Wiley's ExamXpert Quantitative Aptitude (QA) Simplified for CAT 2019 9788126588503, 9788126579679

Table of contents :
Cover
Title Page
Copy right
Acknowledgements
Brief Contents
Table of Contents
Introduction
About the Common Admission Test (CAT) and Other MBA Exam
Common Myths
Key Features of the Book
Analysis and Strategy to Prepare for MBA Exams
Common Admission Test (CAT)
Xavier Aptitude Test (XAT)
Indian Institute of Foreign Trade (IIFT)
Symbiosis National Aptitude Test (SNAP)
Chapter 01: Number System – I Factors, Multiples and Exponents
Introduction
Types of Numbers
Natural numbers
Whole numbers
Integers
Rational numbers
Irrational numbers
Complex numbers
Even/Odd Numbers, Prime and Composite Numbers
Even and Odd Numbers
Prime Numbers
Composite Numbers
Property of Prime Numbers
Checking Whether a Number Is Prime or Not
Co-prime
VBODMAS Rule
Divisibility Rules
Divisibility Rule of 2, 4, 8, 16
Divisibility Rule of 2
Divisibility Rule of 4
Divisibility Rule of 8
Divisibility Rule of 3 and 9
Divisibility Rule of 3
Divisibility Rule of 9
Divisibility Rule of 5, 25 and 125
Divisibility Rule of 5
Divisibility Rule of 25
Divisibility Rule of 125
Divisibility Rule of 10, 50 and 100
Divisibility Rule of 10
Divisibility Rule of 50
Divisibility Rule of 100
Divisibility Rules for Composite Numbers
General Rule for Composite Numbers
Divisibility Rule for 11
Divisibility Rule for 7 and 13
General Divisibility Rule for 11
Special Rule
Divisibility Rule for 27 and 37
Surds and Indices
Surds
Laws of Surds
Indices
Factors of Composite Numbers
Factors
Unique Factorization Theorem
How to Find Number of Factors of a Composite Number
Number of Ways in Which a Composite Number Can Be Expressed As a Product of Two Factors
Number of Ways in Which a Composite Number Can Be Expressed As a Product of Two Co-prime Factors
How to Find Sum and Product of Factors of a Composite Number
LCM and HCF
Various Methods of Finding the LCM
Various Methods of Finding the HCF
Division Algorithm
LCM and HCF of Fractions
LCM and HCF of More Than Two Fractions
Vedic Maths
Base Method of Squaring
Squares of Numbers Ending With 5
Squares of Numbers Consisting of 9s only
Squares of Numbers Consisting of 1s only
Practice Exercise
Answer Key
Answers and Explanations
Chapter 02: Number System – II Remainders, Factorials and Base System
Basic Concepts
Remainders
Fermat’s Theorem
To Find the Last Digit of Numbers
Rule of Cyclicity
Thumb Rule for Finding the Unit’s Digit of abc
To Find the Last Two Digits of Numbers
Factorial
To Find the Highest Power of a Prime Number in N!
To Find the Highest Power of a Composite Number in N!
To Find the Number of Zeros at the End of N!
Wilson’s theorem
Base System and Conversions
What is Base System?
Conversion From Base X to Base 10 and From Base 10 to Base X
Binary Operation of Addition, Subtraction and Multiplication
Practice Exercise
Answer Key
Answers and Explanations
Chapter 03: Commercial Math
Basic Concepts
Percentage
Concept of Base
Population Formula
Profit and Loss
Faulty weights
Simple and Compound Interest
Difference Between Simple Interest (SI) and Compound Interest (CI)
Compounding Period
Doubling Time Period
Future Value (FV) and Present Value (PV)
Compounded Annual Growth Rate (CAGR) and Simple Annual Growth Rate(SAGR)
Practice Exercise
Answer Key
Answers and Explanations
Chapter 04: Ratio, Proportion and Mixtures
Introduction
Ratio
Proportion
More Results
Important Result
Continued Proportion
Variation
Direct Variation
Inverse Variation
Partnership
Averages
Arithmetic Mean (A.M.)
Weighted Arithmetic Mean
Geometric Mean
Weighted Geometric Mean
Harmonic Mean
Weighted Harmonic Mean
Relationship between A.M., G.M. and H.M.
Mixtures and Solutions
Alligation
Replacements
Practice Exercise
Answer Key
Answers and Explanations
Chapter 05: Time and Work
Basic Concepts
Time and Work
Pipes and Cisterns
Practice Exercise
Answer Key
Answers and Explanations
Chapter 06: Time, Speed and Distance
Introduction
Average Speed
Applications of Ratio and Proportion
Relative Speed
Upstream and Downstream Motion
Circular Motion
Clocks
Practice Exercise
Answer Key
Answers and Explanations
Chapter 07: Polynomials, Algebraic Formulae and Linear Equations
Basic Concepts
Polynomials
Division of a Polynomial by Another Polynomial
Remainder Theorem and Factor Theorem
Remainder and Factor Theorem for Such a Case
Algebraic Formulae
Important Results
Polynomial Equations
Linear Equation
Practice Exercise
Answer Key
Answers and Explanations
Chapter 08: Quadratic Equation
Introduction
Nature of Roots
Relationship Between Roots and Coefficients of Equation
AM, GM, HM and Roots of Equations
Continued Fractions
Common Roots
Graphical Interpretation of Quadratic Function
Practice Exercise
Answer Key
Answers and Explanations
Chapter 09: Functions and Graphs
Basic Concepts
Open and Closed Interval
Domain and Range
Common Functions and their Graphs
Trigonometric Functions: sin x and cos x
Logarithmic Functions: logax
Exponential Functions: ex
Greatest Integer Functions: [x]
Fractional Functions: {x}
Algebraic Functions: x, x2, x3 and 1/x
Even and Odd Functions
Symmetricity of Even and Odd Function
Inverse of a Function
Composite Functions
Recursive Functions
Rotation of Graphs
Practice Exercise
Answer Key
Answers and Explanations
Chapter 10: Inequalities, Maxima and Minima
Inequalities
Fundamental Rules
Linear Inequality
Quadratic and Higher Order Inequalities
Modulus Based Inequality
Logarithmic Inequality
Maxima and Minima
Introduction
Use of AM, GM and HM
Practice Exercise
Answer Key
Answers and Explanations
Chapter 11: Logarithms
Basic Concepts
Definition
Rules of Logarithms
To Find the Number of Digits in a Number
Practice Exercise
Answer Key
Answers and Explanations
Chapter 12: Progressions
Introduction
Definition
Arithmetic Progression (AP)
Geometric Progression (GP)
Harmonic Progression (HP)
Three numbers in AP or GP or HP
Infinite AGP and Sum of Its Terms
Arithmetico-Geometric Progression (AGP)
Special Series
Practice Exercise
Answer Key
Answers and Explanations
Chapter 13: Set Theory and Venn Diagrams
Introduction
Set
Subset
Number of Subsets
To Prove Two Sets A and B to be Equal Sets
Universal Set and Complement of a Set
Operations on Sets
Union of Sets
Intersection of Sets
Difference of Two Sets
Venn Diagrams
Properties
Number of Elements in Sets
Minima and Maxima in Venn Diagrams
Applications of Four Venn Diagrams
Practice Exercise
Answer Key
Answers and Explanations
Chapter 14: Permutation and Combination
Introduction
Fundamental Principle of Counting
Product Rule
Sum Rule
Permutation and Combination
Combination
Permutation
Applications of Fundamental Principle of Counting
Distribution of Objects
When Objects are Distinct
When Objects to be Distributed Are Similar
Number of Whole Number, and Natural Number, Solutions of Equations
Arrangement of Distinct Objects with Conditions
Arrangement of Objects, Some of Which Are Similar
Selection of Objects When Any Number of Them Can be Selected
Circular Arrangements
Applications to Geometrical Figures
Group Formation
Practice Exercise
Answer Key
Answers and Explanations
Chapter 15: Probability
Basic Concepts
Product Rule for Independent Events
Binomial Probability
Conditional Probability
Baye’s Theorem
Geometrical Probability
Practice Exercise
Answer Key
Answers and Explanations
Chapter 16: Plane Geometry
Introduction
Lines and Angles
Basic Proportionality Theorem (BPT)
Triangles
Types of Triangles
Classification of Triangles
Similarity of Triangles
Congruence of Triangles
Centroid
Orthocentre
Circumcentre
Incentre
Angle Bisector Theorem
Area of Triangle
Quadrilaterals
Square
Rectangle
Rhombus
Parallelogram
Trapezium
Practice Exercise
Answer Key
Answers and Explanations
Chapter 17: Polygons and Circles
Basic Concepts
Polygons
Difference Between Convex and Concave Polygons
Interior and Exterior Angles
Area of Regular Polygons
Number of Diagonals
Circles
Arc, Sector and Segment
Properties Related to Circles
Secants and Tangents to a Circle
Practice Exercise
Answer Key
Answers and Explanations
Chapter 18: Mensuration
Introduction
Cube
Cuboid
Sphere and Hemi-Sphere
Right Circular Cylinder
Right Circular Cone
Pyramids
Prisms
Tetrahedron
Surface Area and Volume of Regular Tetrahedron
Frustum
Euler’s Theorem
Practice Exercise
Answer Key
Answers and Explanations
Chapter 19: Coordinate Geometry and Trigonometry
Coordinate Geometry
Distance Formula
Section Formula
Centroid and Incentre of a Triangle
Area of a Triangle
Collinearity
Equations of Straight Line
Point-point Form of Equation of Line
Point-slope Form of Equation of Line
Slope-intercept Form of Equation of Line
General Form of Equation of Line
Intercepts Form of Equation of Line
Important Points
Parallelogram
Area of a Parallelogram
Angle Between Two Lines
Condition for Lines to be Parallel
Condition for Lines to be Perpendicular
Concurrent Lines
Distance between a Point and a Line
Distance between Two Parallel Lines
Position of a Point with Respect to a Line
Trigonometry
Basic Results
Trigonometric Functions of Common Angles
Practice Exercise
Answer Key
Answers and Explanations
Solved Papers
CAT 2017 Solved Paper Slot 1
Quantitative Aptitude (QA)
Data Interpretation and Logical Reasoning (DILR)
Answers and Explanations
CAT 2017 Solved Paper Slot 2
Quantitative Aptitude (QA)
Data Interpretation and Logical Reasoning (DILR)
Answers and Explanations
CAT 2018 Solved Paper Slot 1
Quantitative Aptitude (QA)
Data Interpretation and Logical Reasoning (DILR)
Answers and Explanations
CAT 2018 Solved Paper Slot 2
Quantitative Aptitude (QA)
Data Interpretation and Logical Reasoning (DILR)
Answers and Explanations
Back Cover

Citation preview

WILEY Simplified

for CAT 2019

Ashu Jain

Wiley ExamXpert

QA Simplified for CAT 2019 Copyright © 2019 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks or trade names of their respective holders. Wiley is not associated with any product or vendor mentioned in this book. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W ILI First Edition: 2017 Revised Edition: 2019 ISBN: 978-81-265-7967-9 ISBN: 978-81-265-8850-3 (ebk) www.wileyindia.com Printed at:

Acknowledgements The work on this book started with the sole objective of presenting all the concepts asked in major MBA entrance exams in such a manner that even the students who are weak in Maths would be able to comprehend them easily. I have taken immense care to write in a manner that will help students develop the right skills and strategies required to crack these exams. I cannot help but thank, from the bottom of my heart, the editorial team at Wiley India, who gave me this opportunity to present this book to all the students aspiring for CAT, XAT and other MBA entrance exams. Without their support, the book and its knowledge would have remained with me only, and not have come out in its present form. I also want to thank all my students over the years, for the immense experience I have received from training them for close to two decades and which I have tried to incorporate in the present book, to the best of my ability. Last but not least, I would like to thank my family members: my mother (Mrs Sushila Jain), father (Mr M.K. Jain), wife (Ruchi Jain) and both daughters (Shreya and Stuti), without whose emotional support, this work would not have been possible. Thanks to all of you, for your unwavering support and encouragement for the successful completion of this book!

Ashu Jain

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Brief Contents Chapter 1: Number System – I: Factors, Multiples and Exponents   1 Chapter 2: Number System – II: Remainders, Factorials and Base System   71 Chapter 3: Commercial Math  102 Chapter 4: Ratio, Proportion and Mixtures   144 Chapter 5: Time and Work   185 Chapter 6: Time, Speed and Distance   207 Chapter 7: Polynomials, Algebraic Formulae and Linear Equations   245 Chapter 8: Quadratic Equation  276 Chapter 9: Functions and Graphs   303 Chapter 10 Inequalities, Maxima and Minima   351 Chapter 11: Logarithms  384 Chapter 12: Progressions  395 Chapter 13: Set Theory and Venn Diagrams   426 Chapter 14: Permutation and Combination   455 Chapter 15: Probability  488 Chapter 16: Plane Geometry  508 Chapter 17: Polygons and Circles   560 Chapter 18: Mensuration  597 Chapter 19: Coordinate Geometry and Trigonometry   622 Solved Papers: CAT 2017 and 2018 Solved Papers   P-1

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Table of Contents Acknowledgements.......................................................................................................................................... iii Brief Contents.................................................................................................................................................... v Introduction...................................................................................................................................................... xi

1. Number System – I: Factors, Multiples and Exponents...................................... 1

Types of Numbers...................................................................................................................................1



VBODMAS Rule..................................................................................................................................... 11



Divisibility Rules................................................................................................................................... 12



Surds and Indices................................................................................................................................ 23



Factors of Composite Numbers........................................................................................................... 30



LCM and HCF........................................................................................................................................ 37 Vedic Maths.......................................................................................................................................... 50

Practice Exercise..................................................................................................................................... 60

2. Number System – II: Remainders, Factorials and Base System....................... 71 Remainders...........................................................................................................................................71

To Find the Last Digit of Numbers...................................................................................................... 79



To Find the Last Two Digits of Numbers............................................................................................. 82

Factorial................................................................................................................................................ 84

Base System and Conversions............................................................................................................ 91

Practice Exercise..................................................................................................................................... 98

3. Commercial Math...............................................................................................102

Percentage......................................................................................................................................... 102 Profit and Loss.................................................................................................................................... 112 Simple and Compound Interest........................................................................................................122

Practice Exercise...................................................................................................................................133

4. Ratio, Proportion and Mixtures.........................................................................144 Ratio.................................................................................................................................................... 144 Proportion........................................................................................................................................... 145 Variation.............................................................................................................................................. 149 Partnership......................................................................................................................................... 151 Averages............................................................................................................................................. 152 Mixtures and Solutions...................................................................................................................... 164 Practice Exercise................................................................................................................................... 174

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5. Time and Work....................................................................................................185

Time and Work...................................................................................................................................185



Pipes and Cisterns............................................................................................................................. 197

Practice Exercise................................................................................................................................... 201

6. Time, Speed and Distance.................................................................................207

Average Speed................................................................................................................................... 207



Applications of Ratio and Proportion................................................................................................ 210



Relative Speed................................................................................................................................... 212



Upstream and Downstream Motion.................................................................................................. 217



Circular Motion................................................................................................................................... 224

Clocks.................................................................................................................................................232 Practice Exercise................................................................................................................................... 236

7.

Polynomials, Algebraic Formulae and Linear Equations.................................245 Polynomials........................................................................................................................................ 245

Algebraic Formulae............................................................................................................................252



Polynomial Equations........................................................................................................................ 257



Linear Equation..................................................................................................................................258

Practice Exercise................................................................................................................................... 266

8. Quadratic Equation............................................................................................. 276

Nature of Roots.................................................................................................................................. 276



Relationship Between Roots and Coefficients of Equation............................................................. 281



AM, GM, HM and Roots of Equations...............................................................................................284



Continued Fractions...........................................................................................................................286



Common Roots...................................................................................................................................288



Graphical Interpretation of Quadratic Function...............................................................................290

Practice Exercise................................................................................................................................... 297

9. Functions and Graphs........................................................................................303

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Open and Closed Interval..................................................................................................................303



Domain and Range............................................................................................................................304



Common Functions and their Graphs............................................................................................... 307



Even and Odd Functions....................................................................................................................323



Inverse of a Function.........................................................................................................................326

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Composite Functions.........................................................................................................................329



Recursive Functions........................................................................................................................... 331



Rotation of Graphs.............................................................................................................................336

Practice Exercise................................................................................................................................... 340

10. Inequalities, Maxima and Minima.....................................................................351 Inequalities............................................................................................................................................ 351

Fundamental Rules............................................................................................................................ 351



Linear Inequality.................................................................................................................................352



Quadratic and Higher Order Inequalities..........................................................................................353



Modulus Based Inequality................................................................................................................. 357



Logarithmic Inequality.......................................................................................................................362

Maxima and Minima.............................................................................................................................. 364 Introduction........................................................................................................................................364

Use of AM, GM and HM.....................................................................................................................364

Pratice Exercise..................................................................................................................................... 378

11. Logarithms..........................................................................................................384

Definition............................................................................................................................................384



Rules of Logarithms...........................................................................................................................384



To Find the Number of Digits in a Number.......................................................................................389

Practice Exercise................................................................................................................................... 391

12. Progressions.......................................................................................................395

Definition............................................................................................................................................395



Arithmetic Progression (AP)...............................................................................................................395



Geometric Progression (GP)..............................................................................................................399



Harmonic Progression (HP)...............................................................................................................406



Three numbers in AP or GP or HP.....................................................................................................406



Arithmetico-Geometric Progression (AGP)........................................................................................406



Special Series.....................................................................................................................................408

Practice Exercise................................................................................................................................... 416

13. Set Theory and Venn Diagrams.........................................................................426 Set.......................................................................................................................................................426 Subset.................................................................................................................................................428

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Universal Set and Complement of a Set..........................................................................................429

Operations on Sets............................................................................................................................430



Venn Diagrams...................................................................................................................................430

Properties...........................................................................................................................................432 Practice Exercise...................................................................................................................................446

14. Permutation and Combination..........................................................................455 Fundamental Principle of Counting..................................................................................................455

Permutation and Combination..........................................................................................................459



Applications of Fundamental Principle of Counting.........................................................................462



Circular Arrangements.......................................................................................................................469



Applications to Geometrical Figures................................................................................................. 470



Group Formation................................................................................................................................ 472

Practice Exercise................................................................................................................................... 474

15. Probability...........................................................................................................488 Product Rule for Independent Events...............................................................................................495

Binomial Probability........................................................................................................................... 497



Conditional Probability.......................................................................................................................498



Baye’s Theorem..................................................................................................................................500



Geometrical Probability.....................................................................................................................500

Practice Exercise................................................................................................................................... 502

16. Plane Geometry..................................................................................................508 Lines and Angles................................................................................................................................508 Triangles.............................................................................................................................................509 Quadrilaterals.....................................................................................................................................529 Practice Exercise................................................................................................................................... 541

17. Polygons and Circles..........................................................................................560 Polygons..............................................................................................................................................560 Circles.................................................................................................................................................566 Practice Exercise...................................................................................................................................584

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18. Mensuration........................................................................................................597 Cube.................................................................................................................................................... 597 Cuboid.................................................................................................................................................598

Sphere and Hemi-Sphere..................................................................................................................599



Right Circular Cylinder.......................................................................................................................599



Right Circular Cone............................................................................................................................600

Pyramids.............................................................................................................................................608 Prisms.................................................................................................................................................608 Tetrahedron........................................................................................................................................ 610 Frustum............................................................................................................................................... 611

Euler’s Theorem................................................................................................................................. 612

Practice Exercise................................................................................................................................... 616

19. Coordinate Geometry and Trigonometry...........................................................622 Coordinate Geometry............................................................................................................................ 622

Distance Formula...............................................................................................................................623



Section Formula................................................................................................................................. 624



Centroid and Incentre of a Triangle.................................................................................................. 627



Area of a Triangle...............................................................................................................................628

Collinearity..........................................................................................................................................628

Equations of Straight Line................................................................................................................. 631

Parallelogram.....................................................................................................................................634

Angle between Two Lines..................................................................................................................635



Concurrent Lines................................................................................................................................636



Distance between a Point and a Line............................................................................................... 637



Distance between Two Parallel Lines............................................................................................... 637



Position of a Point with Respect to a Line........................................................................................638

Trigonometry.......................................................................................................................................... 640

Basic Results......................................................................................................................................640



Trigonometric Functions of Common Angles.................................................................................... 641

Practice Exercise................................................................................................................................... 652



Solved Papers ..................................................................................................... P-1 CAT 2017 QA and DILR Tests Slot 1 .......................................................................................................... P-3 CAT 2017 QA and DILR Tests Slot 2 ........................................................................................................ P-24 CAT 2018 QA and DILR Tests Slot 1 ........................................................................................................ P-46 CAT 2018 QA and DILR Tests Slot 2 ........................................................................................................ P-69

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Introduction ABOUT THE COMMON ADMISSION TEST (CAT) AND OTHER MBA EXAMS The Common Admission Test (CAT) is an aptitude test that is conducted annually by the Indian Institutes of Management (IIMs). Each year, one of the IIMs sets and conducts the CAT exam. This year, the CAT will be conducted by Indian Institute of Management, Lucknow. The CAT scores are accepted not only by the 20 IIMs in India but also by over 100 B-schools across the country. Apart from the IIMs, some other well established institutions that accept the CAT score are: Faculty of Management Studies (FMS), University of Delhi S P Jain Institute of Management and Research (SPJIMR), Bharatiya Vidya Bhawan, Mumbai Management Development Institute (MDI), Association of MBAs, Gurgaon National Institute of Industrial Engineering (NITIE), Mumbai Jamnalal Bajaj Institute of Management Studies (JBIMS),University of Mumbai Institute of Management Technology (IMT), Ghaziabad IIT School of Management (IIT SoMs), Across India MICA (formerly known as Mudra Institute of Communications, Ahmedabad) Around 4,000 seats are offered through the 20 IIMs and approximately 2 lakh MBA aspirants take the test every year.  The CAT changed from a paper-based exam to an online exam in the year 2009. As can be ascertained from the data shared above, one needs to obtain a high score in order to gain admission to IIMs and/or other top B-schools. However, getting a high score is not sufficient. After getting a call from a particular B-school, one needs to go through the second stage of the selection process, that is, Group Discussion/Personal Interview/Written Aptitude Test round. This round is also an extremely challenging one and demands close attention and focussed preparation. Here, we would like to share insights regarding the entire selection process which comprises: 1.

Written Test

2.

Group Discussion/Personal Interview/Written Aptitude Test

Before discussing about the written test stage, we would like to mention that, in order to get admission in a renowned, or even a good B-school, the CAT is not the only exam. MBA aspirants usually focus too much on the CAT preparation and ignore other exams. This should not be the approach adopted by a candidate. Some of the major non-CAT exams worth considering are: XAT, IIFT, TISS, SNAP, NMAT and so on. So, while preparing for MBA entrance exams, one must focus not only on the CAT, but also on non-CAT exams as well. In fact, there are some common misconceptions in the minds of many students regarding the CAT. Let’s discuss some of these: Quantitative Aptitude Simplified for CAT 2019

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COMMON MYTHS .All B-schools accepting the CAT score are top B-schools. Though it does appear to be the case, an analysis of the list of all B-schools that accept the CAT score reveals .that quite a large chunk of these institutes are not worth considering. CAT is extremely tough, whereas non-CAT exams are not that tough. We have been taking the actual CAT for more than a decade (and have analysed earlier CAT papers), and we are of the view that, despite variations in test structure over the years, the CAT exam is NOT A TOUGH EXAM, though it is generally perceived by many to be so. Quite often, students feel that the CAT is tough mainly because of the uncertainty surrounding its pattern. And, this generates fear in the minds of students. The key to getting great scores is not so much about learning more and more advanced concepts as it is about getting rid of the fear of the CAT. And here comes the ultimate guide to preparing for the CAT and other MBA entrance exams. To get great scores, all you need is to get rid of the fear of the CAT exam, besides developing comfort with the major areas asked in the exam. .Preparing only for the CAT will automatically prepare one for non-CAT exams as well. Nothing can be farther from the truth. Nowadays, many exams have Current Awareness and General Knowledge section (asked in XAT, IIFT, SNAP and so on), which are not asked in the CAT. The XAT exam has one full section on Decision Making, which is totally absent in the CAT. Certain areas of Logical Reasoning are not asked in the CAT, but are extensively tested in IIFT, SNAP, NMAT, and so on. Even if the topics tested in the CAT and other exams are the same, the kind of questions asked is quite different. So, though there are some common testing areas between CAT and other exams, every examination adopts its own unique testing and selection process. Hence, if one intends to prepare for non-CAT exams as well, then one has to devote special attention to the kind of questions asked in these exams. Besides, the strategies required to succeed in the CAT are somewhat different from those required to succeed in the other competitive exams. I.f I learn a lot of concepts, especially the advanced ones, my chances of success in the exams will increase tremendously. This belief is so strongly etched in the minds of the aspirants that they spend the entire time for the CAT preparation learning more and more concepts. By the time they finish learning these concepts, they are hardly left with time for anything else. The flaw in this preparation strategy is that they have not focussed on taking tests similar to the CAT and other exams. And because of this lack of practice of tests, even if they have learnt loads of concepts, they are not able to secure even average marks in the exams. In short, the aspirants have not focussed on learning TEST TAKING SKILLS, rather they have ignored them and; therefore, despite knowing so many concepts and shortcuts, they are not able to score well. The above ideas should not be understood in isolation. Rather all the points mentioned above are connected to each other in some way or the other. Hence, if one learns not only smart techniques of solving questions, but also Test Taking Skills, the apprehension faced by a candidate with respect to the CAT and other competitive exams reduces drastically, resulting in enhanced scores and finally, success. All this requires careful planning and focus. While beginning preparation for the CAT and other MBA entrance exams, one should start by reviewing concepts of various sections, that is, Quantitative Aptitude, xiv

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Data Interpretation, Reasoning, Verbal Ability, and so on. One should also start taking practice tests after having learnt a couple of topics. This method of regular testing enables one to identify one’s strengths and weaknesses and accordingly adjust the exam preparation. The present book aims to ensure that students learn all the concepts relevant for Quantitative Ability section of the CAT in the simplest possible manner. Note that the syllabus covered in the CAT is not just restricted to what is taught in class IX and X, but also includes elements of what is taught in class XI and XII. For example, topics such as Permutation and Combination, Probability, Sequence and Series, Set Theory, Functions and Graphs, Inequalities, Maxima and Minima, Quadratic Equations and their graphs are the major areas of Algebra, which carries the highest weightage in the CAT and which is from the syllabus of class XI and XII. The way all the topics are covered in this book will make even those students who consider themselves weak in Maths, understand the concepts easily. As you begin with this book, you can safely assume that your preparation process has already started. This book has been specially designed and developed for the candidates who wish to streamline their preparation with respect to the CAT. The purpose behind this book is two-fold: To provide conceptual clarity with respect to the different elements that are tested and help you revise the core topics. To assist in the development of an effective and targeted strategy geared for the admission process. The CAT aims at assessing a candidate’s ability to demonstrate accuracy in a limited period of time. Although this book caters mainly to the CAT, since the syllabus of Quantitative Ability of major MBA entrance tests is almost same, this book can also be used for learning the topics asked in other MBA entrance exams, such has XAT, IIFT, SNAP, NMAT and so on. However, there are some differences in the kind of questions asked as well as the Sections tested (for example, in the XAT, there is a section that tests you on Decision making but the same is not the case with the CAT. Also, GK is tested in the XAT as well as the IIFT exams). Every examination varies with respect to the approach followed and the elements tested. Therefore, a thorough understanding of the demands of a particular examination, combined with a customized and targeted strategy will help a candidate gain entrance into his or her desired management institute.

KEY FEATURES OF THE BOOK There are certain unique features of the book which are mentioned below: Focus on key concepts: The book contains all the relevant concepts asked in the key MBA exams. The theoretical concepts are explained in a simple manner which enhances understanding. Moreover, the explanations are extensive and cover all relevant aspects. Solved Examples: The theory is interspersed with a number of solved examples strategically placed within the text matter so as to enhance learning and application. In order to build conceptual clarity, it is important to explain concepts with examples in a coherent and cohesive manner. This book aims to do just that by providing theoretical concepts followed by relevant examples. Practice Exercises: Each chapter is followed by a Practice Exercise, which contains questions asked in past year CAT, XAT and IIFT exams from the topic at hand. The purpose is to make students realize that the concepts provided in each chapter are all that is required to solve each and every question asked in the said exams. Moreover, alternative or smarter solutions are provided for these questions wherever required.

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Videos: The book contains exclusive ‘expert speak’ videos that will provide you tips and strategies for test preparation and will also serve as doubt clearing resources where you’ll find answers to many of your questions related to quant-related concepts and theories. These videos are accessible through QR codes that are strategically placed throughout the book. The objective of providing these free videos with the purchase of this book is to bring alive the classroom experience for the learners who should find these useful in their exam preparation strategy.

ANALYSIS AND STRATEGY TO PREPARE FOR MBA EXAMS In this section, we present an analysis of the Quantitative Aptitude Section of some key MBA entrance exams such as CAT, XAT, IIFT, SNAP, and so on. An understanding of the structure of these exams will help you devise a customised strategy targeted for each examination.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to analyse a Mock Test and use the strategies to improvise upon one’s score in these tests.

Now, let us examine the recent trends in these examinations with respect to the structure and content.

Common Admission Test (CAT) CAT is an online examination. The sectional breakup for CAT 2018 is provided below: Section Name

Questions

Time limit

Verbal Ability and Reading Comprehension

34

60 min

Data Interpretation and Logical Reasoning

32

60 min

Quantitative Ability

34

60 min

TOTAL

100

180 min

Sectional cut-offs are applicable to the B-schools accepting the CAT score, for example, the IIMs, FMS, MDI, IMT Ghaziabad, IMI, and so on.

Xavier Aptitude Test (XAT) Xavier Aptitude Test (XAT) is one of the most prestigious and difficult MBA entrance exams. Conducted by XLRI Jamshedpur, XAT is held on the first Sunday of January every year. Till 2017 XAT remained a paper based test. With effect from 2018, XAT has become an online computer based test. 

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Part

Part 1

Part 2

Section Name

Number of Questions 2019

2018

2017

2016

Verbal and Logical Ability

26

26

24

26

Decision Making

21

21

21

23

Quantitative Ability and Data Interpretation

27

27

27

29

TOTAL

74

74

72

78

General Knowledge

25

25

25

25

Essay Writing

–

--

--

--

GRAND TOTAL

99

99

97

103

XAT 2019 exam was of 3 hours duration instead of 3.5 hours. Candidate were allowed to go to Part 2 in XAT 2019 after completing the time duration of 165 minutes in Part 1. The total time allotted to Part 2 was 15 minutes. There is negative marking for questions not attempted by students, which creates additional concerns for a candidate. The score obtained in Part II is not considered in the first stage of selection, but is considered in the final process.

Indian Institute of Foreign Trade (IIFT) Indian Institute of Foreign Trade (IIFT) schedules an Entrance Test for those candidates who wish to gain admission in the MBA (International Business) program at its New Delhi and Kolkata campuses. This is a national level entrance examination which is attempted by thousands of students every year. It is a paper-based exam. Sections

Part

Number of Questions

Time Limit

2018

2017

2016

2015

Logical Reasoning

20

20

22

20

Data Interpretation

20

20

20

18

Reading Comprehension

16

16

16

16

Verbal Ability

20

20

20

20

3

General Knowledge

18

18

25

28

4

Quantitative Ability

20

20

20

22

114

114

123

124

1 2

TOTAL

120 min

Symbiosis National Aptitude Test (SNAP) Symbiosis National Aptitude (SNAP) Test is a national-level management entrance examination, conducted by Symbiosis International University (SIU). SNAP test scores are valid for admission to 15 institutes under SIU. The scores are also valid for admission to various other MBA or equivalent courses. SNAP has duration of two hours and is conducted in 30 cities across India. Till 2016 SNAP remained a paper based test. With effect from 2017, SNAP has become an online computer based test. Quantitative Aptitude Simplified for CAT 2019

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Section

Number of Questions 2018

2017

2016

2015

Verbal Ability and RC

35

40

40

40

QA and DI

35

40

40

40

Logical Reasoning

35

40

40

30

General Awareness

25

30

30

40

TOTAL

130

150

150

150

Exam pattern and scoring pattern was changed in SNAP 2018. Total number of questions was reduced to 130 from earlier 150. Questions were divided into MCQs (Normal Questions) and Non-MCQ (Special questions) just like TITA questions in CAT. Having understood the pattern of the key MBA entrance exams, we will now analyse the Quantitative Aptitude section of these exams. There are basically two ways to analyse any test: Identifying the number of questions asked from each topic in a section over the past few years, thereby ascertaining the important topics and focussing on them; Segregating questions on the basis of difficulty level and devising strategies for preparation in accordance with the distribution of questions. If we adopt the first approach for analysis, then we might identify a particular set of topics as being more important, due to their high frequency. This will lead students to focus exclusively on these topics, at the cost of other topics. However, this strategy entails a risk because most competitive exams refrain from following a set pattern and are constantly looking for ways to challenge students. Therefore, it is not prudent to ignore a topic because it is not asked frequently. Also, the difficulty level of the questions based on a frequently asked area might be high or they might involve extensive calculations. If any or both of these scenarios occur, the students will be impacted negatively with respect to the number of attempts and accuracy, which will ultimately result in a low score.

EXPERT SPEAK Scan this QR Code to watch a video that explains the exam preparation strategy.

In the second approach, we generally segregate questions into three types:

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1.

Type I: Questions which are easy to understand, and equally easy to solve (assuming that students are aware of the basic concepts of the topics from which the questions are asked).

2.

Type II: Questions which are easy to understand, but are lengthy. Such questions will take more than the ‘average time per question’ apportioned to solve.

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3.

Type III: Questions which are difficult to understand and hence might take extra time to grasp thoroughly, even if students are aware of the basic concepts of the topics at hand.

We recommend the second method of analysis for any test. In this method of analysis, preparing for all the topics is a sine qua non. During preparation, a student might realise that there are certain topics that are easier to understand while other topics present greater difficulty. Once the preparation is complete, a student should try to categorize the questions with respect to the level of difficulty and distribution. The strategy should be that during the exam, he/she should first try to identify Type I questions, irrespective of the topic and attempt those questions first. He/she should be careful not to fall prey to the Type II or III questions, and waste time attempting these questions. An examiner, when he/she designs a test, ensures that Type II and Type III questions are sprinkled all over the test between Type I questions. It is also quite possible that first few questions are of Type III (high difficulty level) and when a student observes that he has left the first 5 or 6 questions, it creates psychological pressure and can have a negative impact on motivation and concentration. Please note that these are aptitude tests designed to assess resilience, mental strength, focus and concentration, among other aspects. It is advised to all the students that they must persist and maintain equanimity. Don’t lose heart and continue with the suggested strategy. If the test turns out to be a difficult one, it will be difficult for most of the students and the cut-off will reduce automatically. Once a student has attempted Type I questions, he/she should revisit the entire section or test and start attempting Type II questions. By this time, the student would have gained momentum and a fair understanding of the structure of the exam. The remaining time should be spent on these questions. In the rare case that some time is still left after exhausting Type II questions, a student may proceed to Type III questions. Care should be taken to maintain accuracy while attempting the questions. Otherwise, negative marking will have a detrimental impact on the score, even if the attempts are on a higher side. With respect to the specific exam based strategy; a student should attempt adequate number of mock tests/ practice tests for that exam. For example, if a student is preparing for the CAT, he/she should attempt 12 to 15 mock tests simulating the CAT pattern. Students should also develop the right knowledge and conceptual skills desired to attempt all the sections. For example, students should know that XAT has a section on Decision Making, and developing the conceptual know-how of the section is important to ensure a good number of attempts within the time limit. Similarly, General Knowledge is asked in XAT, IIFT and SNAP and so students should be aware of the elements of General Knowledge and awareness that are assessed in the various exams. In the beginning, students might find it difficult to sit for 3 hours at a stretch, while maintaining a high degree of concentration. This could be due to lack of mental stamina. To overcome this, it is advised that students start taking Section Tests of one-hour duration. Gradually, they should increase the time duration of the tests. So, once they are attuned to sitting for 1 hour, they should now take two such tests consecutively, and develop the mental habit of sitting for 2 hours at a stretch. Over a period of time, students should develop stamina of sitting for 3 hours in one go. Once they develop such stamina, then they should focus on improving their scores in these exams. It is also seen that students have a habit of solving Quantitative Aptitude questions by writing each and every step. Kindly note that the more you write the more time is wasted which has a negative impact on the number of questions that can be attempted. Over a period of time, with practice, students will realize that a lot of mathematical calculations can be performed mentally using short-cuts. If students can enhance this skill, they will be able to save a lot of time which can be utilised to attempt more questions. Of course, this takes immense hard work and rigorous concentration, but the results are worth it!

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One should realize that there are two ways to perform in a test, assuming that students have developed the stamina of sitting for the entire test duration and assuming that students have learnt all the relevant concepts and skills required in a particular test. One way is to focus on accuracy, and not worry too much about the number of attempts. The other is to focus on speed and try to attempt more questions with little focus on accuracy. Both of these are extreme strategies and should be avoided. During the initial days of preparation, one must definitely focus on accuracy. Gradually, one should increase the number of attempts, without diluting the accuracy. And finally, during the last days of preparation, increase the number of questions attempted to such level as is desired to obtain the cut-off scores, without compromising on the accuracy.

EXPERT SPEAK Scan this QR Code to watch a video that is a continuation of the Exam Preparation Strategy.

To conclude this chapter, we would emphasize that preparing for these exams requires a specific set of skills, including time management across all the sections, and one should leave no stone unturned in developing those skills to achieve the desired objective. We wish you the best for your test preparation and hope that you find this book useful in your journey.

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Number System – I: Factors, Multiples and Exponents

Chapt er 01

Number System – I Factors, Multiples and Exponents INTRODUCTION Numbers are an integral part of our life. We encounter them at every stage of our day-to-day functioning. In this chapter, we intend to elucidate certain essential properties and facts about numbers which will prove not only interesting, but also useful to the students who are preparing for competitive exams such as CAT, XAT, IIFT and so on.

Types of Numbers Natural numbers We can have a class of 30 students and the weight of a boy can be 45.6 kg. But, surely, we cannot have 45.6 students in a class, because fractional number of students is strange and impossible. The number of students will be either 45 or 46. Numbers like 45, 46, 30 and so on are called natural numbers. Therefore, natural numbers include 1, 2, 3, 4, 5 up to infinity. The set of natural numbers is denoted by ‘N’ and is represented as below: N = {1, 2, 3, 4, ....} Note: Infinity is not a natural number. In fact, infinity is not a number at all. It is represented as .

1

Quantitative Aptitude Simplified for CAT

The property of closure for natural numbers If a natural number is added to any natural number, we always get a natural number. Therefore, we can say that natural numbers are closed with respect to addition. Similarly, any natural number multiplied with any natural number yields another natural number. Therefore, we can say that natural numbers are closed with respect to multiplication. But, any natural number subtracted from any other natural number does not always yield another natural number. For example, 4 – 3 = 1 is a natural number, but 6 – 9 = –3 is not a natural number. Therefore, natural numbers are not closed with respect to subtraction. Note that natural numbers are also not closed with respect to division, because any natural number divided by any other natural number does not always yield a natural number. For example, 5 divided by 3 does not result in a natural number.

Whole numbers A natural number minus the same natural number is zero. For example, 6 – 6 = 0, which is a Whole Number. Whole numbers are defined as numbers like 0, 1, 2, 3 and so on. The set of whole numbers is denoted by ‘W’ and represented as below: W = {0, 1, 2, 3, ....} The only difference between whole numbers and natural numbers is the inclusion of zero in whole numbers.

The property of closure for whole numbers Like natural numbers, whole numbers are closed with respect to addition and multiplication but not with respect to subtraction (check this yourself).

Integers Integers include all the numbers such as 0, 1, 2, 3 and so on. They also include negative numbers such as –1, –2, –5, and so on. The set of integers is denoted by ‘’ or ‘Z’ and represented as below:  = {..., –3, –2, –1, 0, 1, 2, 3, ...} Therefore, Integers include negative numbers, zero and positive numbers. Zero and positive numbers collectively are known as whole numbers, whereas all positive numbers are natural numbers. From above, we can say that, 

Set of Natural Numbers is a subset of the set of Whole Numbers.



Set of Whole Numbers is a subset of the set of Integers.

Note: Zero is neither positive nor negative.

The property of closure for integers Integers are closed with respect to addition, subtraction and multiplication. We can easily see that neither natural numbers, nor whole numbers, nor integers are closed with respect to division. For example, when 5 is divided by 7, we do not get a natural number, nor whole number, nor integer.

Rational numbers Rational numbers are those numbers which can be expressed in the form of p , where p and q are integers and q

q  0. The set of rational numbers is denoted by Q.

2

Number System – I: Factors, Multiples and Exponents

If p and q are integers, then rational numbers would include numbers like

4 13 22 9 , , , and so on. Therefore, 7 25 7 23

all fractions are covered in this. All integers are also types of rational numbers. Therefore, 3, 5, 6, 28, –4, –9, 0 are all examples of rational numbers. Rational numbers can be classified as integers or fractions. Fractions are of two types: proper and improper. Proper fractions are those fractions in which the numerator is less than the denominator. For example,

5 23 , 7 148

etc. Improper fractions are those fractions in which the numerator is more than the denominator. For example, 7 148 , etc. 5 23

Therefore, reciprocal of proper fractions would be improper fractions and vice versa. Of course, if numerator becomes equal to the denominator, then the fraction reduces to 1, which is not a fraction as such. 7 2 can be written as 1 . In this 5 5 2 mixed fraction form, the fractional part is always a proper fraction, that is, is a proper fraction. Thus, a 5

All improper fractions can be converted to mixed fraction form. For Example,

mixed fraction is a mixture of integer and proper fraction. From above, we can conclude that all natural numbers, whole numbers, integers and fractions are subsets of rational numbers. Subset is denoted by the symbol . Therefore, we can represent the statement made above in the following manner: N  W Q Numbers such as 4.56, 0.2345, 0.3434343434343434...... are also rational numbers. Note that 4.56 can be written as

456 2345 and 0.2345 can be written as . 100 10000

Numbers such as 4.56 and 0.2345 are called terminating decimals. Terminating decimals are those decimals in which there is finite number of digits after the decimal point. Integers such as 2, 4, 67, –1, –26 are also terminating decimals. Numbers such as 0.3434343434343434......, also written as 0.34 , are called recurring decimals. Recurring decimals are those decimals in which a digit or a group of digits repeats infinitely. Here, the digit group ’34’ is repeating infinitely. Therefore, numbers such as 0.2333333...., 0.33333...., 9.345454545...., are all examples of recurring decimals. We can represent 0.233333.... as 0.23 ; 0.33333.... as 0.3 ; 9.345454545.... as 9.345   . Generally, the dots are placed above the first Instead of bar, we can also use dots. Therefore, 9.345  9.345  and last digit of the recurring group. Therefore, 9.34597  9.34597 .

Terminating as well as recurring decimals are types of rational numbers. So, it is possible to convert recurring decimals to p form. q

Conversion of Recurring Decimals to

p q

form

Let us learn to convert 0.273333..... to p/q form. Basic Method: Let x = 0.273

... (1)

3

Quantitative Aptitude Simplified for CAT

Multiply equation Eq. (1) by 100 so that we are left with only the ‘recurring part’ of the number after ‘decimal point’. So, ... (2)

100x = 27.3

Now, multiply Eq. (2) by 10 (because 10 has as many zeros as there are digits in the recurring part). We get, 1000x = 273.3

... (3)

Subtracting Eq. (2) from Eq. (3), we get: 900x = 273 – 27 = 246. So, x =

246 . 900

Shortcut Method: Here 27 is non-recurring whereas 3 is recurring. Write the number as it is (without ‘decimal’ and ‘bar’), that is, 273. Subtract the non-recurring part, that is, 27 from 273. We get 246. Divide 246 by a number having as many 9’s as there are recurring digits, followed by as many 0’s as there are non-recurring digits. So, we divide 246 by 900 to get

246 . 900

We can generalize the above discussion: form of 0.abcdedededede... or 0.abcde =

abcde  abc . 99000

In other words, write the number as it is (without ‘decimal’ and ‘bar’). Subtract the non-recurring part from this and divide the result by a number which consists of as many 9’s as there are digits in the recurring part, followed by as many 0’s as there are digits in the non-recurring part. Let us convert 0.34 . Basic Method: Let x = 0.34

... (1)

Since there are 2 recurring digits, multiply both sides of the equation by 100. We get, 100x = 34.34 .

... (2)

Now, subtract Eq. (1) from Eq. (2), we get: 99x = 34, or x =

34 . 99

Shortcut Method: Write the digit group as it is (without ‘decimal’ and ‘bar’), that is, 34. Divide this by a number having as many 9’s as there are recurring digits. Therefore, 0.34343434.... = 0.34 = Similarly, 0.235235235.... =

34 . 99

235 5467 ; 0.546754675467.... = . 999 9999

Conclusion: In case there is no non-recurring digit after decimal point, then nothing is subtracted in the numerator and no zero appears in the denominator. Example 1 Convert 0.4358989898989.... to p form. q

4

Number System – I: Factors, Multiples and Exponents Solution 0.4358989898989.... =

43589  435 43154  . 99000 99000

Example 2 Convert 12.3425 to p form. q

Solution 12.3425  100  0.123425

= 100 

123425  1234 122191  . 990000 9900

Conclusion: In questions where there are non-zero digits before decimal point, there will be as many zeros in the denominator as there are non-recurring digits after the decimal point. Example 3 Let x = 0.abcdabcdabcdabcd....., where a, b, c and d are non-negative integers such that not all of them are simultaneously zero. By which of the following numbers should x be multiplied to make x a positive integer? A.

9990

B.

999

C.

49995

D.

None of these

Solution: C x = 0.abcd =

abcd . Therefore, x must be multiplied by 9999 or a multiple of 9999 to make it a positive 9999

integer. We observe that 49995 (= 5  9999) is a multiple of 9999. Note: All numbers which are terminating (and non-recurring) decimals or recurring (and Non-terminating) decimals are rational numbers. Moreover, if a number is a rational number, it will be either terminating decimal or a recurring decimal.

Irrational numbers Consider a number like 1.234323456545676545678889765... where the digits after decimal point go on forever without any particular pattern or order. Also consider: 1.10100100010000100000...., which is also neither recurring nor terminating. Such numbers which are neither recurring nor terminating are called as irrational numbers. Obviously, since they are neither recurring, nor terminating, we cannot express them in the form of p . Therefore, irrational

q p numbers are those numbers which cannot be expressed in the form of , where p and q are integers and q

q  0. Some of the examples of irrational numbers are 2, 3, 3 4 , (2 + on.

5

3

),

2

+

3

, , e, e2, log57, log102 and so

Quantitative Aptitude Simplified for CAT

22 , it might be mistaken to be a rational 7 22 22 number. But, actually  is not equal to . The value of  is 3.14159...., whereas the value of is 7 7 22 3.142857142857… = 3.142857 , which is a recurring decimal. We can see that  < . 7

Note: Since  is generally (though incorrectly!) known to be equal to

Example 4 The number 0.318564318564318564....... is: A.

a natural number

B.

an integer

C.

recurring decimal

D.

neither recurring nor terminating

Solution: C 0.318564318564318564....... = 0.318564 . So, the number is recurring. Example 5 Which of the following statements is not true about 3 ? A.

It is an irrational number.

B.

Its value is non-terminating.

C.

Its product with a rational number is always an irrational number.

D.

Its product with an irrational number is always a rational number.

Solution: D 3

is an irrational number and its product with an irrational number may be irrational or rational.

Real numbers It is clear from the above discussion that if a number is rational, it cannot be irrational and if a number is irrational, then it cannot be rational. Therefore, the set of rational and irrational numbers are disjoint sets. The union of these two sets is the set of real numbers, that is, rational and irrational numbers together are called real numbers. Note that between any two rational numbers, we can have infinite rational numbers as well as infinite irrational numbers. Similarly, between any two irrational numbers, we can have infinite rational as well as irrational numbers. We will later see how to introduce ‘n’ rational numbers between two rational numbers. Example 6 How many rational numbers are there between 5 and 6? A.

2

B.

3

C.

4

D.



Solution: D There are infinite rational numbers between any two real numbers.

6

Number System – I: Factors, Multiples and Exponents

Complex numbers Let us briefly discuss about complex numbers. Complex numbers are numbers which can be expressed in the form of a + ib, where ‘a’ and ‘b’ are real numbers, and i =  1 . Here, ‘a’ is the real part whereas ‘ib’ is the imaginary part. Examples of complex numbers are 2 + 3i, 5 – 7i and so on. ‘i' (pronounced as ‘iota’) is the unit of imaginary numbers. If b = 0, then a + ib = a, which is a real number. Therefore, we can say that all real numbers are subsets of complex numbers. Note that if b  0, then such numbers are called as imaginary numbers. Now, since i =  1 , i2 = –1, i3 = (i2)i = –i and i4 = (i2)2 = (–1)2 = 1. In general, i4n = 1, that is, iota raised to the power “multiple of 4” is always equal to 1. Example 7 Find the value of i237. Solution i237 = i236  i = 1  i = i, since 236 is a multiple of 4. Example 8 Find the value of i + i2 + i3 + i4 + ..... + i2005. Solution: We can easily see that i + i2 + i3 + i4 = i – 1 – i + 1 = 0. Therefore, every cluster of 4 terms is always zero. Therefore, (i + i2 + i3 + i4 + .....+ i2004) + i2005 = 0 + i2005 = i.

Conjugate of a complex number If we have a complex number a + ib, then a – ib is the conjugate of a + ib. Therefore, 2 – 5i is the conjugate of 2 + 5i. Important property of the conjugates is that the product of a complex number and its conjugate is always a real number. For example, (2 + 5i)(2 – 5i) = 22 – (5i)2 = 4 – (–25) = 29. In general, (a + ib) (a – ib) = a2 + b2 which is always real. Conjugates are also used to simplify certain fractions in which the denominator is imaginary number. Example 9 Simplify:

2 4  3  5i 5  3 i

Solution 2 should be multiplied and divided by the conjugate of 3 + 5i so that upon simplification, the denominator 3  5i

becomes a real number. Therefore,

Similarly,

2  2   3  5i   2(3  5i )  3  5i =    =    17 3  5i  9  25   3  5i   3  5i 

4  4   5  3i   4(5  3i )  10  6i =    =  25  9   17 5  3i  5  3i   5  3i   

7

Quantitative Aptitude Simplified for CAT

Adding them, we get:

2 4 13  11 i  = . 3  5i 5  3 i 17

Even/Odd numbers, prime and composite numbers Even and odd numbers As we have already seen, natural numbers are numbers such as 1, 2, 3 and so on. Natural numbers can be classified as odd and even numbers. Even numbers are numbers which are divisible by 2, whereas odd numbers are numbers which are not divisible by 2. Note that zero is not an even number though numbers ending with zero are even numbers. When we talk of even or odd numbers, we mean only natural numbers and zero is not a natural number. Note: (Even Number) + (Even Number) = (Even Number) (Even Number) + (Odd Number) = (Odd Number) (Odd Number) + (Odd Number) = (Even Number) Example 10 If x is an even positive integer and y is an odd positive integer, then which of the following statements is true? A.

(x – 1) y is even

B.

x(y – 1) is odd

C.

(x – 1) (y – 1) is odd

D.

x(y – 1) is even

Solution: D For such questions, work with the options. (x – 1) is odd because x is even. Since y is already odd, therefore (x – 1) y is definitely odd. Hence [A] is not the correct option. Likewise, [B] and [C] are incorrect.

Prime Numbers Similarly, natural numbers can also be classified as: prime numbers, composite numbers and the number 1. Prime numbers are numbers which are divisible by one and itself. More specifically, prime numbers have exactly two distinct factors. For example: 2, 3, 5, 7, 11 and so on. From 1 to 100, there are 25 prime numbers. Note that all the prime numbers are odd except 2 which is the only even prime number and the smallest prime number. Since 1 has only one factor, that is, 1, therefore 1 is not a prime number. In fact, 1 is the only natural number which has only one factor.

Composite numbers Composite numbers are numbers having more than 2 factors. For example, 12, 36, 225 and so on. We can easily see that 1 is also not a composite number. Therefore, 1 is neither prime nor composite.

Property of prime numbers If p is a prime number more than or equal to 5, then p can be expressed as either 6n + 1 or 6n – 1. In other words, if p  5, where p is prime, then p = 6n ± 1. For example, 5 can be written as 6 – 1, 7 can be written as 6 + 1, 13 can be written as 6(2) + 1, and so on. But every number of the form 6n ± 1 need not be a prime number. For example, 25 = 6(4) + 1, but 25 is not a prime number. Similarly, 49 = 6(8) + 1, but 49 is not a prime number. The prime numbers 2 and 3 cannot be written in the form of 6n ± 1.

8

Number System – I: Factors, Multiples and Exponents Example 11 If p is a prime number more than or equal to 5, then which of the following is necessarily true? A.

p divided by 6 leaves a remainder 1

B.

p divided by 6 leaves a remainder 5

C.

p divided by 6 leaves a remainder 1 or 5

D.

p divided by 6 leaves a remainder 1, 3 or 5

Solution: C Since p is a prime number more than or equal to 5, it can be written as 6n ± 1, which when divided by 6 leaves remainder 1 or –1. The remainder of –1 when divided by 6 is same as 6 – 1 = 5. Therefore, the remainder is either 1 or 5. Example 12 Find the remainder when p2 is divided by 12, where p is a prime number  5. Solution Method 1: p = 6n ± 1. Therefore, p2 = (6n ± 1)2 = 36n2 ± 12n + 1, which when divided by 12 leaves a remainder of 1. Method 2: Alternatively, take a couple of prime numbers. Checking with 5, 7 and 11, we see that 25, 49 and 121, when divided by 12, leave remainder 1 in each case. Note that method 2 is not a sure shot method because we are not sure if there is a case when the remainder is not 1. But by method 1, we are sure that remainder will be always 1. Example 13 By which largest number is p2 – 1 always divisible, where p is a prime number  5? A.

8

B.

24

C.

12

D.

48

Solution: B Method 1: p2 – 1 = (p – 1) (p + 1). If p is a prime number  5, it is necessarily odd and hence p – 1 and p + 1 are both even numbers. Now, these two are consecutive even numbers. Therefore, one of them is a multiple of 2 and other a multiple of 4 and hence their product will be a multiple of 8. Further, p – 1, p and p + 1 being three consecutive natural numbers, one of them must be a multiple of 3. Since p cannot be multiple of 3 (it being a prime number), either p – 1 or p + 1 must be a multiple of 3. This means that the product of p – 1 and p + 1 will also be multiple of 3. Therefore, the product (p – 1)(p + 1) will be a multiple of 8  3, that is, 24. Method 2: In the previous example, we have seen that p2 = 36n2 ± 12n + 1. Therefore, p2 – 1 = 36n2 ± 12n = 12n (3n ± 1). Now, n can be either even or odd. If n is even, then 12n is a multiple of 24 and hence the result. If n is odd, then 3n ± 1 is even and again the same result. Therefore, p2 – 1 is always divisible by 24.

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Quantitative Aptitude Simplified for CAT

Method 3: Check with some prime numbers. Taking p = 5, 7, 11, we see that p2 – 1 = 24, 48 and 120, each of which is divisible by 24. Example 14 Which of the following statements is always correct? A.

Every prime number can be expressed as 6n + 1

B.

Every prime number can be expressed as 6n – 1

C.

Every prime number can be expressed as 6n ± 1

D.

None of these

Solution: D None of the options [A], [B] or [C] is true because for a prime number to be expressible as 6n + 1 or 6n – 1, the prime number has to be more than or equal to 5, which is not mentioned in the statements. Except for 2, all prime numbers are odd. Therefore, sum of two prime numbers (none of which is 2) must be even. For example, 5 + 7 = 12, which is an even. But the sum can be odd also. For example, 11 + 2 = 13. But for such a case, one of the prime numbers has to be even, and that can be 2 only. Example 15 Let the prime number 34abc3249 be equal to sum of x and y, where x and y are prime numbers. If x < y, find the value of x. A.

3

B.

5

C.

None of [A] and [B]

D.

No such value of x is possible

Solution: C Since the given number is odd, one of x or y has to be equal to 2. Since x < y, x has to be 2 only. Example 16 Express 23 as the sum of two prime numbers. Solution Let 23 = a + b, where a and b are two prime numbers. Since 23 is odd, one of a and b must be 2. But, in such a case, the other number will be 21, which is not a prime number. Therefore, 23 cannot be expressed as the sum of two primes.

Checking whether a number is prime or not Let 437 be a number. We want to check whether it is a prime number or not. Find the square root of 437, which comes out to be approximately 20.9. The integral number is 20. Find all the prime numbers from 1 to 20. Check whether 437 is divisible by any of them or not. The primes from 1 to 20 are 2, 3, 5, 7, 11, 13, 17 and 19. We see that the number is not divisible by any prime number except 19. In fact, 437 = 19  23. Therefore, 437 is not a prime number. Example 17 Check whether 257 is a prime or not?

10

Number System – I: Factors, Multiples and Exponents Solution The integral number in the square root of 257 is 16. The primes up to 16 are 2, 3, 5, 7, 11 and 13. Upon checking, we see that 257 is not divisible by any of these prime numbers. Therefore, the number 257 is a prime number.

Co-prime Two numbers a and b are co-prime if they do not have any common factor other than 1, that is, their HCF is 1. For example, (4, 9) are co-prime since the HCF of 4 and 9 is 1. Co-prime is also known as relatively prime. Therefore (4, 9) are relatively prime also. Based on the above discussion of the various kinds of numbers, we can develop a number tree as below:

Figure 1: Number Tree

VBODMAS Rule What do you think the answer to 2 + 3  5 would be? Is it (2 + 3)  5 = 5  5 = 25 or 2 + (3  5) = 2 + 15 = 17 The correct answer is 17; but why? When you perform a series of mathematical operations, there is a specific order in which you should perform the operations. This order is referred to as the VBODMAS Rule. The rule states that the operations should be done in the following order: – V stands for Vinculum – B stands for Bracket – O stands for Of

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Quantitative Aptitude Simplified for CAT

– D stands for Division – M stands for Multiplication – A stands for Addition – S stands for Subtraction Note that the operation ‘Of’ means multiplication only, but when ‘Of’ is written, then it is given higher priority than when multiplication symbol ‘’ is used. Example 18 Calculate 4 + 70  10  (1 + 2)2 – 1. Solution 4 + 70  10  (1 + 2)2 – 1 = 4 

70 70 2  3 1 = 4   9  1 = 4 + 7 9 – 1 = 4 + 63 – 1 = 67 – 1 = 66 10 10

If the order of precedence is not followed, one might land up with wrong answers. Note: There are three types of brackets: ( ) is parenthesis, { } is curly bracket and [ ] is square bracket, and as such there is no order of priority between these brackets. The preference order between Addition and Subtraction can be interchanged, that is, addition and subtraction can be done in any order as this will not alter the answers at all. For example, 4 + 7 – 2 = (4 + 7) – 2 = 11 – 2 = 9 OR 4 + 7 – 2 = 4 + (7 – 2) = 4 + 5 = 9.

Example 19 What is the value of 1  [1 + 1  {1 + 1  (1 + 1  3)}] A.

11 7

B.

7 11

C.

18 7

D.

7 18

Solution: B 1  [1 + 1  {1 + 1  (1 +

1 4 3 )}] = 1  [1 + 1  {1 + 1  }] = 1  [1 + 1  {1 + }] 3 3 4

7 4   11 7 = 1   1  1   = 1  1   = 1  = . 4 7 7 11  

Divisibility Rules In this section, we will discuss divisibility rules of all the important numbers. Divisibility rules of numbers help us to determine whether any given number is divisible by a certain number or not. For example, to check whether 31245768023546 is divisible by 4 or not, we need not divide the entire number by 4. We only need to check the number formed by unit’s and ten’s digit, that is, 46. If 46 is divisible by 4, then the entire number is divisible by 4. Since 46 is actually not divisible by 4, we can say that 31245768023546 is not divisible by 4.

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Number System – I: Factors, Multiples and Exponents

Divisibility rule of 2, 4, 8, 16 Divisibility rule of 2 If the unit’s digit of the number is 0, 2, 4, 6 or 8, then the number is divisible by 2. For example, 25842, 2540, 638 (unit’s digit is underlined) are divisible by 2 because unit’s digit of these numbers are 2, 0 and 8 respectively, whereas 321 and 4533 are not divisible by 2.

Divisibility rule of 4 If the number formed by the unit’s and ten’s digit of the number is 00 or divisible by 4, then the number is divisible by 4. For example, 4356, 12324, 1500, are all divisible by 4, because the number formed by last two digits (as underlined) are 56, 24 and 00 respectively, which are all divisible by 4. Numbers such as 3454, 43678, 321 are not divisible by 4.

Divisibility rule of 8 If the number formed by the last three digits of the number is 000 or divisible by 8, then the number is divisible by 8. For example, 546000 and 45432 are divisible by 8, whereas 43689 and 45678 are not divisible by 8. Note: Observe that the divisibility rule for 2 is determined by the last digit of the number, for 4 it is determined by the last 2 digits and for 8 it is determined by the last three digits of the number. We can say that the divisibility rule of 2n is determined by the last n digits of the number. Therefore, a number is divisible by 16, if the number formed by the last 4 digits is 0000 or divisible by 16. For example, 23548096 is divisible by 16 because 8096 is divisible by 16. Explanation Let us take the example of 342567. This can be written as 342560 + 7. Since 342560 is divisible by 2, the last digit 7 determines whether the number is divisible by 2 or not. Similarly, 342567 can be written as 342500 + 67. Since 342500 is divisible by 4, the number formed by last two digits 67 determines whether the number is divisible by 4 or not. Similarly, we can explain divisibility rule for 8 or 16 or in general 2n. Important The number 342567 can be written as 342500 + 67. For the number to be divisible by 4, the number 67 should be divisible. But, 67 is not divisible by 4 and upon division by 4 leaves a remainder 3. Therefore, 342567 when divided by 4 also leaves the remainder 3. From this, we can conclude that “the remainder when 342567 is divided by 4 is same as when 67 is divided by 4”. It means that applying divisibility rule of 4 provides information not only about whether the given number is divisible by 4 or not, but also on the remainder obtained when the given number is divided by 4. The same can be said regarding divisibility rule of any number. Example 20 Find the value of p such that (i)

234p48 is divisible by 8.

(ii)

43p456 is divisible by 8.

(iii) abcp4 is divisible by 4. Solution (i)

Applying the rule, only ‘p48’ needs to be checked. Putting the values of p (by hit and trial), we see that p can be 0, 2, 4, 6 or 8.

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Quantitative Aptitude Simplified for CAT

Alternatively, the number can be written as 234000 + p00 + 48. Since 234000 as well as 48 are divisible by 8, p00 also must be divisible by 8 and so p can be 0, 2, 4, 6 or 8. (ii)

Since ‘456’ is divisible by 8, p can take any value. Therefore, p can be anything from 0 to 9.

(iii) Applying the rule, only ‘p4’ needs to be checked. Again, by hit and trial, we see that p can be 0, 2, 4, 6 or 8. Example 21 What should be added to the number 342587 so that the number is divisible by 8? Solution We observe that 587 when divided by 8 leaves remainder 3. Therefore, we must add 5 to the number for it to be divisible by 8. Example 22 What least number must be added or subtracted so that the number 3457 is divisible by 8? Solution 457 = 400 + 56 + 1. Since ‘400 + 56’ is divisible by 8, if we subtract 1 or add 7, the number would become divisible by 8. Since the least number is asked, we should subtract 1 so that the number is divisible by 8. Example 23 What is the least number by which 12342 should be multiplied to make it divisible by 2  4  8? Solution We see that 12342 is divisible by 2 and not by 4 or 8. Therefore, to make it divisible by 4  8 as well, we need to multiply the given number by 4  8, that is, 32 or a multiple of 32. Since the least number is asked, the correct answer would be 32. Example 24 Natural numbers are written one after the other like: 12345678910111213... If this number contains 100 digits, what is the remainder when this is divided by 8? Solution The first 9 digits will be single digit numbers from 1 to 9. Then 10 onwards, each number contributes 2 digits. The next 90 digits would be obtained by 45 numbers (starting from 10). Starting from 10, first number is 10, second number is 11 and so on. Likewise, 45th number will be 54. So, by writing natural numbers from 1 till 54, we have written 99 digits. The 100th digit will be ten’s place digit of the next natural number 55. Therefore, the last three digits of the 100 digit number are 545, which when divided by 8 leaves a remainder of 1.

Divisibility rule of 3 and 9 Divisibility Rule of 3 If the sum of the digits of the given number is divisible by 3, then the number is divisible by 3. For example, 34257 is divisible by 3 because sum of the digits is 3 + 4 + 2 + 5 + 7 = 21, which is divisible by 3.

Divisibility Rule of 9 If the sum of the digits of the given number is divisible by 9, then the number is divisible by 9. For example, 31437 is divisible by 9 because sum of the digits is 3 + 1 + 4 + 3 + 7 = 18, which is divisible by 9.

14

Number System – I: Factors, Multiples and Exponents

Explanation Let us say we have a 6 digit number ‘abcdef’. Now, abcdef = a00000 + b0000 + c000 + d00 + e0 + f = a (100000) + b(10000) + c(1000) + d(100) + e(10) + f = [a (99999) + b(9999) + c(999) + d(99) + e(9)] + [a + b + c + d + e + f] Since [a (99999) + b (9999) + c (999) + d (99) + e (9)] is divisible by 3 as well as 9, we can say that only [a + b + c + d + e + f] needs to be checked for divisibility by 3 or 9. Hence the rule. Further, we can say that if the digit sum is not divisible by 3, then the remainder obtained upon division of the digit sum by 3 will be same as when the original number is divided by 3. Similarly for 9. Example 25 What should be a + b if 3ab544 is to be divisible by (i)

3

(ii)

9

Solution Adding the digits, we see that 3 + a + b + 5 + 4 + 4 = 16 + a + b. (i)

For the number to be divisible by 3, the sum a + b should be 2 or 5 or 8 or 11 or 14 or 17. The sum cannot exceed 18 as the largest value of either of a or b can be 9 and 9 + 9 is 18.

(ii)

For the number to be divisible by 9, the sum a + b should be 2 or 11. Once again, the sum cannot exceed 18.

Example 26 In the previous example 7, how many pairs (a, b) exist for the number to be divisible by 9? Solution From what has been discussed in the previous example, we see that a + b can be 2 or 11. When a + b = 2, then (a, b) can be (0, 2), (1, 1) or (2, 0). When the sum is 11, then (a, b) can be (2, 9), (3, 8), ...., (9, 2). Therefore, a total of 11 pairs of (a, b) can exist for the number to be divisible by 9. Note that (2, 9) is different from (9, 2) because 329544 is different from 392544. Example 27 What is the remainder when 324590 is divided by 9? Solution Digit sum is 23 which when divided by 9 leaves the remainder 5. Therefore, the remainder is 5. Note: While checking for divisibility of a number by 3, when we add the digits, we can ignore the digits which are 3 or multiples of 3 and also ignore those digit groups whose sum is 3. Similarly, while checking for divisibility by 9, when we add the digits, we can ignore the digit 9 whenever it appears and also ignore those digit groups whose sum is 9. Example 28 Find the remainder when 32452170989 is divided by 9.

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Quantitative Aptitude Simplified for CAT

Solution Digit sum of the given number = 3 + 2 + 4 + 5 + 2 + 1 + 7 + 0 + 9 + 8 + 9 = (3 + 2 + 4) + 5 + (2 + 7) + (1 + 8) + 9 + 9 Ignoring 9’s and digit groups whose sum is 9, we are left with 5 only, which is the digit sum of the given number. When the given number is divided by 9, the remainder is 5. Example 29 Find (a, b) if 459a78b is divisible by 4 as well as 9. Solution If the given number is divisible by 4, 8b should be divisible by 4. The possible values of b are 4 and 8. Also, the given number is divisible by 9. So, digit sum = 4 + 5 + 9 + a + 7 + 8 + b = 33 + a + b. When b = 4, a should be 8 so that the number becomes divisible by 9. When b = 8, a should be 4 so that the number becomes divisible by 9. So, there are two sets of value of a and b possible. Therefore, (a, b)  (4, 8), (8, 4).

Divisibility rule of 5, 25 and 125 Divisibility Rule of 5 If the unit’s digit is 0 or 5, then the number is divisible by 5. Therefore, numbers like 345, 5460 and so on are divisible by 5.

Divisibility Rule of 25 If the number formed by last two digits is divisible by 25, then the number is divisible by 25. Therefore, numbers like 325, 5450 and so on are divisible by 25.

Divisibility Rule of 125 If the number formed by last three digits is divisible by 125, then the number is divisible by 125. Therefore, numbers like 17625, 5250 and so on are divisible by 125. Note: In general, like in case of 2n, a number is divisible by 5n if the number formed by last n digits is divisible by 5n.

Divisibility rule of 10, 50 and 100 Divisibility Rule of 10 If the unit’s digit is 0, then the number is divisible by 10. Therefore, numbers like 360, 5400 and so on are divisible by 10. Note that if a number is divided by 5 and the remainder is 3, then we can surely say that the unit’s digit of the number is 3 or 8. In general, if upon division by 5, the remainder is n (where n < 5), then the unit’s digit of the number will be either n or 5 + n. Similarly, if a number is divided by 10 and the remainder is 3, the unit’s digit of the number will be 3. In general, if upon division by 10, the remainder is n (where n < 10), then the unit’s digit of the number will be n.

Divisibility Rule of 50 If the number formed by last two digits is 00 or 50, then the number is divisible by 50.

16

Number System – I: Factors, Multiples and Exponents

Divisibility Rule of 100 If last two digits of the number are 00, then the number is divisible by 100. Example 30 How many four-digit numbers, which are divisible by 25, can be formed by using the digits 2, 3, 5 and 7, using each digit only once? Solution The numbers that are divisible by 25 have 00 or 25 or 50 or 75 as their last two digits. Zero is not one of the four digits given in the question. Hence we only need to consider the numbers ending in 25 and 75. Numbers ending in 25: Four-digit numbers formed with last two digits as 25 and using 3 and 7 as the other two digits are 3725 and 7325. Numbers ending in 75: Similarly, four-digit numbers formed with last two digits as 75 and using 2 and 3 as the other two digits are 2375 and 3275. Hence four numbers, satisfying the given conditions, can be formed using the given digits.

Divisibility rules for composite numbers We will now study the divisibility rules for composite numbers like 6, 12, 15, 18 and so on. Now, 6 = 2  3, where 2 and 3 are co-prime. Then, we can say that a number is divisible by 6 if the number is also divisible by 2 and 3. Similarly, a certain number is divisible by 12, if the number is divisible by 3 and 4, where 3 and 4 are co-prime. Note that we cannot say that a number is divisible by 12 if it is divisible by 2 and 6. This is so because (2, 6) is not a co-prime pair of numbers. So, we see that the two numbers into which the composite number breaks should not have any common factor. We have seen in the previous chapter that such numbers are called co-prime numbers. We now have a general rule for composite numbers.

General rule for composite numbers If a number N is divisible by ‘a’ and ‘b’, where ‘a’ and ‘b’ are co-prime numbers, then the number N is also divisible by ‘a  b’. A number is divisible by 15 if the number is divisible by 3 and 5. For divisibility by 18, the co-prime pair would be (2, 9). For divisibility by 20, the co-prime pair would be (4, 5). For divisibility by 75, the co-prime pair would be (3, 25). Therefore, if the number is divisible by 3 as well as by 25, then the number will be divisible by 75. This way, we can generate divisibility rule for practically any composite number. Example 31 Is 1223334444 divisible by 48?

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Quantitative Aptitude Simplified for CAT

Solution The number should be divisible by 3 and 16. The digit sum is 30 which is divisible by 3. To check for 16, the number formed by last 4 digits should be checked. The number formed is 4444 which is not divisible by 16. Hence the number is not divisible by 48. Example 32 What is the remainder when 1223334444 is divided by 48? Solution The number is divisible by 3. So, remainder upon division by 3 is 0. The remainder when 1223334444 is divided by 16 is same as when 4444 is divided by 16. The remainder is 12. The required remainder will be that smallest number which when divided by 3 leaves remainder 0 and when divided by 16 leaves remainder 12. The smallest number that meets both the criterion is 12 itself. Example 33 To make the number 1223334444 divisible by 48, what should be: (i)

added

(ii)

subtracted

(iii) multiplied to the number? Solution In the previous example, we have seen that the number 1223334444 when divided by 48 leaves the remainder 12. (i)

Therefore, to this number, if we add 36 (because 12 + 36 = 48), the number would then become divisible by 48.

(ii)

If we subtract the 12, the number would become divisible by 48.

(iii) 1223334444 when divided by 48 leaves the remainder 12. When 12 is multiplied by 4, we get 48 which is divisible by 48. So, 1223334444 should be multiplied by 4 to make the resultant number divisible by 48. Example 34 What should be ‘a’ and ‘b’ if 345ab68 is divisible by 12, given that the number ‘ab’ is divisible by 3? Solution The number is divisible by 4. For 3, add the digits which comes out to 26 + a + b. So, a + b can take many values. If ‘ab’ is divisible by 3, then a + b should also be divisible by 3. But in that case 26 + a + b cannot be divisible by 3 as 26 is not divisible by 3. Therefore, no value of a and b would make the number divisible by 12. Example 35 Find the value of a if 254a6 is divisible by 12. Solution The given number should be divisible by 3 as well as 4. For divisibility by 4, the number a6 should be divisible by 4. For this, a can be 1, 3, 5, 7 or 9. For divisibility by 3, sum of digits = 17 + a should be divisible by 3. Therefore, possible values of a are 1, 4, 7. Therefore, a can be 1 or 7.

18

Number System – I: Factors, Multiples and Exponents

Divisibility rule for 11 If the difference of the sum of digits at odd places and sum of digits at even places is 0 or a multiple of 11, then the number is divisible by 11. Let us take an example. To check whether 432576 is divisible by 11 or not: The digits at odd places (from unit’s digit side) are 6, 5, and 3 whose sum is 14. Let us call this sum as S1. The digits at even places (from unit’s digit side) are 7, 2, and 4 whose sum is 13. Let us call this sum as S2. The difference of 14 and 13 is 1, which is neither zero nor a multiple of 11. Therefore, the number is not divisible by 11. So, we should check whether S1 – S2 is 0 or a multiple of 11. Let us take another number, 56432563. Here, S1 = 3 + 5 + 3 + 6 = 17, and S2 = 6 + 2 + 4 + 5 = 17. Since S1 – S2 = 0, the number is divisible by 11. Explanation Let us say we have a 6-digit number ‘abcdef’. Then, abcdef = a(100000) + b(10000) + c(1000) + d(100) + e(10) + f = a(100001) + b(9999) + c(1001) + d(99) + e(11) + (f – e + d – c + b – a) = a(100001) + b(9999) + c(1001) + d(99) + e(11) + [(f + d + b) – (a + c + e)] We can see that a(100001) + b(9999) + c(1001) + d(99) + e(11) is always divisible by 11. Therefore, only [(f + d + b) – (a + c + e)], which is effectively S1 – S2, needs to be checked for divisibility by 11. We can immediately see that [(f + d + b) – (a + c + e)] will contribute to the remainder when the number ‘abcdef’ is divided by 11. Therefore, in case of 432576, S1 – S2 = 1. So, when 432576 is divided by 11, remainder is 1. Example 36 What is the remainder when 564375 is divided by 11. Solution S1 = 6 + 3 + 5 = 14; S2 = 5 + 4 + 7 = 16. S1 – S2 = 14 – 16 = –2. Here, the difference is negative. In such cases, we add the divisor, that is, 11 to get a positive remainder. Therefore, 11 added to –2 gives 9, which is the remainder. Therefore, (S1 – S2) is the remainder whenever the number abcdef is divided by 11, whether S1 – S2 is positive or negative. Example 37 Find the remainder when 32146753 is divided by 11. Solution Here, S1 = 3 + 7 + 4 + 2 = 16; S2 = 5 + 6 + 1 + 3 = 15. Now, S1 – S2 = 16 – 15 = 1, which is not divisible by 11. So, 1 is the remainder when 32146753 is divided by 11. Example 38 Find the remainder when 23417673 is divided by 11.

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Quantitative Aptitude Simplified for CAT

Solution Here, S1 = 3 + 6 + 1 + 3 = 13; S2 = 7 + 7 + 4 + 2 = 20. Now, S1 – S2 = 13 – 20 = –7, which is not divisible by 11. In case we get negative remainder, we should add the divisor (11, in this case) to get the correct remainder. So, –7 + 11 = 4 is the remainder when 23417673 is divided by 11. Example 39 The positive number M39048458N is divisible by 8 and 11 where M and N are single-digit numbers. What are the values of M and N respectively? Solution A number is divisible by 8 if the last three digits of that number are divisible by 8. In this question the last three digits are 58N, and the only number between 580 and 589 that is divisible by 8 is 584  N = 4. For divisibility by 11, S1 – S2 = (M + 9 + 4 + 4 + 8) – (3 + 0 + 8 + 5 + 4) = M + 5 should be divisible by 11  M = 6. Note that M has to be a single digit number only.

Divisibility rule for 7 and 13 Let us understand the rule through an example. Example 40 Check whether 3542356765679863579095643 is divisible by 7 or not. If not, find the remainder. Solution From the unit’s digit side of the number, form groups of 3 digits as (003) (542) (356) (765) (679) (863) (579) (095) (643) From the unit’s digit side, add the alternate groups and call the sum S1. Similarly, add the remaining alternate groups and call the sum S2. So, S1 = 003 + 356 + 679 + 579 + 643 = 2260. S2 = 542 + 765 + 863 + 095 = 2265. S1 – S2 = 2260 – 2265 = –5. If S1 – S2 = 0 or a multiple of 7, then the number is divisible by 7, otherwise not. Here, since the difference is –5, the number is not divisible by 7. Moreover, like in case of divisibility of 11, here also we should add the divisor 7 to get the remainder. Therefore, remainder = –5 + 7 = 2. Explanation Let us say we have a 6 digit number ‘abcdef’. Now abcdef = abc000 + def (quite like 324567 = 324000 + 567) = (abc) (1000) + (def) = (abc) (1001) + [(def) – (abc)]. Now, 1001 = 7  11  13. Therefore, for any value of ‘abc’, abc  1001 will always be divisible by 7.

20

Number System – I: Factors, Multiples and Exponents Therefore, [(def) – (abc)], which is effectively S1 – S2, needs to be checked. Also, S1 – S2 will contribute to the remainder, if not divisible by 7. From the above rule for 7, we can also say that the same rule can be applied for 11 as well as for 13. This is so because 1001 is divisible by 11 and 13 also.

General divisibility rule for 11 In any given number, form groups of ‘n’ digits from unit’s digit side, where ‘n’ is odd. Add alternate groups starting from unit’s digit side and call the sum S1. Add the remaining groups and call the sum S2. If S1 – S2 is 0 or a multiple of 11, then the number is divisible by 11, otherwise not. If S1 – S2 is not divisible by 11, then S1 – S2 will contribute to the remainder. Example 41 Check whether the number 576324589706 is divisible by 11 or not? Solution Rule 1: In the number 576324589706, add the digits at odd places. Therefore, S1 = 7 + 3 + 4 + 8 + 7 + 6 = 35. Add the digits at even places: S2 = 5 + 6 + 2 + 5 + 9 + 0 = 27. S1 – S2 = 35 – 27 = 8, which is not divisible by 11. Rule 2: From the right side, form groups of 3 digits. 576324589706 = (576) (324) (589) (706). S1 = (324) + (706) = 1030; S2 = (576) + (589) = 1165. S1 – S2 = 1030 – 1165 = –135, which is not divisible by 11. Rule 3: From the right side, form groups of 5 digits. 576324589706 = (00057) (63245) (89706). S1 = (00057) + (89706) = 89763; S2 = 63245 S1 – S2 = 89763 – 63245 = 26518, which is not divisible by 11. Note that when the given number is divided by 11, the remainder would be 8.

Special rule Let a digit ‘p’ repeat 6 times. We obtain ‘pppppp’. Now, pppppp = ppp(1001) = p(111)(1001) = p(3  37)(7  11  13). Therefore, if a digit repeats 6 times, the number so formed will be always divisible by 3, 7, 11, 13 and 37. In fact, if the digit ‘p’ repeats ‘6 times’ or ‘a multiple of 6 times’, the resulting number will be always divisible by the same prime numbers, viz., 3, 7, 11, 13 and 37. Therefore, if ‘p’ repeats 12 times, 18 times or 36 times, the number would be divisible by the said prime numbers.

21

Quantitative Aptitude Simplified for CAT

Example 42 The digit 7 repeats 77 times to form a 77 digit number. When this number is divided by 13, what is the remainder? Solution The number = (7777.................) = (7777.................7700000) + 77777 (77 times)

(7 appears 72 times)

Now, the first number is divisible by 13 because the digit 7 appears 72 times, where 72 is a multiple of 6. Therefore, the remainder will be contributed by 77777. To find remainder, we can form groups of 3 digits from right side in the number 77777. Therefore, S1 – S2 = 777 – 77 = 700, which when divided by 13 leaves the remainder 11. Example 43 Find the value of a if 4326a1567321 is divisible by 13. Solution Write the number as 432 6a1 567 321. S1 = 321 + 6a1; S2 = 567 + 432 = 999. Now, S1 – S2 = 321 + 6a1 – 999 = 6a1 – 678, which should be divisible by 13. So, 6a1 – 678 = (600 + a0 + 1) – (600 + 70 + 8) = a0 + 1 – 78. Here, 78 is divisible by 13. So, a0 + 1 = a1 should be divisible by 13 and therefore a = 9.

Divisibility rule for 27 and 37 Let there be a number 423567. From the unit’s digit side, form groups of 3. Add all these groups. If this sum is divisible by 27, then the number is divisible by 27, otherwise not. Here, we have 423 + 567 = 990, which is not divisible by 27. Hence the number is not divisible by 27. The rule of divisibility for 37 is similar to that for 27. Since 990 is not divisible by 37, the given number is not divisible by 37. Explanation Let there be a number ‘abcdef’. Then, abcdef = abc(1000) + def = abc(999) + (abc + def). We see that 999 = 27  37. Hence the rule. In general, if a digit ‘p’ repeats 3 times, we get ‘ppp’ which is always divisible by 3 and 37. Example 44 Check whether 111222333444 is divisible by 37 or not? Solution Applying the rule, we get 111 + 222 + 333 + 444. We see that each group is divisible by 37, therefore, the sum would also be divisible by 37. Example 45 abcabcabcabc is divisible by A.

7

B.

11

22

Number System – I: Factors, Multiples and Exponents C.

13

D.

all of these

Solution: D Grouping the digits results in (abc – abc + abc – abc) = 0, which means that the number is divisible by 7, 11 as well as 13.

Surds and Indices We will now focus our attention on the laws governing surds and indices.

Surds Surds are those numbers which can be expressed as number such that

n

a

Is

3 4

n

a

, where ’a’ is a rational number and n is any real

is not reducible to a rational number. For example, 2, 3 5, 5 12 and so on.

1

In general,

n

1

a  a n . Therefore,

2  22 ;

1 3

5  53 ;

1 5

12  12 5

2 a surd? On the face of it, it seems that a =

surd. This is not true as surd.

3 4

2 can be simplified as

4

12

2

2

and this being irrational, the given number is not a and now 2 being a rational number, the number is a

Note: All surds are irrational numbers, though not all irrational numbers are surds. For example, 3 12 is a surd and also an irrational number, whereas log102 is an irrational number but not a surd. Same goes for , which is irrational but not a surd. Therefore, the set of all surds is a subset of all irrational numbers, but vice versa is not necessarily true.

LAWS OF SURDS Certain laws govern surds and now we will list those laws, all of which can be easily derived using the fact that 1 n

a  a n . In all these rules, please bear in mind that m, n and a are such that none of the expressions reduces

to a rational number, because in that case, the result would no more be a surd. The fact that the expressions reduce to a rational number does not in any way mean that the law would not be applicable. 1

(a)

n m

(b)

n

am 

(c)

n

ab 

(d)

n

a na  b nb

a  m n a  mn a  a mn n m

n

m

a  an anb

Example 46 Simplify: 5

(i)

8 64  8 3 3 27

23

Quantitative Aptitude Simplified for CAT

(ii)

       

 

 a

a3b4c b3c4

3 5 4 7

a



1

 b 4 /3  4

1 /3

2 5 2 3 5  2 c  3 b  3 a   

       

1

Solution: (i) The given expression can be rewritten as 3 5

8 3 27   8 3 64

5

23  3 3 3

8

3  6 22

3 1  3

25  3



1

1

 25

1

3

1 8

4

7

 215  3 8 or

15 4

8

273.

38  23

(ii) The given expression    =     

 

 a

a3b4c b3c4

1 3 3 1    5 10 12

= a7

2

 b7



3 5 4 7

a



1 / 3

 b 4 /3 

1 4 2

5 2 3  2 c  3 b  3 

6 1 1   25 5 3

4

 c 21



5 a  

4 3  15 20

       

1

2



3

4

b c a







221

a3b4c

346

4 7

5 2 3 5   2 c  3 b  3 a     3 5

  a

1 1 / 3

 b 4 /3  4

129

 a 420  b 525  c 420

Square and square root of binomial surd A binomial surd of the type a +



For example, 2  3



2

 22 

b

 3

2

or

a b

can be squared using the basic rule (a + b)2 = a2 + 2ab + b2.

22 3  74 3 .

To find the square root of a binomial surd, we have to reverse the process of squaring. Let us understand the process with the help of the following example. Example 47 Find the square root of 21 + 8 5 . Solution First, ensure that the coefficient of the square root term is 2. So, 21 + 8 5 = 21 + 2  4 5 = 21 + 2 80 This ensures that the coefficient of square root term becomes 2. Next, search for two numbers whose sum is 21 and whose product is 80. The two such numbers are 16 and 5. Therefore, 21  2 80  16  5  4  5 . Example 48 Simplify:

51  14 2  51  14 2

.

51  14 2  51  14 2

Solution Proceeding as above for finding the root of a surd,

24

Number System – I: Factors, Multiples and Exponents 51  14 2  51  2 2  49  49  2  7  2 . Similarly, 51  14 2  51  14 2

Therefore,

51  14 2  51  14 2



51  14 2  7  2 .

 7  2    7  2   14  7  2   7  2  2 2

7 7 2  2 2

Example 49 Find the square root of 9 + 2  6  12  2 2  . Solution Since there are three root terms, we can conjecture that the root of the given surd should be of type.



a b c

2  a  b  c  2

ab  bc  ca

a b c



The given surd should be written as 9 + 2  6  12  8  . Comparing the two, we observe that a = 2, b = 3 and c = 3. 92

Therefore,





6  12  8  2  3  4  2  2  3

Conjugate of a binomial surd and rationalisation A binomial surd is a surd which has two terms, at least one of which is a pure surd. For example,

2  5, 11  2 13, 2  3 5 and so on.

Changing the sign of the coefficient of pure surd term is called the conjugate of the given surd. Therefore, conjugate of

Similarly, conjugate of

2

5

is either 2  5 or 5  2 (as both the terms in this surd are a pure surd).

11  2 13  11  2 13

or

2 13  11

.

If we multiply a surd by its conjugate, the result is a rational number. This fact is used in the process of rationalization. Rationalization means to convert a surd to a rational number. The surd with which the given surd is multiplied is called rationalizing factor. So, rationalizing factor of 3  7 is 3  7 and that of

 3  7  3  7 

5

is

5

, as

= 9  7 = 2 (a rational number), and 5  5  5 (a rational number).

Example 50 Rationalize

2 3 . 4  15

Solution Rationalization of the given expression requires conversion of the denominator to a rational number. This is done by multiplying and dividing the given expression by the conjugate of the denominator.  2  3   4  15  2 3    4  15  4  15   4  15 

(Multiplying and dividing the expression by the same surd does not change the value)

25

Quantitative Aptitude Simplified for CAT

 = 

 2 3

 4 



15   = 8  2 15  4 3  3 5 .  16  15 

 

Example 51 11  4 7  16  6 7

Simplify

2 3  5

.

Solution 11  4 7  2  7 and

16  6 7  3  7







5 2 3  5 11  4 7  16  6 7 2  7  3  7   2 3  5 2 3  5 2 3  5 2 3  5

Therefore,









 10  5 3  5 5  2  4 3  = 10  5 3 2 5 5  10  5 3  5 5  2  3 

=

24 3

5

 2  4 3  2  4 3 

20  40 3  10 3  60  10 5  20 15 40  30 3  10 5  20 15  4  48 44

In the denominator, we could have taken 2 +









3  5 and rationalized further instead of 2  3  5 . The

answer in either case will be the same. Example 52 1 1 2  13  2   2  33   23  6 3  33     Simplify . 1 1 2 2 2 1 1 1 1  13    3 3 3 3 3 3 3 3 3  2  3  5   2  3  5  6  15  10   3  30   

Solution 1

1

1

Let a = 23 , b = 3 3 and c = 5 3 . Then, the expression becomes (a  b)  a2  ab  b2  (a  b  c)  a2  b2  c 2  ab  bc  ca   3abc

=



a3  b3 a  b3  c 3 3

2 3 1  . 235 2

Here, we have used the standard results given in Chapter 7: Polynomials, Algebraic Formulae and Linear Equations. Example 53 Arrange the following surds in ascending order: (i)

3, 3 4, 4 5

(ii)

3  7, 2  6, 5  5

(iii)

4

5, 5 6, 6 7

26

Number System – I: Factors, Multiples and Exponents Solution (i) The given surds are 31/2, 41/3, 51/4. The denominators in the powers are 2, 3 and 4 whose LCM is 12. Raising each surd to the power 12 will eliminate the fractional powers. We get 36, 44 and 53. So, we will now compare 36, 44 and 53, whose values are 729, 256 and 125. Naturally, the largest is 36 and smallest is 53. Therefore, the surds in ascending order are 4 5, 3 4, 3 . (ii) Squaring all the given surds, we get



3 7

2  6  



2

 3  7  2 21  10  2 21

2

5 5

 4  6  2 24  10  2 24



2

 5  5  2 25  10  2 25

Now the surds can be easily compared and the surds in ascending order are

3  7, 2  6, 5  5 .

(iii) Here, if we proceed as in (i), LCM of 4, 5 and 6 would be 60. Raising each surd to such a high power will make it difficult to calculate. We will compare the surds in pairwise manner. First, let us compare

45

5

and

6.

LCM of 4 and 5 is 20. Raising both the surds to the power 20, we get 55 and 64. Now, 55 = 3125 and 64 = 1296. So, 64 < 55  5 6  4 5 Let us now compare

5

6

6 and

(1)

7.

LCM of 5 and 6 is 30 and proceeding as above, we get 66 and 75. ------- (2)

66 = 46656 and 75 = 16807. So, 75 < 66  6 7  5 6 From Eq. (1) and (2), we get:

6

7

5

6

4

5

.

Indices The number an consists of two variables: a and n. The variable ‘a’ is called base and ‘n’ is called the exponent or index or power. The laws governing these are given below. (a)

(am)n = (an)m = amn

(b)

am+n =(am)(an)

(c)

m am–n = a n

a

Therefore, if m = n, then am–m =

am  a0 = 1. But this is true for all values of ‘a’ except when am

a = 0 or  (as the denominator can’t be 0 or ). Note that 00 and 0 are indeterminate forms. (d)

(ab)m = (am)(bm)

(e)

 a    b

(f)

a b   ab 

c

m



am bm c

27

Quantitative Aptitude Simplified for CAT

Example 54  a4 b7 

 a 2 b 5 

2

Simplify  3 5    . 3   a b   (ab)  Solution  a4 b7   a2 b5   3 5    3   a b   (ab) 

2

 a4 b7 

 a 2 b 5 

2

=  3 5    = a4+3–4–6b7+5+10–6 = a–3b16. 3   a b   (ab) 

If there are two distinct natural numbers a and b and if a < b, then ab > ba provided a, b  3. For example, 34 is more than 43, as 3 < 4. In other words, if a and b are both more than or equal to 3, then (smaller number)larger number > (larger number)smaller number Example 55 Which of the two is bigger: 2327 and 2723? Solution Since 23 is less than 27, we can say that 2327 > 2723. Example 56 Arrange the following in increasing order: (i)

7600, 3800, 21000

(ii)

712, 811, 910, 109

Solution (i) 7600 = (76)100 3800 = (38)100 21000 = (210)100 Comparing the given three terms is equivalent to comparing 76, 38 and 210 only. Now, 76 = 493 which is close to 503 which is 125000. 38 = 812 = which is close to 802 which is 6400. 210 = 322 = 1024. Since the values (obtained by approximation) are far and wide enough, we can safely say that 210 < 38 < 76. Hence, 21000 < 3800 < 7600. (ii) 712 = 496 = 24013. Similarly, 811 = 233 = (230)(23) = (10243)(8) = 20483. Therefore, 712 > 811. Further, 910 = 320 = (36)3(32) = (729)3(3)2  (729)3(2)3 = 14583. Therefore, 811 > 910. From the result that ab > ba if a < b (where a, b  3), we can surely say that 910 > 109. Hence, 712 > 811 > 910 > 109. Example 57 2

Find the value of 2 3 .

28

Number System – I: Factors, Multiples and Exponents Solution 2

If we find 23 first, then we get 8. Then, 2 3 = 82 = 64. 2

If we find the value of power, which is 32 = 9, then 2 3 = 29 = 512. The latter method is correct. Simply because, in the question, 2 is raised to the power 32 and not that 23 is raised to the power 2. For, if 23 were raised to the power 2, that would be written as (23)2 which is definitely not 32

equal to 2

. c

In general, we can say that ab  (ab) c. We also know that (ab)c = abc (by law of indices). Example 58 Using the digit 3 four times, how many different numbers can be written without using the symbols of +, –, , and brackets? Write all those numbers in ascending order and hence identify the largest of them. What if the digits to be used are (i)

1

(ii)

2

(iii) 4 (The students must arrive at all these ascending orders themselves for a very good clarity of the concepts of indices and their laws) Solution 3

3

33

33

Following numbers are possible: 3333, 3333, 3333, 333 , 3333, 333 , 33 , 33 . To find the largest number, we have to compare them pairwise. Out of 3333 and 3333, we can easily conjecture that 3333 is more than 3333. Further, 3333 = 33  3332 = 33  (332)16 = 33  (1089)16, which is obviously more than 3333. 3

3333 is more than 333 because base is same and the power in first case is 33 and in the second case is 33 which is 27, which is less than the power 33. Comparing 3333 and 3333, we can say that 3333 < 8133 = 3132, which is surely less than 3333. 3

Out of 3333 and 333 , only the powers need to be compared. The powers are 333 and 333. 3

332 = 1089 which is more than 333, so naturally 333 will be more than 333. Hence, 333 is larger than 3333. 3

33

333 is less than 813, which is equal to 312, which is less than 333. Hence, 333 < 33 . 33

Finally, out of 33

33

33

and 33 , we can write 33

33

3

3

27

27

33

as 33 and so 33 < 33 . 33

3333 < 3333 < 333 < 3333 < 3333 < 333  33  33 33

The largest number is 33 . (i) If the digit to be used was 1, the possible numbers are 1

1

11

11

1111, 1111, 1111, 111 , 1111, 111 , 11 , 11

Proceeding on the lines of previous case, we come to the following order: 1

11

11

1111 = 111  11  11

1

< 111 < 1111 < 1111 < 1111

The value of each of the first 4 terms is 1.

29

Quantitative Aptitude Simplified for CAT

(ii) If the digit to be used was 2, then the numbers are 2

2

22

22

2222, 2222, 2222, 222 , 2222, 222 , 22 , 22

22

2

2

The increasing order is: 2222 < 2222 < 22 < 222 < 2222 < 2222 < 222  2 2

22

(iii) If the digit to be used was 4, then the numbers are 4

4

44

4444, 4444, 4444, 44 4 , 4444, 444 , 4 4 , 4 4

44

4

4

44

44

The increasing order is: 4444 < 4444 < 4444 < 4444 < 444  444  44  44

Factors of Composite Numbers In this section, we will gain an understanding of concepts pertaining to factors, multiples and factorization process. This will help us understand all about least common multiple, that is LCM and highest common factor, that is HCF. HCF is also called GCD (greatest common divisor). We have seen earlier that composite numbers are those numbers which have more than or equal to 3 distinct factors, For example, 18, 24, 28, 343, and so on are all examples of composite numbers.

Factors Factors of a number are those numbers by which the given number is divisible. For example, 2 is a factor of 18 because 18 is divisible by 2. Here we can see that 3 is also a factor of 18. Let us list out all the factors of 18. They are: 1, 2, 3, 6, 9 and 18. So, we see that the number is a factor of itself. Moreover, it is not hard to understand that 1 is a factor of all natural numbers. Similarly, factors of 12 are: 1, 2, 3, 4, 6 and 12.

Unique factorization theorem According to this theorem, every composite number can be uniquely written in terms of its prime factors.

For example, 18 = 2  3  3 = 2  32 Note: Every composite number can be expressed as a product of its prime factors in a unique way. We can easily determine the factors of the following numbers: 18 = 1, 2, 3, 6, 9, 18 27 = 1, 3, 9, 27 98 = 1, 2, 7, 14, 49, 98 Note that it is easy to count the number of factors of simple composite numbers like 18, 27 and so on by actually writing all the factors and then counting them. But for big numbers like 244, 376, and so on, the process of counting in this manner becomes very tedious. The principles of mathematics help us in counting the number of factors of any composite number without actually writing the factors!

How to find number of factors of a composite number Let C be a composite number which can be prime- factorized and written as below: C = am  bn  cp  ... where a, b, c, ... are prime factors and m, n, p, ...are positive integers. Then, the number of factors N is given by the following formula:

30

Number System – I: Factors, Multiples and Exponents N = (m + 1)(n + 1)(p + 1)...... Therefore, to count the number of factors of 18, we need to prime-factorize it as 18 = 2  32 = 21  32. Here, m = 1 and n = 2. Therefore, the number of factors, N = (m + 1) (n + 1) = (1 + 1) (2 + 1) = 6. Let us try this for bigger numbers like 244. Now, 244 = 4  61 = 22  611. Therefore, the number of factors, N = (2 + 1) (1 + 1) = 3  2 = 6. Example 59 Find the number of factors of 48. Solution 48 = 24  31. Therefore, the number of factors, N = (4 + 1) (1 + 1) = 5  2 = 10. Example 60 Find the number of factors of y = 25  68  73. Solution Here, students would be inclined to do this: m = 5, n = 8 and p = 3. Therefore, N = 6  9  4 = 216. But, this is incorrect as the number needs to be prime-factorized first! Therefore, y = 25  68  73 = 213  38  73. Number of factors, N = 14  9  4 = 504. Of course, we should not write the factors of y and then count them! Explanation Let us take an example to understand the logic behind the rule for finding the number of factors of a composite number. Let us consider the number 12. 12 can be prime factorised as 22  31. Sub-factors of 22 are 20, 21 and 22. Similarly, sub-factors of 31 are 30 and 31. Factors of 12 are 1, 2, 3, 4, 6, and 12. Now, 1 = 20  30 2 = 21  30 3 = 20  31 4 = 22  30 6 = 21  31 12 = 22  31 We can see that the factors of a composite number are nothing but associations of the sub-factors of 22 with sub-factors of 31. In other words, association of 20 with 30 gives us 20  30 = 1. Similarly, association of 21 with 31 gives us 21  31 = 6. We can develop the following associations.

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Quantitative Aptitude Simplified for CAT

Therefore, the line connecting 20 with 31 means 20 multiplies with 31 which gives 3. We see that each subfactor of 22 associates with every -sub-factor of 31 to create a factor of 12. The number of associations is, therefore, the number of factors of the number 12. Since there are 3  2 = 6 associations, we can say that it has 6 factors. Observe that there are 3 sub-factors of 22, that is, 20, 21, 22. In general, if we have am, then there will be (m + 1) sub-factors of am, that is, a0, a1, a2, a3,..., am. Similarly, in bn, there will be (n + 1) sub-factors. Hence, the number of associations between am, bn, cp, and so on, will be (m + 1) (n + 1)(p + 1).... and hence the result. So, the idea is to count the number of associations because each association will yield each factor. Suppose we want to count those factors of 23  34  73, which are odd numbers. Once again, we make use of ‘associations’. To create an odd factor, the power of 2 in the sub-factors should be zero, because 20 = 1 which when associated with any sub-factor of 34 and/or 73 would always yield an odd factor. The following network diagram gives us the total number of factors.

= 80 Factors We only want odd factors. Therefore, any association with 21 or 22 or 23 is forbidden. So, total number of desired associations is given by the following network diagram (left diagram):

Number of Odd Factors = 20

Number of Even Factors = 60

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Number System – I: Factors, Multiples and Exponents Therefore, number of factors which are odd = 1  5  4 = 20. To find number of even factors, refer to the above network diagram (right diagram) Therefore, number of even factors = 3  5  4 = 60. Important Learning All perfect squares have odd number of factors. Converse is also true, viz., if a number has odd number of factors, then it must be a perfect square. For example, 4 has three factors: 1, 2, 4. 9 has three factors: 1, 3, 9. 16 has five factors: 1, 2, 4, 8, 16. 36 has nine factors: 1, 2, 3, 4, 6, 9, 12, 18, 36. In these examples, we can see that the “number of factors” is odd. Explanation If C is a composite number, then it can be written as C = am  bn  cp  ... If C is a perfect square, then each of the power m, n, p, ...and so on must be an even number. Then, number of factors, N = (m + 1) (n + 1) (p + 1)....... We observe that each of the terms (m + 1), (n + 1), (p + 1)...... is odd and hence, the number of factors, that is, N has to be odd. Therefore, we obtain the said result. Of course, the same logic can explain the converse of the result. Example 61 Find the number of factors of the following perfect squares: (i)

332

(ii)

24  92

(iii) 2401 Solution (i)

332 = 32  112. Therefore, number of factors = 3  3 = 9.

(ii)

Number of factors of 24  92 is 5  3 = 15.

(iii) 2401 = 74. Therefore, it has 5 factors. Example 62 For the number x = 23  35  57, (a)

Find the number of factors.

(b)

Find the number of factors which are even, and also the number of factors which are odd numbers.

(c)

Find the number of factors which have 2 factors. Also find the number of factors which have 3 factors.

(d)

How many factors are there which have 4 factors?

(e)

How many factors have even number of factors?

(f)

Find the number of factors which are perfect cubes.

(g)

Find the number of factors which are divisible by 4 but not by 81.

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Solution (a)

Number of factors = 4  6  8 = 192

(b)

To find even factors, we will not consider the sub-factor 20 of the factor 23. Therefore, number of associations = 3  6  8 = 144 Number of odd factors = 1  6  8 = 48

(c)

Note that prime numbers have 2 factors! Therefore, there are 3 factors which have 2 factors, viz., 2, 3 and 5. Numbers which are p2 type will always have 3 factors, where ‘p’ is a prime number. For example, 22 has 3 factors, 32 has 3 factors and similarly, 52 has 3 factors. Therefore, there are 3 factors which have 3 factors.

(d)

Numbers which are p3 type will have 4 factors. But, number like p  q types will also have 4 factors. Numbers of p3 type are 23, 33 and 53. Numbers of p  q type are 2  3, 3  5 and 2  5. Therefore, there are 6 factors of x which have 4 factors.

(e)

To find factors having even number of factors, let us find those factors which have odd number of factors, which we know are perfect squares. To find perfect square factors, we need to associate perfect square sub-factors of the number only.

= 24 Factors See the above network diagram. The number of factors which are perfect squares are 2  3  4 = 24 Therefore, there are 24 factors which have odd number of factors. Therefore, there are 192 – 24 = 168 factors which have even number of factors. (f)

To find perfect cube factors, the sub-factors should be perfect cubes themselves. The number of perfect cubes are = 2  2  3 = 12.

(g)

If factors are to be divisible by 4, we should take the sub-factors 22 and 23. Similarly, if the factors are not to be divisible by 81, we should not take 34 and 35. So, the required number of factors = 2 × 4 × 8 = 64.

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Number System – I: Factors, Multiples and Exponents

Number of ways in which a composite number can be expressed as a product of two factors Let us take a composite number 24. Write all its factors. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24 Now, 24 can be expressed as product of 2 factors in following ways: 1  24; 2  12; 3  8; 4  6. We observe that this is nothing but forming pairs of factors. Therefore, the number of ways in which a composite number could be expressed as a product of two factors would be number of pairs of factors, that is, half of the number of factors! So, if we have a composite number, C = am  bn  cp  ..., then the number of ways in which it can be expressed as a product of two factors is given by

1 [(m + 1) (n + 1) (p + 1)....] 2

What will happen in case of “perfect square composite numbers” where there are odd number of factors and hence we cannot halve the number of factors? In cases of perfect squares, if N is the number of factors, then number of ways is given by

1 (N + 1). 2

Let us take an example. Example 63 In how many ways can we express 36 as a product of two factors? Solution Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36. Following are the ways in which 36 can be expressed as desired: 1  36; 2  18; 3  12; 4  9; 6  6. Therefore, number of ways =

1 1 (N + 1) = (9 + 1) = 5. 2 2

Number of ways in which a composite number can be expressed as a product of two co-prime factors In this case, the number of ways in which a composite number can be expressed as a product of two co-prime factors would be 2m – 1, where m is the number of prime factors of the given number. Example 64 With the help of previous example, elucidate the statement that the number of ways in which a composite number can be expressed as a product of two co-prime factors is 2m1, Solution For the number 36, we have seen that the total number of ways is given below: 1  36; 2  18; 3  12; 4  9; 6  6. Of these, only 1  36 and 4  9 are the cases where the two factors are co-prime. Using the formula as well, we can see that since 36 = 22  32, there are 2 prime factors and hence m = 2. Therefore, number of ways = 22 1 = 2.

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How to find sum and product of factors of a composite number Let C be a composite number which can be prime-factorized and written as C = am  bn  cp  ... Then, the sum of the factors of C is given by the following formula:  am1  1   bn1  1   c p 1  1 

Sum =     ...  a1   b1   c 1  1

N

Product = C 2 , where C is the composite number and N is the number of factors of C. Example 65 Find the sum and product of the factors of 18. Solution Factors of 18 are 1, 2, 3, 6, 9, and 18. Sum = 1 + 2 + 3 + 6 + 9 + 18 = 39.  22  1   33  1 

Using the formula, we get =    = 39.  2 1   3 1  Product of factors = 18(6/2) = 183. Example 66 Find the sum and product of the factors of x = 23  35  57. Solution  24  1   3 6  1   5 8  1    .  2 1   3 1   5 1 

The sum = 

Number of factors of x = (3 + 1) (5 + 1)(7 + 1) = 192 Product of factors = (23  35  57)192/2 = (23  35  57)96. Example 67 Find the sum of factors of 28. Solution The required sum = 1 + 2 + 4 + 7 + 14 + 28 = 56. In the last example, the summation is twice the number. Such numbers whose sum of factors is twice the number itself are called perfect numbers. In other words, if we add all the factors of the perfect number excepting the number itself, we would get the original number. Smallest perfect number is 6, next is 28, and so on. Example 68 Let a1, a2, a3, ... an are factors of P, a perfect number. Find the value of

1 1 1 1    ... . a1 a2 a3 an

Solution a1 + a2 + a3 + ... + an = 2P. Dividing both sides by P, we get

36

a 1 a1 a2 a3 a    ... n = 2. But, 1  , because a1 P an P P P P

Number System – I: Factors, Multiples and Exponents

= 1 and P = an. Similarly,

1 1 1 1 1 a2     ... and likewise. Therefore, we get = 2. P an1 a1 a2 a3 an

Note that this is a standard result for all perfect numbers! Example 69 The product of factors of a composite number is 29162. Find the sum of factors of the composite number. Solution 1

N

Product = 29162 = C 2 . If the composite number C is 2916 = 4 × 729 = 22 × 36, then number of factors = 3 × 9 = 27, which does not match with the given information. Now, 2916 = 542 and so product of factors can also be written as 544. If the composite number C is 54 = 2 × 27 = 2 × 33, then number of factors = 2 × 4 = 8 and so product of factors will be 54 8/2 = 544, which matches  22  1   34  1 

with the given information. So, the composite number is 54 and its sum of factors =    =3×  2 1   3 1  40 = 120.

LCM and HCF Let us say we have two numbers 12 and 18. Write some multiples of 12 and 18 as below. 12:

12,

24,

36,

48,

60,

72,

18:

18,

36,

54,

72,

90,

...

84,

...

Write only those multiples of these numbers which are common to them. Therefore, common multiples of 12 and 18 are: 36, 72, 108, and so on. The least amongst these is 36, so 36 is the Least Common Multiple (LCM) of 12 and 18. Similarly, to learn about HCF, write all the factors of 12 and 18 as below. 12:

1,

2,

3,

4,

6,

12

18:

1,

2,

3,

6,

9,

18

Write only the common factors of 12 and 18. So, the common factors of 12 and 18 are: 1, 2, 3 and 6. The highest amongst these is 6, and so 6 is the Highest Common Factor (HCF) of 12 and 18. Observe that 6 is the highest of the common factors of 12 and 18. Hence, we can say that 6 is the highest common factor, that is HCF (or GCD) of 12 and 18. This method of finding LCM or HCF is tedious for big numbers and hence we must learn better methods of finding the LCM or HCF. Still, the above method clarifies the basic meaning of the terms LCM and HCF. From above, we can conclusively say that LCM of numbers is always a multiple of HCF of the same numbers. For example, LCM and HCF of 12 and 18 are 36 and 6, and we know that 36 is a multiple of 6.

Various methods of finding the LCM Let us find the LCM of 12, 14, 18 and 21. Method 1: Divide these numbers by factors successively till all the numbers reduce to 1.

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Quantitative Aptitude Simplified for CAT

The divisors used are multiplied to get the LCM. Therefore, LCM = 2  2  3  3  7 = 252. Method 2: This method is known as Factorization Method. In this method, we factorize the given numbers and highest power of every prime factor appearing across all the numbers is taken. 12 = 22  3 = 22  31  70 14 = 2  7 = 21  30  71 18 = 2  32 = 21  32  70 21 = 3  7 = 20  31  71 Taking the highest powers of 2, 3 and 7, we get LCM = 22  32  71 = 252. Example 70 Find the LCM of 23  512  75 and 32  53  79 112. Solution 23  512  75

= 23  30  512  75  110.

32  53  79  112

= 20  32  53  79  112.

Therefore, LCM

= 23  32  512  79  112.

Example 71 Find the LCM of 1212  93  105 and 610  97  208. Solution First of all, prime factorize the numbers as shown below: 1212  93  105 = 2(2×12)+5  312+(2×3)  55 = 229  318  55. 610  97  208 = 210+(2×8)  310+(2×7)  58 = 226  324  58. LCM = 226  324  58. Method 3: In this method, we use the rule that LCM of the numbers given is either the biggest of those numbers or a multiple of this biggest number. So, LCM of 12 and 18 is either 18 or a multiple of 18. Let us find the LCM of 12 and 18 by this method. Start with the assumption that 18 is the LCM of 12 and 18. But for this to be true, 18 should be divisible by 12, which is not the case. So, we check the next multiple of 18, which is 36 and which is divisible by 12. So, 36 is the LCM of 12 and 18. Similarly, we can find the LCM of 12, 14, 18, 21. Start with the assumption that 21 is the LCM of the given numbers. Since 21 is not divisible by the given numbers, we check the next multiple of 21. But, checking each multiple can be tedious. We realize that 21 consists of one 3 and one 7. The next number 18 consists of one 2 and two 3’s. So, 21 should be multiplied by those factors which are missing in 21 but available in 18. We get 21 × 3 × 2. The next number 14 consists of

38

Number System – I: Factors, Multiples and Exponents one 2 and one 7, both of which are available in 21  3  2. Now, 12 consists of two 2’s and one 3. Since 21  3  2 already consists of one 2 and two 3, and so the deficiency of an extra 2 needs to be provisioned for. Therefore, LCM of the given numbers = 21  3  2  2 = 252.

Various methods of finding the HCF Method 1: This is factorization method. In this method, we factorize the given numbers and the lowest power of every prime factor appearing across all the numbers is taken. 12 = 22  3 = 22  31  70 14 = 2  7 = 21  30  71 18 = 2  32 = 21  32  70 21 = 3  7 = 20  31  71 Taking the lowest powers of 2, 3 and 7, we get HCF = 20  30  70 = 1. Example 72 Find the HCF of 23  512  75 and 32  53  79 112. Solution 23  512  75

=

23  30  512  75  110.

32  53  79  112

=

20  32  53  79  112.

Therefore, HCF

=

20  30  53  75  110 = 53  75.

Example 73 Find the HCF of 1212  93  105 and 610  97  208. Solution First of all, prime factorize the numbers as shown below: 1212  93  105 = 2(2×12)+5  312+(2×3)  55 = 229  318  55. 610  97  208 = 210+(2×8)  310+(2×7)  58 = 226  324  58. HCF = 226  318  55. Method 2: This method can also be called as method of differences. For example, to find the HCF of 12 and 18, we take the difference of the numbers, and check whether the difference obtained is the HCF or not. Difference of 12 and 18 is 6. Here, 6 is the factor of 12 as well as 18, and hence is the HCF of 12 and 18. Let us find the HCF of 12 and 20. The difference is 8, which is not the factor of 12 (nor that of 20). Check the next factor of 8, which is 4 and which is the factor of both 12 as well as 20. So, 4 is the HCF of 12 and 20. Example 74 Find the HCF of 12, 18 and 22. Solution 18 – 12 = 6 and 22 – 18 = 4. The smaller of 4 and 6 is 4. Now, 4 is not the HCF, as it does not divide 18 and 22 (even if it divides 12). So, we check the next factor of 4, that is 2, which divides all the three numbers, and hence is the HCF of these numbers. The method can also be applied in a slightly different format. To find the HCF of 12 and 18, we can take the HCF of the smaller number, 12 and the difference of the two numbers, 6.

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Quantitative Aptitude Simplified for CAT

So, HCF (12, 18) = HCF (12, 6) The process can be continued till both the numbers in the brackets become same, and then that number is the HCF. So, HCF (12, 18) = HCF (12, 6) = HCF (6, 6) = 6 HCF (12, 20) = HCF (12, 8) = HCF (8, 4) = HCF (4, 4) = 4 Example 75 Find the HCF and LCM of 244 and 248. Solution Using the method of differences, we see that 248 – 244 = 4, which divides 244 as well as 248, and hence 4 is the HCF. For LCM, if we use factorization method, then it will be very tedious. Recall the following theorem regarding LCM and HCF of 2 numbers: If we have two numbers (a, b) and L and H are their LCM and HCF resp., then a  b = L  H that is, product of the numbers = product of their LCM and HCF. But this result is true only for the case of two numbers and not for more than 2 numbers. Using this result, we can find LCM by first finding HCF (which is much easier to find), which is 4. So, LCM of 244 and 248 =

244  248 = 15128 4

The method of difference to find the HCF can be improvised further. Suppose we want to find the HCF of a and b (where b > a) Then, HCF (a, b) = HCF (a, r), where r is the remainder when b is divided by a. For example, HCF (95, 35) = HCF (35, 25) = HCF (25, 10) = HCF (10, 5) = HCF (5, 0) = 5 Example 76 Consider the fraction A.

12n  1 for some natural number n. Then, 30 n  2

The given fraction can be reduced to simpler form by cancelling some common factors

B.

The fraction is already in its simplest form for any value of n

C.

The fraction may or may not be reduced to simpler form depending upon the value of n

D.

The fraction can be reduced to simpler form only for some specific prime number values of n

Solution: B All we need to check is whether the HCF of 12n + 1 and 30n + 2 is 1 or not. Now, HCF (12n + 1, 30n + 2) = HCF (12n + 1, 6n) = 1 [Note that when 30n + 2 = 2(12n + 1) + 6n, and so 6n is the remainder] Example 77 Check whether the result “Product of the numbers = Product of their LCM and HCF” is true or not for the following numbers. (i)

a = 18, b = 24

(ii)

a = 14, b = 18, c = 20.

40

Number System – I: Factors, Multiples and Exponents Solution (i)

L (18, 24) = 72; H (18, 24) = 6 Product of numbers = 18  24 = 432 Product of L and H = 72  6 = 432 The result is true in this case.

(ii)

L (14, 18, 20) = 1260; H (14, 18, 20) = 2 Product of numbers = 5040 Product of L and H = 2520 The result is not true for this case, because we have three numbers.

Can the result ever be true for 3 or more than 3 numbers? Of course, the result can be true in case of 3 or more numbers, provided each pair of numbers is co-prime to each other. So, if the 3 numbers are a, b and c, then a  b  c = LCM  HCF, only if each pair (a, b), (b, c) and (a, c) is a co-prime pair. For example, for the numbers 12, 25 and 77, LCM = 12  25  77 = 23100, and HCF = 1. So, product of all the three numbers = Product of their LCM and HCF. This is so because each pair (12, 25), (25, 77) and (12, 77) is co-prime pair. Note: If L (a, b, c, ....) = H (a, b, c, ....), then a = b = c = ...., That is, if LCM of some numbers = HCF of those numbers, then the numbers are equal. Note that vice versa is also true.

Division algorithm If a number N is divided by D, and if quotient is Q and remainder R, then N=DQ+R This is known as division algorithm. Therefore, Dividend = Divisor  Quotient + Remainder Therefore, if 18 is divided by 7, the quotient is 2 and the remainder is 4. It can also be written as 18 = (7  2) + 4. Here, 18 is called the dividend, 7 is called the divisor, 2 is called the quotient and 4 is called the remainder. This will be found useful in the following examples. Example 78 Find the number which when divided by 12 leaves quotient 9 and remainder 4. Solution Dividend = Divisor  Quotient + Remainder = 12  9 + 4 = 112 Example 79 A number when divided by 42 leaves remainder 17. Find the remainder when the same number is divided by 7.

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Quantitative Aptitude Simplified for CAT

Solution The given number = 42x + 17, where x is the quotient (as we do not know the value of the quotient). Now, 42x when divided by 7 does not leave any remainder as 42 is divisible by 7 (and this does not depend upon the value of x). So, remainder will be obtained on account of 17 which when divided by 7 leaves remainder 3. Example 80 Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor? Solution Let the two numbers be A and B and divisor be d. Then, A = dx + 431; B = dy + 379. Therefore, A + B = dx + dy + (431 + 379) = dz + 810 = (dz + 599) + 211. Since 211 is the remainder, 599 is divisible by d and hence d should be 599 or a factor of 599. But, d cannot be less than any of the remainders, that is 431 and 379. So, d = 599. Example 81 Find the smallest number which is divisible by 24 and 45. Solution Basically, we are looking for the smallest number which is a multiple of both 24 and 45. Conceptually, we understand that such a number would be the LCM of 24 and 45.

Factorizing the given numbers, we get 24 = 23  3 45 = 32  5 Therefore, LCM = 23  32  5 = 360. So, 360 is the smallest number which is divisible by 24 and 45. Example 82 Find the least number which when divided by 24 and 45 will leave the remainder 2. Solution The LCM of 24 and 45 is 360 (from previous example). The general number when divided by 24 and 45 would leave the remainder 2 if the number will be of the form 360k + 2, because 360 is divisible by 24 as well as by 45. In order to obtain the smallest number, take k as 1. Hence, the smallest number = 360 + 2 = 362. Example 83 Find the smallest number which when divided by 5, 7, 8 and 9 would leave the remainders 2, 4, 5 and 6 respectively. Solution In this case, the remainder is not the same. But, we notice that the difference between the divisor and respective remainder is constant. In this example, this difference is 3. To solve such questions, we subtract this difference (3) from the LCM of the divisors. LCM of divisors = 5  7  8  9 = 2520. Therefore, the required smallest number is 2520 – 3 = 2517.

42

Number System – I: Factors, Multiples and Exponents Explanation When the number N is divided by 5, the remainder is 2. This means the number is 2 more than a multiple of 5, which also means the number is 3 less than the next multiple of 5. For example, if 27 is 2 more than 25 (a multiple of 5), then it is also 3 less than 30 (next multiple of 5). So, remainder of 2 upon division by 5 is equivalent to remainder of –3. Similarly, the number when divided by 7 leaves remainder 4 or –3; when divided by 8 leaves remainder 5 or–3; when divided by 9 leaves remainder 6 or –3. Therefore, the question effectively becomes: “what is the smallest number which when divided by 5, 7, 8, 9 leaves the remainder of –3?” This question is similar to the previous example. LCM of 5, 7, 8, 9 = 2520. Therefore, N = 2520k + (–3). For smallest number, take k as 1 and so the smallest number = 2520 – 3 = 2517. Example 84 Find the smallest three digit number which when divided by 5 leaves remainder 3 and when divided by 3 leaves remainder 1. Solution This is similar to the previous example, as the difference of divisor and respective remainder is same, that is 5 – 3 = 2 and 3 – 1 = 2. So, general number N = [LCM (5, 3)] k – 2 = 15k – 2. Since the smallest 3 digit number is asked, N  100 or 15k – 2  100  15k  102  k  102/15 = 6.8. So, the smallest (integer) value of k = 7 and hence the smallest three digit number = 15(7) – 2 = 103. Example 85 What is the largest three digit number which when divided by 5 leaves a remainder 3 and when divided by 6 leaves a remainder 5? Solution In this example, the difference of divisors and respective remainders is not constant, and so the approach has to be changed! Let us first find a general number meeting the requirements, and then find the largest three digit number. Start with the biggest divisor. We know that the number when divided by 6 leaves the remainder 5. This means N = 6q + 5 = (5q + q) + 5 = (5q + 5) + q. We represented 6q as 5q + q, so that 5q becomes divisible by 5, our next divisor. Now, (5q + 5) + q when divided by 5 leaves remainder 3. Since (5q + 5) is divisible by 5, the remainder will be q. The smallest value of q which when divided by 5 leaves remainder 3 is 3. So, q = 3, and hence the smallest such number = 6q + 3 = 6(3) + 5 = 23. The general number, N = [LCM (5, 6)] k + 23 = 30k + 23. For largest 3 digit number, 30k + 23  999  k  976/30 = 32.53. So, the largest value of k = 32, and the largest 3 digit number = 30(32) + 23 = 983.

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Quantitative Aptitude Simplified for CAT

Example 86 When a number is divided successively by 5, 7 and 8, we get the remainders 2, 3 and 7 respectively. What will be the remainders when the same number is divided successively by 8, 7 and then 5, that is, when the order of division is reversed? Solution When the number N is divided by 5, the remainder is 2. Let us say that the quotient is x. This x when divided by 7 leaves the remainder 3 and quotient y and when y is divided by 8, it leaves the remainder 7 and quotient z. This can be written as below: N = 5x + 2

(i)

x = 7y + 3

(ii)

y = 8z + 7

(iii)

Putting (iii) in (ii), x = 7(8z + 7) + 3 = 56z + 52

(iv)

Putting (iv) in (i), N = 5(56z + 52) + 2 = 280z + 262 Smallest such number is 262, when z = 0. Therefore, N = 262. Dividing 262 by 8, the quotient is 32, remainder is 6. Dividing 32 by 7, the quotient is 4, remainder is 4. Dividing 4 by 5, the quotient is 0, remainder is 4. Therefore, when the order of division is reversed, the remainders are 6, 4 and 4 respectively. Example 87 Find the smallest number which when divided by 3, 5, 6 and 7 leaves the remainder 2 but is divisible by 11. Solution LCM of 3, 5, 6 and 7 is 210. Therefore, number is of the form 210k + 2. It is given that 210k + 2 is divisible by 11. To find the number, we can put the values of k, one by one, and then check. But this is going to be tedious. So, a better approach would be: N = 210k + 2 = [(11  19) + 1] k + 2 = (11  19)k + (k + 2). Since (11  19) k is always divisible by 11, we just need to find k so that k + 2 is also divisible. Obviously, k has to be 9 so that 9 + 2 = 11 is divisible by 11. Therefore, the smallest number is 210  9 + 2 = 1892. Example 88 Find the product of two numbers whose LCM is 25 and HCF is 11. A.

275

B.

25

C.

250

D.

None of these

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Number System – I: Factors, Multiples and Exponents Solution: D If a and b are two numbers, then ab = lh, where l and h are their LCM and HCF. Seemingly, the product is 25  11 = 275, but this is incorrect because LCM must always be a multiple of HCF, whereas 25 is not a multiple of 11. Therefore, no such two numbers exist! Example 89 We have four concentric circles, each circle made of blinking lights. The outermost circle lights blink every 3 sec, the next circle lights blink every 5 sec, the next one every 6 sec and the innermost one every 8 sec. At 12 o’clock, lights of all the four circles blink together. By 1 o’clock, how often would they have blinked together? Solution The LCM of the blinking times is the first time they all blink together. LCM (3, 5, 6, 8) = 120 sec. Therefore, every 120 sec or 2 min, all the four circles blink together. Therefore, in 1 hour, they would have blinked 30 times. Example 90 In a morning walk, three persons start off together. Their steps measure 80 cm, 85 cm and 90 cm, respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps? Solution Minimum distance = LCM of 80, 85 and 90 = 12240 m. Example 91 Aman and Chaman met at Vasant Vihar after a long time. Aman resides in Vivek Vihar and Chaman resides in Gurgaon. After having met, they reached the Vasant Vihar bus-stop to return to their respective residences. They got to know that a bus had left for each of their respective destinations just then. Neither of the two wanted to leave the other one alone at the bus stop, and decided that they would board the buses to their respective destinations simultaneously. If the frequency of buses from Vasant Vihar to Gurgaon was one in 7 minutes and that from Vasant Vihar to Vivek Vihar was one in 11 minutes, how long would they wait at the bus stop? How many buses going to their destinations would each one decide not to board? Solution Since 7 and 11 are co-prime, the earliest that they can simultaneously board a bus is 7 Now, in 77 minutes: Total number of buses for Gurgaon that would have left Vasant Vihar is

x

11 = 77 minutes.

77  11 7

Total number of buses for Vivek Vihar that would have left Vasant Vihar is

77 7 11

 Aman would not board 6 buses and Chaman would not board 10 buses. Example 92 When an army is arranged in rows of five jawans each, we have two jawans extra. When arranged in rows of six jawans each, we have three jawans extra and when arranged in rows of nine jawans each, we have six jawans extra. (i)

When least number of such jawans is arranged in rows having 12 jawans in each row, how many jawans we would be left with?

(ii)

What least number of jawans should be added/removed to form them into a solid square?

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Quantitative Aptitude Simplified for CAT

Solution (i)

Let N be the number of jawans. When N is divided by 5, 6 and 9, we are left with remainders 2, 3 and 6 respectively. The difference of divisor and remainder is same. LCM of 5, 6 and 9 is 90. Therefore, least number of such jawans is 90 – 3 = 87. When these 87 jawans are arranged in rows, each having 12 jawans, we would be left with 3 jawans, because 87 divided by 12 leaves the remainder of 3.

(ii)

To find the least number of jawans to be added or removed to form a solid square, the total number of jawans must be a perfect square. Therefore, 81 and 100 are the perfect squares. But 81 being the nearest perfect square to 87, we need to remove 6 jawans. To make the number 100, we would need to add 13 jawans. But 13 is larger than 6. Therefore, 6 must be removed.

Example 93 Find the LCM and HCF of 281 – 1 and 227 – 1. Solution In such questions, we should find the HCF of the powers of 2, that is HCF of 81 and 27, which is 27. Let us assume that x = 227. Then, the given numbers become x3 – 1 and x – 1. The HCF of these can be obtained by factorizing x3 – 1 as (x – 1) (x2 + x + 1). Therefore, HCF of the two numbers is x – 1, that is 227 – 1. Example 94 Find the HCF of 284 – 1 and 254 – 1. Solution The HCF of 84 and 54 is 6. Let x = 26. Then, HCF (284 – 1, 254 – 1) = HCF (x14 – 1, x9 – 1). x14 – 1 = (x – 1)(x13 + x12 + x11 + … + 1) x9 – 1 = (x – 1)(x8 + x7 + x6 + … + 1) So, HCF = x – 1 = 26 – 1. Example 95 Find the largest number by which 24, 33 and 45 would be always divisible. Solution Here, we are basically asking for the HCF of the numbers. Therefore, HCF (24, 33, 45) = 3. Example 96 Find the largest number by which when 37, 72 and 107 are divided, the remainder is always 2. Solution If N is the required number, then 37 divided by N would leave the remainder 2. Therefore, 37 – 2 = 35 is divisible by N. Similarly, 72 – 2 = 70 and 107 – 2 = 105 are also divisible by N. The required number = HCF of 35, 70 and 105 = 35, which is the number. Example 97 When 73 is divided by a number, the remainder is 1. When 106 is divided by that number, the remainder is 2 and when 131 is divided by that number, the remainder is 3. What is the largest such number? Solution We would have to find the HCF of (73 – 1), (106 – 2) and (131 – 3), that is, HCF (72, 104, 128) = 8.

46

Number System – I: Factors, Multiples and Exponents Example 98 Find the largest number by which when 41, 125 and 167 are divided; we are left with the same remainder. Also find that remainder. Solution Let the number be N and the remainder be r. Then, (i)

41 = Nq1 + r

(ii)

125 = Nq2 + r

(iii) 167 = Nq3 + r Subtracting (i) from (ii) and (ii) from (iii), 84 = N(q2 – q1) and 42 = N(q3 – q2). From this, it is obvious that N is a factor of 84 and 42. Since N has to be the largest number, N would be the HCF of 42 and 84, which is 42. Therefore, largest such number is 42. When 41 is divided by 42, remainder = 41. Example 99 Find the length of the largest stick which can measure the lengths 60’, 75’ and 135’ completely. Solution The length of the stick should be such a number which divides all the given numbers. Since it should also be the largest number, we need to find the HCF of these numbers. HCF (60’, 75’, 135’) = 15’. Therefore, a stick of length 15’ can measure all the given lengths completely. Example 100 What is the minimum number of equal sized square pieces into which a rectangular piece of cloth measuring 20’  35’ can be cut without being left with any piece of cloth? What if the squares do not need to be of equal size? Solution Let the size of the square be x’. Then, x should be such that it divides the length of 20’ as well as the length of 35’ so that we are not left with any piece of cloth. And of course x should be as large as possible for the number of squares to be minimum. Therefore, x = HCF (20’, 35’) = 5’.

Therefore, number of squares along the length = 7 and those along the breadth = 4. Therefore, number of square pieces = 7  4 = 28. Alternatively, we can divide the total area of the rectangular piece by the area of each square piece. Therefore, number of square pieces =

20  35 = 28. 55

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Quantitative Aptitude Simplified for CAT

If the squares need not be of equal size, then first of all, the largest sized square will be 20’  20’. After this, we will be left with a rectangular strip of dimensions 15’  20’ from which a 15’  15’ square can be cut out. Finally, we will be left with a strip of dimensions 5’  15’, from which exactly 3 squares of dimensions 5’  5’ will be cut and now we are left with no strip. Therefore, in all we could cut 1 + 1 + 3 = 5 square pieces.

Example 101 In a class, there are 100 students where the ratio of girls to boys is 3 : 2. For an examination, what is the minimum number of class rooms required so that girls and boys do not sit together and equal number of students is present in each classroom? Solution The number of girls and boys is 60 and 40. Let there be x students in each classroom. The value of x must be such that x divides 60 as well as 40 (so that girls and boys do not sit together) and x should be as large as possible (so that number of classrooms required is minimized). Therefore, x = HCF (60, 40) = 20. That is, in each classroom 20 students should be seated, either boys or girls, so that the number of classrooms required is 100/20 = 5. Note the similarity in the last two examples. The square size in the rectangular strip is conceptually similar to the number of students in each classroom, and the number of squares is same as the number of classrooms required!

LCM and HCF of fractions To find LCM and HCF of fractions, we need to use the following formulae: LCM of Fractions =

LCM of Numerators HCF of Numerators ; HCF of Fractions = HCF of Denominators LCM of Deno min ators

But, care must be taken that when the fractions are taken for finding LCM or HCF, the fractions should be in their simplest form, that is, the numerator and denominator must not have any common factor. Example 102 Find the LCM and HCF of

21 12 and . 35 33

Solution Before we find LCM or HCF, the fractions must be reduced to their simplest form. So, LCM (3,4) 12 21 12 3 4  Therefore, LCM  ,  = LCM  ,  = = 12. 35 33 5 11 HCF (5,11) 1     HCF (3,4) 1 21 12 3 4  HCF  ,  = HCF  ,  = . LCM (5,11) 55  35 33   5 11 

48

21 3 12 4  and  . 35 5 33 11

Number System – I: Factors, Multiples and Exponents Let us see what happens if the fractions are not expressed in their simplest form, using the above example.  21 12 

LCM (21,12)

84

 LCM  ,  = = 84, which is incorrect. Similarly, HCF (35,33) 1  35 33  HCF (21,12) 21 12  HCF  ,  =  35 33 

LCM (35,33)

3 1  , which is also incorrect. 1155 385

Also note that in case of two fractions, f1 and f2, if their LCM and HCF are l and h, then f1  f2 = l  h, that is, Product of two fractions = Product of their LCM and HCF The result is true only for the case of two fractions and not for more than two fractions. Let us prove that Product of two fractions = Product of their LCM and HCF. Let f1 =

a c and f2 = , and for the sake of simplicity assume that these fractions are in their simplest forms, b d

that is there is no common factor between a and b, and also between c and d. Then, LCM (a, c) HCF (a, c) a c a c and HCF  ,  = LCM  ,  = HCF (b, d) LCM (b, d)  b d  b d LCM (a, c) HCF (a, c) LCM (a, c)  HCF (a, c) a c a c So, LCM  ,   HCF  ,  =  = b d b d HCF ( b , d ) LCM ( b , d ) LCM (b, d)  HCF (b, d)    

We already know that for two numbers a and c, LCM (a, c)  HCF (a, c) = a  c and similarly for b and d. So, a c a c ac a c LCM  ,   HCF  ,  =   = f1  f2. bd b d  b d  b d

LCM and HCF of more than two fractions The result can also be true for more than two fractions, under certain conditions only. For example, if the three fractions are

a c e , and (and assume that all the fractions are in their simplest forms), then the product of b d f

fractions will be equal to the product of their LCM and HCF, provided each pair of numerators, that is (a, c), (c, e) and (a, e) is co-prime, and each pair of denominators, that is (b, d), (d, f) and (b, f) is co-prime. Example 103 Arrange the fractions

2 3 5 , and in the ascending order of their magnitude. 15 10 21

Solution Factorizing the denominators of the given fractions: 15 = 3  5; 10 = 2  5; 21 = 3  7. Hence, LCM of denominators of fractions = 2  3  5  7 = 210  The given fractions can be rewritten as

28 63 50 , and . 210 210 210

Now, we can arrange the fractions simply by comparing the magnitude of their numerators (since they have identical denominators) 

2 5 3   . 15 21 10

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Quantitative Aptitude Simplified for CAT

Example 104 Two persons A and B are running around a circular track at the speeds of 22 m/s and 30 m/s. The length of the track is 100 m and they start at the same time from the same place in the same direction. When would they meet for the third time at the starting point after they start running? Solution A is at the starting point every

100 50 100 10  sec and B is at the starting point every  sec. 22 11 30 3

Therefore, they both will be at the starting point after time t sec, where t is the multiple of They both will be at the starting point for the first time when t is the LCM of

50 10 as well as . 11 3

50 10 sec and sec. 11 3

So, LCM of the given fractions = 50 sec. Hence, they will meet for the third time after 50  3 = 150 sec. [We will revisit such examples when we discuss in details the concepts of relative speed and its application to circular motion in the chapter of “Time, Speed and Distance”.] Example 105 In the house of Santa Singh, there are two pendulum clocks. First clock strikes 12 o’clock in 8 other one strikes 12 o’clock in 8

1 sec and the 4

4 sec. Both the clocks develop some mechanical snag, because of which 5

upon striking 12 o’clock, the pendulums of the clocks continue to make the sound at their respective rates without stoppage for 1 minute. How often does it happen that the pendulums of both the clocks gong together, if the first gong occurs together? Solution When it strikes 12 o’clock, there are 11 time gaps which takes 8 sec =

1 33 4 sec = sec for the first clock and 8 4 4 5

44 sec for the second clock. 5

For the first clock, if 11 time gaps take

33 1 33 3 sec, each time gap will take   sec. 4 11 4 4

For the second clock, each time gap will take

1 44 4   sec. 11 5 5

3 4 12 So, the time taken for them to gong together for the first time = LCM  ,   = 12 sec. 1  4 5

Therefore, the two clocks gong together every 12 sec. In 1 minute, they would gong 5 times after the start.

Vedic Maths Presented below are certain techniques for finding squares of natural numbers. These techniques are very simple and applicable to virtually all numbers. It has been observed that students find a lot of difficulty in performing simple mathematical functions like multiplication, squaring, division and so on, which has a detrimental impact on their performance. The following techniques will prove useful in helping the student develop a quick understanding of these basic

50

Number System – I: Factors, Multiples and Exponents principles. It is important to note that one must possess an acute sense of observation to be able to grasp the methods completely, which will go a long way in developing intuitive logic – an important skill!

Base method of squaring Example 106 Find the square of 13. Solution Identify a ‘base’. ‘Base’ is a number which must satisfy only two conditions: it must be a very simple number, and it must be close to the given number. Thus we choose our base to be 10. The difference between number 13 and base 10 is +3. Since the difference between the given number and the base chosen is ‘positive’, we call this difference ‘surplus’. Add (+3) to 13 to obtain 16. Multiply 16 by the chosen base 10. We get 160. To the number 160 so obtained, add the square of the surplus number, that is, 32 = 9. We get 160 + 9 = 169, which is the square of 13. Example 107 Find the square of 18. Solution As mentioned in the previous example, we can choose the base to be 10 or 20. Let’s choose the base to be 10. The difference between number 18 and base 10 is +8, which is surplus. Add (+8) to 18 to obtain 26. Multiply 26 by the base 10 chosen. We get 260. To the number 260 so obtained, add the square of the surplus number, that is, 82 = 64. We get 260 + 64 = 324, which is the square of 18. If we choose the base to be 20, let’s observe the process now. The difference between number 18 and base 20 is 2, which is deficit. Add (2) to 18 to obtain 16. Multiply 16 by the base 20 chosen. We get 320. To the number 320 so obtained, add the square of the deficit number, that is, (2)2 = 4. We get 320 + 4 = 324, which is the square of 18. Note that the process is faster when base 20 is chosen as compared to when base 10 is chosen. Example 108 Find the square of 19. Solution Base = 20. Deficit = 1. 19 – 1 = 18. 18 × 20 = 360. Now, 360 + (1)2 = 361, which is square of 19.

51

Quantitative Aptitude Simplified for CAT

Example 109 Find the square of 23. Solution Base = 20. Surplus = +3. 23 + 3 = 26. 26 × 20 = 520. Now, 520 + (+3)2 = 529, which is square of 23. Example 110 Find the square of 24. Solution Base = 20. Surplus = +4. 24 + 4 = 28. 28 × 20 = 560. Now, 560 + (+4)2 = 576, which is square of 24. Example 111 Find the square of 28. Solution Base = 30. Deficit = 2. 28 – 2 = 26. 26 × 30 = 780. Now, 780 + (2)2 = 784, which is square of 28. Example 112 Find the square of 34. Solution Base = 30. Surplus = +4. 34 + 4 = 38. 38 × 30 = 1140. Now, 1140 + 42 = 1156, which is square of 34. Example 113 Find the square of 39.

52

Number System – I: Factors, Multiples and Exponents Solution Base = 40. Deficit = 1. 39 – 1 = 38. 38 × 40 = 1520. Now, 1520 + (1)2 = 1521, which is square of 39. Example 114 Find the square of 48. Solution Base = 50. Deficit = 2. 48 – 2 = 46. 46 × 50 = 2300. Now, 2300 + (2)2 = 2304, which is square of 48. In this process, we can skip the process of multiplication of 46 by 50. Instead of multiplying 46 by 50, we can multiply 46 by 100 and then divide by 2, which is 2300. Students would find this method easier than multiplying 46 by 5 and then putting a zero at the end of it. So, we should always try to take the base as 50 as much as possible to make use of this benefit. Note that to find the square of 39 (discussed above) where we took the base as 40, we will now take base as 50 as described below. Base = 50. Deficit = 11. 39 – 11 = 28. 28 × 50 = 1400 (by multiplying 28 by 100 and dividing by 2). Now, 1400 + (11)2 = 1521, which is square of 39. Example 115 Find the square of 43. Solution Base = 50. Deficit = 7. 43 – 7 = 36. 36 × 50 = 1800. Now, 1800 + (7)2 = 1849, which is square of 43. Example 116 Find the square of 53.

53

Quantitative Aptitude Simplified for CAT

Solution Base = 50. Surplus = 3. 53 + 3 = 56. 56 × 50 = 2800. Now, 2800 + 32 = 2809, which is square of 53. Example 117 Find the square of 61. Solution Base = 50. Surplus = +11. 61 + 11 = 72. 72 × 50 = 3600. Now, 3600 + (11)2 = 3721, which is square of 61. Example 118 Find the square of 94. Solution Base = 100. Deficit = 6. 94 – 6 = 88. 88 × 100 = 8800. Now, 8800 + (6)2 = 8836, which is square of 94. Example 119 Find the square of 97. Solution Base = 100. Deficit = 3. 97 – 3 = 94. 94 × 100 = 9400. Now, 9400 + (3)2 = 9409, which is square of 97. Example 120 Find the square of 104. Solution Base = 100.

54

Number System – I: Factors, Multiples and Exponents Surplus = +4. 104 + 4 = 108. 108 × 100 = 10800. Now, 10800 + (4)2 = 10816, which is square of 104. Example 121 Find the square of 112. Solution Base = 100. Surplus = +12. 112 + 12 = 124. 124 × 100 = 12400. Now, 12400 + (12)2 = 12544, which is square of 112.

Squares of numbers ending with 5 Example 122 Find the square of 65. Solution The number 65 can be divided into two parts: 6 / 5 The number to the left of 5 is 6. The natural number which comes after 6 is 7. Multiply 6 by 7 to obtain 42, which forms the left part of the answer. The unit’s digit 5 is squared to obtain 25, which forms the right part of the answer. So, square of 6 / 5 = 42 / 25 = 4225. Example 123 Find the square of 45. Solution The number 45 can be divided into two parts: 4 / 5 The number to the left of 5 is 4. The natural number which comes after 4 is 5. Multiply 4 by 5 to obtain 20, which forms the left part of the answer. The unit’s digit 5 is squared to obtain 25, which forms the right part of the answer. So, square of 4 / 5 = 20 / 25 = 2025. Example 124 Find the square of 75. Solution The number 75 can be divided into two parts: 7 / 5 The number to the left of 5 is 7. The natural number which comes after 7 is 8. Multiply 7 by 8 to obtain 56, which forms the left part of the answer.

55

Quantitative Aptitude Simplified for CAT

The unit’s digit 5 is squared to obtain 25, which forms the right part of the answer. So, square of 7 / 5 = 56 / 25 = 5625. Example 125 Find the square of 105. Solution The number 105 can be divided into two parts: 1 0 / 5 The number to the left of 5 is 10. The natural number which comes after 10 is 11. Multiply 10 by 11 to obtain 110, which forms the left part of the answer. The unit’s digit 5 is squared to obtain 25, which forms the right part of the answer. So, square of 1 0 / 5 = 110 / 25 = 11025. Example 126 Find the square of 135. Solution The number 135 can be divided into two parts: 1 3 / 5 The number to the left of 5 is 13. The natural number which comes after 13 is 14. Multiply 13 by 14 to obtain 182, which forms the left part of the answer. The unit’s digit 5 is squared to obtain 25, which forms the right part of the answer. So, square of 1 3 / 5 = 182 / 25 = 18225. In this example, it becomes difficult to multiply 13 by 14. To do this, an easier method will be to find square of 13 and add 13. So, 169 + 13 = 182. Rest of the calculations can be done as above. Example 127 Find the square of 165. Solution 16/5 16 × 17 = 162 + 16 = 256 + 16 = 272. 52 = 25. So, 1652 = 27225. Example 128 Find the square of 625. Solution 62/5 62 × 63 = 622 + 62 = 3844 + 62 = 3906. 52 = 25. So, 6252 = 390625. In this example, it becomes difficult to multiply 62 by 63, even by doing 622 + 62. To solve such questions, let us learn the technique of finding squares of numbers ending with 25 taking the same example above.

56

Number System – I: Factors, Multiples and Exponents Divide 625 into two parts, with 25 on right side and the rest of left side. So, 625 = 6 / 25 Find the square of 6 and add half of 6 to it. We get 62 + 3 = 39. Multiply the result by 10 to get 39 × 10 = 390. This forms the left part the answer. The right part of the answer is square of 25, that is 625. So, 6252 = 390 / 625 = 390625. Example 129 Find the square of 825. Solution 8 / 25 82 + 4 = 64 + 4 = 68, 68 × 10 = 680, 252 = 625, 8252 = 680 / 625 = 680625. Example 130 Find the square of 925. Solution 9 / 25 92 + 4.5 = 81 + 4.5 = 85.5, 85.5 × 10 = 855, 252 = 625, 9252 = 855 / 625 = 855625. Example 131 Find the square of 1125. Solution 11 / 25 112 + 5.5 = 121 + 5.5 = 126.5, 126.5 × 10 = 1265, 252 = 625, 11252 = 1265 / 625 = 1265625.

Squares of numbers consisting of 9s only Example 132 Find the square of 9999.

57

Quantitative Aptitude Simplified for CAT

Solution To find the square of this number which consists of 9’s only, the last digit 9 should be replaced by 8, keeping the rest of the digits same. We get 9998, which consists of one less number of 9’s, that is, it consists of three 9’s. Now, place as many zeroes at the end of it as the number of 9’s in it, followed by 1. We get 99980001, which is the square of 9999. Example 133 Find the square of 9999999. Solution To find the square of this number which consists of 9’s only, the last digit 9 should be replaced by 8, keeping the rest of the digits same. We get 9999998, which consists of one less number of 9’s, that is, it consists of six 9’s. Now, place as many zeroes at the end of it as the number of 9’s in it, followed by 1. We get 99999980000001, which is the square of 9999999. Example 134 Find the square of 999999999. Solution 9999999992 = 999999998000000001.

Squares of numbers consisting of 1s only Example 135 Find the square of 111. Solution The number consists of three 1’s. So, we write natural numbers one after the other starting from 1 up to 3. We get 123. Now, write the natural numbers in reverse order, without repeating the biggest number 3. We get 12321, which is square of 111. Example 136 Find the square of 11111. Solution 111112 = 123454321. Example 137 Find the square of 11111111. Solution 111111112 = 123456787654321. Example 138 Find the square of 111111111111.

58

Number System – I: Factors, Multiples and Exponents Solution In such cases, we will get two digit numbers also while writing natural numbers. The ten’s place digit of each such natural number is carried over to the previous digit. So, 1 2 3 4 5 6 7 8 9 ‘10’ ‘11’ ‘12’ ‘11’ ‘10’ 9 8 7 6 5 4 3 2 1 12345678901210987654321 11111 1 2 3 4 5 6 7 8 ‘10’ 1 2 3 2 0 9 8 7 6 5 4 3 2 1 12345678012320987654321 1 12345679012320987654321 So, square of 111111111111 is 12345679012320987654321. To master these techniques, one must practice each method extensively and try new questions on one’s own. As one practices, it is important that one gradually develops the ability to perform the entire calculation orally without any pen and paper. Only then one would develop the ability of viewing numbers differently and working with them effectively.

EXPERT SPEAK Scan this QR Code to watch a video that explains some useful methods of finding squares of numbers.

59

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE A:

Direction for questions 1 – 3: Answer the questions on the basis of the information given below.

n has a perfect integer-valued divisor which is greater than 1 and less than n

B:

The seven basic symbols in a certain numeral system and their respective values are as follows:

n has a perfect integer-valued divisor which is greater than n but less than n

Then,

I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000

A.

Both A and B are false.

B.

A is true but B is false.

In general, the symbols in the numeral system are read from left to right, starting with the symbol representing the largest value; the same symbol cannot occur continuously more than three times; the value of the numeral is the sum of the values of the symbols. For example, XXVII = 10 + 10 + 5 + 1 + 1 = 27. An exception to the left-to-right reading occurs when a symbol is followed immediately by a symbol of greater value; then, the smaller value is subtracted from the larger. For example, XLVI = (50  10) + 5 + 1 = 46.

C.

A is false but B is true.

D.

Both A and B are true. CAT 2003 (R)

5.

If a, a + 2, and a + 4 are the prime numbers, then the number of possible solutions for a is: A.

one

B.

two

C.

Three more than three

D. CAT 2003 (R) 1.

2.

3.

4.

CAT 2003 (R)

The value of the numeral MDCCLXXXVII is: A.

1687

6.

Let x and y be positive integers such that x is prime and y is composite. Then

B.

1787

A.

y  x can’t be even integer

C.

1887

B.

xy can’t be even integer

D.

1987

C.

The value of the numeral MCMXCIX is

A.

1999

B.

1899

C.

1989

D.

1889

D.

xy can’t be even integer. x

None of these CAT 2003 (R)

7.

Which of the following can represent the numeral for 1995?

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B − A is perfectly divisible by 7, then which of the following is necessarily true?

(a)

MMLXXV

(b)

MCMXCV

A.

100 < A < 299

(c)

MVD

(d)

MVM

B.

106 < A < 305

A.

only (a) and (b)

C.

112 < A < 311

B.

only (c) and (d)

D.

118 < A < 317

C.

only (b) and (d)

D.

only (d)

CAT 2005 8.

Let n ( > 1) be a composite integer such that is not an integer. Consider the following statements:

60

For a positive integer n, let pn denote the product of the digits of n, and sn denote the sum of the digits of n. The number of

Number System – I: Factors, Multiples and Exponents integers between 10 and 1000 for which pn + sn = n is

B.

25

C.

41

A.

81

D.

67

B.

16

E.

73

C.

18

D.

9

CAT 2006 13. The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be

CAT 2005 9.

Let S be a set of positive integers such that every element n of S satisfies the conditions i

1000 ≤ n ≤ 1200

ii.

every digit in n is odd Then how many elements of S are divisible by 3?

A.

9

B.

10

C.

11

D.

12

A.

101 : 88

B.

87 : 100

C.

110 : 111

D.

85 : 98

E.

97 : 84 CAT 2006

14. When you reverse the digits of the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?

CAT 2005 10. If x = –0.5, then which of the following has the smallest value?

A.

5

A.

21/x

B.

6

B.

1 x

C.

7

D.

8

C.

1 x2

E.

10

D.

2x

E.

CAT 2006

1

15. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

x

CAT 2006 11. Which among 21/2, 31/3, 41/4, 61/6 and 121/12 is the largest?

A.

3

B.

2

A.

21/2

C.

4

B.

31/3

D.

0

C.

41/4

E.

1

D.

61/6

E.

121/12

CAT 2007 16. A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which

CAT 2006 12. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers? A.

21

61

Quantitative Aptitude Simplified for CAT

of the following best describes the value of x? A.

2≤x≤6

B.

5≤x≤8

C.

9 ≤ x ≤ 12

D.

11 ≤ x ≤ 14

E.

13 ≤ x ≤ 18

A. B. C. D. E.

CAT 2008 20. Find the sum: 1 1 1 1 1  2  2  1  2  2  ... 1 2 2 3

CAT 2008 17. The integers 1, 2, … , 40 are written on a blackboard. The following operation is then repeated 39 times: In each operation, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b − 1 is written. What will be the number left on the board at the end? A.

820

B.

821

C.

781

D.

819

E.

780

1≤m≤3 4≤m≤6 7≤m≤9 10 ≤ m ≤ 12 13 ≤ m ≤ 15

 1

1 1  2 2007 20082

A.

2008 

1 2008

B.

2007 

1 2007

C.

2007 

1 2008

D.

2008 

1 2007

E.

2008 

1 2009

CAT 2008

CAT 2008

18. Suppose, the seed of any positive integer n is defined as follows:seed (n) = n, if n < 10 = seed(s(n)), otherwise, where s(n) indicates the sum of digits of n. For example, seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc. How many positive integers n, such that n < 500, will have seed(n) = 9? A. 39 B.

72

C.

81

D.

108

E.

55

21. There is a deck of 52 playing cards. All the cards are placed face down in a straight line. First person comes and flips the position of all the cards, i.e. flips them face up. The second person comes and flips the position of all the cards which are placed at positions which are multiples of 2, i.e. if the card is face up, he will flip it face down, and if face down, he will flip is face up. The third person comes and flips the position of all the cards which are placed at positions which are multiples of 3, and so on. So, the nth person comes and flips the position of all cards which are placed at positions which are multiples of n. After all the 52 persons have flipped the positions of the cards, one after the other, which all cards will be face up?

CAT 2008

A. B. C. D.

19. Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

7 51 15 45 CAT 2012

22. If a number is same as 8 times the sum of digits and added to 1, and is same as 13

62

Number System – I: Factors, Multiples and Exponents times the difference and added to 2, how many such numbers are there? A. B. C. D.

27. If n = 2x  3y, find the number of factors of n.

0 1 2 3

A.

xy

B.

(x – 1)(y – 1)

C.

(x + 1)(y + 1)

D.

xy + 1 CAT 2015

CAT 2012 28. If m =

23. Let M be a 100 digit number a99a98a97.......a2a1a0, where a99, a98 etc are the digits of the number. Also, N is another 100 digit number a0a1a2......a98a99. Then, |M  N| is A.

Always divisible by 9, but not necessarily by 99.

B.

Always divisible by 99

C.

Always divisible by 999

D.

No such divisibility rule can be established

integer values of m is n a positive integer? A.

B.

3a = 2b

C.

a=b

D.

None of these

B.

2

C.

1

D.

8

29. Four Prime Numbers are written in ascending order. The product of first three is 7429 and the product of last three is 12673. Find the sum of first and last number?

24. It is given than 5a = 26, 125b = 676. Then, 2a = 3b

4

CAT 2015

CAT 2013

A.

76  n , then for how many positive n1

A.

40

B.

46

C.

42

D.

Cannot be determined CAT 2016

CAT 2014

30. Find the number of factors of 10800 which are divisible by 12 but are not divisible by 36?

25. Sum of two numbers is 96 and ratio of the same two numbers is 5 : 7. Find the LCM. A.

140

A.

B.

280

B.

12

C.

420

C.

18

D.

Cannot be determined

D.

24

9

CAT 2014

CAT 2016

26. 3x + 2y and 2x + 3y when divided by 5 leave remainders of 2 and 3 respectively. Find the remainder when x  y is divided by 5 (where x > y).

31. A commander was trying to arrange his soldiers in the form of a square, but 63 soldiers were left. So he arranges them in form of 2 squares and there were no soldiers left. Find the minimum number of soldiers?

A.

1

B.

2

C.

3

A.

D.

4

B.

88

C.

72

D.

Cannot be determined

CAT 2014

79

CAT 2016

63

Quantitative Aptitude Simplified for CAT

32. The leftmost digit of a six-digit number is 1. If the first digit of this number is shift to the end of the number leaving the rest of the digits in exactly the same sequence, the resulting number is thrice the original number. What is the sum of the digits of the number? A.

21

B.

27

C.

24

D.

18

37. A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is A.

6912

B.

864

C.

288

D.

240 IIFT 2012

38. If k is an integer and 0.0010101 × 10k is greater than 1000, what is the least possible value of k?

CAT 2016 33. Sum of the squares of 3 consecutive integers can be equal to

A.

4

A.

654002

B.

5

B.

151877

C.

6

C.

234104

D.

7

D.

None of these

IIFT 2012 CAT 2016

34. The highest number amongst A. B. C. D.

3

4

3

39. The equation 7x – 1 + 11x – 1 = 170 has

4

2, 3, 4 is

A.

no solution

2

B.

one solution

3

C.

two solutions

D.

three solutions

4

IIFT 2012 All are equal 40. If N = (11p + 7)(7q – 2)(5r + 1)(3s) is a perfect cube, where p, q, r and s are positive integers, then the smallest value of p + q + r + s is :

IIFT 2016 35. The simplest value of the expression  p  14  2  2p 4  p  2 2 

1 /p

    

is:

A.

5

B.

6

4

C.

7

B.

8

D.

8

C.

4p

E.

9

D.

8p

A.

XAT 2017 IIFT 2015

41. For two positive integers a and b, if (a + b)(a +b) is divisible by 500, then the least possible value of a × b is:

36. If the product of the integers a, b, c and d is 3094 and if 1 < a < b < c < d, what is the product of b and c? A.

26

B.

91

C.

133

D.

221

A. B. C. D. E.

8 9 10 12 None of the above XAT 2016

IIFT 2013

64

Number System – I: Factors, Multiples and Exponents 42. An ascending series of numbers satisfies the following conditions: i. ii.

A. B. C. D. E.

II.

When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2. When divided by 11, the numbers leave no remainder. The 6th number in this series will be: 242 2882 3542 4202 None of the above XAT 2015

A. B. C. D. E.

45. p and q are positive numbers such that pq = qp, and q = 9p. The value of p is A. B. C. D. E.

43. A three-digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three-digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions? A. B. C. D. E.

The arithmetic mean of the numbers is greater than 4. Which of the following statements would be sufficient to determine the sum of the four numbers? Statement I Statement II Statement I and Statement II. Neither Statement I nor Statement II. Either Statement I or Statement II. XAT 2014

9 6

9

3

9

9

9

8

9

XAT 2013 46. A number is interesting if on adding the sum of the digits of the number and the product of the digits of the number, the result is equal to the number. What fraction of numbers between 10 and 100 (both 10 and 100 included) is interesting?

5 6 7 8 None of the above

A. B. C. D. E.

XAT 2015 44. Consider four natural numbers: x, y, x + y, and x – y. Two statements are provided below:

0.1 0.11 0.16 0.22 None of the above XAT 2013

I.

All four numbers are prime numbers.

ANSWER KEY 1.

B

2.

A

3.

C

4.

B

5.

9.

Z

10. B

11. B

12. C

13. E

14. B

15. A

16. B

17. C

18. E

19. A

20. A

21. A

22. B

23. A

24. A

25. B

26. D

27. C

28. A

29. B

30. A

31. C

32. B

33. B

34. B

35. B

36. B

37. B

38. C

39. B

40. E

41. B

42. C

43. B

44. A

45. E

46. E

65

A

6.

D

7.

A

8.

D

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS

Take x = 13, y = 4, so [B] is eliminated Take x = 13, y = 39, so [C] is eliminated 7.

1.

Answer: B

Explanation: Let A = 100x + 10y + z and B = 100z + 10y + x and z > x because B > A. B – A = 99z – 99x is divisible by 7  (z – x) is divisible by 7. So, (x, z) can be (1, 8) and (2, 9). Therefore, 100’s place digit will be 1 or 2 and so 100 < A < 299.

Explanation: Value of number MDCCLXXXVII = 1000 + 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1 = 1787. 2.

Answer: A Explanation: MCMXCIX = 1000 + (1000 – 100) + (100 – 10) + (10 – 1) = 1999.

3.

8.

Answer: C

Answer: B Explanation: For checking whether a number is prime or not, we take its square root and divide by all the numbers less than its square root. Since n is a composite number, it will definitely have a perfect integer-valued divisor which is greater than 1 and less than n .

5.

a=

9b (bc  99)

The maximum value of b × c can be 9 × 9 = 81 and so denominator is always negative which is not possible. Therefore, n cannot be a 3-digit number. Therefore, there are only 9 possible values of n.

Answer: A Explanation: a, a + 2, a + 4 are the prime numbers. The only values which satisfy the given condition is 3, 5, 7. Note that the given 3 numbers are consecutive odd numbers or consecutive even numbers. Since they are supposed to be odd numbers, so they are consecutive odd numbers. In such a case, one of the three numbers will definitely be a multiple of 3, and so cannot be a prime number. Therefore, no other set of prime numbers is possible.

6.

Answer: D Explanation: Number can be two digit or 3-digit number. If it is a 2-digit number, say ab, then according to question, a × b + a + b = 10a + b  a × b + a = 10a  a × b = 9a  a(b – 9) = 0  a = 0 or b = 9 Since a can’t be zero, therefore b = 9. Numbers can be 19, 29, 39, … 99. So, 9 such numbers exist. If it is 3-digit number, say abc, then a × b × c + a + b + c = 100a + 10b + c  a × b × c = 99a + 9b  a × b × c = 9(11a + b)

Explanation: (A) starts with MM which means number will be more than 2000. So (A) is not possible. MCMXCV = 1000 + (1000 – 100) + (100 – 10) + 5 = 1995 MVM = 1000 + (1000 – 5) = 1995. 4.

Answer: A

9.

Answer: A Explanation: Let the four-digit number be 1abc. Since n ≤ 1200 and a cannot be even, so n < 1200 and so a = 1. As it is divisible by 3, a + b + c = 5, 8, 11, 14, 17, 20, 23, 26. As all the numbers are odd, sum can’t be even.  a + b + c = 5, 11, 17 or 23. Also, since a = 1, b + c = 4, 10, 16 or 22. But b + c cannot be more than 9 + 9 = 18. Therefore, b + c can be 4, 10 or 16. Various possibilities are: (1, 3), (1, 9), (3, 7), (5, 5), (9, 7).

Answer: D Explanation: Use options. Take x = 13, y = 9, so [A] is eliminated.

66

Number System – I: Factors, Multiples and Exponents 15. Answer: A

Therefore, numbers are: 1113, 1131, 1119, 1191, 1137, 1173, 1155, 1197 and 1179.

Explanation: Let the four digit number be XXYY i.e. 1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) is a perfect square. Therefore, number is a multiple of 11. Since it is a four-digit number, it must be square of numbers more than or equal to 33. Checking with the squares of 33, 44, 55, 66, 77, 88 and 99, we get 882 = 7744. There is only one such number possible. Alternatively, Since the last two digits are same, those can be 00 or 44. If the number is XX00, then XX must also be a perfect square. No such case is possible. Therefore, the last two digits of the perfect square must be 44, which is also the last two digits of square of 12. Therefore, required number will be of the type = 50k  12. Coupled with the fact that the number must be more than 32 and also a multiple of 11, the only possibility is 100 – 12 = 88.

10. Answer: B Explanation: 1 is the only value which is negative for x x

= –0.5. Other options yield positive values. 11. Answer: B Explanation: Comparing 21/2 and 31/3, we can raise them to the power 6 so that we get 23 and 32. Out of these, 32 is larger. Similarly, 34 > 43; 36 > 63; 312 > 123. So, 31/3 is the largest of them all. Moreover, 3m > m3 for all m > 3. 12. Answer: C Explanation: Let the 4 numbers be 2n – 3, 2n – 1, 2n + 1, 2n + 3 whose sum = 8n. The sum is a multiple of 8 and ends with 0. Possible sums are 40, 80, 120, 160, 200, and so on. Since the sum is divided by 10 to yield a perfect square, the sums can be 40, 160 and so on. The sum cannot be 40 because sum of 4 consecutive two-digit odd numbers will be more than 40. For 160, the 4 numbers will be 37, 39, 41 and 43. So, 41 is one of the possible odd numbers.

16. Answer: B Explanation: In such questions, we should start in reverse order of distribution of rice. So, after giving to the third customer, we are left with no rice. The half a kg rice given to the third customer, as a last step, is half of 1 kg. So, before giving to the third customer, we had 1 kg of rice. Before giving to the second customer, we had (1 + 0.5)  2 = 3 kg and before giving to the first customer, we had (3 + 0.5)  2 = 7 kg rice. So, the value of x = 7 kg.

13. Answer: E Explanation: Adding the two values in each of the ratios, we get: 189, 187, 221, 183 and 181, out of which 181 is the only number which is a prime number. So, 97 : 84 is a possible ratio.

17. Answer: C Explanation: For simplicity, lets perform the operation in sequential order. After first operation, we get (1 + 2 – 1). After second, we get (1 + 2 – 1) + (3 – 1), and so on. In all, there will be 39 steps and so 1 will be subtracted 39 times. That is, 39 is to be subtracted from the sum of first 40 natural

14. Answer: B Explanation: Let the number be ‘ab’, which is equal to 10a + b. Reversing the digits, we get 10b + a. Now, (10b + a) – (10a + b) = 18, or b – a = 2. The possible values of (a, b) are: (1, 3), (2, 4), …., (7, 9). There are 7 in all and other than 13, there are 6 numbers only.

numbers. We get = 39 = 820 – 39 = 781.

67

n(n  1) 40  41  39  – 2 2

Quantitative Aptitude Simplified for CAT

18. Answer: E

If we take 10x + y = 13(y – x) + 2 or 23x – 13y = 2 or 23(7y + 1)/2 – 13y = 2 or 161y + 23 – 26y = 4 which gives us negative value of y. So, this case is not possible. Therefore, only one number

Explanation: seed(n) will be 9 for all those numbers which are divisible by 9. The number of numbers less than 500 which are multiples of 9 500

  = [55.55…] = 55, where [x] is the =   9  greatest integer function.

23. Answer: A Explanation: M = a99(1099) + a98(1098) + a97(1097) + … + a0 N = a0(1099) + a1(1098) + a2(1097) + … + a99 |M – N| = a99(1099 – 1) + a98(1098 – 10) + a97(1097 – 102) + … + a0(1 – 1099) Each term in the bracket is divisible by 9, but not necessarily by 99.

19. Answer: A Explanation: Let us assume three consecutive positive integers to be 1, 2, 3. Then, 1 + 22 + 33 = 32, which is not a perfect square. Similarly, 2 + 32 + 43 = 75, which is also not a perfect square. Further, 3 + 42 + 53 = 144, which is a perfect square. So, smallest value is 3.

24. Answer: A Explanation: 125b = 676  53b = 676  (5a)3b/a = 676 

20. Answer: A

(26)3b/a = 262. Therefore,

Explanation: First term = 1 

25. Answer: B

1 1 9 3 1     2 . 12 22 4 2 2

Explanation: Since ratio is 5 : 7, the numbers will be 5x and 7x and hence 5x + 7x = 96 or x = 8, Therefore, numbers are 40 and 56. Their LCM = 280.

3 49 8 1 First two terms =    3  . And so 2 36 3 3

on. So, sum of all the terms = 2008 

3b = 2 or 3b = 2a. a

1 . 2008

26. Answer: D 21. Answer: A Explanation: 3x + 2y = 5k + 2; 2x + 3y = 5m + 3. Subtracting the equations, we get: x – y = 5 (k – m) – 1, which when divided by 5 leaves remainder –1 or 4.

Explanation: If a card is flipped odd number of times, it will be face up. E.g., card number 9 will be flipped up by first person, then flipped down by third person and then flipped up by ninth person. Card number 6 will be flipped by 1st, 2nd, 3rd and 6th persons and so it will be face down. We also know that if a number is a perfect square, it will have odd number of factors. Hence all perfect square card numbers will be face up. From 1 to 52, there are 7 perfect squares: 12, 22, 32, …, 72.

27. Answer: C Explanation: Number of factors = (x + 1)(y + 1), as the number is already in prime factorized form. 28. Answer: A Explanation:

22. Answer: B

m=

Explanation: Let the number be a two-digit number xy. Then, 10x + y = 8(x + y) + 1 or 2x – 7y = 1. Also, 10x + y = 13(x – y) + 2 or 14y – 3x = 2. Solving these two equations, we get x = 4, y = 1.

n  1  77 77 1 . If m and n both are n1 n1

to be positive integers, n – 1 should be a factor of 77. The number of factors of 77, i.e. 71 × 111 is (1 + 1)(1 + 1) = 4. Therefore, there are 4 positive integer values of m.

68

Number System – I: Factors, Multiples and Exponents 29. Answer: B

= (a – 1)2 + a2 + (a + 1)2 = 3a2 + 2. Now use options. If 3a2 + 2 = 654002, then a2 = 218000, which is not a perfect square. If 3a2 + 2 = 151877, then a2 = 50625 = 2252.

Explanation: 7429 = 17 × 19 × 23; 12673 = 19 × 23 × 29. Therefore, the four prime numbers are 17, 19, 23 and 29. Therefore, the sum is = 17 + 29 = 46.

34. Answer: B

30. Answer: A

Explanation:

Explanation: First of all, we will factorize the given number. So, 10800 = 24 × 33 × 52. Since the factor is divisible by 12, it must have at least two 2’s and at least one 3. But, it is also given that the factor is not divisible by 36. So, the factor cannot have more than one 3. There is no restriction on the number of 5’s. Therefore, we will take combinations of (22, 23, 24), (31), (50, 51, 52). So, number of factors will be = 3 × 1 × 3 = 9.

1

1

1

The given numbers are: 22 , 33 , 44 . LCM of denominators of fractional powers = LCM (2, 3, 4) = 12. Raising each number to the power 

1



12



1



12



1



12, we get:  2 2  ,  3 3  ,  4 4  









12

or 26, 34,



43. The highest among them is 34. Therefore, the highest number is 3 3 . 35. Answer: B Explanation: Given expression

31. Answer: C

 p 1 p  4 4  22  p  2 2 

Explanation: Let the number of soldiers be x2 + 63. Now, x2 + 63 = a2 + b2. Using hit and trial, the minimum value of x is 9 and a = b = 6. So, there are 72 soldiers.



1 p 1 p 2p    1 2 2 2 2

= 2

32. Answer: B



Explanation: Let the original number be 1abcde. After shifting, the number becomes abcde1, whose value is thrice that of 1abcde. 3 × e should be result in a number ending with 1. Therefore, e must be 7 and so carryover is 2. Original and new numbers become: 1abcd7 and abcd71. Now, 3 × d + 2 (carryover) should be result in a number ending with 7. Therefore, d must be 5. Original and new numbers become: 1abc57 and abc571. And the process can be carried further. Finally, the original and new numbers are 142857 and 428571. Therefore, sum of digits = 1 + 4 + 2 + 8 + 5 + 7 = 27.

  

1 /p

    

 2p  1 ( p 1) / 2 2 2 2  p /2 22  

1 /p

    

1/p

  23 p 

1 /p

= 8.

36. Answer: B Explanation: 3094 = 2  7  13  17 = a  b  c  d. As 1  a  b  c  d, a = 2, b = 7, c = 13, d = 17. Therefore, b  c = 7  13 = 91. 37. Answer: B Explanation: Let the length of the rod be 3x. Then, length of each part is x. The first part of length x is divisible by 12, the second part is divisible by 18 and the third part divisible by 32. Therefore, x = LCM of 12, 18 and 32 = 288. Therefore, length of rod = 3x = 288 × 3 = 864.

33. Answer: B

38. Answer: C

Explanation: Let the three consecutive numbers be (a – 1), a, (a + 1). Sum of their squares

Explanation: We need 0.0010101 × 10k > 1000. Therefore, decimal place should shift so that

69

Quantitative Aptitude Simplified for CAT

given number becomes 1010.1. For this, k must be at least 6.

The number will surely end with 0. Let the number be xy0. Upon changing the places the digits, the number becomes yx0 (as the number is still divisible by 10). Now, (100x + 10y) – (100y + 10x) = 90(x – y), which is divisible by 40 or (x – y) is divisible by 4. So, x – y = 4 or 8. Therefore, (x, y) = (5, 1), (6, 2), (7, 3), (8, 4), (9, 5), (9, 1).

39. Answer: B Explanation: 7x – 1 + 11x – 1 = 170. Each term on left side is exponential term, which is a continuously increasing function. So, this equation can have only one solution. By taking values of x as 1, 2, 3, … etc, we easily hit upon the value of x as 3 which satisfies the given equation.

44. Answer: A Explanation: Considering Statement I:

40. Answer: E Explanation: In order to be a perfect cube, every prime number should be a power of 0, 3, 9 … Also, each of p, q, r and s must be minimum possible, as well as more than 0. Therefore, p = 2, q = 2, r = 2 and s = 3. Therefore, the smallest value of p + q + r + s = 2 + 2 + 2 + 3 = 9.

x

y

x+y

x-y

5

2

7

3

Since this is the only possible solution, statement I is sufficient. Considering Statement II:

41. Answer: B Explanation: The least no. of the form mm divisible by 500 is 1010. Therefore, a + b = 10. Hence, minimum value of a × b = 1 × 9 = 9.

x

y

x+y

x-y

5

3

8

2

6

4

10

2

Since no unique solution statement II is not sufficient.

42. Answer: C

is

possible,

45. Answer: E

Explanation: LCM (3, 4, 5, 6) = 60. Therefore, numbers are of the form 60k + 2, which should be divisible by 11. Given number can be written as: 60k + 2 = 55k + 5k + 2. So, k should be such that 5k + 2 is divisible by 11. By hit and trial, we get k = 4. Therefore, smallest such number = 60 × 4 + 2 = 242. General number = 660m + 242. For 6th such number, m = 5 and so number is 660 × 5 + 242 = 3300 + 242 = 3542.

Explanation: pq = qp and q = 9p  p9p = (9p)p  p9 = 9p  p8 = 9  p = 8 9 . 46. Answer: E Explanation: Let us say the number is ab. Then, according to question, a × b + a + b = 10a + b  a × b + a = 10a  a × b = 9a  a(b – 9) = 0  a = 0 or b = 9 Since a can’t be zero, therefore b = 9. Numbers can be 19, 29, 39, … 99. So, 9 such numbers exist.

43. Answer: B

Fraction of numbers =

Explanation:

70

9 = 0.099. 91

Chapt er 02

Number System – II Remainders, Factorials and Base System BASIC CONCEPTS Remainders We all know that when a number such as 235 is divided by 7, we are left with some remainder which can be anything from 0 to a number less than 7 (the divisor). In this case, the remainder is 4. Therefore, 235 = 7n + 4. Let us understand how to find remainders in more complicated cases and see the underlying principle. When we learnt divisibility rules, we saw that if a number upon division by 11 leaves a remainder of –4, then the correct remainder is obtained by adding the divisor, that is, 11 in this case. Therefore, the correct remainder is –4 + 11 = 7. Therefore, for the divisor 11, the remainder of –4 is same as the remainder of +7. For the divisor 7, the remainder of –3 is same as the remainder of +4. Details: When divisor is 7 and remainder is –3, the number can be written as 7n – 3, which is same as 7n – 7 + 4 = 7m + 4. Hence, we can say that the remainder of –3 is same as +4 and similarly, the remainder of –1 would be same as the remainder of +6 (when the divisor is 7). That’s the way negative remainders can be converted to positive ones and vice versa. This rule will be extremely useful while finding the remainders. Example 1 Find the remainder in each of the following processes: (i)

23 is divided by 7

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Quantitative Aptitude Simplified for CAT

(ii)

45 is divided by 7

(iii) 23 + 45 is divided by 7 (iv) 23  45 is divided by 7 Solution (i)

We can easily see that 23 divided by 7 leaves the remainder 2.

(ii)

We can easily see that 45 divided by 7 leaves the remainder 3.

(iii) Instead of adding the numbers 23 and 45 and then finding the remainder, we will find the respective remainders and add those remainders. So, 23 and 45 when divided by 7 leave the remainders 2 and 3 respectively. Hence, 23 + 45 divided by 7 leaves the same remainder as 2 + 3 divided by 7. So, remainder is 2 + 3 = 5. (iv) Just like in case of addition, here also we will multiply the remainders to obtain the final remainder. So, find the individual remainders when each multiplicand is divided by 7. We find that the remainders when 23 and 45 are divided by 7 are 2 and 3 respectively. Multiply these remainders to obtain 6, which when divided by 7 leaves the remainder 6 itself. Therefore, the remainder is 6. Now, of course if we are asked to find the remainder when 2354  6574 is divided by 7, we can easily do so by finding the remainders when 2354 and 6574 are separately divided by 7. When 2354 is divided by 7, the remainder is 2. When 6574 is divided by 7, the remainder is 1. Therefore, when 2354  6574 is divided by 7, the remainder is 2  1, that is, 2. Important Result: When x is divided by n, let the remainder be r1. When y is divided by n, let the remainder be r2. When z is divided by n, let the remainder be r3. Then, when x  y  z is divided by n, the remainder will be r1  r2  r3, provided this product is less than n. If this product is more than n, then divide the product again by n to find the final remainder. Explanation When x is divided by n, since the remainder is r1, therefore x = nq1 + r1, where q1 is some quotient. Similarly, y = nq2 + r2 and z = nq3 + r3. Therefore, x  y  z = (nq1 + r1)(nq2 + r2)(nq3 + r3). When we open the brackets, we see that there are 8 terms, of which first 7 terms are each a multiple of n and the last term is r1  r2  r3. These 7 terms are divisible by n and hence the last term contributes to the remainder. Hence the result. Example 2 What is the remainder when 324 × 431 × 635 is divided by 11? Solution When 324 is divided by 11, remainder is 5; when 431 is divided by 11, remainder is 2 and when 635 is divided by 11, remainder is 8. Therefore, remainder will be 5 × 2 × 8 = 80, which when again divided by 11 leaves the remainder 3.

72

Number System – II: Remainders, Factorials and Base System Alternatively, we can use negative remainders. So, 635 divided by 11 leaves the remainder 8 or 3. So, remainder will be 30 which when divided by 11 leaves remainder 8 and so remainder is 3. Example 3 Find the remainder when 41 × 42 × 43 × 44 × 45 is divided by 46. Solution Remainder = (5)(4)(3)(2)(1) = 120 which when divided by 46 leaves remainder 28 to which if we add 46, we get 18. Or adding 46 three times, we get 120 + 138 = 18, which is the remainder. Alternatively, We can cancel the factor of 2 from dividend and divisor. So, now we have to find the remainder when 41 × 21 × 43 × 44 × 45 is divided by 23. Remainder = (5)(2)(3)(2)(1) = 60, which when divided by 23 leaves the remainder 9. However, it is important to note that 9 should not be treated as the remainder. The obtained remainder of 9 should be multiplied by the cancelled factor of 2 to get the correct remainder 18. Now, we will learn to solve questions of the type when xa is divided by n and the remainder is asked. The first step is to find the remainder when x is divided by n. If remainder is r, then the remainder when xa is divided by n is ra. If ra is more than n, then ra is processed further as described in the examples below. Example 4 What is the remainder when 5023 is divided by 7? Solution 5023 = 50  50  50  ... 23 times

50 divided by 7 leaves the remainder 1. Each 50 would leave the remainder of 1. Therefore, the remainder would be equal to 1  1  1  ..... = 1. Hence, 1 is the remainder. Note: If x divided by n leaves the remainder of r, then xa when divided by n would leave the remainder as ra which would be further divided by n to get the final remainder. Example 5 What is the remainder when 4823 is divided by 7? Solution 4823 = 48  48  48  ..... (23 times) 48 divided by 7 leaves the remainder 1. Each 48 would leave the remainder of 1. Therefore, the remainder would be equal to (1)  (1)  (1)  ..... (23 times) = 1. Hence, the remainder is 1 or 1 + 7 = 6. Example 6 What is the remainder when 6726 is divided by 17?

73

Quantitative Aptitude Simplified for CAT

Solution 6726 divided by 17 leaves the remainder = (1)26 = 1 (because 26 is an even number). Example 7 Find the remainder when 5145 is divided by 7. Solution 51 divided by 7 leaves the remainder 2. Therefore, the remainder would be same as when 245 is divided by 7. Now, 245 = (23)15 = 815. Since 8 divided by 7 leaves a remainder of 1, therefore 8 15 divided by 7 leaves the same remainder as 145 would leave, and, that is, 1 only! In the above example, we saw that in one of the steps 245 was written as (23)15. The objective of doing this is that 2 should be raised to its lowest power such that the result is “one more” or “one less” than the divisor. Therefore, 2 was raised to the power 3 so that the result 8 is one more than the divisor 7. This process is done intuitively or by hit and trial. Example 8 Find the remainder when 5137 is divided by 7. Solution 5137 = 237 (as the divisor is 7). Now, 237 = 2 × 236 = 2 × (23)12 = 2 × 812 = 2 × 112 = 2. Alternatively, Since 2 must be raised to the power 3 to get a result which is one more than the divisor, the power of 2 must be a multiple of 3. Since the power is 37 which is not a multiple of 3, divide 37 by 3 and obtain the remainder 1. Now, 2 should be raised to this remainder to obtain the result. Therefore, 237 = 21  remainder is 2.

EXPERT SPEAK Scan this QR Code to watch a video that explains the basic concepts involved in finding Remainders.

Example 9 Find the remainder when 51 37

42

is divided by 7.

Solution 51 37 7

42

42

2 37  7

.

Now, 2 should be raised to the power 3, which gives us 8 which is one more than 7. Now, 3742 when divided by 3 leaves remainder 1.

74

Number System – II: Remainders, Factorials and Base System 42

Therefore, 2 37

 21  2  2 is the remainder when 51 37

42

is divided by 7.

Example 10 Find the remainder when 549 is divided by 8. Solution Method 1: 549 = (–3)49 (as the divisor is 8) = –(3)49. Here, we see that 3 raised to the power 2 is 9 which is one more than the divisor 8. Therefore, –(3)49 = –3(3)48 = (–3)(9)24 = (–3)(1) = –3  remainder is 5. Method 2: 549 = 5(5)48 = 5(25)24. But, 25 divided by 8 leaves the remainder 1. Therefore, 5(25)24 = 5(1)24 = 5, which is the remainder. So, whether you deal with negative remainders (and later convert them to positive remainders), or with positive remainders, it does not make any difference! We should somehow be able to get the remainder. In the last two examples, we saw that 2 raised to the power 3 yields 8 which when divided by 7 leaves the remainder 1 (and we neglected the term in which 2 was raised to the power “multiple of 3”). Also, when 5 raised to the power 2 is divided by 8, the remainder is again 1. But, there are cases where we are left with a remainder other than 1. In such cases, one must be extremely careful and proceed with caution. Refer to the following example to understand this more clearly. Example 11 Find the remainder when 3637 is divided by 17. Solution 3637 = 237 (because divisor is 17). Now, 237 = 236  2 = (24)9  2 = 169  2. 16 divided by 17 leaves the remainder of 16 or –1 (recall the process of conversion of positive remainders to negative and vice versa). Note that here we cannot neglect this. Therefore, 169  2 = (–1)9  2 = –2, which means the remainder is 15 because divisor is 17. Example 12 Find the remainder when 36 37

38

is divided by 17.

Solution 36 37

38

 2 37

38

(as the divisor is 17). Once again, 2 should be raised to the power 4.

But, power is 3738, which is not a multiple of 4. To make it a multiple of 4, divide 3738 by 4. The remainder is 1. Therefore, 24N+1 = 16N  2 = (–1)N  2. Now, we need to know whether N is an even or an odd number.. To overcome this difficulty, we need to raise 2 to the power 8 instead of 4. We observe that 2 when raised to the power 8 is 256 which when divided by 17 leaves the remainder 1. In this case, divide 3738 by 8. When 3738 is divided by 8, we get 538 = 2519 = 119 = 1 (as divisor is 8)

75

Quantitative Aptitude Simplified for CAT

Therefore, 36 37 divided by 17.

38

 2 37

38

= 28N+1 = 256N  2 = 1N  2 = 2, which is the remainder when the given number is

Example 13 11

Find the remainder when 11 11

is divided by 7.

Solution 11

11 11

11

 411 .

Now, 4 raised to the power 3 is 64 which when divided by 7 leaves the remainder 1. Therefore, the power 1111 should be divided by 3 to get the remainder 211 = (–1)11 = –1 = +2. 11

Therefore, 411

= 43N+2 = 42 = 16 which leaves the remainder 2 when divided by 7.

Example 14 11

11 Find the remainder when 11

is divided by 9.

Solution 11

11 11

11

 211 .

Now, 26 = 64 which when divided by 9 leaves the remainder 1. But, 1111 when divided by 6 leaves the remainder –1 and hence +5. 11

Therefore, 211

= 26N+5 = 25 = 32  5.

Example 15 Find the remainder when 296 is divided by 96. Solution Divisor is 96 which can be factorized as 25  3. In cases where the dividend and divisor have some common 296 291 factors, we can delete the common factors. Therefore,  . 96 3 Now, 291 = (–1)91 = –1 = +2 (as the divisor is 3) A very common mistake students commit is to think that the remainder is finally 2, which is absolutely wrong! 2 is the remainder when 291 is divided by 3 and not when 296 is divided by 96. In cases where the common factors were deleted to find the remainder, the correct remainder is obtained by multiplying the obtained remainder with the common factor deleted. In the current example, the common factor deleted is 32 and, hence, the obtained remainder 2 is to be multiplied with 32 to get 64, which is the correct remainder. The question is why do we have to multiply like this. The reason is simple: We have seen that 291 when divided by 3 leaves the remainder 2. This can be written in the form of division algorithm as 291 = 3q + 2 where q is some quotient. Multiply both sides of the equation with 32 (the deleted factor), 291  32 = 3q  32 + 2  32 296 = 96q + 64

76

Number System – II: Remainders, Factorials and Base System which tells us that when 296 is divided by 96, the remainder is 64 (and not 2). Example 16 Find the remainder when 7100 is divided by 2402. Solution We know that 74 = 2401. Therefore, 7100 = (74)25 = 240125 = (–1)25 = –1  remainder is 2401. Example 17 Find the remainder when 31 ! 31 !

31 !

is divided by 91.

Solution 31! = 1  2  3  4  ..... 30  31. We see that 91 = 13  7. Since both the factors: 13 and 7, are present in 31!, obviously the remainder will be zero. Example 18 What is the remainder when 2431500 – 1 is divided by 1001? Solution We know that 1001 = 7  11  13. Let us check for the given number’s divisibility by 7, 11 and 13 separately. 2431500 (2)1500 (2 3 ) 500 8 500    1 7 7 7 7 Therefore, when 2431500 is divided by 7, the remainder is 1. Hence, 2431500 – 1 when divided by 7 would leave no remainder. Similarly,

2431500 (1)1500   1 and, hence, 2431500 – 1 when divided by 11 leaves no remainder. 11 11

2431500 (4)1500 (64) 500    1 and, hence, 2431500 – 1 when divided by 13 also leaves no remainder. 13 13 13 Therefore, when 2431500 – 1 is divided by 1001 it would leave no remainder. Example 19 For the number N = 1! + 2! + 3! + ...... + 99! (a)

find the remainder when N is divided by 5. Also find the same for division by 4 and 6.

(b)

find the smallest number n > 3 that will always divide the given number.

(c)

find the remainder when NN is divided 4, 5 and 6 separately.

Solution (a)

We see that 5! onwards, every number ends with 0 and hence when divided by 5 leaves no remainder. Therefore, the remainder is because of 1! + 2! + 3! + 4!, whose value is 33 which when divided by 5 leaves the remainder 3.

To check for remainder when divisor is 4, we observe that 4! onwards, all the numbers are divisible by 4. Hence, remainder would be same when 1! + 2! + 3! is divided by 4 or when 9 is divided by 4. Therefore, remainder is 1.

77

Quantitative Aptitude Simplified for CAT

For 6, we see that 3! onwards, all the numbers are divisible by 6. Therefore, remainder is 1! + 2! = 3. (b)

Since 4, 5 and 6 separately do not divide the given number, check for 7, then 9 and so on. We see that 9 is the next such number by which N is divisible. To check for 9, we see that 6! = 720 which is divisible by 9 and hence 6! onwards, every number is divisible by 9.

Moreover, 1! + 2! + ... + 5! = 1 + 2 + 6 + 24 + 120 = 153, which is also divisible by 9. Therefore, N is divisible by 9. (c)

N divided by 4 leaves the remainder 1. Therefore, when NN is divided by 4 it leaves the remainder 1N = 1.

N divided by 5 leaves the remainder 3. Therefore, NN = 3N. Now, N = 4x + 1 (because when N is divided by 4, we get 1 as remainder) Therefore, 3N = 34x+1 = 34x  31 = 81x  3 = a number which ends with 3. We know that in case of division by 5, the remainder is same as the unit’s digit of the number. Therefore, when NN is divided by 5, the remainder is again 3. N when divided by 6 leaves the remainder 3. Therefore, N = 6p + 3. Therefore, NN = (6p + 3)N which will always be odd and a multiple of 3. Therefore, the number upon division by 6 must leave the remainder of 3. This is so because any odd number can be written as either 6n + 1 or 6n + 3 or 6n + 5. But 6n + 1 and 6n + 5 are not multiples of 3 and hence cannot be the value of (6p + 3)N. Example 20 Using the five fingers of your hand, you start counting, starting with thumb as 1, then index finger as 2, and so on and when you reach the smallest finger counting it as 5, you move to the finger adjacent to it counting it as 6, then next to it counted as 7, next to which is index finger counted as 8 and then thumb counted as 9. You keep counting till you count 2002. Which finger are you on? If you use both the hands (all the ten fingers), when you count 2002, which finger will you be on? Solution First cycle of counting ends when you count 8 (and you are on index finger). Next cycle ends at 16. Therefore, each cycle ends when you count a number which is a multiple of 8. This means that when you count a number which is a multiple of 8, you land up at the index number (towards the end of the cycle) and tend to go towards the thumb and return. 2002 when divided by 8 leaves a remainder of 2. Therefore, when you count 2000, you are at the index finger. You will count 2001 at the thumb and 2002 back at the index finger. When both the hands are used, the cycle completes when you count 18 (and you are on the index finger). Therefore, 2002 divided by 18 leaves the remainder 4. Therefore, on counting 2002, you will be on the ring finger (next to the smallest finger) of the hand with which we started counting.

Fermat’s Theorem If p is a prime number, then np–1 divided by p leaves remainder 1 if n and p are co-prime. For example, 522 divided by 23 leaves remainder 1 since 23 is a prime number and 5 and 23 are co-prime. In general, n(p–1)k when divided by p leaves a remainder 1. So, 1284 divided by 29 leaves remainder 1 because 1284 = 1228×3 and we know that 1228 divided by 29 leaves remainder 1. Note that 34–1 – 1 = 33 – 1 = 26 is not divisible by 4 since 4 is not a prime number.

78

Number System – II: Remainders, Factorials and Base System Example 21 Find the remainder when 296 is divided by 97. Solution Since 97 is a prime number, 296 – 1 is divisible by 97 and hence 296 when divided by 97 leaves a remainder of 1. Example 22 Find the remainder when 81000 is divided by 29. Solution 81000 = 8(28×35+20) = 8(28×35) × 820 = 1 × 820 = 6410 = 610 = 365 = 75 = 2401 × 7. Since 2401 when divided by 29 leaves remainder 23, the remainder is 23 × 7 = (6) × 7 = 42 = 16.

To Find the Last Digit of Numbers We have already learnt how to find the remainder when a particular number is divided by another number. The concepts and ideas developed there will be found useful here. The rule of cyclicity is used to know what will be the unit’s digit of any number raised to the power any other number. Therefore, we can find the unit’s digit of numbers such as 2354657890.

Rule of cyclicity Let us take an example, say 34745. Note that the unit’s digit of 34745 will be same as unit’s digit of 4745 as also unit’s digit of 745. We can say that the unit’s digit of 34745 would depend upon the unit’s digit of the base number along with the value of the exponent. In the unit’s digit of any number, we can have any one of the digits in the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Also note the following results: If a number ends with 0, then this number raised to the power any natural number always ends with 0”. The same is true for a number ending with 1, 5 or 6. Therefore, (.....0)n = (......0) (.....1)n = (......1) (.....5)n = (......5) (.....6)n = (......6) Therefore, the unit’s digit repeats in case the unit’s digit in the original number is 0 or 1 or 5 or 6. Example 23 Find the unit’s digit of (a)

26534

(b)

54005467

(c)

43212345

Solution (a)

Since the unit’s digit of 265 is 5, we can say that the unit’s digit of 26534 will also be 5.

(b)

Unit’s digit being 0, the unit’s digit of 54005467 will also be 0.

(c)

Similarly, the unit’s digit of 43212345 will also be 1.

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Quantitative Aptitude Simplified for CAT

Example 24 In the previous example, part (b), find the number of zeros at the end of the number. Solution 5400 ends with 2 zeros. Therefore, 54005467 ends with 2  5467 = 10934 zeros! Now, the following rules pertain to the numbers ending with 2, 3, 4, 7, 8 and 9. (any odd number)4n = (………1) (any even number)4n = (………6) So, any odd number, that is number ending with either 3 or 7 or 9, when raised to the power which is a multiple of 4, will always end with 1. So, for example, 2348 ends with 1. Similarly, 4268 ends with 6. What if the power is not a multiple of 4? Let us take the same example to elaborate this. 34745 = (.....7)45 = (.....7)44+1 = (.....7)44 × (.....7)1 = (.....1) × (.....7) = (.....7). Therefore, in such cases, we will break the power into two parts: the first part being a multiple of 4 and the other will be the rest. In the current case, 45 was written as 44 + 1. Then, we applied the law of indices and proceeded accordingly. Alternatively, divide the power 45 by 4 and we are left with the remainder 1. Place the remainder in the power of the unit’s digit of the base number. So, 34745 = (…7)1 = (…7). Example 25 Find the unit’s digit of 12345677654314. Solution Power divided by 4 leaves the remainder 2. So, 12345677654314 = (...........7)2 = (.......9). Note: From the foregoing discussion, we can generalize that (a)

If unit’s digit of a number is 0 or 1 or 5 or 6, the unit’s digit repeats no matter what the power of that number is.

(b) For other digits, apply the following rule (any odd number)4n = (.........1) (any even number)4n = (.........6) If the power is not a multiple of 4, then divide the power by 4 and keep the remainder only in the power of the unit’s digit of the base number.

EXPERT SPEAK Scan this QR Code to watch a video that explains the basic concepts involved in finding Unit's Digit of any number.

Thumb rule for finding the unit’s digit of ab

c

c

The thumb rule is that while finding the unit’s digit of the number ab , we check – unit’s digit of the base number, that is, a.

80

Number System – II: Remainders, Factorials and Base System – whether the immediate power, that is, b is a multiple of 4 or not. If not, then bc should be divided by 4 and the remainder should be found. This remainder then becomes the power of ’a’. Example 26 Find the unit’s digit of (a) 37 47 (e) 2 3

57

54

(b) 37 47

 23

45

58

(c) 37 46

57

(d) 23

1516

17

(g) 1314

(f) 1315

45

(h) 7!7!

 54

32

7!

Solution (a)

The power 4757 is not a multiple of 4, so we will divide 4757 by 4. We see that the remainder is (1)57 = 1 = 3. Therefore, unit’s digit = (....7)3 = (....3). Therefore, the number ends with 3.

(b)

When 4758 is divided by 4, the remainder is (1)58 = 1. Therefore, unit’s digit = (....7)1 = (....7).

(c)

In the number, we see that the power is 4657, which upon close examination, is a multiple of 4. This is so because 4657 = (46 × 46) × (4655). The number in the first bracket is a multiple of 4. Hence the result. In general, (any even number)n = multiple of 4, where n > 1. Therefore, unit’s digit = (.....7)4n = (.....1).

(d)

A single 2 and a single 5 multiplied with each other will result in a number which ends with zero. In the current question, there are large number of 2’s and 5’s. Therefore, the unit’s place will naturally be 0.

(e)

To find the unit’s digit of 2 3

54

4

, we divide the power by 4 and find the remainder. So, 3 5 divided by 4

4 leaves the remainder (1)5 = 1 = 3. So, unit’s digit of the number = 23 = (…8) To find the unit’s digit of 2 3 the remainder (1) Note that 2 3 23

54

 23

45

45

45

45

, we divide the power by 4 and find the remainder. So, 3 4

5

divided by 4 leaves

= 1. So, unit’s digit of the number = 21 = (…2)

is bigger than 2 3

54

because 45 > 54. Therefore,

54   45    23  2 3  = [(….2) – (….8)] = (….4).  

So, unit’s digit is 4. (f)

1517 divided by 4 leaves the remainder –1 or +3. Therefore, unit’s digit = (....3)3 = (....7)

(g)

Since 1415

(h)

Since 7! ends with zero, 7!7!

16

is a multiple of 4, unit’s digit = (....3)4n = (....1). 7!

also ends with zero.

Example 27 x

If x = 1! + 2! + 3! + 4! + ....+ 1000!, find unit’s digit of x, xx and x x . Solution We have seen earlier that x ends with 3. To find unit’s place digit of xx, we must know that x = 1! + 2! + 3! + (4! + ......+ 1000!) and the term in the bracket is a multiple of 4 (each term being a multiple of 4). Therefore, x = 1 + 2 + 6 + 4n = 1 + 8 + 4n = 4m + 1. Therefore,

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Quantitative Aptitude Simplified for CAT

xx = (....3)4m+1 = (....3)1 = (....3). x

For x x , the power is xx which is equal to (4m + 1)4m+1, which when divided by 4 leaves the remainder 1. Therefore, unit’s digit = (.....3)1 = (.....3). Example 28 Find the unit’s digit of 13125 + 23125 + 33125 + 43125 + .... + 993125. Solution 3125 divided by 4 leaves the remainder 1. Therefore, required unit’s place will be same as the unit’s place of (1 + 2 + 3 + ....9) + (1 + 2 + 3 + ....9) + ....+ (1 + 2 + 3 + ....9), where (1 + 2 + 3 + ....9) appears 10 times. This is equal to 45 × 10 = 450. Therefore, required unit’s place is 0. Example 29 Find the remainder when 2367 is divided by 10. Solution Note that remainder upon division by 10 is same as the unit’s digit of 2367. The unit’s digit of the given number would be 7.

To find the last two digits of numbers Let us now find out the technique of finding the 10’s place digit of the numbers such as1234128. First of all, the students should know that the 10’s place digit of the number would be dependent upon the unit’s and ten’s place digit of the base, that is, 34. To begin with, let us find the last two digits of a number, say 422. To find that, find the difference between 42 and 50, which is 8. Square of 8 ends with the same last two digits as the square of 42 ends with, that is 64. Similarly, to find the last two digits of 822, find the difference between 82 and 100, which is 18. Square of 18 ends with the same last two digits as the square of 82 ends with, that is 24 (since 182 = 324). So, we can say that to find the last two digits of x2, find the difference of x from 50 or 100 (whichever is closer to x) and find the square of this difference. To find last two digits of 682, we find the difference between 50 and 68, which is 18. Now, 182 ends with 24 and hence 682 also ends with 24. To find last two digits of 1342, we find the difference between 150 (because this is a multiple of 50 nearest to 134) and 134, which is 16. Since 162 ends with 56, therefore 1342 also ends with 56. Alternatively, the digit 1 in 134 is redundant as only last two digits are important. So, 1342 ends with same last two digits as 342 ends with. So, gap from 50 is 16 whose square ends with 56 and so 1342 ends with 56. The logic for this is that any natural number can be expressed as 50k ± a, where 0  a  25. Therefore, (50k ± a)2 = 2500k2 ± 100ak + a2. So, only a2 contributes to the last two digits. Therefore, the last two digits of any perfect square will always be same as the last two digits of one of the natural numbers from 1 to 25. The ‘last two digits’ of the squares of natural numbers from 1 to 25 are: 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25. Therefore, a perfect square can only end with

82

Number System – II: Remainders, Factorials and Base System 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96. This would help in finding the last two digits of the numbers like 34128. 34128 = (....56)64 = (.....36)32 = (.....96)16 = (.....16)8 = (.....56)4 = (.....36)2 = (.....96). Here, we did squaring successively, each time making use of the above technique of taking gap from 50 or 100. Example 30 Find the remainder when 2367 is divided by 100. Solution Like in one of the previous examples, we can understand that here, the last two digits are asked. We need to do successive squaring to find the answer. 2367 = (23)66  (23) = (....29)33  (23) = (....29)32  (23)  (29) = (....41)16  (67) = (....81)8  (67) = (....61)4  (67) = (....21)2  (67) = (....41)  (67) = (....47) Hence, the last two digits of the number are 47. We observe that the above process can become lengthy if the power is too large. To avoid lengthy calculations, refer to the following rules: 

(any odd number)20n = (…… 01)

(except those numbers ending with 5)



(any even number)20n = (…… 76)

(except those numbers ending with 0)

Therefore, 1740 = (…. 01) and 42160 = (…. 76). In case the power is not a multiple of 20, we split the powers as we did in case of finding unit’s digit.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to find Ten's Place digit of a number.

Example 31 Find the last two digits of 2645. Solution 2645 = 2640 × 265 = (....76) × 264 × 26 = (....76) × 762 × 26 = 76 × 26 = 26. Example 32 Find the last two digits of 561000. Solution Since power is a multiple of 20, we get 561000 = (....76). Example 33 Find the last two digits of 37255.

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Quantitative Aptitude Simplified for CAT

Solution 37255 = 37240 × 3715 = (....01) × 3714 × 37 = 697 × 37 = 696 × 69 × 37 = 613 × (… 53) = 612 × 61 × (… 53) = 21 × 61 × 53 = (….93) If a number ends with 5, then the rule is: (…..a5)n = (…. 75) , if both a and n are odd = (…. 25) , otherwise where a is the ten’s place digit of the base number. Example 34 Find the last two digits of 1351235. Solution Since both ten’s place digit, 3 as well as power 1235 are odd, the last two digits are 75.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to apply the concept of finding Ten's Place digit to more cases.

Example 35 18

Find the last two digits of 1416 . Solution Since power is not a multiple of 20, we need to split it. For this, we divide the power 1618 by 20 and find the remainder. Remainder upon division by 20 can be obtained if we know the last two digits of 1618. So, 1618 = (… 56)9 = (….56)8 × (… 56) = (… 36)4 × (…56) = (… 96)2 × (… 56) = (…16) × (…56) = (….96). So, when this is divided by 20, remainder is 16. Therefore, given number = 1416 = (…96)8 = (…16)4 = (…56)2 = (…36).

Factorial Factorial of a number is symbolically written as n! or n . We will use the symbol n! to denote factorial of a number n. So, we have factorials such as 5!, 7!, 100! and so on. 5! means 1  2  3  4  5 7! means 1  2  3  4  5  6  7 100! means 1  2  3  4  5  ......  99  100 In general, n! means 1  2  3  ......  (n – 1)  n. So, n! is the product of the first n natural numbers. Therefore, 1! = 1 2! = 1  2 = 2 3! = 1  2  3 = 6

84

Number System – II: Remainders, Factorials and Base System 4! = 1  2  3  4 = 3!  4 = 24 5! = 1  2  3  4  5 = 4!  5 = 120 6! = 1  2  3  4  5  6 = 5!  6 = 720 Note that 5! can be written as 4!  5. In general, n! can also be written as (n – 1)!  n. Also, the value of 0! is 1 (by definition). It is important to know that factorial of negative numbers or fractions do not exist. So, in n!, n must be a whole number. Therefore, we cannot have cases such as (2.5)!, (–3)! and so on. Example 36 Find the LCM and HCF of 5!, 12! and 25! Solution Since 25! is a multiple of 12! as well as 5!, we can say that 25! is the LCM. Since 5! is a factor of 12! as well as 25!, we can say that 5! is the HCF. Example 37 Find the value of

12! 10!

Solution 12! = 12  11! = 12  11  10!. So,

12! 12  11  10!  = 12  11 = 132. 10! 10!

Example 38 Find the value of

1 1 1 1    ...  . 1! 2! 3! 6!

Solution 6! is the LCM of the denominator terms in the given expression. So,

1 1 1 1 720  360  120  30  6  1 1237    ...   = 1! 2! 3! 6! 6! 720

To find the highest power of a prime number in N! Let us take 10! 10! = 1  2  3  4  5  6  7  8  9  10 = 2  3  (2  2)  5  (2  3)  7  (2  2  2)  (3  3)  (2  5) = 28  34  52  71, which is the prime factorization of 10!. Note that the highest power of 2 in 10! is 8. Similarly, the highest powers of 3, 5 and 7 in 10! are 4, 2 and 1 respectively. We can ask the following questions 

If 10! is divisible by 2a, what is the largest value of a?



If we start dividing 10! by 2 successively, how often can we continue to divide 10! by 2 without being left with any remainder?

The answer to all these questions is 8, same as the highest power of 2 in 10!. 10! being a relatively small number, we could write it as the product of first 10 natural numbers and subsequently find the number of 2’s, number of 3’s, and so on. But such an observation based method is not

85

Quantitative Aptitude Simplified for CAT

going to be very successful if we have factorials of much bigger numbers. For example, if we were to find the number of 2’s in 100!, the above approach completely fails. We need to have some method for such cases. To find the highest power of 3 available in 50! The number 50 in 50! is taken and divided by 3 successively disregarding the remainders each time. Therefore, 50 divided by 3 gives the quotient 16 and remainder 2. Now, 16 when divided by 3 gives quotient 5, which when further divided by 3 gives the quotient 1. Now, since the quotient 1 is less than the divisor 3, the process stops here. Add all the quotients: 16 + 5 + 1 = 22. So, 22 is the highest power of 3 in 50!. Therefore,

50 16 5 = 16, = 5 and = 1 (disregarding remainders each time) 3 3 3

The highest power of 3 in 50! = 16 + 5 + 1 = 22. Example 39 Find the highest power of 7 by which 105! is divisible. Solution 105 15 = 15; = 2 . Therefore, 15 + 2 = 17 is the highest power of 7 by which 105! is divisible. 7 7

To find the highest power of a composite number in N! The above method (of successive division and adding the quotients) is applicable only when the divisor is a prime number. If the divisor is a composite number, then this method cannot be applied. What happens if the divisor is not a prime number? We have seen that the highest power of 2 in 10! is 8. But, if we are required to find the highest power of 4 in 10!, then we should first convert 4 to 22. Now, 28 = 224 = (22)4 = 44. So, if the highest power of 2 in 10! is 8, then the highest power of 4 in 10! is 4. In 10!, what would be the highest power of 8? Again, 8 = 23. So, 28 = 232+2 = (23)2 22 = 82 22. The highest power of 8 in 10! is 2. The other factor of 22 does not yield another 8, and hence is redundant. Let us say we want to find the highest power of 6 in 25!. Express 6 as a product of co-prime pair of 2 numbers. We know that 6 = 2  3, where (2, 3) are co-prime. Now, we will count the highest power of 2 and 3 in 25!. Applying the method, we find that the highest power of 2 in 25! = 12 + 6 + 3 + 1 = 22, and the highest power of 3 = 8 + 2 = 10. Therefore, 25! = 222  310  .... = 212  (210  310)  .... = 212  610  .... Hence the highest power of 6 in 25! is 10. Effectively, the highest power of 6 in 25! is same as the smaller of the powers of 2 and 3. Since 10 (the power of 3) is less than 22 (the power of 2), the highest power of 6 in 25! = 10. Let us now find the highest power of 72 in 25!. We know that 72 = 8  9, where (8, 9) are co-prime. To find the highest power of 8, we need to first find the power 2, and to find the highest power of 9, we need to find the power of 3. The highest power of 2 in 25! = 22. Now, 222 = 237+1 = (23)7 21 = 87 21. So, the highest power of 8 in 25! = 7. The highest power of 3 in 25! = 10. Now, 310 = 325 = (32)5 = 95. So, the highest power of 9 in 25! = 5.

86

Number System – II: Remainders, Factorials and Base System Now, 5 (the highest power of 9) is less than 7 (the highest power of 8). So, the highest power of 72 in 25! is 5 (the smaller of the powers of co-prime pair). Example 40 Find the highest power of 12 in 70!. Solution 12 = 22  3. Therefore, to find 12’s, we need to find 2’s and 3’s in 70!. Number of 2’s in 70! = 35 + 17 + 8 + 4 + 2 + 1 = 67. Number of 3’s in 70! = 23 + 7 + 2 = 32. Therefore, 70! = 267  332  .... = 2  266  332  .... = 2  (433  332)  .... = 2  (4  432  332)  .... = 2  4  1232  .... Therefore, the highest power of 12 in 70! is the smaller of the powers of 4 and 3. Since the powers of 4 and 3 are 33 and 32 respectively, the required highest power of 12 is 32. Example 41 In the number 50!, find the highest power of 2, 3, 6, 12, 18, 24 and 72. Solution The highest power of 2 in 50! = 25 + 12 + 6 + 3 + 1 = 47. The highest power of 3 in 50! = 16 + 5 + 1 = 22. Since 6 = 2  3, the highest power of 6 will be the smaller of the powers of 2 and 3. Therefore, the highest power of 6 in 50! = 22. Since 12 = 4  3, we need to first find the highest power of 4. Now, 247 = 2223+1 = (22)23 21 = 423 21. So, the highest power of 4 is 23, and hence the highest power of 12 is 22. For 18, the co-prime pair is (2, 9). Now, 322 = 3211 = (32)11 = 911. So, the highest power of 18 is 11. For 24, the co-prime pair is (8, 3). Now, 247 = 2315+2 = (23)15 22 = 815 22. So, the highest power of 24 is 15. For 72, the co-prime pair is (8, 9). We have already found the highest powers of 8 and 9, which are 15 and 11 respectively. So, the highest power of 72 in 50! = 11. From the above example, we can make two observations: 



If a composite number c = a  b, where a and b are prime numbers and a < b, then the highest power of c in n! is same as the highest power of b, the bigger prime number. For example, to find the highest power of 6, it is sufficient to find the highest power of 3, the bigger of the two prime numbers 2 and 3. This is simply because in n!, there will be less number of 3’s than number of 2’s, and we are only interested in finding the smaller power of the co-prime pair.

However, the above is not necessarily applicable when at least one of the two numbers in the co-prime pair is a composite number. While finding the highest power of 12, the co-prime pair is (3, 4), where 4 is a composite number. Even if 4 is more than 3, the power of 4 is more than the power of 3 in 50!, and so the power of 3 determines the power of 12 in 50!.

To find the number of zeros at the end of N! Recall that 1! = 1

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Quantitative Aptitude Simplified for CAT

2! = 1  2 = 2 3! = 1  2  3 = 6 4! = 1  2  3  4 = 3!  4 = 24 5! = 1  2  3  4  5 = 4!  5 = 120 6! = 1  2  3  4  5  6 = 5!  6 = 720 Note that 5! = 120, which ends with a single zero. What is the reason of the presence of this zero? Zero at the end of a number means it is a multiple of 10. In fact, 120 can be written as 12  10 = 12  101. So, we can say that the number of zeros in 5! (= 120) is same as the highest power of 10 in 12  101. Consider the number 143600000. This number ends with 5 zeros. The number can also be written as 1436  105. So, the number of zeros is same as the highest power of 10 in 1436  105. We can generalize that to find the number of zeros at the end of n!, we only need to find the highest power of 10. To find the highest power of 10, we write 10 = 2  5, where (2, 5) is co-prime pair. Hence, to find the highest power of 10 in n!, we should find the highest power of 2 as well as 5, and whichever power is the smaller, that is the power of 10. But, we have already mentioned that the power of bigger prime number will be less than the power of the smaller prime number. So, the power of 5 will be always less than the power of 2 in n!. Therefore, the highest power of 5 will determine the number of zeros at the end of n!. We can now say that, the number of zeros at the end of n! is same as the highest power of 5 in n!. Example 42 Find the number of zeros at the end of 100!. Solution To find number of zeros, we need to find the highest power of 10 and that means 2  5. Number of 2’s = 50 + 25 + 12 + 6 + 3 + 1 = 97. Number of 5’s = 20 + 4 = 24. Therefore, number of 10’s in 100! is 24 and, hence, the number of zeros at the end of 100! is also 24. Obviously, there was no need to find the highest power of 2, but it is shown here so that we can observe that the power of 2 is more than that power of 5. The highest power of 5 in 5! = 1, and so 5! ends with 1 zero. The highest power of 5 in 6! = 1, and so 6! ends with 1 zero. The highest power of 5 in 7! = 1, and so 7! ends with 1 zero. It is important to note that each of the factorials 5!, 6!, 7!, 8! and 9! ends with 1 zero only. The reason is that 6! = 6  5!, which means 5! is multiplied by 6. Multiplying 5! by 6 does not increase the number of 5’s and hence does not increase the number of zeros. The moment we reach 10!, the introduction of 10 leads to increase in one 5, and hence the total number of 5’s becomes 2, that is the total number of zeros becomes 2. So, the highest power of 5 in 10! = 2, and so 10! ends with 2 zeros. It should now be obvious that each of 10!, 11!, 12!, 13! and 14! ends with 2 zeros; factorials from 15! to 19! end with 3 zeros; factorials from 20! to 24! end with 4 zeros; factorials from 25! to 29! end with 6 zeros. Note that the number of zeros suddenly jumped from 4 to 6 the moment we moved from 24! to 25!. The reason is that 25 contains two 5’s and multiplying 24! with 25 leads to introduction of two 5’s and hence two zeros. It is now obvious that for any natural number n, n! can never end with exactly 5 zeros.

88

Number System – II: Remainders, Factorials and Base System Similar jumps in number of zeros will happen whenever the multiplying factor has more than one 5, that is when n = 50, 75, 100, 125, and so on. For example, 49! ends with 9 + 1 = 10 zeros, whereas 50! ends with 10 + 2 = 12 zeros, that is jump of two zeros. Example 43 If n! ends with 20 zeros, find n. Solution 100! ends with 24 zeros. Therefore, n has to be less than 100. Let’s try 75!. Now, 75! ends with 15 + 3 = 18 zeros. Therefore, 80! ends with 19 zeros and hence 85! ends with 20 zeros. But, 86!, 87!, 88! and 89! also end with 20 zeros. So, n can take 5 values, that is 85, 86, 87, 88 and 89. Example 44 If n! ends with exactly 30 zeros, find the value of n. Solution Factorial of 100 to 104 will have 24 zeros. Factorial of 105 to 109 will have 25 zeros. Continuing like this, we see that factorial of 120 to 124 will have 28 zeros. Now, number of zeros in 125! = 25 + 5 + 1 = 31 zeros. Therefore, there is no number n, whose factorial will ever end with 30 zeros!, Also, n! cannot end with exactly 29 zeros. Example 45 Find the number of zeros at the end of 10!  20!  30!  40!  50!. Solution We need to find the highest power of 5 in each of the factorial terms. The highest power of 5 in 10! = 2. The highest power of 5 in 20! = 4. The highest power of 5 in 30! = 6 + 1 = 7. The highest power of 5 in 40! = 8 + 1 = 9. The highest power of 5 in 50! = 10 + 2 = 12. Since all the terms are multiplied, the number of zeros in each will be added to get the total number of zeros, that is 2 + 4 + 7 + 9 + 12 = 34. Example 46 A person starts multiplying consecutive positive integers starting from 20. How many integers should he multiply to get a number which will end with 3 zeroes? Solution A number will end in 3 zeroes when it is multiplied by three 10’s. Factorizing 10, you get, 10 = 5  2  He should multiply till he gets three 5’s and three 2’s in the numbers being multiplied. The number 20 already has one 5. If we multiply 20 with positive integers till 24, number of 2’s increases, but number of 5’s does not increase. The moment we multiply with 25 (which has two 5’s), the number of 5’s becomes 3, whereas number of 2’s has already gone beyond 3. So, he has to multiply from 20 to 25 that is 6 numbers. Example 47 Find the number of zeros at the end of

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Quantitative Aptitude Simplified for CAT

(i)

11  22  33  .....  100100.

(ii)

1!1!  2!2!  3!3!  .....  10!10!.

(iii) 1!1! + 2!2! + 3!3! + ..... + 10!10!. Solution (i)

The number of zeros is same as the number of fives. Five appears in terms like 55, 1010, 1515, and so on. Moreover, terms like 2525 have 5 appearing 50 times and not 25 times. Therefore, we will count them as 25 + 25. Same goes for 5050, 7575 and 100100. Therefore, number of fives = (5 + 10 + 15 + 20 + .....+ 90 + 95 + 100) + (25 + 50 + 75 + 100) = 1050 + 250 = 1300.

(ii)

In this case, the number of zeros will be because of 5’s. We see that 5’s appear only in 5!5! and 10!10!. Therefore, number of 5’s = 5! + 2(10!).

(iii) Here, the terms are getting added. Now, 1!1! + 2!2! + 3!3! + ..... + 10!10! = 1 + 4 + (....6) + (....6) + (....0) + ...... + (....0). This implies that the unit’s digit of 1!1! + 2!2! + 3!3! + ..... + 10!10! is 7 and hence it does not end with any zero. Example 48 When 20! is divided by 100000, what is the remainder? Solution We know that 20! ends with 4 zeros because it contains four 5’s. These 4 zeros will cancel with the 4 zeros of 100000 and we would have only single zero. The question therefore, effectively is to find the remainder when 20! (after removing all its zeros) is divided by 10. And this means to find the unit’s digit of 20! after removal of the 4 zeros. Prime factorizing 20!, we get 20! = 218 × 38 × 54 × 72 × 111 × 131 × 171 × 191. Removing 4 zeros means removing 4 twos and 4 fives. Therefore, we are left with 214 × 38 × 72 × 111 × 131 × 171 × 191. Finding unit’s place digit of this, we get 214 × 38 × 72 × 111 × 131 × 171 × 191 = (....4) × (....1) × (....9) × (....1) × (....3) × (....7) × (....9) = (....6) × (....1) × (....9) = (....4) Since unit’s place is 4, we see that the remainder when 214 × 38 × 72 × 111 × 131 × 171 × 191 is divided by 10, the remainder is 4 and hence when 20! is divided by 100000, the remainder is 4 × 10000 = 4 × 104. Recall that the common factor 104, which was removed, must be introduced back to get the correct remainder!

Wilson’s theorem If n is a prime number, then (n – 1)! + 1 is divisible by n. e.g., (2 – 1)! + 1 = 2 which is divisible by 2, because 2 is a prime number. (3 – 1)! + 1 = 3 which is divisible by 3, because 3 is a prime number. (4 – 1)! + 1 = 7 which is not divisible by 4 because 4 is not a prime number. (5 – 1)! + 1 = 25 which is divisible by 5, because 5 is a prime number. Likewise, 16! + 1 is divisible by 17 because 17 is a prime number. Obviously, (n – 1)! is divisible by all natural numbers from 1 to (n – 1) and hence (n – 1)! + 1 is not divisible by any natural number from 2 to (n – 1), every number being divisible by 1.

90

Number System – II: Remainders, Factorials and Base System Example 49 What is the remainder when 22! is divided by 23. A.

1

B.

22

C.

23

D.

2

Solution: B Since 22! + 1 is divisible by 23, we can say that 22! + 1 = 23n. Therefore, 22! = 23n – 1 which when divided by 23 leaves a remainder of –1 which is same as the remainder of 22. Hence the remainder is 22. Example 50 Find the remainder when 21! is divided by 23. Solution From the previous example, we know that 22! when divided by 23 leaves the remainder 22. Now, 22! = 22 × 21!. Let remainder when 21! is divided by 23, be r. So, 22! = 22 × 21! divided by 23 leaves the remainder 22 × r. But 22! leaves the remainder 22. So, r should be such that when 22 × r is divided by 23, we are left with remainder 22. Therefore, r can be nothing else but 1. We can now generalize that (n – 2)! when divided by n leaves a remainder 1, if n is a prime number. Example 51 What is the remainder when 298! + 30! is divided by 299? Solution Note that 299 is not a prime number. Factorizing, we get 299 = 13  23. Therefore, 298! is divisible by 13 as well as 23. Similarly, 30! is divisible by 13 as well as 23. Therefore, the remainder is 0.

Base System and Conversions We will study (i)

how to convert a number from a decimal system to any other base system and vice versa

(ii)

addition, subtraction and multiplication of the numbers in bases other than 10.

What is base system? We have natural numbers such as 2, 43, 4567, and so on which use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 only. Let us take the number 4567. 4567 = 4000 + 500 + 60 + 7 = (4 × 103) + (5 × 102) + (6 × 101) + (7 × 100). This 10 whose powers are used, is called the Base of the number and the number is said to be in the Base of 10. Base system of 10 is called in common parlance as Decimal System. Therefore, we can see that every number has to have some base which is generally mentioned especially if the number is in some base other than decimal system. We can have Base System of 2, 3, 4, 8, 9 10, 12, 16 and so on In fact, any of the natural numbers can act as a Base. More commonly used Bases are 2, 8, 10, 12 and 16.

91

Quantitative Aptitude Simplified for CAT

Base 2 system is called Binary System. Base 8 system is called Octal System. Base 16 system is called Hexadecimal System. BASE

DIGITS USED

Base 2

=

0, 1

Base 3

=

0, 1, 2

Base 4

=

0, 1, 2, 3

Base 8

=

0, 1, 2, 3, 4, 5, 6, 7

Base 10

=

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Base 12

=

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B

... ... ...

Note that in base 12, after the digit 9, we do not have any other digit and hence we make use of other characters to represent bigger numbers. Therefore, 10 is represented as A and 11 is represented as B. Base 16

=

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

A means 10, B means 11, C means 12, D means 13, E means 14 and F means 15. In general, Base n

=

0, 1, 2, 3, ....., (n – 1)

Therefore, numbers in different bases are written in the following manner: (100111001)2, (2356)8, (100432)7, (324A21)12, (3AB5E)16. The numbers in the subscript represent the Base System used. Also note that numbers like (32568)8 is invalid number, as digit 8 or more cannot be used in base 8.

Conversion from base X to base 10 and from base 10 to base X Let us say we have a number in base x, such as (32146)x, (assuming x > 6). Then, (32146)x = 3x4 + 2x3 + 1x2 + 4x1 + 6x0. The number in the RHS so obtained is the number in base of 10. Moreover, the power of x is always in decreasing order such that the last term (in the expansion) carries the power of x as zero. Therefore, if we have a 5 digit number, the power of x will start from 4 and if we have a 10 digit number, the power of x will start from 9. Example 52 Convert (1011011)2 to the base 10. Solution (1011011)2 = (1 × 26) + (0 × 25) + (1 × 24) + (1 × 23) + (0 × 22) + (1 × 21) + (1 × 20) = 64 + 0 + 16 + 8 + 0 + 2 + 1 = (91)10. Example 52 Convert (2314)5 to the base 10.

92

Number System – II: Remainders, Factorials and Base System Solution (2314)5 = (2 × 53) + (3 × 52) + (1 × 51) + (4 × 50) = 250 + 75 + 5 + 4 = (334)10. Example 54 (3AD5)16 = (.........)10. Solution (3AD5)16 = (3 × 163) + (A × 162) + (D × 161) + (5 × 160) = (3 × 163) + (10 × 162) + (13 × 161) + (5 × 160) = 12288 + 2560 + 206 + 5 = (15059)10. To convert from base 10 to, say, base 2, we will divide the given number successively by 2. Let us take an example. Example 55 Convert (43)10 to base 2. Solution Successive division is shown like this: 2

43

2

21

1

2

10

1

2

5

0

2

2

1

2

1

0

0

1

Firstly, 43 is divided by 2 (written to the left of 43) and we get quotient 21 and remainder 1. Quotient is written below 43 and remainder to the right of 21. Next, we divide quotient 21 by 2 (written to the left of 21) and we get quotient 10 and remainder 1. The process is continued till the quotient becomes 0. The first remainder becomes the unit’s digit of the number in binary. The second stage remainder becomes the ten’s digit of the number. This goes on and the last remainder becomes the leftmost digit of the number in binary. So, the number in binary is (101011)2. Example 56 (647)10 = (.......)8. Solution The successive division is shown below: 8

647

8

80

7

8

10

0

8

1

2

0

1

Therefore, (647)10 = (1207)8.

93

Quantitative Aptitude Simplified for CAT

Example 57 (2142)10 = (......)16. Solution The successive division is shown below: 16

2142

16

133

14

16

8

5

0

8

Therefore, (2142)10 = (85E)16.

Binary Operation of Addition, Subtraction and Multiplication Sometimes, we are required to add two (or more) numbers which are given in base other than decimal system. Addition is done as explained below. Example 58 (35216)8 + (14657)8 = (.........)8 Solution There is no need to convert each of them to base 10 and do the addition process. Instead, we realize that the process of addition is exactly similar to the process if the numbers were in decimal system. Add the unit’s digit numbers 6 and 7 as usual and obtain 13. Divide 13 by the base 8 as 13 = 8 × 1 + 5

(division algorithm)

where 1 is the quotient and 5 is the remainder. The remainder is placed as the unit’s digit of the answer and the quotient is the carry-over. The next step is to add the digits at the ten’s place and also add the carry-over to get 1 + 5 + 1 = 7, which when divided by 8 leaves the remainder 7 and quotient 0. Hence no carry-over. Continuing like this, we get (3 5 2 1 6)8 + (1 4 6 5 7)8 (5 2 0 7 5)8 Example 59 Add: (34AB7)16 and (EF871)16. Solution 11110 (3 4 A B 7)16 + (E F 8 7 1)16 (1 2 4 3 2 8)16 And hence the result of addition is (124328)16.

94

Number System – II: Remainders, Factorials and Base System For subtraction, we need to take care of the fact that when we borrow a number from the previous digit, it means that the base number is borrowed. Let us elaborate this through the following examples. Example 60 Solve the following: (36432)9 – (14378)9. Solution The unit’s digit of the larger number is 2 and that of the smaller number is 8. Since 8 cannot be subtracted from 2, we will borrow from the previous place. Borrowing 1 means borrowing 9 because base is 9. Therefore, 2 + 9 = 11 and 11 – 8 = 3 which is the unit’s digit of the answer. Continuing like this, we get (3 6 4 3 2)9 - (1 4 3 7 8)9 (2 2 0 4 3)9 Process: Let A = (36432)9 and B = (14378)9. Unit’s digit of A is 2 and that of B is 8. Since 2 < 8, borrow 1 which means 9 (the base). Therefore, 2 + 9 – 8 = 3. Ten’s digit of A is 3 which now becomes 2. Again borrow 1 which is 9. Therefore, 2 + 9 – 7 = 4. Hundred’s digit of A is 4 which now becomes 3. Now, 3 – 3 = 0. 6 – 4 = 2. 3 – 1 = 2. Hence the answer is (22043)9. Example 61 Solve the following: (DA34B)16 – (ABCDE)16. Solution Working on similar lines as in previous example, we get (D A 3 4 B)16 - (A B C D E)16 (2 E 6 6 D)16 Process: B < E. Therefore, borrow 1 which means 16 (the base). Therefore, 11 + 16 – 14 = 13 = D. 4 becomes 3. Now, 3 < D. Therefore, borrow 1 which is 16  3 + 16 – 13 = 6. 3 becomes 2. Now, 2 + 16 – 12 = 6. A becomes 9. Now, 9 + 16 – 11 = 14 = E. D becomes C. C – A = 2. Therefore, the answer is (2E66D)16. As regards multiplication, the process is exactly same as addition because multiplication only means adding the same thing certain number of times. Let us take the following example.

95

Quantitative Aptitude Simplified for CAT

Example 62 Solve: (43)8  (65)8. Solution The process is explained below. Process: First, we multiply 43 by 5, in base 8. 3  5 = 15. Divisor being 8, Q = 1, R = 7. (5  4) + 1 = 21, Q = 2, R = 5. Therefore, (43)8  (5)8 = (257)8. Similarly, (43)8  (60)8 = (3220)8. Finally, (257)8 + (3220)8 = (3477)8. Example 63 Solve: (AB)16  (3C)16. Solution The multiplication is done as shown below Process: B  C = 11  12 = 132, Q = 8, R = 4. (A  C) + 8 = (10  12) + 8 = 128, Q = 8, R = 0. Therefore, (AB)16  (C)16 = (804)16. Similarly, (AB)16  (30)16 = (2010)16. Finally, (AB)16  (3C)16 = (804)16 + (2010)16 = (2814)16. Example 64 In a certain base system x, 3214 + 5434 = 13052. Find x. Solution 4 + 4 = 8 which when divided by x should leave the remainder 2. That is possible only when x = 6. Upon further checking, we find the entire result is correct once we assume that x = 6. Therefore, the value of x is 6. Example 65 In a certain base x, (10)  (10) = (100). Find x. Solution It is natural on the part of students to think immediately that x must be 10 only. But, the fact of the matter is that x is any number more than or equal to 2. Here is how. (10)x  (10)x = (100)x or, (x + 0)  (x + 0) = (x2 + 0 + 0) or, x  x = x2, which is always true irrespective of the value of x. Hence the result.

96

Number System – II: Remainders, Factorials and Base System Example 66 Upon converting 2100 – 1, from base 10 to base 2, how many 1’s and 0’s would you write? Solution When we write 21 – 1, that is, 1 in base 2, we write (1)2. When we write 22 – 1, that is, 3 in base 2, we write (11)2. When we write 23 – 1, that is, 7 in base 2, we write (111)2. Therefore, when we write 2100 – 1, in base 2, we write (11.......)2, where 1 appears 100 times. There is no zero in the number when converted to base of 2. Example 67 Convert 335 from base 10 to base 9. Solution Dividing 335 by 9, Q = 333, R = 0. Dividing 333 by 9, Q = 331, R = 0. Continuing like this, we see that the conversion looks like this: (300000.......0)9, where 3 appears only once and 0 appears 17 times. Example 68 When (757)10 is converted to base 8, what is the unit’s digit? Solution We observe that when 757 is divided by 8, the first remainder so obtained is placed at the unit’s digit place. Therefore, unit’s digit required is same as the first remainder when 757 is divided by 8. The remainder when 757 is divided by 8 = (1)57 = 1 = 7. Example 69 When we convert (100!)10 to base 8, how many zeros occur at the end? Solution We observe that the number of zeros at the end is same as the highest power of 8 in (100!). The highest power of 2 in 100! is 97. Therefore, 100! = 297 × …. = (23)32 × 21 × … Therefore, there will be 32 zeros at the end of the number upon the said conversion.

EXPERT SPEAK Scan this QR Code to watch a video that explains the applications involved in the Base Conversion process.

97

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

What is the remainder when 496 is divided by 6?

B.

61

C.

01

A.

0

D.

41

B.

2

E.

81

C.

3

D.

4

CAT 2008 7. CAT 2003 (R)

2.

The remainder when (1523+ 2323) is divided by 19, is A.

4

B.

15

C.

0

D.

18

For a sequence of terms, first two terms are 2 and 2. Third term onwards, every term is product of previous two terms. Find the unit's digit of 2007th term. A.

2

B.

4

C.

6

D.

8 CAT 2012

CAT 2004 3.

8.

If x = (163 + 173 + 183 + 193), then x divided by 70 leaves a remainder of

1  2  3  4  ...  20 Find the minimum value 6n

A.

0

B.

1

of n so that the given expression is not an integer.

C.

69

A.

D.

35 CAT 2005

4.

Consider the expression

Let n! = 1 × 2 × 3 × … × n for integer n  1. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of

8

B.

9

C.

18

D.

19 CAT 2014

9.

The unit digit in the product of (8267)153 × (341)72 is A.

1

A.

10

B.

0

B.

2

C.

7

C.

7

D.

1

D.

9 IIFT 2012

CAT 2005 5.

10. Z is the product of first 31 natural numbers. If X = Z + 1, then the number of primes among X + 1, X + 2, ..., X + 29, X + 30 is

The rightmost non-zero digit of the number 302720 is A.

1

B.

3

C.

7

D.

9

A.

CAT 2005 6.

B.

2

C.

Cannot be determined

D.

None of the above IIFT 2012

What are the last two digits of 72008? A.

30

11. Two numbers in the base system B are 2061B and 601B. The sum of these two

21

98

Number System – II: Remainders, Factorials and Base System numbers in decimal system is 432. Find the value of 1010B in decimal system.

Expression

A.

110

B.

120

x What would be the remainder if x is 13!

C.

130

D.

140

E.

150

13

1

n

can also be written as

n1

divided by 11?

XAT 2016 12. Amitabh picks a random integer between 1 and 999, doubles it and gives the result to Sashi. Each time Sashi gets a number from Amitabh, he adds 50 to the number, and gives the result back to Amitabh, who doubles the number again. The first person, whose result is more than 1000, loses the game. Let ‘x’ be the smallest initial number that results in a win for Amitabh. The sum of the digits of ‘x’ is:

A.

2

B.

4

C.

7

D.

9

E.

None of the above XAT 2014

15. Consider the expression: (xxx)b = x3, where b is the base, and x is any digit of base b. Find the value of b. A.

5

B.

6

C.

7

A.

3

D.

8

B.

5

E.

None of the above

C.

7

D.

9

E.

None of these

XAT 2013 16. Please read the following sentences carefully. XAT 2014

13. Two numbers, 297B and 792B, belong to base B number system. If the first number is a factor of the second number, then the value of B is:

(I)

103 and 7 are the only prime factors of 1000027

(II)

6

6!  7 7!

A.

11

B.

12

(III) If I travel one half of my journey at an average speed of x km/h, it will be impossible for me to attain an average speed of 2x km/h for the entire journey.

C.

15

A.

D.

17

B.

Only Statement II is correct

E.

19

C.

Only statement III is correct

D.

Both statements I and II are correct

E.

Both statement I and III are correct XAT 2013

XAT 2014 14. Read the following instruction carefully and answer the question that follows:

All the statement are correct

ANSWER KEY 1. D

2. C

3. A

4. D

5. A

6. B

7. B

8. B

9. C

10. D

11. C

12. C

13. E

14. D

15. E

16. C

99

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS 1.

6.

Explanation: Dividing the power by 20, we get remainder as 8. Therefore, last two digits of 72008 will be same as those of 78. Therefore, 78 = (49)4 = (…01)2 = (…01).

Answer: D Explanation: 496 2192 2191 (1)191     1 = 2. 6 6 3 3 Therefore, correct remainder is 2 × 2 = 4.

2.

3.

7.

Explanation: 15 when divided by 19 leaves remainder 4 and 23 when divided by 19 leaves remainder 4. So, remainder when (1523 + 2323) is divided by 19, is (4)23 + 423 = (4)23 + 423 = 0. Alternatively, Use the formula xn + yn = (x + y)(xn–1  xn–2y + xn–3y2  … + xyn–1), if n is odd. In other words, xn + yn is divisible by (x + y). Here, 1523 + 2323 is divisible by (15 + 23) = 38, which is divisible by 19. So, given expression is divisible by 19.

8.

Answer: A

9.

Answer: B Explanation: We first need to find the higher power of 6 in 20!. This means the highest power of 3, which is 6 + 2 = 8. So, the highest power of 6 in 20! is 8. Therefore, minimum value of n so that the given expression is not an integer is 9. Answer: C Explanation: Dividing the power by 4 for first term, we get: (8267)153 × (341)72 = (…7)1 × (…1) = (…7).

10. Answer: D Explanation: Z = 31!. Now, Z is divisible by all numbers less than 32. X = 31! + 1 X + 1 = 31! + 2, will be divisible by 2, X + 2 = 31! + 3 will be divisible by 3, X + 3 = 31! + 4 will be divisible by 4 and so on. Hence none of the numbers will be prime.

Answer: D Explanation: Note that 10 × 10! = (11 – 1) × 10! = 11! – 10!. Similarly, 9 × 9! = 10! – 9!, and so on. Therefore, p = (11! – 10!) + (10! – 9!) + (9! – 8!) + … + (2! – 1!) = 11! – 1!. Now, p + 2 = 11! + 1, which when divided by 11! leaves a remainder of 1.

5.

Answer: B Explanation: Unit’s digits of the terms are: 2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 2, 6, … We observe that after every 6 terms, the pattern repeats. So, unit’s digit of 2004th term (where 2004 is a multiple of 6) will also be 6 and so the next few terms (only unit’s digit) will be 2, 2, 4. So, unit’s digit of 2007th term = 4.

Answer: C

Explanation: x = (163 + 173 + 183 + 193) x = (163 + 193) + (173 + 183) x = (16 + 19) (162 + 192 – 16 × 19) + (17 + 18) (172 + 182 – 17 × 18) x = 35[162 + 192 – 16 × 19 + 172 + 182 – 17 × 18] x = 35 × (Even Number) Therefore, x is divisible by 70. 4.

Answer: C

11. Answer: C Explanation: 2061B = (2B3 + 6B + 1)10; 601B = (6B2 + 1)10. As per the question, 2B3 + 6B2 + 6B + 2 = 432  B3 + 3B2 + 3B + 1 = 216  (B + 1)3 = 216  B = 5. Therefore, (1010)B = (1010)5 = (130)10.

Answer: A Explanation: 302720 = 32720 × 102720 The rightmost non-zero digit of the given number will be same as that of 32720. Since power is a multiple of 4, unit’s digit = 1.

100

Number System – II: Remainders, Factorials and Base System 13

Important: Note that there is an error in this question, because digit 6 is not allowed to be used if base is 5.

1

x

 n  13! . Therefore, x n1

13! 13! 13! 13! 13! 13!    ...    1 2 3 11 12 13 All the terms in x are divisible by 11 except 13! . 11 Now, 13! 1  2  3  ...  11  12  13 10!11  12  13   11 11 11 = 10! × 12 × 13, which when divided by 11 leaves the remainder = (1)(1)(2) = 2 = 9.

=

12. Answer: C Explanation: Let the smallest number be ‘X’ Amitabh

Sashi

Step 1:

2X

2X + 50

Step 2:

4X + 100

4X + 150

Step 3:

8X + 300

8X + 350

Step 4:

16X + 700

16X + 750

Step 5:

32X + 1500

32X + 1550

15. Answer: E Explanation: From the given equation, bx2 + bx + x = x3  bx + b + 1 = x2  x2 – bx – (b + 1) = 0  x

b  b2  4(b  1) b  (b  1)2  2 2 b  (b  1)  1 1   or b  2 2 2 both of which is not possible. So, no solution exists.

In step 5, Amitabh will definitely lose. So, for Amitabh to win, step 4 should be the last step and for that the number with Sashi must be more than 1000. So, 16X + 750 > 1000 or the least value of X = 16. The sum of digits of 16 is 7.



13. Answer: E

16. Answer: C

Explanation: 297B = 2B2 + 9B + 7; 792B = 7B2 + 9B + 2 = 3(2B2 + 9B + 7) + (B2 – 18B – 19). As per the question, (B2 – 18B – 19) = (B – 19)(B + 1) should be divisible by (2B2 + 9B + 7) = (2B + 7)(B + 1), which is possible when B = 19.

Explanation: Statement I – Prime factors of 1000027 are 7, 19, 73, 103. Hence, the statement is false. Statement III – Average speed = HM of the individual speeds if distances are equal. So, 2( x )(v ) or v = x + v, which is impossible. 2x = x v Hence it is true statement. Note that we need not check for statement II.

14. Answer: D Explanation:

101

Quantitative Aptitude Simplified for CAT

Chapt er 03

Commercial Math BASIC CONCEPTS Percentage The concept of “Percentages” lays the foundation for the study of a lot of topics, for example, Profit and Loss, Simple Interest, Compound Interest, Stocks and Shares, Partnerships, Ratios and Proportions, and many more. These topics are generally asked in CAT and other MBA entrance examinations. We will understand these applications later. For the moment, let us concentrate on studying and strengthening the basic concepts of “Percentages”. The word “percent” means “for every hundred”. Therefore, 25% means 25 out of 100; 35% means 35 out of 100. Now, r% means r out of hundred or (in formula form) r r% = 100 Therefore, 25% =

25 1 35 7 = and 35% = .  100 4 100 20

We see that every percentage can be converted to fraction form using the above result. The students are advised to remember the important conversions from percentage to fraction and vice versa. Note that 25% is

1 1 1 th and hence half of that, that is, 12.5% is half of th, that is, th. Therefore, 4 4 8 1 = 50%; 2

1 = 12.5%; 8

1 = 25%; 4

3 = 37.5%; 8

5 = 62.5%; 8

3 = 75%; 4

7 = 87.5%; 8

1 = 16.66%; 6

1 = 33.33%; 3

2 = 66.66%; 3

5 = 83.33%. 6

1 = 20%; 5

2 = 40%; 5

3 = 60%; 5

4 = 80%; 5

Conversion from percentage to fraction and vice versa helps us do the calculations quickly.

102

Commercial Math Example 1 Express

15 as a percentage. 42

Solution 15 15   100 = 35.71%. 42 42

Example 2 Express 45

3 % as a fraction. 8

Solution 3 45 %  8

3 45 % 8  363 100 800

Example 3 Find the value of 320% of 25. Solution We note that 320% of 25 is same as 25% of 320, because 320% of 25 = 320. Therefore, the value = 25% of 320 =

320 25  25   320 = 25% of 100 100

1  320 = 80. 4

Therefore, we can conclude that a% of b = b% of a Example 4 25% of 37.5% of 87.5% of 66.66% of y = ky. Find k. Solution 1 3 7 2 7 7 25% of 37.5% of 87.5% of 66.66% of y =      y   y k= . 4 8 8 3 128 128  

Example 5 A man spends 37.5% of his income on food and clothing. Of the remaining, he spends 20% on entertainment and the remaining part is divided equally between his wife and 2 sons. What percent of his income is the share of each son? Solution If x is his income, then 37.5%, that is (3/8)th of x is spent on food and clothing. He is left with (5/8)th of x. If 80 5 50 20% of (5/8)th is spent on entertainment, then 80% of (5/8)th = = 50% of x is remaining which   100 8 100 is divided equally between his wife and 2 sons. Therefore, each son’s share = (1/3)rd of 50% = 16.66% of his income. Alternatively, since he spends 37.5% of his income on food and clothing, and 37.5% is 3/8th, we can assume that his income is Rs.80 (which is a multiple of 8), to avoid fractions. Now, 3/8th of 80 = 30 and so he is left with Rs.50. Now, he spends 20% of the remaining, which is Rs.10 and so is left with Rs.40, which is 50% of Rs.80. The remaining 50% is divided equally between 3 persons and so each gets 16.66%.

103

Quantitative Aptitude Simplified for CAT

Concept of Base Base is one of the most important concepts to be learnt. What is a Base? Base is any quantity which undergoes a certain change. For example, a person’s weight is 60kg and because of overeating, his weight now becomes 75 kg. We can say that the weight has increased by 15 kg. This increase of 15 kg is usually expressed as a percentage of 60. This is done as 15  100 = 25% 60

Therefore, we can also say that the weight has increased by 25%. But, 25% of what? It is 25% of 60. This 60 is the base. Every percentage has to have some Base. In more general terms, we can say that if x is any quantity and it increases by y units, then the percentage increase, r% is given by the formula r% =

y  100 x

In other words, r% of x = Upon increase, the quantity becomes x + y = x 

r x =y 100

r r   x  x  1   . 100 100  

In general, if any quantity x increases by r%, the new quantity will be r   x1   100   20   Hence, if a quantity x increases by 20%, the new quantity is x  1   or 1.2x. We can say that the original 100   quantity gets multiplied by 1.2.

If a quantity increases by 25%, the quantity becomes 1.25x. 120  If a quantity increases by 120%, the quantity becomes x  1   = 2.2x. 100   350   If a quantity increases by 350%, the quantity becomes x  1   = 4.5x. 100   2257  If a quantity increases by 2257%, the quantity becomes x  1   = 23.57x. 100   200  If a quantity increases by 200%, the quantity becomes x  1   = 3x. 100  

If a quantity becomes 1.5x, the quantity has increased by 50%. If a quantity becomes 2.75x, the quantity has increased by 175%. If a quantity becomes 13.45x, the quantity has increased by 1245%. If a quantity becomes 4x, the quantity has increased by 300%. One should understand these fundamentals very clearly.

104

Commercial Math Example 6 A quantity increases by 255% and becomes 7100. What is the quantity? Solution If x is the original quantity, then it becomes 3.55x. Therefore, 3.55x = 7100 or x = 710000/355 = 2000. Example 7 20 is what percent of 50? Solution 20 is

20  100 = 40% of 50. 50

Example 8 If Ram’s income is 25% more than Shyam’s income, by how much percent is Shyam’s income less than Ram’s income? Solution In such cases, it is best to take Shyam’s income as Rs 100. Now, if Shyam’s income is Rs 100, then Ram’s income will be Rs 125. Therefore, Shyam’s income is Rs 25 less than Ram’s income of Rs 125. Therefore, to calculate the corresponding percentage, we get 25  100  20 % 125

Hence, Shyam’s income is 20% less than that of Ram’s income. Example 9 If Ram’s income is 25% less than Shyam’s income, by how much percent is Shyam’s income more than Ram’s income? Solution Once again, we should take Shyam’s income to be Rs 100. Therefore, Ram’s income = Rs 75. Percentage by which Shyam’s income is more than Ram’s income =

25 1  100  33 % . 75 3

There is no need to create a formula for every such variety of questions, a common tendency amongst the students. Formula based approach should be avoided as far as possible. Example 10 If Ram’s salary is 20% more than Rahim’s and Shyam’s salary is 40% of Rahim’s salary, how much % is Ram’s salary that of Shyam’s salary? Solution The trick in this question lies in the words! Shyam’s salary is 40% of Rahim’s salary and not 40% “less” than Rahim’s.

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Therefore, if Rahim’s salary is Rs 100, then Ram’s salary is Rs 120 and Shyam’s is Rs 40. Therefore, Ram’s 120 salary is  100 = 300% of Shyam’s salary. 40 We can also say that Ram’s salary is 200% more than Shyam’s salary. Example 11 If Ram’s and Shyam’s salary got raised by 5%, who got a better deal? Solution It depends upon the current salaries of each. If both had the same salary initially, then obviously both are on equal platform. If Ram’s initial salary is Rs 20,000 and Shyam’s initial salary is Rs 6000, then increase in the salary of Ram is Rs 1000, whereas increase in Shyam’s salary is Rs 300 and hence Ram got a better deal.

Population Formula r  Pn  P0  1   100  

n

where, r = rate of growth (in %) n = number of time periods (generally in years) P0 is the population in the beginning, and Pn is the population in the end. Example 12 If the population today is 10,000 and increases at the rate of 5% per annum, what was the population 4 years ago? Solution Note that, in this example, Pn = 10,000, r = 5%, n = 4 years and Po is to be calculated.

5   Therefore, 10000  P0  1   100  

4

 Po » 8227.

If the population is decreasing, rate of growth will be taken as negative and

r  Pn  P0  1   100  

n

Further, if every year, the population increases at a different rate, then r  r  r   Pn  P0  1  1   1  2   1  3  100   100   100  

Note that if population increases in the first year at the rate of 5%, then decreases at the rate of 7% and in the subsequent year also it decreases at the rate of 2%, but in the fourth year, population increases at the rate of 4%, then 5  7  2  4   Pn  P0  1   1   1   1   100   100   100   100  

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Commercial Math Example 13 If a bacteria population increases at the rate of 6% in the first 10 minutes, and then at the rate of 10% in the next 10 minutes, then what is the overall percentage increase in the population? Solution 6  10  Pn  P0  1   1   = 1.166P0. Therefore, there is a 16.6% increase. 100   100  

Alternatively, (1.06) (1.10) = (1.166). Hence, 16.6% increase. Example 14 If a quantity increases by 20% and then decreases by 10%, then what is the net increase/decrease %? Solution If we take the quantity to be x, then x (1.2) (0.9) = 1.08x. Therefore, there is a net increase of 8%. Alternatively, if we take the quantity to be 100, then after the 20% increase, the quantity becomes 120 and a decrease of 10% on this will mean a decrease of 12. Therefore, the quantity becomes 108 and hence results in a net increase of 8%. Example 15 In the above example, if the order of percentage change is reversed, will the answer change? Solution If the quantity is decreased by 10% and then increased by 20%, the result will be as below: x (0.9)(1.2) = 1.08x. It is important for students to realize that the order in which the changes happen does not really matter! In the above example, we note that the two changes happen successively, that is, one after the other. Let there be a quantity x which undergoes two successive changes of a% and b%. Then, the quantity becomes a  b  a b ab    x1     1   x1   100 100 100 100 10000     

If the overall percentage change is represented by r%, the quantity should become r   x1   100  

Comparing the two, we get: ab  r% =  a  b  % 100  

As we can see that the order in which changes happen is not important:  a  b  ab  %   b  a  ba  %     100  100   

This result will be found extremely useful in many cases. Let us apply this in the above two examples. In the first example, the successive increments are 6% and 10%. Therefore, a = 6% and b = 10%.

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ab  (6)(10)   Applying the result, overall percentage change, r% =  a  b   % =  6  10   % = 16.6% 100  100    (20)(10)  Applying the result for the next example, overall percentage change =  20  10   % = 8%. 100  

In the last case, since one of the changes was a decrement, the sign gets changed from positive to negative. Therefore, we can see that the result is true in cases of decrements also. Example 16 A shopkeeper offers a discount of 20% on certain goods. Upon constant insistence by the customer, the shopkeeper agrees to a further discount of 20%. What is the overall discount % offered by the shopkeeper? Solution This is a case of successive discounts. If the goods are marked at Rs 100, then after the first discount, the price becomes Rs 80. Second discount of 20% is on the base of Rs 80. Therefore, the discount is 20% of Rs 80, which is Rs 16 and hence the final price is Rs 64. Overall discount is 36%. ab  Alternatively, we can apply the above stated result of successive changes  a  b  %. 100  

Therefore, overall percentage change =  20  20 

(20)(20) = 36%. 100

The sign of 36% is negative which only shows that overall there is a discount. Example 17 The price of an article is first raised by 10% and then a discount of 10% is offered. Which of the following statements is correct? (a)

There is no profit nor loss to the seller

(b)

The profit or loss depends upon the price of the article

(c)

There is a net loss of 1%, no matter what the price is

(d)

None of these

Solution Since the percentage value is same, one may think that there is no profit no loss to the seller. But this is not the case. Let us say that the price of an article is Rs 100. When raised by 10%, the price becomes Rs.110 which when discounted by 10% leads to reduction of Rs 11 and hence the person loses Re 1 on the cost price of Rs 100. There is a net loss of 1%. Correct answer is (c). ab  (10)(10) Alternatively, we can apply the result as  a  b  = 1%.  % = 10  10  100  100 

Alternatively, if the original price is Rs x. Then, the price after these changes becomes x(1.1)(0.9) = 0.99x, which implies 1% loss. Let us analyze why there is a net loss. This is happening because of the concept of base. When there is an increase of 10%, the base has increased from 100 to 110. The amount of increase is 10 units. But, when 10%

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Commercial Math is now reduced, there is a decrease of 11 (which is more than the earlier increase of 10 units). That is why there is a net decrease. What happens if there are three percentage changes or four percentage changes? Do we have another formula? And does that mean that we have to remember a new formula every time something new happens? In case there are three changes, we can still apply the same formula twice. Let us see how. Example 18 Price of a share increases by 10% on day 1, 20% on day 2 and 30% on day 3. What is the overall percentage increase in the share price? Solution By the end of day 2, the price would have increased by 10  20 

(10)(20) = 32% 100

Over this, a further increase of 30% means overall change of 32  30 

(32)(30) = 71.6%. 100

Therefore, there is an overall increase of 71.6%. And likewise, we can tackle cases of 4 or more percentage changes, irrespective of increment or decrement. It is important to note the following. Let there be three variables, P, Q and R such that P = Q × R. Also, if Q changes by a%, R changes by b%, then ab  percentage change in P is given by successive changes formula, that is  a  b  %. 100   So, thumb rule is that if one variable is the product of the other two variables, then we can apply the formula of successive percentage changes as described above. Example 19 A quantity is raised by 20%. By what percentage should the new quantity be decreased so that the original quantity is reached? Solution Let the initial quantity be 100. Then 20% increase means the quantity becomes 120. There must be a decrease of 20 in this so that 100 is obtained back. This means percentage change of

20  100 = 16.66%. 120

Alternatively, if r% should be the new change, then

20  r 

(20)(r ) 20 = 0 or r = –  100 = –16.66%. 100 120

The negative sign only indicates the decrement. Therefore, in such cases also, we can apply the successive increments formula. Example 20 If price of petrol increases by 10%, what should be the reduction in its consumption so that (i)

the expenses remain the same as earlier,

(ii)

the expenses can increase by 5%.

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Solution (i) Let the price of petrol be Rs 100. After the increase, it becomes Rs 110. To decrease the consumption so that expenses remain the same, we must ensure that we spend only Rs 100 on petrol. Therefore, percentage 10 1 decrease in the consumption is given by  100  9 %. 110 11 Alternatively, We observe that Expense = Price × Quantity and so we can apply the formula of successive percentage change. If expense is to remain same, percent change in expense will be taken as 0%. Also, percent change in price = a% = 10%, and we need to find percent change in quantity, that is b%. Using the formula, we get 0% = 10 + b +

10b 10 1 or b =   9 % (negative sign means percent decrease). 100 1.1 11

(ii) If expenses increase by 5%, the expenses become Rs 105. Therefore, we should decrease petrol prices from Rs. 110 to Rs. 105, which is Rs 5 over Rs 110. Percentage decrease in consumption =

5 6  100  4 %. 110 11

Alternatively, we can use successive percentage change formula. 5% = 10 + b +

10b 5 6 or b =   4 % (negative sign means percent decrease). 100 1.1 11

Example 21 If price of Maruti Swift is raised by 40% which leads to doubling of revenue, what is the percentage change in the number of units being sold? Solution Revenue = Price  Number of units sold, or R = P  n. Let the number of units change by r%. Then, r  r   2R = (1.4P)  n 1    2 = (1.4)  1    r = 42.85%. 100 100    

Therefore, number of units sold increases by 42.85%. Alternatively, Revenue = Price × Quantity sold. Since revenue has doubled, percentage increase in revenue is 100%. So, 100 = 40 + r +

40r 60 or r = = 42.85%. 100 1 .5

EXPERT SPEAK Scan this QR Code to watch a video that explains the applications of Percentages to typical problems.

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Commercial Math

EXPERT SPEAK Scan this QR Code to watch a video that explains the applications of Successive Percentage Change formula in various situations.

Example 22 Reduction in price of sugar by 20% allows a household to buy 45 kg more for Rs 450. Find the original price of the sugar. Solution If original price = p, then new price = 0.8p. Therefore,

450 450   45 or p = Rs. 2.5/kg. 0.8 p p

Alternatively, The question means that if earlier, we were able to buy x kg of sugar for Rs.450, we can now buy (x + 45) kg of sugar because of reduction in the price. Therefore, 20% of 450 (which is Rs.90) allows the household to buy 45 kg of sugar (at the reduced price rate). Reduced price rate of the sugar = Rs 90/45 = Rs 2 per kg. Therefore, original price of the sugar =

2 = Rs 2.5 per kg.  1  20    100  

Example 23 In a fraction, if the numerator increases by 10% and the denominator decreases by 20%, the fraction A.

increases by 30%

B.

increases by 10%

C.

decreases by 10%

D.

None of these

Solution Let the fraction be

x (1.1) x x   (1.375) . Therefore, the fraction increases by . Then, the new fraction is y (0.8) y y

37.5%. Alternatively, let z =

x or x = yz. Here, x increases by 10% and y decreases by 20%. We can use formula of y

successive percentage changes, where percentage change in x denotes overall percentage change. Therefore, 10 = 20 + b +

(20)(b) 30 or b =  37.5% 100 0.8

Example 24 10% of a person’s earnings are donated. Of the remainder, 50% goes towards his children’s education, entertainment and clothing. Whatever is left is divided into 9 parts, out of which 5 parts are distributed among the 5 children. If he had only 4 children, what % more would he save than he does now?

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Quantitative Aptitude Simplified for CAT

Solution Let the earnings of the person be Rs 100. After donation, he is left with Rs 90. After the said expenses, he is left with Rs 45. This is divided into 9 parts; therefore each part is Rs 5,making the total Rs 25. Hence, he saves Rs 20. If he had only 4 children, he would save Rs 5 more. Therefore, he saves 25% more than Rs 20. Note that if the question were “what % would he save than he does now”, then the answer would have been 5%. Observe how base was chosen. Example 25 In an election between two candidates, one candidate who gets favor from 40% of the voters (in that constituency) wins by 200 votes. If there were 5000 persons in the constituency, how many did not vote? Solution 40% of 5000 = 2000. Therefore, loser gets 2000 – 200 = 1800. Total votes polled = 3800. Therefore, there were 5000 – 3800 = 1200 persons who did not vote. Example 26 From a 40 litres salt solution of 10% strength, how much water must evaporate to increase the strength to 15%? Solution Let x be the quantity of water that must evaporate. The evaporation keeps the salt content constant. The initial salt content = 10% of 40, that is, 4 units. Therefore, initially we have 36 litres of water. Since x litres now 4 evaporates, we have 36 – x litres of water. Therefore, % strength of the solution is given by  100 = 15 36  x 1 or x = 9 litres. 3 Example 27 Price of sugar increases by 200% and then decreases by 300%, what is the net % change? Solution The price of sugar cannot decrease by more than 100%. Therefore, such a case is impossible!

Profit and Loss “Profit, Loss and Discount” is one of the applications of the concept of percentages which we touched upon in the previous section. We will understand the application of the concepts of percentages in cases of “profit and loss” in this section. Whenever any article is bought or sold, two parties are involved: the customer (or buyer) and the shopkeeper (or seller). This seller buys goods from some wholesaler at a certain price which is called the cost price (CP) for the seller. Obviously, the seller will not sell that article for the same price. He will sell it for more than the cost price at which he bought from the wholesaler. The price at which he is able to sell the article to the customer is called the selling price (SP). The difference between the selling price and cost price is called profit (P). Generally, the selling price is more than the cost price and the seller usually gains. But, the selling price can be less than the cost price and hence the seller may lose. The difference between the selling price and cost price in such cases is called loss (L).

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Commercial Math Therefore, SP – CP = P CP – SP = L Profit or loss is usually expressed in percentage terms. We know that for percentage, we require some base. As a standard, cost price is the base for profit as well as loss. Therefore, Profit % =

Profit SP  CP  100 =  100 . Cost Price CP

(i)

From this, we can obtain the following formula: P%  SP = CP  1   100  

Similarly, Loss % =

(ii) Loss CP  SP  100 =  100 . Cost Price CP

(iii)

From this, we can obtain following the formula: L%  SP = CP  1   100  

(iv)

Use formulae (i) or (iii) to find profit% or loss%. Use formulae (ii) or (iv) to find SP or CP, given the other data. Example 28 If a shopkeeper sells certain goods for Rs 125 that were sold to him by a wholesaler for Rs 110, find the profit or loss %. Solution Wholesaler sold the goods to the shopkeeper for Rs 110 and therefore, Rs 110 becomes the cost price for the shopkeeper. Profit = Rs (125 – 110) = Rs 15 Profit % =

15 7  100  13 %. 110 11

Example 29 A sells to B for Rs 120 and makes a profit of 25%. What is the cost price for A? Solution P%   Use the formula SP = CP  1   . Therefore, 120 = CP (1.25) or CP = Rs.96. 100  

Example 30 A sells to B and makes a profit of 10%. B sells to C and makes a profit of 20%. What is the overall profit % made? Solution If CP for A = Rs.100, then SP for A = 100(1.1) = Rs.110 = CP for B. So, SP for B = 110(1.2) = Rs.132. Therefore, overall profit% = 32%. Alternatively, we realize that this is similar to the case of successive increments. Therefore, overall profit % = 10  20 10 + 20 + = 32%. 100

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Quantitative Aptitude Simplified for CAT

Example 31 In the above example, at what loss % should C sell to D so that overall there is no profit no loss? Solution 32 + r +

32 800 8 32  r or r =   24 %. 1.32 33 33 100

Therefore, loss % should be 24

8 %. 33

Alternatively, if cost price for A is Rs 100, then cost price for C is Rs 132. To obtain a cost price of Rs 100 32 8 again, we need to decrease it by Rs 32 and hence loss % =  100  24 %. 132 33 Example 32 Ram buys a watch for Rs 600 which he sells to Shyam at a profit of 20%. Shyam sells that watch to Radha earning the same profit percentage, whereas Radha sells it to Sita at a profit percentage which is double of the profit percentage earned by Shyam. What is the price at which Radha sold the watch to Sita? Solution CP of the watch for Ram = Rs 600. CP of the watch for Shyam = 600 × 1.20. CP of the watch for Radha = 600 × 1.20 × 1.20 = 600 (1.2)2. Profit % for Radha = 40%. Therefore, SP for Radha = 600(1.2)2(1.4) = 600(1.44)(1.4) = Rs 1209.6. Example 33 A trader sells certain goods and earns profit which is 25%, but he being illiterate, calculates it on the selling price. What is the true profit percentage? Solution If SP is Rs 100, then profit = 25% of SP = Rs 25 and hence CP = Rs 75. True profit percentage =

25 1  100  33 % . 75 3

Example 34 A trader sells goods at a profit % of 20%. If he had sold it for Rs 50 more, he would have gained 30%. What is the selling price now? Solution If original cost price is Rs x, then SP = 1.2x New SP = 1.2x + 50 which should be equal to 1.3x (because profit % now is 30%). Therefore, x = Rs 500 and selling price now is Rs 650. Alternatively, if CP = Rs.100, then SP = Rs.120. New SP = Rs.130. Difference in the 2 SP’s = Rs.10. Applying unitary method, If difference is Rs.10, then CP = Rs.100; If difference is Rs.50, then CP = Rs.500.

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Commercial Math Quite often, in such transactions, shopkeepers tend to give some discounts to the customers so that the buyer feels that the article is now cheap enough and hence buys the article. But, shopkeepers raise the cost price so that even after offering discount(s), he ends up making a profit. Therefore, a shopkeeper will buy goods from some wholesaler for some price (called cost price) and raise the cost price by certain amount. This act of raising the price of articles is called marking-up, which is usually expressed as a percentage of cost price. Therefore, cost price is the base for mark-up percentage. The price so obtained after marking-up is called marked price (MP). This is also sometimes known as list price or tag price. This is the price which we see in the shops attached as a tag on the articles such as T-shirts, Trousers, Television, and so on. Cost Price + Mark-up = Marked Price or CP + M = MP r   If mark-up % is r%, then, MP = CP  1   100  

If the trader is able to sell the articles at the marked price, the marked price becomes the selling price. But, as usually happens, customer asks for some discount, generally in the form of percentage. You would hear people asking for discounts like 20%, 35% and so on. This percentage, again, must have some base and that base is always the marked price. Marked Price – Discount = Selling Price or MP – D = SP d%  If discount % is d%, then SP = MP  1   100  

After discount amount is subtracted from the marked price, the article is sold for that price which is called selling price, which if higher than the cost price leads to earning of profits. The following table lists some terms, and their corresponding bases. TERMS

BASE

Profit %

Cost Price

Loss %

Cost Price

Mark-up%

Cost Price / Price

Discount %

Marked Price

Example 35 A trader gains 20% despite a discount of 20%. What is the mark-up %? Solution Let CP = Rs 100. Then, SP = Rs 120. 20  If MP (marked price) = x, then price after discount = x  1   = 0.8x = SP. 100  

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Therefore, 0.8x = 120 or x = Rs 150. This suggests that the trader has marked-up by Rs 50, and hence mark-up % =

Mark  up 50  100   100 = 50%. Cost Pr ice 100

In this example, we have only assumed that cost price is Rs 100. This is done only for convenience of calculations and the student is advised not to mistake it for the actual cost price. The fact is that the mark-up % will be 50% no matter what value of cost price is taken! Example 36 A man would lose 15% by selling his furniture for Rs 510. To what sum should he mark up the price to convert loss into profit? Solution SP = 510. Loss% = 15%. Therefore, CP = 600. Therefore, markup should be Rs 90. Marked price must be Rs 690. Example 37 A trader offers a discount of 20%. A customer comes to buy an article but wants a further discount of 30%. What is the overall discount % offered by the trader? Solution Let the MP be Rs 100. A discount of 20% means Rs 20. Price becomes Rs 80. Further discount of 30% means 30% of Rs 80, that is, Rs 24. Therefore, the price becomes Rs 56. Therefore, there is an overall discount of Rs 44 on the price of Rs 100. Therefore, the discount % is 44%. Alternatively, this is a case of successive discounts and hence the formula of successive increments can be (20)(30) applied. This gives us – 20 – 30 + = –44%. Negative sign only indicates discount. 100 Example 38 The price of an article is first raised by 10% and then a discount of 10% is offered. Which of the following statements is correct? A.

There is no profit no loss to the seller

B.

The profit or loss depends upon the price of the article

C.

There is a net loss of 1%, no matter what the price is

D.

None of these

Solution: C Since the percentage value is same, students tend to think that there is no profit no loss to the seller. But this is not the case! Let us say that the price of an article is Rs 100. When raised by 10%, the price becomes Rs 110 which when discounted by 10% leads to reduction of Rs 11 and hence the person loses Re 1 on the cost price of Rs 100. There is a net loss of 1%. Alternatively, we can apply the result as below: 10 – 10 +

(10)(10) = –1%. 100

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Commercial Math Alternatively, if the original price is Rs x; then, the price after these changes becomes x (1.1)(0.9) = 0.99x, which implies 1% loss. Example 39 A salesman in a shop is told by the owner of the shop to write off a discount of 25%. The stupid salesman wrote off the discount but on the cost price instead of the marked price! The owner quickly realized the mistake committed by the salesman. What should be the mark-up percentage which the owner should apply to ensure that he earns a profit of 25% on the true cost price? Solution Let the CP be Rs 100. Then, the salesman wrote off the discount of 25% or Rs 25 on Rs 100 and hence reduced the price to Rs 75. The owner would raise this price of Rs 75 by such an amount so that he gains 25% which means SP should be Rs.125. Therefore, marking-up should be of the amount Rs (125 – 75), that is, Rs.50. Mark-up % =

50 2  100  66 % 75 3

Example 40 If a shopkeeper sells two identical items whose selling prices are same and he earns profit of 10% on one, whereas loss of 10% on the other, then the combined profit or loss % will be A.

1% loss

B.

1% profit

C.

no profit, no loss

D.

Cannot be determined

Solution: A Let SP of both the items be Rs 100. Then, CP of one where profit is earned = CP of one where loss is incurred = Total CP =

100 . 1.1

100 . 0.9

100 100 + and total SP = Rs 200. 1.1 0.9

Since CP is more than the SP, there is net loss.

100 100   200  100 = 1%. Loss % = 1.1 0.9 100 100  1.1 0.9 Important Learning If SP of two items be Rs x each, earning r% on one and losing r% on the other, then there is a net loss and Loss r2 %= %. 100 Example 41 In the above example, if CPs are same, then what is the overall profit or loss?

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Solution There will be no profit, no loss, because amount of profit will be offset by the same amount of loss. Example 42 What % of water must be added to pure milk so that by selling the mixture at cost price, the milkman earns a profit of 20%? (a)

Assuming water is free of cost.

(b)

Assuming the cost of water is 20% of the cost of milk.

Solution (a)

Let us say that we have 100 Lt. of milk @ Re 1 per litre so that the cost of milk is Rs 100. To earn 20% profit, SP should be Rs 120, for which we must have 120 litres of mixture. Therefore, we must add 20% water to milk, or 20 lt.

Note that addition of water which is r% of the amount of milk, will lead to the profit of r% only. (b)

Since 100 lt of water would cost Rs 20 (as the CP of water is 20% of the CP of milk),

20 lt of water would cost only Rs 4. Therefore, total cost price of the solution = Rs 100 + Rs 4 = Rs 104. Total SP of the solution = Rs 120. Therefore, profit % =

16  100 = 15.38%. 104

Example 43 On sale of 3 items, a shopkeeper offers 1 item free to a customer. What discount is being offered? Solution Let price of each item be Re.1. He offers a customer 4 items (marked price = Rs.4) whereas he charges him for 43 3 items (selling price = Rs.3). Therefore, discount % =  100 = 25%. 4 Example 44 A shopkeeper gives two items free to a customer if the customer buys 8 items. If the shopkeeper does not give it free, then on each item, he earns a profit of 20%. What is the profit/loss percentage when he introduces the scheme of giving two items free? Solution Let the CP of each item be Rs 10. Then, CP of these 10 items is Rs 100. CP of 8 items is Rs 80. SP of these 8 items when the scheme is not introduced is 80 + 20% of 80 = Rs 96. When the scheme is in force, then he receives Rs 96 for all the 10 items. But the CP of these 10 items was Rs 100. Therefore, there is a net loss of Rs 4 or 4%. Example 45 At what profit should he have been selling before introduction of scheme so that introduction of scheme wipes off all his profits and he incurs no loss?

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Commercial Math Solution Before scheme, sale of 8 items (whose CP is Rs 80) should fetch him Rs 100, so that upon introduction of the 20 scheme, he makes no profit and no loss. Therefore, profit % should have been  100 = 25%. 80 Example 46 A person buys a clock for Rs 600, sells it for Rs 700, buys it again for Rs 800 and sells it back for Rs 900. Find the amount of loss/profit. Solution The common mistake students commit is: There is a profit of Rs 100 in the first transaction. Loss of Rs 100 in the second transaction does away with the profit in the first transaction. In the last transaction, there is a profit of Rs 100 and hence there is an overall profit of Rs 100. This is wrong calculation. There are many ways to do this. Method 1: Let us assume that the person initially had Rs 1000. On buying the clock, he is left with Rs 400. On selling, the person now has Rs 1100. On buying the second time, he is left with Rs 300 and finally, upon selling, he has Rs 1200, viz., Rs 200 more than he started with. Therefore, he gains Rs 200, and not Rs 100. Method 2: Total cash outflow is 600 + 800 = 1400 and total cash inflow is 700 + 900 = 1600. Therefore, there is a net profit of Rs 200. Method 3: Alternatively, we can say that the shopkeeper bought another watch for Rs 800 and sold it for Rs 900. Therefore, he earned Rs 100 from the first watch and another Rs 100 from the second watch! Method 4: The base of Rs 600 must not be disturbed. When the watch was sold for Rs 700, there was a profit of Rs 100. When it was bought back for Rs 800, there was a loss of Rs 200, or net loss of Rs 100. When it was finally sold for Rs 900, there was a profit of Rs 300, which will offset the loss incurred up till now and thus there will be net profit of Rs 200. Example 47 There is an interesting story about a shopkeeper. In a market area, there were two shopkeepers A and B, selling bicycles. One day, a customer came to A and asked for a bicycle. The shopkeeper quoted the price of the bicycle at Rs 400, though he knew that it had cost him only Rs 300. The customer offered him a Rs 500 note but the shopkeeper did not have change. So, he came to B and asked for change. A gave B the Rs 500 note given by the customer and received the change of the same in exchange. Then, he returned Rs 100 note to the customer and the customer gladly went away from the market. The next day, B came running to A saying that the note of Rs 500 was a counterfeit note and hence A should return the money to B. Quietly, A gave B the real Rs 500 note, and sat down calculating the amount of loss in the entire transaction. Can you help A in doing the calculation? Solution The shopkeeper B made no losses as he got his Rs 500 back. The loss was incurred only by A. Shopkeeper A lost the bicycle worth Rs 300. Since we are calculating the loss incurred by A, we will look at what was the cost A had to bear for the bicycle and not the expected profit. Therefore, shopkeeper A lost a bicycle worth Rs 300 and Rs 100 which A gave to the customer in exchange for the Rs 500 note given by the customer. Therefore, the shopkeeper lost a total of Rs 400! When we come across a question wherein only the articles are mentioned and no value of the articles is given, we can assume some convenient price for the articles. Generally, we should take the cost price of the articles to be Re 1 per article.

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Quantitative Aptitude Simplified for CAT

Example 48 If selling price of 9 articles is same as cost price of 12 articles, then what is profit/loss%? Solution In such cases, always assume that cost price of one item/article is Re 1. In this example, if this assumption is taken, then SP of 9 articles = CP of 12 articles = Rs 12. Since CP of 9 articles is Rs 9, therefore, the trader earns profit of Re 3 on sale of 9 articles. Profit % =

3 1  100 = 33 %. 9 3

Example 49 Upon selling 10 items, a trader gains the selling price of 2 articles. What is the profit %? Solution Let SP of each item be Rs 1. Then, SP of 10 articles = Rs 10. Gain = Rs 2. Therefore, CP = Rs 8. Therefore, profit % = 25%.

Faulty weights Faulty weights are also a means adopted by shopkeepers to trick the customers into thinking that they are getting the right amount. If a trader cheats the customers by using faulty weights, then he gains not only through marking-up of the cost price, but also because of faulty weights. Toothpaste tubes marked as having 100 gm of toothpastes might not have an actual weight of 100 gm, but 98 gm. Therefore, the manufacturer is able to save 2 gm on sale of 98 gm (actual weight) but charges for 100 gm. If in a year, he is able to sell 1 lakh such tubes, he will save 2 lakh grams of paste or 200 kg of paste from which he can create more such tubes and sell! Example 50 A company sells toothpaste in tubes on which the weight of the paste is written as 100 gm. The actual weight of the paste was found to be 96 gm. (a)

What is the gain percent if the company claims to sell at cost price of the paste (assume that the tube is free of cost)?

(b)

What is the gain percent if the company would have gained 20% had it sold the complete amount?

(c)

If the company gains 20%, what would be the gain % if the weight of the paste were 100 gm?

Solution (a)

Let us convert grams to rupees. Let 1 gm paste cost Re 1. Then, 100 gm would cost Rs 100. If a tube is sold, the price claimed is Rs 100 whereas only paste worth Rs 96 is given to the customer.

Therefore, CP = Rs 96 and SP = Rs 100. Therefore, gain % =

1 4  100 = 4 % . 96 6

(b)

Had it sold the complete amount of paste, then CP = Rs 100 and SP = Rs 120. But, actual CP = Rs 24 96. Therefore, P% =  100 = 25%. 96

(c)

Company gains 20% means that the SP is 20% more than the actual CP of Rs 96. Therefore, SP = Rs 115.20.

Hence, if complete paste is given, then profit % is only 15.2%, because CP = Rs 100 instead of Rs 96.

120

Commercial Math Example 51 A trader buys 1 quintal of wheat from wholesaler and sells in the market. He uses a weight of 1 kg whose actual weight is 980 gm. He starts selling in the morning and by evening he is able to sell the entire stock. If he marks up the price by 10% only, what is the actual gain %? Solution Using beam balance, if he puts the weight of 1 kg on one balance, and wheat on the other, the actual weight of the wheat put is 980 gm only. Actual cost is Rs 980, whereas the cost shown is Rs 1000. Marked price = Rs 1100 (10% more than the cost price). Gain % =

120  100 = 12.24%. 980

Example 52 If a trader’s weighing balance shows 950 grams for every kilogram, will there be loss or profit to the shopkeeper who claims to sell the goods at cost price? How much is the profit or loss %? Solution There will be loss to the shopkeeper. Let the cost price of each item be Re 1. Then the shopkeeper actually gives away goods worth Rs 1000 whereas what he and the customer see on the weighing balance is 950 grams and accordingly the shopkeeper asks for Rs 950. (For something that cost him Rs 1000.) So, to the shopkeeper, there is a net loss of 50 grams for every kilogram (actual) sold. CP to the shopkeeper = Rs 1000. SP = Rs 950. Loss % =

50  100 = 5%. 1000

Example 53 During summers, the meter scale of a cloth merchant expands by 2%. If he claims to sell at cost price, what is the gain/loss%? Solution Since the meter scale expands, he would actually sell 102 cm for every 100 cm visible on the scale! Therefore, there is a clear cut case of loss. CP (actual) = Rs 102 and SP = Rs 100. Loss % =

2  100 = 1.96%. 102

Example 54 While buying from whole-seller, a trader uses a 1 kg weight which weighs 1100 gm. While selling, he uses a 1 kg weight which weighs 940 gm. If he claims to sell at cost price, what is the gain %? Solution He buys goods worth Rs 1100 from the whole seller for Rs 1000. He sells goods worth Rs 940 to the customers for Rs 1000. Therefore, he would sell goods worth Rs 1100 for Rs. 940. Therefore, his cost price = Rs 1000 and SP = Rs 1170.21.

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Quantitative Aptitude Simplified for CAT

Gain % = 17.021%. Alternatively, the seller is selling goods worth Rs 940 for Rs 1100 and hence gains 17.021%. Example 55 A trader gains 10% while buying and gains 10% while selling. What is the overall gain%? Solution For the trader, what goes out is cost price and what comes in is the selling price. While buying, if he gives Rs 100 and gets Rs 110 (or worth of goods), he would gain 10%. This Rs 110 worth of goods is sold at a profit of 10%. Therefore, SP = 1.1  110 = 121. Profit % = 21%. Example 56 A trader defrauds to the extent of 10% while buying and 10% while selling. What is the gain%? Solution This is a completely different example from the previous example! When he defrauds while buying, it means that if he buys goods whose weight is 100 kg, he shows it to be 90 kg. This means that he pays Rs 90 for 10 1 something which actually costs Rs 100. Profit % in such a case would be  100  11 % and not 10%. 90 9 Therefore, students are advised to note the difference between the previous and this example. Now, once he gets 100 kg, he sells it for Rs 110. Therefore, profit % =

110  90 2  100  22 % . 90 9

Example 57 Shreya purchased a house for Rs 204782. She wants to sell it for a profit of exactly 15%. However, she does not want to sell it herself. She wants a Real Estate Agent to sell it for her. The Agent must make a commission of exactly 5.8% (in addition to Shreya’s 15% profit). How much must the Agent sell the house for? Solution Purchase Cost + Shreya’s Profit + Agent’s Commission = Selling Price 204782 +

15 5.8 (204782) + S = S, 100 100

where S is the Selling Price in Rs, or S = 250000. Note that the commission is to be calculated on the sale price and not the cost price.

Simple and Compound Interest The concept of “interest” comes whenever you keep certain amount of money with someone for a certain period of time; or offer to lend some amount of money to someone for certain period of time. The amount of “interest” would depend upon (i)

the amount of money kept with that person,

(ii)

the rate of interest chargeable and

(iii) duration for which the money was kept.

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Commercial Math The rate of interest depends upon the use the borrower would make of that money. The higher the risk involved in the use of the money by the borrower, the more the interest rate the lender would charge. There are two ways to calculate the rate of interest depending upon whether compounding is done or not. If not, then it is called simple interest whereas if compounding is done, it is called compound interest.

Difference between Simple Interest (SI) and Compound Interest (CI) Let A offer to lend some money to B. This money lent is called the principle. When B returns this principle to A after some time, the money returned is higher than the principle. This money retuned is called the amount. The extra money paid is called interest. Therefore, Amount = Principle + Interest Or A=P+ This interest could be simple or compound. If the money is lent at simple interest, say r% per annum (p.a.), then after one year, interest would be r% of P. Therefore, SI =

r Pr of P = . 100 100

In the second year, to calculate the interest, principle is not changed (in case of simple interest) and it remains the same. Therefore, interest in the second year will also be r% of P and hence interest amount will also remain the same. In general, in t years, the interest would be would be SI =

Pr each year and hence, total simple interest in all these years 100

Pr t 100

As a result of this, the amount becomes, A = P +

rt  Pr t  A = P  1   100 100  

If the same money is lent at compound interest, say r% p.a., then after one year, interest would be r% of P. Therefore, CI =

r Pr of P = 100 100

Here we can see that the interest at the end of first year is same, irrespective of it being simple or compound. This interest accrues to the principle at the end of the first year and hence leads to the increase in the principle. Therefore, the amount at the end of first year becomes the principle for the second year. r  Amount at the end of first year = P + r % of P = P  1   100  

This is the principle for the second year and hence amount at the end of second year

r  r   r   = P  1    1  = P 1  100   100  100   

2

t

r   Continuing like this, we see that the amount at the end of t years becomes, A = P  1   . 100  

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Quantitative Aptitude Simplified for CAT

This is one of the most important formulas in commercial math. This formula is not just useful to calculate the interest you would receive from a bank where some of your money is deposited in savings account, but also useful to calculate the EMI (equated monthly instalment) when you take a loan from some bank. It is also to helpful in calculating the amount you would receive if you open a deposit account with a bank! Since A = P + , interest  is given by: =A–P t

r   In case of compound interest, CI = A – P = P  1   –P=P 100   We have already seen that simple interest, SI =

t   r   1   1 . 100    

Pr t 100

It is important to understand that in the first year, interest value is same irrespective of it being simple or compound. The difference starts coming after first year, that is, second year onwards. Since in the second year, principle remains the same in case of simple interest whereas increases in case of compound interest, we observe that at the same rate of interest r, compound interest will be higher than simple interest. Not just this, the value of compound interest will keep on rising higher and higher for subsequent years, whereas the value of simple interest will remain the same. The gap between CI and SI keeps increasing as years go by! We can also prove this using the formulae. t   r  CI = P   1   1 = P 100    

  Pr t Pr 2 t(t  1)  rt t(t  1) r2     .....  1   ....   1    100 2 1002 2(100)2    100 

which is obviously more than SI =

Pr t for t > 1 years. 100

Example 58 A sum of Rs 6000 is deposited in a bank which offers 6% rate of interest p.a. (compounding being done). Find (i)

the amount after 3 years.

(ii)

compound interest in first, second and third year.

If this sum is deposited where simple interest is offered, what is the SI in first, second and third year? Solution (i)

3  6   Amount after 3 years = 6000   1    = 6000 (1.06)3  Rs 7146. 100    

(ii)

CI in first year = 6% of 6000 = Rs 360.

(Note that CI in first year is same as SI in first year) To find CI in second year, there are many methods. Method 1: CI in second year = Amount at the end of second year – amount at the end of first year = 6000(1.06)2 – 6000(1.06) = Rs 381.6. Method 2: CI in 2nd year = 6% of (amount at the end of first year) = 6% of 6360 = Rs 381.6. Method 3: CI in 2nd year = CI in first year + 6% of CI in first year = 360 + 6% of 360 = Rs 381.6.

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Commercial Math From method 3 above, we must note that difference in SI and CI in 2 years is because of the interest on the interest in the first year. Interest in the first year is Rs 360 and interest on this Rs 360 is Rs 21.60 (@ 6% p.a.). Hence, the result. SI in all the years is same and is equal to the CI in the first year, that is, Rs 360. From the foregoing discussion, we can say that

r   CIn = CIm  1   100  

nm

, where

CIn = CI in the nth year (and not CI in n years), CIm = CI in the mth year (and not CI in m years), r = rate of interest applicable in the question. Example 59 If CI in third year is Rs.360 and that in fifth year is Rs.435.6, find rate of interest and also Principle. Solution

r   CI5 = CI3  1   100  

53

r    435.6 = 360  1   100  

Therefore, P (1.1)2 – P = 360 or P = Rs.

2

r  121   1  . Therefore, r = 10%.   100  100 

12000 . 7

Example 60 Difference in SI and CI in the first two years is Rs 50. Find the principle if rate of interest charged is 5%. Solution 2   2Pr r  Pr 2 1   Method 1: CI (in 2 years) – SI (in 2 years) = P   1   Rs 50.   2 100   100 100  

Therefore, P =

1002  50 = Rs 20,000. 52

Method 2: As seen in the above example (ii), we know that the difference in SI and CI in two years = interest on the interest in first year = 5% of , where  = interest in first year. Therefore,

100   = Rs 50. Hence,  = Rs 1000. 5

Now,  = 5% of Principle  Principle = Rs 20,000. Example 61 Difference in SI and CI in the first three years is Rs 100. If rate of interest is 5%, find the principle. Solution 3   3Pr r  Pr 3 3 Pr 2 Pr 2  r  CI (in 3 years) – SI (in 3 years) =   1     3  1   3 2 2  100 100 100 100 100 100      

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Quantitative Aptitude Simplified for CAT

= Rs.100 P =

100 1002  2 = Rs.13114.75. 3.05 5

Example 62 Amount at the end of 2 years is Rs 3000 and that at the end of 3 years is Rs 3450. Find the rate of interest (compounding being done). Solution Amount at the end of 2 years becomes the principle for the third year. Therefore, rate of interest =

3450  3000  100  15% 3000

Example 63 CI in the second year is Rs 11 and CI in third year is Rs 12.1. Find the rate of interest. Solution

r   12.1 = 11  1   100  

32

 r = 10%.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to approach problems based on Simple and Compound Interest.

Compounding period We have seen that in case of compound interest, the interest gets accrued to the principle and amount so obtained becomes the new principle for the next year. So far, we considered the cases where the compounding was done annually. This may not be the case always. Sometimes, compounding is done half-yearly, in which case, the interest gets accrued to the principle after 6 months instead of 1 year. If compounding is done on a quarterly basis, the interest is accrued to the principle after 3 months. Therefore, we can have various compounding periods. We have seen that the amount in case of compounding is given by

r   A = P 1  100  

t

where r is the rate of interest p.a. and compounding is done annually. In fact, if compounding period is not mentioned, we will assume it to be annual. If compounding period is different, we use the following formula to find out the amount.

r   A = P 1  c  100  

tc

where rc = rate of interest per compounding period tc = number of compounding periods in t years.

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Commercial Math In appearance, the above formula is same as the amount formula mentioned earlier. In fact, this is the most general formula to be used for all circumstances. Therefore, if compounding is done half-yearly, then rate of interest per half-year becomes

r   compounding periods in t years becomes 2t, that is A = P  1   200   r   and if compounding is done quarterly, then A = P  1   200  

r and number of 2

2t

4t

and likewise,

r   If compounding is done daily, then A = P  1   36500  

365t

Let us say that we deposit Rs 100 in a bank which offers 10% p.a rate of interest and where compounding is done annually. There is another bank in the same locality which offers 10% p.a. rate of interest and here the compounding is done half-yearly. At the end of 1 year,

10   Amount (bank 1) = 100  1   200  

1

= Rs 110.

2

5   Amount (bank 2) =  1   = Rs 110.25 200   We see that the amount in the second case is higher than in the first case. If compounding is done quarterly, then 4

2.5   Amount = 100  1    Rs 110.38 100   Can we say that the shorter the compounding period (all else remaining same), the higher the amount? Of course yes. The reason for this is not difficult to find. The shorter the compounding period, the earlier the interest gets accrued and changes the value of the principle for the next compounding period. This leads to higher amount if the compounding is done for a shorter period. The remaining period earns interest on the interest also along with the principle and this is the real reason for higher amounts in case of shorter compounding period. Therefore, amount is higher if compounding is done on quarterly basis than if it is done on half-yearly basis which will be higher than if it is done on an annual basis. Obviously, daily compounding yields better returns than quarterly compounding. In the above example, we can see that Rs 100 becomes Rs 110 if interest rate is 10% and compounding is done annually and it is Rs 110.25 if compounding is done half-yearly. We will get the same amount of Rs 110.25 if rate of interest is 10.25% p.a., compounding done annually. We say that “10.25% p.a. compounding done annually” is the effective rate of interest for “10% p.a. compounding done half-yearly”. Thus, if rate of interest is 6% p.a. compounded half-yearly, the effective rate of interest is 6.09% p.a. compounded annually, because (1.03)2 = 1.0609. To find the effective rate of interest in case of half yearly compounding, we can also use the formula for successive percentage changes. So, if rate of interest is 6% p.a. compounded half-yearly, then rate of interest 33 per compounding period is 3%. So, effective (annual) rate of interest = 3 + 3 + = 6.09%. 100

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Quantitative Aptitude Simplified for CAT

Example 64 A sum is divided into two equal parts and each part invested in two instruments, one offering 8% p.a. compounding done annually and the other offering 8% p.a. compounding done half-yearly. If the difference in the income after one year is Rs 10,000, what is the value of the sum? Solution Let P be divided equally into two parts. Therefore, each part is P/2. P P P 10,000 (1.04)2 – (1.08) = 10,000  = = Rs 62,50,000. 0.0016 2 2 2

Therefore, value of the sum = Rs 1, 25, 00,000. Example 65 If a bank offers 5%, 7% and 4% p.a. (compounding done annually) in three consecutive years respectively, what is the overall rate of return? Solution (1.05)(1.07)(1.04) = 1.16844  overall rate of return = 16.844%. Alternatively, use successive percentage change formula.

Doubling time period If we have some amount of money, we would like to know in how many years this sum will double itself. Naturally, this depends only upon the rate of interest offered and not on the sum invested. The higher the rate of interest, the earlier the doubling will happen. Let P be the initial sum and rate of interest offered is r%. Then, if compounding is done, amount would double in n years as given below:

r   2P = P  1   100  

n

or

r   2 = 1   100  

n

As stated earlier, doubling period depends only upon the rate of interest and nothing else. If simple interest is calculated, then =

Pr t 100

Since amount doubles, therefore, interest = principle (as A = 2P = P + ). Hence, P=

Pr t 100 or t = . 100 r

In case of SI, if money is to become n times, then t =

(n  1)  100 r

Example 66 A sum of money doubles in 5 years. In how many years will the money become 8 times, if compounding is done?

128

Commercial Math Solution 5

r   Since compounding is done,  1   = 2. 100   3

5  r   Cubing both sides of the equation,   1    = 23  = 8. 100    

Therefore, in 15 years, the money would become 8 times. Example 67 In the above example, what will happen if simple interest is calculated? Solution Doubling means interest = principle. Therefore, P =

Pr(5) . 100

If money is to become 8 times, interest would be 7 times principle. Multiplying the above equation by 7, we get: 7P =

Pr(5) Pr(35) 7  . 100 100

This means that the money would become 8 times in 35 years, if simple interest is calculated. Alternatively, if money doubles in 5 years, 5 = t=

100 . For money to become 8 times, r

(n  1)  100 (8  1)  100 7  100 =   35 years. r r r

Example 68 In how many years will Rs 10,000 become Rs 20,000 if rate of interest is 8%, provided (i)

simple interest is calculated.

(ii)

compounding is done.

Solution (i)

In case of simple interest, the formula for doubling is t =

100 . r

Since r = 8%, t = 12.5 years. n

(ii)

r   If compounding is done, then doubling would happen as per the formula 2 =  1   . 100  

Since r = 8%, then (1.08) n = 2 n = 9 years. Now, solving this is going to be tough in the exam. For doubling in case of compounding, use the empirical formula: t =

72 , where 5% ≤ r ≤ 12%. r

If we restrict ourselves in this range of r, the error rate would be less than 1%. If we go beyond the range, the error would be high. The farther we go beyond this range, the greater would be the percentage error. Solving the above example using this formula, we get t =

72 72  9 years.  r 8

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Quantitative Aptitude Simplified for CAT

Future Value (FV) and Present Value (PV) If I borrow Rs 50,000 from one of my friends and return Rs 50,000 to him 20 years hence, he would not like it! That’s because the value of money changes with time. The rate of interest prevalent determines the speed with which this value changes. In fact, n

r   FV = PV  1   , 100   where FV = Future Value, PV = Present Value, r = rate of interest applicable and n = number of years. Therefore, if rate of interest applicable is 10% and Rs 5000 is borrowed to be paid after 5 years, then the amount of money which will pay off the debt is given by 5

10   FV = 5000  1   = Rs 8052.55 100   Example 69 A is to pay B Rs 12000 after 3 years. If A wants to pay after 1 year, what amount should be paid; provided the rate of interest is 5% p.a.? Solution Note that since A wants to pay after 1 year, time = 2 years will be taken. Let x be the amount payable 1 year from now. Then, 12000 = x (1.05)2. Therefore, x  Rs 10,884. Example 70 Ranjeet borrows Rs 1 lakh from Ravi and agrees to pay after 2 years with 8% CI. But after 2 years, he is not able to pay the money. He borrows from Rajan the amount due, and pays Ravi. Ranjeet agrees to pay Rajan 3 years hence the amount payable at the new rate of interest of 10%. What is the amount Ranjeet is required to pay to Rajan? Solution Amount payable to Ravi = 100000(1.08)2. Amount payable to Rajan = 100000(1.08)2  (1.1)3 = Rs 155247.84. The concept of FV and PV can be used to calculate the EMI (equated monthly instalment) if a loan is taken. In such a case, the borrower agrees to pay a fixed amount on a monthly basis for a predetermined period of time. The value of EMI depends upon: the amount of loan, rate of interest applicable and duration of the loan. Similarly, we can apply these principles in case we are saving through recurring deposit, where monthly or quarterly payment is made. Observe the following examples. Example 71 A person opens a recurring deposit account with a bank and invests an amount of Rs 500 per month for one year at the rate of interest of 12% p.a. compounding being done monthly. He pays the first instalment on 1st January, 2002. Next instalment is paid on 1st February, 2002 and likewise last instalment on 1st December, 2002. He will receive lump sum amount on 1st January, 2003. How much money is he entitled to receive?

130

Commercial Math Solution The first instalment of Rs 500 is kept in the bank for exactly one year and hence the rate of interest is 1% per 12 1   month (because r = 12% per annum). This Rs 500 will become 500  1   . The second instalment was 100   with the bank for only 11 months. Continuing like this, we observe that the last instalment was kept with the bank for 1 month only. Therefore, total amount he is entitled to receive should be

1   500  1   100  

12

1   500  1   100  

11

1   500  1   100  

10

1   ....  500  1   100  

1

12 11 10 1  1  1  1  1      = 500   1    1    1    ....   1    100  100  100  100       

This is a geometric progression. Let 1.01 = a. Then, we have 500(a + a2 + a3 + ..... + a12) =





500a a12  1 500(1.01)(1.01)12  1  = 6404.6. a  1 0.01

Example 72 A person is under a debt of Rs 6000, which he can’t pay today. He agrees to pay in monthly instalments where the rate of interest payable is 12% p.a. (compounding done monthly). What equal monthly instalment will pay off the debt in one year? Solution Let EMI be Rs x. The instalment is paid at the end of every month. x (1.01)11 + x(1.01)10 + .... + x (1.01)0 = 6000(1.01)12 x=

6000(1.01)12 = Rs.533.  1.01 12  1     1.01  1 

Example 73 If the above person is to receive Rs 10,000 after 5 years from a third person, can he pay off the debt charged to him @ 12% p.a. CI, compounding done annually? Solution 5

12   Future value of Rs 6000 @ 12% after 5 years will be 6000  1   = 10,574. 100   Therefore, the person will not be able to pay off the debt. He will fall short of money by Rs 574. 12  5  If the lender agrees to payment by simple interest, then the amount will be 6000  1   or 9600. In 100   this case, he can pay off the debt.

Compounded Annual Growth Rate (CAGR) and Simple Annual Growth Rate (SAGR) The concept of CAGR and SAGR is a hallmark concept for students aspiring for MBA and which is also used extensively in Data Interpretation questions.

131

Quantitative Aptitude Simplified for CAT

Let us assume that a quantity grows at the rate of 10% in the first year and at the rate of 20% in the subsequent year. We know that the overall growth rate is 32% (applying the concept of successive increments). What is the annual growth rate? The immediate question should be: are we doing compounding or not? 2

r   If compounding is done, then the rate r% is determined by  1    1.32 . 100   2p  and if simple interest is being calculated, then the rate p% is determined by  1    1.32 . 100  

This r is called CAGR and p is called SAGR. Solving, we get r = 14.89% and p = 16%. Therefore, if simple interest is being calculated, the overall growth rate (for two years), that is, 32% is divided by number of “growth” years (that is, 2) to get SAGR, which is 16%. Calculating CAGR is quite demanding. Let us see the calculation technique employed. 2

r  1  r   1.32 1+ =   100  100 

1.32 .

Now, to find the square root of 1.32, we know that 1.44 and 1.21 are squares of 1.2 and 1.1. Therefore, square root of 1.32 will lie between 1.1 and 1.2. The square of 1.15 = 1.3225, which is very close to 1.32 and more than 1.32. Therefore, rate will be only slightly less than 15% and hence the value should be approximately 14.9%. Example 74 If a population grows at the rate of 10%, 15% and then 25% in the three successive years, then find CAGR and SAGR. Solution 3

r  10   15   25    If CAGR is r%, then  1    P1   1  1  100  100   100   100    3

r   or  1   = 1.1 ×1.15×1.25=1.58125 100   Therefore, 1+

r  3 1.58125  1.165 .Therefore, r = 16.5%. 100

If SAGR is p%, then p =

1 3

rd

of 58.125% = 19.375%.

Note that CAGR is always less than SAGR for same overall % change.

132

Commercial Math

PRACTICE EXERCISE 1.

The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

B.

36 % in option B and 64% in option C

C.

64 % in option B and 36% in option C

D.

1/3 in each of the three options

A.

remain the same

E.

B.

decrease by 13.64%

30 % in option A, 32% in option B and 38% in option C

C.

decrease by 15%

D.

decrease by 18.75%

E.

decrease by 30%

Directions for questions 4 and 5: Read the following information and answer the questions that follow: Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is: 240 + bx + cx2 where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 2 66 % . However, an increase in daily 3 production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

CAT 2006 Directions for questions 2 and 3: Read the following information and answer the questions that follow: Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others. Option A: Invest in a public sector bank. It promises a return of +0.10%. Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of –3%.

CAT 2007

Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of –2.5%, while a fall will entail a return of +2%.

4.

CAT 2007

2.

3.

The maximum guaranteed return to Shabnam is A.

0.25%

B.

0.10%

C.

0.20%

D.

0.15%

E.

0.30%

What strategy will maximize guaranteed return to Shabnam? A.

5.

the

100 % in option A

133

How many units should Mr. David produce daily? A.

130

B.

100

C.

70

D.

150

E.

Cannot be determined

What is the maximum daily profit, in rupees, that Mr. David can realize from his business? A.

620

B.

920

C.

840

D.

760

E.

Cannot be determined

Quantitative Aptitude Simplified for CAT

5.5 years. Which of her friends offers the MOST appropriate option?

Directions for questions 6 and 7: Answer the questions on the basis of the information given below:

A. B. C. D.

In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks. CAT 2004 6.

7.

8.

CAT 2013 10. A farmer grows some variety of crop every year in his farm. Because of technological innovation, a better yielding crop of the same variety is available to him which improves the productivity by 90%. It also reduces the fertilizer consumption by 5%. However, the new high yielding variety requires doubling the space between two plants compared to older variety. Assume that he incurs cost only on fertilizer, and the new varieties are available free of cost. Also assume that he is able to sell that he produces. If he opts for the new variety, then

If group B contains 23 questions, then how many questions are there in group C? A.

1

B.

2

C.

3

D.

Cannot be determined

If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B? A.

11 or 12

B.

12 or 13

C.

13 or 14

D.

14 or 15

A. B. C. D.

30.9%

B.

29.6%

C.

44%

D.

25%

11. If discount is 40%, profit is 20%. If discount is 25%, what is the profit %? A.

25%

B.

40%

C.

50%

D.

0% CAT 2014

12. If income of Meena is 20% more than that of Ena and 20% less than that of Tina, and income of Reena is 50% of that of Tina, then by how much percentage should the income of Reena be increased to make it equal to that of Ena?

CAT 2012 9.

He loses 5% No profit no loss He loses 10% Data insufficient CAT 2013

If mark-up % was 60%, and trader gives discount of 10% and also bears 10% sales tax, what is overall profit / loss %? A.

Naveen Namita Mohan Shreya

Nikita wants to double her money for which she has the following options: Naveen offers 8% per annum compounding done annually for at least 6 years; Namita offers 12% for a time period not exceeding 5 years; Mohan offers 15% for exactly 4 years; whereas Shreya offers 10% for at most

A. B.

25% 1 33 % 3

C.

50%

D.

20% CAT 2015

134

Commercial Math 13. If a sum of Rs. 64,00,000 becomes Rs. 3,24,00,000 in 12 years, then how much money will become Rs.2,25,00,000 in 18 years, compounding done annually? A.

Rs. 1975308.6

B.

Rs. 2375128.6

C.

Rs. 2576308.6

D.

Rs. 1813218.6

soap at the same price, the approximate percentage by which average cost price is lower or higher than the selling price is A. B. C. D.

10.5 higher 12.5 lower 14.5 lower 8.5 higher IIFT 2016

CAT 2016

17. The pre-paid recharge of Airtel gives 21% less talktime than the same price pre-paid recharge of Vodafone. The post-paid talktime of Airtel is 12% more than its pre-paid recharge, having the same price. Further, the post-paid talktime of same price of Vodafone is 15% less than its pre-paid recharge. How much percent less / more talktime can one get from the Airtel post-paid service compared to the post-paid service of Vodafone?

14. A sum of money becomes Rs. 40,000 in 3 years at 8% per annum compounding done annually. Then, what will the same sum of money become in the same number of years if simple interest was applicable? A.

Rs. 39374

B.

Rs. 37950

C.

Rs. 40000

D.

Rs. 37657 CAT 2016

15. An agent agreed to buy goods from the seller for Rs. 24 lakhs, but the seller said that the agent has to remove all the goods from his godown within the next 30 days, failing which he will charge the agent Rs. 2000 per day for each day beyond 30 days. The cost incurred by the agent for the removal of goods was Rs. 14000 per day. The agent is able to sell all the goods for Rs. 37.8 lakh, thereby earning a profit of Rs. 3.2 lakhs. In how many days was he able to remove all the goods from the premises of the seller? A.

40 days

B.

50 days

C.

70 days

D.

60 days

A.

3.9% more

B.

4.7% less

C.

4.7% more

D.

2.8% less IIFT 2015

18. As a strategy towards retention of customers, the service centre of a split AC machine manufacturer offers discount as per the following rule: for the second service in a year, the customer can avail of a 10% discount; for the third and fourth servicing within a year, the customer can avail of 11% and 12% discounts respectively of the previous amount paid. Finally, if a customer gets more than four services within a year, he has to pay just 55% of the original servicing charges. If Rohan has availed 5 services from the same service centre in a given year, the total percentage discount availed by him is approximately:

CAT 2016 16. In a local shop, as part of promotional measures, the shop owner sells three different varieties of soap, one at a loss of 13 percent, another at a profit of 23 percent and the third one at a loss of 26 percent. Assuming that the shop owner sells all three varieties of

A. B. C. D.

16.52 20.88 22.33 24.08 IIFT 2015

135

Quantitative Aptitude Simplified for CAT

19. Sailesh is working as a sales executive with a reputed FMCG Company in Hyderabad. As per the Company’s policy, Sailesh gets a commission of 6% on all sales upto Rs. 1,00,000 and 5% on all sales in excess of this amount. If Sailesh remits Rs. 2,65,000 to the FMCG company after deducting his commission, his total sales were worth

C.

5.5 percent (loss)

D.

None of the above IIFT 2014

22. If decreasing 70 by X percent yields the same result as increasing 60 by X percent, then X percent of 50 is A. B. C. D.

A.

Rs. 1,20,000

B.

Rs. 2,90,526

C.

Rs. 2,21,054

3.84 4.82 7.10 The data is insufficient to answer the question

D.

Rs. 2,80,000

IIFT 2013 23. Three years ago, your close friend had won a lottery of Rs. 1 crore. He purchased a flat for Rs. 40 lakhs, a car for Rs. 20 lakhs and shares worth Rs. 10 lakhs. He put the remaining money in a bank deposit that pays compound interest @ 12 percent per annum. If today, he sells off the flat, the car and the shares at certain percentage of their original value and withdraws his entire money from the bank, the total gain in his assets is 5%. The closest approximate percentage of the original value at which he sold off the three items is

IIFT 2015 20. Eight years after completion of your MBA degree, you start a business of your own. You invest INR 30,00,000 in the business that is expected to give you a return of 6%, compounded annually. If the expected number of years by which your investment shall double is 72/r, where r is the percent interest rate, the approximate expected total value of investment (in INR) from your business 48 years later is: A.

2,40,00,000

B.

3,60,00,000

C.

4,80,00,000

D.

None of the above

A. B. C. D.

60% 75% 90% 105%

IIFT 2014

IIFT 2013

21. A pharmaceutical company manufactures 6000 strips of prescribed diabetic drugs for Rs. 8,00,000 every month. In July 2014, the company supplied 600 strips of free medicines to the doctors at various hospitals. Of the remaining medicines, it was able to sell 4/5th of the strips at 25% discount and the balance at the printed price of Rs.250. Assuming vendor’s discount at the rate of a uniform 30 percent of the total revenue, the approximate percentage profit / loss of the pharmaceutical company in July 2014 is: A.

5.5 percent (profit)

B.

4 percent (loss)

24. The annual production in cement industry is subject to business cycles. The production increases for two consecutive years consistently by 18% and decreases by 12% in the third year. Again in the next two years, it increases by 18% each year and decreases by 12% in the third year. Taking 2008 as the base year, what will be the approximate effect on cement production in 2012? A. B. C. D.

136

24% increase 37% decrease 45% increase 60% decrease

Commercial Math IIFT 2012

oranges at Rs. 10 apiece and marks 25% profit. If she gets Rs.653 after selling all the apples and oranges, find her profit percentage.

25. Mr. Mishra invested Rs. 25,000 in two fixed deposits X and Y offering compound interest @ 6% per annum and 8% per annum respectively. If the total amount of interest accrued in two years through both fixed deposits is Rs. 3518, the amount invested in Scheme X is A.

Rs. 12,000

B.

Rs. 13,500

C.

Rs. 15,000

D.

Cannot be determined

B.

Loss of Rs. 210

C.

Loss of Rs. 250

D.

None of the above

B.

17.4%

C.

17.9%

D.

18.5%

E.

19.1%

29. The Maximum Retail Price (MRP) of a product is 55% above its manufacturing cost. The product is sold through a retailer, who earns 23% profit on his purchase price. What is the profit percentage (expressed in nearest integer) for the manufacturer who sells his product to the retailer? The retailer gives 10% discount on MRP.

26. Rohit bought 20 soaps and 12 toothpastes. He marked-up the soaps by 15% on the cost price of each, and the toothpastes by Rs. 20 on the cost price each. He sold 75% of the soaps and 8 toothpastes and made a profit of Rs. 385. If the cost of a toothpaste is 60% the cost of a soap and he got no return on unsold items, what was his overall profit or loss? Loss of Rs. 355

16.8%

XAT 2016

IIFT 2012

A.

A.

A.

31%

B.

22%

C.

15%

D.

13%

E.

11% XAT 2015

30. In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs. 10,000 and in the beginning of 2010, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principle amount is:

IIFT 2012 27. A shop, which sold same marked price shirts, announced an offer - if one buys three shirts then the fourth shirt is sold at a discounted price of `100 only. Patel took the offer. He left the shop with 20 shirts after paying `20,000. What is the marked price of a shirt? A.

`1260

A.

Rs. 16,000

B.

`1300

B.

Rs. 18,000

C.

`1350

C.

Rs. 20,000

D.

`1400

D.

Rs. 25,000

E.

`1500

E.

None of the above XAT 2015

XAT 2017

31. The tax rates for various income slabs are given below.

28. Rani bought more apples than oranges. She sells apples at Rs.23 apiece and makes 15% profit. She sells

137

Quantitative Aptitude Simplified for CAT

each when they reach the age of 21. He is looking for plan that will give him a simple interest per annum. The rates of interest of the plans for his younger son and elder son should be

Income Slab (Rs.)

Tax rate

≤ 500

Nil

> 500 to ≤ 2000

5%

> 2000 to ≤ 5000

A.

5% and 7.5% respectively

10%

B.

8% and 12% respectively

> 5000 to < 10000

15%

C.

10% and 15% respectively

D.

15% and 22.5% respectively

E.

20% and 30% respectively

There are 15 persons working in an organization. Out of them, 3 to 5 persons are falling in each of the income slabs mentioned above. Which of the following is the correct tax range of the 15 persons? (E.g. If one is earning Rs. 2000, the tax would be: 500 × 0 + 1500 × 0.05) A.

1350 to 7350, both excluded

B.

1350 to 9800, both included

C.

2175 to 7350, both excluded

D.

2175 to 9800, both included

E.

None of the above

XAT 2013 33. Ramesh bought a total of 6 fruits (apples and oranges) from the market. He found that he required one orange less to extract the same quantity of juice as extracted from apples. If Ramesh had used the same number of apples and oranges to make the blend, then which of the following correctly represents the percentage of apple juice in the blend?

XAT 2015 32. Mr. Mehra is planning for higher education expenses of his two sons aged 15 and 12. He plans to divide Rs 15 lakhs in two equal parts and invest in two different plans such that his sons may have access to Rs. 21 lakhs

A.

25%

B.

33.3%

C.

60%

D.

60.6%

E.

None of the above XAT 2013

ANSWER KEY 1.

E

2.

C

3.

B

4.

B

5.

9.

A

10. B

11. C

12. B

13. A

14. A

15. D

16. A

17. A

18. B

19. D

20. C

21. C

22. A

23. C

24. C

25. C

26. A

27. B

28. B

29. D

30. C

31. A

32. E

33. E

138

D

6.

A

7.

C

8.

A

Commercial Math

ANSWERS AND EXPLANATIONS 1.

2 2 the % rise in cost is 66 % = . At 20 3 3 units, the cost = 240 + 20b + 400c. At 40 units, the cost = 240 + 40b + 1600c. 2 5 As the cost increases by , (240 + 3 3 20b + 400c) = 240 + 40b + 1600c …(1) Similarly, When units increase from 40 to 60, the % rise in cost is 50% 3 (240 + 40b + 1600c) = 240 + 60b 2 +3600c …(2) By solving equations (1) and (2), we get 1 b = 10, c = . The cost equation 10 1 2 becomes 240 + 10x + x . The 10 selling price per unit is Rs.30. Therefore the profit we can obtain by selling x units 1 2 is, P(x) = 30x – (240 + 10x + x) 10 1 2 x. = –240 + 20x – 10 x Therefore, P’(x) = 20 – = 0 or x 5 = 100. At x = 100, P(100) = Rs (–240 + 2000 – 1000) = Rs 760. Hence correct option for question 4 is [B] and for question 5 is [D].

Answer: E Explanation: Let the length, breadth and height be 3, 2 and 1. New length, breadth and height is 6, 1 and 0.5. Total area of the four walls (initial) = 2  1 (3 + 2) = 10 units. Total area of the four walls (final) = 2  0.5 (6 + 1) = 7 units. So, the area of the four walls decreases by 30%.

Common explanation for questions 2 and 3: Let investment in option A, B and C be x, y and z. Then, in rising market, the returns will be 0.001x + 0.05y – 0.025z and in a falling market, the returns will be 0.001x – 0.03y + 0.02z. For maximum guaranteed returns, 0.001x + 0.05y – 0.025z = 0.001x – 0.03y + 0.02z, or 64y = 36z. Assuming Shabnam has Rs 100 to invest, she should invest Rs 36 in option B and 64 in option C, so that her returns = 0.05(36) – 0.025(64) = 1.8 – 1.6 = 0.2%. Since her combined guaranteed returns from option B and C is more than that possible from option A, she should not invest any money in option A. Now,

2.

4.

Explanation: Mr. David produce should produce 100 units daily.

5.

Answer: D Explanation: The maximum daily profit is Rs 760.

Answer: C Explanation: The maximum guaranteed return to Shabnam is 0.20%

3.

Answer: B

6.

Answer: A Explanation: Group A carries 1 mark, Group B carries 2 marks and Group C carries 3 marks. Now group A carries atleast 60%. If group B contains 23 questions, it carries 46 marks. If C contains 1 question, then A contains 100 – 23 – 1 = 76 questions. So, percentage marks for A 76 section =  100 = 60.8%. If 76  46  3

Answer: B Explanation: 36 % in option B and 64% in option C will maximize the guaranteed return to Shabnam.

Common explanation for questions 4 and 5: The cost is given as 240 + bx + cx2. When the units increase from 20 to 40,

139

Quantitative Aptitude Simplified for CAT

10. Answer: B

C contains 2 questions, then percentage marks for A section 75 =  100 = 59.055% < 75  46  6 60%. Therefore, C cannot contain more than 1 question. 7.

Explanation: Production earlier = 100. Production after improvement in productivity = 190. But since only half of the plants can be sown, production now = 95 (in the same area of land). Also, cost has become Rs. 95 from earlier cost of Rs. 100. Per unit cost earlier = Rs. 100 / 100 = Rs. 1 per unit. Per unit cost now = Rs. 95 / 95 = Rs. 1 per unit. So, no profit no loss.

Answer: C Explanation: Group C contains 8 questions that is 24 marks. If group B contains 11 questions, then percentage marks of B section 22 =  100 = 17.32%. So, A 81  22  24 option is ruled out. If group B contains 12 questions, then percentage marks of B section 24 =  100 = 18.75%. So, B 80  24  24 option is ruled out. If group B contains 13 questions, then percentage marks of B section 26 =  100 > 20%. 79  26  24

8.

11. Answer: C Explanation: Let CP = Rs. 100. Then, SP = Rs. 120. Since discount is 40%, Marked Price = 120/0.6 = Rs. 200. If discount is 25%, then SP = Rs. 150. Then, profit % = 50%. 12. Answer: B Explanation: Let income of Ena = Rs. 100. Then, income of Meena = Rs. 120, which is 20% less than that of Tina. So, income of Tina = 120/0.8 = Rs. 150. Income of Reena = 50% of that of Tina = Rs. 75. Now, percentage by which income of Reena should be increased to make it equal to that of Ena 100  75 1 =  100  33 % . 75 3

Answer: A Explanation: [A] Let CP = Rs. 100. Then Marked Price = 160. Price after discount of 10% = Rs. 160 – 16 = Rs. 144. If SP is Rs. x, then after including tax, amount becomes 1.1x, which is equal to 144. So, x = 144/1.1 = Rs. 130.9. Therefore, profit % = 30.9%.

9.

Answer: A

13. Answer: A

Explanation: If Nikita uses offer of Naveen, then amount ≥ P(1.08)6 = 1.586P If Nikita uses offer of Namita, then amount ≤ P(1.12)5 = 1.76P If Nikita uses offer of Mohan, then amount = P(1.15)4 = 1.749P If Nikita uses offer of Shreya, then amount ≤ P(1.10)5(1.05) = 1.691P The only option she has is to use the offer of Naveen.

Explanation: Let the money grow by r% per annum, compounding done annually. Then, 12 r   64,00,000  1   100   = 3,24,00,000. If we assume that 6 r  k   1   , then equation 100   becomes 64,00,000k2 = 3,24,00,000 324 81 9 or k2 = or k =  64 16 4 Now, if the required sum of money is P, then Pk3 = 2,25,00,000 or P = Rs. 1975308.6.

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Commercial Math 14. Answer: A

80.1 (= 70.488) and 55. Therefore, total amount paid = 395.588 or Rs. 395. Therefore, overall discount % availed 500  395 =  100 = 21%. 500

Explanation:

40,000 . Therefore, (1.08) 3 amount in case of SI 40,000  38 =  1   = Rs. 39374. 3 100  (1.08)  Principle, P =

19. Answer: D Explanation: Let total sales = x. Then, total commission = 6% of 100000 + 5% of (x – 100000), which is equal to x – 265000. Therefore, 6000 + 0.05(x – 100000) = x – 265000 or x = 280000.

15. Answer: D Explanation: Cost incurred by the agent = 37.8 – 3.2 = Rs. 34.6 lacs, of which cost of goods = Rs. 24 lacs. So, cost incurred due to delay in removing goods from the godown = 34.6 – 24 = Rs. 10.6 lacs. If he removed the goods in total x days, then cost incurred by him due to delay = 14000x + 2000(x – 30) = 10,60,000 or x = 70 days.

20. Answer: C Explanation: Investment doubles in 72/6 = 12 years So, in 48 (= 12 × 4) years, the investment becomes 24 times. Therefore, total value of the investment (in INR) = 16 × 3000000 = 4,80,00,000.

16. Answer: A Explanation: Let the S.P. of each type be Rs. 100. 100 ≅ Rs. 115. Then, CP of first type = 0.87 100 ≅ Rs. 81 and CP of second type = 1.23 100 CP of third type = ≅ Rs. 135. 0.74 Therefore, total CP = Rs. 331. Total SP 31 = Rs. 300. Therefore, CP is  100 300 = 10.5% higher than SP.

21. Answer: C Explanation: 600 strips were given free to doctors. Of 4 × 5400 = 4320 strips 5400 strips, 5 were sold at 25% discount. Revenue generated from these strips = 250 × 0.75 × 4320 = Rs.8,10,000. Revenue generated from remaining (5400 – 4320 = 1080) strips = 250 × 1080 = Rs.2,70,000 Total revenue = Rs.10,80,000. Vendor’s discount = 30% of the total revenue. Therefore, net profit = 70% of 1080000 = Rs.756000. So, there is net loss of 800000 – 756000 = Rs.44000. Therefore, loss% 44000 =  100 = 5.5%. 800000

17. Answer: A Explanation: If Vodafone pre-paid talktime is 100, then Airtel pre-paid talktime = 79 and Airtel post-paid talktime = 79(1.12) = 88.48. Vodafone post-paid talktime = 85. So, required percentage 88.48  85 =  100 = 4.09%. 85

22. Answer: A Explanation: X  X    70  1    60  1   or 10 100  100    70 X 60 X 10 = or X = % = 7.69%  100 100 1 .3

18. Answer: B Explanation: Let original service charges be Rs.100. Then, amount paid by the customer in the 5 services = 100, 90, 80.1, 0.88 x

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Therefore, X% of 50 = 50% of X = half of 7.69 = 3.84.

Profit from sale of 75% of soaps and 8 toothpastes = 15% of (75% of 20)x + 20 × 8 = 385  x = Rs.100. Overall revenue from soaps = (75% of 20)(1.15)x = 17.25x = Rs. 1725. Similarly, revenue from toothpastes = 8 × (0.6x + 20) = Rs. 640. Total revenue = Rs. 2365. Total cost = 20x + 12 (0.6x) = 2000 + 720 = Rs. 2720. Net loss = Rs. 355.

23. Answer: C Explanation: Gain in assets = 5% of 1 cr = Rs. 5 lacs. If he gets r% of the asset value from sale of those assets, then, 105 = r% of 70 + 62.87 30(1.12)3 or r =  100 = 90% 70 (approx.). 24. Answer: C

27. Answer: B

Explanation: In the time period, there are 3 increases of 18% each and a decrease of 12%. So, there will be net increase and so options [B] and [D] are eliminated. Further, approximate increase % will be 18 × 3 – 12 = 42%. So, nearest option is [C].

Explanation: Since he gets 20 shirts, he buys 15 shirts at Rs x per shirt and 5 shirts at Rs. 100 each. Therefore, 15x + 500 = 20,000 or x = Rs.1300. 28. Answer: B Explanation: If she sells x apples and y oranges, then 23x + 10y = 653  23x = 653 – 10y = (23 × 28 + 9) – 10y = 23 × 28 + (9 – 10y). Now, 9 – 10y should be a multiple of 23. Also note that 9 – 10y will always end with 1 and will be a negative number. Since 23 × 7 ends with 1, 9 – 10y = 23 × 7 = 161. Therefore, y = 17 and so x = 21, which is in line with the fact that Rani bought more apples than oranges. CP of each apple = Rs. 20 and so total cost of all apples = 20 × 21 = Rs. 420 CP of each orange = Rs. 8 and so total cost of all oranges = 8 × 17 = Rs. 136. Total cost of all the fruits = Rs. 556. 653  556 Overall profit % =  100 556 = 17.44%.

25. Answer: C Explanation: Overall returns in 2 years 3518 =  100 = 14.072%. 25000 Overall returns in first investment @ 6% 66 = 6 + 6 + = 12.36%. Similarly, 100 overall returns for second investment 88 =8+8+ = 16.64%. 100 Using Alligation, ratio of investments of first to second = (16.64 – 14.072) : (14.072 – 12.36) = 2.568 : 1.712 = 3 : 2. Therefore, investment in scheme X 3 =  25000 = Rs. 15000. 5 Alternatively, (1.06)2x + (1.08)2(25000 – x) = 25000 + 3518 Or, x = Rs. 15000.

29. Answer: D Explanation: Let Manufacturing Cost be Rs.100. Therefore, MRP = Rs. 155. Since discount is 10%, SP (for retailer) = 155 × 0.9 = Rs.139.5. SP for manufacturer = CP for retailer 139 .5 = = Rs. 113. 1.23

26. Answer: A Explanation: Let cost price of soap be Rs. x. Also, MP = SP as no discount is mentioned. CP of toothpaste = 0.6x.

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Commercial Math Therefore, profit % for manufacturer = 13%.

3). For maximum tax outgo, this list will be (3, 3, 4, 5). Tax in first case > 5 × 0 + 4 × 0 + 3 × 75 + 3 × 375 = Rs.1375. Tax in first case < 3 × 0 + 3 × 75 + 4 × 375 + 5 × 1125 = Rs.7350.

30. Answer: C Explanation: Notice that the time gap between 2004 and 2007 is same as that between 2007 and 2010, that is 3 years. If the money grows by some percent in the first 3 years, the growth rate in the next 3 years will be same. Also, compound interest in the last 3 years also grows by same percent as in the first 3 years. The interest accumulated in the last 3 years = 25000 – 10000 = Rs. 15000, which is 50% more than that in the first 3 years. So, growth rate is 50% every 3 years. In the first 3 years, if money grew by 50% and amount of increase is Rs. 10000, principle will obviously be Rs. 20000.

32. Answer: E Explanation: For younger child, interest in 9 years = 21 – 7.5 = 13.5. So, interest per year 13.5 = 1.5. = 9 1.5 So, rate of interest =  100 7.5 = 20%. We need not calculate for elder one because there is only one option with 20% rate of interest for younger child. However, we can still find for elder one. Interest in 6 years = 13.5 and so interest per year = 2.25. Rate of interest 2.25 =  100 = 30%. 7.5

31. Answer: A Explanation: Let x be the income of any person. If x ≤ 500, tax outgo = Rs.0 If 500 < x ≤ 2000, then tax outgo = 0 + 5% of (x – 500), and so on. For minimum overall tax outgo of all 15 persons, maximum number of persons should be in the lower slabs. If lowest slab has 5 persons, then second slab cannot have 5 persons because then remaining 5 persons have to be distributed between the remaining two slabs and minimum in each slab is 3 persons. So, second slab would then have 4 persons and remaining slabs would have 3 persons each. So, number of persons in the slabs will be (5, 4, 3,

33. Answer: E Explanation: Since one less orange was required to extract the same quantity of juice as extracted from the apples, the juice extracted from 3 apples should be equal to the juice extracted from 2 oranges. LCM of 3 and 2 = 6. So, 3 apples produce 6 units of juice and 2 oranges also produce 6 units of juice. Therefore, 1 apple produces 2 units of juice and 1 orange 3 units of juice. Therefore, percentage of apple juice in the blend 2 =  100 = 40%. 23

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Chapt er 04

Ratio, Proportion and Mixtures INTRODUCTION Ratio, proportion and variation find application in various domains such as, ‘time, speed and distance’, ‘time and work’ problems as well as ‘geometry’.

Ratio What is a ratio? A ratio is nothing but comparison of two or more quantities expressed in the same units. Therefore, we can compare quantity of milk with water and say that in a vessel, milk and water are in the ratio of 3: 2. This only means that out of total 3 + 2, that is, 5 parts, 3 parts is milk and 2 parts is water. This also 3 2 3 th is milk and th is water. In other words, milk is  100  60% of the total quantity of solution; 5 5 5 2 and water is  100 = 40% of the total quantity of solution. Therefore, ratio can be expressed as a fraction, 5

means that

which can be converted to a percentage. Example 1 Ratio of milk to water in a solution is 2: 5. (i)

What is milk as a percentage of water?

(ii)

What is water as a percentage of milk?

(iii) What is the quantity of milk if water is 30 litres? Solution (i)

2 5

Milk as a percentage of water =  100  40% .

Note that milk as a percentage of water should not be taken as

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2 4  100  28 % . 7 7

Ratio, Proportion and Mixtures

(ii)

Water as a percentage of milk =

5  100  250% . 2

(iii) If water is 5 litres, milk is 2 litres. If water is 30 litres, milk is 12 litres. Example 2 In a mixture of whisky, soda and water, if ratio of whisky to soda is 4: 5 and ratio of soda to water is 6: 7, what is the ratio of whisky to water to soda? Solution W: S = 4: 5 and S: Wt = 6: 7. In the first case, soda is 5 units while in the latter case, it is 6 units. To make them same, multiply the first ratio by 6 and second by 5 and then combine as shown below. W: S = 24: 30 and S: Wt = 30: 35. Therefore, W: S: Wt = 24: 30: 35 and hence W: Wt: S = 24: 35: 30. Example 3 Ratio of incomes of Ram to Shyam is 2: 3 and that of Shyam to Rahim is 4: 7. Rahim’s income is how much percent more than that of Ram? Solution Ram: Shyam = 2: 3 and Shyam : Rahim = 4 : 7. Therefore, Ram : Shyam : Rahim = 8 : 12 : 21. Percentage by which Rahim’s income is more than that of Ram =

21  8  100  162.5%. 8

Alternatively, Percent by which Shyam’s income is more than Ram’s =

32  100  50% . 2

Percent by which Rahim’s income is more than Shyam’s =

74  100  75% . 4

To find the percent by which Rahim’s income is more than Ram’s, we will use successive increments. Therefore, 50 + 75 +

50  75  162.5% . 100

Example 4 Ratio of angles in a triangle is 2 : 3 : 5. Is the triangle acute or obtuse? Solution The angles are 2x, 3x and 5x. Therefore, 2x + 3x + 5x = 180°  x = 18  angles are 36°, 54° and 90°. The triangle is neither acute nor obtuse! It is a right angled triangle.

Proportion When two ratios are equated, we get what is called as a proportion. Therefore, if a : b = c : d, then a, b, c and d are said to be in proportion. This is also written as a : b : : c : d.

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If a : b = c : d, then

a c   ad = bc. b d

This is stated as: product of means = product of extremes. Moreover, d =

bc . Here, d is called fourth proportional to a, b and c. a

More results If

a c  , then b d

(i)

Adding 1 on both sides, we get

a c ab c d  1   1 or  b d b d

This rule is known as rule of componendo. (ii)

Subtracting 1 from both sides, we get

a c ab c  d  1   1 or  b d b d

This rule is known as rule of dividendo. (iii) Combining the above two rules, if we divide the two results obtained, we get ab cd b  d  ab  cd ab cd ab cd b d .

This rule is known as rule of componendo and dividendo. Example 5 If a = 3, b = 7, c = 9 and d = 21, then (i)

are a, b, c and d in proportion?

(ii)

are a, c, b and d in proportion?

(iii) are a, d, c and b in proportion? (iv) to make a, c, d and e in proportion, what should be the value of e? Solution (i)

a 3 c 9 3 a c  and   . Therefore,  and hence they are in proportion. b 7 d 21 7 b d

(ii)

For a, c, b and d to be in proportion,

a b should be equal to which is true because c d

a c a b    b d c d .

(iii)

a c a c  does not imply  . Therefore, a, d, c and b are not in proportion. b d d b

(iv) For a, c, d and e to be in proportion, Therefore,

a d  . c e

3 21   e = 63. 9 e

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Ratio, Proportion and Mixtures Example 6 If

x 3  12x 63 , find the value of x.  6x 2  8 62

Solution Applying Componendo and Dividendo, we get

x x

3

 12x    6x2  8 

3

 12x    6x  8  2



63  62 (x  2)3 125 (x  2)    5x  3.  63  62 (x  2)3 1 (x  2)

Important result If a, b, c, d, e, f, .... are in proportion, then 1

 pan  qcn  ren  .....  n a c e    ……and each of the ratios is also equal to  n  . n n b d f  pb  pd  rf  .... 

where p, q, r, ... and so on are such that not all of them become zero simultaneously. Moreover, n cannot be zero because then the power of the expression, which is

1 , would not be defined! n

When p = q = r = .... = n = 1, then a c e a  c  e  ....    ....  b  d  f  .... b d f

Example 7 If

a c e 2 5 a 3  12c 3  20e 3    , find the value of . b d f 5 5b 3  12d 3  20 f 3

Solution 1

a c e 2  5a3  12c 3  20e 3  3    =   , where n = 3 and p, q and r are 5, 12 and –20 respectively. b d f 5  5b 3  12d 3  20f 3  3

Therefore, the value of

5 a 3  12c 3  20e 3 2 8 is equal to    . 5b 3  12d 3  20 f 3 5 125  

Example 8 If a, b, c, d, e, f, g and h are in proportion, and (i)

4 a  3c  2e  g 4b  3d  2f  h

(ii)

4 a  3c  2e  g b  2d  3f  4h

a  2c  3e  4 g 4  , find the value of b  2d  3f  4h 7

Solution (i)

From the above important result, we can say that

a c e g a  2c  3e  4 g 4a  3c  2e  g      b d f h b  2d  3f  4h 4b  3d  2f  h .

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Therefore, (ii)

4a  3c  2e  g 4  . 4b  3d  2f  h 7

Since the coefficients of a, c, e and g do not match with those of the corresponding terms in the denominator, the value cannot be found.

Example 9 If

p r x 4 p .    , find the value of q s y 5 y

Solution Since only the ratios are equal, the values of terms like p, q, r, ... and so on is not known and hence

p cannot y

be found.

Continued proportion The terms a, b, c, d, e, ... and so on are said to be in continued proportion if

a b c d     ..... b c d e

Since the ratios are continuously equal, the terms are also in geometric progression. This is so because if each of the ratios is equal to

1 , then 

b = a, c = b = a2, d = c = a3, ... and so on. Taking the first equality, we get

a b  or b2 = ac  b = b c

ac .

The term b is called mean proportional to a and c. Here, the sequence does not matter. b2 The term c is called third proportional to a and b. Here, sequence matters because third a b2 a2 whereas third proportional to b and a is . proportional to a and b is a b

Moreover, c =

Example 10 If third proportional to 2 and b is 32, find mean proportional to b and d, where d is third proportional to 7 and e, where e is fourth proportional to 3, 6 and 7 (take all numbers to be positive). Solution 2 b   b = 8. b 32

d=

3 7 e2 , where  . 7 6 e

Therefore, e = 14 and hence, d =

14 2  28 . 7

Therefore, mean proportional to b and d is bd  8  28  4 14 . Example 11 If a, b, c and d are such that they are in proportion, and c is third proportional to a and d whereas b is mean proportional to a and c, find the relationship between a, b, c and d.

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Ratio, Proportion and Mixtures Solution a c d2  and c  and b  ac . b d a

Then,

d2  d2 a c   a   ad = bc  ad = ac  a  a b d

Therefore, a2 = d2  a = d. But c 

 d2 d3   .  a a 

d2  c = d. a

Moreover, b  ac = a. Therefore, a = b = c = d.

Variation Variation deals with the nature of variation of one variable with respect to the nature of variation of the other variable. Let there be two variables, x and y.

Direct variation If x and y are so related that an increase in the value of x leads to corresponding increase in the value of y, and a decrease in the value of x leads to corresponding decrease in the value of y, then we can say that y is directly proportional to x, expressed as y  x. The proportionality sign ‘’ can be converted to equality sign ‘=’ by introducing a constant known as proportionality constant. Therefore, we can also write y  x as y = kx, where k is proportionality constant. For example, circumference of a circle, C = 2r. Since 2 is a constant, we can say that C  r. y = k (a constant), that is, the ratio of y to x is constant. Therefore, if y is directly x y y proportional to x, then 1  2 . x 1 x2

Note that y = kx implies

Similarly, the ratio of the circumference of a circle to its radius is also constant, that is,

C  2 = constant. r

Inverse variation If x and y are so related that an increase in the value of x leads to a decrease in the value of y, and a decrease in the value of x leads to an increase in the value of y, then we can say that y is inversely proportional to x, expressed as y 

1 k , or y = . x x

Note that since y =

k , xy = k (a constant). Therefore, if y is inversely proportional to x, then x1y1 = x2y2. x

For example, density = that is, density 

mass . If mass of a body is constant, then density is inversely proportional to volume, volume

1 . volume

Another common example is the relationship speed = proportional to time, that is, speed 

distance . If distance is constant, then speed is inversely time

1 . time

This relationship (between speed, distance and time) will be found extremely useful when we discuss the concepts of “Time, Speed and Distance”.

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Example 12 Value of a coin is directly proportional to its radius and also to the square of its thickness. If radius doubles, and thickness trebles, then find the ratio of the new value to the old value of the coin. Solution Let V be the value of the coin, d the thickness and r the radius. Then, V  d2r or V = kd2r, where k is the constant of proportionality. V1 = kd12r1. V2 = k(3d1)2(2r1) = 18  kd12r1 = 18V1. Therefore, the ratio is 1:18. Example 13 Pressure of a gas is inversely proportional to the volume and directly proportional to the temperature of the gas. The pressure is 20 psi when volume is 230 units at a temperature of 100K. What will be the pressure when the volume of the gas is 400 units and temperature is 150K? Solution P 

1 PV and P  T. Therefore, combining these two results, we get = constant. V T

Therefore,

P1 V1 P2 V2 20  230 P2  400  or  or P = 17.25 psi. T1 T2 100 150

Example 14 Price of a diamond is directly proportional to the square of its weight. A piece of diamond breaks into two pieces whose ratio of weights is 2 : 3. What is the percentage profit/loss due to breakage? Solution Let the weights of the pieces now be 2x and 3x. Therefore, the weight earlier was 5x. Price, P  w2, where w is the weight. Earlier price, P = kw2 = k(5x)2 = 25kx2. The price now, P’ = k[(2x)2 + (3x)2] = 13kx2. This means that there is a loss of 12kx2. Loss % =

12kx 2  100  48% . 25kx 2

SHORTCUT: Let the weights be 2 gm and 3 gm. Then the earlier weight is 5 gm. The price before breakage = 52 = Rs 25, (assuming that the constant of proportionality is 1. Note that this does not affect the required percentage) The price now = 22 + 32 = 13. Therefore, loss % = 48%. Example 15 In the previous question, what should be the ratio in which the piece of diamond should break so that loss % is the maximum? What is the loss %?

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Ratio, Proportion and Mixtures Solution Let the ratio be a : b. Then the prices earlier and now are (a + b)2 and a2 + b2. Loss = (a + b)2 – (a2 + b2) = 2ab. Note that the weight of the unbroken diamond is constant and that means that a + b is constant. Therefore, 2ab is maximized when a = b. Therefore, the required ratio is 1 : 1. Loss % in such a case =

2ab 2  100  2  100  50% . (a  b)2 2

Example 16 The reduction in the speed of an engine of a goods train is directly proportional to the square of the total number of bogies attached. The speed is 99 km/hr when 4 bogies are attached and the speed is 72 km/hr with 5 bogies attached. What is the maximum number of bogies that can be attached so that the train can just move to reach its destination? Also find the corresponding speed of the train at which it can move with maximum number of bogies attached. Solution Let the speed of the train without any bogie be v0 and the speed with certain number of bogies attached be v. Then, v0 – v  n2, where n is the number of bogies attached. Therefore, v0 – v = kn2. Now, v0 – 99 = k (4)2 = 16k v0 – 72 = k (5)2 = 25k Solving, we get v0 = 147 and k = 3. So, v = v0 – kn2 = 147 – 3n2. For maximum number of bogies, v should be 0. Therefore, 0 = 147 – 3n2 or n =

147  49  7 . 3

With 7 bogies, the speed of the train will be zero, or the train will come to a halt. Therefore, the maximum number of bogies that can be attached is 1 less than 7, that is, 6, so that the train can just move. The speed in such a case will be 147 – 3(6)2 = 39 km/hr. This is the “maximum speed” with which the train will move with 6 bogies.

Partnership In cases of partnerships, 2 or more persons come together and set up a business activity by investing a certain amount of money for a certain period. The persons are called “partners” and the money invested “capital”. The duration and start time of investment may vary from investor to investor. Therefore, A and B may enter into a partnership where A invests Rs 10,000 for a period of 12 months and B joins A with a sum of Rs 15,000 after 3 months and stays with him for the rest of the year. Therefore, duration of investment for B would be 9 months. After a year is complete, the profits earned by the partnership firm is distributed to the investors in proportion to the amount of investments keeping in mind the duration of investment.

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In the above case, the ratio in which A and B would distribute the profits will be (10,000  12): (15,000  9), that is, 8 : 9. So, in case the profit earned by them is Rs 1700, A would get

8  1700  Rs. 800 and B would get 17

9  1700  Rs. 900. 17

Example 17 Let A and B start a partnership firm by investing Rs 24,000 and Rs 36,000 respectively. B withdraws after 4 months. Another partner C joins A 2 months later with Rs 20,000. If B joins back in the 12th month, what is the ratio in which the profit earned in 1 year would be distributed? Solution A’s contribution = 24,000  12 B’s contribution = 36,000  (4 + 1) C’s contribution = 20,000  6. The required ratio = (24  12) : (36  5) : (20  6) = 24 : 15 : 10. Example 18 A, B and C enter into partnership with capitals in the ratio of 2 : 5 : 3. B joins a few months later than A and C joins 2 months later than B and withdraws from the business some time before the year ends. If the ratio in which profits are distributed is 8 : 10 : 3, (i)

how many months later does B join, and

(ii)

how many months earlier does C withdraw

Solution (i)

Let B join x months later and C withdraw y months earlier. Amounts of investment are 2r, 5r and 3r.

A’s contribution = 2r  12 = 24r B’s contribution = 5r  (12 – x) C’s contribution = 3r  (12 – x – 2 – y) = 3r  (10 – x – y) Since the ratio of profits for A and B is 8 : 10, therefore, 24r : 5r(12 – x) = 8 : 10 or or

24 8 or x = 6.  5(12  x) 10

Therefore, B invested 6 months after A started. (ii)

Similarly,

24r 8 24r 8     y = 1. 3r(10  x  y ) 3 3r(10  6  y ) 3

Therefore, C withdraws 1 month before the end of the year, that is, at the end of 11 months.

Averages Average of some given terms is nothing but a single value which represents the entire set of terms and which lies somewhere in the middle of the given terms. So, if some terms are given as a1, a2, a3, ....., an, then simple average of all these terms is given as below:

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Ratio, Proportion and Mixtures

Average =

a1  a2  a3  ....  an n

where n is the number of terms. Sometimes, the need for such a central value, also called a ‘measure of central tendency’, arises for comparison between 2 or more sets of values. For example, if we want to compare two sections of 10th standard in a school with respect to their heights and the data provided to us is as follows: Class A has 10 students, the ages of those students is 14, 15, 15, 13, 14, 16, 14.5, 15.5, 13.5, 13 Class B has 12 students, the ages of those students is 15, 16, 14, 14, 15, 15.5, 14.5, 14.5, 13, 13.5, 16, 15. In order to compare the ages of the two classes, a central value is needed which represents the ages of all the students in each class. Now, it is quite possible that that central value found may not be equal to any of the values given in that set. That’s okay as long as it is the representative value. The average age of students in class A =

14  15  15  13  14  16  14.5  15.5  13.5  13 143.5   14.35. 10 10

As we have mentioned, the average age (14.35 years) is not equal to any of the values given. The average age of students in class B =

15  16  14  14  14  15  15.5  14.5  14.5  16  15 176   14.67. 12 12

We can say from the above that the average age of the students in class A is less than the average age of the students in class B. Moreover, if some students get transferred from A to B and some from B to A, the agecomposition of the class will get disturbed. So, if we are given a set of persons, then some feature of that set, for example, age, height, weight, and so on are the observed values whence we can find the corresponding average value which represents that particular feature in a concise manner. Census of population which is done every decade in India contains huge amounts of data regarding people, which shows the demographics. But, every minute detail of every Indian would be of no use to the Prime Minister of India, who has to make some policy decisions based on this. For example: (i)

Average height of all the people of India is of much more value than the individual heights of all the persons.

(ii)

Per capita income is defined as: Total income of all the persons, which is same as GDP of India, divided by the size of population. This per capita income is of much greater value to the policy makers than the individual incomes of the people (which is of more value to Income Tax authorities).

(iii) The average age of the entire population of India gives us a rough idea of relative distribution of people of different ages. For example., if average age of the population in India is 55 years and the same for China is 61 years, then we can say to some degree of accuracy that ‘more’ number of people in China are older than the people in India. (iv) The average calorie intake per capita is also a useful parameter in making policy decisions. Having learnt the importance of ‘average values’, it is essential for a student to understand that the average or mean is of three basic types: Arithmetic Mean, Geometric Mean and Harmonic Mean. A type of Arithmetic

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Quantitative Aptitude Simplified for CAT

Mean is Weighted Arithmetic Mean, which will also be discussed. The type of mean to be used varies from case to case. Let us understand these cases.

Arithmetic Mean (A.M.) If there are n terms, a1, a2, a3, a4, ...., an, then, the arithmetic mean of these n terms is defined as A.M. =

a 1  a 2  a 3 ....  an n

Now, this is used to find the average of values like age, height, weight and so on. While finding out ‘per capita income’ also we can use A.M. Use of A.M. in finding average speed is given in the chapter on “Time, Speed and Distance”. Example 19 Find the average height (in cm) of a class of 15 students: 130, 135, 134, 145, 150, 132, 144, 146, 141, 152, 151, 157, 139, 140, 144. Solution Sum of all the heights = 2140. Average height =

2140 2  142 cm. 15 3

Example 20 Let the average weight of 30 residents in a residential area be 56.5 kg. (i)

If 3 residents whose weights are 40 kg, 46 kg and 75 kg leave the area, what is the average weight of the remaining residents?

(ii)

If the 3 residents (in (i)) are replaced by 3 new persons whose average weight is 53

2 kg, what is the 3

average weight now? Solution (i)

Total weight of 30 residents = 30  56.5 = 1695 kg.

Weight of 27 residents (after 3 residents leave) = 1695 – (40 + 46 + 75) = 1534. The average of these 27 residents = 56.81 Important observation: The total weight of residents leaving the area is (40 + 46 + 75) = 161 kg and hence the average of these 3 persons = 53.67 kg. Reduction of three persons of weights 40, 46 and 75 is equivalent to going out of three persons each of whose weight is 53.67 kg (this is precisely the meaning of averages!) which means that persons with weights lower than the average weight are leaving which means that the average weight of the remaining people would have to be higher than the given average of 56.5 kg. Similarly, if there were three persons of the same weights, that is, 40, 46 and 75 who joined the residential area of 30 persons (leading up to 33 persons in all), this would have meant that persons of with weights lower than the average weight are being added to the existing group of 30 persons, in which case, the average weight of these 33 persons would be lower than the given average of 56.5 kg. We can verify this. The new average = (ii)

1695  161  56.24 kg. 33

After the said replacement, the average weight

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Ratio, Proportion and Mixtures

=

1695  (40  46  75)  (3  53.67) 1695   56.5 kg. 30 30

The above “important observation” clearly explains why the average weight here remains the same! Also note that in total there are three cases possible: (i)

some person(s) leave a group

(ii)

some person(s) join a group

(iii) some person(s) is/are replaced by same number of person(s). All this leads to change in the average figure. We can develop the below mentioned formula. Change in the average figure =

Difference between average and new figure . New number of members

EXPERT SPEAK Scan this QR Code to watch a video that explains how to approach problems based on Averages.

Example 21 If average age of a group of 9 persons is 18 years and one of the persons whose age is 22 years leaves the group, what is the new average age? Solution Here, difference between average and new figure = 22 – 18 = 4. Therefore, change in average figure =

4 8

= 0.5 years. (we have taken 8 because new number of members is now 8). Since the age of the person leaving the group is more than that of the average, the average age of the remaining persons will decrease. So, new average age = 18 – 0.5 = 17.5 years. Using the formula also, we get the same answer:

9  18  22 = 17.5 years. 8

Example 22 The temperatures in a week from Monday to Sunday are recorded and it is given that the average temperature in the first four days is 34.3°C and that in the last four days is 41.25°C. If the temperature on Thursday is 40°, what is the average temperature of the week? Solution T (Mon) represents “temperature on Monday” and so on. Therefore, T (Mon) + T (Tue) + T (Wed) + T (Thurs) = 34.3  4 = 137.2°C T (Thurs) + T (Fri) + T (Sat) + T (Sun) = 41.25  4 = 165°C Adding the two equations, [T (Mon) + T (Tue) + T (Wed) + T (Thurs)] + [T (Thurs) + T (Fri) + T (Sat) + T (Sun)] = 137.2 + 165 = 302.2

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Quantitative Aptitude Simplified for CAT

 T (Mon to Sun) + T (Thurs) = 302.2  T (Mon to Sun) + 40 = 302.2  T (Mon to Sun) = 302.2 – 40 = 262.2  Average Temperature from Monday to Sunday =

262.2  37.46 oC . 7

Example 23 Rajesh found the average of a few terms to be 35 but misread 14 as 41 and hence his average increased by 0.9 units. Find the number of terms which he added. Solution Let there be n terms. Then, sum of terms which he got = 35n. The correct sum of terms = 35n – 41 + 14 = 35n – 27 The correct average =

35n  27 . n

Since the average increased by 0.9 because of misreading by Rajesh, 27  35n  27    . n n n  

 

Alternatively, change in average = 0.9 =

41  14 or n = 30. n

Example 24 Prove that the average of n terms which are in Arithmetic Progression and whose first and last term is a1 and an, is given by

a 1  an n1 . Also, prove that the average of first n natural numbers is . 2 2

Solution If n terms are in AP, whose first and last term is a1 and an, then sum of terms =

n  a1  an  . 2

n   2  a1  an   a  a n Therefore, the average of those n terms =  1 . n 2

In first n natural numbers, first term is 1 and last term is n. Therefore, average of first n natural numbers =

a 1  an 1  n  . Hence proved. 2 2

Example 25 Average age of class A having 30 students is 20 years and that of class B having 40 students is 23 years. (i)

Find the average age of both the classes combined.

(ii)

How many minimum number of students each of age 30 years should be included so that the average is not less than 25?

Solution (i)

Total age of all the students in class A = 30  20 = 600 years.

Total age of all the students in class B = 40  23 = 920 years.

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Ratio, Proportion and Mixtures Total age of combined class = 600 + 920 = 1520 years. Average age of the combined class = (ii)

1520  21.71 years. 70

Let the minimum number of students each of age 30 years be n.

The new average after the inclusion of n students =

1520  30n  25 70  n

 1520 + 30n  1750 + 25n  5n  230  n  46 Therefore, minimum number of students each of age 30, to be included is 46. Example 26 There are 5 balls of weights w1, w2, ...., w5. The balls can be paired in 10 ways. The average weights of such pairs are 45, 46, 49, 50, 51, 51, 53, 55.5, 60 and 61 kg. Find the average weight of all the 5 weights. Solution w1  w2 w  w3 = 45  w1 + w2 = 90 kg; 1 = 46  w1 + w3 = 92 kg; and so on. 2 2

Adding all these 10 equations, we observe that w1 appears 4 times, w2 appears 4 times, w3 appears 4 times and so on. Therefore, 4(w1 + w2 + w3 + w4 + w5) = 2(45 + 46 + 49 + 50 + 51 + 51 + 53 + 55.5 + 60 + 61)  (w1 + w2 + w3 + w4 + w5) = (45 + 46 + 49 + 50 + 51 + 51 + 53 + 55.5 + 60 + 61)/2 = 260.75 kg. Therefore, the average weight of all the 5 balls = 260.75/5 = 52.15 kg.

Weighted arithmetic mean Weighted Arithmetic Mean is given by n

w1 a1  w2 a2  ....  wn an  w1  w2  ....  w n

w a

i i

i 1 n

w

i

i 1

where a1, a2, ... are the values of the observation and w1, w2, ..... are the weights attached. Let us see the following example. Example 27 Let there be three classes, A, B and C which have 20, 25 and 40 students and the average weights of the students in these classes (respectively) are 51 kg, 54 kg and 49.5 kg. If all the three classes are combined to form one class of 85 students, what is the average weight of the combined class? Solution Total weight of all the students together = (20  51) + (25  54) + (40  49.5) = 4350 kg. The average weight =

4350  51.17 kg. 85

As we can see in the above example that the number of students in each class behaves like the weights attached to the values of the observation (which is the weight of the students). Using the formula of weighted arithmetic mean, we get

157

Quantitative Aptitude Simplified for CAT (20  51)  (25  54)  (40  49.5) 4350   51.17 kg 85 85

We can say that “arithmetic mean” is a specific case of “weighted arithmetic mean”, where every weight attached is of value 1. As you will see later, weighted arithmetic mean forms the basis of the concept of Alligation. Example 28 Mahesh travels 3 hours by bus which moves at a speed of 40 km/hr, another 2 hours by train which moves at a speed of 60 km/hr and finally 4 hours by ship which moves at a speed of 45 km/hr. What is the average speed of Mahesh? Solution Average Speed =

Total Distance Total Time

Therefore, Average Speed =

(3  40)  (2  60)  (4  45) 420   46.67 km/hr. 324 9

(Here, the time of travel acts as the weight attached) Example 29 In the previous example, if the time taken in every part of the travel be 1 hour, then what will be the average speed? Solution Average Speed would then be =

(1  40)  (1  60)  (1  45) 145   48.33 km/hr. 1 1 1 3

The last example is just to show that if weights attached become 1 each, the formula for weighted arithmetic mean would reduce to simple arithmetic mean. Example 30 Average age of a group of students is 20 years and that of another group of students is 30 years, whereas the average age of all the students together is 23 years. What is the ratio of number of students in the two groups? Solution Let there be m students in first group and n students in second group. Then, 20m  30n = 23  20m + 30n = 23m + 23n  3m = 7n  m : n = 7 : 3. m n

Geometric mean If there are n terms, a1, a2, a3, ...., an, then the geometric mean of these n terms is defined as G.M. =

n

a1 a2 a3 ....an

So, G.M. of 2 and 8 is

2

3 1  2 3  3 36  9 . 2  8  4 and that of 3, 9 and 27 is 3 3  9  27  3

Geometric Mean is extremely useful in finding the average growth rates. Let a quantity grow by 10% in the first year, by 20% in the next year and by 30% in the third year. What is the average growth rate?

158

Ratio, Proportion and Mixtures  

If x is the original quantity, then the quantity after 1 year becomes  1 

10   x , that is, 1.1x. Similarly, the 100 

quantity after second year becomes (1.1)(1.2)x and finally, the quantity after 3rd year becomes (1.1)(1.2)(1.3)x. If r is the average growth rate, then it means that if the quantity x had grown by a constant rate of growth in these three years, the output would have been same. Therefore, x  1  

3

r   = (1.1)(1.2)(1.3)x. 100 

 Let  1  

r   = R. Then, 100 

R3 = (1.1) (1.2)(1.3) R = 3 (1.1)(1.2)(1.3)  1.1972  r = 19.72%. Therefore, if the original quantity had grown by 19.72% every year consistently for 3 years, the quantity would have become 1.716 times, which is the same result when the same quantity grew by 10%, then 20% and then 30% in the respective years. Mathematically, we observe that (i)

R is the geometric mean.

(ii)

The geometric mean of 1.1, 1.2 and 1.3 is taken, and not of 10%, 20% and 30%, in finding the average growth rate.

Geometric mean of only positive terms can be taken as the nth root of negative term is not real, if n is even. Therefore, in general, the terms should all be positive. But, the application of geometric mean extends to negative growth rates. Example 31 If a price index grows by 25%, –13%, –20%, 30% and 12% respectively in 5 successive years, what is the average growth rate in the index? Solution Taking cue from the above example, R = 3 (1.25)(0.87)(0.8)(1.3)(1.12)  1.048  r = 4.8%. So, a uniform rate of 4.8% every year leads to the same result. Students must note that “average growth rate” calculated above is nothing but “compounded annual growth rate”, CAGR. Refer to CAGR concept introduced earlier for a complete understanding of the geometric mean. Example 32 If the price index decreases by 10%, 20% and then 30% in respective years, then what is the average decrease rate? Solution R=

3

(0.9)(0.8)(0.7) = 0.7958

average decrease rate = (1 – 0.7958) × 100 = 20.42%. Example 33 Find the geometric mean of 1, 2, 22, 23, 24, ..... 2n–1.

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Quantitative Aptitude Simplified for CAT

Solution Geometric Mean =

n

 (n1)n  1  2  n

2123....(n1)  2 

2

n1 . 2

.

Weighted geometric mean If G1 is the geometric mean of n1 terms and G2 is the geometric mean of n2 terms, then Product of n1 terms = G1n and that of n2 terms = G2n . 1

2

Hence, product of all the (n1 + n2) terms =  G2 n

1

 G  n2

2

.

Combined geometric mean, G of all the terms (n1 + n2) =  n1 n2  G1 n1 G2n2 . Taking logarithms on both sides, log G =

n1 log G1  n2 log G2 n1  n2

Example 34 If geometric mean of first 20 terms is 10 and next 10 terms is 80, find the geometric mean of all the 30 terms. Solution G = 30 (20)20 (80)10  30 (10)30 (2)30  20 .

Harmonic mean If there are n terms, a1, a2, a3, ...., an, then the harmonic mean of these n terms is defined as H.M. =

1 1 1 1 1 1    ...    n  a1 a2 a3 an 

So, H.M. of 10, 20 and 40 is



n  1 1 1 1     ...    an   a1 a2 a3

.

3 120 1   17 . 1 1 1 7 7   10 20 40

Example 35 Let there be two quantities one of which is 10. The other quantity is such that the harmonic mean of the two quantities is 20. Find the other quantity. Solution If the other quantity is x, then HM = 20 =

2(10)x  10 + x = x  10 = 0, which is absurd. 10  x

Therefore, such a case is impossible. Let us prove it. If the two terms are a and b and their harmonic mean is h, then h=

2 . If b approaches , then h approaches 2a. Therefore, h  2a is the limiting value of h in case b 1 1  a b

approaches the largest value, .

160

Ratio, Proportion and Mixtures Example 36 There are three varieties of oranges priced at 5 for a rupee, 6 for a rupee and 10 for a rupee. If equal number of each variety is sold, what is the average price realized per rupee? Solution Let the number of each variety sold be x. Then, total of 3x is sold. x 5

Price realized from first variety =

Price realized from second variety = Price realized from third variety =

x 6

x . 10

Therefore, average price realized per rupee = =

Total number of organges sold Price of all the oranges

3x 3  x x x 1 1 1     5 6 10 5 6 10

(which is the harmonic mean of “quantity per rupee”) =

45 3  6 number of oranges per rupee. 7 7

From the above example, we realize that the harmonic mean can be applied to cases where the quantities involved under question are equal, for example, equal number of oranges. Harmonic Mean is also useful in finding average speed in certain cases which is mentioned in the chapter of “Time, Speed and Distance”.

Weighted harmonic mean If H1 is the harmonic mean of n1 terms and H2 is the harmonic mean of n2 terms, then Sum of reciprocal of n1 terms =

1 1 n   ....n1 terms = 1 . a1 b1 H1

Sum of reciprocal of n2 terms =

n 1 1   ....n2 terms = 2 . . a2 b2 H2

Sum of reciprocal of n1 + n2 terms  1 1   1 1  n n n  n2   ....n1 terms      ....n2 terms   1  2  1 a b a b H H H  1 1   2 2  2 2

= 

 the combined harmonic mean, H of all the terms (n1 + n2) is given by H=

n1  n2 n1 n1  H1 H2

This is used in cases where unequal amount of the quantities is involved. Example 37 Using the data of the previous example, if 5 kg of first variety, 10 kg of second variety and 15 kg of the third variety is sold, what is the average number of oranges sold per rupee?

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Quantitative Aptitude Simplified for CAT

Solution Using the formula for weighted harmonic mean, we get H=

5  10  15  7.2. 5 10 15   5 6 10

Therefore, 7.2 oranges per rupee are sold.

Relationship between A.M., G.M. and H.M. Let us first take the case of two terms, a and b. AM of these two terms, A =

ab . 2

GM of these same terms, G = HM of these terms, H =

ab .

2 2ab  1 1 ab  a b

 a  b   2ab 

From this, we can conclusively say that AH =    = ab = G2.  2   a b Therefore, G2 = AH is a standard result. In other words, Geometric Mean of two terms a and b is the Geometric Mean of Arithmetic Mean and Harmonic Mean of the same two terms a and b. But this is true only in case of two terms a and b. If we have n terms, then this is not true. Example 38 Let AM of two terms be 5 and GM of the same two terms be 4.5. What is the HM of those two terms? Also find the two terms and confirm the answer. Solution Using the result G2 = AH, we get 4.52 = 5H  H = 4.05. Let the two terms be a and b. Then,  a + b = 10

(i)

and

(ii)

ab  4.5  ab = 20.25

Eliminating b, we get a(10 – a) = 20.25  a2 – 10a + 20.25 = 0 a= If a =

10  100  81 10  19  . 2 2 10  19 10  19 , then b = 10 – a = . 2 2

Therefore, the two terms are

10  19 10  19 and . 2 2

162

Ratio, Proportion and Mixtures  10  19   10  19  2     2 2 2 ab    100  19  81  4.05 . HM of these two terms =   a  b  10  19   10  19  20 20      2 2    

While solving such questions, we have to be careful as to the correctness of the data. This is so because of the following inequality regarding AM, GM and HM. AM GM  HM This is the an extremely useful inequality (which helps us solve lots of questions on “maxima and minima” as well). Moreover, this is not only true for 2 terms a and b, but also for any number of terms. Therefore, for n terms, a1, a2, a3, ....., an, a1  a2  a3  ....  an n n  a1 a2 a3 ...an  1 1 1 1 n    ...  a1 a2 a3 an

.

Irrespective of the number of terms, all the terms must necessarily be positive. If any term is negative, then GM is not defined and if any term is zero, then HM is not defined. Example 39 Prove that AM, GM and HM are equal to each other only when all the n terms are equal. Solution If all the terms are equal, then a1 = a2 = a3 = ..... = an = a, say. Then, AM of n terms =

a1  a2  a3  ....  an a  a  a  ...n time an =   a. . n n n

Similarly, GM of n terms HM of n terms =

n

a1 a2 a3 ....an  n a.a.a.a......n time  n an  a .

n n n   a .  1 1 1 1  1  1  1  ....n time (n/a)    ....    a a a an   a1 a2 a3

Therefore, when all the n terms are equal, then AM = GM = HM = a (that is, the term itself) This is a standard result (which will be found extremely useful when we learn concepts of Maxima and Minima) Moreover, it is not possible that “AM and GM are equal to each other but not equal to HM”. This means that “either all the three are equal” or “all the three are unequal”. Conclusion: (i)

If all the terms are equal, then AM = GM = HM = the term itself.

(ii)

If all the terms are unequal (some of the terms may be equal, but not all), then AM > GM > HM.

Example 40 If x = 23  25  27  29  31  33 and y = 286, then which of the two is bigger? Solution Let there be 6 terms: 23, 25, 27, 29, 31 and 33. AM of these 6 terms =

23  25  27  29  31  33  28 6

GM of these 6 terms =

6

23  25  27  29  31  33 .

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Quantitative Aptitude Simplified for CAT

Since AM is more than GM (if terms are unequal), 28 > 6 23  25  27  29  31  33 . 286 > 23  25  27  29  31  33 Hence y is bigger than x.

Mixtures and Solutions When two or more types of liquids are mixed with each other in a vessel, we get what is called as a “mixture”. For example, milk and water can be mixed to get a mixture. Mixtures are identified by the name of the liquids and their ratio. So, we can have a mixture in which milk and water are in the ratio of 4 : 7. This means, as we have discussed earlier, 4 parts is milk and 7 parts is water. Mixture can be taken to be same as solution. Note that when we mix two or more liquids, there may be some chemical reaction between the liquids. But, for our current discussion, we will assume that no such reaction takes place. So, when some acid (say, HCL) is mixed with water, some chemical reaction actually takes place, but these things would not be an area of concern in the current study. In mixtures and solutions, we deal with two concepts: (a)

Alligation, and

(b)

Replacements

Let us discuss them in greater detail.

Alligation When we mix two or more solutions, we get another solution in which the constituents are in a new ratio. Let us take an example. Let us say we have two acid solutions, A and B, in which concentration of acid is 10% and 20% respectively. If we mix 2 litres of A with 6 litres of B, the concentration of the resultant solution is obtained thus: Fraction of acid in the resultant solution =

Total acid (10% of 2 lt)  (20% of 6 lt) 0.2  1.2 1.4     0.175 , Volume of solution 2  6 8 8

This means that the resultant solution contains 17.5% acid. Similarly, if we are given that the two solutions, A and B contain 10% and 20% acid concentration, and that upon mixing them, the resultant solution has 17.5% concentration, what is the ratio in which A and B were mixed? Let the ratio be m : n. Then, we can assume that m litres of 10% conc. and n litres of 20% conc. are mixed to get m + n litres solution. Therefore, 10m  20n  17.5 mn

 10m + 20n = 17.5m + 17.5n  7.5m = 2.5n or m : n = 2.5 : 7.5 = 1 : 3. This means that if A and B are mixed in the ratio of 1 : 3, the resultant solution will always contain 17.5% concentration. This also means that if 1 litres of A is mixed with 3 litres of B, or 2 litres of A with 6 litres of B, or even 30 litres of A with 90 litres of B, we will get the same concentration, 17.5%.

164

Ratio, Proportion and Mixtures Observations: (a)

If the ‘relative’ amounts of the two solutions are changed, the concentration of the resultant solution will also change.

(b)

We can also conjecture that the resulting concentration will always lie between 10% and 20% and by no means can exceed this limit. That is, the resulting concentration can never be less than 10% or more than 20%, no matter what the relative amounts are.

A certain resultant concentration is obtained by a fixed “relative amount” of solutions A and B. This “relative amount” can also be obtained using the Rule of Alligation. Let us find the ratio in which A and B are mixed, using the rule of Alligation. According to the rule of Alligation, the concentrations of A, B and the resultant are written in the form as shown below:

The values 2.5 and 7.5 are obtained by subtracting “17.5 from 20” and “10 from 17.5” resp. Taking their ratios, we get the required ratio as 1 : 3. The rule of Alligation allows us to visually place the concentration values in the diagram and quickly obtain the ratio of volumes in which the solutions were mixed. Reason: The logic of the rule of Alligation is not difficult to find. Let us use the same example. We have seen that if two solutions having 10% and 20% concentration are mixed to obtain 17.5% resultant concentration, we can find the ratio of mixing by the following equation 10m  20n  17.5 mn

Upon rearranging all the terms, we note 10m + 20n = 17.5m + 17.5n, or m(17.5 – 10) = n(20 – 17.5), or m : n = (20 – 17.5) : (17.5 – 10)

(i)

which implies that the ratio is 2.5 : 7.5 or 1 : 3. Compare equation (i) with the above diagram. It is now easy to understand the “genesis” of the rule. The larger question is to identify when to use the rule of Alligation. First of all, from whatever has been discussed so far, we can say that the rule of Alligation is nothing but another way of looking at the weighted arithmetic mean, where the ratio of the weights attached, like m or n (in the examples above), is obtained as the result of Alligation. It is important to understand that rule of Alligation is useful to find the ratio in which the solutions were mixed, whereas weighted arithmetic mean formula is useful when the ratio is given but the concentration of the resultant solution is asked. But, we need to also understand that when we apply the rule of Alligation, we get the ratio of which quantities. Following example will help clarify this.

165

Quantitative Aptitude Simplified for CAT

EXPERT SPEAK Scan this QR Code to watch a video that explains how and where to use the technique called 'Alligation'.

Example 41 Arjun goes from Delhi to Agra at a constant speed of 60 km/hr and Agra to Jaipur at a constant speed of 80 km/hr. The overall average speed of Arjun in going from Delhi to Jaipur via Agra comes out to be 65 km/hr. (i)

What is the ratio of times taken to travel these respective distances

(ii)

What is the ratio of distances “Delhi-Agra” to “Agra-Jaipur”.

Solution We can use Alligation to find the ratio, but is this ratio the ratio of “distances” or “time”? (i) 65 =

Let t1 and t2 be the times taken in the respective distances. Then,

60t1  80t2 t1  t2

We observe that this formula is similar to the weighted arithmetic mean formula, which has been shown to be the foundation of the rule of Alligation. We also observe that the weights attached are the times, t1 and t2. Therefore, the ratio which we would obtain using the rule of Alligation will be the ratio of times. Hence,

or the ratio of times is 3 : 1. (ii)

From the ratio of times, we can easily find the ratio of distances because distance = speed  time. Therefore, ratio of distances = s1t1 : s2t2 = 60  3 : 80  5 = 9 : 20.

This example clearly explains where to use Alligation. Therefore, one should use Alligation where one would otherwise have used the weighted arithmetic mean. Example 42 Two mixtures of cement and sand having cement and sand in the ratio of 3 : 2 and 7 : 3 are available. We want to get cement and sand in the ratio of 2 : 1. What is the ratio in which the available mixtures be mixed to get the desired ratio? Solution To get the desired ratio of cement and sand, the mixing ratio can be found using the rule of Alligation.

166

Ratio, Proportion and Mixtures

This implies the required ratio is 1 : 2. Example 43 If two varieties of sugar costing Rs 8 per kg and Rs 13 per kg are mixed in the ratio of 4 : 5, what is the cost price of the resultant variety of sugar? Solution Here the ratio is given and the price of the resultant is asked. We should use the weighted average formula: 8  4  13  5 7  10 45 9

The cost price of the resultant variety is Rs. 10

7 9

Example 44 Two varieties of rice costing Rs 10 per kg and Rs 18 per kg are mixed and sold at Rs 20 per kg thereby gaining 66%. What is the ratio in which the two varieties were mixed? Solution Since selling price of the mixture is Rs 20 per kg and profit is 66

2 %, cost price of the mixture is 3

20  Rs. 12 . 66.66   1   100  

Using Alligation,

The required ratio is 3 : 1. Example 45 There are two mixtures of whisky in soda, A and B, where A contains 15% whisky. When A and B are mixed in a certain ratio, the resultant contains 20% whisky. Now, if we mix this 20% whisky with mixture B in the same ratio as A and B were mixed, the resultant contains 23

1 % whisky. What is the ratio in which A and B are mixed 3

and what is the strength of whisky in B? Solution Let the strength of whisky in B be x%. Then,

167

Quantitative Aptitude Simplified for CAT

Now, 20% is mixed with x% in the ratio of (x – 20) : 5 to get 23

1 % whisky. Applying weighted arithmetic mean 3

formula, 20(x-20)  5x 1  23  x  30% . (x-20)  5 3

Hence B contains 30% whisky and the ratio in which A and B are mixed is 2 : 1. Example 46 What is the ratio in which two milk water solutions having milk and water in the ratio of 2 : 3 and 7 : 3 respectively, should be mixed to get a milk water solution having milk and water in the ratio of 4 : 1? Solution 4  100  80% which is more than the 5 2 7 percentage of any of the mixing solutions, which are  100  40% and  100  70% This means that 40% 5 10

This is a tricky question in that the resultant solution has the milk % as

solution cannot be mixed with 70% solution in any manner to get 80% solution! Example 47 To 30% acid solution whose volume is 30 litres, how much water must be added so that the solution gets diluted by (i)

5 percent

(ii)

5 percentage points

Solution (i)

When 5 percent is mentioned, it means 5% of 30%, which is 1.5%. Therefore, the resultant solution contains 30 – 1.5 = 28.5% acid. Moreover, it is important to note that pure water contains no acid. So, it is taken to contain 0% acid. Using alligation,

The ratio is 57 : 3 or 19 : 1. This means that 1 litre of water must be added to 19 litres of solution to get the desired result. Therefore,

30 11 1  1 litres of water must be added to 30 litres of solution to dilute the given solution by 5 19 30

percent.

168

Ratio, Proportion and Mixtures (ii) When 5 percentage points is mentioned, it simply means that acid strength decreases from 30% to 25%. Proceeding as usual,

Therefore, the ratio is 5 : 1. So, to 30 litres of solution, we must add

30  1  6 litres of water to dilute the given 5

solution by 5 percentage points. Example 48 Three jars of 1 litre each contain milk and water in the ratio of 2 : 3, 4 : 7 and 5 : 3. All the contents of the three jars are poured into a fourth jar of 3 litres. What is the ratio of water to milk in this big jar? Solution 2 3 4 litre milk and litre water. Similarly, 1 litre of second and third jar contains 5 5 11 4 3 litre milk and lt water; and litre milk and litre water respectively. 11 8

1 litre of first jar contains

2 4 5 176  160  275   611 440  Milk-water ratio = 5 11 8  . 3 7 3 264  280  165 709   5 11 8 440

But the question is to find the water : milk ratio, which is 709 : 611. Example 49 If three varieties of rice costing Rs 8 per kg, Rs 13 per kg and Rs 20 per kg are mixed to get Rs.18 per kg variety, what should be the ratio of mixing them? Solution We cannot get Rs 18 per kg variety by mixing Rs 8 and Rs 13 per kg varieties, but we can get that by mixing either Rs 8 with Rs 20 per kg variety or Rs 13 with Rs 20 per kg variety. Let us mix them separately and find the ratios using the rule of Alligation.

The above diagrams suggest that Rs 8 be mixed with Rs 20 in the ratio of 1 : 5 to get Rs 18 variety. Similarly, Rs 13 be mixed with Rs 20 in the ratio of 2 : 5 to get Rs 18 variety. If we mix these new mixtures each of which is Rs 18 per kg variety, the result will obviously be Rs 18 per kg variety. Therefore, “1 kg of Rs 8 mixed with 5 kg of Rs 20” be mixed with “2 kg of Rs 13 mixed with 5 kg of Rs 20” to get a mixture which contains 1 kg of Rs 8, 2 kg of Rs 13 and 5 + 5 = 10 kg of Rs 20. Therefore, to get Rs 18 variety, we can mix the three given varieties in the ratio of 1 : 2 : 10.

169

Quantitative Aptitude Simplified for CAT

It is important to realize that this ratio is not a unique ratio. This means that there are many other ratios possible in which the three given varieties can be mixed to get Rs 18 per kg variety rice. For example, the ratio of 1 : 5 (first diagram) can also be written as 3 : 15. Therefore, 3 kg of Rs 8 can be mixed with 15 kg of Rs 20 to get Rs 18 variety. Therefore, the other ratio can be 3 : 2 : (5 + 15), that is, 3 : 2 : 20. As a general result, we can say that there are infinite ratios possible. But, in the CAT exam, there will be 4 options to choose from. The best strategy in such a case would be to use the options and apply weighted average formula to confirm which option is the correct option.

Replacements In cases of replacements, we deal with instances where, let us say, we have a cask full of wine and some portion of it is taken out and replaced with same amount of water (or for that matter any other liquid). The solution now obtained is a mixture of wine and water. Some portion of this solution is again taken out and replaced with same amount of water. This process is done n number of times. After all such replacements, we are required to find the ratio of wine to water, or the fraction of wine. To solve such cases, we should use the following formula, which will be found very helpful: b  F  f 1  a 

n

where, a = initial amount of wine in the cask; b = amount of wine taken out from the cask; f = initial fraction of wine in the cask; F = final fraction of wine in the cask; n = number of times the replacement is carried out. Note that f and F represent the fraction of liquid which is getting diluted due to replacement cycle. Here, since we are replacing wine with water, wine gets diluted and so f and F represent initial and final fraction of wine. Let us understand the use of this formula with the help of the following example. Example 50 Let us say we have a 40 litres cask containing milk and water in the ratio of 4 : 9. 5 litres of milk-water solution is taken out and replaced with 5 litres of water. This process is repeated. What is the final milk water ratio in the cask? Solution Here, a = 40 lt; b = 5 lt. Since initial milk water ratio is 4 : 9, we can say that the initial fraction of milk is . Since the process is done twice, n = 2. Using the above formula, we can find out the final fraction of milk: n

2

b 4  5  4 49 49  . F  f 1      1    a 13  40  13 64 208 

This means that the ratio of milk to water after the replacement = 49 : (208 – 49) = 49 : 159. Note that the given formula can be used only if in all the replacements, same amount ‘b’ is taken out and replaced with. If in different cycles, different amounts are taken out (but replaced with that amount only), then we use the more general formula:

170

Ratio, Proportion and Mixtures b   F  f 1  1  a 

b2   1   a 

b3    1  a  ......  

where b1 is the amount taken out and replaced with in the first stage, b2 is the amount of replacement in the second stage, and so on. This is so because the fraction of milk at the end of stage 1 of replacement, that is b   f  1  1  becomes the initial fraction for stage 2 of replacement. a 

Let us take an example. Example 51 We have a 40 litres cask containing milk and water in the ratio of 4 : 9. 5 litres of this solution is taken out and replaced with 5 litres of water. Now, 8 litres of the solution is taken out and replaced with water. Finally, 10 litres of solution is taken out and replaced with water. What is the final amount of milk in the cask? Solution Using the above formula, b  b  b  4  5  8  10   F  f 1  1  1  2  1  3   1  1   1   a a a  13  40   40   40  

=

4  7   4   3  21      13  8   5   4  130

The amount of milk =

21 6  40  6 litres. 130 13

EXPERT SPEAK Scan this QR Code to watch a video that explains the formula and applications of the concept of Replacement.

Example 52 We have 40 litres of 2 : 3 milk water solution. (i)

5 litres are taken out and replaced with 5 litres of water. 5 litres are again taken out and replaced with 5 litres of milk. What is the milk water ratio now?

(ii)

5 litres of water is added to the solution and then 5 litres of the solution is taken out. This process is repeated once more. What is the milk water ratio now?

(iii) 5 litres are taken out and replaced with 5 litres of milk. This process is repeated till the percentage of milk in the solution exceeds 60%. How many times should the replacement cycle be repeated for this to happen? Solution (i)

In this, there are two stages of replacement. In the first stage, replacement is done with water whereas in the second stage, the replacement is done with milk. Here, it is important to mention that when replacement is done with water, f and F in original formula represent the ‘initial’ and ‘final’ fraction of milk, whereas if replacement is done with milk, the same variables f and F would represent ‘initial’ and ‘final’ fraction of water (and not milk). In short, f and F represent the fraction of liquid which is getting diluted due to replacement cycle.

171

Quantitative Aptitude Simplified for CAT

For stage 1, F = final fraction of milk. b 2  5  2 7 7  F  f 1    1  .     5 40 a     5 8 20

This

7 is the final fraction of milk at the end of stage 1. 20

For stage 2, since we are replacing with milk, we need to take the initial fraction of water. The initial fraction of water = 1 

7 13  . 20 20

Therefore, final fraction of water = This

13  5  13 7 91   . 1   20  40  20 8 160

91 is the final fraction of water. Hence the ratio of water to milk is 91 : (160 – 91) = 91 : 69 and hence 160

the milk to water ratio is 69 : 91. (ii)

Addition of 5 litres of water would change the ratio of milk and water. In the original solution, the ratio of milk to water is 2 : 3. This means that amount of milk available is

2  40  16 lt and hence 5

the amount of water = 40 – 16 = 24 lt. Adding 5 litres of water would raise the amount of water to 29 litres. Therefore, the milk water ratio in the now obtained 45 litres of solution is 16 : 29. Now 5 litres of solution is taken out and then water is added. For this portion, we can use the ‘replacements formula’ as shown below: n

b 16  5  16 7 14  (as n = 1) F  f 1      1    a 45 40     45 8 45

The ratio of milk to water in the solution now is 14 : 31. Taking out 5 litres of solution would not affect the ratio of milk to water and hence the final ratio is 14 : 31. (iii) Since we are replacing with milk, the fraction of water would be considered (as discussed in (i)). For milk concentration to exceed 60%, the water concentration should go below 40%. n

3 5  2 1   40%  5 40  5 n

7 2 or    3  8

We calculate that less than

7 2 = 87.5% and (0.875)3 = 66.99 which is more than rd whereas (0.875)4 = 0.59 which is 8 3

2 rd. Therefore, the process should be repeated at least 4 times. 3

Making a slight modification to this, it can be asked what amount of solution should be taken out after 3 stages of replacements by 5 litres of milk, so that the milk concentration becomes exactly 60%. To solve this, let the amount taken out after 3 stages be x litres. Then, 3

2 3  7  x  2 (as the fraction of water after this would be th)   1    5  8  40  5 5 3

  2  8    1024  200  or x  1-1        40   1    40  

 3  7   



1029 

1029

lt.

172

Ratio, Proportion and Mixtures A further modification to the question: what should be the amount of replacement so that after 3 stages, the concentration of milk becomes 60%? Now, for this, let us assume that the amount of replacement is x litres. Then, 3 3  x  2 2 3 x  2  or  1     x  40  1  3  = 5.057 (approximately) 1   40 3 3 5 40  5   

Note that the answer has to be slightly more than 5 litres as with 5 litres, we have seen that after three stages of replacements, the concentration is slightly less than 60%.

173

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

a 1 b c 1 d e 1  ,  2,  ,  3,  , then what b 3 c d 2 e f 4 abc ? is the value of def

If

 a c ,    36 e 

C.

 a bd   ,   12 18 

B.

27 8

D.

C.

3 4

 a c  ,   6 d

D.

27 4

E.

1 4

CAT 2003 (R) 5.

a b c If   = r, then r cannot take bc ca ab

A.

1 2

B.

1

C.

1 or 1 2

D.



6.

1 or 1 2

2:3

B.

3:2

C.

1:3

D.

3:4

A.

4:5

B.

29 : 13

C.

13 : 29

D.

None of these

There are two containers, one having oil A and the other having oil B. Now, 60% oil transferred from A to B, then 50% from B to A. Now, ratio of quantity of A to B is 4 : 5. Find the original ratio of quantity of A to B. A.

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture., he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk? A.

Ratio of two areas is 2 : 3. Ratio of rice to wheat in the whole area is 3 : 7. If first area has ratio of rice to wheat 2 : 5, find the ratio in the second one.

CAT 2014

CAT 2004

4:7

B.

5:9

C.

1:2

D.

2:3 CAT 2014

7.

CAT 2004 4.

B.

3 8

any value except

3.

 a b ,    27 e 

A.

CAT 2006 2.

A.

Let a, b, c, d, and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer?

Shyam divides a cake in the ratio of 1 : 2 : 5 and gives the largest piece to Ram. Ram divides the piece received by him in the ratio of 1 : 3 : 6 and gives the smallest piece to Raja. If Raja gets 20 gm of cake, what is the total weight of cake? A.

160 gm

B.

320 gm

C.

80 gm

D.

300 gm CAT 2015

174

Ratio, Proportion and Mixtures 8.

There are two vessels of same volume, one containing water and the other acid. From each vessel, 2 litres are drawn and then poured in the other vessel. Again, 2 litres from each vessel is drawn and poured in the other vessel. Now, the amount of acid in both the vessels become same. Find the original volume of the vessel. A.

8

B.

4

D.

6

D.

5

12. Company ABC starts an educational program in collaboration with Institute XYZ. As per the agreement, ABC and XYZ will share profit in 60 : 40 ratio. The initial investment of Rs.100,000 on infrastructure is borne entirely by ABC whereas the running cost of Rs.400 per student is borne by XYZ. If each student pays Rs.2000 for the program find the minimum number of students required to make the program profitable, assuming ABC wants to recover its investment in the very first year and the program has no seat limits. A. B. C. D. E.

CAT 2015 9.

Ratio of incomes of A, B and C is 3 : 4 : 5 last year. Income of A increases by 33.33% this year, that of B increases by 25% and that of C increases by 12.5%. If total income of all three this year is Rs. 819000, then what is the difference in the incomes of A and C this year? A.

Rs. 90000

B.

Rs. 91000

C.

Rs. 98000

D.

102000

XAT 2016 13. Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?

CAT 2015 10. Ratio of incomes of A, B and C is 3 : 5 : 8. A saves 20% of his income, B saves 30% and C saves 25% of his income. Total savings is what percent of the total income? A.

25%

B.

28%

C.

25.625

D.

25.25%

63 84 105 157 167

A. B. C. D. E.

328 units 368 units 392 units 616 units None of the above XAT 2015

CAT 2015 14. Consider the formula, S 

11. 100 kg and 500 kg of two varieties of rice cost Rs. 500 / kg and Rs. 1000 / kg. The two varieties were mixed and three-fourth of the mixture was sold at Rs. 1200 / kg and the rest was sold at Rs. 800 / kg. Find the profit %.

* , where   *

all the parameters are positive integers. If ⍵ is increased and ⍺, τ and ρ are kept constant, then S: A.

increases

A.

20%

B.

decreases

B.

25%

C.

increases and then decreases

C.

10%

D.

decreases and then increases

D.

15%

E.

cannot be determined XAT 2014

CAT 2016

175

Quantitative Aptitude Simplified for CAT

15. There are two alloys P and Q made up of silver, copper and aluminium. Alloy P contains 45% silver and rest aluminum. Alloy Q contains 30% silver, 35% copper and rest aluminium. Alloys P and Q are mixed in the ratio of 1 : 4.5. The approximate percentages of silver and copper in the newly formed alloy is:

16 kg. One piece each of equal weight was cut off from both the alloys and first piece of first alloy was alloyed with the second piece of second alloy and the second piece of first alloy was alloyed with the first piece of the second alloy. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

A.

33% and 29%

B.

29% and 26%

A.

3.33 kg

C.

35% and 30%

B.

4.67 kg

D.

None of the above

C.

5.33 kg

D.

None of these

IIFT 2015

IIFT 2013

16. X and Y are the two alloys which were made by mixing Zinc and Copper in the ratio 6 : 9 and 7 : 11 respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form another alloy Z, what is the ratio of Zinc and Copper in the new alloy Z? A.

6:9

B.

59 : 91

C.

5:9

D.

59 : 90

19. The duration of the journey from your home to the College in the local train varies directly as the distance and inversely as the velocity. The velocity varies directly as the square root of the diesel used per km, and inversely as the number of carriages in the train. If, in a journey of 70 km in 45 minutes with 15 carriages, 10 litres of diesel is required, then the diesel that will be consumed in a journey of 50 km in half an hour with 18 carriages is

IIFT 2014 17. A milk vendor sells 10 litres of milk from a can containing 40 litres of pure milk to the 1st customer. He then adds 10 litres of water to the milk can. He again sells 10 litres of mixture to the 2nd customer and then adds 10 litres of water to the can. Again he sells 10 litres of mixture to the 3rd customer and then adds 10 litres of water to the can and so on. What amount of pure milk will the 5th customer receive? A.

510 litres 128

B.

505 litres 128

C.

410 litres 128

D.

405 litres 128

A.

2.9 litres

B.

11.8 litres

C.

15.7 litres

D.

None of the above IIFT 2013

20. The Howrah-Puri express can move at 45 km/hour without any wagons attached, and the speed is diminished by a constant that varies as the square root of the number of wagons attached. If it is known that with 9 wagons, the speed is 30 km/hour, what is the greatest number of wagons with which the train can just move? A.

63

B.

64

C.

80

D.

81 IIFT 2012

IIFT 2014

21. A 10 litre cylinder contains a mixture of water and sugar, the volume of sugar being 15% of total volume. A few litres of the mixture is released and an equal amount of

18. Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs

176

Ratio, Proportion and Mixtures water is added. Then the same amount of the mixture as before is released and replaced with water for a second time. As a result, the sugar content becomes 10% of total volume. What is the approximate quantity of mixture released each time? A.

1 litres

B.

1.2 litres

C.

1.5 litres

D.

2 litres

B.

Rs. 288

C.

Rs. 216

D.

Rs. 384

B.

1

C.

n 2

D.

(n  1) 2n

E.

2008

Mark (D) if the question cannot be answered even after using A and B together. 25. The average weight of a class of 100 students is 45 kg. The class consists of two sections, I and II, each with 50 students. The average weight, WI of Section I is smaller than the average weight, WII of Section II. If the heaviest student, say Deepak, of Section II is moved to Section 1, and the lightest student, say Poonam, of Section I is moved to Section II, then the average weights of the two sections are switched, that is, the average weight of Section I becomes WII and that of Section II becomes WI. What is the weight of Poonam?

B.

22 years

A:

WII  WI = 1.0

B:

Moving Deepak from Section II to I (without any move from I to II) makes the average weights of the two sections equal. CAT 2007

26. Average salary of nursing and admin staff is Rs. 300. If average nursing staff earns Rs. 50 more than the admin staff, and there are 300 nursing and 200 admin staff, find the average salary of nursing staff.

24. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eightmember joint family is nearest to 23 years

24 years

Mark (C) if the question can be answered by A and B together, but not by either A or B alone.

CAT 2007

A.

E.

Mark (B) if the question can be answered using B alone but not using A alone.

23. Consider the set S = {2, 3, 4,...., 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y? 0

25 years

Mark (A) if the question can be answered using A alone but not using B alone.

IIFT 2012

A.

D.

Directions for question 25: The question is followed by two statements A and B. Indicate your responses based on the following directives:

22. A sum of Rs. 1400 is divided amongst A, B, C and D such that A’s share : B’s share = B’s share : C’s share = C’s share : D’s share = 3 : 4. How much is C’s share? Rs. 72

21 years

CAT 2007

IIFT 2012

A.

C.

A.

Rs.280

B.

Rs.270

C.

Rs.250

D.

Can’t be determined CAT 2014

177

Quantitative Aptitude Simplified for CAT

27. Sum of ages of A, B and C 10 years ago was 30 years. If after 5 years, average age of A and C is 1.5 years more than the age of B, what is the ratio of A's to C's age? A.

3:2

B.

2:3

C.

20 : 21

D.

Cannot be determined

B.

increase by 1

C.

increase by 1.5

D.

increase by 2 IIFT 2013

29. In 2011, Plasma – a pharmaceutical company – allocated Rs. 4.5 × 107 for Research and Development. In 2012, the company allocated Rs, 60,000,000 for Research and Development. If each year the funds are evenly divided among 2 × 102 departments, how much more will each department receive this year than it did last year?

Rs. 2.0 × 105

B.

Rs. 7.5 × 105

C.

Rs. 7.5 × 104

D.

Rs. 2.5 × 107

Both Statement I and Statement II are required.

D.

Neither Statement I nor Statement II is sufficient.

E.

Either Statement I or Statement II is sufficient.

A.

6

B.

7

C.

8

D.

9

E.

None of these XAT 2014

IIFT 2012 A.

C.

31. Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student can get an A grade in the course if the average of her scores is more than or equal to 90. Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he will score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?

28. The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall remain unchanged

Statement II only.

XAT 2015 CAT 2014

A.

B.

32. A teacher noticed a strange distribution of marks in the exam. There were only three distinct scores: 6, 8 and 20. The mode of the distribution was 8. The sum of the scores of all the students was 504. The number of students in the most populated category was equal to the sum of the number of students with lowest score and twice the number of students with the highest score. The total number of students in the class was:

30. The median of 11 different positive integers is 15 and seven of those 11 integers are 8, 12, 20, 6, 14, 22, and 13. Statement I: The difference between the averages of four largest integers and four smallest integers is 13.25.

A.

50

B.

51

Statement II: The average of all the 11 integers is 16.

C.

53

D.

56

Which of the following statements would be sufficient to find the largest possible integer of these numbers?

E.

57

A.

XAT 2014

Statement I only.

178

Ratio, Proportion and Mixtures 33. The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is A.

B.

28

C.

30

D.

32

E.

34

26

XAT 2013

ANSWER KEY 1. A

2. C

3. A

4. D

5. C

6. A

7. B

8. B

9. B

10. C

11. A

12. E

13. B

14. A

15. A

16. B

17 D

18. C

19. B

20. C

21. D

22. D

23. B

24. E

25. C

26. B

27. D

28. C

29. C

30. E

31. D

32. E

33. D

179

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS 1.

Combining both ratios, a : b : c : d : e = 108 : 18 : 9 : 4 : 3. Only

Answer: A

5.

Explanation: abc  a   b   c       def  d   e   f   a  b  c   b  c  d   c  d  e      b c d c d e d e f 

Answer: C Explanation: a b c abc which will    b  c c  a a  b 2( a  b  c )

1 if a + b + c ≠ 0. If a + b + c = 0, 2 c then a + b = c or  1 . Therefore, r ab 1 cannot take any value except or 1. 2

be equal to

3.

6.

Answer: A Explanation: Let original quantities of A and B be x and y. When 60% is transferred, quantity of oil A and B in the container having B is 0.6x and y. Now, when 50% is transferred from B to A, then quantities of oil A and B transferred are 0.3x and 0.5y. So, oil A and B in first container becomes (0.4x + 0.3x) and 0.5y. Therefore, ratio = 0.7x : 0.5y, which is given to be 4 : 5.

Answer: A Explanation: Out of total 100 litres of mixture, there is 20 litres of water and 80 litres of milk. When he sells 1 th part of mixture, that is 25 litres, water 4 will be 15 litres and 60 litres of milk in total 75 litres of mixture. When he adds 25 litres water in it, total water will be 25 + 15 = 40 litres and milk will be 60 liters. So the required ratio is 40 : 60 = 2 : 3. Alternatively, use the formula of replacement. We get b n 4 1 4 3 3 F  f  1     1      . a 5 4 5 4 5  Therefore, ratio of milk to water = 3 : 2 or ratio of water to milk = 2 : 3.

4.

Answer: C Explanation: Sums of terms in the given ratios are 2 + 3 = 5; 3 + 7 = 10 and 2 + 5 = 7. LCM of 5, 10 and 7 = 70. 2 Let the area of two regions be  70 = 28 5 and 42. Since first area has ratio of rice to wheat as 2 : 5, amount of rice and wheat in this area = 8 and 20. Ratio of overall rice to wheat = 3 : 7. Therefore, overall rice = 21 and overall wheat = 49. Therefore, rice and wheat in the second area = 13 and 29. Therefore, ratio of rice to wheat in the second area = 13 : 29.

1 1 1 1 1 3 =   2    2   3    3    . 2  2 4 8 3 2

2.

c cannot be integer. d

So, 0 .7 x  4  x  4 . Therefore, required 0 .5 y

5

y

7

ratio = 4 : 7. 7.

Answer: B Explanation: Sum of terms in the second ratio is 1 + 3 + 6 = 10. Since 5 parts from 1st ratio is getting divided, we should multiply first ratio by 2. So, the ratio 1 : 2 : 5 becomes 2 : 4 : 10. Now, this 10 parts is divided in the ratio of 1 : 3 : 6. So, smallest part is 1, whose weight is 20 gm. Therefore, 10 parts will be 200 gm and so total parts = 2 + 4 + 10 = 16  320 gm.

Answer: D Explanation: a = 6b = 12c  a : b : c = 12 : 2 : 1 2b = 9d = 12e  b : d : e = 18 : 4 : 3

8.

Answer: B Explanation: After the said operations, if amount of acid in both the vessels becomes same, then

180

Ratio, Proportion and Mixtures 12. Answer: E

amount of water also becomes same and so ratio of acid to water in the two vessels becomes 1 : 1. If original volumes of vessels were 8 litres, then after first transfer, volume of acid in second vessel becomes 6 litres and after the second operation, since we are not transferring pure acid, the amount of acid cannot become 4 litres making equal amounts of water and acid. So, [A] option is ruled out. If original volume were 4 litres, then amount of acid after first transfer becomes 2 litres. So, ratio of acid to water in both the vessels becomes 1 : 1. The second operation would then have no effect on the ratio, and so ratio after second operation remains the same. Therefore, original volume is 4 litres. 9.

Explanation: Let total number of students be x. Then, total expenditure = 100000 + 400x. Revenue = 2000x. Therefore, total profit = 2000x – (100000 + 400x). Profit share with company ABC = 60% of total profit, which should be equal to 100000, to recover the cost. Therefore, 60% of [2000x – (100000 + 400x)] = 100000 (at break-even).  x = 166.7. Therefore, minimum number of students = 167. 13. Answer: B Explanation: 5  864 = 480 units; Y 9 = 864 – 480 = 384 units. 3 B in 480 units of X =  480 = 360 units; B 4 2 in 384 units of Y =  384 = 256 units 3 Total units of B = 360 + 256 = 616, which is 50% of 1232. This 1232 units contains 864 units of M. Therefore, water = 1232 – 864 = 368 units.

In 864 units of M, X =

Answer: B Explanation: Let the incomes be 24, 32 and 40 (multiplying by 8 because income of C 1 th . New increases by 12.5%, that is 8 incomes = 32, 40 and 45. Total = 32 + 40 + 45 = 117.) If total income is 117, difference of incomes of A and C = 13; If total income is 819000, difference 13 =  819000 = Rs. 91000. 117

14. Answer: A Explanation: Divide the numerator and denominator by .  We get, S = . If  increases, then /    decreases and so S increases.

10. Answer: C Explanation: Let the incomes be 3, 5 and 8. Savings of A = 20% of 3 = 0.6; of B = 30% of 5 = 1.5; of C = 25% of 8 = 2. Total savings = 4.1 Required 4.1 percentage =  100 = 25.625%. 16

15. Answer: A Explanation: 1 unit of P has 0.45 units of silver. Now, 4.5 units of Q has 30% silver, that is 1.35 units of silver. So, total silver in the mixture = 1.8 units. P does not contain any copper. Copper in 4.5 units of Q = 35% of 4.5 = 1.575. Therefore, percentage of silver in the new 1.8 alloy =  100 = 33% (approximately). 5.5 Percentage of copper in the new alloy 1.575 =  100 = 28.6% (approximately). 5 .5

11. Answer: A Explanation: Total cost of the mixture = 100 × 500 + 500 × 1000 = Rs. 550,000. Total revenue from sale 3 1 =  600  1200   600  800 4 4 = Rs. 660,000. 11 Therefore, profit % =  100 = 20%. 55

181

Quantitative Aptitude Simplified for CAT

16. Answer: B

20. Answer: C

Explanation:

Explanation: Let speed of train with n wagons attached be v. Then, with n wagons, speed decreases by 45 – v, which is directly proportional to square root of number of wagons attached. So, 45 – v  n  45 – v = k n (where k is proportionality constant). Now, with 9 wagons attached, speed is 30 km/hr. Therefore, 45 – 30 = k 9  k = 5. For maximum number of wagons to be attached, v = 0. So, 45 = k n = 5 n  n = 81. But, when 81 wagons are attached, speed becomes exactly zero and so train would not move. So, we need to attach 1 wagon less. So, maximum number of wagons that can be attached is 80.

6 7 118 Zinc in alloy Z =  40   60  15 18 3 9 11 182 Copper in alloy Z = .  40   60  15 18 3 Therefore, ratio of zinc to copper in the new alloy Z = 118 : 182 = 59 : 91.

17. Answer: D Explanation: Use the formula of “replacements”. a n 10  4 F = f  1    1  1   b 40    (we will take n = 4, because 5th customer gets after 4 cycles) Therefore, amount of pure milk received by 3 4 405 litrers. 5th customer =    10  4 128  

21. Answer: D

18. Answer: C

Explanation: Using the formula of “replacements”, we get 10 = 15

Explanation: Let weight of the cut-off piece = x kg. Let percentage of aluminium in 8 kg and 16 kg alloy be r% and q% respectively. (8  x )r  xq xr  (16  x )q Therefore,  8 16  16r – 2rx + 2qx = rx + 16q – qx  16 (r – q) = 3rx – 3qx  16(r – q) = 3x(r – q)

2

2 x  x  1 x 1    10 3 10  

litres = 1.84 litres  2 litres. 22. Answer: D Explanation: A : B = 3 : 4, B : C = 3 : 4. Therefore, A : B : C = 9 : 12 : 16. Also, C : D = 3 : 4. Therefore, A : B : C : D = 27 : 36 : 48 : 64.

16 x= = 5.33 kg (since r ≠ q). 3

19. Answer: B

Therefore, C’s share =

Explanation: t 

x (where t = time, x = distance, v v

= speed). Also, v 

48  1400 = Rs. 384. 175

23. Answer: B Explanation: From 2 to 2n + 1, there are 2n terms: half of them are even and the other half are odd numbers. Now, X – Y = difference of the average of odd numbers and even numbers = average of the difference of odd and even numbers. Since each difference of consecutive odd and (previous) even number is 1, there will be 1 appearing n times which when divided by n gives 1. OR, X – Y =

d (where d = diesel n

consumed per km and n = number of carriages in the train). Combining the two, we get t 

 2  1   10 3  

xn . Therefore, d

d2 d2 t1 x1n1 45 70  15 50       d2 t2 30 50  18 d1 x2n2 10 70

= 11.8 litres.

3  5  7  ...  (2n  1) 2  4  6  ...  2n  n n

182

Ratio, Proportion and Mixtures



(a  5)  (c  5) = 1.5 + (b + 5)  a + c 2

(3  2)  (5  4)  (6  5)  ...((2n  1)  (2n)) n  1 n n

= 2b + 13 – 10 = 2b + 3. Also, a + b + c = 60. So, 2b + 3 = 60 – b or b = 19. Therefore, a + c = 41. Since we don’t know the age of a or c, we cannot determine the required ratio.

24. Answer: E Explanation: Ten years ago, the sum of the ages of 8 members in the family = 231. After 3 years, the age of every member increases by 3; hence the total age increases by 8  3 = 24. The total age is 231 + 24 = 255. A member of age 60 years died = 255 – 60 = 195. After another 3 years, sum of the ages of all members = 195 + 24 = 219. Again a member of age 60 died = 219 – 60 = 159. We are left with 4 years to reach current year. Hence the sum increases by 8  4 = 32. Sum of the ages of 8 members in the present year = 159 + 32 = 191. Average age =

28. Answer: C Explanation: We can take first 7 natural numbers whose average is 4. If we include next 3 numbers, we get first 10 natural numbers whose average is 5.5, which is 1.5 more than 4. 29. Answer: C Explanation: Difference in average funds of the two years

191 = 23.8 = 24 8

=

(approximately).

(6  4.5)  107 = 75000 = 7.5 × 104. 200

30. Answer: E

25. Answer: C

Explanation: The given numbers in increasing order are: 6, 8, 12, 13, 14, 20, 22. Since 15 is median of 11 integers, there will be 5 numbers on either side of 15. As we can see, the 5 numbers less than 15 are already given to us. Let the 3 unknown numbers be x, y and z. Now, using statement I alone, sum of largest 4 numbers = 13.25 × 4 + (6 + 8 + 12 + 13) = 92. For largest possible integer, the two unknown integers must be lowest possible above 15, so they can be 16 and 17. Therefore, largest possible number = 92 – 22 – 20 – 17 = 33. Statement I alone is sufficient. Using statement II alone, sum of 11 integers = 16 × 11 = 176. Therefore, x + y + z = 176 – (6 + 8 + 12 + 13 + 14 + 15 + 20 + 22) = 66. Once again, for largest possible integer, the other two unknown must be 16 and 17. Therefore, largest possible integer = 66 – 16 – 17 = 33. Statement II alone is also sufficient.

Explanation: We are given the average weight of a class of 100 students which is 45 kg. Also, WI < WII Now, WI + WII = 90 … (1) Moreover, if Deepak weighs d and Poonam p, then 50WI + d – p = 50WII, Or d – p = 50 (because WII – WI = 1). Also, 50WII – d + p = 50WI, or d – p = 50. Therefore, statement A alone is insufficient to solve the question. Using Statement B, (50WI + d)/51 = (50WII – d)/49 This statement alone is also insufficient. Combining both the statements, we can get the weight of Poonam. 26. Answer: B Explanation: Let average salary of admin staff be Rs.x. Then, average salary of nursing staff = Rs. x + 50. Now, 300(x + 50) + 200x = 500 × 300  x = 270. 27. Answer: D

31. Answer: D

Explanation: Sum of ages today = 30 + 30 = 60. Sum of ages after 5 years = 60 + 15 = 75. Also, if a, b and c are the ages of A, B and C today, then

Explanation: Total exams given = x; Total score of previous exams = y.

183

Quantitative Aptitude Simplified for CAT

33. Answer: D

If he needs minimum 97 to get A grade, that is average of 90, then y + 97 = 90x. Similarly, if he scores 70 to just manage a score of 87, then y + 70 = 87x. Therefore, 90x – 97 = 87x – 70 or x = 9.

Explanation: Let the numbers in increasing order be a, b, c, d, e and f. Now, sum of these numbers = 15 × 6 = 90. The sum of c and d = 18 × 2 = 36, because 18 is the median. To maximize f, c and d should be close to each other. Also, a and b should be minimum positive integers = 1, 1. So, sum of first 4 numbers = 1 + 1 + 36 = 38. Sum of biggest two numbers = 52. If c = d = 18, then the data will have two modes, where we can have only one mode. So, c = 17 and d = 19 and so e = 20. Therefore, f = 32.

32. Answer: E Explanation: Let the number of students scoring 6, 8 and 20 be x, y and z respectively. So, 6x + 8y + 20z = 504 (i) x + 2z = y (ii) From (i) and (ii), we get 14y + 8z = 504 or, 7y + 4z = 252 By hit and trial we get y = 32 and z = 7. Therefore, x = 18. Therefore, total number of students = 32 + 7 + 18 = 57.

184

Time and Work

Chapt er 05

Time and Work BASIC CONCEPTS Time and Work Majority of the questions and concepts from this topic are applications of ratios, direct and inverse variation, and unitary method. We will understand these concepts and their applications and also identify more efficient methods of solving the same questions. Let there be a farmer who ploughs his field in certain days with the help of some fellow farmers. Every day, same number of farmers work for same number of hours per day; and thus the ploughing of the field is finished in a few days. We can easily conjecture that if we change the number of farmers, or number of hours worked per day or the amount of work, the number of days in which the work will get finished will change. So, we can say that the number of days taken depends upon the number of people working, number of hours worked per day, amount of work, and of course the efficiency with which people work. Let us formalize this relationship. If M = number of men doing a certain job, D = number of days required to do the job, H = number of hours worked per day, W = amount of work required to be done, Then following proportionalities exist: M

1 D

This means that the number of men is inversely proportional to the number of days required for the job. So, if more number of days are allowed for a certain job, working for a fixed number of hours per day and doing fixed amount of work, then we would require less number of men, and vice versa. Similarly,

185

Quantitative Aptitude Simplified for CAT

M

1 H

This means that the number of men is inversely proportional to the number of hours worked per day. So, if more number of hours can be worked per day, working for fixed number of days and doing fixed amount of work, then we would require less number of men, and vice versa. Finally, MW This means that the number of men is directly proportional to the amount of work. So, if more amount of work is required to be done, working for fixed number of days and hours per day, then we would require more number of men, and vice versa. Combining these results, we get M 

W W MDH  M  k    constant. DH DH W

This is an important result. The useful formula with regard to this is M1D1H1 M2D 2H2  W1 W2

Let us take an example to elaborate its use. Example 1 If 20 men can complete a certain work in 30 days, working 8 hours a day, then in how many days would 25 3 men complete th of the work working for 12 hours per day? 4 Solution M1 = 20, D1 = 30, H1 = 8, W1 = 1 units, M2 = 25, H2 = 12, W2 = Therefore,

3 units. 4

M1D1H1 M2D 2H2 MD H W (20)(30)(8 )(3/4)   D2 = 1 1 1 2   12 days. W1M2H2 (25)(12)(1 ) W1 W2

MDH = constant, we can say that the units of “work” will be same as the units of “MDH”, that is, manW days-hours. In cases where number of days is not mentioned, the units of work become man-hours. So, the common language used by management in an office is that on a particular day 4000 man-hours amount of work has been done. This could mean: 400 men worked for 10 hours or 500 men for 8 hours, and so on.

Since

Example 2 In an office, the amount equivalent of 2500 man-hours is done on a particular day. The next day, 20 more persons join the office who work for same number of hours as the existing staff. If the number of persons has increased by 10% because of this addition, what is the total man-hours clocked on this day? Solution Method 1: Since the number of persons has increased by 10% which is 20 persons, we can say that the number of persons earlier was 200 and hence the number of hours worked per day =

2500  12.5 hrs. 200

186

Time and Work Therefore, the number of man-hours clocked today = 220  12.5 = 2750 man-hours. Method 2: Since the number of men has increased without changing the number of hours per day, we can say that “man-hours” done also increases by same percentage, that is, 10%. Therefore, total man-hours clocked on this day = 2500 + 10% of 2500 = 2750 man-hours. This is based on the assumption that everyone is working with the same efficiency. If efficiency of the first set of M1 persons is different from that of the other set of M2 persons, then the following formula (which is an extension of the earlier one) can be used: M1D1H12 MD H  2 2 2 2 W1 W2

where  is the efficiency of M1 number of persons and  is the efficiency of M2 number of persons. Example 3 Ram and Shyam are two contractors who employ 20 and 25 persons under them respectively. The people under Shyam are lazy and are 80% as efficient as those under Ram. Both are supposed to construct a building, each occupying the same amount of land but Shyam’s building is 25% taller than Ram’s building. People under 1 Ram finish th of the work in 1 day. If all the persons work for the same number of hours per day, then in 10 how many days would Shyam’s building be ready? Solution M1 = 20, M2 = 25, D = 10, H1 = H2, W1 = 1 unit, W2 = 1.25 units. Therefore, (20)(10)(H1 )(100%) (25)(D 2 )(H2 )(80%) M1D1H12 MDH    2 2 2 2  (1) (1.25) W1 W2

 D2 = 12.5 days. Example 4 27 men can do a certain job in 30 days. They all started work together. After 20 days, 10 men left. The remaining people finished the job in x days. Find x. In how much time did the entire project get completed? How much was the delay in the project? Solution Here, work is constant. So, 27 × 30 = 27 × 20 + 15x  x = 18 days. Therefore, the entire project will get completed in 20 + 18 = 38 days. The delay in the project will be 20 + 18 – 30 = 8 days. Example 5 5 men can do as much work as 8 women can do. If 10 men and 12 women together can do a certain work in 15 days, in how many days can 25 men alone do the same work?

187

Quantitative Aptitude Simplified for CAT

Solution 8 women  5 men  12 women  7.5 men (Though fractional number of men is not possible, this is acceptable for calculation purpose) Therefore, 10 men + 12 women  10 men + 7.5 men = 17.5 men. So, M1D1 = M2D2 or (17.5)(15) = (25)(D2)  D2 =

17.5  15  10.5 days. 25

Example 6 5 men can do as much work as 8 women can do and 2 women can do as much work as 3 children can do. If 5 men, 16 women and 9 children can finish a certain work in 100 days, then how many more men are required to finish the work in 75 days, if these extra men added are 25% more efficient than the men already available? Solution 5 men  8 women  12 children. 5 men + 16 women + 9 children = 5 men + 10 men + 3.75 men = 18.75 men. Using the result M1D1 = M2D2, M2 =

M1D1 18.75  100   25 . D2 75

We already have 18.75 men. So, 25 – 18.75 = 6.25 men are required. 25  If x is the number of men of higher efficiency, then  1   x  6.25  x  5  100  

Example 7 If 7 men and 5 women can do a certain work in 10 days and 5 men and 7 women can do the same work in 8 days, then 10 men and 15 women can do the same work in how many days? Solution If m denotes the number of men and w denotes the number of women, then (7m + 5w)  10 (5m + 7w)  8  70m + 50w  40m + 56w  30m  6w  5m = w. Therefore, 7 men + 5 women = 7 men + 25 men = 32 men. Moreover, 10 men + 15 women = 10 men + 75 men = 85 men. If 32 men can do the work in 10 days, then 85 men can do the work in

32 13  10  3 days. 85 17

Example 8 In an army camp, there is provision for food for 200 army men for a month. If after 20 days, war breaks out and 50 men die in the war, how many more days would the food last than it would have lasted previously? Solution Amount of food = 200  30 = 6000 man-days. (a month would always mean 30 days) Amount of food consumed in 20 days = 200  20 = 4000 man-days. If the number of days food lasts now is x, then

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Time and Work

6000 = 4000 + 150x  x =

2000 1  13 days. 150 3

Therefore, the food would last 3

1 1 days more than it would have lasted. So, the answer is 3 days. 3 3

Another type of question that is usually asked is described below: Let there be two persons A and B who can finish a certain task in 10 and 15 days respectively. Since A does the job in less days than B would take, we can say that A is more efficient than B. By how much % is A more efficient that B? Before we discuss this, please understand that the amount of work done per day defines the efficiency. There are two approaches to solve this. Approach 1: In 1 day, amount of work done by A =

1 1 and that done by B = . 10 15

Therefore, percentage by which A is more efficient than B =

w A  wB (1 / 10)  (1 / 15)  100   100  50% . wB (1/15)

Hence, A is 50% more efficient that B. Some students find efficiency by the following (wrong) method. 15  10 1  100  33 %, taking the base to be 15. Using the concept 15 3 of days like this will never give the correct efficiency. To understand why, learn the second approach.

% by which A is more efficient than B =

Approach 2: In this approach, we take the LCM of days given and assume that to be the work to be done. LCM (10, 15) = 30 = amount of work to be done. Then, in one day, the amount of work done by A = units; work done by B (in one day) =

30 3 10

30  2 units. 15

Therefore, percentage by which A is more efficient than B =

w A  wB 32  100   100  50% . wB 2

If C is 50% more efficient than A, then in how many days would C complete the job? Again there are two approaches. Approach 1: Let C take x days to finish the job. Then in one day, amount of work done by C = Amount of work done by A =

1 . x

1 . Now, as per the question, 10

(1/x) - (1/10) 2  100  50%  x = 6 days. (1/10) 3

Approach 2: In one day, work by A =

1 which is same as 10%. 10

So, if total work is 100 units, then A will do 10 units in 1 day, whereas C would do 15 units (as C is 50% more 100 2 efficient than A). Hence, number of days taken by C =  6 days. 15 3 Therefore, essentially, there are two methods which are employed to tackle cases like these:

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Quantitative Aptitude Simplified for CAT

(i)

Either to deal with fractional amounts of work, like

1 1 th or th and so on. In this, we assume 10 15

that total work done is 1 unit. Or, (ii)

take the LCM of the individual number of days and assume that to be the amount of work.

The second approach avoids fractions and hence is better in many instances. Example 9 A and B can finish a job working all alone in 12 day and 18 days respectively. In how many days will they be able to finish the job if they work together? Solution Approach 1: If the total amount of work is 1 unit, then in 1 day the amount of work done by A= B=

1 units, and the amount of work done by 12 1 units. 18

So, in 1 day, the amount of work done by A and B together = If

1 1 32 5    units. 12 18 36 36

5 1 36 1 units is done in 1 day, then 1 unit of work is done in   7 days. 36 5/36 5 5

Approach 2: In this approach, we will take the LCM of 12 and 18 which is 36. Let the amount of work required to be done is 36 units. In 1 day, the amount of work done by A =

36 units. 12

In 1 day, the amount of work done by B =

36  2 units. 18

Therefore, in 1 day the total amount of work done = 3 + 2 = 5 units. If 5 units is done in 1 day, 36 units will be done in

36 1  7 days. 5 5

In general, if A does a job in ‘a’ days working alone and B in ‘b’ days also working alone, then together they can ab complete the job in days. ab Example 10 In the previous example, what will happen if they work on alternate days? Does it depend on ‘who starts’? If yes, how? Solution Let us take two cases: in case 1, A starts the work and in case 2, B starts the work. Case 1: Let A start the work. Then on 1st day,

1 th of the work is finished. 12

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Time and Work

On 2nd day, B works and further

1 th of the work is finished. 18

On subsequent days, the same cycle will repeat and whatever amount of work is finished in first two days, the same amount of further work will be finished on 3rd and 4th day together and so on. Therefore, Work done in first two days =

1 1 5   .. 12 18 36

Work done in next two days =

5 . 36

5 is done and so on. We realize that after 7 such cycles, the total 36 5 5 5 5 5 5 5 35 amount of work done would be =        . 36 36 36 36 36 36 36 36

In the next cycle, another

35 1  . 36 36

Now, the amount of work left to be done =

7 cycles mean 14 days are over. Now, on 15th day, A would work (as we can see that A works on odd days).k 1 1 th of the work whereas only th work is required to be done, which A would take 12 36 1/36 1 1  rd of the day to finish. So, the work is finished in 14 days. 1/12 3 3

On 15th day, A can do

The trick is to find the largest natural number such that upon multiplication, the result is another fraction which is just less than 1 (or may be equal to 1). In the current example, this natural number was 7. We observe that a cyclical pattern emerges which will repeat till fraction of work done is just less than 1 or equal to 1. And this is where the whole trouble lies. Now we will take up case 2 in which B starts the work. Case 2: Let B start the work. Then on 1st day,

1 th of the work is finished. 18

On 2nd day, A works and further

1 th of the work is finished. 12

Going as usual, work done in first two days = Work done in next two days =

1 1 5   . 18 12 36

5 . 36

5 is done and so on. Once again, we realize that after 7 such cycles, the total 36 5 5 5 5 5 5 5 35 amount of work done would be =        . 36 36 36 36 36 36 36 36

In the next cycle, another

Now, the amount of work left to be done = 1 

35 1  . 36 36

7 cycles mean 14 days are over. Now, on 15th day, B would work (as we can see that B works on odd days).

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Quantitative Aptitude Simplified for CAT

1 1 th of the work whereas only th is required to be done, which B would take 18 36 1/36 1 1  of the day to finish. So, the work is finished in 14 days. 1/18 2 2

On 15th day, B can do

We can solve the above example (both the cases) by using LCM approach. LCM of 12 and 18 = 36, which is assumed to be the amount of work required to be done. So, work done by A in 1 day = 3 units; work done by B in 1 day = 2 units. If A starts the work, then work done on 1st day = 3 and work done on 2nd day = 2. So, work done in first two days = 3 + 2 = 5. Work done on 3rd and 4th day is repetition of work done on 1st and 2nd day, that is 3 + 2 = 5. So, work done = (3 + 2) + (3 + 2) + (3 + 2) + …, till 36 units of work is done. But, 36 is not a multiple of 5. So, the cycle of 5 units of work will repeat 7 times, which means 35 units of work is done. Repetition 7 times means 14 days. So, on 15th day, we are left with 1 unit of work which will be done 1 1 by A in rd of the day. So, total time taken = 14 days. 3 3 If B starts the work, then work done on 1st day = 2 and work done on 2nd day = 3. So, work done in first two days = 2 + 3 = 5. Work done on 3rd and 4th day is repetition of work done on 1st and 2nd day, that is 2 + 3 = 5. So, work done = (2 + 3) + (2 + 3) + (2 + 3) + …, till 36 units of work is done. But, 36 is not a multiple of 5. So, the cycle of 5 units of work will repeat 7 times, which means 35 units of work is done. Repetition 7 times means 14 days. So, on 15th day, we are left with 1 unit of work which will be done 1 1 by B in of the day. So, total time taken = 14 days. 2 2 Observation: The amount of work done in each cycle and the number of cycles is the same in both the cases. Even the amount of work done on the last day is also same. The answers are different because of the different efficiencies of the persons to do the work on the last day. It is important to realize that if the work done in “each cycle” is a factor of the total work required to be done, then after “natural number of cycles”, the complete work would be done no matter who starts the work. Observe the following example. Example 11 Let there be two persons A and B who can do a certain job in 10 and 15 days respectively . They work on alternate days. In how many days would the job be completed? Solution Let’s again take two cases. Now, we will discuss only the LCM approach. Case 1: In this case, A starts the work. Proceeding as in the previous example, LCM (10, 15) = 30 units of work is required to be done. Work done by A and B per day is 3 and 2 units. Work done in first two days = 3 + 2 = 5 Work done in next two days = 3 + 2 = 5, and so on. To complete 30 units of work, exactly 6 cycles or 12 days will be taken (since 30 is a multiple of 5).

192

Time and Work Case 2: In this case, B starts the work. Proceeding as in the previous example, Work done in first two days = 2 + 3 = 5 Work done in next two days = 2 + 3 = 5, and so on. To complete 30 units of work, exactly 6 cycles or 12 days will be taken (since 30 is a multiple of 5). So, the answer doesn’t depend upon who starts the work. This is so because the total work required to be done (that is, 30 units of work) is a multiple of the work done by the two person in 1 cycle of 2 days (that is 5 units).

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Example 12 Let there be three persons A, B and C who can finish a certain job in 10, 12 and 18 days. They work on alternate days starting with A, then B, then C and then again A and so on. When will the job be completed? Solution LCM (10, 12, 18) = 180. Work done by A, B and C per day = 18, 15 and 10 units respectively. Work done in 1 cycle (3 days) = 18 + 15 + 10 = 43. After 4 cycles, that is 12 days, 43 × 4 = 172 units of work will be done. On 13th day, A will do 18 units in full day and so 180 – 172 = 8 units in So, time taken for the complete work = 12

8 4  th of the day. 18 9

4 days. 9

If we follow fractional approach, then Work done on day 1 =

1 1 1 units; Work done on day 2 = units; Work done on day 3 = units. 10 12 15

This cycle repeats. Work done in these 3 days =

1 1 1 18  15  10 43     units. 10 12 18 180 180

After 4 such cycles, that is 12 days, the amount of work done = 4  The amount of work left to be done =

43 172  . 180 180

8 2  . 180 45

1 2 units on that day, but the requirement is only units. 10 45 2/45 20 4 Therefore, this much work would be done by A in   th of the day. 1/10 45 9

On 13th day, it is A’s turn to work and he can do

Therefore, the entire work is finished in 12

4 days. 9

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Quantitative Aptitude Simplified for CAT

It should now be clear that LCM approach is better than fractional approach. Example 13 Let A be able to construct a building in 10 days which B can demolish in 20 days. In how many days would the building be constructed if (i)

they both work together

(ii)

they work on alternate days

Solution (i)

LCM (10, 20) = 20 units of work is required to be done.

Work done by A and B per day = +2 and 1 (we are taking negative value for demolition work). Work done in 1 day = 2 – 1 = 1 unit. So, 1 unit of work is done in 1 day; 20 units of work are done in 20 days. Hence, time taken to complete the work = 20 days. (ii)

The common mistake committed in such a question is the rationale that if 1 unit of work is done in 2 days (since they work on alternate days), 20 units would be done in 40 days. This is incorrect.

First of all, we need to realize that work will be started by the construction man only, that is A and also that he will work on the last day as well. Therefore, A works on day 1 and A also works on the last day. So, work done in first two days = 2 – 1 = 1 unit; work done in next two days = 2 – 1 = 1 unit, and so on. On the last day, the maximum work that can be done is 2 units (because A works on the last day). So, the minimum work that must be done by the end of penultimate day will be 20 – 2 = 18 units. If 1 unit is done every 2 days, then 18 units will be done in 36 days. On 37th day, A comes and does 2 units of work and completes the entire work. Therefore, the whole work is done in 37 days. In such cases, always remember that the work is started by the one who does positive work and also ended by the one who does positive work. Example 14 Let there be two persons A and B who can construct and demolish a building respectively in 10 and 30 days respectively. They work on alternate days. When half the work is “first” done, B increases his efficiency to double. In total, how many days are taken to finish the construction work? Solution LCM (10, 30) = 30 units of work is required to be done. Work done per day by A and B = +3 and 1. In first cycle of two days, starting with A, work done = 3 – 1 = 2 units. After 6 cycles, that is 12 days, work done = 12 units. Half the work means 15 units. On 13th day, A comes and does 3 units of work, completing 15 units of work. Now, B increases his efficiency to double and so can demolish 2 units in 1 day. On 14th day, work demolished = 2 units and so work left to be done = 15 + 2 = 17 units. From 15th day onwards, work done in 2 days = 3 – 2 = 1 unit. Since A will work on last day, so the maximum work that will be done on last day = 3 units. Therefore, the minimum work required to be completed by the end

194

Time and Work of penultimate day = 17 – 3 = 14 units, which will be done in 14 cycles of 2 days each. So, time taken = 28 days. After 28 days, work completed = 14 units and work left = 3 units, which will be done by A in 1 day. Therefore, total time taken = 14 + 28 + 1 = 43 days. Example 15 Let there be three persons A, B and C such that A constructs a building in 10 days working alone, B demolishes the same in 15 days and C demolishes in 20 days. On first day, A works, on second day B works, on third day A works and on fourth day C works. The same cycle repeats. In how many days is the work completed? Solution LCM (10, 15, 20) = 60 units of work is required to be completed. Per day work of A, B and C = 6, 4, 3. Work done in first 4 days = (6 – 4 + 6 – 3) = 5 units. On the last day, the maximum work that will be done = 6 units. So, the minimum work done by the end of 54 = 10.8 cycles. penultimate day = 60 – 6 = 54. Now, 5 After 10 cycles, that is 40 days, work done = 50 units. In next 4 days, work done = 50 + 6 – 4 + 6 – 3 = 55 units. Now work remaining = 5 units which will be done by A in So, total time taken for the work = 44

5 th of a day. 6

5 days. 6

Example 16 A monkey wants to climb a 20 m long pole. In the first minute, it climbs 3 m whereas in the next minute it slips by 2 m. This process goes on till the monkey reaches the top of the pole. How much time is taken by the monkey to reach the top of the pole? Solution This is a problem similar to the ones discussed earlier in which one was a construction man and the other a demolition man. Hence, we will follow the same process employed in the previous questions. This means that in the last minute; the monkey will climb 3 m. Hence, in the minute prior to that, it should have climbed by 20 – 3 = 17 m. In the first two minutes, the monkey is able to climb 3 – 2 = 1 m. Since 1 m is climbed in 2 minutes, 17 m would be climbed in 34 minutes. Finally, in the 35th minute, the monkey would climb 3 m and hence would reach the top. Therefore, the monkey takes a total of 35 minutes. Example 17 A certain field can be ploughed by a man in 20 days which his father would plough in 30 days and his grandfather in 60 days, all working alone. They all set out to plough the field. After 5 days, the father of the man leaves for the city to buy good variety of seeds but returns after the entire field is ploughed. The grandfather leaves ploughing 2 days before the field actually gets ploughed. When did the field actually get ploughed?

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Quantitative Aptitude Simplified for CAT

Solution Let the number of days in which the field got ploughed be x. The man works throughout, the father works only for 5 days, whereas the grandfather works for (x – 2) days. Therefore, adding the fractional contributions of each person should yield 1. x 5 x-2    1 4x + 8 = 60  x = 13 days. 20 30 60

Alternatively, LCM of 20, 30 and 60 = 60. Therefore, let us assume that total amount of work is 60 units. In 1 day, work done = 3 + 2 + 1 = 6 units. So in 5 days, the work done = 6 × 5 = 30 units. Then, father leaves. In the last 2 days, the man alone works. In 1 day, he can do 3 units of work. So, in 2 days, he will do 6 units of work. Work which is required to be done by man and grandfather is 60 – 30 – 6 = 24 units. Man and grandfather together do 3 + 1 = 4 units in 1 day. So, to do 24 units of work, time required =

24 = 6 4

units. Therefore, total time taken = 5 + 6 + 2 = 13 days. Alternatively, If total time taken is x days, then man works for x days, father works for 5 days and grandfather for (x – 2) days. Therefore, 3x + 2 × 5 + 1(x – 2) = 60  x = 13 days. The last one is the easiest and fastest approach. Example 18 In the previous example, after father left for city, the man and grandfather continue working together. The grandfather leaves exactly 2 days before the field would have been ploughed completely, had he not left. In how many days will the entire field be ploughed now? By how many days does the work get delayed compared to if nobody had left in between? Solution The father of the man leaves after 5 days, by which time, 30 units of work is over. The remaining work will be 30 done by the man and grandfather in = 7.5 days. 4 But, the grandfather leaves exactly 2 days before the work would have been completed. Therefore, the number of days the grandfather works now is 5.5. In these 5.5 days, the work done = (3 + 1)5.5 = 22 units. Work left = 8 units, which will be done by the man alone in Therefore, the work gets done in 5 + 5.5 + 2

8 2  2 days. 3 3

2 1  13 days. 3 6

If all three worked together and did not leave, the work would have been over in

196

1  10 days. 1 1 1   20 30 60

Time and Work

So, the work is delayed by 3

1 days. 6

EXPERT SPEAK Scan this QR Code to watch a video that explains the further applications of the LCM approach of solving problems based on Time and Work.

Pipes and Cisterns The cases of two or more persons doing a certain work can also be applied to the cases of pipes and cistern. A cistern is a tank of some capacity which has some inlet and some outlet pipes through which water (or some other liquid) flows. You may be asked to find the time taken to fill the completely empty cistern or empty a completely filled tank. A pipe being able to fill a cistern in 8 hours is similar to a man being able to construct a building in, say, 15 days. Let us see some of the problems below. Example 19 A cistern can be filled by two inlet pipes in 6 hours whereas it can be filled by one of the pipes in 10 hours. If the second pipe alone is open, then in how many hours would the cistern be full? Solution Let the second pipe take x hours. Then, 1 1 1 1 1 1 1       x 10 6 x 6 10 15

 x = 15 hours. Alternatively, LCM (6, 10) = 30 units of work is required to be done. Work done by one pipe per hour = 3 units. Work done by both the pipes per hour =

30 = 5 units. 6

So, work done by the second pipe = 5 – 3 = 2 units. Time taken by second pipe =

30 = 15 hours. 2

Example 20 An inlet pipe can fill a tank in 15 hours which an outlet pipe can completely empty in 60 hours. (i)

If both are open simultaneously, in how many hours would the tank be filled?

(ii)

If both are open at alternate hours and initially the tank is completely empty, in how many hours would the tank be filled?

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Quantitative Aptitude Simplified for CAT

Solution (i)

In 1 hour, the fraction of tank filled =

1 1 1   . 15 60 20

So, in 20 hours, the entire tank will be filled. Alternatively, LCM (15, 60) = 60 units of work is required to be done. The per hour work done by the inlet and outlet pipes = +4 and 1 respectively. 4 – 1 = 3 units of work is done in 1 hour; 60 units of work is done in 20 hours. (ii)

In the last hour, the inlet pipe will be working. So, by the hour prior to that, the fraction of tank that 1 14 56 should be filled = 1    . 15 15 60

In first 2 hours, fraction of tank filled =

1 1 3   . 15 60 60

3 54 th is filled in 2 hours, 18 cycles would mean 36 hours in which th would be filled. In the next two 60 60 4 1 3 hours, fraction of work done =   , 60 60 60

Since

Fraction of work done so far = Remaining = 1 

54 3 57   . 60 60 60

57 3 1 1 / 20 3   , which would be filled by inlet pipe in  hours. 60 60 20 1 / 15 4

Therefore, the entire tank can be completely filled in 38

3 hours. 4

Alternatively, LCM (15, 60) = 60 units. Per hour work by inlet and outlet pipes = +4 and 1 unit. By the penultimate hour, minimum 60 – 4 = 56 units of work should be over. In first cycle of 2 hours, work done = 4 – 1 = 3 units. So, in 18 cycles, that is 36 hours, work done = 54 units. Work done in later hours = 54 + 4 – 1 = 57 units, which will take 2 more hours. Now, work left = 3 units, which will be done by inlet pipe in Therefore, total time taken to fill the tank = 38

3 hours. 4

3 hours. 4

Example 21 Two inlet pipes can fill a completely empty tank in 8 hours and 10 hours respectively. An outlet pipe can completely empty the same tank in 12 hours. Initially, only the inlet pipes are open till the tank is half full. Then, the outlet pipe is also opened along with the inlet pipes. In total, how much time is taken to completely fill the tank, if initially the tank is completely empty?

198

Time and Work Solution In the first hour, the fraction of tank filled = Since

1 1 9   . 8 10 40

9 1 1 /2 20 2 is filled in 1 hour, would be filled in  2 hours. 40 2 9 / 40 9 9

Now, as per the question, in 1 hour, the fraction of tank filled = Since

1 1 1 15  12  10 17     . 8 10 12 120 120

17 1 1 /2 60 is filled in 1 hour, would be filled in  hours. 17 / 120 17 120 2

The total time taken to fill the completely empty tank =

20 60 880 115   5 hours. 9 17 153 153

Alternatively, LCM (8, 10, 12) = 120 units of work is required to be done. Per hour work of 2 inlet and 1 outlet pipes = 15, 12 and 10. When only inlet pipes are open, time taken to do 60 units of work =

60 60  . 15  12  10 17

Now, outlet pipe is also opened. Hence, time taken = Total time taken =

60 60  . 15  12 27

60 60 20 60 880     hours. 27 17 9 17 153

Example 22 In the previous example, let the inlet pipes be initially opened for some time t. After this, the outlet pipe is also opened alongside. The tank is filled in total 2t hours. When was the outlet pipe opened? Solution In the first t hours, the fraction of tank filled = In the next t hours, the fraction of tank filled =

t t  . 8 10 t t t   . 8 10 12

30 8 t t   t t t  Therefore,    2 hours.      1 or t = 11 11  8 10   8 10 12 

So, outlet pipe was opened after 2

8 hours. 11

Alternatively, Work = 120 units. Inlets pipes work for total 2t hours and outlet pipe works for t hours. So, 15 × 2t + 12 × 2t – 10 × t = 120 or t = So, outlet pipe was opened after 2

120 30 8  2 hours. 44 11 11

8 hours. 11

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Quantitative Aptitude Simplified for CAT

Example 23 Two tanks whose ratio of volumes is 1: 2 are supplied by 1 and 3 inlet pipes respectively. The pipe in the smaller tank can fill the tank completely in 10 hours. The radii of the 3 pipes in the larger tank are 1: 2: 3, where the smallest pipe is of same dimensions as the one in the smaller tank. If all the three pipes are opened, in how many hours would the larger tank be completely filled? Solution Since the smallest pipe of larger tank is of same dimensions as the pipe in the smaller tank, we can say that this smallest pipe of the larger tank would be able to completely fill the tank in 20 hours (as the tank is of double capacity). The ratio of radii = 1: 2: 3  Ratio of cross sectional areas = 1: 4: 9. It is very important for the student to note that the rate of flow of water through the pipes is directly proportional to the cross sectional areas and not to the radius. Since the second pipe is 4 times as large, it will take pipe will take

1 20 th as much time, that is, hours and the largest 4 4

20 hours to completely fill the tank. 9

Together, in 1 hour, the fraction of tank filled =  The entire tank can be filled in

1 4 9 14    20 20 20 20

20 3 1 hours. 14 7

Example 24 If 40 cows can graze grass in a pasture in 40 days and 30 cows can do it in 60 days, then 20 cows can graze the grass in how many days? Solution Cow-days in the first case is 40 40 = 1600 and in the second case is 30  60 = 1800. Since the cow-days is not the same, the amount of grass is changing and hence it is natural for us to assume that the grass is growing at a certain constant rate every day. Let the initial amount of grass on the land be ‘a’ and the growth rate of the grass per day be ‘b’. Then, 40  40 = a + 40b and 30  60 = a + 60b Solving these equations, we get: a = 1200 and b = 10. In case of 20 cows, if the number of days they take to graze is x, then 20x = 1200 + 10x  x = 120 days.

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Time and Work

PRACTICE EXERCISE 1.

A chemical plant has four tanks (A, B, C, and D), each containing 1000 litres of a chemical. The chemical is being pumped from one tank to another as follows:

4 hours

D.

5 hours CAT 2014

4.

From A to B @ 20 litres/minute From C to A @ 90 litres/minute From A to D @ 10 litres/minute From C to D @ 50 litres/minute From B to C @ 100 litres/minute

If a certain sum of money is just sufficient to pay A for 21 days, and the same amount of money is just sufficient to pay B for 28 days, then the same money is just sufficient to pay A and B together for how many days? A.

12 days

B.

7 days

From D to B @ 110 litres/minute

C.

14 days

Which tank gets emptied first, and how long does it take (in minutes) to get empty after pumping starts?

D.

15 days

A.

A, 16.66

B.

C, 20

C.

D, 20

D.

D, 25

CAT 2015 5.

CAT 2005 2.

C.

In Nuts And Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts. Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts? A.

130 minutes

B.

135 minutes

C.

170 minutes

D.

Amar takes twice as much time as Akbar and Anthony together; Akbar takes as much time as Amar and Anthony together. If Anthony takes 10 days to do the work, then how much time will Amar and Akbar will take together to do the same work? A.

1 day

B.

4 days

C.

3 days

D.

2 days CAT 2016

6.

In the marketing management course of an MBA programme, you and your roommate can complete an assignment in 30 days. If you are twice as efficient as your roommate, the time required by each to complete the assignment individually is A.

180 minutes

45 days and 90 days

B.

30 days and 60 days

C.

40 days and 120 days

D.

45 days and 135 days

CAT 2004 IIFT 2016 3.

If pipe A can fill a tank in 3 hours, pipe B in 2 hours and pipes A, B and C together in 1 hour, then pipe C alone can fill the tank in how many hours? A.

8 hours

B.

6 hours

7.

201

A tank is connected with both inlet pipes and outlet pipes. Individually, an inlet pipe can fill the tank in 7 hours and an outlet pipe can empty it in 5 hours. If all the pipes are kept open, it takes exactly 7 hours for a completely filled-in tank to empty. If the

Quantitative Aptitude Simplified for CAT

11. 12 men can complete a work in 10 days. 20 women can complete the same work in twelve days. 8 men and 4 women started working and after nine days 10 more women joined them. How many days will they now take to complete the remaining work?

total number of pipes connected to the tank is 11, how many of these are inlet pipes? A.

2

B.

4

C.

5

D.

6 IIFT 2015

8.

Three carpenters P, Q and R are entrusted with office furniture work. P can do a job in 42 days. If Q is 26% more efficient than P and R is 50% more efficient than Q, then Q and R together can finish the job in approximately: A.

11 days

B.

13 days

C.

15 days

D.

17 days

33.33%

B.

50%

C.

66.67%

D.

None of these

B.

5 days

C.

8 days

D.

10 days

12. Four two-way pipes A, B, C and D can either fill an empty tank or drain the full tank in 4, 10, 12 and 20 minutes respectively. All four pipes were opened simultaneously when the tank is empty. Under which of the following conditions the tank would be half filled after 30 minutes?

A mother along with her two sons is entrusted with the task of cooking Biryani for a family get together. It takes 30 minutes for all three of them cooking together to complete 50 percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after 1 hour and the younger son cooks for further 3 hours. If the mother needs 1 hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour? A.

2 days

IIFT 2012

IIFT 2015 9.

A.

A.

Pipe A filled and pipes B, C and D drained

B.

Pipe A drained and pipes B, C and D filled

C.

Pipes A and D drained and pipes B and C filled

D.

Pipes A and D filled and pipes B and C drained

E.

None of the above XAT 2017

13. A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 8 hours while an outlet pipe can empty the full tank in 12 hours. If all pipes are left open simultaneously, it takes 6 hours to fill the empty tank. What is the relationship between M and N?

IIFT 2013

A. 10. Capacity of tap Y is 60% more than that of X. If both the taps are opened simultaneously, they take 40 hours to fill the tank. The time taken by Y alone to fill the tank is A.

60 hours

B.

65 hours

C.

70 hours

D.

75 hours

M:N=1:1

B.

M:N=2:1

C.

M:N=2:3

D.

M:N=3:2

E.

None of the above XAT 2016

14. Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty cone in 8

IIFT 2013

202

Time and Work hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

15. Albela, Bob and Chulbul have to read a document of 78 pages and make a presentation next day. They realize that the article is difficult to understand and they would require team work to finish the assignment. Albela can read a page in 2 minutes, Bob in 3 minutes, and Chulbul in 4 minutes. If they divide the article into 3 parts so that all three of them spend the equal amount of time on the article, the number of pages that Bob should read is

A.

Pipe A was open for 19 hours

B.

Pipe A was open for 19 hours 30 minutes.

A.

24

B.

25

C.

Pipe B was open for 19 hours 30 minutes

C.

26

D.

27

D.

Pipe C was open for 19 hours 50 minutes.

E.

28

E.

The situation is not possible.

XAT 2013 XAT 2015

ANSWER KEY 1. C

2

C.

3. B

4. A

5. D

6. A

7. D

9. B

10. B

11. A

12. A

13. E

14. C

15. A

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8. B

Quantitative Aptitude Simplified for CAT

5.

ANSWERS AND EXPLANATIONS 1.

Explanation: Let time taken by Amar and Akbar be x and y days. Then, 20 y  10 y  x = 2 .   10  y 10 y  

Answer: C Explanation: Tank A, (90 – 20 – 10) = 60 litres/min fill Tank B, (110 – 100 + 20) = 30 litres/min fill Tank C, (100 – 90 – 50) = − 40 litres/min fill or 40 litres/min. empty. Tank D, (10 + 50 − 110) = − 50 litres/min. fill or 50 litres/ min. empty. Tank D gets emptied first. Then, total emptying time of D =

2.

Similarly,  20 y  10    10  y   200 y  20 y  100  30 y 10     10  y  10  100 + 30y = 200  y = and x 3 10  20    3  5. = 10 10  3 Time taken by Amar and Akbar together 10  5   =  3   2 days. 10 5 3 10 x  y =     10  x 

1000  20 min. 50

Answer: C Explanation: First machine produces 1000 nuts in 10 + 5 = 15 min or 2000 nuts in 30 min. Second machine produces 1500 bolts in 20 + 10 = 30 min. Total 1500 nuts in 30 min. So, 9000 nuts in 180 – 10 = 170 min (we have to subtract last 10 min of rest). In these 170 min, we would have produced more than 9000 bolts. So, 9000 pairs of nuts and bolts are produced in minimum 170 min.

3.

6.

Answer: A Explanation: Use options. Option [A]: LCM (45, 90) = 90 units. So, work done per day by them = 2 units and 1 unit. Therefore, work done together = 2 + 1 = 3 90 = 30 units. So, time taken for 90 units = 3 days. Alternatively, If “you” are twice as efficient as your roommate, the roommate will take double the time taken by you. Hence, options 3 and 4 can be eliminated. Further, since time taken by two persons is 30 days, no one can alone do the work in 30 days. So, [B] option is also ruled out.

Answer: B Explanation: LCM (3, 2, 1) = 6 units of work is required to be done. Per hour work done by A, B and “all three together” = 2 units, 3 units and 6 units. So, work done per hour by C = 6 – 2 – 3 = 1 unit. Therefore, time taken by pipe C alone to fill the tank is 6 hours.

4.

Answer: D

Answer: A Explanation: Let the amount of money = LCM (21, 28) = Rs.84. Money charged by A per day = Rs.4 and that charged by B = Rs.3. When they work together, charges per day = 4 + 3 = Rs.7. So, number of days for which payment can be 84 made to A + B = = 12 days. 7

7.

Answer: D Explanation: Let there be x inlet and y outlet pipes. Therefore, x + y = 11. LCM (5, 7) = 35 units of work is required to be done. When all the pipes are open, work is done in 7 hours. So, work done per hour = 5 units.

204

Time and Work Work done by each inlet pipe = 5 units and that done by outlet pipe = 7 units. So, work done by x inlet and y outlet pipes = 7y – 5x (assuming emptying the tank is positive work) Therefore, 7y – 5x = 5  7(11 – x) – 5x = 5  x = 6. 8.

=

11. Answer: A Explanation: 12m × 10 = 20w × 12  1m = 2w. Now, 8m + 4w = 8m + 2m = 10m. In 9 days, work done = 10m x 9 = 90m. Remaining work = 120m – 90m = 30m. Since 10 women joined, effectively 5 men joined. So, number of men now is 15. If they take x days to complete the remaining work, 30m = (15m)x  x = 2 days.

Answer: B Explanation: Let the amount of work required to be done is 42 units. So, work done by P in 1 day = 1 unit. That done by Q in 1 day = 1.26 and that done by R = 1.5 × 1.26 = 1.89. When Q and R work together, work done in 1 day = 1.26 + 1.89 = 3.15. Time taken to complete the work 42 40 1   13 days. = 3.15 3 3

9.

260  40 = 65 hours. 160

12. Answer: A Explanation: Let the capacity of the tank = LCM (4, 10, 12, 20) = 60. So, work done by A, B, C and D = 15, 6, 5, 3 respectively. Option [A]: Work Done in a minute = 15 – 6 – 5 – 3 = 1. So, after 30 minutes, the tank will be half filled.

Answer: B Explanation: Let total work be 100 units. Since mother is faster than elder son, and in 1 hour, the work is completed when all 3 work together, percentage of work done by mother can’t be 33.33%. If mother does 50% work in an hour, then she can do 100% work in 2 hours. So, elder son would take 3 hours to complete the work alone. When both the sons are working, elder one works for only 1 hour and he does 33.33% work. The remaining 66.66% work is done by younger son in 4 hours. So, younger son can do complete work alone in 6 hours. If mother, elder son and younger son can do the work alone in 2 hours, 3 hours and 6 hours, then they can complete the work together in 1 hour. This matches with the information given.

13. Answer: E Explanation: Let the total work required to fill the tank be 24 units. The work done by an inlet pipe and an outlet pipe per hour is 3 units and 2 units respectively. Now, 6(3M – 2N) = 24  3M – 2N = 4. This equation has infinite solutions. When N = 1, M = 2; when N = 4, M = 4; and so on. 14. Answer: C Explanation: Let the capacity of the tank be LCM (8, 12, 4) = 24 units. Work done per hour by the three pipes = 3 units, 2 units and 6 units. When height remains 50%, radius also 24 becomes 50% and so volume becomes 8 = 3 units and so in 20 hours, 21 units of work was done. Now use options. Option [A]: If A was open for 19 hours, then B and C was open for 20 hours. Work done by them = 19 × 3 + 20 × 2 – 20 × 6 = –23. This is not possible.

10. Answer: B Explanation: Tap X does 100 units of work in 1 hour. So, tap Y does 160 units of work in an hour. When they work together, total work done in 1 hour = 260 units. So, in 40 hours, total work done = 260 × 40 units. So, time taken by Y alone to fill the tank

205

Quantitative Aptitude Simplified for CAT

15. Answer: A

Option [B]: If A was open for 19.5 hours, then B and C was open for 20 hours. Work done by them = 19.5 × 3 + 20 × 2 – 20 × 6 = –21.5. This is not possible. Option [C]: If B was open for 19.5 hours, then A and C was open for 20 hours. Work done by them = 20 × 3 + 19.5 × 2 – 20 × 6 = –21, that is emptying work of 21 units is done.

Explanation: LCM (2, 3, 4) = 12. In 12 minutes, number of pages read by Albela, Bob and Chulbul are 6, 4 and 3. So, together they have read = 6 + 4 + 3 = 13 pages. So, to read 78 pages, they would need 12 × 6 = 72 min. Therefore, in 72 72 min, number of pages read by Bob = 3 = 24 pages. .

206

Time, Speed and Distance

Chapt er 06

Time, Speed and Distance INTRODUCTION In this chapter, we will deal with all the aspects of “time, speed and distance” including the concept of “clocks”, which is an application of “circular motion”. We all know the basic formula of speed, that is, v=

d t

where, v = speed of an object, d = distance travelled by that object, and t = time taken in this process. Therefore, the units of speed depend upon the units chosen for distance and time. Usually, the units of speed are expressed in km/hr or m/s. Students are expected to know how to convert one system of units to the other system of units. Therefore, 1 km/hr means that in one hour, a distance of 1 km is travelled. In other words, 1 km/hr =

1 km 1000m 5   m/s . 1 hour 60  60 sec 18

This means that 18 km/hr = 5 m/s or more importantly, 36 km/hr = 10 m/s. Therefore, 72 km/hr = 20 m/s.

Average Speed The concept of average speed is a fundamental concept that should be clear to students. Let us say, Shyam travels from city A to city B at a certain speed, and then travels from city B to C at another speed. In doing so, he takes some time. If instead he had gone from A to B to C at a “constant” speed such that the total time taken was the same in all cases, then that “constant” speed would have been called average speed. Therefore, average speed is defined as the total distance travelled divided by he total time taken, that is, Average Speed =

Total Distance Total Time

If Shyam goes from A to B at a speed of u and B to C at a speed of v, where

207

Quantitative Aptitude Simplified for CAT

distance AB = x and distance BC = y, then average speed =

xy . x y  u v

There are numerous scenarios possible in such cases. If distances in different parts of the journey are equal, then average speed becomes =

2x 2uv ,  uv  x  x   u v

which is the harmonic mean of individual speeds. It is important to realize that the average speed here is ‘independent’ of the distance AB (or BC). We can say that “if distance in two parts of the journey is same, then average speed is the harmonic mean of the individual speeds in the two parts”. We can extend this case to more scenarios. For example, if Shyam goes from A to B to C to D, where AB = BC = CD, at the respective speeds of u, v and w, then average speed is obtained as Average Speed =

Total Distance 3x 3   . x x x 1 1 1 Total Time     u v w u v w

Thus, in the general case, “if distance in n different parts of the journey is same, then average speed for the whole journey is the harmonic mean of the n individual speeds in these n parts”, that is, Average Speed =

n  1 1 1    .....n terms u v w

An important aspect to be careful about here is that Shyam should not stop at any of the cities! Otherwise, the concept of harmonic mean cannot be applied, as stoppage at any of the cities will increase the total travel time. Example 1 If Shyam goes from his home to office at a speed of 40 km/hr and immediately returns home at a speed of 60 km/hr, what is the average speed? Solution Since distance between home and office is constant for both parts of the journey, we can apply the concept of harmonic mean. Therefore, Average Speed =

2uv 2  40  60   48 km/hr. u v 40  60

Example 2 There is a race between Chintu and Pintu. They have to run around a park (which is square shaped) and come back at the starting point. Chintu runs AB at a speed of 5 m/s, BC at a speed of 10 m/s, CD at a speed of 8 m/s and DA at a speed of 4 m/s. Pintu maintains a constant speed of 7 m/s. Who wins the race? Solution For Chintu, since AB = BC = CD = DA (as the park is square shaped), we should find out the average speed of Chintu using the concept of Harmonic mean. Average speed of Chintu =

4 160  m/s. 1 1 1 1 27    5 10 8 4

208

Time, Speed and Distance Since the speed of Pintu is 7 m/s, which is more than the average speed of Chintu, we can say that Pintu won the race. Now, if Shyam goes from A to B at a speed of u and B to C at a speed of v, where time taken from A to B = time taken from B to C, then average speed =

Total Dis tan ce ut  vt u  v   , Total Time tt 2

which is the Arithmetic Mean of the individual speeds. Here, the average speed is independent of times taken in the individual parts of the journey. We can say that “if time taken in two parts of the journey is same, then average speed is the arithmetic mean of the individual speeds in the two parts”. We can extend this case to more scenarios. For example, if Shyam goes from A to B to C to D, where TAB = TBC = TCD, at the respective speeds of u, v and w, then average speed is obtained as Average Speed =

Total Distance ut  vt  wt u  v  w   . Total Time 3t 3

Thus, in the general case, “if time taken in n different parts of the journey is same, then average speed for the whole journey is the arithmetic mean of the n individual speeds in these n parts”, that is, Average Speed =

u  v  w  .....n terms . n

Example 3 Shyam goes from home to office at a speed of 40 km/hr. From office, he goes to the airport at a speed of 60 km/hr to see off his boss. From the airport, he goes back home at a speed of 45 km/hr. He does not stop either at the office or at the airport. Upon reaching home, he realizes that he took as much time to reach airport from office as from home to office, and as much time to reach home from airport as from office to airport. At what constant speed could he have travelled throughout such that he reaches home at the same

time as he reached in the given case? Solution In this, the time taken in the different parts of the journey is same, and hence we can apply the concept of Arithmetic Mean of the speeds. Therefore, the average speed =

u  v  w 40  60  45 145 1    48 km/hr. 3 3 3 3

Example 4 Chintu and Pintu are having a race in which they have to go from Delhi to Chandigarh and come back without stopping anywhere at all. Karnal is exactly midway between Delhi and Chandigarh. The average speed of Chintu from Delhi to Karnal is 30 km/hr, from Karnal to Chandigarh is 40 km/hr and from Chandigarh to Delhi is 45 km/hr. The average speed of Pintu from Delhi to a restaurant (situated somewhere between Delhi and Chandigarh) is 25 km/hr (Pintu doesn’t stop here) and from here to Chandigarh to Delhi (non-stop) is 50 km/hr, taking double the previous time. Who won the race? Solution Let distance between Delhi to Karnal be x. Then, average speed of Chintu for the whole journey =

4x = 38.9 km/hr. x x 2x   30 40 45

209

Quantitative Aptitude Simplified for CAT

If time taken by Pintu from Delhi to that restaurant is t, then the time taken for the rest of the journey is 2t. Therefore, the average speed of Pintu is =

25t  50(2t) 125 2   41 km/hr. t  2t 3 3

Therefore, Pintu won the race. If distance in the different parts of the journey is not same, and the time taken in these parts is also not same, then neither of the above 2 concepts can be applied. But, the fundamental concept of average speed can always be applied, that is, Average Speed =

Total Distance Total Time

Therefore, if Shyam goes from A to B to C to D at speeds of u, v and w resp., taking times t1, t2 and t3 resp., then the average speed is given by Average Speed =

ut 1  vt 2  wt 3 t1  t 2  t 3

whereas if instead of times, distances d1, d2 and d3 are mentioned, then Average Speed =

d1  d2  d 3 d1 d2 d 3   u v w

Applications of Ratio and Proportion d . If time is constant, then speed is directly proportional to t d  1 . d2

As we have seen in the beginning that speed, v = distance, that is, v  d. This also means that

v1 v2

If instead distance were kept constant, then speed becomes inversely proportional to time, that is, v 

v t 1 . This means that 1  2 or v1t1 = v2t2. v 2 t1 t

Example 5 A man goes from home to office every day. One fine day, due to heavy traffic, his average speed of going to office reduces by 20%. By how much percent will his travel time increase? Solution Method 1: Let v be the usual speed and t be the usual time of going to office.

20   The new speed =  1   v = 0.8v. 100   r   The new time =  1   t , where r is the percent increase. 100   r   Since distance is constant, v1t1 = v2t2 or vt = (0.8v)  1  t . 100  

210

Time, Speed and Distance

 1  1   100  25% Therefore, r =  .   0.8  Method 2: This is a situation where the distance is constant (from home to office). Therefore, speed is inversely proportional to time. Since speed decreases by 20%, that is,

1 th , we can say that 5

1 4  the speed becomes  1   th, that is, th. 5 5  This means that time taken would become

5 4

of the usual time. Therefore, it takes

th

time, which implies his travel time increases by

1 4

th

1 4

th

more than the usual

or 25%.

Example 6 Payal travels from home to hospital every day to see her husband who is hospitalized because of an accident. On one of the occasions, her speed decreases by 25% and hence she is late to the hospital by 20 minutes. What is the time Payal usually takes on days when she is not late? Can you find the speed with which she usually travels? Solution This example is somewhat similar to the previous example in the sense that the distance is constant. So, once again, we can make use of the result that the speed is inversely proportional to the time taken. Since speed 3 4 1 decreases by 25%, the speed becomes th, the time taken would become rd. This means that she takes 4 3 3 rd of the usual time over and above the usual time. Thus,

1 3

rd

of the usual time is the time by which she is late, which is given to be 20 minutes. Therefore, if ‘t’ is

the usual time, then

1 t = 20 or t = 60 minutes. 3

The usual travel time of Payal is 60 minutes from home to hospital. This also means that since today she is late by 20 minutes, she reaches the hospital in 1 hour 20 minutes. The usual speed of Payal cannot be found out, since the distance from home to hospital is not given. Example 7 One day Payal has to reach the hospital in 45 minutes. How much faster than usual should she drive to reach in time? Solution Now, the usual time is 60 minutes and today, the time allowed is 45 minutes. This means that the time should 1 3 decrease by 15 minutes or th of the usual. That is, the time taken now is th of the usual. 4 4 Therefore, the speed should be

4 3

rd

of the usual, that is,

1 3

rd

extra. So, Payal should drive 33

1 % faster than 3

usual to reach in time. In the above examples, we have seen that ratios and fractions make the process very convenient. In the example below, the above method as well as the traditional method have been given. The students are advised to learn both the methods so that depending upon the needs and requirements of the question, they may

211

Quantitative Aptitude Simplified for CAT

apply the appropriate method in the time limit given to them. Some students learn only one of the methods and become comfortable with it. This limits the scope and reduces the speed of the students. Example 8 Payal (from the above example) has to reach the hospital in 45 minutes but has to pick up her children from school which lies midway between her home and the hospital. She travels 20% faster than usual, from home to school and picks up the kids. How much faster than the usual speed should she drive for the rest of the distance so as to be in time? Solution Method 1: If she drives at normal speed, she would reach the school in 30 minutes. But, since her speed is 6 5 20% faster than normal, her speed becomes th of the normal and hence the time taken becomes th of the 5 6 1 th normal time, that is, less. Therefore, she reaches the school in 25 min.Now she will reach the hospital in 6 30 minutes if she travels at normal speed, but she has to reach in 20 minutes, so that overall time taken is 45 20 th minutes (assuming that the time of picking up the kids is nil). Reaching in 20 minutes means taking , that 30 2 is, rd of the normal time. 3 3 1 time of the normal speed, that is, of the normal speed more than the normal 2 2 speed. Therefore, she should travel at a speed 50% faster than normal speed so that she reaches the hospital in time.

So, the speed should be

Method 2: Let v = usual speed and t1 = time taken from home to school. For the journey between home to school, 1 (v  60) = (1.2v)t1 or t1 = 25 minutes. 2

For the journey between school to hospital, 1 (v  60) = 2

1  r    v  (45 – 25)  r = 50%. 100  

Relative Speed The concept of relative speed is one of the most important concepts to be learnt. If two persons are moving in the same or opposite direction, then the speed of one of the persons as recorded (or observed) by the other person is referred to as ‘relative speed’ of one with respect to the other. The value of relative speed depends upon the relative directions in which the persons are moving. So, if A and B are moving in the same direction at the speeds of v1 and v2 respectively, then speed of B with respect to that of A would be v2 – v1. Let the speed of A be 20 km/hr and that of B be 30 km/hr, and they are moving in the same direction, then the speed of B relative to that of A is 30 – 20 = 10 km/hr. What will be the speed of A relative to B? It should be 20 – 30 = –10 km/hr. Though theoretically, negative speeds have their own meanings, for all practical purposes, we will consider only those cases where relative speed of faster person with respect to that of the slower one is asked. If the two persons are moving in opposite directions, then two cases arise:

212

Time, Speed and Distance (i)

both are moving towards each other, or

(ii)

both are going away from each other.

Observing more closely, we can infer that these are not different cases. This is so because when they are moving towards each other, then upon meeting if they continue in their directions of motion, they would be seen to be moving away from each other. So, when two persons are moving in opposite directions (whether towards or away), the relative speed is given by v1 + v2. In the above example, if speed of A is 20 km/hr and that of B is 30 km/hr and they are moving in opposite directions, then the relative speed between them is given by 20 + 30 = 50 km/hr. Let us try to interpret the meaning of the relative speed. When we say that the relative speed between A and B is 50 km/hr, it only means that the gap between A and B is changing at the rate of 50 km every hour. Let us say that police running at a speed of 30 km/hr chase a thief who is running at a speed of 20 km/hr. Then, the relative speed between them is 30 – 20 = 10 km/hr. This speed of 10 km/hr means that the gap between the police and thief is decreasing at the rate of 10 km every hour. Same is true if they were moving in opposite direction. Therefore, when they are moving in opposite direction, the relative speed would be 20 + 30 = 50 km/hr and the gap between them would be increasing at the rate of 50 km every hour. Observe the following example. Example 9 A thief escapes a jail and runs at a speed of 40 km/hr. After 2 hours, the police come to know about the escape and starts chasing the thief at a speed of 50 km/hr. (i)

How long did the thief run before he gets caught?

(ii)

How far from the jail is the thief caught?

(iii) If the thief increases his speed to 45 km/hr after having run for 6 hours, when would the thief be caught from the time he increased his speed? (iv) At what speed should the police chase to nab the thief in 5 hours. (v)

After how much time should the police realize the escape so as to nab the thief in 5 hours?

Solution Since the thief runs for 2 hours, the gap between the thief and police = 2  40 km/hr = 80 km. (i)

Police start chasing the thief at the speed of 50 km/hr. Since the police and thief are moving in the same direction, the relative speed between the two would be = 50 – 40 = 10 km/hr.

The gap of 80 km would be covered at the relative speed of 10 km/hr in

80 km  8 hours 10 km/hr

Since the police runs for 8 hours, the thief would have run for 10 hours (including the first 2 hours during which only thief was running). To make matters easier, it is best to make the following assumption if we want to find the ‘time’ of meeting in case two persons are running. Assumption: In this assumption, we make the slower person rest and make the faster one move at the relative speed. Here, slower one is thief (speed = 40 km/hr). So, we will assume that the thief is not moving at all after the gap of 80 km is created, that is at the time when both police and thief are running. And we make the police move at the speed of relative speed, that is 50 – 40 = 10 km/hr. Now, if we interpret the question again, we realize that the police are moving towards thief at the speed of 10 80 km km/hr and thief is 80 km away, not moving at all. So, the police will catch the thief in  8 hours 10 km/hr

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(ii)

When the thief is caught, the police would have run for 8 hours @ 50 km/hr and hence the distance travelled by police is 8  50 = 400 km. This is the distance between the jail and the place where the thief is caught.

(iii) When the thief has run for 6 hours, he would have travelled a distance of 6 40 = 240 km. The police would have run for 4 hours and the distance the police travel is 4  50 = 200 km. This means that the gap between police and thief is 240 – 200 = 40 km. The speed of thief now changes to 45 km/hr and hence the relative speed becomes 50 – 45 = 5 km/hr. The time taken by the police to catch the thief =

40 km  8 hours . 5 km/hr

Therefore, the time taken by the police to catch the thief after the thief has run for 6 hours is 8 hours. (iv) Let the speed of the police be v. Then, the relative speed between them is (v – 40) km/hr. The gap between the police and the thief is 80 km and the time to be taken is 5 hours. Therefore, (v)

80  5 v = 16 + 40 = 56 km/hr. v  40

Let the police realize the escape after t hours. Then, the thief would have travelled a distance of 40t. Now,

Time of meeting = 5 hours =

40t or t = 1.25 hours = 1 hour 15 min. 10

EXPERT SPEAK Scan this QR Code to watch a video that explains how to understand and solve problems based on Relative Speed.

Example 10 Ram and Shyam are standing on the opposite ends of a swimming pool of length 100 m. They start swimming towards each other at the respective speeds of 4 m/s and 6 m/s. When and where would they meet? Solution The gap between the persons swimming is 100 m. The relative speed = 4 + 6 = 10 m/s (as they are moving in opposite directions). Now make the slower one (Ram) rest and faster one (Shyam) move at the relative speed of 10m/s. For Shyam to meet Ram, he has to travel a distance of 100 m at a speed of 10 m/s. Therefore, the time of meeting =

100 m  10 sec. . 10 m/s

In these 10 sec, Ram would have travelled 4  10 = 40 m and Shyam would have travelled 6  10 = 60 m. So they meet at a place which is 40 m from where Ram started or 60 m from where Shyam started. The ratio of the speeds of Ram and Shyam is 4 : 6 = 2 : 3; and the ratio of the distances they travel (respectively) is 40 : 60 = 2 : 3. Note that the ratio of speeds is same as the ratio of distances because time of

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Time, Speed and Distance travel for both the persons is same. Recall that if time is constant, then speed is directly proportional to v d distance, that is, 1  1 . v 2 d2 When Ram meets Shyam, each would have travelled for the same time, that is, they both travel for 10 2 seconds. Since ratio of distances travelled is 2 : 3, and total distance is 100 m, Ram travels  100 = 40 m 5 3 and Shyam travels  100 = 60 m. It is important to be understand that this is true if the start time of both 5 the parties is same. If it is not the case, then the one who starts earlier is allowed to move till the other person starts moving. So, in this method, we do not require travel time to find the distance travelled by the persons. Example 11 Ram and Shyam start from A and B respectively. and move at the speeds of 20 m/s and 15 m/s. If A and B are separated by a gap of 40 m, after how much time will they have a gap of 320 m, if they are moving away from each other? Solution Make the slower one (Shyam) rest and faster one (Ram) move at the relative speed, that is 20 + 15 = 35 m/s. They already have a gap of 40 m. To create a gap of 320 m, Shyam has to run 320 – 40 = 280 m. So, time taken =

280 280   8 sec . 20  15 35

Example 12 Let Ram and Shyam be 70 km apart and they are moving in same direction at speeds of 40 km/hr and 30 km/hr, with Ram chasing Shyam. In how much time will the gap between them be 10 km? Solution Make Shyam rest and make Ram move at the relative speed = 40 – 30 = 10 km/hr. To create a gap of 10 km, 60 = 6 hours. Ram has to run 60 km. So, time taken = 10 Note that this is not the only answer! The gap will also be 10 km when Ram has met Shyam and further travelled a distance of 10 km. In that case, distance travelled by Ram = 70 + 10 = 80 km (with Shyam not moving at all). In that case, time taken =

80 = 8 hours. 10

Example 13 Relative speed of Ram with respect to Shyam is 30 m/s and that of Shyam with respect to Rahim is 25 m/s. What is the relative speed of Ram with respect to Rahim? Solution Since direction of motion of the persons is not given, the required relative speed cannot be found out. Let us take two cases to understand this. Case 1: If all are moving in the same direction and the speeds of Ram, Shyam and Rahim are v1, v2 and v3, then

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v1 – v2 = 30 and v2 – v3 = 25. Therefore, v1 – v3 = (v1 – v2) + (v2 – v3) = 30 + 25 = 55 m/s. Case 2: If Ram and Rahim are moving in the same direction, but Shyam is moving in a direction opposite to that Ram and Rahim, then v1 + v2 = 30 and v2 + v3 = 25. Therefore, v1 – v3 = 30 – 25 = 5 m/s. We can apply the concept of relative speeds to many cases such as: train vs train, train vs platform, train vs pole, train vs man, man inside a train, boat in river, aero plane flying in the air, and so on. If we are dealing with cases of trains, care must be taken to consider the length of the train if it is substantial compared to the distances involved in a particular question. Let us study the following examples to understand concepts discussed above. Example 14 Two trains of length 250 m and 150 m are moving towards each other on parallel tracks such that in the beginning, their engines are drawn level. If the speeds of the trains are 36 km/hr and 54 km/hr respectively, then (i)

in how much time would the two completely pass each other?

(ii)

in how much time would the mid-points of the two just pass each other?

How will your answers change if the trains were moving in the same direction and the engine of the faster train is parallel to the tail end of the slower train? Solution The speeds of the trains should first be converted to the units of m/s. 36 km/hr = 10 m/s; and 54 km/hr = 15 m/s. (i)

The trains travel a total distance which is the sum of the lengths of the trains, when they completely pass each other.

Therefore, the total distance travelled = l1 + l2 = 250 + 150 = 400 m. Relative speed = 10 + 15 = 25 m/s. The time taken to completely pass each other = (ii)

400  16 sec . 25

Here, the distance travelled by the trains would become

Therefore, time taken =

1 (250 + 150) = 200 m. 2

200 = 8 sec. 25

If the two trains were running in the same direction, the only change that would happen is in relative speed, which would now be 15 – 10 = 5 m/s. Therefore, the answer to (i) would be and the answer to (ii) would be

400  80 sec 5

200  40 sec. 5

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Time, Speed and Distance Example 15 If the engines of the above trains were initially separated by a gap of 0.5 km, then (assuming that they are moving in opposite direction, (i)

in how much time would they completely pass each other?

(ii)

in how much time would they collide if they were on the same track?

Solution (i)

For them to completely pass each other the distance travelled by them would increase by 500 m, that is, the distance travelled now becomes 400 + 500 = 900 m.

Time taken to completely pass each other = (ii)

900  36 sec. 25

For them to collide they only have to travel 500 m and the lengths of the trains would definitely not be 500 added here. Therefore, time taken =  20 sec. 25

Example 16 If train A takes 10 sec to completely pass a 200 m long platform and 8 sec to completely pass a man who is running at a speed of 36 km/hr in the same direction as the train, then how long will the train take to completely pass another train B whose length is 250 m and which is running at a speed of 72 km/hr in the opposite direction as the first train? Solution Let the length of the train be x. When it passes the platform, the speed of the train comes out to be x  200 m/s. 10 x  200 – 10, because the speed 10 of the man is 36 km/hr = 10 m/s, and train and man are moving in same direction.

When it passes a man, then relative speed between the man and the train =

Therefore,

x  200 x x  , as is also the relative speed between the man and the train. 10 8 8

 8x + 800 = 10x or x = 400 m. Hence the speed of the train A is 60 m/s. The speed of train B is given as 72 km/hr = 20m/s. Since the two trains are moving in opposite directions, time taken for them to completely pass each other =

400  250 1  8 sec. 60  20 8

Upstream and Downstream Motion This is an application of the concept of relative speed, though it is different from the cases discussed so far. This involves cases like a boat moving in a river, aero plane flying in air, man running inside the train or even a man running up an escalator which is also moving. Let us take a simple case of a boat in a river and let us assume that the river water is stationary. The boat has an attached engine with the help of which it can move at a speed of v. If the river water now starts running at a certain speed, say u, then the speed of the boat will get altered depending upon the direction (with respect to water) in which it moves. If it moves along the river water, the speed of the boat would be aided by that of the

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water, whereas if it were moving against the flow of river, the speed of the boat would be reduced and it just might happen that the boat would be seen to be not moving at all if the speed of the boat becomes same as that of water and the movement of the boat is against that of the river! All this can be summed-up as: If the boat moves along the river water, the motion is called as downstream motion and the speed of the boat becomes v + u. If the boat moves against the river water, the motion is called as upstream motion and the speed of the boat becomes v – u. So, downstream speed, D = v + u and upstream speed, U = v – u. Here, we will assume that v is more than u. Otherwise, the boat will be seen to be moving along the stream though it is trying to move upstream. Adding the two equations, we get D + U = (v + u) + (v – u) = 2v  v =

DU . 2

Similarly, subtracting the second equation from the first, we get D – U = (v + u) – (v – u) = 2u  u =

DU . 2

So, if we know the speed of the boat in still water (v) and that of the stream (u), then we can find out the downstream (D) and upstream speed (U), whereas if we know the downstream and upstream speed, then we can find out the speed of the boat in still water as well as that of the stream. The concept discussed above can be equally applied to “aero plane in the air”, “man running in the train”, and so on. Example 17 In a river, there are 3 points along its length: A, B and C such that A is somewhere between C and B. It is also given that CA = 10 km and AB = 20 km. A boat starts from A and reaches B in 4 hours, whereas it goes from B to C in 5 hours. Find the speed of boat in still water and also speed of the stream? Solution Speed of boat from A to B =

20 30 = 5 km/hr and speed from B to C = = 6 km/hr. 4 5

Therefore, we can interpret that A to B direction is upstream direction and B to C is downstream direction. So, D U 6 5 D U 6 5 speed of boat in still water =  = 5.5 km/hr, and speed of stream  = 0.5 km/hr. 2 2 2 2 Example 18 A swimmer can swim in running water from A to B in 5 hours and B to A in 8 hours. (i)

If the water in the river were stationary, then how much time will the swimmer take for the round trip?

(ii)

In the above example, if there is a paper boat placed at A, in how much time will it reach B?

(iii) If we increase the speed of stream by some amount (the increased speed still being less than that of the swimmer), then will the overall trip time of the swimmer increase or decrease or remain the same?

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Time, Speed and Distance Solution (i)

Since the time taken from A to B is less than that taken from B to A, we can conclusively say that the direction A to B is downstream whereas B to A is upstream. Let the distance between A and B be x. Then, x x Downstream Speed, D = and Upstream Speed, U = . 5 8 If t is the time taken by the swimmer in going from A to B in still water, then x Speed of the swimmer in still water = . t x x  x 5 8 2  5  8 80 DU   t  . Therefore, v = 2 t 2 58 13 160 4 For the round trip, the time taken would be twice of that, that is,  12 sec. 13 13

(ii)

In cases of paper boat or log of wood or even a cap, the object would move at the speed of the stream, as it cannot have any speed of its own. So, in this example, what is needed is to calculate the speed of the stream and hence the time. If the time taken by the paper boat in going from A to B is t, then

x x  x DU 5 8 u   t 2 2 2  8  5 80 2 or t =   26 hours. 85 3 3 (iii) Let the speed of swimmer in still water and that of stream be v and u. If x is the distance between A and B, then overall trip time, T will be x x 2vx T=   2 . v  u v  u v  u2 We observe that if u increases, the denominator decreases and so T increases. So, we can say that the faster the speed of stream, the more will the overall trip time be. As we will keep on increasing the speed of stream, there will come a time when the speed of the stream becomes equal to that of the swimmer, and in that situation, the swimmer cannot move upstream at all, that is he will take infinite time to come back, which is also evident from the above formula. Example 19 A boat travels a distance of 28 km downstream in 4 hours. Once the boat starts the upward journey, the water in the river flows at double the speed because of which the boat is able to reach the place from where it started, in 14 hours. What is the speed of the stream now? Solution Downstream speed =

28  7 km/hr. 4

If the speed of the boat in still water is v, and that of the stream is u, then v + u = 7 and v – 2u =

28  2 km/hr. 14

Eliminating v in the above equations,

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3u = 7 – 2 = 5  u =

5 2 1 . 3 3

Therefore, the speed of the stream now = 2 

5 10 1   3 km/hr. 3 3 3

Example 20 A train of length 550 m is running eastwards at a speed of 108 km/hr. Two persons are standing inside the train, one near the engine and the other at the tail end of the train. They start moving towards each other in the running train at the speeds of 40 m/s and 15 m/s with respect to a man standing on the platform. In how much time will they meet each other? Solution This is an excellent application of the concept of relative speed. The speed of the man at the tail end with respect to the man on the platform is 40 m/s. But, the train itself is running at a speed of 30 m/s with respect to the same man on the platform. Therefore, the speed of the man (at the tail end) with respect to the train = 40 – 30 = 10 m/s. And the speed of the man (in the engine) with respect to the man on the platform = 15 m/s and therefore, the relative speed of the man (in the engine) with respect to the train = 30 + 15 = 45 m/s. The relative speed between the two men in the train = 10 + 45 = 55 m/s, as they are moving towards each 550 other. Therefore, the time taken by them to meet =  10 sec. 55 Note that while calculating the relative speed of the men, the train would be considered to be stationary! Also note that the relative speed between the men with respect to train is same as the relative speed of the men with respect to the man on the platform. Example 21 An aero plane moves against the wind at the speed of 550 km/hr from A to B, and travels double this distance in the same direction further, but the speed of the wind also gets doubled. If the speed of the plane (without the wind blowing) is 750 km/hr, what should be the speed of the plane on the return trip so as to reach the starting position in half the time, with the new speed of the wind? Solution Let the distance between A and B be x and the speed of the wind be u. If the speed of the plane in air when no wind is blowing, is v. Then, v – u = 550  u = 750 – 550 = 200 km/hr. Time taken for travelling the distance x is given by

x x  hrs. v - 4 550

Similarly, time taken for travelling the further distance is given by 2x 2x 2x x    v  2u 750  400 350 175

 x  x  Total time taken =   hours.  550 175  If the speed of the plane on the return trip is w, then time taken to return would be

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Time, Speed and Distance

3x 1 x x      w  400 2  550 175  

3 1 1 1  550  175 725        w  400 2  550 175  2  500  175 192500

w=

192500  3 287500 11500 16  400    396 km/hr. 725 725 29 29

Example 22 Raju and Ramu are moving up a moving-up escalator. The ratio of speeds of Raju to that of Ramu is 2 : 1. Raju has to walk 40 steps to reach to the top whereas Ramu takes 30 steps to reach to the top. If the escalator were turned off, how many steps would be needed to be climbed by any of them? CAT 2002 Solution It is interesting to note that the speed of Ramu is less than that of Raju, yet he takes less number of steps to reach the top. On some hard thinking, the students should be able to understand why: since Ramu travels slowly, much of his walking effort is taken up by the escalator. As an extreme case, consider that Ramu is extremely lazy and takes only 2 steps to reach the top. He would definitely reach the top as almost all the effort is done by the escalator! There are two ways to solve the question. Method 1: First of all, let’s assume that the escalator has x steps in all. Now Raju reaches the top after having taken 40 steps. This means in the same time, the escalator moves through (x – 40) steps. Since the speed of Ramu is half that of Raju, we can say that in the same time in which Raju walks 40 steps, Ramu would take only 20 steps. Therefore, when Ramu has taken 20 steps, the escalator would have moved only x – 40 steps. So, when Ramu walks 30 steps, the escalator moves through

x  40  30  1.5(x  40) steps. 20

But, by now Ramu has reached the top. So, total number of steps taken by Ramu and the escalator together is 30 + 1.5(x – 40), which should be same as x. So, 30 + 1.5(x – 40) = x or x = 60. Therefore, the number of steps in the escalator is 60, which is what Raju or Ramu would have to walk to be able to reach the top. Method 2: Let us assume that Raju reaches the top in 1 minute. Then, the speed of Raju = 40 steps/min. Therefore, speed of Ramu = 20 steps/min. Let the speed of the escalator be u. Then, the downstream speed of Raju = (40 + u) steps/min. From this, we can also say that the total number of steps in the escalator is 40 + u. Downstream speed of Ramu = (20 + u) steps/min. Ramu walks 20 steps in 1 minute. Therefore, he would walk 30 steps in 1.5 minutes. In 1.5 minutes, Ramu (along with escalator) would have moved through 1.5(20 + u), which is the total number of steps in the escalator.

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Hence, 1.5(20 + u) = (40 + u)  u = 20 steps/min  Total number of steps in the escalator = 40 + 20 = 60 steps.

RACES Races can be linear or circular. As for the linear races, a few terms need to be understood. The following type of language is used in races. For instance, if it is given that: In a race of 1 km, A beats B by 30 m. This means that in the time A travels 1000 m, B travels only 970 m and hence loses by 30 m. Therefore, ratio of speeds of A to B = 1000 : 970 = 100 : 97. If, for instance, it is given that: In a race of 1 km, A gives B a start or head start of 30 m. This means that at the beginning of the race, A allows B to run for 30 m. Whether A or B wins the race, depends upon the speeds. It may also be given that: In a race of 1 km, A beats B by 5 sec. It means when A reaches the finish line, B will take 5 more seconds to reach the finish line. So, B is so far from the finish line that running at his speed, he will take 5 sec to reach the finish line. Similarly, in a 1 km race, if A gives B a start or head start of 5 sec, it means that at the beginning of the race, A allows B to run for 5 sec. Therefore, A can beat B in some race by some distance which can also be expressed in terms of the time. Note that whenever a race begins, all the participants will be assumed to start the race together, unless mentioned otherwise. If all the racers reach the finish line at the same time, the race is said to end in a ‘dead heat’. Example 23 In a 1 km race, if A beats B by 50 m or 5 sec, what is the speed of B and A? Solution When A runs 1000 m, B runs 950 m. After this, B would take 5 sec to travel the rest of the distance of 50 m. This implies that the speed of B is

50  10 m/s. 5

When A runs 1000 m, B runs 950 m. This means that ratio of speed = ratio of distance, So, if the speed of A and B is v1 and v2, then v1 : v2 = d1 : d2 = 1000 : 950 Since v2 = 10 m/s, v1 =

1000 10  10  10 m/s. 950 19

Example 24 In a km race, if A beats B by 50 m and B beats C by 50 m, by what distance does A beat C? Solution Method 1: Ratio of distance travelled by A to that travelled by B = 1000 : 950 = 20 : 19 = (20  20) : (19  20) Ratio of distance travelled by B to that travelled by C = 1000 : 950 = 20 : 19 = (20  19) : (19  19) Therefore, ratio of distance travelled by A to that travelled by B to that travelled by C = (20 20) : (19  20) : (19  19) = 400 : 380 : 361. When A travels 400 m, C travels 361 m.

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Time, Speed and Distance

Therefore, when A travels 1000 m, C travels

1000  361  902.5 m. 400

Therefore, A beats C by 1000 – 902.5 = 97.5 m. Method 2: When A travels 1000 m, B travels 950 m, and when B travels 1000 m, C travels 950 m. So, when B travels 950 m, C travels

950  950  902.5 m . 100

Therefore, A beats C by 1000 – 902.5 = 97.5 m. Method 3: When A travels 1000 m, B travels 950 m. So, speed of B is 5% less than that of A. Similarly, when B travels 1000 m, C travels 950 m. So, speed of C is 5% less than that of B. So, to find the percentage by which speed of C is less than that of A, we can use the formula of successive percentage changes. So, –5–5+

(5)(5) = –9.75%. So, speed of C is 9.75% less than that of A. 100

Therefore, when A runs 1000 m, C runs 9.75% less, that is, 97.5 m less. So, A beats C by 97.5 m. Example 25 In a kilometer race, A and B are running at speeds where the speed of A is 20% more than that of B. After some time, B increases his speed so that his speed now is 20% more than that of A. The race ends in a dead heat. How far is B from the starting point when he increases his speed? Solution Let the place where B changes his speed be x m from the starting point. If speed of B originally is v, then the speed of A = 1.2v. The speed of B later on becomes 1.2(1.2v) = 1.44v. Method 1: Since the race ends in a dead heat, time taken by A for the entire race is same as that for B. Time taken by A =

1000 1.2v

Time taken by B =

x 1000  x  v 1.44v

Therefore,

1000 x 1000  x    1200 = 1.44x + (1000 – x) 1.2v v 1.44v

 200 = 0.44x x =

5000  454.54 m. 11

Method 2: When B is x units away, then A is 1.2x units away. A has yet to travel 1000 – 1.2x and B has to travel 1000 – x. The ratio of distances B has yet to travel to that which A has yet to travel should be 1.2, because ratio of speed of B to A now is 1.2 : 1. Therefore,

1000  x 6  1.2  x  454 = 454.54 m. 1000 - 1.2x 11

Method 3: We can use allegation also, where the speeds of B are v and 1.44v, and the average speed of B is the speed of A, that is, 1.2v.

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The ratio comes out to be 24 : 20 = 6 : 5. Therefore, ratio of distances in the two parts of the journey = 6(v) : 5(1.44v) = 5 : 6. Therefore, x =

5  1000 = 454.54 m. 11

Example 26 In a 500 m race, A beats B by 100 m. In a 2 km race, A beats C by 150 m. In a km race between B and C, who wins the race and by how much distance? Solution In a 500 m race, A beats B by 100 m. So, in a km race, A beats B by 200 m. Similarly, in a 2 km race, A beats C by 150 m. So, in a km race, A beats C by 75 m. So, when A has travelled 1000 m, B has travelled 800 m and C has travelled 925 m. Naturally, C wins the race. When C has travelled 925 m, B has travelled 800 m. Therefore, when C has travelled 1000 m, B would have travelled

800  1000  864.86 m. 925

C beats B by 1000 – 864.86 = 135.14 m. What we have discussed is linear races. What follows now is circular races, commonly called as circular motion.

Circular Motion Circular motion deals with those cases where the persons (or objects) involved are moving in a circular fashion, whether in the same direction or in the opposite direction. The concepts of relative speed discussed in one of the earlier sections are applicable to the case of circular motion as well. Let us first discuss the case of circular motion in which there are only two persons running, and later we will take up the case of 3 or more persons on the circular track. This is best discussed with the help of an example. Let there be two persons - A and B - standing at the same location on a circular track. They start running in the same direction at the same point of time at the speeds of 10 m/s and 7 m/s. They keep on running on the track non-stop. The length of the track is 1 km. There are many questions that can be asked from this case. Some of them are listed below: (i)

When would A and B meet for the first time anywhere on the circular path?

(ii)

Find the place on the circular path when they meet for the first time anywhere on the path.

(iii) When they meet for the first time anywhere on the path, how many rounds have A and B each made? (iv) When would A and B meet at the starting point for the first time? (v)

When they meet at the starting point for the first time, how many rounds have A and B each made?

(vi) After the start, how often have they met before meeting again at the starting point? How would the answer of all these questions change if A and B were running in the opposite direction? Let us take them up one by one.

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Time, Speed and Distance (i)

The moment the race starts, A is just ahead of B. Another point of view can be that A is behind B by 1000 m. So, for A to meet B, he has to cover this gap of 1000 m. The relative speed between them is 10 – 7 = 3 m/s.

So, if t is the time when they meet for the first time anywhere on the path, then t=

1000 sec. 3

So, we can generally say that the time of their meeting for the first time anywhere on the path is given by Time taken for the first meeting =

Length of the track Relative Speed

There is another way to understand this. Distance travelled by A = 10t and that travelled by B = 7t. If they are meeting for the first time, we can surely say that the difference of the distance travelled by them would be the length of the track. Therefore, 10t – 7t = 1000 or t =

1000 1  333 sec. 3 3

Recall that in case of linear motion, we had mentioned that we make the slower person rest and the faster one move at the relative speed. The same thing can be applied here too. Since B is the slower one, he is made to rest, that is his speed is assumed to be 0 m/s, and A moves at the relative speed, that is 3 m/s. Now, for A to meet B, A has to cover a distance of 1000 m at the speed of 3 m/s and so time of meeting =

1000 1  333 sec . 3 3

(ii)

In

1000 1000 10000 seconds, distance travelled by A = 10   m 3 3 3

1 1 10000 1  3333   3000  333  m. This means that A is 333 m away from the starting point in 3 3  3 3 their direction of motion. Now,

Also, B would have travelled 7 

1000 7000  m. 3 3

7000 1 1 1  2333   2000  333  m. This means that B is also 333 m away from the starting point 3  3 3 3 in their direction of motion. Now,

We can easily observe that the difference of the distances travelled by A and B is 1000 m, which is nothing but the length of the track and it is at this time that they are at the same location, that is, they are meeting. 10000 m. Since the length of one round is 1000 m, we can say that the 3 10000 10 1 1 number of rounds made by A =  1000   3 . So, A makes 3 complete rounds and rd of 3 3 3 3 a round further.

(iii) We have seen that A travels

7000 m. Since the length of one round is 1000 m, we can say that the number of rounds 3 7000 7 1 1 made by A =  1000   2 . So, B makes 2 complete rounds and rd of a round further. 3 3 3 3

Similarly, B travels

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Quantitative Aptitude Simplified for CAT

It is important to note that the difference of the number of rounds made by A and B is 1 round. This is bound to be if they are to meet anywhere on the path for the first time. (iv) To find the time when they meet at the starting point, a totally different approach is followed. Here, we observe that A will be at the starting point after every 1000 m of travelling. This means that A is at the 1000  100 sec. starting point every 10 Similarly, B will be at the starting point every

1000 sec. 7

For both of them to be together at the same time t, we understand that t has to be the smallest multiple of 100 1000 1000 sec. This means nothing but LCM of 100 and . sec as well as 7 7 LCM of 100 sec and

LCM of 100 and 1000 1000 1000   1000 sec. sec = HCF of 1 and 7 1 7

(Refer Chapter 1: Number System to learn how to find LCM and HCF of fractions). (v)

Total distance travelled by A = 10  1000 = 10000 m. This means 10 rounds.

Total distance travelled by B = 7  1000 = 7000 m. This means 7 rounds. (vi) First meeting takes place after be after

1000 2000 sec. Second meeting will be after sec and third meeting will 3 3

3000 sec, that is, 1000 sec. Therefore, before meeting at the starting point, they have met 3

twice. Note that the first meeting creates a difference in the number of rounds by 1. Second meeting would happen when they have a difference in the number of rounds by 2. And the third meeting would happen when they have a difference in the number of rounds by 3, as we can see above that the difference in the number of rounds when they meet at the starting point = 10 – 7 = 3 rounds. This can be explained logically also, as when they meet for the first time, this location where they are meeting becomes the new starting point for further running. And hence the process goes on and on. Let us answer all these questions if A and B were running in the opposite direction. (i)

A and B would meet anywhere on the path after

1000 1000  sec, where 10 + 7 = 17 m/s is the 10  7 17

relative speed between the two. (ii)

1000 4 4  588 m. So A meets B at a distance 588 m away 17 17 17 4 7000 13 from the starting point in the direction in which A was running, or 1000  588   411 m 17 17 17 away from the starting point in the direction in which B was running.

In this much time, A has travelled 10 

(iii) They meet for the first time while making the first round itself. A has made

10000 10 7  1000  rounds. Similarly, B has made rounds. 17 17 17

(iv) For A and B to meet at the starting point, we have to know how often A and B are available at the starting 1000 point. We know that A is at the starting point every 100 sec and B is at the starting point every sec. 7 Therefore, LCM of these two times is the time when both would be at the starting point at the same point of time.

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Time, Speed and Distance

1000   LCM  100,   1000 sec. 7   Note that this answer is the same as in case they move in the same direction. This is so because the frequency of their being at the starting point is independent of the direction in which they are moving. (v)

Number of rounds made by A when they meet at the starting point =

The number of rounds made by B when they meet at the starting point = (vi) First meeting happens after Second meeting happens after Third meeting happens after

(10m/s)  (1000 sec)  10 . (1000 m)

(7m/s)  (1000 sec) 7 . (1000 m)

1 000 sec. 17

2000 sec. 17

3000 sec. 17

and so on. 17th meeting happens after =

17000 1000 sec, which is when they meet at the starting point. 17

So, before meeting at the starting point, they have met 16 times. From this, we can say that if T is the time when A and B meet at the starting point for the first time and t is the time when they meet anywhere on the path for the first time, then the number of times they have met before T meeting at the starting point is given by  1 . t Whatever has been discussed in the context of 2 persons, can be applied to the case of 3 or more persons also. Let there be 3 persons, A, B and C, running on a circular track of length 1 km, at the speeds of 10 m/s, 7 m/s and 5 m/s resp. in the same direction starting from the same place at the same point of time. (i)

When would all three meet for the first time anywhere on the circular path?

(ii)

Find the place on the circular path when they meet for the first time anywhere on the path.

(iii) When they meet for the first time anywhere on the path, how many rounds have each of them made? (iv) When would they all meet at the starting point for the first time? (v)

When they meet at the starting point for the first time, how many rounds have each of them made?

(vi) After the start, how often have they met before meeting again at the starting point? (i)

For all three to meet, we can make A meet B and B meet C. Then, taking LCM of these two times is the time when all three meet.

From the previous discussion, we can say that A meets B every B meets C every

1000 1000   500 sec. 75 2

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1000 1000  sec. 10  7 3

Quantitative Aptitude Simplified for CAT

 1000 ,500  So, LCM   = 1000 sec. We can say that after 1000 sec, A will be meeting B and B will be meeting  3  C and hence all three will be meeting each other. (ii)

In 1000 sec, A would have travelled 10  1000 = 10000 m, which means that A is at the starting point. This also means that all the three will always meet at the starting and at no other place. But this result is only because of the data and just a coincidence. It just may happen that all the three would meet somewhere other than the starting point if we change the values of their speeds. (Check this out if the speeds of A, B and C are 13 m/s, 11m/s and 7m/s and all are moving in the same direction)

(iii) When they meet for the first time, A would have made Similarly, B would have made and C would have made

10000  10 rounds. 1000

7000  7 rounds. 1000

5000  5 rounds. 1000

(iv) For them to meet at the starting point for the first time, we will find the time each of them takes to be at the starting point and then take the LCM of those individual times. Therefore, A is at the starting point every

1000  100 sec. 10

B is at the starting point every

1000 sec. 7

C is at the starting point every

1000  200 sec. 5

1000  , 200  = 1000 sec. LCM  100, 7   As was expected, A, B and C will meet for the first time at the starting point after 1000 sec, as was seen in (ii). (v)

Refer to explanation (iii).

(vi) Since they all meet only at the starting point, the answer to this question would be: never. Making a slight variation in the question, if A and B run in clockwise direction whereas C moves in counterclockwise direction, then A meets B every

1000 1000 250 sec and B meets C every  sec. 3 12 3

Therefore, all three would meet anywhere on the path for the first time after the time which is the LCM of 1000 250 1000 and , which comes out to sec. 3 3 3 Whereas the time when all the three are at the starting point is 1000 sec (as this does not depend upon the direction of motion). Observe the following examples. Example 27 Let Chintu and Pintu run on a circular path starting from the same place and at the same time but moving in opposite directions at the respective speeds of 10 m/s and

228

Time, Speed and Distance 15 m/s. When they meet for the first time anywhere on the path, they reverse their direction of motion and also exchange their speeds, that is, Chintu now runs at a speed of 15 m/s and Pintu at 10 m/s. Every time they meet, they change their direction and speed as mentioned. The length of the circular track is 2.5 km. (i)

How far are they along the track from the starting point when they meet for the 4th time?

(ii)

What is the total distance travelled by Chintu upon meeting for the third time?

(iii) When would they meet at the original starting point? Solution Let Chintu runs in the clockwise direction and Pintu in counter-clockwise direction. The starting point is O. (i)

Upon meeting for the first time at P, they exchange their speeds and also change their directions. So, now Chintu is running in counter-clockwise direction at a speed of 15 m/s and Pintu is running in clockwise direction at a speed of 10 m/s, and so on.

2500  100 sec and hence the place where they meet is 10  100 = 1000 m 10  15 along the direction in which Chintu runs, that is, clockwise direction.

First meeting will happen in

We can easily analyze that the second meeting happens 1 km further in the clockwise direction starting from the first meeting place. And going like this, we can say that the fourth meeting happens 4 km from the starting place in the clockwise direction. But, the length of the track being 2.5 km, it means that the fourth meeting place is 1.5 km in the clockwise direction from the starting point, or 1 km in the counter-clockwise direction. Important analysis: Since ratio of speeds of Chintu to Pintu is 2 : 3, we can divide the entire track length into 5 equal parts. Let the starting point be called as A, and other points (in clockwise direction) be called as B, C, D and E. When Chintu runs 2 units in clockwise direction, he reaches C and Pintu runs 3 units in anticlockwise direction, and he also reaches C, and so they meet there. When they exchange their directions of motion as well as speeds, Chintu now runs in anticlockwise direction at higher speed and Pintu runs in clockwise direction at lower speed. We can now say that whosoever runs at lower speed always moves in clockwise direction and whosoever runs at higher speed always runs in anticlockwise direction. Once we grasp this idea, we can say that first meeting happens at C, that is 2 units in clockwise direction, second meeting at E, which is further 2 units along clockwise direction. So, every meeting happens 2 units further along the clockwise direction. So, 4th meeting happens 8 units along clockwise direction. But there are total 5 points only. So, 8 units means 3 units from A along clockwise direction, that is at D. (ii)

Every meeting happens after every 100 sec. Therefore, total distance travelled by Chintu upon meeting for the third time = 100  (10 + 15 + 10) m = 3.5 km.

(iii) First meeting will take place 1 km away from the starting point in clockwise direction. Second meeting will take place 2 km away from the starting point in clockwise direction. Third meeting will take place 3 km away from the starting point in clockwise direction. Fourth meeting will take place 4 km away from the starting point in clockwise direction. Fifth meeting will take place 5 km away from the starting point in clockwise direction. We observe that when they meet for the 5th time, they are meeting at the starting point. We can understand this as we are looking for that distance which is a multiple of 2.5 km (the length of the track) and which should also be a multiple of 1 km (the place of first meeting).

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Quantitative Aptitude Simplified for CAT

Example 28 Ram and Shyam are racing on a circular track starting from the same place with the ratio of their speeds being 1 2 : 3. The length of the track is km. When the faster one has travelled a total distance of 8 km, how many 2 times have they meet on the track? Solution Let the speeds of Ram and Shyam be 2v m/s and 3v m/s. Since it is a race, they are obviously running in the same direction. The first meeting happens after

500 500  sec. 3v  2v v

In this time, Shyam (the faster one) would have travelled

500  3v  1500 m. v

So, when Shyam has travelled 1.5 km, they meet for the first time. When Shyam has travelled 8 km, they would have met

8 1  5 times. 1.5 3

But since the number of meetings cannot be fractional, we can logically conclude that they would have met 5 times. Example 29 Ram and Shyam are standing on the diametrically opposite ends of a circular track of length 1 km. Both start running simultaneously at the speeds of 12 m/s and 15 m/s in the same direction. (i)

When would they meet for the first time anywhere on the path?

(ii)

When would they meet for the first time at the place from where Shyam started?

(iii) How will you answer (ii) if Shyam’s speed is 7 m/s. Solution (i)

The first time when they meet anywhere on the path is given by

500 500  sec, as the gap 15 - 12 3

between them is 500 m. (ii)

500  12  2000 from where Ram started. When they meet, 3 that becomes the starting point for subsequent meeting points. The first meeting happens at the place where Ram started.

The location of the first meeting is

The second meeting will happen after

1000 sec after the first meeting. By then, Ram would have travelled 3

1000  12  4000 m. 3

Therefore, whenever they meet, they always meet at Ram’s starting place. So, they will never meet at the place from where Shyam started. 500 (iii) Method 1: When Shyam’s speed is 7 m/s, then first meeting happens after  100 sec, in which 5 time, Ram would have travelled 100  12 = 1200 m. So, the first meeting happens 200 m away from Ram’s place of start, in the direction in which Ram was running.

230

Time, Speed and Distance

The second meeting will happen after a further time of

1000  200 sec. 5

Therefore, Ram would have travelled 12  200 = 2400 m to meet the second time, which is 400 m away from the first meeting point. So, when they meet for the second time, they are 600 m away from Ram’s place in the direction of motion of Ram. Third meeting is further 400 m away, that is, at the place of start of Ram. Fourth meeting is 400 m away from Ram’s place. Fifth meeting is 800 m away from Ram’s place. Sixth meeting is 200 m away from Ram’s place, which is the place where they had met the first time. This means that they would never meet at Shyam’s place. But doing this may be tedious at times. A better method is shown below. Method 2: We can see that first meeting happens when Ram has travelled 1200 m. Second meeting happens when Ram has travelled 2400 further, that is, total of 3600 m. Third meeting when Ram has travelled 2400 m further, that is, 6000 m. This distance should someday be an odd multiple of 500. Only then would they ever meet at Shyam’s place of start. This is so because when the distance travelled is an even multiple of 500, For example, 1000 m, 2000 m, etc., they are meeting at Ram’s place and odd multiples of 500 means 500 m, 1500 m, 2500 m, and so on, which would never happen going by the trend of the terms in the series. Therefore, they would never meet at Shyam’s place of start. Method 3: First meeting happens 200 m away from Ram’s place of start. Let us call this point A. This becomes the new starting point.

 1000 , 1000  Now, the next meeting at A would be after the time which is the LCM of   or LCM of 7   12 1000  250 , 1000  .   , which is 7  1  3 Before meeting at A, they would meet after

1000 =200 sec, 400 sec, 600 sec, and so on., which means 2400 5

m, 4800 m, and so on. This means, starting from A, every time they meet, they are 400 m further away from the previous meeting point, till they once again meet at A. In none of these cases are they meeting at Shyam’s place. Hence the result.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to apply the concept of Relative Speed to Circular Motion and Clocks.

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Quantitative Aptitude Simplified for CAT

Clocks A clock measures the rotation of earth about its axis. The earth completes one rotation about its axis in 24 hours. Clocks usually have 12 marks on it depicting each hour and two rounds of the clock by the hour hand signifies completion of one day. The concept of clocks is a direct application of the circular motion discussed earlier. Let us first of all learn the basics of clocks. In a normal analog clock, there are three hands - hour hand, minute hand and second hand. The analog clock carries on its face numbers from 1 to 12 and it is divided into 60 equal parts, each part referring to 1 minute. So, one full round of the minute hand corresponds to 60 minutes. Therefore, 60 minutes corresponds to 360° and hence 1 minute corresponds to 6°. Minutes is sometimes shown as ’ and seconds as ”. So, 20 minutes is written as 20’ and 45 seconds as 45”. Therefore, 35’ 50” means 35 minutes and 50 seconds. Notation of time: In case of clocks, if we have to write the time of 15 minutes 12 seconds past 3 o’clock, the notation used is 03:15:12. AM and PM are used to denote the time interval between 12 midnight to 12 noon and 12 noon to 12 midnight. So, 3:00 PM denotes the time between 12 noon to 12 midnight. Moreover, 12 AM means 12 o’clock midnight and 12 PM means 12 o’clock noon time. Relative speed: In 60 minutes, the hour hand travels a certain distance. This distance is expressed in units of ‘degrees’ and not centimeter, metre and so on. Therefore, in 60 minutes, the hour hand moves through 30° 30 o  1     / min . and hence the speed of the hour hand is 60 min  2  In 60 minutes, the minute hand moves through 360° and hence the speed of the minute hand is 360 o = 6/min. 60 min

Similarly , the speed of second hand =

360 o = 360/min 1 min

Since all the three hands move in the same direction, that is, clockwise, the relative speed between hour and 1 1 minute hand is 6 o    5  / min  5.5 / min . 2 2 The meaning of the relative speed here is that minute hand travels 5.5° more than that travelled by hour hand, in one minute. In other words, the gap between the hands is changing at the rate of 5.5°/min. Similarly, the relative speed between minute hand and seconds hand = 360° – 6° = 354°/min. Useful interpretation of the relative speed is that: the hour hand can be thought of as not moving at all and the minute hand can be thought of as moving at the speed of 5.5°/min. The ratio of speed of minute hand to that of hour hand is 6° :

1  = 12 : 1. Therefore, minute hand travels 12 2

times faster than hour hand. All these facts will be found useful while solving questions on clocks. Example 30 At what time between 5 o’clock and 6 o’clock do the minute hand and the hour hand meet?

232

Time, Speed and Distance Solution If you thought the answer to be 5:25, think again. At 5:25, the minute hand is at 5 and hour hand has progressed to some place between 5 and 6. So, at 5:25, they are not meeting. Let us use the concept of relative speed to calculate this. At 5 o’clock, the minute hand is exactly at 12 and the hour hand is exactly at 5. The angle between them is the gap between the hands, which is 150°. Assuming that the hour hand is fixed at 5 and only the minute hand moves at the relative speed of 5.5/min, time taken by minute hand to meet the hour hand = 150 o 300 3 3   27 min . So, the two hands meet 27 min past 5 o’clock. o 11 11 5.5 / min 11 So, we see that the formula used in circular motion is even useful for handling cases of clocks. Example 31 At what time between 3 and 4 o’clock do the two hands enclose an angle of 30°? Solution At 3 o’clock, the angle between the hands is 90°. Going by the interpretation of relative speed given above (where the hour hand can be thought of as being at rest), we can easily see that the minute hand is required to move a distance of 90° – 30° = 60° or 90° + 30° = 120° in order for the hands to enclose an angle of 30°. So, the angle of 30° will be enclosed twice in this gap of 1 hour. The time when the hands enclose this angle the first time = The second time this happens after

60 120 10   10 min. 5.5 11 11

120 240 9   21 min. 5.5 11 11

Therefore, between 3 and 4 o’clock, the two hands will enclose the said angle 10 also 21

10 11

minutes past 3 and

9 minutes past 3. 11

Example 32 What time between 7 and 8 o’clock are the minute and hour hand equidistant from 8? Solution Let the time be x minutes. At this time, the minute hand will be between 8 and 9 and the hour hand will be between 7 and 8. In x minutes, the number of degrees travelled by the minute hand = 6x°, as the speed of the minute hand is 6°/min. Similarly, the hour hand would have travelled So, the hour hand would be (30 –

1 x°. 2

1 x)° away from 8. 2

Similarly, the minute hand would be (6x – 240)° away from 8. As per the question, (30 –

1 x)° = (6x – 240)° 2

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Quantitative Aptitude Simplified for CAT

 270 = 6.5x  x =

270 540 7   41 minutes. 6.5 13 13

Example 33 Prove that in a day of 24 hours, the two hands (minute and hour hand) meet each other completely 22 times. Solution You may think that they should meet 24 times. Method 1: Between 1 and 2 o’clock, the two hands meet once. Between 2 and 3 o’clock, the two hands meet once. Between 3 and 4 o’clock, the two hands meet once. This goes on, but between 11 and 1 o’clock, the two hands meet only once. This is the only gap of 2 hours during which the hands are meeting only once. So, in one full circle of the clock, that is, 12 hours, the hands meet 11 times. Therefore, in a day of 24 hours, the two hands meet 2  11 = 22 times. Method 2: At 12 o’clock, the two hands are together. The minute hand has to cover a gap of 360° to meet the hour hand (as we have seen in the previous example). So, after 12 o’clock, the two hands meet in

360 o 720  min . o 11 5.5 / min

Once they meet, the minute hand has to cover a gap of 360° to meet the hour hand again. This process goes 720 on. We can say that In min, the number of times the two hands meet = 1; 11 In 24 hours or 24  60 min, the two hands meet =

24  60 60  11   22 times. 720/11 30

Remember this as a standard result. Example 34 In a day of 24 hours, how often do the three hands meet? Solution This is like the problem of 3 persons starting from the same place at the same time in the same direction. Proceeding likewise, the minute hand meets the hour hand every The seconds hand meets the minute hand every

360 720  min . 5.5 11

360 60  min . 354 59

 720 , 60   720 min . LCM    11 59  So, all the three hands meet once every 720 min. In 24 hours, or 24  60 min, the three hands would meet

24  60  2 times. 720

Intuitively also, we can understand that three hands will meet every time it strikes 12 o’clock. In a day of 24 hours, 12 o’clock would strike twice and hence the answer. In the following, we will deal with cases where the clock is faulty.

234

Time, Speed and Distance Example 35 A clock loses a minute every hour. In how many minutes will it strike an hour? Solution When 60 minutes elapse in the correct clock, only 59 minutes elapse in the faulty clock, as it loses a minute in an hour. Therefore, when 60 minutes elapse in the faulty clock, the correct clock would show 60 1  60  61 min  59 59

Example 36 There are two clocks, one of which loses 2 minutes in a day and the other gains 3 minutes in a day. Both the clocks are set right at 3 o’clock. (i)

When will they show the same time again?

(ii)

What time are the clocks showing?

Solution (i)

In a day, the gap between the clocks is 5 minutes. The gap of 5 minutes is created in 1 day. The gap of 12 hours will be created in [(12  60)  1]  5 = 144 days.

(ii)

After 1 day, the first clock loses 2 minutes.

Therefore, after 144 days, the first clock would lose 288 minutes, which means = 4 hours 48 minutes. This means that the time this clock would show is 10:12. Similarly, the other clock would have gained 3  144 = 432 minutes =

432 . 60

= 7 hours 12 minutes. This clock would also show 10:12. Example 37 In the above example, if the clocks are set right at 3:00 PM instead of 3 o’clock, how will you answer these questions? Solution If the clock is showing 3:00 PM, it only means that the clock is digital and not analog. So, after 144 days, the losing clock would show 10:12 am and the gaining clock would show 10:12 pm. Therefore, for them to show the same time, the gap should be of 24 hours and not of 12 hours. Therefore, the job is done in 288 days. The time shown would be 5:24 am.

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Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE stop flights between A and B. All the times indicated are local and on the same day.

Directions for questions 1 and 2: Answer the questions on the basis of the information given below.

Departure

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

City

Time

City

Time

B

8:00 am

A

3:00 pm

A

4:00 pm

B

8:00 pm

Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west at 50 km per hour.

CAT 2005 1.

Arrival

At what time do Ram and Shyam first meet each other?

CAT 2007

2.

3.

A.

10 A.M.

B.

10:10 A.M.

C.

10:20 A.M.

D.

10:30 A.M.

4.

At what time does Shyam overtake Ram?

What is the time difference between A and B? A.

1 hour and 30 minutes

B.

2 hours

C.

2 hours and 30 minutes

A.

10:20 A.M.

D.

1 hour

B.

10:30 A.M.

E.

Cannot be determined

C.

10:40 A.M.

D.

10:50 A.M.

5.

Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? A.

3

B.

3.5

C.

4

D.

4.5

E.

5

6.

CAT 2006 Directions for Questions 4 and 5: Answer the questions on the basis of the information given below: Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-

236

What is the plane's cruising speed in km per hour? A.

700

B.

550

C.

600

D.

500

E.

Cannot be determined

Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to A.

6:15 am

B.

6:30 am

Time, Speed and Distance

7.

C.

6:45 am

D.

7:00 am

E.

7:15 am

during next 2 minutes, r/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round? CAT 2008

A.

4

Two boats, traveling at 5 and 10 km/hr, head directly towards each other. They begin at a distance of 20 km from each other. How far apart are they (in km) one minute before they collide?

B.

8

C.

16

D.

32

A. B. C. D.

CAT 2004

1 12 1 6 1 4 1 3

11. 2 boats A and B go upstream at a speed of 10 km/hr, where stream's speed is 5 km/hr. At noon, a snag develops in the engine of B and so it floats along the stream. A realizes the snag in B when the gap between them is 5 km. Then, A turns back and meets B. When would A meet B? A. CAT 2004

8.

Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race? A.

Karan and Arjun reach the finishing line simultaneously.

B.

Arjun beats Karan by 1 metre.

C.

Arjun beats Karan by 11 metres.

D.

Karan beats Arjun by 1 metre.

B.

1:30 pm

C.

2 pm

D.

3 pm CAT 2012

12. Three persons A, B and C start from the same place in the same direction. A and B start at 10 am and C at 12 pm. C overtakes A at 2 pm and then he doubles his speed. C overtakes B at 3 pm. Find the ratio of speeds of A and B.

CAT 2004 9.

1 pm

If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

A.

2:3

B.

5:6

C.

3:4

D.

None of these CAT 2014

13. A train crosses a platform in 30 sec. Speed of train is 60 km/hr. Length of train is 200 m. Find the length of platform.

A.

11 km/hr

A.

300 m

B.

12 km/hr

B.

200 m

C.

13 km/hr

C.

500 m

D.

14 km/hr

D.

100 m

CAT 2004

CAT 2014

10. A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is r during the first 30 seconds, r/2 during next one minute, r/4

14. Rahim and Shyam start at 12 noon from P and Q and move towards Q and P respectively. They meet at 3 pm the same day. Rahim takes 2.5 hours less than Shyam

237

Quantitative Aptitude Simplified for CAT

to reach the respective end. Find the time at which Shyam reaches P. A.

5:30 pm

B.

6:30 pm

C.

7:30 pm

D.

8:30 pm

Vivekananda and return back to Kanyakumari just in time before the ferry sinks? (Current of the sea water from Rock of Vivekananda to Kanyakumari is 2 km per hour.)

CAT 2015 15. A and B are 440 km apart. P and Q start from A at 7:30 am and 8:45 am at the speeds of 50 km/hr and 40 km/hr and R starts from B at 11 am at a speed of 60 km/hr? At what time will P be equidistant from Q and R? A.

11:55 am

B.

12:05 pm

C.

12:30 pm

D.

1:00 pm

B.

30% greater

C.

37.5% greater

D.

50% less

B.

41 km/hr towards the Rock & 38 km/hr while returning to Kanyakumari

C.

42 km/hr towards the Rock & 36 km/hr while returning to Kanyakumari

D.

35 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

18. It was a rainy morning in Delhi when Rohit drove his mother to a dentist in his Maruti Alto. They started at 8.30 AM from home and Rohit maintained the speed of the vehicle at 30 Km/hr. However, while returning from the doctor’s chamber, rain intensified and the vehicle could not move due to severe water logging. With no other alternative, Rohit kept the vehicle outside the doctor’s chamber and returned home along with his mother in a rickshaw at a speed of 12 Km/hr. They reached home at 1.30 PM. If they stayed at the doctor’s chamber for the dental check-up for 48 minutes, the distance of the doctor’s chamber from Rohit’s house is

16. A chartered bus carrying office employees travels every day in two shifts  morning and evening. In the evening, the bus travels at an average speed which is 50% greater than the morning average speed; but takes 50% more time than the amount of time it takes in the morning. The average speed of the chartered bus for the entire journey is greater/less than its average speed in the morning by: 18% less

40 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

IIFT 2014

CAT 2016

A.

A.

A.

15 Km

B.

30 Km

C.

36 Km

D.

45 Km IIFT 2013

IIFT 2015

19. It takes 15 seconds for a train travelling at 60 km/hour to cross entirely another train half its length and travelling in opposite direction at 48 km/hour. It also passes a bridge in 51 seconds. The length of the bridge is

17. A ferry carries passengers to Rock of Vivekananda and back to Kanyakumari. The distance of Rock of Vivekananda from Kanyakumari is 100 km. One day, the ferry started for Rock of Vivekananda with passengers on board, at a speed of 20 km per hour. After 90 minutes, the crew realized that there is a hole in the ferry and 15 gallons of sea water had already entered the ferry. Sea water is entering the ferry at the rate of 10 gallons per hour. It requires 60 gallons of water to sink the ferry. At what speed should the driver now drive the ferry so that it can reach the Rock of

A.

550 m

B.

450 m

C.

500 m

D.

600 m IIFT 2012

238

Time, Speed and Distance 20. Arup and Swarup leave point A at 8 AM to point B. To reach B, they have to walk the first 2 km, then travel 4 km by boat and complete the final 20 km by car. Arup and Swarup walk at a constant speed of 4 km/hr and 5 km/hr respectively. Each rows his boat for 30 minutes. Arup drives his car at a constant speed of 50 km/hr while Swarup drives at 40 km/hr. If no time is wasted in transit, when will they meet again? A.

At 9.15 AM

B.

At 9.18 AM

C.

At 9.21 AM

D.

At 9.24 AM

E.

At 9.30 AM

C. D. E.

6 hours 23 minutes 20 seconds

D.

7 hours 17 minutes 20 seconds

E.

7 hours 23 minutes 20 seconds

Scenario I: Devanand walks East at a constant speed of 3 km per hour and Pradeep walks South at a constant speed of 4 km per hour. Scenario II: Devanand walks South at a constant speed of 3 km per hour and Pradeep walks East at a constant speed of 4 km per hour. Scenario III: Devanand walks West at a constant speed of 4 km per hour and Pradeep walks East at a constant speed of 3 km per hour.

1 7 3 9 7 16 7 16 9

A.

Scenario I only

B.

Scenario II only

C.

Scenario III only

D.

Scenario I and II

E.

None of the above XAT 2015

24. Prof. Mandal walks to the market and comes back in an auto. It takes him 90 minutes to make the round trip. If he takes an auto both ways it takes him 30 minutes. On Sunday, he decides to walk both ways. How long would it take him? XAT 2016

22. Each day on Planet M is 10 hours, each hour 60 minutes and each minute 40 seconds. The inhabitants of Planet M use 10 hour analog clock with an hour hand, a minute hand and a second hand. If one such clock shows 3 hours 42 minutes and 20 seconds in a mirror what will be the time in Planet M exactly after 5 minutes? A.

C.

23. Devanand’s house is 50 km West of Pradeep’s house. On Sunday morning, at 10 a.m., they leave their respective houses. Under which of the following scenarios, the minimum distance between the two would be 40 km?

21. Pradeep could either walk or drive to office. The time taken to walk to the office is 8 times the driving time. One day, his wife took the car making him walk to office. After walking 1 km, he reached a temple when his wife called to say that he can now take the car. Pradeep figured that continuing to walk to the office will take as long as walking back home and then driving to the office. Calculate the distance between the temple and the office.

B.

6 hours 22 minutes 20 seconds

XAT 2016

XAT 2017

A.

B.

A.

100 minutes

B.

120 minutes

C.

140 minutes

D.

150 minutes

E.

None of the above XAT 2013

25. The taxis plying in wasseypur have the following fare structure: Rs 20 for the first two kilometers, Rs 5 for every km in excess of 2 km and up to 10 km, and Rs 8 for every

6 hours 18 minutes 20 seconds

239

Quantitative Aptitude Simplified for CAT

26. Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram’s car at the rate of 25 kmph, while Shyam walked at 5 kmph. After sometime. Hari got off and started walking at the rate of 5 kmph and Ram went back to pick up Shyam. All three reached the destination simultaneously. The number of hours required for the trip was:

km in excess of 10 km. Bullock carts on the other hand charge Rs 2 per km. Sardar Khan takes a taxi from the Wasseypur railways station to his home. On the way, at a distance of 14 km from the railway station, he meets Faizal Khan, and gets down from the taxi to talk to him. Later he takes a bullock cart to reach his home. He spends a total of Rs 102 to reach his home from the railway station. How far is his home from the railway station? A.

17

B.

18

C.

19

D.

20

E.

21

A.

8

B.

7

C.

6

D.

5

E.

4 XAT 2013

XAT 2013

ANSWER KEY 1. B

2. B

3. C

4. D

5. B

6. B

7. C

8. D

9. B

10. C

11. A

12. B

13. A

14. C

15. C

16. B

17. C

18. C

19. A

20. D

21. C

22. B

23. A

24. C

25. C

26. A

240

Time, Speed and Distance

ANSWERS AND EXPLANATIONS 1.

550 – 50 = 500 km/hr and along A to B will be 550 + 50 = 600 km/hr. Time taken from B to A = 3000/500 = 6 hours and that from A to B = 3000/600 = 5 hours and hence total trip time comes out to be 11 hours. Since flying time from B to A is 6 hours and the plane flew from B at 8:00 am, it would reach A at 2 pm (as per local time of B, not A). But at that instant, local time at A is 3 pm. Therefore, time difference between A and B is 1 hour.

Answer: B Explanation: At 10 am, Ram reaches B, whereas Shyam reaches the midpoint C. Now, gap between them is 2.5 km and relative speed between them is 5 + 10 = 15 km/hr. So, time taken to 2.5 2.5 hrs   60 min = 10 min meet = 15 15 after 10 am. Therefore, Ram and Shyam meet first at 10:10 am.

2.

Answer: B

4.

Answer: D

5.

Answer: B

6.

Answer: B

Explanation: At 10:15 AM, Shyam reaches at B and Ram would have travelled a distance of 5 × (0.25 hrs) = 1.25 km. Now, they are moving in same direction and so relative speed = 10 – 5 = 5 km/hr. So, time taken for Shyam to overtake 1.25 Ram = hrs = 0.25 hours = 0.25 x 60 5 min = 15 min after 10:15 am. Therefore, Shyam overtakes Ram at 10:30 am. 3.

Explanation: Since ABC = 60 and BAC = 30, ACB = 90. Therefore, BC = 500 cos 60 = 250 km and AC = 250 3 km. Time taken for the train to reach C 250 km = = 5 hours, that is at 1 pm. 50 km / hr Time taken for Rahim to reach the station 250 3 = that is 6 hours 11.15 min. Adding 70 the margin of 15 minutes, time is 6 hours 26.15 min. So, Rahim should start from A at 6:30 am. If he starts at 6:45 am, he will not be able keep a margin of 15 minutes, though he will reach the station approximately 15 min before the arrival of the train.

Answer: C Explanation: Distance traveled by Arun in 2 hours = 60 km. Time taken for Barun to meet Arun 60 = = 6 hours. 40  30 In 6 hours, distance traveled by Barun = 240 km. This distance will be traveled by Kiranmala in 240/60 = 4 hours. So, time gap between start of Arun and Kiranmala = 2 + 2 = 4 hours.

7.

Answer: C Explanation: Their relative speed is 15 km/hr. In 1 min, 15 1 they would travel ×1= km. So, they 60 4 1 would be km apart 1 minute before 4 collision.

Common explanation for questions 4 and 5: In this case, we should solve question # 5 first. Total travel time from B to A and back = 7 + 4 = 11 hours (though we cannot say that time taken from B to A is 7 hours, but going back and forth automatically adjusts the effects due to difference in time zone). If plane’s cruising speed is 550 km/hr, then plane’s effective speed along B to A will be

8.

Answer: D Explanation: When Karan travels 100 m, Arjun travels 90 m. So, in the second race, when Karan travels 110 m, Arjun would travel 90% of that distance = 99 m and is still 1 m behind the finish line. So, Karan beats Arjun by 1 m.

241

Quantitative Aptitude Simplified for CAT

9.

12. Answer: B

Answer: B

Explanation: Distance travelled by A in 4 hours is same as that travelled by C in 2 hours. So, ratio of speeds of A to C = 2 : 4 = 1 : 2. Distance travelled by B in 5 hours is same as that travelled by C in 3 hours. So, ratio of speeds of B to C = 3 : 5. Therefore, ratio of speeds of A : B A /C 1 /2 5 =   . Therefore, required ratio B /C 3 /5 6 = 5 : 6.

Explanation: Since the distance covered is same, therefore speeds will be inversely proportional to time. So the ratio of times taken in the two cases will be 15 : 10 = 3 : 2. In this ratio, time difference is 1, whereas in actual fact, time difference is 2 hour. Therefore, ratio would be 6 : 2 and so time taken in first case will be 6 hours, which means the man started cycling at 7 am. Now that his speed is 12 km/hr, ratio of speeds is 10 : 12 and so ratio of times will be 12 : 10 = 6 : 5. Since his actual travel time @ 10 km/hr is also 6 hours, his actual travel time at 12 km/hr will be 5 hours and so he will reach the place at 12 pm. Alternatively, once we know that he started cycling at 7 am, we can say that distance travelled = 10 × 6 = 60 km which he will be travel at 12 km/hr in 5 hours and so he will reach the place at 12 pm.

13. Answer: A Explanation: Let the length of platform be x. Then, x + 200 5  =  60    30  x = 300 m. 18   14. Answer: C Explanation: Let Ram and Shyam meet at M. Let speeds of Rahim and Shyam be v and u. Then, PM = 3v and QM = 3u. Now, v 3v 3u 3   2.5  3k   2.5 , where k = u u v k Solving the equation, we get: 3k2 – 2.5k – 3 = 0  k = 2.5  6.25  36 2.5  42.25  6 6 2.5  6.5 9    1.5 6 6 (rejecting the negative answer). Therefore, v : u = 3 : 2 and so PM : QM = 3 : 2. If Rahim covers 3 units distance (= PM) in 3 hours, he will cover 2 units distance (= MQ) in 2 hours and so he will reach Q at 5 pm and so Shyam will reach P at 7:30 pm.

10. Answer: C Explanation: Radius is r meters, so he will cover one circumference in 30 + 60 + 120 + 240 = 450 seconds in first round and in second round he will take 480 + 960 + 1920 + 3840 = 7200 7200 = 16. seconds. So the required ratio is 450 11. Answer: A Explanation: Speed of the boats in still water = 10 km/hr. Since they are going upstream, the upstream speeds of the boats = 10 – 5 = 5 km/hr. After t hours after the snag develops, the gap between the two boats = (5 + 5)t. But this is given as 5 km. So, 10t = 5 or t = 0.5 hours. Now, the boat A takes a U-turn and its downstream speed becomes 15 km/hr. To meet B, it has to cover a distance of 5 km. So, 5 time taken to meet B = = 0.5 hours. 15  5 So, time when A meets B is 1 pm.

15. Answer: C Explanation: At 11 am, distance travelled by P = 50 × 3.5 = 175 km and that by Q = 40 × 2.25 = 90 km. So, gap between P and Q = 85 km and that between P and R = 440 – 175 = 265 km. In the next 1 hour, by 12 pm, distance between P and Q will increase by 10 km, so it becomes 95 km whereas distance between P and R decreases by 110 km, and so it

242

Time, Speed and Distance 5  51 = 850. 18 Therefore, y = 550 m.

becomes 155 km. By 12:30, in next half an hour, gap between P and Q = 95 + 5 = 100 km and gap between P and R = 155 – 55 = 100 km. So, at 12:30 pm, P will be equidistant from Q and R.

(300 + y) = 60 

20. Answer: D Explanation: At 4 km/hr, Arup walks 2 km in 0.5 hours and at 50 km/hr, he covers 20 km in 0.4 hours. At 5 km/hr, Swarup walks 2 km in 0.4 hours and at 40 km/hr, he covers 20 km in 0.5 hours. Thus, time taken by each of them to reach B is equal. Time = 30 + 30 + 24 = 84 minutes. So, they meet again at 9:24 AM.

16. Answer: B Explanation: Let the speed of the bus in morning is 10 km/hr and time taken is 4 hours. Then, speed and time taken in evening is 15 km/hr and 6 hours. So, distances travelled in morning and evening = 40 km and 90 km. Average speed for the whole journey 40  90 = 13 km/hr, which is 30% more = 10 than the average speed in the morning.

21. Answer: C Explanation: Let his walking speed be x km/hr. Then, speed by car = 8x km/hr. Also, let distance from temple to office = y km. As per the question, 1 1y y 9  or y = .  8x x x 7

17. Answer: C Explanation: It will take 4.5 hours to enter 45 more gallons into the ferry. Now, the distance yet to be travelled is 70 km towards the rock and 100 km back to Kanyakumari. Use options. Only [C] option ensures the ferry reaches Kanyakumari just in time.

22. Answer: B Explanation: In the given clock, 10 and 5 are opposite to each other. When the mirror shows 3 hours 42 minutes and 20 seconds, the hour hand is between 3 and 4, the minute hand is between 42 minute mark and 43 minute mark and the second hand is at 5. So, hour hand must be between 6 and 7, the minute hand must be between 17 minute mark and 18 minute mark. Also, the position of the second hand will be same at 5. So, the actual time must be 6 hours 17 minutes and 20 seconds So, after five minutes, the time will be 6 hours 22 minutes and 20 seconds.

18. Answer: C Explanation: 48 4  60 5 hours. Therefore, time of travel both ways 4 21 =5–  . If x is the distance between 5 5 Doctor’s chamber and Rohit’s house, then

Time of stay at Doctor’s chamber =

x x 21    x = 36 km. 30 12 5

19. Answer: A Explanation: Length of the longer and shorter trains are x x x   x 2   15 m and m. Therefore, 5 2 (60  48)  18  x = 300 m. If length of bridge is y m, then

Alternatively, The sum of the time visible in the mirror and the actual time should be equal to 10:00:00 hours. 10:00:00 – 03:42:20 = 06:17:20 hours. Thus, actual time is 6 hours 17 minutes and 20 seconds. So, after 5 minutes the time will be 6 hours 22 minutes and 20 seconds.

243

Quantitative Aptitude Simplified for CAT

23. Answer: A

Since he paid Rs.102, he spent Rs.10 on bullock cart and so distance travelled on bullock cart = 5 km. Therefore, distance between his home and railway station = 14 + 5 = 19 km.

Explanation: In scenario II and III, the gap between them is increasing. So, we need to check scenario I only. Let us assume that the gap between them is 40 km after x hours. Then, distance travelled by Devanand = 3x and by Pradeep = 4x. Therefore, (50 – 3x)2 + (4x)2 = 402 or x = 6 hours. So, after 6 hours, the gap will be 40 km.

26. Answer: A Explanation: Ratio of speed of car to speed of walking = 25 : 5 = 5 : 1. Let starting point be A, end point be B, Hari got off at C and Ram picks up Shyam at D. (AC + CD) : AD = 5 : 1 So, if AD = x, then AC + CD = 5x or (AD + DC) + CD = 5x or x + 2CD = 5x or CD = 2x. Further, (CD + DB) : CB = 5 : 1. So, if CB = y, then CD + DB = 5y or CD + (DC + CB) = 5y or 2x + 2x + y = 5y or x = y. Therefore, AD = x, DC = 2x and CB = x. Therefore, AB = 4x = 100. The car travels a total distance of AC + CD + DB = (x + 2x) + (2x) + (2x + x) = 8x = 200 at the speed of 25 km/hr. So, total time taken = 8 hours.

24. Answer: C Explanation: Walking to market and coming back by auto takes 90 minutes. If he takes auto both ways, then time taken is 30 min. So, time taken by auto for one way is 15 min. Therefore, walking one way = 90 – 15 = 75 min. So, walking both ways = 150 min. 25. Answer: C Explanation: Taxi fare for the first 14 km = 20 + 5 x× 8 + 8 × 4 = Rs. 92.

244

Polynomials, Algebraic Formulae and Linear Equations

Chapt er 07

Polynomials, Algebraic Formulae and Linear Equations BASIC CONCEPTS Polynomials A polynomial is an algebraic expression of the form a0xn + a1xn–1 + a2xn–2 + ..... + an–1x + an where every power of the variable (or variables) is a whole number. Therefore, the expression 2x3 + 5x2 – 4x + 3 is a polynomial, because every power of the variable, x is a whole number. In this definition, we have mentioned the term ‘algebraic expression’, which needs to be understood. Algebraic expressions are those expressions where the power of the variable may not be a whole number, For example, 1 1 x 2  2  5x  7 is not a polynomial but an algebraic expression, as the power of x in the term 2 is –2 which x x 3 is not a whole number. Similarly, x + 3x x – 5x + 3 is an algebraic expression but not a polynomial, because 3 the power of x in one of the terms is , which is not a whole number. 2 We can say that “all polynomials are algebraic expressions, but all algebraic expressions need not be polynomials”. In the above definition of polynomials, we have also stated that a polynomial can have 2 or more variables. Therefore, expressions like x2 + 2xy + y2, x3 – 3x2y + 5xy2 – y3 are all examples of polynomials in 2 variables. Example 1 Identify which of the following are polynomials and which are not.

245

Quantitative Aptitude Simplified for CAT

If an expression is a polynomial, mark [A]. If an expression is an algebraic expression but not a polynomial, mark [B]. If an expression is neither algebraic expression nor polynomial, mark [C]. (i)

5x2 – 3x + 2

(ii)

3 x  3x  7

(iii)

x 3  2x 2  1 2x 2  5 x  2

(iv)

sinx + 3cos2x – logx

(v)

x3 x 2 3 2 y y

(vi)

4x + 3

2

(vii) 4x (viii) 4 (ix)

0 3 2

(x) Solution (i)

Since every power is a whole number, it is a polynomial. Therefore, the answer is [A].

(ii)

It is definitely not a polynomial. It is an algebraic expression. Therefore, the answer is [B].

(iii) It is also an algebraic expression like (ii) above. Therefore, the answer is [B]. (iv) The expression is not algebraic because of the presence of functions like trigonometric and/or logarithmic functions. The expressions which contain such functions are not algebraic as they are not expressed in powers of x. As such, the expression is a function of x, but not algebraic expression. Therefore, the answer is [C]. (v)

The answer is [B].

(vi) to (x) are all polynomials because every power of the variable is a whole number. Therefore, all constants are polynomials. Even (x) is also a polynomial though the number is an irrational number. The nature of the constant doesn’t determine whether an expression is a polynomial or not. 1 2 5 4 Similarly, expressions like x  x  are also polynomials even though none of the coefficients 7 9 2 is a whole number. Hence, the answer is [A]. The highest power of the variable in a polynomial expression is called degree of the polynomial. Therefore, the degree of f(x) = 2x3 + 5x2 – 4x + 3 is 3 and that of g(x) = x2 + 2xy + y2 is 2. Note that degree of the polynomial f(x) = 3 is 0, because 3 can be written as 3x0. We can say that degree of all constant polynomials is 0. However, degree of zero polynomial, f(x) = 0 is not 0. In fact, degree of zero polynomial is not defined. Example 2 Let there be two polynomials P(x) and Q(x). Which of the following is necessarily true?

246

Polynomials, Algebraic Formulae and Linear Equations (i)

P(x) + Q(x) is another polynomial whose degree is same as the degree of P(x) or Q(x) if P(x) and Q(x) have same degree.

(ii)

P(x) – Q(x) is another polynomial whose degree is same as the degree of P(x) or Q(x) if P(x) and Q(x) have same degree.

(iii) P(x)  Q(x) is another polynomial whose degree is more than the degree of P(x) or Q(x). (iv) P(x) Q(x) is another polynomial whose degree is sum of the degrees of P(x) and Q(x). Solution (i)

is not necessarily true as coefficient of highest degree term in P(x) and Q(x) can be opposite in sign and hence the terms may cancel each other. For example, if P(x) = 2x3 + 4x2 – x + 7 and Q(x) = –2x3 + 3x2 + 4x – 5, then P(x) + Q(x) has degree 2.

(ii)

Using same logic as above, the statement is not necessarily true.

(iii) If P(x) is a constant polynomial, then the product will be another polynomial whose degree will be same as the degree of Q(x). So, the statement is not necessarily true. (iv) If degree of P(x) is m and that of Q(x) is n, then the degree of their product is definitely m + n. Therefore, this is a true statement.

Division of a polynomial by another polynomial If we divide a polynomial P(x) (called dividend polynomial) by another polynomial D(x) (called divisor polynomial) whose degree is same or less than that of P(x), then we get Q(x) (called quotient polynomial) and R(x) (called remainder polynomial). These 4 polynomials are connected by the following relationship. P(x) = D(x)  Q(x) + R(x) This relationship is similar to Dividend = (Divisor)(Quotient) + Remainder which we have studied earlier. This is known as division algorithm. Example 3 What should be divided by (x2 – 3x + 2) so that the remainder is 9 and quotient is x3 + 5x2 – 8x + 13? Solution Dividend = (Divisor)(Quotient) + Remainder  Dividend = (x2 – 3x + 2)(x3 + 5x2 – 8x + 13) + 9  Dividend = x5 + 2x4 – 21x3 + 47x2 – 55x + 35. There are two ways to divide a polynomial by another polynomial. One is the long division method and the other is known as “synthetic division” method. It is assumed that the student knows the long division method. Below, we present the synthetic division method. Let us learn to divide x4 + 3x3 – 7x2 + 5x + 2 by x – 2, by synthetic division process. Step 1: As a first step, we will write all the coefficients of dividend polynomial in the following manner (separated by some gap) 1

3

–7

5

2

Note that the coefficients must be written in the decreasing order of the power of x starting with the highest power. If some power of x is missing, then the coefficient 0 is written in its place.

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Quantitative Aptitude Simplified for CAT

Step 2: The divisor is x – 2, where the constant term is the lowest degree term, –2. Equate this to zero and get the value of x. So, x – 2 = 0  x = 2. This number 2 can be called as divisor constant. Write this in the leftmost place of the row obtained in step 1, separated by some vertical line as shown below 2

1

3

–7

5

2

Step 3: The first coefficient of dividend is 1 which is always written as it is, below the horizontal line vertically below the coefficient itself. Multiply this 1 (which is written below the horizontal line) by 2, the divisor constant, and obtain 2. Write this 2 below the next coefficient of dividend, that is, 3 and above the horizontal line, as shown below. Add the second column and obtain 5. 2

1

3

–7

5

2

2 1

5

Step 4: Multiply 5 by 2 (divisor constant) once again and obtain 10. Write this below next column. Add the column to obtain 3. Proceeding like this, we obtain the following. 2

1

1

3

–7

5

2

2

10

6

22

5

3

11

24

Separate the last column by another vertical line (as shown above). The numbers below the horizontal line in the row 1

5

3

11

are the coefficients of the quotient polynomial and the last number 24 is the remainder when the dividend polynomial is divided by the divisor polynomial. Therefore, when we divide x4 + 3x3 – 7x2 + 5x + 2 by x – 2, the quotient polynomial is x3 + 5x2 + 3x + 11 and remainder is 24, or x4 + 3x3 – 7x2 + 5x + 2 = (x – 2)(x3 + 5x2 + 3x + 11) + 24. The students must observe this process of finding quotient and remainder carefully. Example 4 Divide 2x5 + 6x4 – 7x3 + 12x + 3 by x + 3 and obtain the quotient and remainder. Solution Proceeding as explained above, the divisor constant will be –3. Also note that the term x2 is missing. So, we supply zero in its place as a correction. All this is mentioned in the process below. –3

2

2

6

–7

0

–6

0

21

– 63

153

0

–7

21

– 51

156

248

12

3

Polynomials, Algebraic Formulae and Linear Equations Therefore, quotient is 2x4 + 0x3 – 7x2 + 21x – 51 and remainder is 156. The quotient can also be written as 2x4 – 7x2 + 21x – 51. This method of synthetic division is very convenient and quick and should be always used whenever required. Example 5 Divide 2x + 7 by x – 5. Solution Again using synthetic division method: 5

2

7 10

2

17

Therefore, quotient is 2 and remainder is 17. Note that, 2x + 7 = (x – 5)2 + 17 Note: Synthetic division can also be applied if the divisor is quadratic, cubic or even higher degree. But those concepts are beyond the context of the exam requirement.

Remainder theorem and factor theorem Let P(x) be a polynomial. If P(x) is divided by x – a, then P(a) is the remainder. This is known as remainder theorem. If P(x) is divisible by x – a, then P(a) = 0. This is known as factor theorem. Both the theorems are basically similar. Let us prove remainder theorem from which factor theorem can be easily proved. Proof: If P(x) upon division by x – a leaves the quotient Q(x) and remainder R(x), then using division algorithm, P(x) = (x – a)Q(x) + R(x) Putting x = a, we get: P(a) = (a – a)Q(a) + R(a)  P(a) = R(a). This means that the remainder is P(a). This is remainder theorem. As for factor theorem, if P(x) is ‘divisible’ by x – a, then the remainder has to be zero. But remainder theorem states that the remainder is P(a). This means that P(a) = 0. This is factor theorem. If P(x) = x3 – 2x2 + 5x + 3 is divided by x – 1, then the remainder is P(1), as a = 1. Therefore, remainder = P(1) = 13 – 2(1)2 + 5(1) + 3 =1 – 2 + 5 + 3 = 7. (the students may verify this by using synthetic division method) Example 6 Find the remainder when 2x5 + 6x4 – 7x3 + 12x + 3 is divided by x + 3.

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Quantitative Aptitude Simplified for CAT

Solution Remainder = 2(–3)5 + 6(–3)4 – 7(–3)3 + 12(–3) + 3 = 156. Please see that the answer is same as obtained in one of the previous examples. Example 7 When we divide x3 + 5x2 – 2x – k by x + 1, the remainder is 5. Find k. Solution By remainder theorem, (–1)3 + 5(–1)2 – 2(–1) – k = 5  k = 1. Example 8 If x4 – 4x3 + 2x2 + kx – 5 is divisible by x – 3, find k. Solution (3)4 – 4(3)3 + 2(3)2 + k(3) – 5 = 0  k =

14 . 3

Example 9 If f(x) is divided by (x – ), then p is the remainder and if divided by (x – ), then q is the remainder. If f(x) is divided by (x – )(x – ), then what is the remainder? Solution If f(x) is divided by (x – ), then the remainder is f() = p (given) If f(x) is divided by (x – ), then the remainder is f() = q (given) Let R(x) be the remainder when f(x) is divided by (x – )(x – ). Since divisor is quadratic, the remainder will be either linear or a constant. Let R(x) = ax + b. Then, f(x) = (x – )(x – )Q(x) + (ax + b)  f() = ( – )( – )Q() + (a + b) and f() = ( – )( – )Q() + (a + b)  p = (a + b) and q = (a + b)  a( – ) = p – q  a =

pq . 

 p  q  q   p Putting this value in any of the equations, we get: b = p      .      pq q   p p  x     q ( x   ) Therefore, remainder is R(x) = ax + b =   x            

Example 10 Find the quotient and remainder when x4 + x3 – 3x2 – 2x + 5 is divided by 2x – 6. Solution The divisor has coefficient of x as 2. Therefore, 2x – 6 = 2(x – 3). Let us first divide the given polynomial by (x – 3). 3

1

1

1

–3

–2

5

3

12

27

75

4

9

25

80

250

Polynomials, Algebraic Formulae and Linear Equations Therefore, x4 + x3 – 3x2 – 2x + 5 = (x – 3)(x3 + 4x2 + 9x + 25) + 80

9 25  1 = (2x – 6)  x 3  2x 2  x   + 80 2 2 2   Hence, Q(x) =

1 3 9 25 x  2x 2  x  and R = 80. 2 2 2

b  The above example illustrates that if P(x) is divided by ax + b, then ax + b should be written as: a  x   a  b  and then the polynomial should be divided by  x   by synthetic division method. The quotient so obtained a  is then divided by ‘a’ to get the actual quotient when the divisor is ax + b, whereas the remainder is kept as it is.

Remainder and factor theorem for such a case  b  b If P(x) is divided by ax + b, then remainder is P    . If P(x) is divisible, then P    = 0.  a  a Example 11 What is the remainder when x64 + x32 + x16 + x8 + x4 + x2 + 1 is divided by x2 – 1. Solution The divisor is x2 – 1 = (x – 1)(x + 1). When f(x) is divided by x – 1, remainder = 7; When f(x) is divided by x + 1, remainder = 7. Using the result obtained in one of the previous examples, remainder obtained when f(x) is divided by x2 – 1 =

p  x     q( x  ) 7 x  1   7( x  1 ) = 7. (taking  = 1 and  = –1)   1  (  1)

Example 12 What is the value of the expression given below? (23)7 – 24(23)6 + 26(23)5 – 71(23)4 + 45(23)3 + 25(23)2 – 50(23) + 100. Solution At first, the question looks daunting. On closer examination, we observe that the given expression is nothing but the remainder when x7 – 24x6 + 26x5 – 71x4 + 45x3 + 25x2 – 50x + 100 is divided by (x – 23). If we use remainder theorem, we will get back the same expression. So, remainder theorem is of use only to the extent of knowing that the given expression is the remainder. Now, we will use synthetic division to find the remainder. 23

1

1

–24

26

–71

45

25

–50

100

23

–23

69

–46

–23

46

–92

–1

3

–2

–1

2

–4

8

Therefore, the remainder is 8 and hence the value of the given expression is also 8.

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Quantitative Aptitude Simplified for CAT

EXPERT SPEAK Scan this QR Code to watch a video that explains how to use Remainder and Factor Theorems.

Algebraic Formulae In this section, we will provide very important and special results which the students must learn without fail. These results are as given below: (i)

(a + b)2 = a2 + 2ab + b2

(ii)

(a – b)2 = a2 – 2ab + b2

Therefore, (iii) (a + b)2 + (a – b)2 = 2(a2 + b2), and (iv) (a + b)2 – (a – b)2 = 4ab Moreover, (v)

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(vi) (a – b)3 = a3 – 3a2b + 3ab2 – b3 (vii) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac (viii) a3 + b3 = (a + b)(a2 – ab + b2) (ix) a3 – b3 = (a – b)(a2 + ab + b2) (x)

a2 – b2 = (a – b)(a + b)

(xi) a4 – b4 = (a2 – b2)(a2 + b2) = (a – b)(a + b)(a2 + b2) = (a – b)(a3 + a2b + ab2 + b3) = (a + b)(a3 – a2b + ab2 – b3) Example 13 If x2 + (i) (ii)

1  7, find x2

1 x 1 x– x

x+

(iii) x3 +

1 x3

(iv) x3 –

1 . x3

Solution 2

(i)

x2 +

1 1 1   x    2  7   x   2 x x x   

2

9x

252

1  3 . x

Polynomials, Algebraic Formulae and Linear Equations 2

1 1 1   x    2  7  x    5 2 x x x 

(ii)

x2 +

(iii)

x 1   x 

3

 x3 

Therefore, if x + If x +

1 1 1 = ± 3.  3  x   Now, from (i) in this example, x + x x x3 

1 1 = 3, then x 3  3  3 3  3(3)  18 . x x

1 1 1 = –3, then x 3  3  (3)3  3(3)  18 . Therefore, x 3  3  18 . x x x

Alternatively,

1  1 1   x    x 2  2  1   ( 3)(7  1)  18 3 x x x  

x3 

(iv)

x1   x 

3

 x3 

1 1  3  x   3 x x 

Now, from (ii) in this example, x  Therefore, if x  If x 

1  5. x

1 1  5 , then x 3  3  x x

1 1   5 , then x 3  3   5 x x



 5 3  3

3  3

5 8 5.

5  8 5 . Therefore, x 3 

1  8 5 . x3

Alternatively,

x3 

1  1 1   x    x 2  2  1    5  (7  1)   5 . 3 x  x x  





Example 14 Simplify the expression

2.253  1.753 . 2.252  1.752  (2.25)(1.75)

Solution Here, if a = 2.25 and 1.75 = b, then

2.253  1.753 a3  b 3  a  b (using the result (ix)) = 2.252  1.752  (2.25)(1.75) a2  b2  ab = 2.25 – 1.75 = 0.5. Example 15 Simplify the expression

4 4  24 . 4 3  4 2  2  4  22  2 3

Solution

4 4  24 a4  b 4 = = a – b (using the result (xi)) 3 2 a  a  b  a  b2  b3 4 3  4 2  2  4  22  2 3 = 4 – 2 = 2.

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Quantitative Aptitude Simplified for CAT

Some other important results (xii) a2 + b2 + c2 – ab – bc – ca =

1 [(a – b)2 + (b – c)2 + (c – a)2] 2

Therefore, a2 + b2 + c2 – ab – bc – ca ≥ 0. Also, a2 + b2 + c2 – ab – bc – ca = 0  [(a – b)2 + (b – c)2 + (c – a)2] = 0  a = b = c. (xiii) a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) =

1 (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2] 2

Deductions from this formula: A)

If a + b + c = 0, then a3 + b3 + c3 = 3abc

B)

If a3 + b3 + c3 = 3abc = 0, and a + b + c ≠ 0, then a = b = c.

(xiv) a4 + a2b2 + b4 = (a2 + ab + b2)(a2 – ab + b2)

EXPERT SPEAK Scan this QR Code to watch a video that explains how to use important Algebraic Formulas.

Example 16 If a2 + b2 + c2 ≤ ab + bc + ca, and abc = 512, then find the value of a2b + b2c + c2a. Solution a2 + b2 + c2 ≤ ab + bc + ca  a2 + b2 + c2 – (ab + bc + ca) ≤ 0. But a2 + b2 + c2 – (ab + bc + ca) cannot be less than 0. So, a2 + b2 + c2 – (ab + bc + ca) = 0  a = b = c. Therefore, abc = 512  a3 = 512. Hence, a2b + b2c + c2a = 3a3 = 3  512 = 1536. Example 17 If a3 + b3 + c3 = 3abc and a + b + c  0, then A.

a=b=c=1

B.

a b  c

C.

a=b=c

D.

a+b+c=1

Solution a3 + b3 + c3 = 3abc 

1 (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2] = 0 2

 either a + b + c = 0 or [(a – b)2 + (b – c)2 + (c – a)2] = 0. Since a + b + c  0, [(a – b)2 + (b – c)2 + (c – a)2] = 0  a = b = c. Hence correct option is [C].

254

Polynomials, Algebraic Formulae and Linear Equations Example 18 Simplify the expression

4 44  33 3555 334 5 . 44 3355 43 3553

Solution

a3  b 3  c 3  3abc 4 44  33 3555 334 5 = 2 44 3355 43 3553 a  b2  c2  ab  bc  ca = a + b + c = 4 + 3 + 5 = 12. Example 19 4

4

1  1   1  1  1     3 3  . Simplify the expression  2 2 1  1   1  1  1     3 3   Solution 4

4

1 1  1   1  1  1  2 1  2 1  a4  4  1  a  2  1   a  2  1      3 3 a a         a2  1  1  a =    2 2 1 1 2  a2   a2   1  1   1  1  1 a   1  1   2     a a2   3 3   2

2

16 9 256  81  144 193  4  3  1   =      1  . 9 16 144 144  3  4 Example 20 Find the value of the expression (a – b)2 + (b – c)2 + (c – a)2 + 2[(a – b)(b – c) + (b – c)(c – a) + (c – a)(a – b)] Solution If x = a – b, y = b – c and z = c – a, then the given expression becomes x2 + y2 + z2 + 2xy + 2yz + 2zx, which is equal to (x + y + z)2. But, x + y + z = 0. Therefore, the given expression is zero. Example 21 The prime numbers by which 343 + 213 – 553 is always divisible are A.

3, 5, 7, 11, 13

B.

3, 5, 7, 17 only

C.

3, 5, 7, 11, 17

D.

2, 3, 5, 7, 11, 17 CAT 2003 (C)

Solution 343 + 213 – 553 = a3 + b3 + c3. But, a + b + c = 34 + 21 – 55 = 0. Therefore, a3 + b3 + c3 = 3abc  the given expression 343 + 213 – 553 = 3(34)(21)(–55) = –(2  32  5  7  11  17).

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Quantitative Aptitude Simplified for CAT

Therefore, the given expression is always divisible by 2, 3, 5, 7, 11 and 17. Hence correct option is [D]. Example 22 If a2 + b2 + c2 = 1, then find the maxima and minima of ab + bc + ca. Solution We know that a2 + b2 + c2 – ab – bc – ca =

1 [(a – b)2 + (b – c)2 + (c – a)2]. 2

Since RHS is always summation of the squares, we can say that the expression a2 + b2 + c2 – ab – bc – ca is always positive, that is, a2 + b2 + c2 – ab – bc – ca  0, or a2 + b2 + c2  ab + bc + ca Since a2 + b2 + c2 = 1, therefore 1  ab + bc + ca. Moreover, (a + b + c)2  0. Therefore, a2 + b2 + c2 + 2(ab + bc + ca)  0, or ab + bc + ca  –

1 2 (a + b2 + c2). 2

Therefore, ab + bc + ca  – Hence, –

1 , as a2 + b2 + c2 = 1. 2

1  ab + bc + ca  1. 2

Maxima and minima of ab + bc + ca is 1 and –

1 . 2

Important results Following results are equally important to understand and remember. an – bn = (a – b)(an–1 + an–2b + ....+ abn–2 + bn–1) for all n, even or odd an + bn = (a + b)(an–1 – an–2b + ....– abn–2 + bn–1) for n odd only an – bn = (a + b)(an–1 – an–2b + ....+ abn–2 – bn–1) for n even only For example, a3 + b3 = (a + b)(a2 – ab + b2) a3 – b3 = (a – b)(a2 + ab + b2) a5 + b5 = (a + b)(a4 – a3b + a2b2 – ab3 + b4) a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4) a2 – b2 = (a – b)(a + b) a4 – b4 = (a – b)(a3 + a2b + ab2 + b3), or = (a + b)(a3 – a2b + ab2 – b3)

256

Polynomials, Algebraic Formulae and Linear Equations and likewise for higher powers. Note that a2 + b2, a4 + b4, and so on cannot have a factor of (a + b) The above can also be written as follows: an – bn is divisible by a – b for all values of n, even or odd. an – bn is divisible by a + b only for even values of n. an + bn is divisible by a + b only for odd values of n. an + bn is not divisible by a + b for even values of n. Example 23 2335 + 1535 is divisible by A.

19

B.

17

C.

3

D.

7

Solution Since power is 35, which is odd, therefore 2335 + 1535 is divisible by 23 + 15 = 38 and hence 19. Hence correct option is [A]. Example 24 Which of the following statements is true? A.

an – bn is divisible by a – b only when n is odd.

B.

an + bn is divisible by a – b when n is odd.

C.

an + bn is divisible by a + b for all n, even or odd.

D.

an – bn is divisible by a + b only when n is even.

Solution From what has been discussed above, we can easily see that correct option is [D].

Polynomial Equations We have already discussed about polynomials. We have also seen that a polynomial is a function of one or more variables. When a polynomial function, f(x) is equated to zero, we get what is commonly called as a polynomial equation, f(x) = 0. If the polynomial is of degree 1, the corresponding equation is linear equation. If the polynomial is of degree 2, the corresponding equation is quadratic equation. If the polynomial is of degree 3, the corresponding equation is cubic equation. and so on and so forth. Quadratic and higher degree equations will be discussed in the next chapter.

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Quantitative Aptitude Simplified for CAT

Linear Equation Linear equations are those equations where the degree of the polynomial is 1. The number of variables can be 1, 2 or more. Therefore, 2x + 3 = 7; 4x – 9y = 12; x + 4y + 6z = 8 are examples of linear equations with one, two and three variables, respectively. Let us look at these equations in some more detail. 2x + 3 = 7 is a linear equation with one variable. If we solve this equation for the value of x, we get: x=

73 2 . 2

So, we see that we get a single value of x and there is no other value of x which satisfies the given equation. We can say that the equation 2x + 3 = 7 has a unique solution. The other equation is 4x – 9y = 12, which can be rewritten as x =

9 y  12 . 4

So, the value of x depends upon the value of y. If we take some value of y, it will give a corresponding value of x. So, the value of x is not unique. For example, if y = 0, x =

9(0)  12 3 4

if y = 1, x =

9(1)  12  5.25 4

if y = 2, x =

9(2)  12  7.5 4

and so on. Therefore, there are infinitely different ordered pairs of (x, y) possible. We say that the equation 4x – 9y = 12 has infinite solutions. Similarly, the third equation is x + 4y + 6z = 8, which can be rewritten as x = 8 – (4y + 6z), and whose value also depends upon on y and z. By providing values to y and z, we get corresponding values of x. We can easily see that this equation also has infinite solutions. From the discussion, we can conclude that: ”a linear equation of 1 variable has a unique solution whereas a linear equation of more than 1 variables will always have infinite solutions”.

SYSTEM OF SIMULTANEOUS EQUATIONS Linear equations can also occur in pairs, triplets, and so on. If we have two equations in 2 variables, or 3 equations in 3 variables, then they are called system of simultaneous equations. So, a1x + b1y = c1 a2x + b2y = c2 is an example of a system of simultaneous equations. Let us discuss the system for two variables in some more detail. If we have a system of simultaneous equations, then the system can have unique solution, infinite solutions or no solution. There are certain conditions which determine how many solutions exist.

258

Polynomials, Algebraic Formulae and Linear Equations

If

a1 b1  , then the system has a unique solution. a2 b2

If

a1 b1 c1   , then the system has infinite solutions. a2 b2 c 2

If

a1 b1 c1   , then the system has no solution. a2 b2 c 2

For example, if the system of equations is 2x + 5y = 7 and 4x + 9y = 12, then the system has a unique 2 5 solution, because  . 4 9 Similarly, if the system of equations is 2x + 5y = 7 and 4x + 10y = 14, then the system has infinite solutions, 2 5 7 because   . 4 10 14 Finally, if the system of equations is 2x + 5y = 7 and 4x + 10y = 13, then the system has no solution, because 2 5 7   . 4 10 13 We also observe in case of “infinite solutions” that the second equation 4x + 10y = 14 is obtained by multiplying the first equation 2x + 5y = 7 by some factor (which is 2 in the current case). We can generalize that whenever one equation is a multiple of the other equation, then the system of equations will always have infinite solutions; whereas if only the coefficients of the variables (but not the constant term) in one equation are a multiple of those in the other equation, then the system of equations has no solution. Example 25 For what value of k will the following system of equations have infinite solutions? 5x + 3y + 9 = 0; 15y + kx + 45 = 0 Solution Rearranging the equations, we get: 5x + 3y = –9 kx + 15y = –45 Applying the condition for “infinite solutions”, we get: 5 3 9    k = 25. k 15  45

Example 26 For what value of k will the following system of equations have no solution? 3x + 5y = 9; 4x + ky = 12 Solution For “no solution”, 3 5 9   4 k 12

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Quantitative Aptitude Simplified for CAT

Solving

3 5 20 5 9 20  , we get k = and solving  , we get k  . 4 k 3 k 12 3

So, for no value of k will the system of equations have no solution. This system of equations will either have unique solution or infinite solutions. Example 27 State true or false: Let there be a system of equations a1x + b1y = c1; a2x + b2y = c2. a b 1. If 1  1 , then there will be infinite solutions. a2 b2 2.

If

a1 c1  , then there will be infinite solutions. a2 c2

3.

If

a1 c1  , then there will be no solution. a2 c2

4.

If

a1 c1 b1   , then there will be unique solution. a2 c 2 b2

Solution 1.

2.

3.

4.

a1 b1  , then we cannot surely say that there will be infinite solutions, because the said ratio a2 b2 a b c may not be equal to 1 . Therefore, the statement is false. If 1  1 , then at best we can say that c2 a2 b2 there will be either infinite solutions or no solution. Unique solution would never exist. a c If 1  1 , then also we cannot surely say that there will be infinite solutions, because the said a2 c2 b ratio may not be equal to 1 . At best we can say that the case of “no solution” would never prevail. b2 So, (2) is also a false statement. a c a b a b If 1  1 , then either 1  1 or 1  1 . The former is a case of “no solution” and the latter is a2 c2 a2 b2 a2 b2 a case of “unique solution”. So, (3) is not necessarily correct. a b a If 1  1 , then there will definitely be a unique solution, irrespective of whether 1 is equal to a2 a2 b2 c1 or not. Therefore, this is a correct statement. c2 If

Example 28 Solve the following system of three equations 2x + 5y = 7 4x – 3y = –12 10x + 7y = –1 Solution In such a case, we will solve any two equations. If the solution obtained satisfies the third equation, then

260

Polynomials, Algebraic Formulae and Linear Equations that is the solution of the given system of equations. Solving the first two equations, 4x + 10y = 14 4x – 3y = –12 Subtracting, we get: 13y = 26  y = 2  x = –

3 . 2

Observe that the solution obtained satisfies the third equation. Hence this is the solution of the given system of equations.

CONSISTENCY AND INCONSISTENCY OF EQUATIONS We have discussed above that if we have a system of simultaneous equations, then we can have either one solution, infinite solutions or no solution. Consistency simply means existence of solution and inconsistency means absence of solution. So, if a system of equations has one solution or infinite solutions, then the system of equations is said to be consistent and if it has no solution, then the system of equations is inconsistent. Example 29 Check for consistency and inconsistency of the following system of equations: (i)

3x + 5y = 12; 4y – 2x = 7

(ii)

x + 6y + 7 = 0; 3x + 18y = 21

(iii) 6x + 9y = 17; 4x + 3y = 10y – 7x Solution (i)

In this,

b a1 3 5   and 1  . The system has one solution and hence is consistent. a2 2 b2 4

c a1 1 b1 1   , but 1   , so the system has no solution. Therefore, the system is c2 3 a2 3 b2 inconsistent. a 6 b 9 (iii) Since 1  and 1   , the system is consistent. a2 11 b2 7 (ii)

Here,

DEPENDENT AND INDEPENDENT EQUATIONS Let us say we have two equations 2x + 3y = 5 and 6x + 9y = 15. The second equation is obtained by multiplying the first equation by 3. Such a system of two equations is called dependent equations. If the equations are 2x + 3y = 5 and 6x + 7y = 12, then the second equation is not a multiple of first equation and such a system is called independent equations. It is very important to understand that we talk of dependent and independent equations only in case the system of equations is consistent. If the system is inconsistent, then we cannot talk of dependent and independent equations. Note that independent equations will always have a unique solution and dependent equations will always have infinite solutions. Let us take the case of three variables. We have a system of three equations 2x + y + 3z = 3

(a)

261

Quantitative Aptitude Simplified for CAT

3x – 2y + 4z = 7

(b)

13x – 4y + 18z = 27

(c)

We observe that third equation is obtained by a linear combination of first two equations (2A + 3B = C). This effectively means that we have only two independent equations and not three equations (as it would appear otherwise). Such a system of three equations forms a system of dependent equations and hence such a system will have infinite solutions.

Conclusion If we have n ‘independent’ equations in n variables, then we will always have a unique solution. In other words, in a given system of simultaneous equations, we need minimum as many independent equations as the number of variables in order to be able to get a unique solution. Example 30 Check whether the following system of equations is dependent or independent. (i)

2x + 4y = 7; 8x + 16y = 28

(ii)

x – 7y + 6 = 0; 3x – 20y = 9

(iii) 2x + 3y + z = 9; x – 5y + 3z = 3; 5x + y + 5z = 10 Solution (i) (ii)

Since second equation is 4 times the first equation, the system of equations is dependent. 1 7  . Hence the system of equations is independent. 3  20

(iii) For the system to be dependent, third equation should be a “linear combination” of the first two. In other words, we need to check whether or not m(2x + 3y + z – 9) + n(x – 5y + 3z – 3) = 5x + y + 5z – 10. Comparing the coefficients of x and y, we realize that 2m + n = 5; 3m – 5n = 1  13m = 26  m = 2 and n = 1. Comparing the coefficient of z, we get m + 3n = 5, which is satisfied by the value of m and n obtained. Comparing the constant terms, we get –9m – 3n = –10, which is not satisfied by the value of m and n obtained. Therefore, m(2x + 3y + z – 9) + n(x – 5y + 3z – 3) = 5x + y + 5z – 10 is not possible for any value of m and n. In other words, 2(first equation) + (second equation)  third equation. Hence the system of equations is neither dependent nor independent. The system has no solution (or is inconsistent). Example 31 For what value of  will the following system of equations have a unique solution. x – y + 2z = 3; 4x + 3y + 2z = 5; 3x + y – 2z = 3. Solution For unique solution, the three equations must be independent equations. Therefore, x – y + 2z = 3

(A)

262

Polynomials, Algebraic Formulae and Linear Equations 4x + 3y + 2z = 5

(B)

3x + y – 2z = 3

(C)

We need to check whether mA + nB = C, or not. m(x – y + 2z – 3) + n(4x + 3y + 2z – 5) = (3x + y – 2z – 3) Comparing the coefficients of x and y, we get: m + 4n = 3

(i)

–m + 3n = 1

(ii)

2m + 2n = –2

(iii)

Solving (i) and (ii): 7n = 4  n =

4 5  m = 3 – 4n = . 7 7

If this value of m and n satisfies (iii), then the system will have infinite solutions. So, 2×

5 4 +2× = –2   = –2.2. 7 7

So, if  = –2.2, then the system will be dependent and hence have infinite solutions. Therefore,  should not be equal to –2.2 so that the equations become independent and hence will have a unique solution. Example 32 Discuss the number of solutions of the system of equations: 2x + 3y + 4z = 9 and 4x + 6y + 5z = 15. Solution We observe that the coefficients of x and y in the second equation are multiple of those in first equation. Taking twice of the first equation and subtracting second from first, we get: (4x + 6y + 8z) – (4x + 6y + 5z) = 18 – 15 = 3 3z = 3 z = 1. Putting this value of z in the given equations, we get: 2x + 3y = 5

(A)

4x + 6y = 10

(B)

In these equations,

a1 b1 c1   . Thus, the system of equations (A) and (B) have infinite solutions. a2 b2 c 2

From the above, we can conclude that the given system of two equations in three variables have infinite solutions, though we can uniquely solve for the value of z. The last example above was just to elaborate that even though the number of equations are less than the number of variables, one of the variables may have a unique solution. Example 33 Find the value of  for which the following system of equations does not have unique solution. Check whether the system has infinite solutions or no solution. 2x + 5y + 2z = 3 x – 2y + 7z = 4 5x – y + (3 + 20)z = 5

263

Quantitative Aptitude Simplified for CAT

Solution To find  for no unique solution, m(2x + 5y + 2z) + n(x – 2y + 7z) = 5x – y + (3 + 20)z 2m + n = 5; 5m – 2n = –1; 2m + 7n = (3 + 20) 9m = 9  m = 1  n = 3. It should satisfy 3rd equation for no unique solution. Therefore, 2 + 21 = 3 + 20   = 1. So, the three equations become 2x + 5y + 2z = 3

(A)

x – 2y + 7z = 4

(B)

5x – y + 23z = 5

(C)

We observe that coefficients of (A), (B) and (C) are connected by (A) + 3(B) = (C), but the constant terms do not follow that and hence the system has no solution. Example 34 If 2 chocolates, 3 candies, and 5 cakes are priced at Rs 45, and 4 candies, 2 cakes and 3 chocolates are priced at Rs 56, then what is the price of 16 cakes, 13 chocolates and 18 candies? Solution Let x, y and z represent the price of each chocolate, candy and cake, respectively. Then, 2x + 3y + 5z = 45 3x + 4y + 2z = 56 Since we are given two equations and three variables, we cannot find the values of the variables uniquely. It is very important to note that we must not conclude at this stage that the answer cannot be found out! What is asked is the value of 13x + 18y + 16z. If this expression is some linear combination of (2x + 3y + 5z) and (3x + 4y + 2z), then the price required can be found out. Let m(2x + 3y + 5z) + n(3x + 4y + 2z) = (13x + 18y + 16z) 2m + 3n = 13; 3m + 4n = 18; 5m + 2n = 16. Solving the first two equations in m and n, we get: n = 3 and m = 2, which also satisfies third equation. Therefore, (13x + 18y + 16z) = 2(2x + 3y + 5z) + 3(3x + 4y + 2z) = 2 × 45 + 3 × 56 = Rs 258. Example 35 A two-digit number is such that when its digits are reversed and added to the original number, the result is 46 more than if the two numbers were subtracted. If the ten’s place digit in the original number is 50% more than the digit in unit’s place, find the number. Solution Let the number be ‘ab’ (this is not ‘a’ multiplied by ‘b’). Value of ab = 10a + b. On reversing the digits, the number obtained = ba = 10b + a. Adding them, we get 11(a + b), whereas subtracting them, we get 9(a – b). As per the given condition, 11(a + b) – 9(a – b) = 2a + 20b = 46. It is also given that: a = 1.5b  2a = 3b. Putting this in the first equation, 3b + 20b = 46  b = 2 and hence a = 3. The number is 32.

264

Polynomials, Algebraic Formulae and Linear Equations Example 36 Age of a man is sum of the ages of his wife and the only son. After 10 years, the sum of the man’s age and wife’s age is 2 less than 5 times the son’s age. If after further 10 years, the age of the son is 2 years more than that of the man when the son was born, what is the age of the man when his son was born? Solution Let the age of man, wife and son be x, y and z. As per the question, x=y+z (x + 10) + (y + 10) + 2 = 5(z + 10) (x – z) + 2 = (z + 20) The equations can be simplified as x–y–z=0 x + y – 5z = 28 x – 2z = 18 Putting x = 18 + 2z in the first two equations, we get: y – z = 18 y – 3z = 10  2z = 8  z = 4 x = 26 and y = 22. When man is 26 years old, son is 4 years old. Therefore, man was 22 years old when the son was born. Example 37 Remuneration paid by a rich man to his servant is Rs 400 per month. If the servant completes a year, he will also be given a turban. The servant’s mother (who lives in a village) dies 9 months later and therefore, the servant has to leave the rich man’s house. The rich man pays him Rs 3550 and the turban. What is the price of the turban? Solution Let the price of the turban be x. Then, Annual remuneration = 12  400 + x = 4800 + x. Then, remuneration for 9 month would be =

3 (4800 + x), which is equal to 3550 and a turban. Therefore, 4

3 1 (4800 + x) = 3550 + x  x  3600 – 3550 = 50  x = Rs 200. 4 4

The price of the turban is Rs 200.

265

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

If R = A. B. C. D.

3065  2965 , then 3064  2964

6.

0 < R  0.1 0.1 < R  0.5 0.5 < R  1.0 R > 1.0 CAT 2005 1

2.

If

4  24

1

22  3 c

 

 43

4

2  32  c

1

4  c4

2 3c2

 64 , find

Consider a function f(x) = x4 + x3 + x2 + x + 1, where x is a positive integer greater than 1. What will be the remainder if f(x5) is divided by f(x)? A.

1

B.

4

C.

5

D.

a monomial in x

E.

a polynomial in x

the value of c. A. B. C. D.

XAT 2013

4 4 5 5

7.

CAT 2013 3.

(x + y + z)(xy + yz + zx)  xyz = A.

(x + y)(y + z)(z + x)

B.

(x  y)(y  z)(z  x)

C.

(x + y + z)2

D.

(x + y + z)3 CAT 2014

4.

If x2 + 3x – 10 is a factor of 3x4 + 2x3 – ax2 + bx – a + b – 4, then the closest approximate values of a and b are A.

25, 43

B.

52, 43

C.

52, 67

D.

None of the above

8.

If a, b and c are 3 consecutive integers between –10 to +10 (both inclusive), how many integer values are possible for the expression

A.

15

B.

14

C.

12

D.

10 CAT 2005

IIFT 2013 5.

A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs.250 and Rs.300 per day respectively. In addition, a male operator gets Rs.15 per call he answers and a female operator gets Rs.10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?

What are the values of x and y that satisfy both the equations? 20.7x × 3–1.25y =

a3  b 3  c 3  3abc (a  b  c )2

8 6 27

40.3x × 90.2y = 8 ×

5

A.

x = 2, y = 5

A.

0

B.

x = 2.5, y = 6

B.

1

C.

x = 3, y = 5

C.

2

D.

x = 3, y = 4

D.

3

E.

x = 5, y = 2

E.

4

81

CAT 2006

XAT 2016

266

Polynomials, Algebraic Formulae and Linear Equations 9.

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x ≤ y is:

A.

Over Rupees 13 but less than Rupees 14

B.

Over Rupees 7 but less than Rupees 8

A.

7

C.

B.

13

Over Rupees 22 but less than Rupees 23

C.

14

D.

D.

18

Over Rupees 18 but less than Rupees 19

E.

20

E.

Over Rupees 4 but less than Rupees 5

13. How many pairs of positive integer m, n

CAT 2006

satisfy

Directions for questions 10 and 11: Solve the questions on the basis of the information given below:

integer less than 60?

An airline has a certain free luggage allowance and charges for excess luggage at a fixed rate per kg. Two passengers, Raja and Praja have 60 kg of luggage between them, and are charged Rs.1200 and Rs.2400 respectively for excess luggage. Had the entire luggage belonged to one of them, the excess luggage charge would have been Rs.5400.

B.

25 kg

C.

30 kg

D.

35 kg

E.

40 kg

6

B.

4

C.

7

D.

5

E.

3

14. The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?

10. What is the weight of Praja’s luggage? 20 kg

A.

CAT 2007

CAT 2006

A.

1 4 1 , where n is an odd   m n 12

11. What is the free luggage allowance?

A.

May 21

B.

April 11

C.

May 20

A.

10 kg

D.

April 10

B.

15 kg

E.

June 30

C.

20 kg

D.

25 kg

E.

30 kg

CAT 2007 15. The total number of integer pairs (x, y) satisfying the equation x + y = xy is

12. A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Shailaja noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount?

A.

0

B.

1

C.

2

D.

None of the above CAT 2004

16. If x and y are integers, then the equation 5x + 19y = 64 has:

CAT 2007

267

A.

no solution for x < 300 and y < 0

B.

No solution for x > 250 and y > 100

Quantitative Aptitude Simplified for CAT

C.

a solution for 250 < x < 300

D.

A solution for –59 < y < 56

21. Two farmers were cultivating wheat on their respective agricultural land in a village. Farmer A had an average production of 20 bushels from a hectare. Farmer B, who had 15 hectares of more land dedicated to wheat cultivation, had an output of 30 bushels of wheat from a hectare. If farmer B harvested 530 bushels of wheat more than farmer A, how many bushels of wheat did farmer A cultivate?

CAT 2003 (R) 17. ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0, have a solution, where a, b and c are distinct, if A.

a+b+c=0

B.

a2 + b2 + c2 = ab + bc + ca

C.

(a + b + c)2 = ab + bc + ca

D.

None of these

A. B. C. D.

CAT 2012 18. The number of Rs.2 coins is 4 times as many as the number of Re.1 coins, and the ratio of the number of Rs.2 coins to that of Rs.5 coins is 7 : 2. If total number of coins lies between 200 and 250, what is the total value of all the coins? A.

Rs 415

B.

Rs 515

C.

Rs 510

D.

Rs 410

50 80 160 200 IIFT 2016

22. In 2004, Rohini was thrice as old as her brother Arvind. In 2014, Rohini was only six years older than her brother. In which year was Rohini born? A. B. C. D.

1984 1986 1995 2000

CAT 2013

IIFT 2015

19. Cost of a pen is twice the cost of a pencil. A man wanted to buy certain number of pens and 16 pencils. But by mistake, the number of pens and pencils got exchanged and so the man ended up paying 50% more. How many pens did he want to buy?

23. A firm is thinking of buying a printer for its office use for the next one year. The criterion for choosing is based on the least per-page printing cost. It can choose between an inkjet printer which costs Rs. 5000 and a laser printer which costs Rs. 8000. The per-page printing cost for an inkjet is Rs. 1.80 and that for a laser printer is Rs. 1.50. The firm should purchase the laser printer, if the minimum number of a pages to be printed in the year exceeds

A.

8

B.

16

C.

4

D.

10

A. B. C. D.

CAT 2015 20. The ages of A, B and C are in AP. After 2 years, the age of the eldest is thrice that of the youngest. After further 4 years, the age of the eldest becomes twice that of the youngest. Find the sum of their current ages. A.

12 years

B.

18 years

C.

20 years

D.

15 years

5000 10000 15000 18000 IIFT 2015

24. The value of x for which the equation 4 x  9  4 x  9  5  7 , will be satisfied, is A. B. C. D.

CAT 2015

1 2 3 4 IIFT 2015

268

Polynomials, Algebraic Formulae and Linear Equations 25. Ravindra and Rekha got married 10 years ago. Their ages were in the ratio of 5 : 4. Today Ravindra’s age is one sixth more than Rekha’s age. After marriage, they had 6 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3 : 2 : 1. What is the largest possible value of the present total age of the family?

and mother was 70 years. The sum of the ages of four family members, at the time of Gouri’s birth, was twice the sum of ages of Hari’s father and mother at the time of Hari’s birth. If Chari is 4 years older than Gouri, then find the difference in age between Hari and Chari. A.

5 years

B.

6 years

A.

79

B.

93

C.

7 years

C.

101

D.

8 years

D.

107

E.

9 years

IIFT 2014

XAT 2017

26. Mrs. Sonia buys Rs.249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavored candy priced at Rs.3.30 per candy; each boy receives a chocolate flavored candy priced at Rs.2.90 per candy. How many candies of each type did she buy?

29. In a True/False quiz, 4 marks are awarded for each correct answer and 1 mark is deducted for each wrong answer. Amit, Benn and Chitra answered the same 10 questions, and their answers are given below in the same sequential order. AMIT

T T F F T T F T T F

BENN

T T T F F T F T T F

37, 51

CHITRA

T T T T F F T F T T

27, 51

If Amit and Benn both score 35 marks each, then Chitra’s score will be:

A.

21, 57

B.

57, 21

C. D.

IIFT 2013 27. At a reputed Engineering College in India, total expenses of a trimester are partly fixed and partly varying linearly with the number of students. The average expense per student is Rs.400 when there are 20 students and Rs.300 when there are 40 students. When there are 80 students, what is the average expense per student? A.

Rs.250

B.

Rs.300

C.

Rs.330

D.

Rs.350

A.

10

B.

15

C.

20

D.

25

E.

None of the above XAT 2017

30. In an examination, two types of questions are asked: one mark questions and two marks questions. For each wrong answer of one-mark question, the deduction is

1 4

th

of

a mark and for each wrong answer of two-

IIFT 2012

marks question, the deduction is

28. Hari’s family consisted of his younger brother (Chari), younger sister (Gouri), and their father and mother. When Chari was born, the sum of the ages of Hari, his father

mark. Moreover,

1 3

rd

of a

1 of a mark is deducted 2

for any unanswered question. The question paper has 10 one-mark questions and 10

269

Quantitative Aptitude Simplified for CAT

two-marks questions. In the examination, students got all possible marks between 25 and 30 and every student had different marks. What would be the rank of a student, who scores a total of 27.5 marks? A. B.

C. D. E.

7 8 None of the above XAT 2015

5 6

ANSWER KEY 1. D

2. D

3. A

4. C

5. C

6. C

7. D

8. E

9. B

10. D

11. B

12. D

13. E

14. C

15. C

16. C

17. A

18. B

19. C

20. B

21. C

22. C

23. B

24. D

25. D

26. B

27. A

28. E

29. A

30. A

270

Polynomials, Algebraic Formulae and Linear Equations

ANSWERS AND EXPLANATIONS 1.

5.

Explanation: Substituting this in the expression we get the 2x 2  1 expression . Only x = ±1 gives an 3x integer.

Answer: D Explanation: R=

3065  2965 (30  29)(3064    2964 ) = 64 64 30  29 3064  2964

6.

3064  2964 . Obviously numerator is 3064  2964

=

Answer: D Explanation: 1 24 22  3 c

4 

 4 

 4

23 23c

23 33 c3 23c

1 2 32  c

 

 43

4

  4

33 23c

1 c 4 2 3c2

4 

4 

c3 23c

 64

 64 

7.

3

23  33  c 3  3  23 + 33 + c3 23c = 3 (2 × 3 × c)  2 = 3 = c or 2 + 3 + c = 0. Since 2 ≠ 3, the only possibility is 2 + 3 + c = 0 or c = 5.

Answer: A Explanation: (x + y + z)(xy + yz + zx)  xyz = (x2y + xyz + x2z + y2x + y2z + xyz + xyz + z2y + z2x) – xyz = (x2y + xyz + x2z + y2x + y2z + xyz + z2y + z2x) = x2(y + z) + xy(y + z) + yz(y + z) + xz(y + z) = (y + z)(x2 + xy + yz + xz) = (x + y)(y + z)(z + x) Alternatively, Put x = y = z = 1. Given expression becomes = (1 + 1 + 1)(1 + 1 + 1) – (1)(1)(1) = 8. Putting in option [A], we get (1 + 1)(1 + 1)(1 + 1) = 8. In no other option, we get 8.

4.

Answer: D Explanation: Total calls per day = 1000 Each male can handle 40 calls per day. Each female can handle 50 calls per day Each male earns per day = Rs.(250 + 15× 40) = Rs.850 Each female earns per day = Rs.(300 + 10 × 50) = Rs.800 It is obvious, that 1000 = 40x + 50y, where x and y are number of male and female operators. Then two possible values of (40x, 50y) are (400, 600) and (600, 400) (because at least 7 female operators are compulsory) If 40x = 400 and 50y = 600 x = 10, y = 12 If 10 male and 12 female operators are working, then total cost = (850 × 10 + 800 × 12) = Rs.18100 If 12 male and 10 female operators are working, then total cost = (850 × 12 + 800 × 10) = Rs.19150. Least cost is Rs.18100, when 10 male employees are working.

 64  4 .

Therefore,

3.

Answer: C Explanation: x5 – 1 is divisible by x4 + x3 + x2 + x + 1. Therefore, x5 when divided by x4 + x3 + x2 + x + 1 leaves remainder 1. Now, f(x5) = (x5)4 + (x5)3 + (x5)2 + (x5) + 1. Each term when divided by x4 + x3 + x2 + x + 1 leaves remainder 1. So, f(x5) when divided by x4 + x3 + x2 + x + 1 leaves remainder = 1 + 1 + 1 + 1 + 1 = 5.

more than denominator. 2.

Answer: C

Answer: C Explanation: x2 + 3x – 10 = (x + 5)(x – 2). If f(x) = 3x4 + 2x3 – ax2 + bx – a + b – 4, then f(2) = 0 and f(5) = 0. f(2) = 48 + 16 – 4a + 2b – a + b – 4 = 0  5a – 3b – 60 = 0; f(–5) = 1875 – 250 – 25a – 5b – a + b – 4 = 0  26a + 4b – 1621 = 0. Solving, we get 20a + 78a – 240 – 4863 = 0  a = 52.07  52.

8.

Answer: E Explanation: 8 6 = 23.5 × 3–2.5  0.7x = 27 3.5, –1.25y = –2.5  x = 5, y = 2.

20.7x × 3–1.25y =

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Quantitative Aptitude Simplified for CAT

13. Answer: E

This solution also satisfies the second equation. 9.

Explanation: 1 4 1    m = m n 12

Answer: B

The given equation is:

Explanation: As per the equation, y must be an even integer. The maximum value of y can be 38 and minimum value 2. These are 19 values. However, x ≤ y. So, minimum possible value of y is 14. So, removing these 6 values, the number of solutions is 13.

12n . So, 48 < n < 60 and n is odd. So, n n  48 can be 49, 51, 53, 55, 57, 59, out of which only 49, 51, and 57 will make m to be an integer. So, only three such values of n will satisfy the given conditions.

14. Answer: C

Common explanation for questions 10 and 11:

Explanation: Prices never become equal before 100 days (using options). Price of Darjeeling tea after 100 days = 100 + 0.1  100 = 110. Therefore, 110 = 89 + 0.15n or n = 140  20th May.

Let the free luggage allowance be A and excess luggage for Raja be E. Since Praja pays double the amount paid by Raja for excess luggage, excess luggage for Praja is 2E. When combined, total luggage = 2A + 3E. Excess luggage when combined = A + 3E. Since E corresponds to Rs 1200, A corresponds to 5400 – 3600 = Rs 1800. If E = 2x, then A = 3x. Total luggage of Praja = 7x. Combined total luggage = 12x = 60 kg. So, x = 5. Therefore, weight of Praja’s luggage = 35 kg. Free luggage allowance, A = 3x = 15 kg.

15. Answer: C Explanation: Given equation is x + y = xy. By observation, only two pairs satisfy this equation: (0, 0) and (2, 2). 16. Answer: C Explanation: The given equation can be rewritten as 5x = 64 – 19y. For this, y must be so chosen that RHS becomes a multiple of 5. We observe that y = 1, x = 9, which is one of the possible solution of this given equation. The values of x will be arithmetic progression series with coefficient of y as the common difference. So, x = 9, 28, 47, etc. will give integer values of y. In other words, the gap between consecutive solutions of x will be 19. So, there will definitely be a solution of x between 250 and 300. Similarly, values of y will also an AP with coefficient of x as the common difference. So, y = 1, 4, 9, 14, 19, … etc will give integer values of x. So, [D] option is not possible.

10. Answer: D Explanation: The weight of Praja’s luggage is 35 kg. 11. Answer: B Explanation: The free luggage allowance is 15 kg. 12. Answer: D Explanation: Let cheque amount be Rs X and Y paise. The confused bank teller cashed Rs Y and X paise. It is given that after buying a toffee for 50 paise, Shailaja is left with 3 times of the amount on the cheque. Therefore, Y   X 50  3  X     Y    300X + 100   100 100   3Y = 100Y + X – 50  299X = 97Y – 50 Now use options. When X = 18, then Y = integer.

17. Answer: A Explanation: We can immediately observe that x = 1, y = 1 will satisfy all the equations. So, a + b + c = 0. Alternatively,

272

Polynomials, Algebraic Formulae and Linear Equations Solve the equations: ax + by + c = 0, bx + cy + a=0 (ax + by + c = 0) × b (bx + cy + a = 0) × a a2  bc b2y – acy + bc – a2 = 0  y = 2 x b  ac  a2  bc    c  b  2 b  ac  c 2  ab   2 . Putting this = a b  ac solution in third equation: cx + ay + b = 0, we get:  c2  ab   a2  bc  3 3   a 2  c  2   b  ac  + b = 0  a + b  b  ac    + c3 – 3abc = 0  either a + b + c = 0 or a = b = c. Since a ≠ b ≠ c, a + b + c = 0.

(a + d) + 2 = 3(a – d + 2)  2a – 4d + 4 = 0  a – 2d + 2 = 0 Also, (a + d) + 6 = 2(a – d + 6)  a – 3d + 6 =0 Solving these equations, we get d = 4 and a = 6. So, sum of ages = 3a = 18 years. 21. Answer: C Explanation: If farmer A had x hectares, farmer B would have had (x + 15) hectares. Then, as per the question 30(x + 15) – 20x = 530  x = 8 hectares. So, number of bushels of wheat cultivated by A = 8 × 20 = 160 bushels. 22. Answer: C

18. Answer: B

Explanation: Difference in their ages will always remain 6 years. So, if age of Rohini’s brother in 2004 = x. Then Rohini’s age = x + 6. But, Rohini’s age in 2004 is thrice of her brother’s age. So, x + 6 = 3x  x = 3. So, Rohini’s age in 2004 = 9 and so she was born in 1995.

Explanation: The number of Rs.2 and Rs.5 coins are 7x 7x . and 2x. So, number of Re.1 coins = 4 Therefore, total value of all the coins 103  7x   1 x; =  + (7x) × 2 + (2x) × 5 =  4  4  43 x . If total and total number of coins = 4 number of coins lies between 200 and 250, then x = 20. Therefore, total number of coins = 215. Hence total value of coins = Rs.515. Alternatively, 103 x . So, value must Total value of coins = 4 be a multiple of 103, which is 515 only.

23. Answer: B Explanation: Let number of pages printed be x. Then, total cost due to inkjet printer = 5000 + 1.8x; total cost due to laser printer = 8000 + 1.5x. As per the conditions given in the question, 8000 + 1.5x < 5000 + 1.8x  x > 10000. 24. Answer: D Explanation: Use options. The equation is satisfied for x = 4.

19. Answer: C Explanation: Number of pencils is 16. Let number of pens he wanted to buy = n. Also, let the cost of each pencil = x, so cost of each pen = 2x. Total cost = 16x + 2nx. Because of the mistake, total cost = 16(2x) + nx, which is 50% more than 16x + 2nx. So, 32x + nx = 1.5(16x + 2nx) = 24x + 3nx  2n = 8  n = 4.

25. Answer: D Explanation: The ages of Ravindra and Rekha at the time of their marriage are 5x and 4x. As on today, their ages are 5x + 10 and 4x + 10. As per the question, 7 5x + 10 = (4x + 10)  x = 5. Therefore, 6 their ages at the time of their marriage = 25 and 20 years. To maximize the present total age, triplet should have been born at the end

20. Answer: B Explanation: Let the ages of A, B and C be a – d, a, a + d. Then,

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Quantitative Aptitude Simplified for CAT

of the first year. So, present age of triplets = 9 years each. Age of each twin = 6 years and age of 6th child = 3 years. So, present total age = 35 + 30 + 3 × 9 + 2 × 6 + 3 = 107 years.

Hari’s age be h. So, father’s and mother’s age becomes (f + h) and (m + h). Since Chari is 4 years older than Gouri, we can say that Gouri was born 4 years later. When Gouri was born, ages of father and mother are (f + h + 4) and (m + h + 4) and that of Hari is (h + 4). As per the question,

26. Answer: B Explanation:

(f + h + 4) + (m + h + 4) + (h + 4) + 4 = 2(f + m)  f + m – 3h = 16.

Let the number of strawberry flavor and chocolate flavor candies be x and y. Then,

Also, (f + h) + (m + h) + h = 70  f + m + 3h = 70.

3.3x + 2.9y = 249. Now, use options.

Solving these two equations, we get 6h = 54  h = 9, which is the age of Hari when Chari was born and hence 9 years is the age difference between Hari and Chari.

27. Answer: A Explanation: Let f be fixed cost the total and v be the variable cost (per student). Then,

29. Answer: A

400 × 20 = f + 20v; 300 × 40 = f + 40v.

Explanation:

Solving, we get v = Rs.200 and f = Rs.4000.

The correct answer fetches 4 marks and wrong answer fetches 1 marks. Amit and Benn scored 35 marks each in this 10 questions.

So, total cost for 80 students = 4000 + 80 × 200 = 20000. So, average expense per 20000 = Rs.250. student = 80

4x – y = 35; x + y = 10  x = 9, y = 1 which means 9 correct answers and 1 wrong answer.

Alternatively, When number of students goes up from 20 to 40, that is number of goes up by 20, the cost escalates by Rs.4000. So, per student variable cost = Rs.200. Also, when there are 20 students, average cost per head is 400. So, 400 – 200 = Rs.200 per head is the fixed cost per student. When number of students

Note that except for the third and fifth questions, answers of Amit and Benn are same. So, answers given by Amit and Benn for the remaining questions are correct. Comparing with Chitra’s answers, it is clear that she had answered 3 questions (that is, 1st, 2nd and 9th) correctly.

increases to 80, the fixed cost per head will 1 become th of Rs.200 = Rs.50. So, average 4 cost per head = 200 + 50 = Rs.250.

Not counting score in 3rd and 5th question, she will score 4 × 3 – 1 × 5 = 7 marks. For the third and the fifth questions, her answers match with Benn. So, she will score 4 – 1 = 3 marks. Thus, Chitra will score 7 + 3 = 10 marks.

28. Answer: E Explanation: When Hari was born, let the age of father and mother be f and m. When Chari was born, let

274

Polynomials, Algebraic Formulae and Linear Equations 30. Answer: A Explanation: Following scores are possible under certain situations. Number of correct attempts

Number of incorrect attempts

Number of non-attempts

Score

1 mark

2 marks

1 mark

2 marks

1 mark

2 marks

10

10

0

0

0

0

30

9

10

0

0

1

0

28.5

9

10

1

0

0

0

28.75

10

9

0

0

0

1

27.5

10

9

0

1

0

0

27.67

If we write the scores in descending order, the score of 27.5 is at rank number 5.

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Quantitative Aptitude Simplified for CAT

Chapt er 08

Quadratic Equation INTRODUCTION This chapter is a continuation of the “theory of equations”, in which we will not only discuss the aspects relating to quadratic equations, but also those relating to higher degree equations, like cubic, biquadratic (also known as quartic). An equation of the form ax2 + bx + c = 0, where a, b and c are unknown constants and a  0, is a quadratic equation. The equation can be rewritten as x 2 

b c x 0 a a

2 2 b  c b2 b  b2 c b 2  4 ac      0  x        x     2a  a 4a 2 2a  4 a2 a 4 a2  

b  b 2  4 ac b 2  4 ac     x     2a  2a 4a 2  x=

 b  b2  4 ac . 2a

 b  b 2  4 ac  b  b 2  4 ac and . These values of x are called the roots of the 2a 2a quadratic equation. Roots of a quadratic equation are those values of x which will satisfy the given equation. Roots are generally represented by  and . Therefore,

We have two values of x:

=

 b  b 2  4 ac  b  b 2  4 ac and  = . 2a 2a

Note that the number of roots of any polynomial equation is equal to the degree of the polynomial. So, if the equation is quadratic, then there will be two roots. Cubic equation will have three roots and quartic will have four roots.

Nature of Roots The nature of roots is dependent upon the term b2 – 4ac, which is called the discriminant.

276

Quadratic Equation (i)

If b2 > 4ac, then the roots are real and distinct. This is so because b2 – 4ac is positive and hence its square root will be a non-zero real number.

(ii)

If b2 = 4ac, then the roots are real and equal. This is so because b2 – 4ac is zero and hence b ==  . 2a

(iii) If b2 < 4ac, then the roots are imaginary and conjugate. This is so because b2 – 4ac is negative and hence its square root is not defined. It is important to note that whenever imaginary roots occur, they always occur in conjugate pairs. For example, if one of the roots of the quadratic equation is 2 + 3i, then the other root will be 2 – 3i. Example 1 Discuss the nature of roots of (i)

4x2 + 3x + 7 = 0

(ii)

x2 + 5x + 1 = 0

(iii) 3x2 – 5x + 2 = 0 Also find the roots of these equations. Solution (i)

b2 – 4ac = 32 – 4(4)(7) = 9 – 112 = –103, which is negative. Therefore the roots will be imaginary. 4x2 + 3x + 7 = 0 x =

 3  32  4(4)(7)  3   103  3  i 103   . 2(4) 8 8

Therefore, the two imaginary roots are conjugate. (ii)

b2 – 4ac = 52 – 4(1)(1) = 25 – 4 = 21, which is positive. Therefore, the roots will be real and distinct.

 5  21 , as discriminant is 21. 2 (iii) b2 – 4ac = (–5)2 – 4(3)(2) = 1. The roots are real and distinct. x2 + 5x + 1 = 0  x =

3x2 – 5x + 2 = 0  x =

5 1 2 2  1, . Therefore, the roots are 1 and . 6 3 3

Example 2 Find the range of values of k so that the roots of (i)

3x2 + 4x + k = 0 are real and distinct

(ii)

2x2 – 3kx + 5 = 0 are real and distinct

(iii) x2 + kx + 1 = 0 are imaginary Solution (i)

For roots to be real and distinct, b2 > 4ac. Therefore, 16 > 4(3)(k)  k
4(2)(5)  k2 >

k >

40 9

2 10  2 10 or k < 3 3

(iii) For roots to be imaginary, b2 < 4ac  k2 < 4(1)(1)  k2 < 4  –2 < k < 2.

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Quantitative Aptitude Simplified for CAT

Example 3 Discuss the nature of the roots of 3x2 – 2(a + b + c)x + (ab + bc + ca) = 0. Solution Discriminant = [2(a + b + c)]2 – 12(ab + bc + ca) = 4(a2 + b2 + c2 + 2ab + 2bc + 2ca) – 12(ab + bc + ca) = 4(a2 + b2 + c2 – ab – bc – ca) ≥ 0. Therefore, roots are real. Roots may be distinct or equal, but they will definitely be real. Example 4 If one of the roots of the equation x3 + 5x2 – 2x – 4 = 0 is 1, then find the other roots. Solution If x = 1 is one of the roots of the given equation, then (x – 1) is a factor of the expression x3 + 5x2 – 2x – 4. Dividing x3 + 5x2 – 2x – 4 by (x – 1) using synthetic division, 1

1

1

5

–2

–4

1

6

4

6

4

0

We get the quotient as x2 + 6x + 4. So, x3 + 5x2 – 2x – 4 = 0  (x – 1)(x2 + 6x + 4) = 0  x = 1 or x2 + 6x + 4 = 0. Solving x2 + 6x + 4 = 0, we get x=

 6  36  16  3  5 . 2

Therefore, the three roots of the cubic equation are 1, 3  5 and 3  5 . If a quadratic equation is given, then we can obtain the roots of the equation. If roots are given, then we can obtain the corresponding equation. We have seen that the roots of the quadratic equation are denoted by  and . If  and  are the roots of the equation, then x =  and x =   (x – ) (x – ) = 0  x2 – ( + )x +  = 0 which is the required equation. The above can be rewritten as x2 – (sum of the roots)x + (product of roots) = 0 But this equation must be same as the equation ax2 + bx + c = 0 whose roots are  and . Comparing the coefficients, we get:  +  = 

b c and  = . a a

This is an important result which can also be obtained as =

 b  b 2  4 ac  b  b 2  4 ac and  = . Adding them, we get: 2a 2a

278

Quadratic Equation  b  b 2  4 ac  b  b 2  4 ac b + =  . Similarly, 2a 2a a

+=

2

 =

2

 b  b  4 ac  b  b  4 ac  = 2a 2a

  b 2   b2  4ac   4 a2

2





b 2  b2  4ac c  . a 4a2

EXPERT SPEAK Scan this QR Code to watch a video that explains how to use Quadratic Formula for roots of equation and also Nature of Roots.

Example 5 Find the equation whose roots are 3 and 5. Solution Sum of roots = 3 + 5 = 8 and product of roots = 3  5 = 15. Therefore, the equation is x2 – 8x + 15 = 0. Example 6 If  and  are the roots of the equation 2x2 + 5x – 3 = 0, then find the value of the expression (i)

3 + 3

(ii)

        

2

(iii)  Solution (i)

 3 5 Since  =  and  =  , the expression becomes 2 2 = 

(ii)

2 3  5 3  25 111  3  .  3         2       2   2  2 2 4 8     

        

2

 2   2      2  2              

2

2

2  (5 / 2)2  3  37  1369  =        36  6    (3 / 2) 

25 25  1225 (iii) (2 – 2)2 = ( – )2( + )2 = [( + )2 – 4]( + )2 =   6    .   16  4   4 

We may not always be able to find the value of such expressions. Only those expressions can be found where replacing  by  and  by  would not change the expression at all. For example, in the first expression 3 + 3, if we perform the said replacement, we get 3 + 3 which is essentially same as the original expression.

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Quantitative Aptitude Simplified for CAT

Example 7 Raju, while finding the roots of a quadratic equation misreads the coefficient of x and hence obtains the roots (4, –3), whereas Mahesh while finding the roots of the equation, misreads the constant term and hence obtains the roots (3, 5). Find the correct roots. Solution If roots are (4, –3), then the equation is x2 – x – 12 = 0. But coefficient of x is misread. So, the actual equation is x2 + ax – 12 = 0. Similarly, for the roots (3, 5), the equation would be x2 – 8x + 15 = 0. But the constant term is misread. So, actual equation is x2 – 8x + b = 0. In the first equation, the constant term is correct whereas in the second equation, the coefficient of x is correct. Therefore, the correct equation would be x2 – 8x – 12 = 0. The roots of the equation would be

8  64  48 8  112   4 2 7 2 2





Therefore, roots are 4  2 7 , 4  2 7 . Example 8 Find the quadratic equation if one of the roots of a quadratic equation is 2 + i. Solution If one of the roots is 2 + i, then the other root will be 2 – i (as imaginary roots occur in conjugate pairs). Therefore, Sum of roots = 4 and product of roots = (2 + i)(2 – i) = 5. The corresponding equation is x2 – 4x + 5 = 0. Example 9 Find (i)

minimum value of x2 – 5x + 9

(ii)

maximum value of 4x – 3x2 + 7

Solution (i)

2 2 5 25  5 11 x2 – 5x + 9 can be rewritten as  x    9   x   . 2 4  2 4 

This is known as perfect square form. 2

5  Since the first term is  x   , which is a perfect square, the minimum value of this will be 0, in which case 2  11 the minimum value of the whole expression will be 0 + . 4

Note that this happens when x = (ii)

5 . 2

2 2  2 25 4 7 2 7 4 4x – 3x2 + 7 =  3  x 2  x    3   x      =  3  x    . 3 3 3 3 3 3 9       

280

Quadratic Equation Since the first term is always negative, the maximum value of the whole expression will occur when the first 25 term is 0, in which case the maximum value is . 3

Relationship Between Roots and Coefficients of Equation Ifand are the roots of a quadratic equation, ax2 + bx + c = 0, then += 

b c and  = . a a

Similar relationships exist for higher degree equations. If ,and  are the roots of a cubic equation, ax3 + bx2 + cx + d = 0, then    

b d c    . This is also written as a a a

b d c     a a a

The notation  means “summation of roots taking one at a time”. Similarly,  means “summation of roots taking two at a time”, and so on. If  and  are the roots of a biquadratic equation, ax4 + bx3 + cx2 + dx + e = 0, then  

b d c e      a a a a

Observe that there is a pattern in all this and this pattern is followed no matter what the degree of the polynomial equation is. Recall that in case of quadratic equation, if roots are known, then quadratic equation can be obtained as x2 x . The same can be said of higher degree polynomials. For cubic, if the roots are ,  and , then x = , x = , x =   (x – )(x – )(x – ) = 0  x3 – x2 + x –  = 0 Comparing the coefficients of this with that of the equation ax3 + bx2 + cx + d = 0, we can get the above relationships between the roots and coefficients. Example 10 For the equation 2x3 + 5x2 – 3x + 7 = 0, find the sum of the reciprocal of its roots. Solution If roots are ,  and , then we have to find

Therefore,

1 1 1        which is equal to .    

     (3 / 2) 3 1 1 1   .   =  (7 / 2) 7   

Example 11 In the equation x3 – 2x2 – 9x + 18 = 0, if one root is negative of the other, find all the roots.

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Solution If the roots are , – and , then  =  + (–) +  = 2   = 2. Moreover,  = ––– –2 = –9  = ±3. The three roots are 3, –3 and 2. Example 12 If the roots of x3 – 15x2 + 71x – 105 = 0 are in Arithmetic Progression, then find the roots. Solution Let the roots be a – d, a, a + d. Then, Sum of roots = (a – d) + (a) + (a + d) = 15  a = 5. Product of roots = (a – d)(a)(a + d) = 105  (5 – d)(5)(5 + d) = 105  (5 – d) (5 + d) = 21 25 – d2 = 21  d = ±2. Therefore, the roots are 3, 5 and 7. Example 13 State whether True or False: (i)

It is possible that no real number satisfies a cubic equation with real coefficients.

(ii)

For a polynomial equation with even degree, at least one of the roots must be real.

Solution (i)

This is a false statement. If the given statement were true, that would mean all the roots are unreal, which would mean that the equation would have odd number of imaginary roots. No equation with real coefficients can have odd number of imaginary roots.

(ii)

This is also a false statement. The number of imaginary roots must be even. At least one of the roots being real allows for the possibility that only one root is real and that would mean other roots are imaginary, which is not possible.

Example 14 For what value of k would the equation x4 – 2x3 + k2x2 + 3kx – 3 = 0 never have 1 as its root? Solution If one of the roots is 1, then it should satisfy the given equation. Therefore, (1)4 – 2(1)3 + k2(1)2 + 3k(1) – 3 = 0 k2 + 3k – 4 = 0  k = 1, –4 The value of k for the given equation to have 1 as a root is 1, –4. Therefore, k must not be 1 or –4 so that the given equation never has 1 as its root. Example 15 The roots of the equation ax4 + bx3 + cx2 + dx + e = 0 are x1, x2, x3 and x4. Find the value of (1 – x1)(1 – x2)(1 – x3)(1 – x4)

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Quadratic Equation Solution If x1, x2, x3 and x4 are the roots of the equation, then (x – x1)(x – x2)(x – x3)(x – x4) = 0, which should be same as ax4 + bx3 + cx2 + dx + e = 0. Dividing this equation by ‘a’, we get x4 +

b 3 c 2 d e x + x + x+ = 0. Therefore, a a a a

(x – x1)(x – x2)(x – x3)(x – x4) = x4 +

b 3 c 2 d e x + x + x+ . a a a a

Putting x = 1, we get: (1 – x1)(1 – x2)(1 – x3)(1 – x4) = 1 +

b c d e    . a a a a

Example 16 Find an equation whose roots are reciprocal of the roots of the equation x3 + 2x2 – 5x + 7 = 0. Solution If the roots of the given equation are , , , then roots of the equation required are

    

     5 5   .  7 7

Sum of roots taking two at a time =

        

Sum of roots =

1

Product of roots =

1

1

1

1

1

1

1

1 1 1 , , .   

     2 2   .  7 7

1 1  .  7

Therefore, the required equation is x3 –

5 2 2 1 x + x+ = 0 or 7x3 – 5x2 + 2x + 1 = 0. 7 7 7

The last example is about transforming an equation to another equation with some required roots relationships. Such cases are called as transformation of equation. If we are given a polynomial equation in x and we want to transform it to another equation in X, then x is replaced by X in a certain manner. If we want to transform an equation P(x) = 0 into another whose roots are reciprocal of the roots of P(x) = 0, 1 1 then the new equation would be P(X) = 0, where X  = 0. Therefore, x should be replaced by . The X x equation P(X) = 0 then becomes P = 0. Let us solve the last example by this method. Replacing x by 3

1 3 , x + 2x2 – 5x + 7 = 0 becomes x

2

 1  1  1    2    5    7  0 , or 7x3 – 5x2 + 2x + 1 = 0  x  x  x

Observe the following rules regarding transformations. If we want to transform an equation P(x) = 0 into another whose roots are (i) (ii)

negative of the roots of P(x) = 0, then X  –x and hence x should be replaced by –X. X k times the roots of P(x) = 0, then X  kx and hence x should be replaced by . k

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Quantitative Aptitude Simplified for CAT

(iii)

1 x times the roots of P(x) = 0, then X  and hence x should be replaced by kX. k k

(iv) h more than the roots of P(x) = 0, then X  x + h and hence x should be replaced by X – h. (v)

h less than the roots of P(x) = 0, then X  x – h and hence x should be replaced by X + h.

Example 17 Form an equation whose roots if decreased by 5 would satisfy the equation x4 – 5x3 + 7x2 + 2x + 12 = 0. Solution Since the required equation’s roots are 5 more than the roots of the given equation, replace x by (X – 5). x4 – 5x3 + 7x2 + 2x + 12 = 0 becomes (X – 5)4 – 5(X – 5)3 + 7(X – 5)2 + 2(X – 5) + 12 = 0  X4 – 20X3 + 150X2 – 500X + 625 – 5(X3 – 15X2 + 75X – 125) + 7(X2 – 10X + 25) + 2(X – 5) + 12 = 0  X4 – 25X3 + 232X2 – 943X + 1427 = 0, which is the required equation. Example 18 If the roots of the equation x3 + 2x2 – 7x + 10 = 0 are  then find an equation whose roots are  Solution For the required equation, Sum of roots =  Sum of two roots taking two at a time =   Product of roots =  ) =  =  Therefore, required equation is x3 + 4x2 – 3x – 24 = 0. Alternatively,  +  +  = –2   +  = –2 – . So, we want to find the equation whose roots are –2 – , –2 –  and –2 – . By using transformation, the required equation is: (–2 – x)3 + 2(–2 – x)2 – 7(–2 – x) + 10 = 0  –8 – 12x – 6x2 – x3 + 2(4 + 4x + x2) + 14 + 7x + 10 = 0  x3 + 4x2 – 3x – 24 = 0.

AM, GM, HM and Roots of Equations Let  and  be the roots of quadratic equation ax2 + bx + c = 0. The equation can be obtained from the roots using the result x2 – ( + )x +  = 0 If A is the AM and G is the GM of  and , then A=

  or  +  = 2A. Similarly, 2

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Quadratic Equation

 or  = G2. Putting these in the above equation, we get

G=

x2 – 2Ax + G2 = 0 This is useful to find the equation, the AM and GM of whose roots are known. For example, if AM of roots of a quadratic equation is 5 and GM of the roots is 4, then the quadratic equation is obtained as: x2 – 10x + 16 = 0. Discriminant, D = b2 – 4ac = (2A)2 – 4G2 = 4(A2 – G2). Obviously, if 4(A2 – G2) > 0, then roots are real and distinct; whereas if 4(A2 – G2) = 0, then the roots are real and equal. Here, A = G. Naturally, if A = G, then the terms are equal (recall the result that if AM and GM of two terms are equal, then the terms are also equal, from the inequality AM  GM). In case of cubic equation, if A, G and H are the AM, GM and HM of the three roots, ,  and , then A=

   + = 3A 3

G=

3

H=

  = G3

3G 3 3       H

The cubic equation x3 – ( +  + x2 + ( +  + )x –  = 0 transforms to x3 – 3Ax2 +

3G 3 x – G3 = 0 H

Example 19 If AM, GM and HM of three roots of a cubic equation are 5, 4, 3, find the cubic equation. Solution x3 – 3Ax2 +

3G 3 3 (4 ) 3 x – G3 = 0  x3 – 15x2 + x – (4)3 = 0 H 3

x3 – 15x2 + 64x – 64 = 0. Example 20 Find the Harmonic Mean of the roots of the equation (i)

x3 – 5x2 + 7x + 8 = 0

(ii)

x2 – 7x + 2 = 0

Solution (i)

Comparing the equation with the standard equation x3 – 3Ax2 + GM =

3

5 3G 3 x – G3 = 0, we see that AM = and H 3

 8  2 .

Comparing the coefficient of x, 3G 3 3(  2) 3 24  H = .  H 7 7

(ii)

Comparing the equation with the standard equation x2 – 2Ax + G2 = 0, we see that

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Quantitative Aptitude Simplified for CAT

AM =

7 and G = 2

2 . Using the result, G2 = AH, H =

G2 2 4   . A 7 /2 7

Example 21 If AM of the roots of a quadratic equation ax2 + bx + c = 0 is 8 and GM is 9, then (i)

Find the equation

(ii)

Discuss the nature of roots

Solution (i)

The equation would be x2 – 2(8)x + 92 = 0 or x2 – 16x + 81 = 0.

(ii)

Discriminant, D = b2 – 4ac = 162 – 4  81 = 256 – 324 < 0.

Therefore, the roots are imaginary. This was bound to be the case because for real numbers, AM should be more than GM, but in this example, AM = 8 which is less than GM = 9. In the above example, we saw that if AM is less than GM, then the roots of the corresponding equation will be imaginary.

Continued Fractions a

Continued fractions are fractions which are of the form

c

b d

,

e f  ....

where a, b, c, ... are integers. Continued fractions can have finite terms or infinite terms in the sequence. Let us see how to simplify such fractions. Example 22 2 1

Simplify the continued fraction 3-

5

. 5 7

Solution This is a case of finite continued fraction, because it has finite terms. In such cases, start with the last fraction, that is,

3

2 1 5

 5 7

5 as shown below. 7

2 80 80 .   7 120  7 113 3 40

Example 23

1

Solve and simplify the fraction

.

1

1 1

1 1

1 .....

286

Quadratic Equation Solution This is a case of infinite continued fraction. In such cases, we observe the pattern which continues and repeats. If the given fraction is x, then

1

x=



1

1

1 . 1x

1

1

1

1 .....

Therefore, x2 + x = 1 or x2 + x – 1 = 0. Solving this quadratic, we get x =

1  5 2

1  1  4 1  5  . But the value 2 2

is rejected because this is negative whereas the given expression is positive. Therefore, the

acceptable value of the given expression is

1  5 1  5 . Therefore, x = . 2 2

Example 24 2

Simplify the expression 1  1

2 1  ...

.

Solution x = 1

2 2 1 1  ...

=1+

2  x2 – x – 2 = 0 or x = 2, –1. But the given expression cannot be x 2

–1. Therefore, 1  1

2 1  ....

2.

In the above examples, observe that (a)

Finite continued fractions will be always rational numbers.

(b)

Infinite continued fractions may be rational numbers or irrational numbers.

Example 25 Simplify the following expressions: 1 (i) 1  . 1 2 1 2 2  ... (ii)

42  42  42  42  ...

Solution (i)

1

The expression

1

2 2

shows some pattern and hence would be solved first. Let this be x.

1 2  ...

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Quantitative Aptitude Simplified for CAT

1

Then, x =



1

2 2

1 . 2 x

1 2  ...

2  44 2  8  = –1 ± 2 . But, the value – 1 – 2 2 because this is negative whereas the x must be positive. Therefore, x = –1 + 2 .

Therefore, x2 + 2x – 1 = 0 or x =

1

So, the given expression = 1 

1

2 2

(ii)

x=

2 is rejected

=1+x= 2.

1 2  ...

42  42  42  42  ...  42  x . Solving for x, we get x = 7.

Example 26 Prove that

2  2  2  2  ...  2 2 2 2...

Solution x=

2  2  2  2  ...  2  x . Solving for x, we get x = 2.

y=

2 2 2 2...  2 y . Solving for y, we get y = 2.

Hence the result.

EXPERT SPEAK Scan this QR Code to watch a video that explains applications of Quadratic Equation to Continued Fractions.

Common Roots Let there be two equations P(x) = 0 and Q(x) = 0. If x =  satisfies both the equations, then x =  is the common root of the two equations. Let us take the case of quadratic equations first. If we have two quadratic equations: x2 – 3x + 2 = 0 and x2 – 5x + 6 = 0, then the roots of the first equation are 1 and 2, whereas the roots of the second equation are 2 and 3. We observe that the root x = 2 is common to both the equations and hence 2 is the common root. There are two ways to find the common root. One is to solve both the equations and then say that a particular root is common (if there is any common root). The other method is to simply subtract the two equations. This will reduce the degree of the equations by 1. Therefore, if we subtract two quadratic equations, we will get a linear equation, and if we subtract two cubic equations, we will get a quadratic equation, and so on. Therefore, (x2 – 3x + 2) – (x2 – 5x + 6) = 0  2x – 4 = 0  x = 2.

288

Quadratic Equation The solution of the linear equation is the common root. A very important point is that this method of taking difference of the equations and proceeding should be applied only when there exists a common root. If there does not exist a common root, then the linear equation obtained by finding the difference will yield some value of x, but that will not be the common root. For example, if the two equations are x2 + 5x + 6 = 0 and x2 + 3x + 9 = 0, then taking their differences yields 3 , which does not satisfy any of the given equations and hence is not the common root. It is 2 now clear that there must be some common root for us to be able to apply this method. Otherwise, we may get wrong answers. The thumb rule is that after obtaining the common root, substitute it in the original equations to verify whether they satisfy the equations or not.

2x – 3 = 0  x =

Example 27 Find the number of roots common to x3 – 4x2 + 5x – 2 = 0 and 2x3 + 7x2 – 3x – 6 = 0. Solution Before taking their differences, we must ensure that the coefficient of x3 is same. Instead of dividing the second equation by 2, we would rather multiply the first equation by 2. We get 2x3 – 8x2 + 10x – 4 = 0. Now, subtracting the second equation from the first equation, (2x3 – 8x2 + 10x – 4) – (2x3 + 7x2 – 3x – 6) = 0  –15x2 + 13x + 2 = 0  (x – 1) (–15x – 2) = 0  x = 1, 

2 . 15

We should now check whether the roots obtained satisfy the original equations or not. We note that x = 1 satisfies both the equations, but the other root does not satisfy the equations. Hence, only one root is common to the two equations. Example 28 Find the value of roots which are common to the equations x3 + 5x2 + 4x + 7 = 0 and x3 + 4x2 + 9x + 1 = 0. Solution Taking the difference of the equations, we get x2 – 5x + 6 = 0, whose roots are 2 and 3. But, we observe that none of the roots satisfy the given equations. Therefore, there is no common root! Example 29 Two equations x3 + 2x2 + px + q = 0 and x3 + x2 + px + r = 0 are given. What is the relationship between p, q and r for them to have 2 common roots? Also find the sum of roots which are not common to the equations. Solution Taking the difference of the equations, we get x2 = r – q x=  r q

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Quantitative Aptitude Simplified for CAT

If the third root of first equation is  and that of second one is , then Sum of roots of the first equation =  r  q  r  q    2  = –2. r  q  r  q    1  = –1.

Similarly, sum of roots of the second equation =





r  q   r  q    2   q

For the first equation, product of roots =  2(r – q) = –q  2r = q.

Sum of roots (for first equation) taking two at a time





=  r q  r q 2 r q 2 r q  p  (r – q) = –p  p + r = q, which is the required relationship. Adding the roots which are not common to the equations, we get (–2) + (–1) = –3. Example 30 Solve for real x: (x2 – 5x + 4)2 + (x2 – 7x + 12)2 = 0. Solution If sum of squares of a few terms is zero, then each term is also zero, provided each term is real. Therefore, x2 – 5x + 4 = 0  x = 1, 4 x2 – 7x + 12 = 0  x = 4, 3. x = 4 is the only value of x which satisfies the given original equation

Graphical Interpretation of Quadratic Function The graph of y = ax2 + bx + c is always parabolic. In general, any equation in two variables, where one of the variables appears in quadratic and the other in linear, is always a parabola. Therefore, the equation x = ay2 + by + c is also a parabola. We will restrict ourselves to the equations of the form y = ax2 + bx + c. To understand parabola, we will take an example. Let us throw some stone from the ground at some angle to the ground (which should not be 90°). The trajectory followed by the stone is the shape of the parabola. We can easily see that the stone reaches some height after which it starts declining. The topmost point reached is called the vertex of the parabola. If we draw a tangent to the parabola at the vertex, then a line perpendicular to this tangent passing through the vertex is called the axis of the parabola. For the equations of the type y = ax2 + bx + c, the axis of the parabola is parallel to y-axis. The general shape of the graph of parabola is shown below:

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Quadratic Equation

a>0

a 0, then the graph opens upwards and if a < 0, then the graph opens downwards, both shown above. When the graph opens upwards, the bottom most point (the vertex) is the minimum value of the curve and when the graph opens downwards, the top most point (the vertex) is the maximum value of the curve. Let us take the case: a > 0. The graph can take three distinct positions in such a case (as shown in the diagram). Position 1: The parabola intersects the x-axis at two distinct points. Position 2: The parabola touches the x-axis and hence meets the x-axis at one point only. Position 3: The parabola does not touch the x-axis at all.

Position 1

Position 2

Position 3

The points where the graph meets x-axis are those points where y = 0. In other words, these refer to the roots of the equation ax2 + bx + c = 0. Position 1 corresponds to the roots being real and distinct. The roots are (, 0) and (, 0). Position 2 corresponds to the roots being real and equal. The repeated roots are (, 0) or (, 0). In this case, xaxis acts as the tangent to the parabola. Position 3 corresponds to the roots being imaginary. No real root exists. From the ongoing discussion, we can surely say that the condition for the curve y = ax2 + bx + c to (i)

intersect x-axis in two distinct points is: b2 > 4ac.

(ii)

touch the x-axis is: b2 = 4ac.

(iii) not at all touch x-axis is: b2 < 4ac. If a < 0, then the only difference would be that the graph would open downwards. The above stated conditions would remain the same. This is shown in the graph below.

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Quantitative Aptitude Simplified for CAT

From the foregoing discussion, we can say that If a > 0 and b2 < 4ac, then ax2 + bx + c > 0; and If a < 0 and b2 < 4ac, then ax2 + bx + c < 0. In other words, If b2 < 4ac, then the signs of a and ax2 + bx + c are same, that is a(ax2 + bx + c) > 0. This will be found a useful result in quadratic inequalities as well. Example 31

EXPERT SPEAK Scan this QR Code to watch a video that explains Graphical Interpretation of Quadratic Functions.

Draw the graph of (i)

y = x2 + 4x + 9

(ii)

y = x2 – 5x + 6

(iii) y = –x2 + 6x – 9 Solution (i)

Observe that if x2 + 4x + 9 = 0, then discriminant = b2 – 4ac = (4)2 – 4(1)(9) = –20 < 0.

Therefore, the roots are imaginary and hence the graph of y = x2 + 4x + 9 does not touch x-axis at all. Since a > 0, the graph will open upwards. Moreover, when x = 0, y = 9. Therefore, the curve passes through (0, 9).

292

Quadratic Equation Finally, x2 + 4x + 9 = (x + 2)2 + 9 – 4 = (x + 2)2 + 5 (completion of squares form). Therefore, the curve attains its minima at x = –2, and the minimum value is 5. Therefore, vertex of the parabola is at (–2, 5). The graph of the function is drawn below.

(ii)

Observe that if x2 – 5x + 6 = 0, then discriminant = b2 – 4ac = (5)2 – 4(1)(6) = 1 > 0.

The roots are 2 and 3 and hence the graph of y = x2 – 5x + 6 intersects x-axis at (2, 0) and (3, 0). Since a > 0, the graph will open upwards. Moreover, when x = 0, y = 6. Therefore, the curve passes through (0, 6). Finally, x2 – 5x + 6 = (x – 2.5)2 + 6 – 6.25 = (x – 2.5)2 – 0.25. Therefore, the curve attains its minima at x = 2.5, and the minimum value is –0.25. Therefore, vertex of the parabola is at (2.5, –0.25). The graph of the function is drawn below.

(iii) Observe that if –x2 + 6x – 9 = 0, then discriminant = b2 – 4ac = (6)2 – 4(–1)(–9) = 0. The roots are repeated and the value is 3 and hence the graph of y = –x2 + 6x – 9 touches x-axis at (3, 0) only. Since a < 0, the graph will open downwards. Moreover, when x = 0, y = –9. Therefore, the curve passes through (0, –9). Finally, –x2 + 6x – 9 = –(x2 – 6x + 9) = –(x – 3)2. Therefore, the curve attains its maxima at x = 3, and the minimum value is 0. Therefore, vertex of the parabola is at (3, 0). The graph of the function is drawn below.

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Quantitative Aptitude Simplified for CAT

Note that completion of squares form of f(x) = ax2 + bx + c helps us in many respects. When we convert this into completed squares form, the equation becomes f(x) = a(x – p)2 + q. 1)

If a is positive, the graph opens upwards and (p, q) is the vertex of the parabola. Moreover, f(x) gets minimized at x = p and the minimum value of f(x) is q.

2)

If a is negative, the graph opens downwards and (p, q) is still the vertex of the parabola. Moreover, f(x) gets maximized at x = p and the maximum value of f(x) is q.

Example 32 What should be p in y = px2 + 3x – 5 for the graph of it to intersect x-axis in 2 distinct points? Solution For the graph of y = px2 + 3x – 5 to intersect x-axis in 2 distinct points, b2 > 4ac. Therefore, (3)2 > 4(p)(–5)  –20p < 9  p > 

9  p > –0.45. 20

Example 33 If the graph of y = ax2 + bx + c passes through (2, 3), (4, –1) and (–3, –5), then find the points where the graph intersects x-axis and also y-axis. Also find the maximum/minimum value of y. Solution Putting the coordinates in the equation, we get three equations: 3 = 4a + 2b + c –1 = 16a + 4b + c –5 = 9a – 3b + c Eliminating c from the first two equations, 4 = –12a – 2b. (A) Eliminating c from the second and third equation, 4 = 7a + 7b.

(B)

From (A), b = (–6a – 2). Putting this in (B), we get 4 = 7a + 7(–6a – 2) = –35a – 14 a= 

18 108 38 b= 2  . 35 35 35

 c = 3 – 4a – 2b = 3 +

72 76 4 101  =3– = . 35 35 35 35

294

Quadratic Equation

Therefore, equation is y = 

18 2 38 101 x  x . 35 35 35

To find the points where the graph intersects x-axis, y = 0  18x2 – 38 – 101 = 0 x=

18  38 2  4(18)(101) 38  8716 19  2179   2(18) 36 18

Therefore, roots are

19  2179 19  2179 and . 18 18

Note that one root is positive and the other is negative. The point where the curve meets y axis is when x = 0. Therefore, when x = 0, y =

101 . Therefore, the point is 35

 0, 101   . 35   Since a < 0, the graph will have a point of maxima. y= 

=

18 2 38 101 18  2 19 101  18 x  x  x x    35 35 35 35  9 18  35

2179 18  19   x  630 35  18 

2  19  101 361   x    18  18 324   

2

 Maxima will occur at x =

19 2179 and the maximum value will be . 18 630

Note: It must be noted that if we draw the coordinates of the points on coordinate axis approximately, we would visually know that the parabola opens downwards and hence ‘a’ must be negative. Moreover, by the location of the points, we can also conclude that the two roots will be opposite in sign and hence the product c of roots will be negative, that is < 0. Since a < 0, so c > 0. Finally, observe that the magnitude of positive a root is sufficiently more than the magnitude of negative root and so the sum of roots will be positive. So, sum b b < 0. Since a < 0, b > 0. Therefore, a < 0, b > 0 and c > 0. of roots =  > 0  a a

EXPERT SPEAK Scan this QR Code to watch a video that is a continuation of the previous video on Graphical Interpretation of Quadratic Functions.

Example 34 What is the condition for the graph of x = 2y2 – ky + 3 to not touch y-axis at all? Solution Note that the given function is quadratic in y and hence the axis of the parabola will be parallel to x-axis. The graph of this curve will not touch y-axis at all only if the roots are imaginary.

295

Quantitative Aptitude Simplified for CAT

Therefore, b2 < 4ac or k2 < 4(2)(3) = 24   2 6  k  2 6 . Example 35 Draw the graph of y = x2 – 3x + 2 and y = x2 – 5x + 6. Verify whether they have any common root or not. Explain the graphical interpretation of common roots. Solution The roots of x2 – 3x + 2 = 0 are (1, 2) and that of x2 – 5x + 6 = 0 are (2, 3). The common root is 2. Drawing their graphs on the same coordinate axes, we get the following:

The curve which is shown in bold is the graph of y = x2 – 5x + 6. We observe that (2, 0) is the place where both the curves have a common point. So, graphically we can say that if there are two curves whose intersection point is a point on x-axis, then that intersection point is the common root. If the second function were y = –x2 + 5x – 6, then also the roots would have been (2, 3), but the curve would open downwards as shown below.

The graphical interpretation helps enhance our understanding of the quadratic expression and the roots of the corresponding quadratic equation. This sometimes helps generate better methods of solving questions.

296

Quadratic Equation

PRACTICE EXERCISE 1.

4.

For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive? x2 − y2 = 0 (x − k)2 + y2 = 1 A.

2

B.

0

5.

−7

B.

−4

C.

2

D.

6

E.

Cannot be determined

What is the value of a + b + c?

2

A.

9

D.

− 2

B.

14

C.

13

D.

37

E.

Cannot be determined

Let

x

4 4 4 4 …

to infinity. 6.

Then x equals A.

3

B.

 13  1    2  

C.

 13  1    2  

D.

13

CAT 2005 3.

A.

C.

CAT 2005 2.

What is the other root of f(x) = 0?

If the roots of the equation x3 − ax2 + bx − c = 0 are three consecutive integers, then what is the smallest possible value of b? A.



B.

−1

C.

0

D.

1

1 3

E.

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

1 3

CAT 2008

A.

–119

7.

B.

–159

If f(x) = x3 − 4x + p, and f(0) and f(1) are of opposite signs, then which of the following is necessarily true?

C.

–110

A.

−1 < p < 2

D.

–180

B.

0 |v|. Then,

10. For what range of values of a is the expression x2  ax + (1  2a2) always positive? A.



4 4  a 9 9

4  a  9 2  a  3

B. C. D.



4 9 2 3

1 11. If |a| > , then solve for x:] 2

 1 

4 a2

B.

u < 0, v > 0

C.

u > 0, v < 0

D.

u > 0, v > 0

15. If x2 + x + p = 0 and x2 + px + 1 = 0 have common real roots, how many values of p exist?

2 2  a 3 3

4 a2  1

u < 0, v < 0

CAT 2014

CAT 2013

 2a   2a 

A.

A.

1

B.

2

C.

0

D.

Can't say

x2 4 x

CAT 2015

 2

x 4 x

 4a

A.

2  5, 3  2

B.

2  5, 2  3

C.

5  2, 3  2

D.

2  5, 5  3

1

16.

2

64

B.

8

C.

27

D.

25

1 3  ....

A.

13  3 2

B.

15  3 2

C.

13  3 2

D.

15  3 2

CAT 2013

A.

1

3

12. If f(x) = x2 + ax + b has minimum at x = 7, f(7) = 16, and p and q are roots of the equation f(x) = 0, then find the value of |p  q|.

=

1

2

CAT 2004 17. If x is real, the smallest value of the expression 3x2 – 4x + 7 is: CAT 2013

298

A.

2/3

B.

3/4

Quadratic Equation IIFT 2012

C.

7/9

D.

None of the above

19. A polynomial “ax3 + bx2 + cx + d” intersects x-axis at 1 and –1, and y-axis at 2. The value of b is:

IIFT 2013 18. The value of A.

1

B.

2

C.

3

D.

4

7  7  7  7  .... is

A.

–2

B.

0

C.

1

D.

2

E.

Cannot be determined XAT 2014

ANSWER KEY 1. C

2. C

3. B

4. B

5. E

6. B

7. B

8. B

9. A

10. D

11. B

12. B

13. B

14. B

15. A

16. D

17. D

18. C

19. A

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Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS 1.

6.

Explanation: Let us assume the roots to be −1, 0 and 1. So, b = (−1)(0) + (0)(1) + (−1)(1) = −1, which is also the smallest given option amongst all the options given. Alternatively, Let the three roots be p – 1, p and p + 1. Then, b = (p – 1)( p) + (p)( p + 1) + (p + 1)( p – 1) = p2 – p + p2 + p + p2 – 1 = 3p2 – 1, whose minimum value will be –1 when p = 0.

Answer: C Explanation: x2 – y2 = 0  x2 = y2  x =  y (x  k )2  y 2  1  x 2  k 2  2kx  x 2  1

 2x 2  2kx  k 2  1 For unique solution (that is roots are repeated roots), discriminant = 0  4k2 – 8(k2 – 1) = 0  –k2 + 2 = 0  k2 = 2  k = ± 2 b 2k k So, x =    . So, k must be 2a 4 2 positive for positive root of x. 2.

7.

Answer: C 8.

9.

Answer: B

Answer: A Explanation: Subtracting one equation from other, we get a quadratic equation x2 – 3x + 2 = 0, which has two roots 1, 2. But none of the roots satisfies these equations.

10. Answer: D

Answer: B

Explanation: Coefficient of x2 > 0. So, for the given expression to be positive, discriminant < 0, that is (–a)2 – 4(1 – 2a2) < 0  9a2 – 4 < 0  a2 4 2 2 <    a . 9 3 3

Explanation: Since f(5) = −3f(2), 25a + 5b + c = −3(4a + 2b + c) Or, 37a + 11b + 4c = 0. Moreover, since 3 is one of the roots, 9a + 3b + c = 0 36a + 12b + 4c = 0. Subtracting the two equations, we get a = b. Sum of roots = −b/a = −1 (because a = b). If the other root is , then  + 3 = −1 or  = −4. 5.

Answer: B Explanation: For real roots, discriminant ≥ 0  b2 – 4a ≥ 0. The solution sets (a, b) satisfying the inequality are: (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (4, 4).

Explanation: f(x) = a(x – p)2 + q = a(x – 1)2 + 3. Since f(0) = 1, a(0 – 1)2 + 3 = 1  a = –2. So, f(10) = –2(10 – 1)2 + 3 = –159. 4.

Answer: B Explanation: f(x) = x3 − 4x + p so f(0) = p and f(1) = p – 3. If f(0) and f(1) are of opposite sign, p(p – 3) < 0  0 < p < 3.

Explanation: The term inside the square root is more than 4, so answer has to be more than 2. So, [B] option is not possible. Moreover, the term inside the second square root is less than 4 and so it is less than 2 and so value of x will be less than 6 . 3.

Answer: B

11. Answer: B Explanation: For the given equation to be true, x2 – 4x should be 1 or –1. Now, x2 – 4x = 1  x2 – 4x –1=0x=

Answer: E Explanation: Since roots are 3 and −4, quadratic equation is x2 + x – 12 = 0. Any multiple of this equation will yield a different set of (a, b, c) and hence no unique answer of a + b + c can be found.

4  4 2  4(1) 4  20  2 5 . 2 2 When x2 – 4x = –1  x2 – 4x + 1 = 0  x

300

Quadratic Equation

=

16. Answer: D

4  4 2  4(1) 4  12  2 3. 2 2

Explanation: Observing the pattern, we can say that 1 3 x x=  1 6  2x  1 2 3 x  2x2 + 7x – x – 3 = 0  2x2 + 6x – 3 = 0 or  6  36  24  6  60 x  2(2) 4 .  6  2 15  3  15   4 2 Since given expression cannot be negative, the value of the given expression  3  15 = . 2

12. Answer: B Explanation: f(x) = x2 + ax + b = (x – m)2 + n (completed squares form). Since f(x) is minimized at x = 7 and f(7) = –16, we can say that m = 7 and n = –16. So, f(x) = (x – 7)2 – 16. Now, f(x) = 0 means (x – 7)2 – 16 = 0  (x – 7)2 = 16  (x – 7) = ±4  x = 7 ± 4 = 11, 3. Therefore, |p – q| = 8. 13. Answer: B Explanation: The equation can be rewritten as: 3x 2  8 x  11 . Now check various values y= (4  x) of x. So, (x, y) = (1, 0), (3, 40). When x = 2, we get non-integer value of y. Since we want only positive integral solutions, (1, 0) will also be rejected.

SHORTCUT: The value of x is less than

1 2

because the denominator is more than 2. Checking with the options, we can reject [A] and [C]. Now, 13  3.6 and 15  3.87. Therefore, option [B] has the 3.6  3 value = 0.3 whereas the value of 2 3.87  3 option (d) is = 0.43. The 2 denominator of the given expression is “2 + (less than 1)”. Therefore, the answer is more 1 than . Hence correct option is [D]. 3

14. Answer: B Explanation: Product of roots = c < 0 (because c > 0). So, one root is negative and the other is positive. So, either u < 0 and v > 0 or u > 0 and v < 0 Sum of roots = b < 0 (because b > 0). Since magnitude of u is more than the magnitude of v, u must be negative and v positive so that their sum is negative.

17. Answer: D Explanation: Write the expression in completed squares form. So, 3x2 – 4x + 7 = 2  4 7 2 7 4 3  x 2  x    3   x      3 3 3 3 9    

15. Answer: A Explanation: Subtracting the equations, we get: x(1 – p) + p – 1 = 0  (1 – p)(x – 1) = 0. If p = 1, then the equations become x2 + x + 1 = 0, which has unreal roots. So, p can’t be 1. If x = 1, then putting this value of x in the equation x2 + x + p = 0, we get 1 + 1 + p = 0 or p = 2. Putting this value of p in the equation x2 + px + 1 = 0, we get x2 – 2x + 1 = 0 or x = 1, 1. So, number of values of p is 1.

2 2  2 17  2 17  = 3   x    .   3 x    3 9  3 3    17 So, minimum value of the expression is . 3

301

Quantitative Aptitude Simplified for CAT

18. Answer: C

19. Answer: A

Explanation: Let x =

Explanation: ax3 + bx2 + cx + d intersects x-axis at 1 and 1. Therefore, a + b + c + d = 0 and –a + b – c + d = 0. Therefore, 2(b + d) = 0 or b + d = 0. Also, ax3 + bx2 + cx + d intersects y-axis at 2. Therefore, 0 + d = 2 or d = 2. Therefore, b = 2.

7  7  7  7  .... . So, x

= 7  7  x . Value of the given expression has to be more than 2 because the term inside square root is 7. We realize that x = 3 satisfies the above equation.

302

Functions and Graphs

Chapter 9

Functions and Graphs BASIC CONCEPTS All polynomials and algebraic expressions we have studied so far are examples of functions. So, f(x)  2x  7 is a function, known as linear function. f(x)  3x2 – 2x  5 is also a function, known as quadratic function. In general, f(x)  a0xn  a1xn–1  a2xn–2  a3xn–3  …  anx0  is polynomial function. All algebraic expressions are also functions, for example, 2x3  5x2 – 3x  6, x2 

1 x

2

, x 2  3x  2,

x 1 and so x5

on are all functions of x. All the trigonometric ratios, like sin x, cos x, tan x and so on are also functions. Similarly, ex, log10x, log10(2x – 1), [x] and |x| are more examples of functions. Functions are generally denoted by lower case letters, like f, g, h, and so on Therefore, f(x)  sin x means sin x is a function of x. Note that f is a function and f(x) is the value of the function for a particular value of x. But for all practical purposes, we can say that f(x) is a function of x. Simply speaking, a function processes the value of x and yields a value which is f(x). In other words, f is the link between x and f(x). For any function, if we assign some value to x, we get some value of f(x). Let y  f(x). Then, f(x)  sin x is written as y  sin x. Here, x is the independent variable and y is the dependent variable. This is because as the value of x changes, the value of y changes. So, the value of y depends upon the value of x, whose value can be chosen independently.

Open and Closed Interval Before we proceed any further, let us understand the different types of brackets used. There are basically three types of brackets:

303

Quantitative Aptitude Simplified for CAT

Parenthesis written as ( ) Square brackets written as [ ] Curly brackets written as { } We generally use curly brackets (or braces) to write the elements of a set. Let a and b be two real numbers. Then, (a, b) means set of values between a and b excluding the values of a and b, that is, a  x  b. This is known as open interval. [a, b] means set of values between a and b including the values of a and b, that is, a  x  b. This is known as closed interval. Therefore, (2, 3) means 2  x  3 [2, 3] means 2  x  3 We can use combination of brackets to represent the range of values. For example, [2, 3) means 2  x  3, that is, 2 is included whereas 3 is excluded. (2, 3] means 2  x  3, that is, 2 is excluded whereas 3 is included. Sometimes inverted square brackets are used to represent open interval otherwise shown by parenthesis, ( ). Therefore, (2, 3) is same as ]2, 3[. Hence, [2, 3) can also be written as [2, 3[ and so on. Also note that  cannot be bound, that is, square brackets cannot be used for . Therefore, x  R can also be written as x  (–, ) or ]–, [. We cannot write this as [–, ]. Let us represent (1, 3), [1, 3], (1, 3] and [1, 3) on real number line.

(1, 3)

[1, 3]

In the first number line, representing (1, 3), there are two unfilled circles at the ends of the strong dark line. The unfilled circles mean that the end points, that is, 1 and 3 are not included (as the open interval suggests). The filled circles at the ends of line in the second number line, representing [1, 3], mean that the end points are included (as the closed interval suggests).

(1, 3]

[1, 3)

Likewise, in the above two numbers lines, first represents (1, 3] and second represents [1, 3).

Domain and Range For any function, the independent variable x can take some values and cannot take some values. The set of values which the variable x can take is called domain. For each value of x in the domain, the function yields some value. Set of all such values of f(x) obtained from the domain values of x is called range. Therefore, the set of values of x is domain and the set of values of y (or f(x)) is range. For example, if f(x)  x2, then f(1)  1, f(2)  4, f(3)  9, and so on.

304

Functions and Graphs Note that x can be any real number whereas f(x) cannot be negative. Therefore, domain is all real numbers, shown as x  R, whereas range is all non-negative real numbers, shown as f(x)  [0, ). In other words, –  x   and 0  f(x) . Example 1 Find the domain and range of the following functions: i.

x3

ii.

sin x

iii.

log x

log2  x  1 

iv. v.

x 1 x 3

Solution i.

Domain is the set of values which x can take. Here, x can take any real number and hence the domain of this function is R, that is, x  R. The range is the set of values which the function x3 can take. Note that x3 can be positive as well as negative depending upon whether x is positive or negative. Therefore, the range is also the set of all real numbers, that is, f(x)  R.

ii.

For the function sin x, x can take any real number. But, sin x cannot go beyond 1 and below –1. Therefore, x  R, f(x)  [–1, 1]

iii.

log x is defined only for positive values of x. Therefore, x can be any positive real number. But, log x can take any value from – to . Hence, x  R or x  (0, ) and f(x)  R. Note that R means all positive real numbers.

iv.

In the function, f(x) 

log2  x  1  , log2(x – 1) should be more than or equal to zero so that its

square root is defined, that is, log2(x – 1)  0 (x – 1)  20  1  x  2. Therefore, domain of the given function is x  2 or x [2, ). The range of the function will be all non-negative real numbers as square root of any non-negative real number is always non-negative real number. Therefore, f(x)  [0, ). v.

In the given function, it is very clear that the denominator cannot be zero. Therefore, x cannot be equal to –3. Domain of the function is x  R – {–3}. Note that R – {–3} means that the element –3 would not be included in the set of real numbers. Rest would be part of domain. The domain can also be written as, x  –3 or x  (––3)  (–3, ). To find range means to find the set of values which the given expression can take. Let us say that y  3y  1 x 1 . Then, xy  3y  x – 1  x  . For real x, y cannot be 1. 1y x 3 But y represents range. Therefore, the range of the given function is, f(x)  1.

Example 2 Find the domain and range of the following functions: i.

(x – 1)(x – 2)

ii.

1 (x  1)(x  2)

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Quantitative Aptitude Simplified for CAT

Solution i.

We can easily see that x can take any real number. Therefore, the domain is the set of all real numbers, that is, x  R. 2

3 1  To find the range, (x – 1)(x – 2)  x2 – 3x  2   x    . 2 4   2

3   x   being a perfect square cannot be less than 0. Therefore, the function cannot be less than 2  1 1 1  1  0 – , that is, – . Therefore, range of the function is f(x)  – or f(x)    , 0  4 4 4  4  ii.

As for domain, x can be all real numbers except 1 and 2. To find range, let the expression be equal to y. Then,

1  y or 1  y(x2 – 3x  2) (x-1)(x-2)  yx2 – 3yx  2y – 1  0. For x to be real, discriminant  0, that is, (3y)2  4y(2y – 1) y2  4y  0  y(y  4)  0  y  0 or  –4. But, y cannot be 0. Therefore, the range of the given expression is, f(x)  (–, –4] (0, ) Example 3 Range of the function f(x)  x2 

1 2

is

x 1

A.

[1, )

B.

[2, )

C.

[, )

D.

None of these

Solution f(x)  x 2 

1 1  x2

 x2 

1  x2 1  x2



x2 1  x2

 x2  1 

x2 1  x2

 x2  x2 x2 2, we can say that x 2  Since f(x)  1   x 2  , where is less than x is always a positive   1  x2  1  x2 1  x2 

number. Hence, 1  x 2 

x2 1  x2

is always more than or equal to 1.

The range of the given function is [1, ). Hence, [A]. Alternatively, Let x2  p and let the given function be y. Then yp

1  p2  p(1 – y)  (1 – y)  0. p 1

For p to be real, discriminant must be  0. Therefore, (1 – y)2 4(1 – y)  (1 – y)(1 – y – 4)  0  (y – 1)(y  3)  0  y  1 or  –3

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Functions and Graphs

But y being a positive number (as it is equal to x2 

1 2

), y  1

x 1

Hence, range of the given function is [1, ).

Common Functions and their Graphs Some of the common functions we encounter are sin x, cos x, log x, ex, |x|, [x], {x}, x, x2, x3 and

1 . We will x

study them and their properties in more details.

Trigonometric functions: sin x and cos x Both the functions are defined for all real x and can take values between –1 and 1, including both. Therefore, domain (sin x or cos x)  R, and range (sin x or cos x)  [–1, 1]. We can see that the curve of y  sin x passes through origin and intersects x-axis at infinite points. The curve of cos x does not pass through origin but intersects x-axis at infinite points. In fact, if the curve of sin x is shifted backwards (in the negative x direction) by 90°, we get the curve of cos x. The graphs of both these functions are repeating the same set of values after fixed interval and this repetition is happening infinite times.

Graph of sin x

Graph of cos x

Logarithmic functions: logax logax is defined only for positive values of x. Therefore, x  0. The value of base, ‘a’ can be either more than 1 or lie between 0 and 1, that is either a  1 or 0  a  1. Obviously a ≠ 1. The graph of log ax are different according to a  1 or 0  a  1. If a  1, then graph of logax is drawn below:

logax (where a  1)

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Quantitative Aptitude Simplified for CAT

As the value of x increases, the value of log x also increases. Moreover, as x approaches 0, log x approaches – . This does not mean that log 0 is –. The curve cuts x-axis at (1, 0). This means that for x  1, the curve is below x-axis, that is, log x  0 and for x  1, log x  0. Therefore, domain (log x)(0, ) and range (log x)  (–, ). If 0  a  1, graph of logax is drawn below:

logax (where 0  a  1) As the value of x increases, the value of log x also decreases. The curve cuts x-axis at (1, 0). This means that for x  1, the curve is above x-axis, that is, log x  0 and for x  1, log x  0. Therefore, domain (log x)(0, ), and range (log x)  (–, ).

Exponential functions: ex ex is defined for all values of x. As the value of x increases, the value of ex also increases. Moreover, as x approaches –, ex approaches 0. Let us now draw the graph of log x. The curve cuts y-axis at (0, 1). The curve is always above x-axis, that is, ex is never less than 0. Therefore, domain (ex)  (–, ), and range (ex) (0, ).

General exponential function is f(x)  ax, where a  1 or 0  a  1. The graph of ax in each case is drawn below.

f(x)  ax (where a  1)

f(x)  ax (where 0  a  1)

So, first graph above could be graph of f(x)  2x or 5x, whereas second graph above could be graph of f(x) x

 1    .  3

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Functions and Graphs

Absolute value functions: |x| |x|, pronounced as “mod x”, is a modulus function and is defined as |x|  x

if x  0

 –x

if x  0

Therefore, |3|  3 (as 3  0) and |–5|  –(–5)  5 (as –5  0) We note that |x| can never be negative. It is defined for all values of x, but |x|  0. Therefore, domain (|x|)  R, and range (|x|)  [0, ). The graph is drawn below.

The graph is V-shaped and the lines of V make angles of 45° with positive and negative directions of x-axis, as shown. The general graph of f(x)  k|x – a|  b will have vertex at (a, b), because function is minimized at x  a and minimum value of f(x) is b (assuming k  0). If k  0, then vertex would still be (a, b) but function would be maximized and maximum value will be b. Refer to the following example. Example 4 Draw the graph of: i.

f(x)  |x – 2|  1.

ii.

f(x)  –2|x  1| – 3

Solution i.

Note that (a, b)  (2, 1). So, vertex is (2, 1). Since k  0, graph would have minima and it would open upwards.

ii.

Here, vertex will be (–1, –3). Since k  0, graph would open downwards and have maxima.

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Quantitative Aptitude Simplified for CAT

Greatest integer functions: [x] The function [x] is called “greatest integer function” and is defined as the greatest integer less than or equal to x. Therefore, [2.46]  2, [0.321]  0 and [–6.358]  –7. In the last example, –7 is “the greatest integer” less than or equal to –6.358. Also, [4]  4, that is, [x]  x if x is an integer. The domain of [x]  R, and range of [x] that is, integers. The graph of [x] is drawn below.

The strong solid lines represent the graph. Note that the dotted lines are not part of the graph. The graph of [x] is a discontinuous curve, as it can only take integer values. Also note that at the beginning of each horizontal line there is a filled circle, whose meaning has been explained earlier. Since the graph looks like the steps of a staircase, this function is also called as step function.

Fractional functions: {x} This is known as “fractional function” and defined as the fractional part of x. For example, {2.35}  0.35 and{– 5.74}  0.26. This is so because –5.74  –6  0.26 and hence the fractional part is 0.26 and not 0.74 (as it appears to be the case). Moreover, {2}  0 Note that [x]  {x}  x or {x}  x – [x]. This is another way to define {x}. The domain of {x}  R, and range of {x}  [0, 1) The graph of this function is drawn below.

310

Functions and Graphs

Algebraic functions: x, x2, x3 and 1/x Domain (x)  R and Range (x)  R. Domain (x2)  R and Range (x2)  [0, ). Domain (x3)  R and Range (x3)  R. 1 Domain    R – {0} and Range  R – {0}.  x

The graphs of each of them are drawn below.

Example 5 Solve for x: i.

|x  3|  2  17

ii.

|3x – 13|  12  7

Solution i.

|x  3|  2  17  |x  3|  15  x  3  ±15  x  15 – 3 or –15 – 3. Therefore, x  12 or –18. More rigorous method of solving questions based on modulus is: Since |x|  x if x  0, and –x if x  0, we can say that |x  3|  (x  3), if x  –3  –(x  3), if x  –3 Therefore, if x  –3, then |x  3|  15  x  3  15  x  12, which is consistent with the fact that x  –3. Similarly, when x  –3, then |x  3|  15  –(x  3)  15 x  –18, which is consistent with the fact that x  –3.

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Quantitative Aptitude Simplified for CAT

When x  –3, |x  3|  15  0  15, which is inconsistent and not possible. Therefore, the answer is x  {12, –18}. ii.

|3x – 13|  12  7  |3x – 13|  –5. But, modulus function can never be negative. Therefore, no solution exists. Doing the same question by more rigorous method: |3x – 13|  3x – 13, if x   –(3x – 13), if x  Therefore, if x 

13 3

13 3

13 , then 3

|3x – 13|  –5  3x – 13  –5  x  , which is inconsistent with the fact that x  Therefore, no solution exists for x 

13 . 3

13 13 . When x  , 3 3

|3x – 13|  –5  –(3x – 13)  –5  x

18 13  6, which is inconsistent with the fact that x  . 3 3

13 13 . Even at x  , the equation is not correct. Therefore, for 3 3 the given equation, no solution exists in the entire real number line.

Therefore, no solution exists for x 

Example 6 Solve for x: |x – 1|  |x – 2|  7. Solution For x  2, |x – 1|  |x – 2|  7 x – 1  x – 2  7  x  5, which is consistent with x  2. For x  2, |x – 1|  |x – 2|  7  2 – 1  2 – 2  7 which is incorrect. For 1  x  2, |x – 1|  |x – 2|  7  x – 1  2 – x  7 which is again incorrect. For x  1, |x – 1|  |x – 2|  7  1 – 1  1 – 2  7 which is also incorrect. For x  1, |x – 1|  |x – 2|  7  1 – x  2 – x  7  x  –2, which is consistent with x  1. Therefore, the solution set of the given equation is {–2, 5}.

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Functions and Graphs

EXPERT SPEAK Scan this QR Code to watch a video that explains the principles involved in the Modulus Function and how to solve equations based on it.

Example 7

1  1 2  1 3  1  1 1000   2  1000    2  1000    2  1000   ...   2  1000   ?         Solution

1  1 2  1 3  1  1 1000   2  1000    2  1000    2  1000   ...   2  1000           1 1  1 2  1 3   1 499     1 500   1 1000         2  1000    2  1000   ...   2  1000     2  1000   .....   2  1000   2 1000      (0  0  0  .....  0)  (1  1  1  1 ....  1)  501 (as there are 501 terms from 500 to 1000) Example 8 Solve for x: [x  1]  [2x – 1]  [3.5x]  1.5, where [x] is greatest integer function. Solution Left Hand Side is sum of two terms both of which are integers as each is greatest integer function. But on the Right Hand Side, there are two terms, the first being an integer and the other is not an integer. Such an equation is therefore, logically incorrect. Hence, no value of x satisfies. Example 9   If f(x)  cos []x  cos [x], where [y] is the greatest integer function of y, then f     2

A.

cos 3

B.

0

C.

cos 4

D.

none of these

Solution   3.1416... Therefore, f(x)  cos []x  cos [x]  cos 3x  cos [x]  2         f    cos 3    cos [   ]  0  cos  .  2  2  2  2 

But  being 3.1416..., 2  10 and hence

2 is less than 5. Therefore, 2

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Quantitative Aptitude Simplified for CAT

f(/2)  cos [4.93...]  cos 4. Hence correct option is [C]. Example 10 If for some value of x, [x]  –3 and {x}  0.85, then find {x3}, where [x] is greatest integer function and {x} is fractional function. Solution We know that x  [x]  {x}  x  –3  0.85  –2.15. Therefore, x3  –9.938375  {x3}  0.061625. Example 11 If for some x, [x2]  12 and {x2}  0.25, find [2x]. Solution x2  [x2]  {x2}  12.25 x  3.5 or –3.5. Therefore, 2x  7 or –7  [2x]  7 or –7. Example 12 X is a variable. [X] is the greatest integer less than or equal to X. [X]  [Y]  [X  Y]. For which of the following relation, the values of X and Y are impossible to determine? A.

XY

B.

[X]  [Y]

C.

X  2Y

D.

Indeterminable

Solution For the result [X]  [Y]  [X  Y] to be true, either i.

X  Y, or

ii.

X and Y have to be integers, or

iii.

{X}  {Y} should be less than 1

Even if we know that X  Y, it is not yet possible to determine the value of X and Y. If we know that [X]  [Y], it is not sufficient to determine the value of X and Y. Similarly, option [C] also does not help in determining the value of X and Y. Therefore, the value of X and Y is indeterminable. Hence correct option is [D] Example 13 Which of the following best fits the below data? x

1

2

3

4

5

y

3

7

13

21

31

A

y  4x  1

B.

y  x3 – x2  x  2

C.

y  x2  x  1

D.

None of these

314

Functions and Graphs Solution For such questions, it is best to use options. Checking with option [A], when x  1, y  5. But this does not match with what is given in the table. Therefore, this is a wrong option. Checking with option [B], when x  1, y  3, which matches. When x  2, y  8, which does not match. Hence, even this is a wrong option. Checking with option [C], when x  1, y  3. When x  2, y  7 and so on. We observe that all the values of x yield those values which match with that given in the table. Hence, the correct option is [C]. We can solve this question without the use of options also. For that, we will find the difference of the consecutive values of y. We obtain 7 – 3  4; 13 – 7  6; 21 – 13  8; 31 – 21  10. The values obtained are: 4, 6, 8, 10. If we find the differences of these differences, we get 6 – 4  2; 8 – 6  2 and 10 – 8  2. We observe that the differences obtained in this case are constant. Since the difference (at the second stage) is constant, we can surely say that original expression is a quadratic expression, provided we know that it was a polynomial expression. Assuming this to be the case, we can say that the general quadratic expression for the given set of values is ax2  bx  c. Therefore, y  ax2  bx  c. Now, when x  1, y  3. Therefore, 3  a  b  c. Similarly, using the values of x as 2 and 3, we get 7  4a  2b  c and 13  9a  3b  c. Solving the system of three equations in three unknowns, we get a  b  c  1. Therefore, y  x2  x  1 is the correct answer. Alternatively, once we know that it will be quadratic function, now use options. We observe that [C] is the only option which is quadratic. If data satisfies this quadratic, then this would be the answer. Else [D] option would be the answer. Example 14 Find the number of times the graph of the function y  ||3x  5|  |2x – 1|| meets the x-axis. Solution For the graph of the given function to meet x-axis, y should become zero. Therefore, ||3x  5|  |2x – 1||  0  |3x  5|  |2x – 1|  0. When |3x  5|  0, then |2x – 1| is not zero and when |2x – 1| is zero, then |3x  5| is not zero. Therefore, both the expressions simultaneously cannot be zero. Hence, for any value of x, their summation can never be zero. The graph never meets the x-axis. Example 15 If x²  y²  0.1 and |x – y|  0.2, then |x|  |y| is equal to A. 0.3 B. 0.4 C. 0.2 D. 0.6 CAT 2000

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Quantitative Aptitude Simplified for CAT

Solution |x – y|  0.2  x – y  ±0.2  (x – y)2  0.04  x2  y2 – 2xy  0.04  2xy  0.1 – 0.04  0.06  xy  0.03. This means that (x, y) is either (0.3, 0.1) or (–0.3, –0.1). In any case, |x|  |y|  0.4. Hence, [B]. Alternatively, (|x|  |y|)2  x2  y2  2|xy|  0.1  0.06  0.16. So, (|x|  |y|)  0.4 (the value can’t be –0.4 because |x| or |y| can’t be negative). Example 16 The area bounded by the three curves |x  y|  1, |x|  1 and |y|  1, is equal to A.

4

B.

3

C.

2

D.

1 CAT 2000

Solution |x  y|  1  x  y  1; x  y  –1. |x|  1  x  1, –1. |y|  1  y  1, –1. Drawing the three curves, we get;

The area of the shaded region  area in quadrant (I  II  III  IV) 

1 1 (1)(1)  (1)(1)  (1)(1)  (1)(1)  3. 2 2

Therefore, correct option is [B]. Example 17 Draw the graph of y  x

1 log10 x

.

316

Functions and Graphs Solution y x

1 log10 x

 xlog x 10  10.

But, this is true only if x is positive (Also, x ≠ 1). Therefore, the graph is drawn below (not drawn to scale). The bold line is the graph on which we have also drawn two unfilled circles to show that the function is not defined for x  0 or 1.

Example 18 Draw the graph of i.

f(x)  |x – 1|  |x – 2|

ii.

f(x)  |x – 1|  |x – 2|  |x – 3|

iii.

f(x)  |x – 1|  |x – 2|  |x – 3|  |x – 4|

iv.

f(x)  (x – 1)(x – 2)(x – 3)

v.

f(x)  x-

vi.

f(x)  [2x]

3 3  2 2

Solution i.

f(x)  |x – 1|  |x – 2| When x  2, f(x)  (x – 1)  (x – 2)  2x – 3 When x  2, f(x)  1 When 1  x  2, f(x)  (x – 1)  (2 – x)  1 When x  1, f(x)  1 When x  1, f(x)  (1 – x)  (2 – x)  3 – 2x. The graph is drawn below:

The graph is minimum in the range of 1  x  2 and the minimum value is 1. Also notice that the graph is “flat” at the base. ii.

f(x)  |x – 1|  |x – 2|  |x – 3| Proceeding as before, f(x)  3x – 6

for x  3

f(x)  3

for x  3

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Quantitative Aptitude Simplified for CAT

f(x)  x

for 2  x  3

f(x)  2

for x  2

f(x)  4 – x

for 1  x  2

f(x)  3

for x  1

f(x)  6 – 3x

for x  1

The graph is drawn below.

The graph is minimum at x  2 and the minimum value is 2. Also notice that the graph is “pointed” at the base. iii.

f(x)  |x – 1|  |x – 2|  |x – 3|  |x – 4| f(x)  4x – 10

for x  4

f(x)  6

for x  4

f(x)  2x – 2

for 3  x  4

f(x)  4

for 2  x  3

f(x)  8 – 2x

for 1  x  2

f(x)  6

for x  1

f(x)  10 – 4x

for x  1

The graph is drawn below:

The graph is minimum in the range of 2  x  3 and the minimum value is 4. Also notice that the graph is “flat” at the base. From the above three graphs, we can conclude that a.

When there are even numbers of terms in modulus function, the graph will be “flat” at the base. Hence, the minima will exist for a range of values of x.

318

Functions and Graphs b. iv.

When there are odd numbers of terms in modulus function, the graph will be “pointed” at the base. Hence, the minima will exist for one value of x only.

f(x)  (x – 1)(x – 2)(x – 3) We observe that the curve intersects x axis at x  1, 2 and 3 and at no other point. When x  0, f(x)  –6. Also note that When x  3, (x – 1), (x – 2) and (x – 3) are all positive and hence f(x) will be always positive; When 2  x  3, (x – 1) and (x – 2) are positive whereas (x – 3) is negative and hence f(x) will be always negative; When 1  x  2, (x – 1) is positive whereas (x – 2) and (x – 3) is negative and hence f(x) will be always positive; When x  1, (x – 1), (x – 2) and (x – 3) are all negative and hence f(x) will be always negative. Therefore, the graph of the function can be drawn as shown below.

v.

f(x)  x-

3 3  2 2

3 3 and the minimum value is f(x)  . Therefore, the graph will 2 2 never intersect x-axis. At x  0, f(x)  3. The graph is drawn below:

The function is minimum at x 

The graph has the same shape as the graph of |x|, with the only difference that the vertex of the  3 3 graph is translated from (0, 0) to  ,  .  2 2 vi.

f(x)  [2x] f(x)  0

for 0  x 

f(x)  1

for

1 2

1  x  1 and so on. 2

The graph is drawn below.

319

Quantitative Aptitude Simplified for CAT

Observe the slight difference between the graph of [x] and [2x]. Example 19 Draw the graphs of log10x, log10|x|, |log10x| and |log10|x||. Solution The graphs are drawn below:

Observe that log |x| and |log|x|| are even functions. Example 20 When the curves y  log10 x and y  x–1 are drawn in the x-y plane, how many times do they intersect for x  1? CAT 2003 A.

Never

B.

Once

C.

Twice

D.

More than twice

Solution Draw the curves y  log10 x and y  x–1 in the same x-y plane, as shown below.

320

Functions and Graphs

The abscissa of the point of intersection of the two curves is ‘a’, and we can say that a  1. Therefore, the number of times the two curves intersect for x  1 is one. Hence, [B]. Example 21 The number of non-negative real roots of 2x – x – 1  0 equals A.

0

B.

1

C.

2

D.

3 CAT 2003

Solution 2x – x – 1  0  2x  x  1, whose solution is same as the intersection points of y  2x and y  x  1. Let us draw the graphs of these two curves.

We see that the curves meet at two points and the coordinates of the point of intersection are (0, 1) and (1, 2). Therefore, there are two non-negative real roots of 2x – x – 1  0, that is, x  0 and x  1. Note that x  0 is nonnegative. Example 22 Consider the following two curves in the x-y plane: y  x3  x2  5 y  x2  x  5 Which of the following statements is true for –2  x  2? A.

The two curves intersect once

B.

The two curves intersect twice

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Quantitative Aptitude Simplified for CAT

C.

The two curves do not intersect

D.

The two curves intersect thrice CAT 2003

Solution To find the number of intersection points, equate the two expressions and solve for x. x3  x2  5  x2  x  5  x3  x  x(x2 – 1)  0  x(x – 1)(x  1)  0  x  0, 1 and –1. Therefore, the two curves intersect each other at three distinct points. Hence correct option is [D]. Example 23 In how many real and distinct points do the graphs of the curve f(x)  x4  6x2  1 and g(x)  4x(x2  1), intersect? A.

1

B.

2

C.

4

D.

infinite

Solution Equating the two expressions, we get x4  6x2  1  4x(x2  1)  x4 – 4x3  6x2 – 4x  1  0  (x – 1)4  0  x  1. Therefore, the number of distinct and real points is 1. Hence correct option is [A]. Example 24 The equation x  ex  0 has A.

No real root

B.

One positive real root

C.

Two positive real roots

D.

One negative real root

Solution x  ex  0 ex  –x. The two curves y  ex and y  –x intersect only once as can be seen graphically.

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Functions and Graphs

Even and Odd Functions A function f(x) is said to be an even function if replacing x by –x does not change the value of the function. In other words, a function f(x) is even if f(–x)  f(x) For example, f(x)  x2 is an even function. This is because f(–x)  (–x)2  x2  f(x). Similarly, other examples of even functions are: cos x, |x|, sin2 x, x4  x2  1, and so on. A function f(x) is said to be an odd function if replacing x by –x changes the sign of the function. In other words, a function f(x) is odd if f(–x)  –f(x) For example, f(x)  x3 is an odd function. This is because f(–x)  (–x)3  –x2  –f(x). Similarly, other examples of odd functions are: sin x, tan x, x3  x, x, and so on.

Symmetricity of even and odd function It is important to note that the graphs of even functions are symmetrical about y axis. Refer to the graphs of cos x, |x| and x2 discussed in the earlier section. We can see that these graphs are symmetrical about y axis. If the graph is symmetrical about y-axis, we can naturally conclude that the height of the curve above x-axis for some value x1 is same as that for –x1. This is precisely why f(–x)  f(x), if the function is an even function. Similarly, the graphs of odd functions are symmetrical about origin. We can also say that the graphs of odd functions are symmetrical in opposite quadrants. Refer to the graphs of sin x, x and x3 discussed in the earlier section. We can see that these graphs are symmetrical in opposite quadrants. If the graph is symmetrical in opposite quadrants, we can naturally conclude that the height of the curve ‘above’ x-axis for some value x1 is same as the height of the curve below x-axis for –x1 and vice versa. This is precisely why f(–x)  –f(x), if the function is an odd function. A function f(x) need not always be even or odd. It may be neither even nor odd function. For example, f(x)  x3  x  1 is neither even nor odd. This is because f(–x)  (–x)3  (–x)  1  –x3 – x  1 which is neither f(x) nor –f(x). Further examples of functions which are neither even nor odd are: log x, ex, [x], {x}, and so on Example 25 Identify whether the following functions are even, odd or neither: i. ii. iii.

4tan3 x  3cosx sec 2 x  xcosec 2 x

log x  |cos x  sin x|

x 3 sinx  5xtanx x2cos2 x  1

iv.

1  x2  x4  x6  ....  x100

v.

x2 sin x

vi.

log x  x 2  1



 323

Quantitative Aptitude Simplified for CAT

Solution i.

f(x) 

4tan3 x  3cosx sec 2 x  xcosec 2 x

Now, f(–x) 

.

4tan3 (  x)  3cos (  x) sec2 (  x)  (  x)cosec2 (  x)

4tan3 x  3cosx



sec2 x  x cosec2 x

, which is neither f(x) nor –f(x).

Hence, it is neither even nor odd. ii.

If x is negative, then log x is not defined. The function is neither even nor odd.

iii.

f(x) 

x 3 sinx  5xtanx x2cos2 x  1

Now, f(–x) 

(  x)3 sin (  x)  5 (  x)tan(  x) 2

2



x 3 sin x  5x tan x

(  x) cos (  x)  1

x2 cos2 x  1

 f(x) .

Therefore, the function is even function. iv.

1  x2  x4  .....  x100  1  (–x)2  (–x)4  .....  (–x)100  1  x2  x4  .....  x100  f(x) Therefore, the function is even function.

v.

f(x)  x2 sin x. Now, f(–x)  (–x)2 sin (–x)  x2 (–sin x)  –x2 sin x  –f(x). Therefore, the given function is an odd function.

vi.





f(x)  log x  x 2  1 .

 Now, f(–x)  log   x  

   log   

  x 2  1  

 log



  1 x2  1  x  log    2   x 1  x 



  x  x2  1      log x  x 2  1   f  x  log    2 2  2  x  1  x x 1  x    

x2  1  x



x2  1  x









Therefore, the given function is odd. Let f(x) and g(x) be two even functions. Also let h(x)  f(x)  g(x). Then, h(–x)  f(–x)  g(–x)  f(x)  g(x)  h(x). Therefore, h(x) is an even function. Similarly, following results can be obtained: Let f1(x), f2(x), f3(x), ...., fn(x) be n functions. Also, let Fn(x)  f1(x)  f2(x)  f3(x)  ....  fn(x) and Gn(x)  f1(x)  f2(x)  f3(x)  ....  fn(x) A.

If all the functions are even, then Fn(x) as well as Gn(x) is an even function, irrespective of n being even or odd.

B.

If all the functions are odd, then Fn(x) is always odd irrespective of n being even or odd. But Gn(x) is odd if n is odd and is even if n is even.

C.

If f(x) is odd and g(x) is even, then i. f(x)  g(x) is neither even nor odd

324

Functions and Graphs ii. f(x) – g(x) is neither even nor odd iii. f(x)  g(x) is odd iv. f(x)  g(x) is odd Example 26 Discuss the nature of f1(x)  g1(x)  f2(x)  g2(x)  f3(x)  g3(x)  f4(x)  g4(x)  ....  fn(x)  gn(x) i.

if n is even

ii.

if n is odd

where f is even and g is odd function. Solution f1(x)  g1(x) is odd and similarly every term is an odd function. Therefore, this is a case of odd functions being added. i.

If even number of odd functions is added, the result is an odd function.

ii.

If odd number of odd functions is added, the result is again an odd function.

Example 27 Discuss the nature of the following: [f1(x)  g1(x)] [f2(x)  g2(x)][f3(x)  g3(x)]  [f4(x)  g4(x)]  ....  [fn(x)  gn(x)] i.

if n is even

ii.

if n is odd

where f and g are odd functions. Solution [f1(x)  g1(x)] is an odd function and similarly every term is an odd function. Therefore, this is a case of odd functions being multiplied. i.

If n is even, then the resulting function is an even function.

ii.

If n is odd, then the resulting function is an odd function.

Example 28 If a function is an odd function, then A.

it is symmetrical about x-axis

B.

it is symmetrical about origin

C.

it is symmetrical about the line y  –x or y  x

D.

both B and C mean the same and are correct options

Solution The correct option is B. Example 29 The graph of the function y  f(x) is symmetrical about the line x  2. Then A.

f(x  2)  f(x – 2)

B.

f(2  x)  f(2 – x)

C.

f(x)  f(–x)

D.

None of these

325

Quantitative Aptitude Simplified for CAT

Solution Let us take an example of a function which is symmetrical about the line x  2. Since f(x)  |x| is symmetrical about x  0, we can say that f(x)  |x – 2| is symmetrical about the line x  2. Checking with option [A], f(x  2)  |(x  2) – 2|  |x|; and f(x – 2)  |(x – 2) – 2|  |x – 4|. We see that |x|  |x – 4| and hence [A] is not the correct option. Checking with option [B], f(2  x)  |x|; and f(2 – x)  |(2 – x) – 2|  |–x|  |x|. We see that f(2  x)  f(2 – x) and hence this is the correct option. Hence correct option is [B].

Inverse of a Function Let us learn to find the inverse of a function f(x). The inverse of a function f(x) is denoted as f –1(x). Let f(x)  2x  7 be a function of x. Then, to find the inverse of f(x), we will express x in terms of f(x). Therefore, f(x)  2x  7  x 

f(x)  7 . 2

Finally, replace x by f –1(x) and f(x) by x. We get, f –1(x) 

x 7 x 7 , and hence inverse of 2x  7 is . 2 2

It is very important to note that every function is not invertible. The function f(x) is not invertible, if 2 or more values of x yield same value of f(x). For example, f(x)  x2. Here, f(1)  f(–1)  1 and hence such a function is not invertible. Whether a function is invertible or not, can also be understood graphically. Let there be any function f(x). Draw its graph and run a line (parallel to x-axis) through the whole curve of f(x). If the line is cut by the curve in more than one point at least once, then the function is not invertible. Otherwise, the function is invertible. For example, if f(x)  x2  3x  2. The graph of f(x) is drawn below along with the line parallel to x-axis.

The line parallel to x-axis is shown dotted at three different positions, and denoted by m. In the lowest position, the line does not touch the curve at all. In the middle position, the line touches the curve at one point. When the line moves further upwards, it starts cutting the curve in two distinct points. And hence the function x2  3x  2 is not invertible. Let us draw the curve of f(x)  2x  7 (the example taken above as a sample example) and apply this thumb rule.

326

Functions and Graphs We can easily imagine that no matter what the position of the dotted line is, this line would never be able to intersect the function 2x  7 in more than one point. Example 30 Find the inverse of the following: i.

x3  9

ii.

5x2 – 2

iii. iv.

e x  e x e x  e x

|x|

Solution i.

Let f(x)  x3  9. Therefore, x 

3 f(x)  9

.

Replacing x by f –1(x) and f(x) by x, we get f –1(x)  ii.

3

x  9 , which is inverse of x3  9.

Let f(x)  5x2 – 2. This is not an invertible function as f(1) and f(–1) are same. Alternatively, this being a quadratic expression, the graph is parabolic and as shown above, the function is not invertible.

iii.

Let f(x) 

e x  e x e x  e x

ex 



1

2x e x  e  1 2x 1 ex   x e  1 e

 [f(x)]e2x – f(x)  e2x  1  e2x  x

f(x)  1  f(x)  1   f(x)  1   2x  log e    In   f(x)  1  f(x)  1   f(x)  1 

1  f(x)  1  In   2  f(x)  1 

Replacing as usual, we get f –1(x)  iv.

1  x 1  In   , which is the inverse of the given function. 2  x 1 

Doing the graphical test on the function |x|, we can easily conclude that the given function is not invertible.

Example 31 Find the inverse of the following functions: i.

x

ii.

1 x

iii.

log10 x

iv.

[x]

v.

x-1 2x  3

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Quantitative Aptitude Simplified for CAT

Solution i.

Let f(x)  x. Then x  f(x). After the replacement, we get f –1(x)  x. Therefore, the inverse of x is x.

ii. Let f(x) 

1 1 1 . Then x  . After the replacement, we get f –1(x)  . f(x) x x

Therefore, the inverse of iii.

1 1 is . x x

Let f(x)  log10 x. Then x  10f(x). After the replacement, we get f –1(x)  10x. Therefore, the inverse of log10 x is 10x.

iv.

The function [x] is not invertible as [2.45]  [2.32]  2, or as can be seen graphically.

v.

Let f(x) 

3f(x)  1 x 1 . Then, 2xf(x)  3f(x)  x – 1  x  1-2f(x) 2x  3

 f –1(x) 

3x  1 . 1  2x

Note that if a function is invertible, then the graph of f(x) and f–1(x) are symmetrical about the line y  x. For example, x Let f(x)  2x. Then, f –1(x)  . Drawing them on the coordinate system, we get 2

x are symmetrical 2 about the line y  x. Check the same for the function log10 x whose inverse is 10x.

The dotted line is y  x. We can easily see that the two curves, y  2x and y 

Example 32 Graph of f(x) and f –1(x) (if f(x) is invertible) is symmetrical about A.

y  –x

B.

y-axis

C.

x-axis

D.

yx

Solution By concept, the graphs are symmetrical about the line y  x. Example 33 Find the inverse of f(x)  x2, where x  0.

328

Functions and Graphs Solution We know that f(x)  x2 as such is not invertible, but if only that portion of the function where x  0 is considered, then the function becomes invertible. f(x)  x2  x 

f ( x) .

Note that we have taken x as Therefore, f –1(x) 

f ( x ) and not  f ( x ) because x cannot be negative.

x , which is the inverse of x2 for x  0.

Composite Functions Let f(x)  cos x and g(x)  log x. Then, fog is the composite function of f on g, defined as f[g(x)]. Now, f[g(x)]  f(log x)  cos (log x). Therefore, fog  cos (log x). We can easily see that the process of finding fog is to write it as f[g(x)] and then in the function f(x), replace x by g(x) to transform cos x to cos (log x). Similarly, gof  g[f(x)]  g(cos x)  log (cos x). We can also find fof and gog. fof  f[f(x)]  f(cos x)  cos (cos x) gog  g[g(x)]  g(log x)  log (log x) Similarly, we can find the value of fofog, which is same as fo(fog). fo(fog)  f[f(g(x))]  f[(cos (log x)]  cos [cos (log x)] It is important to know that we may not always be able to find the composite function. For example, in the above example, gogof would mean g[g(f(x))]  log [log (cos x)], which is not defined for any value of x. This is so because for any value of x, cos x  [–1, 1]. But log cos x is not defined for negative values of cos x. Taking only positive values of cos x, we realize that log (cos x) is always negative or at most zero, whose logarithm is not defined. Therefore, log log cos x is not a defined function and hence gogof cannot be found out. Example 34 Let f(x)  sin (2x  3) and g(x)  log x. Find fog, gof, gog and fof. Solution fog  f[g(x)]  f(log x)  sin ((2log x)  3) gof  g[f(x)]  log (sin (2x  3)) gog  g[g(x)]  log (log x) fof  f[f(x)]  sin [2(sin (2x  3))  3] Example 35 Find fofof if f(x) 

1 . Find the domain of fofof. 1-x

Solution fofof  f[f(f(x))].

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Quantitative Aptitude Simplified for CAT

1

f (f ( x )) 

 1  1   1x



1x x 1  1  x 1 x

Therefore,

1 x  x 1  fofof  f     x.   x  1   x  1  x  (x  1)    x  Note that the domain of fofof is not same as the domain of obtained value ‘x’. To find domain of fofof, we need to consider the domain of f(x), as well as f(f(x)). f(x) is not defined when x  1 and f(f(x)) is not defined when x  0. Therefore, domain of fofof is x  R ~ {0, 1}. Example 36 If f(x)  sin–1x, and g(x)  x2  2, find fog. Solution fog comes out to sin–1(x2  2). On close observation, we find that x2  2 is always more than or equal to 2. But sin–1x is not defined for x more than 1. Therefore, this composite function cannot be found. Example 37 If f(x)  2(x  1)3 and f[g(x)]  g[f(x)], then find g(x). Solution f[g(x)]  g[f(x)] is true only in three cases: i.

When f(x) and g(x) are same function, or

ii.

When g(x)  x, or

iii.

When g(x) is inverse of f(x)

If g(x) is inverse of f(x), then g(x) 

3

x 1 . 2

Example 38 If f(x) 

ax  b , then f(x) is same as f–1(x) if cx  d

A. a  b  0 B. a  c  0 C. a  d  0 D. b  c  0 Solution f(x) 

d  f(x)  b ax  b  ax  b  cx[f(x)]  d[f(x)]  x  . a  c  f(x) cx  d

Replacing x by f –1(x) and f(x) by x, we get: f –1(x) 

dx  b . a  cx

330

Functions and Graphs

If f(x)  f–1(x), then

ax  b dx  b ax  b  dx  b    cx  d a  cx cx  d cx  a

If a  –d, then the equality is satisfied. Hence, a  d  0. Hence correct option is [C]. Let f(x) and g(x) be two functions, both being even functions. Then fog is also an even function. Similarly, if both are odd functions, then fog is also odd function; whereas if one is odd and the other is even, then fog is an even function. We can conclude that if at least one of the functions is even and the rest odd, then the composite function fog and gof are both even functions. In fact, if we have more functions in a series, like fogohokom.... where f(x), g(x), h(x), k(x), m(x), .... and so on are functions of x, and if at least one of the functions is an even function and rest of the functions are odd, then the composite function fogohokom.... is also an even function.

Recursive Functions Consider the series: 1, 1, 2, 3, 5, 8, 13, .... This is a fibonacci series where, starting from the third term, every term is the sum of the preceding two terms. Therefore, 211 312 5  2  3, and so on. In general, an  an–1  an–2, a1  1, a2  1. There can be infinite such relationships to find nth term, each defining a new series. Such series are called recursive series or functions. For example, if an  2an–1  3, a1  1, then the series is 1, 5, 13, 29, ..... Example 39 If a1  1 and an  1  2an  5, n  1, 2….., then a100 is equal to A.

(5  299 – 6)

B.

(5  299  6)

C.

(6  299  5)

D.

(6  299 – 5) CAT 2004

Solution a1  1 a2  2a1  5 a3  2a2  5  2(2a1  5)  5  22a1  2  5  5 a4  2a3  5  2[2(2a1  5)  5]  5  23a1  22  5  25  5  2n1  1   2n–1  5(2n–1) – 5  6(2n–1) – 5 an  2n–1a1  5(1  2  22  ....  2n–2)  2n–1a1  5   2  1   

Therefore, a100  6  299 – 5.

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Quantitative Aptitude Simplified for CAT

Alternatively, use options. If A were correct, then an  5(2n–1) – 6. But, in this case, a1 comes out to –1, which is wrong. If B were correct, then an  5(2n–1)  6. But, in this case, a1 comes out to 11, which is again wrong. If C were correct, then an  6(2n–1)  5. But, in this case, a1 comes out to 11, which is wrong. If D were correct, then an  6(2n–1) – 5. In this case, a1 comes out to 1, a2  7, and so on all of which is correct. Hence correct option is D. Example 40 For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference of squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence? A.

147

B.

76

C.

123

D

Cannot be determined CAT 1999

Solution Let the terms of the fibonacci sequence be a, b, a  b, a  2b, 2a  3b, 3a  5b, .... Then, a6  3a  5b and a7  5a  8b. a72 – a62  (5a  8b)2 – (3a  5b)2  (2a  3b)(8a  13b)  517  11  47 Comparing, we see that a  1 and b  3. Therefore, a6  3a  5b  18 a7  5a  8b  29 Therefore, a8  18  29  47; a9  29  47  76 and a10  47  76  123. Hence correct option is [C]. Example 41 For all non-negative integers x and y, f(x, y) is defined as below: f(0, y)  y  1 f(x  1, 0)  f(x, 1) f(x  1, y  1)  f(x, f(x  1, y)) Find f(2, 1). CAT 2000 Solution f(2, 1)  f(1, f(2, 0))

(i)

Now, f(2, 0)  f(1, 1)  f(0, f(1, 0))

(ii)

Further, f(1, 0)  f(0, 1)  1  1  2

332

Functions and Graphs Using this in (ii), we get: f(2, 0)  f(0, 2)  2  1  3, putting which in (i), we get: f(2, 1)  f(1, 3)  f(0, f(1, 2)) Now, f(1, 2)  f(0, f(1, 1)) f(0, 3)  3  1  4. Therefore, f(2, 1)  f(0, 4)  4  1  5. Example 42 For all non-negative integers x and y, f(x, y) is defined as below: f(0, y)  y  1 f(x  1, 0)  f(x, 1) f(x  1, y  1)  f(x, f(x  1, y)) Find f(2, 2). A.

10

B.

8

C.

9

D.

7

Solution f(2, 2)  f(1, f(2, 1))  f(1, 5) (Using the result from the previous example) f(1, 5)  f(0, f(1, 4)) f(1, 4)  f(0, f(1, 3)) f(1, 3)  f(0, f(1, 2))  f(0, 4)  5.  f(1, 4)  f(0, 5)  6  f(1, 5)  f(0, 6)  7  f(2, 2)  7. Hence correct option is [D]. Example 43 Let x and y be real numbers and let f(x, y)  |x  y|, F(f(x, y))  –f(x, y) and G(f(x, y))  –F(f(x, y)). Based on the above relationships, answer the following: i.

Which of the following statement is true? A.

F(f(x, y))  G(f(x, y))  –F(f(x, y))  G(f(x, y))

B.

F(f(x, y))  G(f(x, y))  –F(f(x, y))  G(f(x, y))

C.

F(f(x, y))  G(f(x, y))  G(f(x, y))  F(f(x, y))

D.

F(f(x, y)  G(f(x, y))  f(x, y)  f(–x, –y) CAT 1999

Solution From the given relationships, we can conclude that F(f(x, y))  –f(x, y)  –|x  y|, and

333

Quantitative Aptitude Simplified for CAT

G(f(x, y))  –F(f(x, y))  |x  y| Option [A] is obviously wrong as RHS contains negative sign with the same expressions on both sides. Checking option [B], –|x  y|  |x  y|  –(–|x  y|)  |x  y|  |x  y|  |x  y| This is wrong as RHS is positive whereas LHS is always negative. Option [C] is again wrong as LHS and RHS are same expressions. Checking option [D], –|x  y|  |x  y|  |x  y|  |(–x)  (–y)|  |x  y|, which is correct. Hence correct option is [D]. ii.

What is the value of f(G(f(1, 0)), f(F(f(1, 2)) , G(f(1, 2))))? A. 3 B. 2 C. 1 D. 0

Solution f(1, 0)  |1  0|  1; f(1, 2)  |1  2|  3. Moreover, G(f(1, 0))  1; F(f(1, 2))  –3 and G(f(1, 2))  3. Therefore, f(G(f(1, 0)), f(F(f(1, 2)) , G(f(1, 2))))  f(1, f(–3 , 3))  f(1, 0)  1. Therefore, (C) is the correct option. iii.

Which of the following expressions yields x2 as its result? A. F(f(x, –x)) G(f(x, –x)) B. F(f(x, x))  G(f(x, x))  4 C. –F(f(x, x)  G(f(x, x)  log2 16 D. f(x, x)  f(x, x)

Solution F(f(x, –x))  G(f(x, –x))  –|x – x|  |x – x|  0 F(f(x, x))  G(f(x, x))  4  –|x  x|  |x  x|  4  16x2. –F(f(x, x)  G(f(x, x)  log2 16  |x  x|  |x  x|  4  x2. Hence correct option is [C]. Example 44 For three distinct real numbers x, y and z, let f(x, y, z)  min (max(x, y), max (y, z), max (z, x)) g(x, y, z)  max (min(x, y), min (y, z), max (z, x)) h(x, y, z)  max (max(x, y), max(y, z), max (z, x)) j(x, y, z)  min (min (x, y), min(y, z), min (z, x)) m(x, y, z)  max (x, y, z) n(x, y, z)  min (x, y, z) Based on this, answer the following examples. i.

Which of the following is necessarily greater than 1?

334

Functions and Graphs

A.

h(x,y,z)  f(x,y,z) j(x,y,z)

B.

j(x,y,z) h(x,y,z)

C.

f(x,y,z) g(x,y,z)

D.

f(x,y,z)  h(x,y,z)  g(x,y,z) j(x,y,z) CAT 2000

Solution For such questions, we should take some value of x, y and z. Let x  1, y  2 and z  3. Then, f(x, y, z)  min(2, 3, 3)  2 g(x, y, z)  max (1, 2, 3)  3 h(x, y, z)  max (2, 3, 3)  3 j(x, y, z)  min (1, 2, 1)  1 m(x, y, z)  max (1, 2, 3)  3 n(x, y, z)  min (1, 2, 3)  1 Hence correct option is [D] as its value is 2, whereas the values of other options are less than or equal to 1. ii

Which of the following expressions is necessarily equal to 1? A.

h(x,y,z)  n(x,y,z) m(x,y,z)  j(x,y,z)

B.

m(x,y,z)  f(x,y,z) g(x,y,z)  n(x,y,z)

C.

j(x,y,z)  g(x,y,z) h(x,y,z)

D.

f(x,y,z)  h(x,y,z) f(x,y,z) CAT 2000

Solution Option [A] 

3 1  1. 3 1

Option [B] 

32 1  . 3 1 2

Option [C] 

13 2  . 3 3

Option [D] 

23 1  . 2 2

335

Quantitative Aptitude Simplified for CAT

Hence correct option is [A] iii.

Which of the following expressions is indeterminate? A.

f(x,y,z)  h(x,y,z) g(x,y,z)  j(x,y,z)

B.

f(x,y,z)  h(x,y,z)  g(x,y,z)  j(x,y,z) j(x,y,z)  h(x,y,z)  m(x,y,z)  n(x,y,z)

C.

g(x,y,z)  j(x,y,z) g(x,y,z)  h(x,y,z)

D.

h(x,y,z)  f(x,y,z) n(x,y,z)  g(x,y,z) CAT 2000

Solution In option [B], we observe that the denominator  1  3 – 3 – 1  0 and hence the expression is indeterminate. Hence correct option is [B]. Important Results Please note some important results related to functions: 1.

1 1 If f(x) is a polynomial function of degree n satisfying f(x)  f    f(x) × f   for all x ≠ 0, then,  x  x n f(x)  1 ± x .

2.

If f(x) satisfies f(x  y)  f(x)  f(y) for all x, y  R, then f(x)  x f(1) for all x  R.

3.

If f(x) satisfies f(x  y)  f(x) × f(y) for all x, y  R, then f(x)  [f(1)]x for all x  R.

Example 45 1 1 If f(x) is a polynomial function of degree n satisfying f(x)  f    f(x) × f   for all x ≠ 0, and f(3)  82,  x  x find f(5).

Solution From the above results, we can say that f(x)  1 ± xn. Now, f(3)  1 ± 3n  82  ±3n  81. We will reject negative sign. So 3n  81 or n  4. So, f(x)  1  x4  f(5)  626.

Rotation of Graphs Now, we will focus our attention on another aspect of graphs of functions which is called rotation of graphs. Let f(x) be a function of x. Then, we can find the value of f(–x), –f(x) as well as –f(–x). Note that i.

the graph of f(–x) is obtained by rotating the graph of f(x) about y-axis.

ii.

the graph of –f(x) is obtained by rotating the graph of f(x) about x-axis.

iii.

the graph of –f(–x) is obtained by rotating the graph of f(x) about both the axes one by one, that is, first about x-axis and then about y-axis. The rotation can also be done first about y-axis and then about x-axis. The order of rotation about the axes does not make any difference.

Therefore, when the sign of independent variable changes, rotation is about y-axis and when the sign of the function changes, the rotation is about x-axis, and finally when the both change their signs, then the rotation is about both the axes one by one (in any order).

336

Functions and Graphs Let us take an example. Let f(x)  x3. Then f(–x)  (–x)3  –x3. The graphs of these functions are drawn below. Note the rotation about y-axis.

Let f(x)  x2. Then –f(x)  –(x)2  –x2. The graphs of these functions are drawn below. Note the rotation about x-axis.

Let f(x)  x for x  0  x2 for x  0. The graph of f(x), f(–x), –f(x) and –f(–x) are drawn below.

337

Quantitative Aptitude Simplified for CAT

We observe that if f(x)  x2, then f(–x)  (–x)2  x2  f(x). Therefore, f(x)  x2 is an even function. But f(–x) means the graph of f(x) is rotated about y-axis. Rotation of f(x)  x2 about y-axis has no impact on the shape and position of the graph. In general, if rotation about y-axis has no impact on the shape and position of the curve, the corresponding function is an even function. Similarly, if f(x)  x3, then f(–x)  –f(x) and hence f(x) is an odd function. But –f(–x), which means double rotation about the axes, would have no impact on the shape and position of the graph. Therefore, if rotation about x-axis and y-axis together has no impact on the shape and position of the curve, the corresponding function is an odd function. Instructions for following examples: Given below are three graphs made up of straight-line segments shown as thick lines. In each case choose the answer as below: A.

If f(x)  3f(–x)

B.

If f(x)  –f(–x)

C.

If f(x)  f(–x)

D.

If 3f(x)  6f(–x) for x  0.

CAT 2000

Example 46

Solution Rotation of the graph about y-axis does not change the shape and position of the graph. Hence correct option is C. Example 47

Solution f(1)  2 and f(–1)  1. We observe that 3f(x)  6f(–x) or f(x)  2f(–x). Hence correct option is D.

338

Functions and Graphs Example 48

Solution Graphically, we can say that the function is an odd function. Hence [B] is the correct option.

339

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

to scale and the same scale has been used on each axis.

If a1  1, an1 – 3an  2  4n for every positive integer n, then a100 equals A.

399 – 200

B.

399  200

C.

3100 – 200

D.

3100  200 CAT 2005

2.

In the X-Y plane, the area of the region bounded by the graph |x  y|  |x – y|  4 is A.

8

Which of the following shows the graph of y against x?

B.

12

A.

C.

16

D.

20 CAT 2005

3.

Let g(x) be a function such that g(x  1)  g(x  1)  g(x) for every real x. Then for what value of p is the relation g(x  p)  g(x) necessarily true for every real x? A.

5

B.

3

C.

2

D.

6

B.

CAT 2005 4.

Consider a sequence where the nth term, tn 

C.

n , n  1, 2, …. The value of t3 t4 t5 n2

…. t53 equals A.

2 495

B.

2 477

C.

12 55

D.

1 1485

E.

1 2970

D.

CAT 2006 5.

The graph of y – x against y  x is as shown below. All graphs in this question are drawn

340

Functions and Graphs E.

9.

Let f(x) be a function satisfying f(x) f(y)  f(xy) for all real x, y. If f(2)  4, then what is the  1

value of f   ?  2 A. B. C. CAT 2006 6.

D. E.

A function f(x) satisfies f(1)  3600, and f(1)  f(2)  ...  f(n)  n2f(n), for all positive integers n  1. What is the value of f(9)?

CAT 2008

A.

80

B.

240

Directions for questions 10 and 11: Read the information given below and answer the questions that follow:

C.

200

f1(x)  x

D.

100

1

x1

E.

120

0

otherwise

CAT 2007 Directions for questions 7 and 8: Read the information given below and answer the questions that follow: Let a1  p and b1  q, where p and q are positive quantities. Define an  pbn–1, bn  qbn–1, for even n  1, and an  pan–1, bn  qan–1, for odd n  1.

8.

1 n1

q(pq) 2

B.

1 n1 qp 2 (p  q)

C.

1n q 2 (p  q)

D.

q 2 (p  q) 2

E.

q(pq) 2

1n

If p 

f3(x)  –f2(x)

for all x

f4(x)  f3(–x)

for all x

A. B. C. D.

(p  q)

0 1 2 3

11. Which of the following is necessarily true? A. B. C. D.

f4(x)  f1(x) for all x f1(x)  –f3(–x) for all x f2(–x)  f4(x) for all x f1(x)  f3(x)  0 for all x

12. Consider the sequence of numbers a1, a2, a3, ... to infinity where a1  81.33 and a2  – 19 and aj  aj–1 – aj–2 for j 3. What is the suzm of the first 6002 terms of this sequence?

1n

(p  q) 2

1 2 and q  , then what is the 3 3

smallest odd n such that an  bn  0.01? A. B. C. D. E.

for all x

f1(x) f2(x), f2(x) f3(x), f2(x) f4(x)?

1n

1 n1

f2(x)  f1(–x)

10. How many of the following products are necessarily zero for every x:

Which of the following best describes an  bn for even n? A.

0x1

CAT 2004

CAT 2007 7.

0 1 4 1 2 1 Cannot be determined

A. B. C. D.

7 13 11 9 15

–100.33 –30.00 62.33 119.33 CAT 2004

341

Quantitative Aptitude Simplified for CAT

Directions for questions 13 to 15: Answer the questions on the basis of the tables given below.

C. D.

CAT 2003 (R)

Graph of log An and n is a straight line passing through (log A, 0) Graph of log An and n is a curved line passing through (0, 0)

Two binary operations  and * are defined over the set {a, e, f, g, h} as per the following tables:

CAT 2012 n

e

f

g

h

*

a

e

F

g

h

a

a

e

f

g

h

a

a

a

A

a

a

e

e

f

g

h

a

e

a

e

f

g

h

f

f

g

h

a

e

f

a

f

h

e

g

g

g

h

a

e

f

g

a

g

e

h

f

C.

h

h

a

e

f

g

h

a

h

g

f

e

D.

A. B.

Thus, according to the first table f  g  a, while according to the second table g * h  f, and so on. Also, let f2  f*f, g3 g*g*g, and so on.

18. g(x  y)  g(x)  g(y), g(4)  8, g(3)  ? A. B. C. D.

4 5 2 3

19. Find the smallest integer more than log3 

A. B. C. D.

log4. A. B. C. D.

e f g h

15. Upon simplification, equals:

B.

1 2 3 4 CAT 2012

{a10

*

(f10



g9)}

e8

20. Identify the relationship between f1(x) and f2(x).

e f g h

16. A person donates money such that to the first person he donates r% of the money; to the next person, he donates r% of the money left and so on. Let An  money left after n donations and A  money initially available with the person. Then, A.

0 3 6 10 CAT 2012

14. Upon simplification, f  [f * {f  (f * f)}] equals: A. B. C. D.

an1  an  an-1 1 an1  an  an-1 2 an1  an  an-1 1 an1  an  an-1 2 CAT 2012

13. What is the smallest positive integer n such that gn  e? A. B. C. D.

n

1 3 1 3     . Then, 2    2 

17. an  

 a

Graph of log An and n is a straight line passing through (0, 0) Graph of log An and n is a straight line passing through (0, log A)

342

Functions and Graphs 26. If |x|  2|y|  10, then find the largest distance between any two points on the given curve.

A.

f1(x)  f2(x)

B.

f1(x)  f2(x)

C.

f1(x)  f2(x)

D.

No relation between the two functions can be established CAT 2013

21. 2|x| ≤ 1.97x. Find the number of distinct real roots. A. B. C. D.

B.

20

C.

40

D.

5

27. If an =

2 0 infinite 1

n , then for what value of n will 100

(102  an)(100  an) will be minimum?

22. ai  ai2  2ai1, a1  1 and a2  3. Find a20. 41 39 37 43

A.

1

B.

100

C.

101

D.

102 CAT 2016

28. If f(x)  ax  b, a and b are positive real numbers and if f(f(x))  9x  8, then a  b  CAT 2014

23. Xi  Xi1  1, X10  k. Find X91. A. B. C. D.

10

CAT 2015

CAT 2014

A. B. C. D.

A.

k k1 k1 1–k

A.

3

B.

4

C.

5

D.

6

E.

None of the above XAT 2017

29. f is a function for which f(1)  1 and f(x)  2x  f(x – 1) for each natural number x  2. Find f(31).

CAT 2014 24. If f(x)  ax  b and f(f(f(x)))  8x  21, find 2a  b.

A.

869

A.

7

B.

929

B.

11

C.

951

C.

8

D.

991

D.

15

E.

None of the above XAT 2016

CAT 2015

30. Find the equation of the graph shown below.

1 1 25. If a1  (where x  1) and an  , 1  an1 1x

find a119.

1 x

A.

1

B.

1 1 x

C.

x

D.

None of these CAT 2015

343

Quantitative Aptitude Simplified for CAT

A.

y  3x – 4

B.

1

B.

y  2x2 – 40

C.

2

C. D. E.

x

2y2

– 40

D.

3

y

2x2

 3x – 19

E.

None of the above

x

2y2 

3y – 19

XAT 2014 XAT 2015

f(x2

x4

7x2

f(x3

31. If – 1)  –  k1 and – 2)  9x3  k2, then the value of (k2 – k1) is

x6

34. The figure below shows the graph of a function f(x). How many solutions does the equation f(f(x))  15 have?

–

A.

6

A.

5

B.

7

B.

6

C.

8

C.

7

D.

9

D.

8

E.

None of the above

E.

Cannot be determined from the given graph

XAT 2015

XAT 2013

32. For a positive integer x, define f(x) such that f(x  a)  f(a × x), where a is an integer and f(1)  4. If the value of f(1003)  k, then the value of ‘k’ will be: A.

1003

B.

1004

C.

1005

D.

1006

E.

None of the above XAT 2015

33. The sum of the possible values of X in the equation |X  7|  |X – 8|  16 is: A.

0

ANSWER KEY 1. C

2. A

3. D

4. A

5. D

6. A

7. A

8. D

9. B

10. C

11. B

12. C

13. A

14. D

15. A

16. B

17. D

18. C

19. C

20. C

21. D

22. B

23. D

24. A

25. A

26. B

27. B

28. C

29. D

30. E

31. C

32. E

33. B

34. C

344

Functions and Graphs

ANSWERS AND EXPLANATIONS 1.

Area of AOB 

Area of ABCD (enclosed area)  4 ×2  8.

Answer: C

3.

Explanation: a1  1. a2  3a1  4(1) – 2  5 a3  3a2  4(2) – 2  21 a4  3a3  4(3) – 2  73. Observing the trend, we see that a1  1  31 – 2 a2  5  32 – 4 a3  21  33 – 6 a4  73  34 – 8 and so on. Therefore, an  3n – 2n and hence a100  3100 – 200. Alternatively, Use options. If [A] option is the answer, then an  3n – 1 – 2n. So, a1  30 – 2(1)  –1. So, [A] is not the correct option. If [B] option is the answer, then an  3n – 1  2n. So, a1  30  2(1)  3. So, [B] is not the correct option. If [C] option is the answer, then an  3n – 2n. So, a1  31 – 2(1)  1; a2  32 – 2(2)  5. Hence correct option is [C]. If [D] option is the answer, then an  3n  2n. So, a1  31  2(1)  5. So, [D] is not the correct option. 2.

g(x  3)  –g(x) So, [B] is also rejected. Replacing x by x  3 in (iii), we get g(x  6)  –g(x  3)  g(x). 4.

(ii)

(iii)

Answer: A Explanation: t3 t4 t5 …. t53  3 4 5 6 53 34 2 .     .....   5 6 7 8 55 54  55 495

5.

Answer: D Explanation: From the given graph, we can say that y – x  k(y  x), where k  1 (as slope of the line can be seen to be more than 1).  1k Therefore, y  x   . Since slope of this  1k line is negative, the correct option cannot be [A], [B] or [C]. Let us take some value of k, say 2. Then slope  –3. So, the line will be closer to y axis than to x axis.

If x  0  2| y |  4  y   2 If y  0  2| x |  4  x   2 Plot the points, we get

6.

Answer: A Explanation: f(1)  f(2)  ...  f(n)  n2f(n) f(1)  f(2)  ...  f(n – 1)  n2f(n) – f(n)  (n2 – 1)f(n) n – 1)2f(n – 1)  (n2 – 1)f(n) (n  1) f(n)  f(n – 1) (n  1)

A 2 2

(i)

g(x  3)  g(x  2) – g(x  1) Using (ii) here, we get

Explanation: | x y|| x y| 4

O

Answer: D Explanation: g(x  1)  g(x) – g(x – 1) Replacing x by x  1 in (i), we get g(x  2)  g(x  1) – g(x) So, [C] option is rejected. Replacing x by x  1 in (ii)

Answer: A

D

1  2 2 2 2

B

2 2 C

Therefore, f(9)

345

Quantitative Aptitude Simplified for CAT

 91   9 1   8 1     f (8)       f (7)  91   9 1   8 1  and so on. 8 7 6 5 4 3 2 1 So, f(9)          f (1) 10 9 8 7 6 5 4 3 2   3600  80 . 90

7.

1  1  f   .f(2)  f  2    f(1)  1. 2  2   1 1 Therefore, f    (as f(2)  4). 4  2 Alternatively, Since f(2)  4, we can conjecture that f(x) should be either x2 or 2x or 2x (as a first estimate). Only x2 fits into the equation f(x).f(y) 1  1  f(xy). So, f    . 4  2

Answer: A Explanation: a1  b1  p  q a2  b2  pb1  qb1  q(p  q) a3  b3  a2(p  q)  pq(p  q) a4  b4  b3(p  q)  pq2(p  q) and so on. Checking with options by putting n  4, we observe that only A] satisfies.

8.

10. Answer: C Explanation: Let us first draw the graph of f1(x).

Answer: D Explanation: a5  b5  a4(p  q)  p2q2(p  q). Looking at the pattern (including from previous solution), we conclude that for n odd, an  bn  p

n1 n1 2 q 2 ( p  q)

f2(x) is obtained by rotation of f1(x) about yaxis. f3(x) is obtained by rotation of f2(x) about x-axis. Finally, f4(x) is obtained by rotation of f3(x) about y-axis. The graphs of these are drawn below.

In the given question, p  q  1. So, n1 n1 p 2 q 2

 2    9

1  0.01     3

n1 2

 2    3

n1 2

 0.01

 0.01 . Now use options.

2 If n  7, then    9 than 0.01.

3

2 If n  9, then    9 than 0.01.

9.

n1 2



8 which is not less 729

4



16 which is less 6561

Answer: B Explanation: When x  1, y  2, given equation becomes f(1).f(2)  f(2), or f(1)  1. 1 Similarly, when x  , y  2, equation becomes 2

346

Functions and Graphs But (a1  a2  a3  a4  a5  a6)  (a1)  (a2)  (a2 – a1)  (–a1)  (–a2)  (–a2  a1)  0. Therefore, a1  a2  a3  ....  a6002  a6001  a6002  a1  a2  81.33 – 19  62.33. 13. Answer: A Explanation: We want gn  e. Check various values of n. When n  2, g*g  h When n  3, g*g*g  g*h  f When n  4, g*g*g*g  g*f  e

From the above graphs, we can surely say that f1(x) is always zero for x  0 and f2(x) is always zero for x  0 and hence for all x, f1(x)  f2(x) is always zero. But f2(x) and f3(x) are both zero only for positive values of x and non-zero for negative values of x. Therefore, the product f2(x)  f3(x) is not always zero. Likewise, f2(x) is always zero for x  0 and f4(x) is always zero for x  0 and hence for all x, f2(x)  f4(x) is always zero.

14. Answer: D Explanation: f  [f *{f  (f * f)}]  f  [f *{f  h}]  f  [f * e]  f  f  h. 15. Answer: A

11. Answer: B

Explanation: {a10 * (f10  g9)} e8 e8  (e*e)4(e*e)2(e*e)  e g9  g*g*g*g*g*g*g*g*g  h*h*h*h*g  e*e*g  e*g  g f10  (f*f)5  h5  (h2)2h  e2h  e*h  h a10  (a2)5  (a)5  a. {a10 * (f10  g9)} e8  {a * (h  g)} e  {a * f}  e  a  e  e.

Explanation: Graphically, –f3(–x)  –f4(x), which means rotating f4(x) about x-axis, which will yield us the graph of f1(x). Algebraically, –f3(–x)  f2(–x)  f1(x) (from the definitions of these functions). 12. Answer: C Explanation:

16. Answer: B

a1  81.33

Explanation: Money left after n donations, An  A

a2  –19 a3  a2 – a1

n

r   1  . Taking log on both sides, 100  

a4  a3 – a2  – a1

r   log An  log A  n log  1  . 100  

a5  a4 – a3  –a1 – (a2 – a1)  –a2 a6  a5 – a4  –a2  a1

When n  0, log An  log A. Therefore, graph of log An and n is a straight line passing through (0, log A).

a7  a6 – a5  (–a2  a1)  a2  a1 Since a7 is same as a1, a8 would be same as a2 and so on. Therefore,

17. Answer: D

a1  a2  a3  ....  a6002  (a1  a2  a3  a4  a5  a6)  (a7  a8  a9  a10  a11  a12)

Explanation: an1 

 .... (a5995  a5996  a5997  a5998  a5999  a6000)  a6001  a6002

347

Quantitative Aptitude Simplified for CAT

1 3   2   

n 1

1 3     2 

n1

Then a1  1; a2  2

n

1 3 1 3        2  2     , a0  1  1  2.

n

1 3 1 3 1 3 1 3             2   2   2   2 

1 3  an  1     2 

1 3     2 

Explanation: If g(x  y)  g(x)  g(y), then g(x)  x g(1). Therefore, g(4)  4 g(1)  8 or g(1)  2. So, g(3)  3 g(1)  3 × 2  6.

n1

19. Answer: C

n

n









Explanation: log3  log4  log12  m, say. Then, 12  m. Since value of   3.14…, 12 lies between 2 and 3 and so 2  m  3. Hence smallest integer more than the given expression is 3.

1 3   2 1 3        2  2    n 2 1 3  1 3       2  2    1 3      2 

n



1 3  3  1     2 



20. Answer: C Explanation: f2(x) is obtained from f1(x) by rotating f1(x) about both the axes one after the other.

n



3 1



21. Answer: D

1 3n 1 3n            2  2      n n  1 3   1 3    3         2  2     So,

Explanation: Clearly, x  0 satisfies the inequality. If x  0, then 2x ≤ 1.97x  x ≤ 0, which is not compatible with x  0. If x  0, then 2x ≤ 1.97x  x ≥ 0, which is not compatible with x  0. Therefore, the only real solution is x  0.

1 an–1 2

22. Answer: B

n n  1 3   1   1  3         2   2  2     n n 1 3  3   1  3         2   2  2     

So, an1 

42 3 42 3 2 4

18. Answer: C

1 3   2  1 3   2             2  1 3   2  1 3  n



Only D option satisfies this. Hence correct option is D.

n n  1 3   1 1 3         2   2  2      n n  1 3   3   1  3         2   2  2      n 1

2

Explanation: a1  1; a2  3; a3  2a2 – a1  5; a4  2a3 – a2  7; and so on. We get a series of consecutive odd numbers. So, an  2n – 1 and so a20  39.

1 an–1  an. 2

23. Answer: D Explanation: X11  1 – X10  1 – k; X12  1 – X11  k; and so on. For odd n, Xn  1 – k and for even n, Xn  k. So, X91  1 – k.

Alternatively, put n  1.

348

Functions and Graphs 24. Answer: A

28. Answer: C

Explanation: f(f(x))  a(ax  b)  b  a2x  ab  b; f(f(f(x)))  a(a2x  ab  b)  b  a3x  a2b  ab  b. Comparing with 8x  21, we observe that a  2 and so a2b  ab  b  4b  2b  b  21  b  3. Hence 2a  b  7.

Explanation: f(f(x)  a(ax  b)  b  a2x  ab  b  9x  8  a  3 and ab  b  8 3b  b  8  b  2. Therefore, a  b  5. 29. Answer: D Explanation: From the given function: f(2)  4  f(1)  5  22  1 f(3)  6  f(2)  6  5  11  32 2 f(4)  8  f(3)  8  11  19  42  3 We can see that f(x)  x2  (x – 1). Hence f(31)  312  30  991.

25. Answer: A Explanation: 1  a2  1  a1





1  1  a2

1  1  1    1 x



x 1 ; a3 x

1  x ; a4  x 1  1    x 

30. Answer: E Explanation: that vertical axis is x-axis, and not y-axis. Moreover, the graph is a parabola whose axis is parallel to x-axis. So, equation will be of the form x  ay2  by  c. This rules out options [A], [B] and [D]. Also, since x-axis itself is not the axis of the parabola, [C] option is ruled out. Alternatively, From the graph, observe that at y  0, x  – 19. Only x  2y2  3y – 19 satisfies the above criteria.

1 1   a1. 1  a3 1  x

So, a5  a2, a6  a3, and so on. Therefore, x 1 1 a119  a2  1 . x x 26. Answer: B Explanation: When x  0, y  5; when y  0, x  10. The graph of |x|  2|y|  10 is drawn below.

31. Answer: C Explanation: Put x  0 in f(x2 – 1)  x4 – 7x2  k1  f(–1)  k1 … (i) Put x  1 in f(x3 – 2)  x6 – 9x3  k2  f(–1)  1 – 9  k2 … (ii) From (i) and (ii), k1  –8  k2  k2 – k1  8.

From the graph, we can easily say that the largest distance between any two points on the curve is 10  10  20.

32. Answer: E Explanation: f(1)  4; f(2)  f(1  1)  f(1 × 1)  f(1)  4; f(3)  f(2  1)  f(2 x 1)  f(2)  4. From this, we observe that f(2), f(3), f(4), … all will have same value of 4. So, f(1003)  4  k.

27. Answer: B Explanation: an  x. Then the given expression becomes (102 – x)(100  x)  x2  2x  10200  –(x2 – 2x – 10200)  –[(x – 1)2 – 10200 – 1]. This expression is minimized when x  1  an  1  n  100.

349

Quantitative Aptitude Simplified for CAT

33. Answer: B

compatible with X  7. So, X  7.5 is one of the solutions. Sum of all solutions  8.5 – 7.5  1.

Explanation: If X  8, then |X  7|  |X – 8|  16  X  7  X – 8  16  X  8.5, which is compatible with X  8. So, X  8.5 is one of the solutions. If 7  X  8, then |X  7|  |X – 8|  16  X  7 – (X – 8)  16  15  16, which is incompatible. So, no solution exists in this range. If X  7, then |X  7|  |X – 8|  16  X  7  (X – 8)  16  X  7.5, which is

34. Answer: C Explanation: f(f(x))  15 will be true when f(x)  4 or f(x)  12. From the graph, f(x)  4 for 4 values of x, while f(x)  12 for 3 values of x. .

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Inequalities, Maxima and Minima

Chapt er 10

Inequalities, Maxima and Minima INEQUALITIES In inequalities, we deal with various types of inequalities, each of which involve specific methods of solving. Solving an inequality means finding the set of values of the variable which satisfies the given inequality. Therefore, we may have linear inequality, quadratic inequality, modulus based inequality, logarithmic inequality, and so on. But, there are three fundamentals rules that help us in solving these inequalities in a logical manner.

Fundamental Rules Rule 1: If a > b, then –a < –b. Therefore, if 3 > 2, then –3 < –2. This implies that whenever both sides of an inequality are multiplied by –1, then the inequality changes direction. In other words, whenever an inequality is multiplied or divided by the same positive real number, the inequality remains unaffected. Rule 2: If a > b, then (i)

1 1  , provided both a and b are of same sign, that is, ab > 0. a b

Therefore, 3 > 2  (ii)

1 1 1 1  and –3 > –4   . 3 2 3 4

1 1  , provided a and b are of opposite signs, that is ab < 0 a b

Therefore, 3 > –2 

1 1  . Here, inequality does not change sign. 3 2

Rule 3: If a > b, then (i)

logca > logcb, if c > 1

(ii)

logca < logcb, if 0 < c < 1

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Quantitative Aptitude Simplified for CAT

This is easy to understand. We know that if a > b, then log a > log b. Dividing both sides by log c (which is positive as c > 1), the inequality log a log b  is not affected. Therefore, > or logca > logcb. log c log c Similarly, if c lies between 0 and 1, then log c is negative and dividing both sides by log c would change the inequality. Hence, (ii).

Linear Inequality Consider the inequality 3x – 5 > 7 This is an example of linear inequality, because degree of the corresponding polynomial is 1. To solve this, we will shift the various terms on similar lines to what we did in case of linear equations. So, 3x – 5 > 7  3x > 12  x > 4. Note that in the last step, we divided both sides of the inequality by 3, which is a positive real number and so the direction of inequality wouldn’t change. Now, consider the following: 2x + 7 < 19 This becomes 2x < 19 – 7  2x < 12. Now, dividing both sides of the inequality by -2, which is a negative real number will change the direction of inequality and so we get 2x < 12  x > 6. Example 1 Solve for x:

3x  7  3 , where x  R. 2x  5

Solution Method 1: We cannot shift the denominator 2x – 5 to the right side of the inequality unless we know the sign of it. But 2x – 5 contains a variable and so the sign of 2x – 5 depends upon the value of x.

5 , then the denominator becomes positive and we can shift it to the right side of the 2 inequality without changing the direction of inequality. We get, If 2x – 5 > 0, that is x >

3x + 7 > 3(2x – 5)  x
. The solution which is common to the solutions x < 3 2

5 5 22 22 and x > is:

when x
7, each of the terms (x – 2), (x – 5) and (x – 7) is positive and hence their product is also positive. So, (x – 2)(x – 5)(x – 7) > 0. When 5 < x < 7, each of the terms (x – 2) and (x – 5) is positive whereas (x – 7) is negative and hence their product is negative. So, (x – 2)(x – 5)(x – 7) < 0. When 2 < x < 5, the term (x – 2) is positive whereas each of the terms (x – 5) and (x – 7) is negative and hence their product is positive. So, (x – 2)(x – 5)(x – 7) > 0. When x < 2, each of the terms (x – 2), (x – 5) and (x – 7) is negative and hence their product is also negative. So, (x – 2)(x – 5)(x – 7) < 0. We note that the sign of (x – 2)(x – 5)(x – 7) changes alternately from positive to negative to positive and so on, as we move from x > 7 to 5 < x < 7 to 2 < x < 5 and so on. Also note that the largest root of the equation f(x) in the above example is x = 7 and when x is more than the largest root, the sign of f(x) is positive and then sign of f(x) will alternate. So, we can generalize that If f(x) = k(x – a)(x – b)(x – c)(x – d)…. (x – n), where a < b < c < d < … < n, then f(x) > 0 if x > n, f(x) < 0 if m < x < n,

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f(x) > 0 if l < x < m, and so on However, one must remember that the above situation of signs changing alternately will happen when the power of each term in f(x) is 1. The following examples will clarify all these points.

EXPERT SPEAK Scan this QR Code to watch a video that explains the basic principles of Linear and Quadratic Inequalities.

Example 2 Solve the inequalities (i)

(x – 2)(x – 5) > 0

(ii)

(x – 3)(x + 1) < 0

(iii) x2 + 5x + 3 < 0 (iv) x2 + 5x + 7 < 0 (v)

x2 + 5x + 6.25 > 0

(vi) 3 + 4x – x2 > 0 Solution (i)

We know that f(x) = (x – 2)(x – 5) will be positive when x > 5 and then sign of f(x) will alternate. So, (x – 2) (x – 5) > 0 when x > 5 or x < 2. This can also be written as x  (–, 2)  (5, ), or x  R – [2, 5].

(ii)

(x – 3) (x + 1) < 0  (x – 3) (x – (–1)) < 0  –1 < x < 3 (as discussed above)

(iii) To solve such inequalities, we first find the roots of x2 + 5x + 3 = 0. Therefore, x=

5  25  12 5  13 . The given inequality can, therefore, be written as  2 2

 5  13   5  13  5  13 5  13 . x  x    x    0  2 2 2 2    (iv) In the inequality x2 + 5x + 7 < 0, we observe that the discriminant = (5)2 – 4(1) (7) = –3. Since discriminant is negative, the graph of x2 + 5x + 7 would never intersect x-axis. Since the graph opens upwards, the graph is completely above x-axis and hence the expression x2 + 5x + 7 will never be less than 0. We can conclude that the inequality is never true. So, no solution exists. Alternatively, 2

2

5 25 5 3   x2 + 5x + 7 < 0   x    7   0   x     0 , which is impossible because 2 4 2 4   2

5 3   x    is always positive. 2 4 

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Inequalities, Maxima and Minima (v)

x2 + 5x + 6.25 > 0  (x + 2.5)2 > 0, which is always true except when x = –2.5, when the expression becomes equal to 0. Therefore, solution set is, x  R – {–2.5} or x  –2.5 or x 

(vi) 3 + 4x – x2 > 0  –x2 + 4x + 3 > 0  x2 – 4x – 3 < 0 The roots of the equation x2 – 4x – 3 = 0 are



4  16  12 4  28  2 7 2 2







Therefore, inequality becomes  x  2  7    x  2  7   0      2  7  x  2  7 . Example 3 Solve the inequality: (i)

(x – 1) (x – 2) (x – 3) > 0

(ii)

(x – 1) (x – 2) (x – 3) (x – 4) < 0

(iii) (x – 1) (x – 2)2(x – 3) > 0 (iv) (x – 1) |x – 2|(x – 3) < 0 Solution (i)

The function f(x) = (x – 1) (x – 2)(x – 3) will be positive when x > 3, and then the sign of f(x) will change alternately. Therefore, f(x) will be positive when x > 3, negative when 2 < x < 3, again positive when 1 < x < 2 and will be negative when x < 1. So, (x – 1)(x – 2)(x – 3) > 0 when x > 3 or 1 < x < 2.

(ii)

f(x) = (x – 1)(x – 2)(x – 3)(x – 4) will be positive when x > 4 and then the sign changes alternately. Therefore, (x – 1) (x – 2) (x – 3) (x – 4) < 0 when 3 < x < 4 or 1 < x < 2.

(iii) The solution of (x – 1) (x – 2)2(x – 3) > 0 is x > 3 or x < 1. Observe that (x – 2)2 has no impact on the solution of the inequality as (x – 2)2 is always positive. So, the inequality effectively becomes (x – 1) (x – 3) > 0 as far as the solution is concerned. (iv) As in the previous example, |x – 2| is always positive, and hence the inequality effectively becomes (x – 1) (x – 3) < 0 whose solution is 1 < x < 3. However, at x = 2, expression becomes 0, which is not less than 0. So, x cannot be equal to 2. So, solution of the given inequality is (1, 2) U (2, 3) Example 4 Solve: (i)

8x5 > x2

(ii)

(x  1)(x  2) 0 (x  3)(x  4)

(iii)

(iv) (v)

(x  1)(x  2) x 2  4x  5

0

(x  1)(x  2) -x2  4x  5 (x  1)(x  2) x 2  4x  2

0 0

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Solution (i)

8x5 > x2  8x5 – x2 > 0  x2(8x3 – 1) > 0 x = 0 does not satisfy the inequality. If x ≠ 0, then x2 > 0. So, x2(8x3 – 1) > 0  (8x3 – 1) > 0  x3 >

1 1 x> . 8 2

Hence solution of the given inequality is x > (ii)

Solution of

1 . 2

(x  1)(x  2)  0 is similar to solution of (x –1)(x – 2)(x – 3)(x – 4) > 0 (x  3)(x  4)

Therefore, solution is x > 4 or 2 < x < 3 or x < 1. (iii) To solve

(x  1)(x  2) x 2  4x  5

 0 , we should first of all study the sign of the denominator.

x2 – 4x + 5 = (x – 2)2 + 1 which is always positive. Hence the sign of the expression (x  1)(x  2)  0 depends only upon the numerator, that is, (x – 1) (x – 2). Therefore, x 2  4x  5

(x  1)(x  2) x 2  4x  5

 0  (x – 1)(x – 2) > 0  x > 2 or x < 1.

(iv) Let us study the sign of the denominator –x2 + 4x – 5. Now, –x2 + 4x – 5 = –(x2 – 4x + 5) = –[(x – 2)2 + 1], which is always negative. Therefore,

(x  1)(x  2) 2

 x  4x  5 (v)

0 

(x  1)(x  2) x2  4x  5

 0  (x – 1)(x – 2) < 0  1 < x < 2.

The denominator is x2 – 4x + 2. The roots of x2 – 4x + 2 = 0 are x=

4  4 2  4(1)(2)  2 2 . 2









Therefore, x2 – 4x + 2 =  x  2  2    x  2  2      The inequality

(x  1)(x  2) 2

 0 becomes

x  4x  2

(x  1)(x  2) 0 x  2 2   x  2 2     

















which effectively means (x – 1)(x – 2)  x  2  2    x  2  2  < 0.     The increasing order of 1, 2, 2  2, 2  2 is 2  2 , 1, 2, 2  2 .









Therefore, the solution of (x – 1)(x – 2)  x  2  2    x  2  2  < 0 is     2 < x < 2  2 , or 2  2 < x < 1. Example 5 Solve for x:

(x  1)(x  2)  2. (x  3)(x  4)

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Inequalities, Maxima and Minima Solution

(x  1)(x  2) (x  1)(x  2)  2   20 (x  3)(x  4) (x  3)(x  4)

x  

2

 

 3x  2  2 x 2  7x  12 (x  3)(x  4)

  0

 x2  11x  22 x2  11x  22 0 0 (x  3)(x  4) (x  3)(x  4)

The roots of x2 – 11x + 22 = 0 are: x =

11  11 2  4(1)(22) 11  33  . Therefore, 2 2

  11  33     11  33    x      x     2 2 x  11x  22       0 0 (x  3)(x  4) (x  3)(x  4) 2

whose solution is

11  33 11  33  x  3, 4  x  2 2

Modulus Based Inequality The basic meaning of |x| is distance of x from origin. Therefore, when we say |x| < 3, it only means that the distance of x from origin is less than 3. Therefore, solution of |x| < 3 is –3 < x < 3. Similarly, |x| > 5 means distance of x from origin is more than 5. Therefore, solution of |x| > 5 is x > 5 or x < –5. In general, (i)

|x| < a  –a < x < a

(ii)

|x| > a  x > a or x < –a

Example 6 Solve the inequality: |2x – 1| < 7 Solution |2x – 1| < 7  –7 < 2x – 1 < 7. Adding 1 on all sides, we get –6 < 2x < 8  –3 < x < 4, which is the solution of the given inequality. Example 7 Solve: (i)

|2x – 3| < |x + 5|

(ii)

|x – 1| + |x – 2| > |x – 4| + 3

Solution (i)

Method 1: Since both the sides are positive, we can square both sides without affecting the inequality. Therefore,

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Quantitative Aptitude Simplified for CAT

(2x – 3)2 < (x + 5)2 4x2 – 12x + 9 < x2 + 10x + 25  3x2 – 22x – 16 < 0  x lies between the roots of 3x2 – 22x – 16 = 0, which are x=

22  222  4(3)(16) 22  26 2   8,  . 6 6 3

Therefore, solution of the given inequality is 

2 < x < 8. 3

Method 2: We know that |2x – 3| = 2x – 3 if x >

3 3 and |2x – 3| = 3 – 2x if x < . 2 2

Similarly, |x + 5| = x + 5 if x > –5 and |x + 5| = –(x + 5) if x < –5. So, we get 3 ranges for x: x > When x >

3 3 , –5 < x < and x < –5. 2 2

3 , the inequality becomes 2

2x – 3 < x + 5  x < 8  solution set is: When x =

0
|x – 4| + 3, we will take the appropriate ranges for x. When x > 4, then the given inequality becomes x – 1 + x – 2 > x – 4 + 3 or x > 2. This implies that x > 4. When 2 < x < 4, the given inequality becomes x – 1 + x – 2 > 4 – x + 3 or x >

10 10 . This implies that < x < 4. 3 3

When 1 < x < 2, the given inequality becomes x – 1 + 2 – x > 4 – x + 3 or x > 6. Therefore, no solution exists in this range. When x < 1, the given inequality becomes 1 – x + 2 – x > 4 – x + 3 or x < –4. This implies that x < –4. At x = 4, |4 – 1| + |4 – 2| > |4 – 4| + 3 or 5 > 3, which is true. This implies that x = 4 is also part of the solution. At x = 2, |2 – 1| + |2 – 2| > |2 – 4| + 3 or 1 > 5, which is false. Therefore, x = 2 is not a part of the solution. At x = 1, |1 – 1| + |1 – 2| > |1 – 4| + 3 or 1 > 6, which is also false. Therefore, x = 1 is also not a part of the solution. Therefore, our final solution to the given inequality is x >

 10  In other words, x   , 4    , .  3  Example 8 Solve the inequality: (i)

|x2 – 7x + 4| < 6

(ii)

|x2 – 7x + 12|  x2 – 7x + 12

(iii) |x2 – 7x + 12|  x2 – 3x + 2 Solution (i)

|x2 – 7x + 4| < 6  –6 < x2 – 7x + 4 < 6.

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10 or x < –4. 3

Quantitative Aptitude Simplified for CAT

This involves two inequalities, x2 – 7x + 4 < 6 and x2 – 7x + 4 > –6. x2 – 7x + 4 < 6  x2 – 7x – 2 < 0 The roots of x2 – 7x – 2 = 0 are x =

7  72  4(1)(2) 7  57  . 2 2

 7  57  Therefore, x2 – 7x – 2 < 0   x   2  

 7  57  7  57 7  57  x .  x    0  2 2 2  

As regards the other inequality, we have x2 – 7x + 4 > –6  x2 – 7x + 10 > 0. But the roots of x2 – 7x + 10 = 0 are imaginary and hence the expression x2 – 7x + 10 is always positive. Therefore, the solution of |x2 – 7x + 4| < 6 is 7  57 7  57  x 2 2 (ii)

x2 – 7x + 12 is more than 0 when x > 4 or < 3; whereas x2 – 7x + 12 is less than 0 when 3 < x < 4. Therefore, |x2 – 7x + 12| = x2 – 7x + 12

when x > 4 or < 3.

|x2 – 7x + 12| = –(x2 – 7x + 12)

when 3 < x < 4.

When x > 4, the given inequality becomes x2 – 7x + 12 x2 – 7x + 12, which is true. When 3 < x < 4, the given inequality becomes –(x2 – 7x + 12)  x2 – 7x + 12  2(x2 – 7x + 12)  0  (x – 3)(x – 4)  0  3  x  4  solution is 3 < x < 4 (Common region between 3 < x < 4 and 3  x  4) When x < 3, the given inequality becomes x2 – 7x + 12  x2 – 7x + 12, which is true. At x = 3 or 4, the given inequality is true. Therefore, the given inequality is always true and hence the solution is: x  R. Such questions are best done using graphs. If we imagine their graphs, we would realize that for any value of x, the expression |x2 – 7x + 12| is always either more than x2 – 7x + 12 or equal to x2 – 7x + 12. If we replace x2 – 7x + 12 by y, then the inequality becomes |y|  y, which is always true irrespective of whether y is positive or negative. (iii) |x2 – 7x + 12|  x2 – 3x + 2  |(x – 3)(x – 4)|  (x – 1)(x – 2) We observe that whenever (x – 1) (x – 2) is negative, the given inequality always holds true because the left hand side is in modulus. This is the case when 1 < x < 2. Therefore, this is always a part of the solution. For x > 4, the inequality becomes x2 – 7x + 12  x2 – 3x + 2 or x 

5 , which is incompatible with x > 4. 2

For 3 < x < 4, the inequality becomes –(x2 – 7x + 12)  x2 – 3x + 2 or x2 – 5x + 7  0 or (x – 2.5)2 + 0.75  0, which is false. Therefore, no solution exists in the range 3 < x < 4.

360

Inequalities, Maxima and Minima For x < 3, the inequality becomes x2 – 7x + 12  x2 – 3x + 2 or x 

5 5 , which is possible. The solution set is x  . 2 2

Therefore, the complete solution for the given inequality is x  (–, 2.5)  (1, 2)  x 

5 . 2

Example 9 Which of the following is the solution for x satisfying |x + 5|  4; 2x + y = 1, where y  2? A.

–9  x  –1

B.



C.

–9  x  

D.



9  x –1 2 1 2

9 1  x   2 2

Solution |x + 5|  4 –4  x + 5  4 –9  x  –1. But, 2x + y = 1 or y = 1 – 2x. Also, y  2  1 – 2x 2  x  

1 . 2

Combining the two results, we get –9  x –1. Example 10 For which of the following equations, –3  x  –2 is the solution set? A.

|x + 4|  1, |x|  3

B.

|x + 4|  2, |x|2

C.

|x + 4|  1, |x|  2

D.

|x + 4|  2, |x|  3

Solution If the solution set is –3  x  –2, we need not check for [B] and [C] options, because –3 doesn’t satisfy |x|  2. Option A: |x + 4|  1  –1  x + 4  1  –5  x  –3. |x| 3  –3  x 3. Combining these, we get x = –3. Therefore, A is not correct option. Hence correct option is D. Alternatively, While checking option A, x = –2 is part of the given solution set but it doesn’t satisfy [A].

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Quantitative Aptitude Simplified for CAT

EXPERT SPEAK Scan this QR Code to watch a video that explains the basic principles of Higher Degree Polynomial Inequalities, Modulus and Logarithm Based Inequalities.

Logarithmic Inequality The inequalities where logarithms are involved are logarithmic inequalities. While solving such inequalities, one must remember the properties related to logarithms. For example, in log x, x must always be positive. Recall the standard result related to logarithmic inequalities mentioned at the beginning of this chapter and is rewritten below: If a > b, then (i)

logca > logcb, if c > 1

(ii)

logca < logcb, if 0 < c < 1

One must remember that logab = c  b = ac. Similarly, logab > c  b > ac (if a > 1), and logab > c  b < ac (if 0 < a < 1) Observe the following examples: Example 11 Solve for x: (i)

log2 (x – 2) > 5

(ii)

log0.2 (x – 2) > 5

(iii) log0.3(x – 1) > log0.09(x – 1) Solution (i)

log2 (x – 2) > 5. First of all, for log (x – 2) to be defined, x must be more than 2. Also, since base is 2 which is more than 1, then shifting of base to the right side does not lead to change in direction of inequality. So, log2 (x – 2) > 5  (x – 2) > 25 or x > 34.

(ii)

log0.2 (x – 2) > 5. Here, the base is 0.2 which lies between 0 and 1. So, shifting to the right side leads to change in direction of inequality. Hence, log0.2 (x – 2) > 5  (x – 2) < (0.2)5  x < 2 + (0.2)5, that is, x < 2.00032. But log (x – 2) is defined only for x > 2. Therefore, the solution is 2 < x < 2.00032.

(iii) log0.3(x – 1) > log0.09(x – 1) 

log(x  1) log(x  1) log(x  1) log(x  1)    . log0.3 log0.09 log0.3 2log0.3

362

Inequalities, Maxima and Minima Since log 0.3 is negative, cancelling it from both sides means we are cancelling a negative term and hence the inequality will change direction. Therefore, log (x – 1)
–1  x < (0.5)–1 or x < 2. Similarly, log0.5x < 1  x > (0.5)1 or x >

1 . 2

1  Therefore, the solution set of the given inequality is x   , 2  . 2  Example 13 Solve the inequality: log(x – 1)(x2 – 4x + 5) < 1 Solution Two cases arise: 0 < (x – 1) < 1 or (x – 1) > 1. Case 1: If 0 < (x – 1) < 1, that is 1 < x < 2, then shifting of base will lead to change in the direction of inequality. So, log(x – 1)(x2 – 4x + 5) < 1  (x2 – 4x + 5) > (x – 1)1 = x – 1  x2 – 5x + 6 > 0  (x – 2)(x – 3) > 0  x < 2 or x > 3. Compatible solution set is: 1 < x < 2. Case 2: If (x – 1) > 1, that is x > 2, then shifting of base will not lead to any change in the direction of inequality. So,

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Quantitative Aptitude Simplified for CAT

log(x – 1)(x2 – 4x + 5) < 1  (x2 – 4x + 5) < (x – 1)1 = x – 1  x2 – 5x + 6 < 0  (x – 2)(x – 3) < 0  2 < x < 3. Compatible solution set is: 2 < x < 3. Moreover, x cannot be equal to 2 as base would become 1, which is not possible. Hence, complete solution set is: x  (1, 2)  (2, 3).

MAXIMA AND MINIMA Introduction When we deal with concepts of maxima and minima, it is commonly believed that the knowledge of calculus (that is, differentiation) is a must to be able to find maximum or minimum value of any given function. As far as CAT and other MBA entrance exams are concerned, we do not see the use of differentiation for finding minima or maxima anywhere. In fact, some of the questions are such that we cannot use differentiation to solve them and hence other techniques are found useful. In the following pages, we will discuss some of these techniques.

Use of AM, GM and HM Let us recall an important and standard inequality (discussed in the chapter on “averages”), that is, AM  GM  HM or a1  a2  a3  ...  an n n  a1 a2 a3 ...an  1 1 1 1 n    ...  a1 a2 a3 an

where AM = arithmetic mean, GM = geometric mean and HM = harmonic mean. We learnt in the chapter on “averages” that the above inequality is true only when all the terms a1, a2, a3, ..., an are positive real numbers. We also learnt that the “equality” holds true when all the terms are equal, that is, when a1 = a2 = a3 = .... = an, then AM = GM = HM =

a  a  a  ...  a na  a . n n n ( a)( a)( a)....( a)

n

 an  a .

n n   a. . 1 1 1 n    ....n times   a a a  a

The fact that “the equality holds true when the terms are equal” gives us a clue to the concept of maxima/minima. Let us take an example to understand this. Example 14 If sum of two positive real numbers, a and b is 10, then find the maximum value of the product ‘ab’. Solution We know that a + b = 10. If a = 1, b = 9, then ab = 1  9 = 9.

364

Inequalities, Maxima and Minima If a = 2, b = 8, then ab = 2  8 = 16. If a = 3, b = 7, then ab = 3  7 = 21. If a = 4, b = 6, then ab = 4  6 = 24. If a = 5, b = 5, then ab = 5  5 = 25. If a = 6, b = 4, then ab = 6  4 = 24. If a = 7, b = 3, then ab = 7  3 = 21; and so on. We observe that as the terms get closer to each other, the value of the product increases till the time when the terms become exactly equal to each other, in which case the product is maximum, that is 25. We can generalize that “if the sum of two positive real numbers is constant, then the product of the terms is maximized when the terms are equal”. Let a = 5 – d and b = 5 + d, where d is some positive real number. Then, a + b = 10 and ab = (5 – d)(5 + d) = 25 – d2, whose maximum value occurs when d2 = 0. This implies that the product is maximum when the terms are equal. We can use the inequality AM  GM to arrive at the above result as well. AM =

ab and GM = 2

ab . Therefore,

ab 10  ab   ab  ab  52 = 25. 2 2 Therefore, the maximum value of the product ‘ab’ is 25, and we can easily verify that this happens when a = b. Through the above example, we could show that if the sum of two terms is constant, the product is maximized when the terms are equal. In fact, this can be extended to more than 2 terms. So, if the sum of n positive real numbers is constant, then the product of these n terms is maximized when the terms are equal. Arithmetic mean of n terms involves summation of n terms and geometric mean of the same n terms involves product of those terms. Since GM  AM, the maximum value of geometric mean is the arithmetic mean of those terms. We can hence conclude that “if the sum (that is AM) is constant, then the product (that is GM) is maximized when the terms are equal”. The rigorous proof of this result is beyond the scope of the present work. The converse of the above is also true, that is if the product of n positive real numbers is constant, then the sum of these n terms is minimized when the terms are equal. The above result and its converse can be applied to geometry in the following manner: Of all rectangles having same perimeter (sum), square has the largest area (product). Of all rectangles having same area (product), square has the least perimeter (sum). Example 15 For n positive terms, if AM  GM, prove that GM  HM. Solution Let a1, a2, a3, .... an, be n positive terms. Then, their reciprocal are also positive terms. Applying the result AM  GM for the reciprocal of these terms, we get

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Quantitative Aptitude Simplified for CAT

1 1 1 1    ....  a a2 a3 an AM = 1 ; GM = n

n

 1 1 1   1       ...   .  a1   a2   a3   an 

1 1 1 1    ....   1 1 1   1 a1 a2 a3 an Now, AM  GM   n       ...   n  a1   a2   a3   an 



n a a a .....a 1 2 3 n



n . 1 1 1 1    ....  a1 a2 a3 an

Hence proved. Example 16 Prove that (i)

a2 + b2 + c2 ab + bc + ca

(ii)

(p + q)(q + r)(r + p)  8pqr (p, q, r > 0)

(iii) a4 + b4 + c4 + d4  4abcd (iv)

x

1 2 x

Solution (i)

We have already seen earlier, that a2 + b2 + c2  ab + bc + ca. We will now use the inequality AM  GM to prove this. If a, b and c are positive real numbers, then AM of a2 and b2 =

a2  b 2 and GM = 2

Therefore, AM  GM 

a2  b2 = ab.

a2  b2  ab  a2 + b2  2ab. 2

Similarly, b2 + c2  2bc and c2 + a2  2ca. Adding all these inequalities, we get a2 + b2 + b2 + c2 + c2 + a2  2ab + 2bc + 2ca or a2 + b2 + c2  ab + bc + ca. (ii)

Since p, q, r > 0, AM  GM 

pq qr rp  pq;  qr and  rp . 2 2 2

 pq  qr   r  p Multiplying all these, we obtain:        ( pq)(qr )(rp)  2   2   2  or (p + q)(q + r)(r + p)  8pqr. (iii) AM  GM 

a4  b 4  c 4  d 4 4 4 4 4 4  a b c d a4 + b4 + c4 + d44abcd. 4

(iv) Let x be a positive real number. Then, for x and

366

1 , x

Inequalities, Maxima and Minima

x

AM =

2

1 x and GM =

1 ( x )    1 . Therefore,  x

1 x 1  x1  2. AM  GM  2 x x

Since x is positive, LHS is positive. To ensure it is always positive, we put modulus. Therefore, x 

1  2. . x

This is an important inequality and can be stated as: “A real number and its reciprocal is either ‘more than or equal to 2’ or ‘less than or equal to –2’”. Example 17 Find the minimum value of 4 tan2  + 9 cot2 . Solution AM  GM 

4 tan2   9 cot 2   2

 4 tan   9 cot    2

2

36  6

 4 tan2  + 9 cot2 12 Therefore, minimum value of 4 tan2+ 9 cot2  is 12. Alternatively, 4 tan2  + 9 cot2  = (2 tan  – 3 cot )2 + 12 which is minimized when 2 tan  = 3 cot . The minimum value is 12. Example 18 Let f(x) = 3sin2 x + 9cosec2 x + 27tan2 x + 81 cot2 x. Then, minimum value of f(x) is A.

4(3)2.5

B.

(3)2.5

C.

4(3)5

D.

120

Solution Using the result: AM  GM, 3sin2 x  9cosec2 x  27tan2 x  81cot 2 x 4  (3sin2 x)(9cosec2 x)(27tan2 x)(81cot 2 x)  4



3sin2 x  9cosec 2 x  27tan2 x  81cot 2 x 4 10  3  32.5 4

3sin2 x + 9cosec2 x + 27tan2 x + 81 cot2 x 4(3)2.5. Hence, option [A] is correct option. Example 19 Find the minimum value of the expression

(p2  p  1)(q2  q  1)(r 2  r  1) , where p, q, r > 0. pqr

367

Quantitative Aptitude Simplified for CAT

Solution

1  1  1  The expression =  p  1    q  1    r  1   . p  q  r  1 1   The minimum value of  p   is 2 (if p is positive). Therefore, minimum value of  p  1   is 3. Similarly, p p   for the other terms. Therefore, the minimum value of the given expression is 3  3  3 = 27. Alternatively, Apply AM ≥ GM. So,

p2  p  1 3 2 q2  q  1 r2  r 1  p  p  1  p . Similarly,  q and r 3 3 3  p2  p  1  Multiply all the inequalities to obtain    3   

 q2  q  1    3  

 r2  r  1     pqr 3  

(p2  p  1)(q2  q  1)(r 2  r  1)  27 . Hence, the minimum value of the given expression is 27. pqr

EXPERT SPEAK Scan this QR Code to watch a video that explains how to use inequality involving AM, GM and HM in solving problems based on Maxima.

Example 20 2

Solve for x: e  sin

x

1

 e

 sin2 x

 cos( e x ) .

Solution The minimum value of LHS is 2 (as it is an expression of the form “a number and its reciprocal”), whereas the maximum value of RHS is 1. Therefore, the equation is inconsistent. Hence, no solution is possible. Example 21 If x, y, z are distinct positive real numbers, then A.

>4

B.

≥6

C.

>6

D.

None of the above

x2 ( y  z)  y 2 ( x  z)  z 2 ( x  y ) would be xyz

CAT 2003 (C)

368

Inequalities, Maxima and Minima Solution The expression =

x2 y  x2 z  y 2  y 2 z  z2 x  z 2 y x x y y z z       xyz z y z x y x

x z  x y  y z =             2  2  2 . z x    y x  z y

Therefore, the expression is more than 6. Hence correct option is [C]. Note that the terms are “distinct” and hence AM will not be equal to GM and so [B] is not the correct option. Example 22 Find the maximum value of

x 2

.

x 1 A.

1 2

B.

3 2

C.

4 3

D.

6 7

Solution For the expression to be maximized, x must be positive. The expression can be rewritten as

1 x

1 x

.

The expression is maximized when denominator is minimized (as the denominator is positive). The minimum 1 value of the denominator is 2 and hence the maximum value of the expression is . 2 Hence correct option is [A]. Example 23 If x + y + z = 13, find the maximum value of (x + 1)(y – 2)(z + 3). Solution x + y + z = 13  (x + 1) + (y – 2) + (z + 3) = 13 + 1 – 2 + 3 = 15. Let X = x + 1, Y = y – 2 and Z = z + 3. Then, we have to find the maximum value of XYZ where X + Y + Z = 15. If sum is constant, the product is maximized when the terms are equal. Therefore, X = Y = Z = 5 and hence XYZ = 53 = 125. Example 24 If 2x + 3y = 7, find the maximum value of x3y4.

369

Quantitative Aptitude Simplified for CAT

Solution 2x + 3y = 7 can be rewritten as

2x 2x 2x 3y 3y 3y 3y       7. 3 3 3 4 4 4 4

 2x 2x 2x 3y 3y 3y 3y  Taking AM and GM of the seven terms:  , , , , , ,  , we get  3 3 3 4 4 4 4 2x 2x 2x 3y 3y 3y 3y 3 4       3 3 3 4 4 4 4  7  2x   3 y      7  3   4   7     7

7

 2x     3 

3

 3y     4 

4

 x3 y 4 

The maximum value of x3y4 =

32 . 3

32 . 3

Alternatively, if sum is constant, then product is maximized when terms are equal. Since 3

2x 2x 2x 3y 3y 3y 3y  2x   3y         7 , that is, sum is constant, the product     3 3 3 4 4 4 4  3   4  maximized when terms are equal. In other words, x3y4 is maximized when terms are equal.

Therefore, for x3y4 to be maximized,

2x 3 y  . 3 4

Solving the system of equations: 2x + 3y = 7 and 3

3 2x 3 y 4  , we get x = , y = . 3 4 2 3

 3  4 Therefore, the maximum value of x3y4 =      2  3

4



32 3

The above example leads to an important result:

x y z   . m n p

If x + y + z = k, then xmynzp is maximized when

This can be extended to any number of variables. More generally, If ax + by + cz = k, then xmynzp is maximized when

ax by cz   , and so on. m n p

Example 25 If 3x + 5y + 7z = 12, then find the maximum value of x3y7z12. Solution For maxima,

3x 5y 7z   . Therefore, 3 7 12

3x + 5y + 7x = 12 becomes 3x + 7x + 12x = 12 or x = 3

6 72 42 ,y= and z = . 11 55 77 7

 6   42   72  Therefore, maximum value of x3y7z12 =        11   55   77 

370

12



246  3 34 57  75  11 22

.

4

would be

Inequalities, Maxima and Minima Example 26 Prove that a4 + b4 + c4  abc(a + b + c). Solution a4 + b4 + c4 = (a2)2 + (b2)2 + (c2)2  (a2)(b2) + (b2)(c2) + (c2)(a2) = (ab)2 + (bc)2 + (ca)2  (ab)(bc) + (bc)(ca) + (ca)(ab) = (abc)(a + b + c). Hence proved. It is to be noted that we have applied the result a2 + b2 + c2  ab + bc + ca, two times to prove the result. Example 27 If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a) (1 + b) (1 + c) (1 + d)? A.

4

B.

1

C.

16

D.

18 CAT 2002

Solution

1a  (1)(a)  a . Similarly, 2 1a 1c 1d  b;  c and  d . 2 2 2 Multiplying all these inequalities, we get

 1  a  1 b  1 c   1 d          abcd = 1  2   2   2  2   (1 + a) (1 + b) (1 + c) (1 + d)  16. Therefore, the minimum value of the product is 16. Hence correct option is [C] Example 28 If the product of n positive real numbers is unity, then their sum is necessarily A.

a multiple of n

B.

equal to n +

C.

never less than n

D.

a positive integer

1 n

CAT 2003 (C) Solution If the product is 1, then GM = 1. Therefore,

371

Quantitative Aptitude Simplified for CAT

AM =

a1  a2  a3  ....  an  1  a1 + a2 + a3 + .... + an  n. n

Therefore, the sum is never less than n. Hence correct option is [C]. Example 29 Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true? A.

The minimum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m + 1

B.

The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1

C.

The maximum possible value of a2 + b2 + c2 + d2 is 4m2 – 2m + 1

D.

The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1 CAT 2003 (C)

Solution Since m is a positive integer, m can take any large value, thereby increasing the value of a + b + c + d, and hence the value of a2 + b2 + c2 + d2. So, maximum value of a2 + b2 + c2 + d2 has no limit. For minimum value, m should be minimum, which is 1. In that case, a + b + c + d = 5. To minimize a2 + b2 + c2 + d2, each of a, b, c and d should have numerically smallest value. So, one of the variables’ values is 2 and the rest have value of 1 each. So, a2 + b2 + c2 + d2 = 12 + 12 + 12 + 22 = 7. Now, 4m2 + 2m + 1 = 7, when m = 1. Hence correct option is [B]. Example 30 If x + y + z = 5; xz + xy + zy = 3, what could be the maximum value of x? A. B.

5 3

19

C.

13 3

D.

None of these CAT 2002

Solution In this question, students generally tend to do the following: x + y + z = 5  (x + y + z)2 = 52  x2 + y2 + z2 + 2xy + 2yz + 2zx = 25 x2 + y2 + z2 = 25 – 2(3) = 19. Now, when x is maximized, y and z should be zero. In such a case, x2 = 19  x = But this is not true because when x =

19 .

19 , y = 0 and z = 0, the given equation x + y + z = 5 is not satisfied.

Even if we realize that x should be slightly less than 19 , the confusion between [C] and [D] option is not yet over. So, this is not the right approach. The proper approach is as explained below: To maximize x, we need to minimize y + z. Now, applying AM ≥ GM on y + z, we get:

372

Inequalities, Maxima and Minima

yz 5 x  yz   3  xy  xz 2 2 

5 x  3  x( y  z)  3  x(5  x) 2

Squaring both sides, we get: (5 – x)2 ≥ 4(3 – 5x + x2)  25 + x2 – 10x  12 – 20x + 4x2  –3x2 + 10x + 13  0  3x2 – 10x – 13  0. Now, roots of 3x2 – 10x – 13 = 0 are x =

10  102  4(3)(13) 10  16 13 .   1, 6 6 3

13  13  Therefore, 3x2 – 10x – 13  0  ( x  1)  x  .   0  1  x  3 3   Therefore, maximum value of x is

13 . Hence correct option is [C]. 3

Example 31 Find the maxima and minima of

2x2  5x  7 x2  6x  2

, for real x.

Solution Let

2x2  5x  7 x2  6x  2

= m. Then, 2x2 + 5x – 7 = mx2 – 6mx + 2m

or (m – 2)x2 – (6m + 5)x + 2m + 7 = 0. For x to be real, discriminant ≥ 0. Therefore, Discriminant = (6m + 5)2 – 4(m – 2)(2m + 7)  0  36m2 + 60m + 25 – 8m2 – 12m + 56  0 28m2 + 48m + 81  0 The roots of 28m2 + 48m + 81 = 0 are imaginary (as discriminant is negative) and hence the inequality 28m2 + 48m + 81  0 is true for all values of m. Therefore, maximum value of m is  and minimum value is –. But m is nothing but the expression 2x2  5x  7 . Therefore, the maximum and minimum value of the expression is  and – x2  6x  2 Example 32 Find the maximum value of the expression

1 2

.

x  5x  9 Solution The denominator is x2 + 5x + 9, which is always positive as the discriminant of x2 + 5x + 9 = 0 is negative. 1 Therefore, 2 is maximized when x2 + 5x + 9 is minimized. x  5x  9 To find the minimum value of x2 + 5x + 9, we rewrite this as below:

373

Quantitative Aptitude Simplified for CAT

(x + 2.5)2 + (9 – 6.25) = (x + 2.5)2 + 2.75. Therefore, minimum value of the expression is 2.75 when x = –2.5. Hence the maximum value of

1 2

x  5x  9 1 4 is  . 2.75 11

In this example, we do not need to equate the given expression to m and proceed as in the previous example simply because the denominator, here, is always positive. In the previous example, the denominator may or may not be positive depending upon the value of x. Example 33 Let g(x) = max (5 – x, x + 2). The smallest possible value of g(x) is A.

4

B.

4.5

C.

1.5

D.

None of the above CAT 2003 (C)

Solution g(x) is the maximum of the two functions: (5 – x) and (x + 2). The function (5 – x) may be more than or less than (2 + x) depending upon the value of x. 5–x>x+2x
1.5, the line y = x + 2 is above the line y = 5 – x and hence y = x + 2 is part of g(x). For x < 1.5, the line y = 5 – x is above the line y = x + 2 and hence y = 5 – x is part of g(x). In other words, g(x) = x + 2

if x > 1.5

g(x) = 5 – x

if x < 1.5

Therefore, g(x) is minimum when x = At x =

3 . 2

3 3 7 3 7 , g(x) = 5 – x = 5 – = . Alternatively, g(x) = x + 2 = +2= . 2 2 2 2 2

374

Inequalities, Maxima and Minima Therefore, the minimum value of g(x) is 3.5, at x = 1.5. Hence correct option is [D]. Example 34 Find the maximum of f(x) = min[(2 + x), (4 – x2)], in the range [–2, 2]. Solution 2 + x > 4 – x2  x2 + x – 2 > 0  (x – 1)(x + 2) > 0  x > 1 or < –2. Similarly, 2 + x < 4 – x2  –2 < x < 1.

y= 4-x 2

The graph of f(x) (shown in bold in the second figure) is drawn below. 2 x+ y=

The intersection points of the curves 2 + x and 4 – x2 are at x = 1 and x = –2. f(1) = 3, f(–2) = 0 and f(2) = min[(2 + x), (4 – x2)] = min(4, 0) = 0. Therefore, the maximum value of f(x) is 3 at x = 1. Graphically also, we can easily verify this. Example 35 Find the minimum and maximum value of (sin x)(cos x). Solution (sin x)(cos x) =

1 1 [2(sin x)(cos x)] = sin 2x. 2 2

The maximum and minimum value of sin 2x is 1 and –1. Therefore, the maximum and minimum value of 2x is

1 1 and – . 2 2

 1 1 In other words, (sin x)(cos x)    ,  .  2 2 Example 36 Find the maxima and minima of

p p and given that |2p – 1| = 3 and |q + 3| = 5. q q

Solution |2p – 1| = 3  2p – 1 = ±3  p = 2, –1. Similarly, |q + 3| = 5  q = 2, –8. Possible values of

p 2 1 2 1 1 1 1  , , , , or 1,  ,  , . q 2 2 8 8 2 4 8

375

1 sin 2

Quantitative Aptitude Simplified for CAT

The maximum value is 1 and minimum value is  Possible values of

1 . 2

1 1 1 p = 1,  ,  , . q 2 4 8

The maximum value is 1 and minimum value is

1 . 8

Example 37 Min(x, y) = Minimum of (x, y), Max(x, y) = Maximum of (x, y), Mod(x) = |x|, where x = –4 and y = –7. Find the value of Max{(x – 1) + Mod(Min(x, y)), Mod[x + Max(Mod(x), Mod(y))]}. A.

–3

B.

–2

C.

3

D.

2

Solution Max{(x – 1) + Mod(Min(x, y)), Mod[x + Max(Mod(x), Mod(y))]} = Max{(–4 – 1) + Mod(Min(–4, –7)), Mod[–4 + Max(Mod(–4), Mod(–7))]} = Max{–5 + Mod(–7), Mod[–4 + Max(4, 7)]} = Max{–5 + 7, Mod[–4 + 7]} = Max{2, 3} = 3. Hence correct option is [C]. Example 38 Function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at x = A.

2.3

B.

2.5

C.

2.7

D.

None of the above CAT 2003 (C)

Solution We have seen in the chapter on “Functions” that if there are three terms in modulus, the graph is minimum at one point and the point is represented by that value of x for which the middle term in modulus is zero. The middle term here is |x – 2.5|, which is zero when x = 2.5. Therefore, the function is minimized at x = 2.5. Hence correct option is [B]. Example 39 Find the minimum value of the function, f(x) = |x – 1| + |x – 2| + |x – 3| + .....+ |x – 20|.

376

Inequalities, Maxima and Minima Solution On lines of the previous question and what we have learnt in the chapter on “Functions”, we can say that when there are even number of terms, the graph is flat at the base. Therefore, in the current example, the function is minimized for 10  x  11. f(10) = 9 + 8 + 7 + ... + 2 + 1 + 0 + 1 + 2 + 3 + 4 + .... + 10 = 100. Example 40 Let um + vm = wm, where u, v, m, w belong to set of integers. A.

m min of u, v, w

B.

m  max of u, v, w

C.

m is always less than min of u, v, w

D.

None of these CAT 2002

Solution Let u = 3, v = 4, w = 5 and m = 2. Then, um + vm = wm  32 + 42 = 52. This is also true when u = –3, v = –4, w = –5 and m = 2, that is, (–3)2 + (–4)2 = (–5)2. When u = 3, v = 4, w = 5 and m = 2, m  min(u, v, w). When u = –3, v = –4, w = –5 and m = 2, m  min(u, v, w). Therefore, nothing definite can be said. Hence correct option is [D].

377

Quantitative Aptitude Simplified for CAT

PRATICE EXERCISE 2

1.

1

4.

What values of x satisfy x 3  x 3  2 ≤ 0? A.

–8 ≤ x ≤ 1

B.

–1 ≤ x ≤ 8

C.

1 0

B.

maximized whenever a > 0, b < 0

C.

minimized whenever a > 0, b > 0

D.

minimized whenever a > 0, b < 0 CAT 2004

CAT 2006 2.

5.

Let f(x) = max(2x + 1, 3 – 4x), where x is any real number. Then the minimum possible value of f(x) is:

If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y? 1

A. B. C.

1 3

A.

23

1 2

B.

23

2 3

C.

24

D.

24

D.

4 3

E.

5 3

2

1

3

CAT 2003 (R) 6.

1 0. If a < 0, then a + |a| = 0. If a = 0, then also a + |a| = 0. Therefore, a + |a|  0 When b > 1, then |a|b2 > |a|  a + |a|b2 > a + |a|  0. Therefore, for all values of a, a – xb  0.

(2) (0.5)

= 4. Therefore, maximum value of w is 4. z When v = –1, w = – . Taking the same u (2) values for z and u, we get w = – = –4. (0.5)

14. Answer: D Explanation: x 7

( x  7)  0 , whose ( x  9)( x  4) x  5 x  36 solution set is x > 7 or 9 < x < 4. So, the smallest integer value of x is 8. Note that x cannot be 9 as the denominator becomes 0 at x = 9. 2

Therefore, –4  w  4. 11. Answer: B Explanation: Dividing the numerator and denominator by cos2, we get: 1 . 2 2 sec   cos ec   tan2   cot 2  To maximize this expression, we need to minimize the denominator. For minimum value of denominator, use the inequality AM ≥ GM. So, 2

2

Explanation: AM ≥ GM  a1  a2  a3  ...  an n  a1 a2 a3 a4 ...an  n nn  n n Therefore, a1 + a2 + a3 + … an ≥ n2. 16. Answer: B

2

Explanation: 4(y – 3)2 + 4(x – 2)2 – 2(x – 3)2 = 4(y – 3)2 + 4(x2 + 4 – 4x) – 2(x2 + 9 – 6x) = 4(y – 3)2 + 2x2 – 2 – 4x = 4(y – 3)2 + 2(x2 – 2x – 1) = 4(y – 3)2 + 2(x – 1)2 – 4 Therefore, the minimum value of the given expression will be –4.

sec2   cos ec2  tan2   cot 2 

sec2   cos ec2  tan2   cot 2  1 4  The minimum value of sec2 + cosec2 + tan2 + cot2 is 4. So, maximum value of the 

given expression is

1 . 4

17. Answer: C Explanation: abc – (a + b + c) = a(bc – 1) – b – c. For maximum value of the expression, b and c both have to be negative with maximum numerical value. So, b = –10 and c = –9. Also, a has to be positive and maximum possible, but a ≠ 9 or 10, because its numerical value cannot be equal to that of b or c. So, a = 8.

12. Answer: B Explanation:

x2  4 x  4

0

15. Answer: D

sec   cos ec   tan   cot   4 4

0

Explanation: The given equation becomes 5y – 3x = 13, whose solution sets are: (1, 2), (4, 5), (9, 8), and so on. We observe that the smallest value of |x – y| occurs when x = 4 and y = 5 and the minimum value is 1.

Explanation: vz z . Let v = 1. Then, w = . w= u u

2

2

13. Answer: A

10. Answer: B

When z = –2 and u = –0.5, then w =

x 2  2x  24

( x  6)( x  4) 0 ( x  6)( x  1) x  7x  6 The solution set of the inequality is: (6, 1)  (4, 6). 

2 x2  4 x  4 2   0 x2  7 x  6 3 x 2  7x  6 3 3( x2  4 x  4)  2( x 2  7x  6)  0 3( x 2  7x  6) 

382

Inequalities, Maxima and Minima

 a2  a  1  b2  b  1  c2  c  1   d2  d  1  e2  e  1 

The value of the expression becomes = 720 – (8 – 9 – 10) = 731. 18. Answer: C Explanation: a  b  c  d a  b  c  d  2d 2d  1 ab c d abcd ab cd .

The given expression is, therefore, maximum when d is maximum and a + b + c is minimum but more than d so that denominator is positive. So, d = 25 and a + b + c = 26. Therefore, value of the given expression 2(25) = 1  51 . 26  25

abcde 1  1    a 1    b 1   a  b  1  1  1   c 1    d 1    e 1   c d e     The minimum value of each term in the bracket is 3. Hence minimum value of the expression = 35 = 243. 20. Answer: C Explanation: As p, q, r are non-negative integers, the given expression will be maximum when the value of the variables is as close to each other as possible. So, p, q, r = 4, 3, 3 (not necessarily in the same order). Therefore, pq + qr + pr + pqr = 33 + 36 = 69.

19. Answer: E Explanation:

383

Quantitative Aptitude Simplified for CAT

Chapt er 11

Logarithms BASIC CONCEPTS Definition If am = b, then m is called logarithm of ‘b’ to the base ‘a’. In other words, m = logab For example, 23 = 8  3 = log28; 54 = 625  4 = log5625.

Rules of Logarithms There are five basic rules of logarithms which will help to solve all the questions based on them. They are (i)

(ii)

log (a b  c  .....) = log a + log b + log c + ....... that is, logarithm of the product of numbers = sum of the logarithms of each number. a = log a – log b log b that is, logarithm of the ratio of numbers = difference of the logarithm of the numerator and denominator.

(iii) log am = m log a log c a (iv) logba = , where c is any positive real number. This is known as base change formula. log c b

 

Note that log bn am  (v)

logam mloga m  log a  m      log b a . Here we have used rule (iii) and (iv). nlogb n  log b  n logbn

aloga N  N. More generally, x log z y  y log z x .

Proof Let aloga N  X. Taking logarithm on both sides (and using rule (iii)), we get (logaN)  (log a) = log x

384

Logarithms Using rule (iv), we get

log N  (log a) = log x log a log N = log x  x = N. Hence the result. Let’s prove the more general rule: x log z y  y logz x . Proof Let k = x log z y . Taking logarithm on both sides, we get

 log y   log x   log x   log y     log y log z x  = log y log z x . log k = log x log z y  log z y log x  =  log z log z    









 k = y log z x . So, x log z y  y logz x . Besides the above rules, also note that log 1 = 0; log 10 = 1; log 100 = 2. logaa = 1 Since log 1 = 0, log1a does not exist, as log1a =

log a , in which case the denominator would be zero. Therefore, log 1

in logarithms, base can never be 1. Example 1 If log 15 = 1.176, find the value of log (0.0015). Solution log (0.0015) = log (15 10–4) = log 15 + log 10–4 = 1.176 + (–4) = –2.824. Example 2 Find the value of (a)

log216

(b)

log27243

(c)

5log5 27

(d)

5log25 27

(e)

25log125 9

(f)

log 1212 log 3 144  log 1728

Solution

log 16 log 2 4 log 2 4 4 log 2     4. log 2 log 2 log 2 log 2

(a)

log 216 =

(b)

log 27243 =

(c)

Using rule (v), we can directly say that 5log5 27  27.

log 243 log 3 5 log 24 5 log 3 5     . log 27 log 2 log 3 3 3log 3 3

385

Quantitative Aptitude Simplified for CAT

(d)

5

log 25 27

log 27 log 25

5

5

log 27 log 52

log 9

 log 125

25 log125 9  5 2

(e)



5

log 27 2 log 5

 log 9 2  3 5  log 5

  

1

 52

log 2 27

 log 9  2   5  3 log 5 



1 /2

 5 log 5 27 2

 53

log 5 9



27 .

 5 log 5 9

2 /3

 3 81

log 1212 12log 12 12log 12 72    . 2 /3 3 /2 3 2 3 13 log 12  log 12   log 144  log 1728    log 12  3 2

(f)

Example 3 Solve the following equations: (a)

log2(x – 3) = 3

(b)

log3(x2 – 4x + 6) = 1

(c)

log2log3(5x – 6) = 1

Solution (a)

log2(x – 3) = 3  (x – 3) = 23 = 8  x = 11.

(b)

log3(x2 – 4x + 6) = 1  (x2 – 4x + 6) = 31  x2 – 4x + 3 = 0  x = 3, 1.

(c)

log2log3(5x – 6) = 1  log3(5x – 6) = 21  (5x – 6) = 32 = 9  x = 3.

Example 4 Arrange in ascending order: (a)

8log4 9 , 27log9 4

(b)

7log343 3 , 11log14641 4 , 5 log5 3125

1

Solution

8

(a)

log4 9

 log 9  3   ,  2  log 4 

log9 4

27

,

 log 4  3   3  log 9 

log9 4

Therefore, 27

7log 343 3 

(b)

3

 22 

log4 9

8

log 3 3 7 log 7



log2 9

3 log3 4 32

1 4

5

5

log 3125 5



 3log3 4

 27

3/2

8

33.

11 log 4641  11 log 11  11 4 1 log 5 3125

3/2

.

1 log 7 3 73

log 4 4

 2log2 9

log 5 5 log 5 5

log11 4 1

 55

44 .

log 55



5

5

So, we have to compare 3 3 , 4 4 , 5 5 . Comparing these surds has been discussed (Chapter 1: Number System). Taking cue from there, we see that the ascending order is 5 5  4 4  3 3 . Example 5 1 1 1 =?   1  log c ab 1  log a bc 1  log b ca

386

Logarithms Solution 1 1 1    1  log c ab 1  log a bc 1  log b ca

1 1 1    log ab   log bc   log ca  1    1    1    log c log a      log b 

log c log a log b log c log a log b      log c  log ab log a  log bc log b  log ca log abc log abc log abc

=

= log abc  1 . log abc

Example 6 1 1 1   ? 1  log ab c 1  log bc a 1  log ca b

Solution 1 1 1    1  log ab c 1  log bc a 1  log ca b

=

1 1 1    log c   log a   log b  1    1    1     log ab   log bc   log ca 

log ab log bc log ca log a2b 2c 2     2. log ab  log c log bc  log a log ca  log b log abc

Example 7 If xn = log tan n°, then find the value of (i)

x1 + x2 + x3 + .... + x89

(ii)

x1 x2 x3 .... x89

Solution (i)

x1 + x2 + x3 + ..... + x89 = log tan 1° + log tan 2° + log tan 3° + .... + log tan 89°

= log [(tan 1°) (tan 2°) (tan 3°)..... (tan 89°)] Now, (tan 1°) (tan 89°) = (tan 1°) (cot 1°) = 1. Similarly, (tan 2°) (tan 88°) = (tan 2°) (cot 2°) = 1. Likewise, product of each pair of terms from the ends = 1. The given expression becomes log (tan 45°) = log 1 = 0. (ii)

x1 x2 x3 .... x89 = (log tan 1°)(log tan 2°)(log tan 3°)....(log tan 89°).

Now, one of the terms in the product is log (tan 45°) = log (1) = 0. Therefore, the entire expression reduces to zero. Example 8 Express log616 in terms of log1227. Solution Let log1227 = x. Then,

387

Quantitative Aptitude Simplified for CAT

x=

log 27 3log 3  log 12 2log 2  log 3

 2x log 2 + x log 3 = 3 log 3 2xlog2 3x

 log 3 =

Now, log 616 =

=

4log2  log2  log3

4log2 4(3  x)log2 4(3  x)   2xlog2  (3  x)log2  2xlog2 (3  x) log2     3x 

4(3 - log 12 27) . (3  log 12 27)

Example 9 Find the value of log11. Solution It would wrong to think that log11 = 1 (using the result logaa = 1). This is so because we know that base can never be 1. The value of log11 cannot be determined. Example 10

 x The solution set (x, y) for the system of equations log2 xy = 5 and log 1   = 1, is y 2 A.

(–4, –8)

B.

(4, 8)

C.

(8, 4)

D.

Both A and B

Solution: D log2 xy = 5  xy = 25 = 32.

 x x 1 log 1   = 1   . y y 2  2 Multiplying the two equations, (xy) 

x 1 = 32   x2 = 16 x = ±4  y = ±8. y 2

Therefore, the pair of values are (4, 8) and (–4, –8). We can also use options to get the answer. Example 11 1x If f(x) = log   , find f(x) + f(y). 1x

A.

f(x + y)

B.

 1y  f    1  xy 

388

Logarithms

C.

 1  (x  y)f    1  xy 

D.

f (x) 

f(y) 1  xy

Solution: B

 1  x   log  1  y   log  1  x  1  y   log  1  x  y  xy  f(x) + f(y) = log    1  y   1  x 1  y   1  x  y  xy  1x       Dividing the numerator and denominator in the brackets by (1 + xy), we get 1 xy  1  xy log  1 xy  1  xy 

   , which is nothing but   

 xy  f   .  1  xy 

To Find the Number of Digits in a Number We know that log101 = 0 and log1010 = 1 and log10100 = 2 and so on. If 1 < x < 10, then log101 < log10x < log1010, that is, 0 < log10x < 1. In other words, if x is a single digit number, then its logarithm will lie between 0 and 1. Similarly, if 10 < x < 100, then log1010 < log10x < log10100, that is, 1 < log10x < 2. In other words, if x is a two-digit number, then its logarithm will lie between 1 and 2, and so on. So, if log of x lies between 3 and 4, then x is a four-digit number. For example, log 23 1.36; log 3245  3.51. In general, to find the number of digits in some number x, follow the following algorithm: Step 1: Find log10x Step 2: Find the integral part of log10x, that is find the value of [log10x], where [.] is the greatest integer function. Let [log10x] = k. Then, the number of digits in x will be k + 1. Example 12 If log 3 = 0.477, find the number of digits in 320. Solution Taking logarithm of 320, we get: 20 log 3 = 20 0.477 = 9.54. [9.54] = 9. Therefore, the number of digits in 320 is 9 + 1, that is, 10. Example 13 If log 5 = 0.699, then the number of digits in the number 264 will be A. 10 B. 64 C. 23 D. 20

389

Quantitative Aptitude Simplified for CAT

Solution: D We know that log 10 = 1. But 10 = 25. Therefore, log (2  5) = 1  log 2 + log 5 = 1 log 2 = 1 – log 5 = 0.301. Now, log10264 = 64 (log102) = 64  0.301 = 19.264. Therefore, 264 has 20 digits. Example 14 If log 12 = 1.079 and log 27 = 1.431, then find the number of digits in 235  325. Solution log 12 = 1.079  2 log 2 + log 3 = 1.079. But log 27 = 1.431  3 log 3 = 1.431  log 3 = 0.477. Using this in first equation, 2 log 2 + (0.477) = 1.079  log 2 =

1 (1.079 – 0.477) = 0.301. 2

Now, log (235  325) = 35 log 2 + 25 log 3 = 35 (0.301) + 25 (0.477) = 10.535 + 11.925 = 22.46. This means that 235  325 has 23 digits.

390

Logarithms

PRACTICE EXERCISE 1.

5.

If x ≥ y and y >1, then the value of the  x  y expression log x    log y   can never be  x  y

A.

−1

B.

−0.5

C.

0

D.

1

A. B. C.

CAT 2005 2.

D

If logyx = (a  logzy) = (b logxz) = ab, then which of the following pairs of values for (a, b) is not possible? A.

  2, 1    2 

B.

(1, 1)

C.

(0.4, 2.5)

D.

 , 1     

E.

6.

A.

no solution for x

B.

exactly one solution for x

C.

exactly two distinct solutions for x

D.

exactly three distinct solutions for x

C. D.

m2 m3 m4  log 2  log 3  .... n n n n 2

A.

 nn1 log  n1 m

   

 mm  log  n   n 

n /2

B.

C.

 m1  n log  1m n

   

D.

 m n 1 log  n1  n

   

n /2

n /2

CAT 2003 (R) 7.

If log10X – log10 X = 2 logX10, then a possible value of X is given by

B.

What is the sum of ‘n’ terms in the series:

(2, 2)

Let u = (log2x)2 – 6log2x + 12, where x is a real number. Then the equation xu = 256, has

A.

9 N 9 N9 = M 3 M3 = N 3 N9 = M

M9 =

log m + log

CAT 2004 4.

1 log3M + 3log3N = 1 + log0.0085, then 3

CAT 2003 (R)

CAT 2006 3.

If

10 1 100 1 1000

If 64x = 48y = 36z, find A.

1 y

B.

2 y

C.

3 y

D.

1 2y

1 1  x z

None of these

CAT 2015 CAT 2003 (R) 8.

22log2 (log10 x )  log 10 x   log 10 x 

range of x. A.

391

x < 10

2

1. Find the

Quantitative Aptitude Simplified for CAT

B.

x = 10

B.

30

C.

x > 10

C.

100

D.

0 2001. Therefore, m = 6, that is, 6 terms should be taken so that the sum is less than 1000. The sum of 6 terms =

1 6 (3  1) = 364. 2

Example 9 Find the sum to infinity of 1 1 1 (i)    .... 2 6 18 2 6 18 (ii)    .... 7 49 343 Solution

(i)

(ii)

1 1 1 a 3 a = , r = . Therefore, S =  2  . 2 3 1r 1 1 4 3 2 2 3 a 1 7 a = , r = . Therefore, S =   . 7 7 1r 1 3 2 7

400

Progressions Example 10 Solve the following: (i)

2 + 2.2 + 2.42 + 2.662 + ......

(ii)

0.1 + 0.03 + 0.009 + ...... 1 1 1    .... (iii) 27 9 3 Solution

2.2  1.1 , which is more than 1. Hence the sum to infinity is . 2 a 0.1 1 (ii) The common ratio is 0.3. S =   1  r 1  0.3 7 1 /9  3 , which is more than 1. Hence the sum to infinity is . (iii) The common ratio is 1 / 27

(i)

The common ratio is

Example 11 When a ball is dropped from a height of 10 m, it rebounds to a height that is

3 4

th

of the height from which it fell

to the ground previously. Every time it falls to the ground, it rebounds to a height that is

3 4

th

of the height from

which it last fell. Find the total distance travelled by the ball before it comes to rest. Solution Before hitting the ground, the distance travelled by the ball = 10 m.

3  Between first and second hit, the distance travelled = 2    10  . 4   3 2  3 3  Between second and third hit, the distance travelled = 2      10    2      10  = ,and so on.  4  4   4  We see that the series formed is an infinite GP. 2 3 3  3 3 Therefore, total distance travelled = 10 + 20         ....   4  4   4  

 3    = 10 + 20   4  = 70 m. 1 3   4  Example 12 Find the sum of the terms: 7 + 77 + 777 + 7777 + ... n terms. Solution 7 + 77 + 777 + 7777 + ... n terms = 7(1 + 11 + 111 + 1111 + ....) =

7 7 (9 + 99 + 999 + 9999 + .... n terms) = [(10 – 1) + (102 – 1) + (103 – 1) + ...+ (10n – 1)] 9 9

401

Quantitative Aptitude Simplified for CAT

=

 7  10(10n  1) 7 7  n   10n1  9n  10 . [(10 + 102 + 103 + ...+ 10n) – n] =  9  (10  1) 9  81





Example 13 If product of 5 terms which are in GP is 243, find the third term. A.

3

B.

2

C.

9

D.

Cannot be determined

Solution Let the terms are a, ar, ar2, ar3, ar4. Then their product = a5r10 = 243. Third term = ar2 =

5

a 5 r 10  5 243  3 .

Alternatively, Take the terms as a=

5

a a , , a, ar , ar 2 , so that their product = a5 = 243. 2 r r

243  3 .

Example 14 Evaluate: 1

1

1

2  2 3  2 9  2 27 .....

Solution

1

1

1

1 1 1 1   .... 3 9 27

2  2 3  2 9  2 27 .....  2



   1     1 1    3  2

 23 / 2  8

Example 15 Find the product of the terms: 1, 2, 4, 8, 16, ..... up to 100 terms. Solution The 100th term = 299. Therefore, the product of 100 terms = 1  2  22  23  24  ....  299 = 2(0 + 1 + 2 + 3 + ... + 99) = 2

99(100) 2

 2 4950

The series in the power of 2 is an AP, whose formula for sum is used here. Example 16 A square has a side of 10 cm. Another square is formed by joining the mid-points of the sides of the given square and this process is repeated infinitely. Find the sum of the areas of all the squares. Solution The largest square has side 10 cm.

402

Progressions

The next smaller square has side

10 cm. 2

The next smaller square has side

10  5 cm, and so on. 2 2

Sum of areas of all these squares =

102

 10  +    52  ....  2

which is an infinite GP with common ratio Sum =

1 . 2

100 = 200 cm2. 1 1 2

Example 17 A cube is inscribed in a sphere of radius 10 cm. A small sphere is inscribed in this cube and in this small sphere, another cube is inscribed. This process of alternately inscribing a cube and a sphere is continued forever. What is the total volume of all the spheres? Also find the ratio of the volume of all the spheres to that of all the cubes? Solution Let us name the spheres S1, S2, S3, and so on in decreasing order of radii. Similarly, cubes are named as C1, C2, C3, and so on in decreasing order of length of edge. Radius of S1 = 10 cm. Diagonal of C1 is the diameter of S1. Therefore, diagonal of C1 = 20 cm. Hence, edge of C1 =

20 cm. 3

Diameter of S2 = edge of C1 =

20 10 cm. Therefore, radius of S2 = cm. 3 3

 20     3   20 Edge of C2 = , and so on. 3 3 We observe that the radii of spheres are in GP and edges of cubes are also in GP. Therefore, total volume of all the spheres is given below:

4 3 (1000) 3  4  10 10 4 1 1        ....    10 3 1   10  3     ....   3 =     1  3 3   3  3 3 27  3   1  3 3



4000 3 cm3 3 3 1

Total volume of all the cubes:

403

Quantitative Aptitude Simplified for CAT

8000  20   20    20   ....   20   1  1  1  ....   3 3  8000         =  cm3.  3 3 3 27 3 3 1  3   1 1   3 3  3  3 3 3

3

3

The required ratio =

3

Total volume of all spheres  3 .  Total volume of all cubes 2

SHORTCUT: We observe that the ratio of volumes of spheres is same as that of cubes. Therefore, to find the required ratio, it is sufficient to find the ratio of the volume of S1 to that of C1. 4 (10)3 Volume of S 1  3 3 The required ratio = .   3 Volume of C1 2  20     3

Example 18 On a chessboard, there are 64 squares numbered 1 through 64. Re 1 is placed on square 1; Rs 2 is placed on square 2; Rs 4 is placed on square 3, Rs 8 is placed on square 4, and so on. When all the squares are filled like this, what is the total amount of money kept on the board? Solution Total money kept = 1 + 2 + 4 + 8 +.... 64 terms =





1 264  1  Rs 264  1 (2  1)





Example 19 Ram owes Shyam some amount of money and agrees to pay Rs 2000 on day 1, Rs 1000 on day 2, Rs 500 on day 3, Rs 250 on day 4, and so on. What is the maximum amount which Ram could possibly owe Shyam? (Shyam does not charge Ram any rate of simple or compound interest) Solution The maximum amount would be payable if Ram goes on paying Shyam forever. Therefore, the maximum amount = 2000 + 1000 + 500 + 250 + .... =

2000  Rs. 4000. 1 1 2

Example 20 Find the number of terms common to the two series (i)

2, 6, 18, 54, ..... 100 terms and 2, 18, 162, .... 100 terms.

(ii)

3, 36, 432, .... 50 terms and 4, 72, 1296, ..... 50 terms.

Solution (i)

The common ratio of first series is 3 and that of second series is 9. The first term in both the series is 2. It is to be observed that third term of first series is obtained by multiplying first term by 3 two times, which is same as multiplying by 9, which is the common ratio of the second series. Therefore, alternate terms in the first series are the terms in second series, starting with 2. Therefore, 1st, 3rd, 5th, 7th, ... terms of first series are same as the terms of the second series.

404

Progressions For the first series, a1 = 2, a2 = 6, a3 = 18, .... a100 = 2  399. For the second series, a1 = 2, a2 = 18, a3 = 162, .... a100 = 2  999 = 2 3198. The new series formed by the common terms will have a common ratio of 9. The two series have common terms (other than the first term) if the common ratio of one is a multiple of the common ratio of the other. The first term of the new series is 2, common ratio is 9 and the last term is  398. Therefore, an = 2  398 = 2   number of terms = 50. Therefore, there are 50 terms common to the two series. (ii)

The first series has common ratio of 12 and second series has common ratio of 18. The first term of the two series is not same. Also, since the common ratio of one is not a multiple of the common ratio of the other, there will be no common term.

Example 21 Find the sum of the series: 3 – 3 + 3 – 3 + ...... Solution Many students would take the answer to be 0, which is not exactly the case! There are two answers possible, depending upon the number of terms in the series. If there are even number of terms, then the sum is zero and if the number of terms is odd, then the sum is 3. That is, 3 – 3 + 3 – 3 + ..... = 0

(if there are even number of terms)

3 – 3 + 3 – 3 + ..... = 3

(if there are odd number of terms)

Such a series is called an oscillating series and the value of such a series oscillates between 0 and 3. Example 22 Find the sum of the series: 1 + 2 – 3 + 1 + 2 – 3 + 1 + 2 – 3 + .... . Is this series same as 2 + 1 – 3 + 2 + 1 – 3 + 2 + 1 – 3 + .... ? Solution 1 + 2 – 3 + 1 + 2 – 3 + ....  = 0

(if the number of terms is 3n)

1 + 2 – 3 + 1 + 2 – 3 + ....  = 1

(if the number of terms is 3n + 1)

1 + 2 – 3 + 1 + 2 – 3 + ....  = 3

(if the number of terms is 3n + 2)

The value of this series oscillates between 0, 1 and 3. For the second series, 2 + 1 – 3 + 2 + 1 – 3 + ....  = 0

(if the number of terms is 3n)

2 + 1 – 3 + 2 + 1 – 3 + ....  = 2

(if the number of terms is 3n + 1)

2 + 1 – 3 + 2 + 1 – 3 + ....  = 3

(if the number of terms is 3n + 2)

The value of this series oscillates between 0, 2 and 3. Since the sum of the second series is not always same as the sum of the first series, we can say that the two series are not same.

405

Quantitative Aptitude Simplified for CAT

Harmonic Progression (HP) If the terms of the sequence are such that the reciprocal of the terms of the series are in arithmetic progression, then the sequence is called Harmonic Progression, or H.P. for short. That is, if a1, a2, a3, .... are in HP, then

1 1 1 , , ,.... are in AP. a1 a2 a3 1 1 1 1 . , ,.... a a  d a  2d a  (n  1 )d

Therefore, terms in an HP can be written as , Where ‘

1 ’ is the first term. a

From above, we can say that the nth term, denoted by an, is given by: an =

1 a  (n  1 )d

Moreover, sum of n terms of an HP is not given by any standard formula.

Three numbers in AP or GP or HP Let there be 3 numbers, a, b and c. If these numbers are in AP, then b – a = c – b  2b = a + c or b =

ac . 2

If these numbers are in GP, then b c  b2 = ac  a b

If these numbers are in HP, then 1 1 1 2 1 1 2 ac are in AP     b  . , , a b c b a c ac

Example 23 If logax, logbx and logcx are in HP, then a, b and c are in A.

AP

B.

GP

C.

HP

D.

Nothing can be said

Solution: B If logax, logbx and logcx are in HP, then logxa, logxb and logxc are in AP. Therefore, 2 logxb = logxa + logxc = logxac  logxb2 = logxac  b2 = ac. Hence, a, b and c are in GP.

Arithmetico-Geometric Progression (AGP) If the terms of the sequence a1b1, a2b2, a3b3, ...., anbn, are such that the terms a1, a2, a3, ...., an are in AP and b1, b2, b3, ...., bn are in GP, then the sequence is called Arithmetico-Geometric Progression, or AGP for short.

406

Progressions Therefore, terms in an AGP can be written as a, (a + d)r, (a + 2d)r2, (a + 3d)r3, ...., [a + (n – 1)d]rn–1 where ‘a’ is the first term, d is the common difference and ‘r’ is the common ratio. From above, we can say that nth term, denoted by an, is given by an = [a + (n – 1)d]rn–1 We would focus on finding the sum to infinity of an AGP.

Infinite AGP and sum of its terms The technique of finding the sum of infinite AGP is best explained through examples, as given below. Example 24 Find the sum to infinity of the series 1 2 3 n  2  3  .... n  .... (i) 3 3 3 3 (ii)

1 3 6 10     .... 7 72 73 74

(iii) 1 + 2x + 3x2 + 4x3 + ..... , where |x| < 1 Solution (i)

Let S =

1 2 3 10 n  2  3  4  .... n  .... . 3 3 3 7 3

Multiply both sides of the above equation by the common ratio of the GP portion of the series, that 1 is, . 3 1 1 2 3 S  2  3  4  .... . 3 3 3 3 Subtracting the two equations, we get:

1 1  2 1   3 2   4 3  S    2  2    3  3    4  4   .... 3 3 3 3  3 3  3 3  1 2 1 1 1 1 1  S   2  3  4  ....  3   1 2 3 3 3 3 3 1 3 3 S= . 4

S–

(ii)

S=

1 3 6 10 1  3  4  5 ....  Since the common ratio of the GP portion is , multiply both sides by 2 7 7 7 7 7

1 . 7 1 1 3 6 10 S  2  3  4  5 .... . 7 7 7 7 7 Subtracting, we get 6 1 2 3 4 5 S   2  3  4  5 .... 7 7 7 7 7 7

(i)

407

Quantitative Aptitude Simplified for CAT

The RHS is an AGP. Multiply both sides by

1 again, 7

16  1 2 3    ....  S  7  7  72 73 74

(ii)

Subtracting (ii) from (i), we get: 6 S 7

1 1 1  6   1 1  7  7 S    7  72  73  74  ....  

1 49 36 1  S  7   S = . 1 6 49 216 1 7

In this question, we had to multiply both sides by the common ratio two times, because the terms 1, 3, 6, 10, and so on are not in AP, but the numbers formed by their differences, that is, (3 – 1), (6 – 3), (10 – 6), and so on are in AP. Example 25 1

1

1

1

Find the value of 2 2 4 4 8 8 16 16 ...... . Solution 1

1

1

1

1

2

3

4

1 2 3 4    .... 4 8 16 .

2 2 4 4 8 8 16 16 ......  2 2 2 4 2 8 2 16 ......  2 2

Now, the power of 2 in the last term is in AGP. Let the sum of this AGP be S. Then, S=

1 2 3 4     ..... . 2 4 8 16

Multiplying both sides by

1 , we get: 2

1 1 2 3 4 S  2  3  4  5  .... . 2 2 2 2 2 Subtracting second equation from first, we get: 1 1 1 1 1 1 1 S  S   2  3  4  5 ....  2  1 1 2 2 2 2 2 2 1 2

 S = 2. Therefore, the given expression = 2S = 22 = 4.

Special Series There are some sequences of terms in which we can easily spot the pattern, but the terms are in none of the standard forms of series: AP, GP, HP or AGP. For example, 1×2+2×3+3×4+4×5+… The pattern in the above series is easy to identify, but this is none of AP, GP, HP or AGP. Such series fall under the category of Special Series. We will now learn how to find the nth term or sum of such series.

408

Progressions There are basically 3 different types of series. Type 1 Following examples come under first type of series: 2 × 5 + 4 × 8 + 6 × 11 + 8 × 14 + … 1×2×3+2×3×4+3×4×5+4×5×6+… Type 2 Following examples come under second type of series: 2 + 5 + 9 + 14 + 20 + 27 + … 2 + 5 + 11 + 23 + 47 + 95 + … Type 3 Following examples come under first type of series: 1 1 1 1 1      ... 1 2 23 34 4 5 56 1 1 1 1     ... 2  5 5  8 8  11 11  14

Let us learn more about each type of special series. Type 1 To find the sum of series of this type, we first need to remember the following 3 formulae: Sum of first n natural numbers, n = 1 + 2 + 3 + .... + n =

n(n  1) . 2

Sum of squares of first n natural numbers, n2 = 12 + 22 + 32 + ..... + n2 =

n(n  1)(2n  1) . 6

n(n  1)  Sum of cubes of first n natural numbers, n3 = 13 + 23 + 33 + ..... + n3 =   2  

2

= (n)2.

In general, to find the sum of the series of this type, we first find out the nth term of the given series and then apply  (summation sign). While finding , we use some of the above given 3 formulae and get the answer. The process will be clearer from the following examples. Example 26 Find the sum the series: 1  2 + 2  3 + 3  4 + .... 20 terms Solution 1  2 + 2  3 + 3  4 + .... = a1  b1 + a2  b2 + a3  b3 + .... where a1, a2, a3, ... forms an AP and b1, b2, b3, ... forms another AP. a1 = 1, a2 = 2, a3 = 3, ...., an = n. b1 = 2, b2 = 3, b3 = 4, ...., an = n + 1. Therefore, nth term of the given series is given by Tn = n(n + 1). Sum of n terms, Sn = Tn = [n(n + 1)] = n2 + n =

n(n  1)(2n  1) n(n  1) n(n  1)(n  2)   . 6 2 3

409

Quantitative Aptitude Simplified for CAT

Put n = 20 in the summation formula to obtain the sum of 20 terms. Sum of 20 terms, S20 =

20  21  22  3080 . 3

Example 27 Find the sum of the series: (1) + (1 + 2) + (1 + 2 + 3) + ..... n terms Solution First term = 1 Second term = 1 + 2 Third term = 1 + 2 + 3, and so on. Therefore, nth term = 1 + 2 + 3 + .... + n =

n(n  1) . 2

Sum of n terms, Sn

n(n  1) 1  2 2

=



=

n(n  1)(n  2) . 6

 n   n  12  n(n  1)(62n  10  n(n2 1)   n(n4 1)  2n3 1  1  2

Example 28 Find the sum of the series: 1  2  3 + 2  4  7 + 3  6  11 + ..... 10 terms Solution This series is a combination of 3 APs. Proceeding as above, an = n; bn = 2n; cn = 4n – 1. Therefore, nth term of the given series is given by (n) (2n) (4n – 1) = 8n3 – 2n2. Sum of n terms, Sn = (8n3 – 2n2) = 8n3 – 2n2  = 8 

2

n(n  1)  n(n  1)(2n  1)  n(n  1)(2n  1)  2 2 .   2   2n (n  1)  2  8 3  

Sum of 10 terms, S10 = 2(10)2(10+1)2 –

10(10  1)(2(10)  1)  24200  770  23430 . 3

Example 29 Find the sum of the series: 212 + 222 + 232 + .... 402. Solution 212 + 222 + 232 + .... 402 = (12 + 22 + 32 + .... 402) – (12 + 22 + 32 + .... 202) =

40(41)(2  40  1) 20(21)(2  20  1)   22140  2870  19270 . 6 6

Example 30 Find the sum of n terms of the series whose nth term is given by 2n2 + 3n + 5.

410

Progressions Solution The sum of n terms of the given series can be found by applying ‘’ to the nth term. Therefore, Required Sum = 2n2 + 3n + 5) = 2n2 + 3n + 5 = 2n2 + 3n +5 = 2(12 + 22 + 32 + .... + n2) + 3(1 + 2 + 3 + .... + n) + (5 + 5 + 5 + .... n terms) Using the above special results, we get:

2  

n(n  1)(2n  1)  n(n  1)(2n  1) 3n(n  1)  n(n  1)    5n .   3  2   5n  6 3 2

Example 31 Find the nth term of the series: (1) + (2 + 3) + (4 + 5 + 6) + (7 + 8 + 9 + 10) + .... n terms. A. B.

n n(n  1 ) 2

C.

n(n2  1) 2

D.

n(2n  1 ) 2

Hence find the sum of the series. n(n  1 ) A. 2 n 3 n  2n2  3n  2 B. 8 n 3 C. n  2n2  n  2 8 n 3 D. n  2n2  n  4 8













Solution The number of terms in first term = 1. The number of terms in second term = 2. The number of terms in third term = 3. Similarly, the number of terms in nth term = n. Total number of terms in all these n terms = 1 + 2 + 3 + .... + n =

n(n  1) . 2

Since the terms in the brackets are consecutive natural numbers starting from 1, we can say that the last term n(n  1) (n  1)n of the nth bracket will be and hence the last term of the (n – 1)th bracket will be . So, the first 2 2 term of the nth bracket =

n(n  1) n2  n  2 . 1  2 2

Therefore, in the nth bracket, a1 =

n(n  1) n2  n  2 and an = . 2 2

The sum of these n terms is the nth term, that is,

411

Quantitative Aptitude Simplified for CAT

an =

  2   2 n  a1  an   n  n  n  2  n(n  1)   n  2n  2   n n2  1 . 2 2 2 2  2 2  2





Alternatively, an = (a1 + a2 + a3 + .... + an) – (a1 + a2 + a3 + .... + an–1)

n(n  1)   n(n  1)   =  1  2  3  .....     1  2  3  .....   2 2       n(n  1)   n(n  1)  1    2 2    =   2  

   n(n  1)   n(n  1)  1     2 2        2    

     

n n = (n  1)(n2  n  2)  (n2  n  2)  n2  1 . 8 2





Alternatively, Use options. Putting n = 1 should give us first term of the series, that is 1. This rules out [D] option. Putting n = 2 should give us the second term of the series, that is (2 + 3) = 5. Only [C] option gives value 5. Now, using nth term, we can find the sum of n terms of the series. Sn = an = =

n

 2 n

2



1 

1 2

 n   n  12   n   n  12  n(n2 1)   n(n2 1)  1  2

3

1 n n(n  1)(n2  n  2)  n3  2n2  3n  2 . 8 8









We can also find the sum of AP series using type 1 approach. Consider the following example. Example 32 Find the sum of n term of the AP series: 2 + 9 + 16 + 23 + .... Solution The nth term of the series = 7n – 5.

 Therefore, Sum = an = (7n – 5) = 7n – 5 = 7  

n(n  1)  7 2 3   5n  n  n . 2  2 2

Type 2 For type 2 questions, the process is elaborated through following examples. Example 33 Find the nth term and hence sum of n terms of the series: 1 + 3 + 6 + 10 + 15 + ..... n terms. Solution Let S = 1 + 3 + 6 + 10 + 15 + .... an–1 + an S = 1 + 3 + 6 + 10 + 15 + .... + an–1 + an Subtracting, we get: 0 = 1 + (3 – 1) + (6 – 3) + (10 – 6) + .... + (an – an–1) – an

412

Progressions 0 = 1 + [2 + 3 + 4 + ......(n – 1) terms] – an 2 (n  1 ) 2(2)  (n  2)1   1  (n  1)( n  2)  n  n . 2 2 2

 an = 1 + Sum, Sn =



n2  n 1  2 2

 n   n  12  n(n  16)(2n  1)  n(n2 1)   n(n  16)(n  2) . 2

Example 34 Find the nth term and hence sum of n terms of the series: 2 + 5 + 9 + 14 + 20 + … n terms Solution Let S = 2 + 5 + 9 + 14 + 20 + .... an–1 + an S = 2 + 5 + 9 + 14 + 20 + .... + an–1 + an Subtracting, we get: 0 = 2 + (5 – 2) + (9 – 5) + (14 – 9) + .... + (an – an–1) – an 0 = 2 + [3 + 4 + 5 + ......(n – 1) terms] – an 2 (n  1 ) 2(3)  (n  2)1   2  (n  1)(n  4)  n  3n . 2 2 2

 an = 2 + Sum, Sn = =



n 2  3n 1  2 2

 n

2

3

1 n(n  1)(2n  1) 3n(n  1)    6 2 

 n   2 

n(n  1)(n  5) . 6

Example 35 Find the nth term and hence sum of n terms of the series: 2 + 5 + 11 + 23 + 47 + 95 + … Solution Let S = 2 + 5 + 11 + 23 + 47 + .... an–1 + an S = 2 + 5 + 11 + 23 + 20 + .... + an–1 + an Subtracting, we get 0 = 2 + (5 – 2) + (11 – 5) + (23 – 11) + (47 – 23) + .... + (an – an–1) – an 0 = 2 + [3 + 6 + 12 + 24 + ......(n – 1) terms] – an The terms in the bracket are in GP.  an = 2 + Sum, Sn =





3 2 n 1  1  2  3 2 n1  1  3  2 n1  1 . 21





n

n

n

n 1

n 1

n 1

 3  2n1  1   3  2n1   1

= 3(20 + 21 + 22 + 23 + … 2n–1) – n =





3 2n  1  n = 3(2n – 1) – n 21

Type 3 For type 3 questions, the process is elaborated through following examples.

413

Quantitative Aptitude Simplified for CAT

Example 36 Find the sum of the series:

1 1 1 1 1 1      ...  1 2 2 3 3 4 45 56 29  30

Solution We split each term into difference of two terms. So, 1 1 1 1 1 1 1 1 1 1 1 1 .   ;   ;   ; …;   1 2 1 2 23 2 3 34 3 4 29  30 29 30

Therefore,

1 1 1 1 1 1      ...  1 2 2 3 3 4 45 56 29  30

 1 1  1 1  1 1  1 1  1  1   1  1  =                 ...       1 2  2 3  3 4  4 5  28 29   29 30  All the intermediate terms will cancel and we get: 1 1 29 , which is the required sum.   1 30 30

Example 37 Find the sum of the series:

1 1 1 1 1     ...  2  5 5  8 8  11 11  14 (3n  1)(3n  2)

Solution

1 1 1 1 1 1 1 1 1 1 1 1  1 1 1 1      ;     ;        ; …; . 2  5 3  2 5  5  8 3  5 8  8  11 3  8 11  (3n  1)  (3n  2) 3  3n  1 3n  2  Therefore,

=

1 1 1 1 1     ...  2  5 5  8 8  11 11  14 (3n  1)(3n  2)

1  1 1  1 1  1 1   1 1  1 1   1 1             ...             3   2 5   5 8   8 11   11 14   (3n  4) (3n  1)   (3n  1) (3n  2)  

All the intermediate terms will cancel and we get:

11 1  n , which is the required sum.     3  2 3n  2  2(3n  2) Example 38 Find the sum of the series:

1 12 1  2 3 1  2  3  ...  n    ...  3 . 1 3 1 3  23 1 3  23  3 3 1  23  33  ...  n3

Solution

n(n  1) 1  2  3  ...  n 2 2   The nth term of the given series = 3 . 3 3 3 2 n ( n  1) 1  2  3  ...  n  n(n  1)     2 

414

Progressions

Therefore, T1 =

2 2 2 2 2 2 ; T2 = ; T3 = , and so on.    2(2  1 ) 2  3 3 (3  1 ) 3  4 1(1  1 ) 1  2

1 1 1 1   1 So, the given series becomes = 2       ...  n(n  1)  1 2 23 34 45 1 1 1 1 1 1 1 1 1   = 2    1                ...     2 2 3 3 4 4 5 n n  1            

1  2n  = 2 1 .   n1  n1 

415

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

1.8

D.

1.7 CAT 2004

5.

A.

3

B.

4

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

C.

5

A.

0

D.

6

B.

30

E.

7

C.

20

D.

Can’t be determined

CAT 2006 2.

C.

Consider the set S = {1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?

CAT 2004 6.

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

A.

3

A.

666

B.

4

B.

676

C.

6

C.

683

D.

7

D.

777

E.

8

CAT 2003 (R) CAT 2006 7.

3.

The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is A.

78

B.

19

C.

20

D.

77

E.

22

1

4 9 16 25     ... 7 72 7 3 7 4

equals: A. B. C. D.

CAT 2008 4.

The infinite sum

On January 1, 2004 two new societies, S1 and S2, are formed, each with n members. On the first day of each subsequent month, S1 adds b members while S2 multiplies its current number of members by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b =10.5n, what is the value of r?

27 14 21 13 49 27 256 147 CAT 2003 (R)

8.

If (a2 + b2)d2  2(a + c)bd + (b2 + c2) = 0, where d has only one real value, then a, b, c are in A.

AP

B.

GP

A.

2.0

C.

HP

B.

9

D.

None of these CAT 2012

416

Progressions 9.

Sum of first 6 terms of an AP is 3. First term is 4 times the third term. Find fifth term.

has numbers 2, 10, 37 as first, third and sixth number, find the eighth number.

A.

–4

B.

–1

C.

2

D.

1

A. B. C. D.

49 64 65 54 CAT 2016

CAT 2014

14. There are two series, A = {2, 8, 14, 20, 26, ...} and B = {2, 6, 10, 14, ...}. If the terms common to the two series are removed from A, what is the sum of first 10 remaining terms of the series A?

10. In a GP, sum of n terms is 2n and sum of 2n terms is n. Find sum of 3n terms of the same GP. A.

3n

B.

6n 3n 2 2n 3

C. D.

A. B. C. D.

CAT 2016 CAT 2014

15. A child, playing at the balcony of his multistoried apartment, drops a ball from a height of 350 m. Each time the ball

11. Find the sum of the series: 3 5 7    ... 1 2 1 2  22 1 2  2 2  3 2 41  2 1  22  32  ...  20 2

rebounds, it rises

B. C. D.

40 7 20 7 40 21 70 3

B.

45

C.

60

D.

35

of the height it has

A.

2530 m

B.

2800 m

C.

3150 m

D.

3500 m

16. Shruti and Krishna left Delhi for Noida at the same time. While Shruti was driving her car, Krishna, an environmentalist by profession, was traveling on his bicycle. Having reached Noida, Shruti turned back and met Krishna an hour after they started. Krishna continued his journey to Noida after the meeting, while Shruti turned back and also headed for Noida. Having reached Noida, Shruti again turned back and met Krishna 30 minutes after their first meeting. The time taken by Krishna to cover the distance between Delhi and Noida is

12. Angles of a quadrilateral are in AP. If the largest angle is 3 times the smallest angle, then what is the value of the smallest angle? 30

th

IIFT 2016

CAT 2015

A.

4 5

fallen through. The total distance travelled by the ball before it comes to rest is

.

A.

540 600 520 620

CAT 2015 13. A sequence is such that difference of consecutive terms are in AP. If the series

A.

2 hours

B.

2.5 hours

C.

3 hours

D.

None of the above IIFT 2016

417

Quantitative Aptitude Simplified for CAT

17. What is the sum of integers 54 through 196 inclusive? A.

28,820

B.

24,535

C.

20,250

D.

17,875

21. Seema has joined a new Company after the completion of her B.Tech from a reputed engineering college in Chennai. She saves 10% of her income in each of the first three months of her service and for every subsequent month, her savings are Rs. 50 more than the savings of the immediate previous month. If her joining income was Rs. 3000, her total savings from the start of the service will be Rs.11400 in

IIFT 2016 18. A multi-storied office building has a total of 17 rows of parking spaces. There are 20 parking spaces in the first row and 21 parking spaces in the second row. In each subsequent row, there are 2 more parking spaces than in the previous row. The total number of parking spaces in the office building is A.

380

B.

464

C.

596

D.

712

A.

6 months

B.

12 months

C.

18 months

D.

24 months IIFT 2015

22. Let P1 be the circle of radius r. A square Q1 is inscribed in P1 such that all the vertices of the square Q1 lie on the circumference of P1. Another circle P2 is inscribed in Q1. Another square Q2 is inscribed in the circle P2. Circle P3 is inscribed in the square Q2 and so on. If SN is the area between QN and PN+1, where N represents the set of natural numbers, then the ratio of sum of all such SN to that of the area of the square Q1 is

IIFT 2016 19. The sum of 4 + 44 + 444 + …. upto n terms, is A.

40 n 5n 8 1  81 9

B.

40 n 4n 8 1  81 9

C.

C.

40 4n 10n  1  81 9

D.

D.

40 5n 10 n  1  81 9





A. B.

 





4 2 2  4   2 2 None of these IIFT 2014





23. In a school, students were called for the Flag Hoisting ceremony on August 15. After the ceremony, small boxes of sweets were distributed among the students. In each class, the student with roll no. 1 got one box of sweets, student with roll number 2 got 2 boxes of sweets, student with roll no. 3 got 3 boxes of sweets and so on. In class III, a total of 1200 boxes of sweets were distributed. By mistake one of the students of class III got double the sweets he was entitled to get. Identify the roll number of the student who got twice as many boxes of sweets as compared to his entitlement.

IIFT 2016 20. If p, q and r are three unequal numbers such that p, q and r are in A.P., and p, r – q and q – p are in G.P., then p : q : r is equal to A.

1:2:3

B.

2:3:4

C.

3:2:1

D.

1:3:4 IIFT 2015

418

Progressions A.

22

B.

24

C.

28

D.

30 IIFT 2014

24. The student fest in an Engineering College is to be held in one month’s time and no sponsorship has yet been arranged by the students. Finally the General Secretary (GS) of the student body took the initiative and decided to go alone for sponsorship collection. In fact, he is the only student doing the fund raising job on the first day. However, seeing his enthusiasm, other students also joined him as follows: on the second day, 2 more students join him; on the third day, 3 more students join the group of the previous day; and so on. In this manner, the sponsorship collection is completed in exactly 20 days. If an MBA student is twice as efficient as an Engineering student, the number of days which 11 MBA students would take to do the same activity, is

B.

2 3

C.

2 3

D.

3 2 IIFT 2014

27. A tennis ball is initially dropped from a height of 180 m. After striking the ground, it rebounds

3 of the height from which it has 5

fallen. The total distance that the ball travels before it comes to rest is A.

540 m

B.

600 m

C.

720 m

D.

900 m IIFT 2013

28. The sum of series, (–100) + (–95) + (–90) + …………+ 110 + 115 + 120, is: A.

0

A.

70

B.

220

B.

80

C.

340

C.

90

D.

450

D.

100

E.

None of the above

IIFT 2014

XAT 2017

25. A bouncing tennis ball is dropped from a height of 32 metre. The ball rebounds each time to a height equal to half the height of the previous bounce. The approximate distance travelled by the ball when it hits the ground for the eleventh time, is:

29. Consider the set of numbers {1, 3, 32, 33,…...,3100}. The ratio of the last number to the sum of the remaining numbers is closest to:

A.

64 metre

B.

96 metre

C.

128 metre

D.

150 metre

A.

1

B.

2

C.

3

D.

50

E.

99 XAT 2016

IIFT 2014

30. What is the sum of the following series? –64, –66, –68,……..….., –100

26. The sum of 1

A.

1 1 1  1 1 5        ... is 6  6 4   6 4 18 

2 3

419

A.

–1458

B.

–1558

C.

–1568

D.

–1664

Quantitative Aptitude Simplified for CAT

E.

None of the above XAT 2015

31. x, 17, 3x – y2 – 2, and 3x + y2 – 30, are four consecutive terms of an increasing arithmetic sequence. The sum of the four number is divisible by: A.

B.

3

C.

5

D.

7

E.

11 XAT 2014

2

ANSWER KEY 1.

D

2.

D

3.

C

4.

A

5.

9.

A

10. C

11. A

12. B

13. C

14. D

15. C

16. A

17. D

18. C

19. C

20. A

21. C

22. A

23. B

24. A

25. B

26. D

27. C

28. D

29. B

30. B

31. A

420

A

6.

B

7.

C

8.

B

Progressions

ANSWERS AND EXPLANATIONS 1.

4.

Explanation: It is given that b = 10.5n. So, number of members of S1 on July 2, 2004 = n + 63n (for 6 months) = 64n. Similarly, number of members of S2 = n × r6. So 64n = n × r6. Hence r = 2.

Answer: D Explanation: Let there be n rows and x children in the front row. The, total number of children = (x) + (x – 3) + (x – 6) + …. + [x – 3(n – 1)] 3(n  1)n = 630 = nx – 2 630 3(n  1) x=  n 2 When n = 6, the value of x is a fraction, which is not possible. For all other values of n, x is an integer.

2.

5.

Answer: A Explanation: 11 2a  10d  and S19 S11 = 2 19 2a  18d  . As per the question, = 2 11 2a  10d  = 19 2a  18d   22a + 2 2 110d = 38a + 342d  16a + 232d = 0  2a + 29d = 0. 30 2a  29d  = 0. S30 = 2

Answer: D Explanation: If number of elements in a particular series is n, then 1000 = 1 + (n – 1)d, where d is the common difference. 999 = (n – 1)d. So, (n – 1) is a factor of 999. Total number of factors of 999 or 33  371 is (3 + 1)(1 + 1) = 8. Out of this, (n – 1) cannot be equal to the smallest factor, 1 (as n must be more than 2). Therefore, there are 7 possible values of n – 1 and hence 7 APs are possible.

3.

Answer: A

6.

Answer: B Explanation: The numbers with the required property are 10, 17, 24, …, 94. If total number of terms is n, then an = a + (n – 1)d  94 = 10 + (n – 1) × 7  n = 13. 13 2  10  12  7 = 676. Sn = 2

Answer: C 7.

Explanation: The first term which is common to the two series is 21. The common difference of the first and second series is 4 and 5. So, the common difference of the series formed from the common terms will be the LCM of the common difference of the individual series, that is, 20. Therefore, terms in the common series will be 21, 41, 61, 81, … The last term will be less than or equal to the smaller of 417 and 466, that is, 417. The last term will naturally be 401 (as terms are 20k + 1 type). Therefore, if there are n terms in the common series, 401 = 21 + (n – 1)20 or n = 20.

Answer: C Explanation: 4 9 16 25 Let x = 1   2  3  4  ... 7 7 7 7 Multiplying both sides by the common ratio of GP portion, we get 1 1 4 9 16 25 x =  2  3  4  5  ... 7 7 7 7 7 7 Subtracting, we get 3 5 7 9 11 6 x = 1   2  3  4  5  ... (i) 7 7 7 7 7 7 Multiplying both sides of the equation (i) 1 again by , we get 7 16   x 77 

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Quantitative Aptitude Simplified for CAT

1 3 5 7 9 11  2  3  4  5  6  ... (ii) 7 7 7 7 7 7 Subtracting (ii) from (i), we get: 66   x 77  1 1 1 1 1 1  = 1  2   2  3  4  5  6  ... 7 7 7 7 7 7   1    1 4 = 1  2 7  = 1  2   1  6 3  1   7  49 Therefore, x = . 27 =

8.

Now, S3n =

3

  1  1     S r 1 3    2  .  3n   n  1 Sn 4 r  1    1 2 3 3n Therefore, S3n = (2n)  . 4 2 11. Answer: A Explanation: 2n  1 2n  1 6   Tn = . 2 n ( n  1 )( 2 n  1 ) n ( n  1)   n   6 Therefore, series becomes 1  1  1  1  ...   = 6 20  21  1 2 2 3 3 4



Answer: B

1 1 1 1 1 1   1 1  6              ...      20 21    1 2  2 3  3 4 1 1  20 40  6      2 7 7  1 21  12. Answer: B Explanation: Let the angles be a, a + d, a + 2d and a + 3d. Now, a + 3d = 3a  2a = 3d (i) Also, sum of all the angles is 360. Therefore, 4a + 6d = 360  2a + 3d = 180 (ii) Solving (i) and (ii), we get a = 45, d = 30. The smallest angle is 45.

Answer: A Explanation: 6 S6 = 2a  5d   3  2a + 5d = 1. 2 Also, a = 4(a + 2d)  3a + 8d = 0. Solving the equations, we get a = 8 and d = 3. Therefore, fifth term = a + 4d = 8 – 12 = –4.

13. Answer: C Explanation: The series is: 2, ___, 10, ___, ___, 37. Now, 2 = 12 + 1; 10 = 32 + 1 and 37 = 62 + 1. So, series becomes 2, 5, 10, 17, 26, 37, 50, 65. So, 8th term is 65. Alternatively, Second term will be less than the average of 2 and 10. So, it will be less than 6. If second term is 5, then the difference between 1st and 2nd terms is 3, that between 2nd and 3rd terms is 5 and so subsequent differences should be 7, 9, 11, and so on. So, 4th and 5th terms should be 10 + 7 = 17 and 17 + 9 = 26. Therefore, T7 = 37 + 13 = 50 and T8 = 50 + 15 = 65.

10. Answer: C Explanation: a rn 1 a r 2n  1  2n ; S2n = n Sn = r 1 r 1 2a r 2n  1  2n . Therefore,  r 1 a r n  1 2a r 2n  1   rn – 1 = 2(r2n – 1). r 1 r 1 Let rn = k. Then, k – 1 = 2(k2 – 1) = 2k2 – 2  2k2 – k – 1 = 0 1  18 1  9 1   1,  . But r ≠ k= 4 4 2 1 1. Therefore, rn =  . 2



















3n

Explanation: The equation is quadratic in variable d. Since d has only one real value, roots of the equation are real and equal. So, [2(a + c)b]2 – 4(a2 + b2)(b2 + c2) = 0  b2(a2 + c2 + 2ac) = a2b2 + a2c2 + b4 + b2c2  2ab2c = a2c2 + b4  b4 – 2ab2c + a2c2 = 0  (b2 – ac)2 = 0  b2 = ac 9.



a r 3n  1 r 1





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Progressions 14. Answer: D

Total time = 60 + 30 + 15 + … =

Explanation: First series has common difference = 6 and second series has common difference = 4. So, common difference of common terms series = LCM (6, 4) = 12. So, if first common term is 2, then the other common terms are 14, 26, 38, and so on. So, remaining terms in A = 8, 20, 32, … Sum of first 10 terms of the series A 10 2  8  9  12  = 620. = 2

60  120 1 1 2

min = 2 hours. 17. Answer: D Explanation: Required sum = (Sum from 1 to 196) – (sum 196  197 53  54 from 1 to 53) =  2 2 = 19306 – 1431 = 17875. 18. Answer: C Explanation: The series = 20 + 21 + 23 + 25 + … 17 terms. Except for first term, the remaining 16 terms are in AP. So, sum = 20 + 16 2  21  (16  1)2 = 596. 2

15. Answer: C Explanation: Distance covered by ball when it touches ground for the first time = 350 m. Distance travelled when it touches for the 4 second time = 2 × × 350 m. 5 Distance travelled when it touches for the 4 4 third time = 2 × × × 350 m. 5 5 Therefore, total distance travelled by the ball will be sum of the infinite series 4 4 4 = 350 + [2 × × 350] + [2 × × × 5 5 5 350] + … 4 = 350 + [2 × × 350] 5 2 3    1  4   4    4   ...   5  5  5  

19. Answer: C Explanation: Taking 4 common and multiplying and dividing by 9, the series becomes 4  9  99  999  9999  ... . 9 The common ratio of the GP inside bracket will be 10. Now use options and check only [C] and [D] options. When n = 1, sum of series should be 4. Value obtained from option [C] = 4. 35 Value obtained from option [D] = . 9 20. Answer: A

   1  4  = 3150 m. = 350 + [2 × × 350]  5 1 4    5 

Explanation: Use options. Checking [A] option, let p = 1, q = 2 and r = 3. Then, r – q = 1 and q – p = 1. So, p, r – q and q – p are in GP.

16. Answer: A

21. Answer: C

Explanation: The first meeting is after one hour, the second is after an additional half an hour, and this process goes on forever. Each time, time interval between subsequent meetings gets halved compared to previous interval of meeting. So, the total time taken by Krishna is sum of infinite GP series.

Explanation: Her savings from third month onwards are in AP. The series is 300 + 300 + (300 + 350 + 400 + 450 + …) n = 600 + 2  300  (n  1)50  = 11400 2  n(50n + 550) = 21600  n2 + 11n – 432 = 0  n = 16, –27. Therefore, n = 16. Hence, the total number of months = 16 + 2 = 18 months.

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1 (n2 + n) 2 1 n(n  1)(2n  1) n(n  1)  n(n  1)  2n  4        2 6 2  4  3  n(n  1)(n  2)  6 Therefore, S20 = 1540. Each MBA student collects Rs. 2 in one day. So, number of days required by 11 MBA 1540 students to collect Rs. 1540 = = 70 2  11 days.

22. Answer: A

Therefore, Sn =

Explanation: Let radius of P1 = r; length of side of Q1 = 2r r 2 r  r 2 ; radius of P2 =  ; length 2 2 2 r  2 = r. of side of Q2 = 2 SN = Ar(Q1 – P2) + Ar(Q2 – P3) + Ar(Q3 – P4) + = Ar(Q1 + Q2 + Q3 + …) – Ar(P2 + P3 + P4 + …) 2   2  r    r 2  r2     ...   2  





2 2   r 2   r  r            ...   2 2 2   2  

2

 r 1 1  2r 2  1    ...    2 4    2

   

25. Answer: B Explanation: The total distance travelled = 32 + 2 × 32 × 1 1 1 + 2 × 32 × × +… 2 2 2 1 1 1  = 32 + 2 × 32 ×  1    ... = 32 + 2 2 4  1 32 × = 96 m. 1 1 2

 1  1  1  ...    2 4

     2    r2  1  =  = (4 – π)r2  2 1 1     2  (4  )r 2 4    . Therefore, required ratio = 2 2r 2

26. Answer: D 23. Answer: B Explanation: 1 1 1 1 1 5  1            ... 6  6 4   6 4 18 

Explanation: Total number of boxes distributed = (1 + 2 + 3 + … + n) + x = 1200, where x is the roll number of the student who got twice as many as compared to his entitlement. n(n  1) Therefore,  1200  x . Now, n(n + 1) 2 should be approximately 2400 but less than 2400. The largest value of n possible is 48. When n = 48, equation becomes 48(48  1)  1200  x or x = 24. 2

The given sum is 5 1 5 21 5    ...    ... 6 24 24  18 24 24  18 = 373   ... 24  18 5 So, the sum will be more than . Also, sum 6 nd rd of 2 and 3 term is negative, sum of 3rd and 4th term is negative and so on. So, sum of the given series will be less than 1.

24. Answer: A

27. Answer: C

Explanation: Let money collected by 1 engineering student in 1 day be Re.1. Then, total money collected in 20 days by all the students = (1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + … 20 terms. This is type 1 problem of Special Series. Now, Tn of the series = (1 + 2 + 3 + … n) n(n  1) = . 2

Explanation: Sum of the series = 180 + 2 × 180 × × 180 ×

3 3 × +… 5 5

= 180 + 2 × 180 ×

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3 +2 5

2  3  3  3 1    ...     5 5  5 

Progressions

= 180 + 216 ×

1 3 1 5

30. Answer: B

= 720 m.

Explanation: Number of terms =

28. Answer: D Therefore, sum =

Explanation: (–100) + (–95) + ….. + 120 = (–100) + (–95) + ….. + 100 + 105 + 110 + 115 + 120 = 105 + 110 + 115 + 120 = 450. The sum of numbers from –100 to 100 will be 0.

19   64  100   1558 . 2

31. Answer: A Explanation: Since terms are in AP, 2 × 17 = x + (3x – y2 – 2)  y2 = 4x – 36. Also, 2(3x – y2 – 2) = 17 + (3x + y2 – 30)  y2 = x + 3. Therefore, 4x – 36 = x + 3 or x = 13. Sum of the given numbers = 7x – 15 = 76, which is divisible by 2.

29. Answer: B Explanation: Sum of the numbers from 1 to 399 1 3100  1 3100  1  = 3 1 2 Required ratio 3100 3100  2   2 assuming that =  3100  1  3100  1   2  



 100  64  1  19 . (  2)



3100 – 1 is almost equal to 3100.

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Quantitative Aptitude Simplified for CAT

Chapt er 13

Set Theory and Venn Diagrams INTRODUCTION It is important to note that questions based on sets are not asked frequently in the MBA entrance exams, whereas those on Venn Diagrams are asked more frequently in Logical Reasoning section than in Quantitative Aptitude section. Hence the student should pay adequate attention to the relevant ideas discussed in this chapter.

Set A Set is a collection of well-defined objects. The objects, also called as elements of the set, are enclosed in curly brackets, { }. Therefore, a set of first 10 natural numbers is written as below: A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} We say that the set A has 10 elements. As per the definition, the set should be a collection and the objects should be well defined. For example, the above set of first 10 natural numbers is a collection of objects, and the objects are well defined. Is the set of 5 most beautiful women, a set? This is not a set because the set is not well defined. Everybody will have his own list of most beautiful women and we cannot come to a common list! Similarly, set of the 10 best cricket players in the world is not a set, as the list is not well defined. But, the set of countries which participated in the last world cup is a set, as it is well defined. Example 1 Which of the following are sets? Also mention how many elements does it have? (i)

{2, 4, 3, 5, 7, 0}

(ii)

{}

(iii)

{2, 3, {4, 7, 9}, {1}}

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Set Theory and Venn Diagrams (iv) Set of first five prime ministers of Independent India (v)

Set of all the countries of the world

(vi) Set of 5 most advanced countries in the world Solution (i)

It is a set and has 6 elements.

(ii)

It is also a set and has no element. It is called as null set.

(iii) It is also a set and has 4 elements. The last two elements are sets themselves. That does not matter at all. (iv) This is also a set as the set is well defined. (v)

This is a well-defined set.

(vi) This is not a set as it is not mentioned as to what is meant by advanced country. Is the advancement in terms of military power, or economic power, or spirituality? Nothing of this sort is mentioned and hence it is not a set. A few points should be noted: (i)

The order in which the elements are written does not make any difference, as a set is simply a collection of terms. For example, {2, 3, 4} is same as {4, 2, 3}.

(ii)

If an element is repeated, the count of number of elements is not changed. For example,

{1, 2, 2, 2, 3, 4, 5, 5, 6, 7, 7, 7, 7} has 7 elements written as {1, 2, 3, 4, 5, 6, 7}. A set can be written in two ways. For example, A = {x|x is a natural number less than or equal to 10} is set builder form, while A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is roster form of writing the same set. Note that x|x is pronounced as “x such that x”. So, the set builder form is pronounced as “x such that x is a natural number less than or equal to 10”. So, in a set builder form, some common property of the elements is stated whereas in roster form, the values of the elements themselves are written as it is. Therefore, {x|x is a vowel in English Alphabets} is set builder form whereas {a, e, i, o, u} is roster form. Two sets A and B are called equal sets if every element of A is in B and that of B is in A, and hence they have same elements. Two sets A and B are called equivalent sets if they have same number of elements irrespective of the elements being same or not. Therefore, if A = {2, 4, 6, 8, 10}; B = {2, 4, 6, 8, 10}; C = {a, e, i, o, u}, then A and B are equal sets, which is written as A = B; whereas A and C are equivalent sets. Of course, all equal sets are equivalent sets, but all equivalent sets may not be equal sets. If a set has finite number of elements, then the set is finite set. If the number of elements is infinite, then the set is called infinite set. For example, the set of prime numbers less than 100 is a finite set, whereas set of natural numbers is an infinite set. More examples of finite sets are: (i)

set of alphabets in English Language,

(ii)

set of persons in Delhi city,

(iii) set of planets in our solar system. The examples of infinite sets are: (i)

set of real numbers,

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Quantitative Aptitude Simplified for CAT

(ii)

set of galaxies in the universe,

(iii) set of points on the circumference of a circle. If there is no element in a set, then such a set is called as empty set or null set. The set of prime numbers less than 2 is a null set. The set of natural number between 2 and 3 is also an example of a null set. A null set is denoted as { } or . Note that {} is not a null set. This is a set having single element which itself is a null set. Such sets which have only one element are called singleton sets. The set of the smallest even number is a singleton set. The set of number of stars in our solar system is also a singleton set (our solar system has only one star, which is sun). Example 2 Identify the type of set: (i)

{1, 2, 3, 4, .....}

(ii)

{set of prime numbers divisible by 2}

(iii) {x|x is the solution of x3 – 4x2 – 3x + 6 = 0} (iv) {x|x lies between 2 and 5 and x  R} Solution (i)

This is the set of natural numbers which is an infinite set.

(ii)

This set contains only 1 element, that is 2, and hence is an example of singleton set.

(iii) This set has three elements, as the cubic will have three roots. This is a finite set. (iv) There are infinite real numbers between 2 and 5 and hence is an infinite set.

Subset If A and B are two sets such that every element of A is in B, then A is called a subset of B, written as A  B. The symbol ‘’ denotes ‘is a subset of’. We can also say that B is a superset of A, which is written as B  A. For example, A = {a, e, i, o, u}, B = {a, b, c, d, e, f, .....x, y, z} We observe that every element of A is in B. Therefore, A B or B  A. If some elements of A are not in B, then A is not a subset of B, which is written as A  B.

Number of subsets Let A = {1}. Then its subsets are: { }, {1} There are 2 subsets. Recall that { } is a null set which is a subset of every set. If A = {1, 2}. Then its subsets are: { }, {1}, {2}, {1, 2} There are 4 subsets. If A = {1, 2, 3}. Then its subsets are: { }, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}. There are 8 subsets.

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Set Theory and Venn Diagrams We observe that if there is 1 element in a set, then there are 2 subsets. If there are 2 elements, there are 4 subsets and if 3 elements, there are 8 subsets. We can generalize that the total number of subsets is given by 2n, where n is the number elements in a set. Proof: Let there be a set having n elements. Then, null set means selecting ‘no’ element from the set, which can be done in nC0 ways. One element from the set can be selected in nC1 ways; Two elements from the set can be selected in nC2 ways; Similarly, n elements can be selected from the set in nCn ways. Therefore, total number of ways = nC0 + nC1 + nC2 + ......+ nCn = (1 + 1)n = 2n, which is the total number of subsets of the set having n elements. Hence proved.

To prove two sets A and B to be equal sets If A  B and B  A, then A and B are equal sets, that is, A = B. Thus, two sets A and B are defined to be equal sets if A  B and B  A. Example 3 Write all the subsets of A = {{2, 3}, 4}. Solution This set has two elements, {2, 3} and 4. Therefore, there will be 22, that is, 4 subsets. They are: { }, {{2, 3}}, {4} and {{2, 3}, 4}. Example 4 Write all the subsets of A = {{ }, {{ }}, {{{ }}}}. Count how many brackets, like { }, are written. Solution This set has three elements each of which is a kind of set. There will be 8 subsets. They are: { }; {{ }}; {{{ }}}; {{{{ }}}}; {{ }, {{ }}}; {{ }, {{{ }}}}; {{{ }}, {{{ }}}}; {{ }, {{ }}, {{{ }}}}. Total number of brackets = 1 + 2 + 3 + 4 + 4 + 5 + 6 + 7 = 32. Example 5 In the set A = {1, 2, 3, ..., 9, 10}, how many subsets are there? Of these subsets, how many subsets are there which have at least 2, 3, and 5 as elements? Solution The number of subsets is obviously 210 = 1024. Of these, if the elements 2, 3 and 5 must be there, then the other 7 elements may be chosen in 27 ways. Therefore, there will 27 subsets.

Universal Set and Complement of a Set Universal set is a set which contains at least all the elements of all the sets under consideration. For example, if we are considering two sets: set of even numbers and set of prime numbers, then the set of natural numbers is a universal set. Note that the universal set is not unique. For example, in the current example, the set of integers could as well be chosen as the universal set, instead of natural numbers.

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Quantitative Aptitude Simplified for CAT

In general, if we have some sets, A, B, C and D, and a set P contains all the elements of the sets A, B, C and D together, then the set P or any set which is a superset of P is a universal set. Let A be a set and U be the universal set. Then, complement of the set A is denoted as A’ or Ac, and is defined as the set of elements which are in U but not in A. For example, A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 8} and universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Now, A’ = {6, 7, 8, 9, 10}, B’ = {1, 4, 6, 7, 9, 10}.

Operations on Sets Union of sets Union of sets is another set which contains all the elements of the given sets. For example, A = {1, 2, 4, 5, 8} B = {4, 5, 7, 9} Then, union of A and B is denoted as A  B, and which is equal to {1, 2, 4, 5, 7, 8, 9}.

Intersection of sets Intersection of sets is another set which contains all the elements which are common to the given sets. For example, A = {1, 2, 4, 5, 8} B = {4, 5, 7, 9} Then, intersection of A and B is denoted as A  B, and which is equal to {4, 5}

Difference of two sets Let there be two sets A and B as below: A = {1, 2, 4, 5, 8} B = {4, 5, 7, 9} Difference of two sets A and B, written as A – B is another set which contains all the elements of A but not of B. Therefore, A – B = {1, 2, 8}. Similarly, B – A = {7, 9}. Now, we can define complement of a set A as difference of U and A, that is, A’ = U – A.

Venn Diagrams Venn diagrams are the diagrammatic representations of sets. A Venn diagram may be elliptical or circular or even triangular. It is only a representation and hence the shape does not matter. Thus, to represent the set A, given by {1, 2, 4, 5, 8}, the Venn Diagram would be

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Set Theory and Venn Diagrams

The dots shown in the diagram may or may not be there. If we have to show another set B = {4, 5, 7, 9}, then the Venn diagram would be

If both the sets are shown together in a Venn diagram, then they would appear as shown below:

The dots are not shown. The elements 4 and 5 are common to the two sets and hence are shown in the common area enclosed by the two circles. Using Venn diagram, we can easily understand the ‘union’, ‘intersection’ and ‘difference’ of the sets, as defined and discussed above. In terms of Venn diagrams, (i)

the complete area of both the circles together is called union, A  B.

(ii)

the area common to the two circles is called intersection, A  B.

(iii) the area in A which is not common to the area of B is called A – B. Similarly, the area in B which is not common to the area of A is called B – A. Diagrammatically,

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Quantitative Aptitude Simplified for CAT

The set A, A’ (or Ac) and U (universal set) are shown below:

Universal sets, by convention, are always shown as rectangles (or squares). The circle inside it is the set A and the area outside the circle and inside the rectangle is Ac.

Properties Note some of the important properties and try to understand them using Venn diagrams. (1) A  Ac = U (2) A  Ac =  (3) De Morgan’s Laws: (i) (A  B)c = Ac  Bc (ii) (A  B)c = Ac  Bc (4) (Ac)c = A (5) c = U and Uc = . Two sets A and B are disjoint sets if they do not have any element common. For example, set of even numbers and set of odd numbers are disjoint sets. Obviously, the intersection of two disjoint sets will be a null set, that is, A  B = . Two disjoint sets are represented by Venn Diagram as shown below:

As we can see, the two circles do not intersect each other and hence A B = . If there are two sets A and B such that some elements of A and B are common and some are not common, then such sets are called intersecting sets. For example,

432

Set Theory and Venn Diagrams A = {1, 2, 3, 4, 5}, B = {3, 5, 6, 8}. Example 6 If (A – B)  (B – A), then which of the following is necessarily correct? A.

AB

B.

BA

C.

(A – B) = (B – A)

D.

(A  B) = A

Solution: A In such cases, we need to verify all the options. If A  B, then (A – B) is a null set, whereas (B – A) is a non-empty set. We know that null set is a subset of every set. If B  A, then (A – B) is a non-empty set, whereas (A – B) is a null set. We know that a non-empty set cannot be a subset of null set. Therefore, [B] is not the correct option. Checking (c), (A – B) = (B – A)  A = B, in which case also (A – B)  (B – A) is a correct statement. Therefore [C] is also a possible answer. Checking [D], (A B) = A  B is a subset of A. This being a similar case to [B] option, we can say that [D] is also incorrect. Note that when A = B, then (A – B)  (B – A), but even when A ≠ B, (A – B) can be subset of (B – A). Therefore, (A – B)  (B – A) does not necessarily imply that A = B. Hence [C] is not always true. Example 7 Find the complement of: (i)

ABC

(ii)

ABC

Solution (i)

(A  B  C)c = [A  (B  C)]c. Now using De Morgan’s law, that is, (A  B)c = Ac  Bc, we get [A  (B  C)]c = Ac  (B  C)c = Ac  (Bc  Cc) = Ac  Bc  Cc

(ii)

On similar lines, (A  B  C)c = Ac  Bc  Cc.

Example 8 If A and B are subsets of X where X contains at least one element which is neither in A nor in B, and where A is not a subset of B, nor disjoint with B, then A.

(A  B)c = 

B.

A  Bc

C.

B  Ac

D.

A  Bc 

Solution: D Drawing the Venn diagram, we get:

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Quantitative Aptitude Simplified for CAT

Now, check each option.

Number of elements in sets If A is a set, then the number of elements in it is denoted by n(A). Let there be two sets A and B (not necessarily disjoint). Then, number of elements in A  B is given by n(A  B) = n(A) + n(B) – n(A B) This is a very important result. Refer to the Venn Diagram below:

Note that the region represented as: ‘a’ means ‘only A’ (which is same as A – B); ‘b’ means ‘only B’ (which is same as B – A); ‘c’ means ‘both A and B’ and ‘d’ means ‘neither A nor B’. Also, a + b + c = A  B and d = (A  B)c. Since a + b + c + d = U, we can say that n(A  B) + d = Universal Set U. This can be extended to the case of three sets, A, B and C. In such a case, A  B  C is given by n(A B  C) = n(A) + n(B) + n(C) – [n(A B) + n(B C) + n(C A)] + n(A B C) Drawing Venn Diagram for this, we get the following:

434

Set Theory and Venn Diagrams

Thus, A  B  C represents the area of all the three circles combined. Note that the region represented as: ‘a’ means ‘only A’; ‘b’ means ‘only B’; ‘c’ means ‘only C’; ‘d’ means ‘A and B but not C’; ‘e’ means ‘B and C but not A’; ‘f’ means ‘A and C but not B’; ‘g’ means ‘all the three A, B and C’; ‘h’ means ‘none of A, B or C’. Likewise previously, here too we can say that n(A B  C) + h = Universal Set U. Example 9 If n(A  B) = 10, n(A) = 5, n(B) = 7, then find the number of elements common to A and B. Solution Using the formula: n(A  B) = n(A) + n(B) – n(A  B), we get: n(A  B) = n(A) + n(B) – n(A  B) = 5 + 7 – 10 = 2, which is the number of elements common to A and B. We can also say that number of elements belonging to only A is 5 – 2 = 3 and those belonging to only B is 7 – 2 = 5. Example 10 In a class of 50 students, 28 pass in Maths, 42 passed in English. If everybody passed in at least one subject, then how many passed in both the subjects? Solution Let M represent the set of students passing in Maths and E represent the set of students passing in English. Then, n(M  E) = n(M) + n(E) – n(M  E) n(M) = 28, n(E) = 42.

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Quantitative Aptitude Simplified for CAT

Since everybody passed in at least one subject, n(M E) = 50. Therefore, n(M  E) = n(M) + n(E) – n(M  E) = 28 + 42 – 50 = 20. Therefore, there are 20 students who passed in both the subjects. Example 11 In a class of 50 students, 25 passed in Maths, 35 passed in English and 4 failed in both the subjects. How many students passed in both the subjects? Solution Let M represent the set of students passing in Maths and E represent the set of students passing in English. Then, n(M  E) = n(M) + n(E) – n(M  E) n(M) = 28, n(E) = 42. Since 4 failed in both the subjects and total number of students is 50, we can say that 50 = 4 = n(M E). Therefore, n(M E) = 46. Now, n(M  E) = n(M) + n(E) – n(M  E) = 25 + 35 – 46 = 14. Therefore, there are 14 students who passed in both the subjects. Example 12 In a class of 70 students, 40 passed in Physics, 35 in Maths and 43 in Chemistry. There are 15 students who passed in Physics and Maths, 18 who passed in Maths and Chemistry and 20 who passed in Physics and Chemistry. (i)

How many passed in all the three subjects, if everybody passes in at least one subject?

(ii)

What is the minimum number of students who should pass in Physics and Chemistry for consistency in data?

Solution Let A = Physics, B = Maths and C = Chemistry. (i)

Since everybody passes in at least one subject, then n(A  B  C) = 70; n(A) = 40; n(B) = 35; n(C) = 43; n(A  B) = 15; n(B C) = 18; n(C  A) = 20 Hence, n(A B  C) = n(A) + n(B) + n(C) – [n(A B) + n(B C) + n(C A)] + n(A B C)  70 = 40 + 35 + 43 – [15 + 18 + 20] + x (where x = number of students who passed in all three subjects)  n(AB C) = 5

(ii)

Let the number of students passing in Physics and Chemistry be x. Then, n(A  B  C) = n(A) + n(B) + n(C) – [n(A B) + n(B C) + n(C A)] + n(A B C)  70 = 40 + 35 + 43 – (15 + 18 + x) + n(A B C)  x = 15 + n(A B C)

For minimum value of x, n(A B C) should be zero. Therefore, minimum value of x is 15. Hence, minimum 15 students must pass in Physics and Chemistry so that the data is consistent.

436

Set Theory and Venn Diagrams Example 13 In the above example, what is the number of students (i)

passing in Maths only

(ii)

passing in Physics only

(iii) passing in Maths and Physics only (iv) failing in Chemistry only (v)

passing in exactly two subjects

Solution Drawing Venn diagram for the above, we get:

We can easily see that there are 7 students who passed in Maths only. There are 10 who passed in Physics only. There are 10 students who passed in Maths and Physics but not Chemistry. These same students failed in Chemistry only. The number of students passing in exactly two subjects = 10 + 15 + 13 = 38.

Minima and maxima in Venn diagrams Let there be a classroom, in which 70% students passed in Maths and 80% passed in Physics. What percentage passed in both? In such questions, assuming that everybody passes in at least one subject is not correct! Therefore, using the formula: n(A  B) = n(A) + n(B) – n(A  B) and obtaining n(A  B) = n(A) + n(B) – n(A  B) = 70 + 80 – 100 = 50% is not the correct approach, as n(A  B) = 100% is a wrong assumption. This is so because it is quite possible that some students failed in both the subjects, or the 70% students passing in Maths are part of the 80% students passing in Physics, leaving the remaining 20% who may have failed in both these subjects! This means that there will be a minimum and maximum number of students passing in both subjects. Let the percentage of students passing in both be x and passing in at least one subject be y. Then, y = 70 + 80 – x = 150 – x  x = 150 – y. x is minimized when y is maximized and vice versa. The maximum value of y is 100%. Therefore, the minimum value of x is 150 – 100 = 50%. Thus, 50% is the minimum percentage of students passing in both the subjects. For maximum value of x, y should be minimized. The minimum value of y (that is percentage of students passing in at least one subject) is 80%. This is so because 80% students have already passed in Physics.

437

Quantitative Aptitude Simplified for CAT

Therefore, x = 150 – 80 = 70%. For these extremum situations, following two Venn Diagrams are drawn:

On one extreme is the case when all those passing in Maths are part of those passing in Physics. In such a case, one set is a subset of the other. On the other extreme is the case when there may be some students who passed in Maths but not in the other and vice versa. In such a case, the two sets are intersecting each other trying to enclose as less common area as possible. In the first case, the percentage of students (a)

passing in at least one subject is 80%

(b)

passing in exactly one subject is 10%

(c)

passing in both the subjects is 70%

Similarly, in the second case, the percentage of students (a)

passing in at least one subject is 100%

(b)

passing in exactly one subject is (80 – 50) + (70 – 50) = 50%

(c)

passing in both the subjects is 50%

Combining both the results, we can say that (a)

minimum and maximum percentage of students passing in at least one subject is 80% and 100%

(b)

minimum and maximum percentage of students passing in exactly one subject is 10% and 50%

(c)

minimum and maximum percentage of students passing in both the subjects is 50% and 70%

Note that when “percentage of students passing in both” is maximized, then “percentage of students passing in at least one” is minimized and vice versa. In the same example, we can get the number of maximum and minimum students failing in (i)

at least one subject

(ii)

exactly one subject

(iii) both the subjects To get that, we should draw the Venn diagram for percentage of students failing in these subjects. So, if there are 80% passing in Physics, surely there are 20% failing in Physics. Similarly, there are 30% failing in Maths. Drawing Venn diagrams for this, we get:

438

Set Theory and Venn Diagrams

We observe that (a)

Minimum and maximum percentage of students failing in at least one subject is 30% and 50%

(b)

Minimum and maximum percentage of students failing in exactly one subject is 10% and 50%

(c)

Minimum and maximum percentage of students failing in both the subjects is 0% and 20%

“Failing” is complement of “Passing”. Therefore, if A represents percentage of students passing in Physics, and B represents percentage of students passing in Maths, then Ac and Bc represent percentage of students failing in Physics and Maths, respectively. Using De Morgan’s law, (A  B)c = Ac  Bc and (A  B)c = Ac  Bc When (A B) is minimized, then (A  B)c is maximized. Since in the current example, minimum value of (A  B) = 80%, maximum value of (A  B)c = 20%. But, (A  B)c = Ac  Bc. Therefore, maximum value of Ac  Bc, that is, maximum percentage of students failing in both the subjects is 20%. This is same as we have obtained above. Similarly for other cases. Let us now extend this to the case of three subjects: Physics, Chemistry and Maths. Let us understand this with the help of an example. Example 14 There are 200 students in a class. 70 students pass in Physics, 80 students pass in Chemistry, 60 students pass in Maths. 35 students pass in Physics and Chemistry, 30 students pass in Chemistry and Maths, 40 students pass in Maths and Physics. (a)

What is the minimum and maximum number of students passing in (i) all the three subjects

(ii) exactly one subject

(iii) exactly two subjects

(iv) at least one subject

(v) at least two subjects

(vi) at most one subject

(vii) at most two subjects (b)

What is the minimum and maximum number of students failing in all the three subjects.

Solution Here, P means Physics, C means Chemistry and M means Maths. The relationship for three sets is given by

439

Quantitative Aptitude Simplified for CAT

n(PC  M) = n(P) + n(C) + n(M) – [n(P C) + n(C M) + n(M P)] + n(P C M) We cannot assume that n(P  C  M) = 200, as it is not mentioned that everybody passes in at least one subject. Therefore, there exists maxima and minina. If n(P  CM) = x and n(P C M) = y, then the equation reduces to x = 70 + 80 + 60 – (35 + 30 + 40) + y x = 105 + y It is important to note here that when “number of students passing in all three” is maximized, then “number of students passing in at least one subject” is also maximized and vice versa. We know that n(P C M) is always a subset of n(P C), n(C M) and n(M P) and hence cannot exceed 35 or 30 or 40. Therefore, maximum value of n(P C M) is the smallest of 35, 30 and 40, which is 30. If n(P C M) = 30, then the following Venn diagram is obtained.

Since nowhere we are getting negative answers (which would have been inconsistent and hence unacceptable), we can surely say that the maximum value of n(P C M) = 30. One way of knowing whether a particular value chosen is maximum or minimum is that there must be at least one region whose value is zero. In the above diagram, the region representing number of students passing in ‘Maths and Chemistry only’ is zero. Now, n(A  B C) = 105 + 30 = 135, which is the maximum value. Number of students failing in all subjects = 200 – 135 = 65. The number of students passing in (i)

all the three subjects = 30 (already calculated)

(ii)

exactly one subject = 25 + 45 + 20 = 90

(iii) exactly two subjects = 10 + 5 + 0 = 15 (iv) at least one subject = 135 (already calculated) (v)

at least two subjects = 5 + 10 + 0 + 30 = 45

(vi) at most one subject = 200 – (number of students passing in at least 2 subjects) = 200 – 45 = 155. Alternatively, at most one subject = passing in exactly 1 subject + failing in all subjects = 90 + 65 = 155. (vii) at most two subjects = 200 – (number of students passing in all 3 subjects) = 200 – 30 = 170. Alternatively, at most two subjects = passing in exactly 1 subject + passing in exactly 2 subjects + failing in all subjects = 90 + 65 + 15 = 170.

440

Set Theory and Venn Diagrams The minimum value of n(P C M) is 0, but before concluding this to be the final answer, we must check whether the data given is consistent with this assumption. For this, we draw another Venn diagram (shown below). We observe that there are two regions where we end up with negative number of students, which is not possible. The largest negative value is –10 which means that the least value of y should be [0 – (–10)], that is, 10. This will automatically take care of the other negative value. The revised Venn diagram for this is also drawn alongside.

Here too, we observe that at least one of the regions is zero. Therefore, minimum value of n(P C M) = 10. Therefore, n(P  C M) = 105 + 10 = 115, which is also the minimum value. Number of students failing in all subjects = 200 – 115 = 85. The number of students passing in (i)

all the three subjects = 10 (already calculated)

(ii)

exactly one subject = 5 + 25 + 0 = 30

(iii) exactly two subjects = 25 + 20 + 30 = 75 (iv)

at least one subject = 115 (already calculated)

(v)

at least two subjects = 75 + 10 = 85

(vi) at most one subject = 200 – (number of students passing in at least 2 subjects) = 200 – 85 = 115. Alternatively, at most one subject = 30 + 85 = 115. (vii) at most two subjects = 200 – (number of students passing in all 3 subjects) = 200 – 10 = 190. Alternatively, at most two subjects = 30 + 85 + 75 = 190. Compiling all that has been discussed, we get The number of students passing in

Minimum

Maximum

(i)

all the three subjects

10

30

(ii)

exactly one subject

30

90

(iii) exactly two subjects

15

75

(iv) at least one subject

115

135

(v)

45

85

at least two subjects

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Quantitative Aptitude Simplified for CAT

(vi) at most one subject

115

155

(vii) at most two subjects

170

190

Note that maxima or minima for all these cases will exist when n(P C M) is maximized or minimized, though when n(P C M) is maximized, it need not mean that all these cases get maximized, and vice versa. (b)

The number of students failing in all the three subjects = 200 – (number of students passing in at least one subject) Therefore, maximum number of students failing in all the three subjects = 200 – (minimum number of students passing in at least one subject) = 200 – 115 = 85 Minimum number of students failing in all the three subjects = 200 – (maximum number of students passing in at least one subject) = 200 – 135 = 65.

Applications of four Venn diagrams Since CAT 2012, there have been questions based on Venn diagrams of four sets. So it is important to learn the basics of the same. If A, B, C and D are four sets then, n(A  B  C  D) = n(A) + n(B) + n(C) + n(D) – [n(A  B) + n(A  C) + n(A  D) + n(B  C) + n(B  D) + n(C  D)] + [n(A  B  C) + n(A  B  D) + n(A  C  D) + n(B  C  D)] – n(A  B  C  D). There are various ways in which we can draw the Venn diagram for 4 sets, but the most convenient way is as drawn below.

To draw this diagram, one needs to draw two horizontally oriented rectangles, and two vertically oriented rectangles. Drawing this way divides the entire region into 15 individual areas marked by lowercase alphabets. a means only A; b means only B; c means only C; d means only D; p means only A and B; q means only A and C; r means only A and D; s means only B and C;

442

Set Theory and Venn Diagrams t means only B and D; u means only C and D; v means only A, B and C; w means only A, B and D; x means only A, C and D; y means only B, C and D; z means all four A, B, C and D; Note that A  B does not mean only A and B. So, A  B is represented by p + v + w + z. To understand how to solve questions based on 4 Venn diagram, it is best to take an example. Example 15 In a competition, 50 members of Gymkhana Club, 63 members of NSCI, 41 members of US Club, and 5 members of WIA participate. 2 of them are members of all the 4 clubs; 5 are members of Gymkhana, NSCI, and US Club; 3 are members of NSCI, US and WIA; 3 are members of Gymkhana, US, and WIA; 25 are members of Gymkhana and US; none is a member of any of the other clubs taken 2 or 3 at a time. How many people participated in the competition? A.

141

B.

140

C.

145

D.

Cannot be determined

Solution: A Suppose Gymkhana = A, NSCI = B, US = C, WIA = D. n(A  B  C  D) = n(A) + n(B) + n(C) + n(D) – n(A  B) – n(A  C) – n(A  D) – n(B  C) – n(B  D) – n(C  D) + n(A  B  C) + n(A  B  D) + n(A  C  D) + n(B  C  D) – n(A  B  C  D) Now, n(A) = 50, n(B) = 63, n(C) = 41, n(D) = 5, n(A  C) = 25, n(A  B) = n(B  C) = n(C  D) = n(A  D) = n(B  D) = 0, n(A  B  C) = 5, n(B  C  D) = 3, n(A  C  D) = 3, n(A  B  D) = 0, n(A  B  C  D) = 2  n(A  B  C  D) = 50 + 63 + 41 + 5 – 25 + 5 + 3 + 3 – 2 = 143. Example 16 In a class, 4 students failed in all 4 subjects: Maths, Physics, Chemistry and English; 12 passed in Maths; 15 in Physics, 13 in Chemistry and 16 in English; 5 passed in Maths and Physics, 5 in Maths and Chemistry, 4 in Maths and English, 6 in Physics and Chemistry, 6 in Physics and English and 5 in Chemistry and English; 3 passed in Maths, Physics and Chemistry; 3 in Maths, Physics and English; 3 in Maths, Chemistry and English; 4 in Physics, Chemistry and English. 2 passed in all 4 subjects. (i)

Find the total number of students who passed in at least one subject.

(ii)

Find the total number of students in the class.

(iii) Find the total number of students who passed in only one subject. (iv) Find the total number of students who passed in at most 2 subjects.

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Quantitative Aptitude Simplified for CAT

Solution (i)

Using the formula for 4 Venn Diagrams, total number of students who passed in at least one subject = n(M  P  C  E) = n(M) + n(P) + n(C) + n(E) – [n(M  P) + n(M  C) + n(M  E) + n(P  C) + n(P  E) + n(C  E)] + [n(M  P  C) + n(M  P  E) + n(M  C  E) + n(P  C  E)] – n(M  P  C  E) = 12 + 15 + 13 + 16  (5 + 5 + 4 + 6 + 6 + 5) + (3 + 3 + 3 + 4)  2 = 36.

(ii)

Total number of students = 36 + those failed in all 4 = 36 + 4 = 40.

(iii) To answer this and next question, let us draw the Venn diagram as below.

Total number of students who passed in only subject = 5 + 6 + 5 + 9 = 25. (iv) Total number of students who passed in at most two subjects = those passed in exactly 2 subjects + those passed in exactly 1 subject + those failed in all 4 subjects = (1 + 1 + 0 + 0 + 1 + 1) + 25 + 4 = 33. Example 17 In a college library, four different business newspapers - Economic Times, Business Standard, Business Line and Financial Express - are available. All students visit the library regularly but 20% of them do not read any business newspaper. The four newspapers given in the above order are read by 230, 180, 180 and 220 students respectively. The number of students reading exactly 2 newspapers for any two newspapers is 20. There are 30 students who read all the four newspapers but there is nobody who reads exactly three out of four newspapers. (i)

How many students do not read any newspaper at all?

(ii)

What percentage of the people reading Business Standard also read at least one other newspaper?

(iii) If all the students in the college including those who do not read any newspaper read at least one newspaper, (Out of the four newspapers above) which they are not reading at present, then what is the least number of students reading all the four newspapers? Solution (i)

We can draw the following diagram with values put in each region.

444

Set Theory and Venn Diagrams

It is given that 230 read ET. Now, 230 = (those reading only ET) + (those reading only ET & one more paper) + (those reading only ET & exactly two more) + (those reading all 4) Or, 230 = (only ET) + (20 + 20 + 20) + (0) + (30), or only ET = 230  90 = 140. Similarly, those reading only Business Standard = 180  (20 + 20 + 20 + 30) = 90. Those reading only Business Line = 180  (20 + 20 + 20 + 30) = 90. Those reading only Financial Express = 220  (20 + 20 + 20 + 30) = 130. So, those reading at least one newspaper = (140 + 90 + 90 + 130) + (20 × 6) + 0 + 30 = 600. Since 20% do not read any newspaper, 80% read at least one newspaper. If total number of persons is x, then 0.8x = 600 or x = 750. So, 750  600 = 150 do not read any newspaper at all. (ii)

Total number of people reading BS = 180. Those reading only BS = 90. So, those reading BS along with at least one more newspaper = 180  90 = 90. Required percentage = 50%.

(iii) Those who earlier read exactly 3 newspapers will now read all 4 newspapers. Since those reading exactly 3 newspapers is 0, the total number of persons reading all 4 now is same as it was earlier, that is 30.

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Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 3.

Directions for questions 1 to 4: Solve the questions on the basis of the information given below: Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, Flood Relief (FR) in Maharashtra, and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project. 



A Maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project. The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.



17 volunteers are involved in the TR project.



The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER project alone.



Ten volunteers involved in the TR project are also involved in at least one more project.

4.

CAT 2005 1.

2.

Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:

After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project. Which of the following statements, then, necessarily follows? A.

The lowest number of volunteers is now in TR project.

B.

More volunteers are now in FR project as compared to ER project.

C.

More volunteers are now in TR project as compared to ER project.

D.

None of the above.

After the withdrawal of volunteers, as indicated in question 3, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers? A.

ER

B.

FR

A.

1

C.

TR

B.

3

D.

Cannot be determined

C.

4

D.

5

5.

Which of the following additional information would enable to find the exact number of volunteers involved in various projects? A.

20 volunteers are involved in FR.

B.

4 volunteers are involved in all the three projects.

C.

23 volunteers are involved in exactly one project.

D.

No need for any additional information.

A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarized information regarding readership in 3 months is given below: Only September: 18; September but not August: 23; September and July: 8; September: 28; July: 48; July and August: 10; none of the three months: 24. What is the number of surveyed people who have read exactly two consecutive issues (out of the three)?

446

Set Theory and Venn Diagrams A.

7

B.

9

C.

12

D.

14

E.

17

60 play Basketball and Volleyball, nobody plays Volleyball and Tennis. How many play Basketball and Badminton? A.

CAT 2006 6.

Consider the sets Tn= {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, …, 96. How many of these sets contain 6 or any integral multiple thereof (that is, anyone of the numbers 6, 12, 18,…)?. A.

80

B.

81

C.

82

D.

83

45

B.

50

C.

70

D.

65 CAT 2013

9.

For 100 ≤ n ≤ 1300, the number of even values of n not divisible by 7 or 9, is A.

441

B.

451

C.

449

D.

459 CAT 2013

CAT 2003 (R) 7.

10. In a class of 100 students, 36 failed in Mathematics, 43 failed in Physics and 50 failed in Chemistry, 26 failed in exactly two subjects and only 1 passed in all three subjects. Find the number of students who failed in all three subjects.

A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular option – air conditioning, radio and power windows – were already installed. The survey found: 15 had AC

A.

2

2 had AC and PW but no radio

B.

1

12 had radio

C.

3

6 had AC & radio but no power windows

D.

Cannot be determined

11 had power windows

CAT 2015

4 had radio and PW Directions for questions 11 to 14: Read the given information and solve the questions that follow:

3 had all three options What is the number of cars that had none of the option? A.

4

B.

3

C.

1

D.

2

In a Business school of 120 students each student has to opt for either one or two or three areas of specializations out of Marketing, Finance and Strategy. The number of students taking exactly two out of three areas of specializations is 92. The number of students taking exactly one area of specialization is six times the number of students taking all specializations. Out of students taking exactly two areas of specializations, those taking Finance and strategy is two times those taking Marketing and Finance. Of those students who opted for only one area of specialization those taking strategy are two more than those taking Marketing and two less than those taking Finance. Also, students opting for strategy specialization is 22 more than those taking marketing

CAT 2003 (R) 8.

In a school, students play Basketball, Tennis, Badminton and Volleyball. A total of 100 students play Basketball, 95 play Tennis, 110 play Badminton and 85 play Volleyball, 20 play Basketball, Badminton and Volleyball, 25 play only Basketball and Badminton, 30 play only Basketball and Tennis, 55 play Basketball and Tennis,

447

Quantitative Aptitude Simplified for CAT

15. 290 students of MBA (International Business) in a reputed Business School have to study foreign language in Trimesters IV and V. Suppose the following information are given

specialization. Following table lists average salary for students opting exactly one, exactly two, exactly three and all students achieved at the placement rounds of the Business School. Specialization

Average Salary (Rs. lacs)

i.

120 students study Spanish

1 Area

10

ii.

100 students study Mandarin

2 Area

15

iii.

3 Area

?

At least 80 students, who study a foreign language, study neither Spanish nor Mandarin

All

14.1

Then the number of students who study Spanish but not Mandarin could be any number from

CAT 2016 11. Number of students opting for Finance as area of specialization is

A.

80 to 170

B.

80 to 100

A.

74

C.

50 to 80

B.

75

D.

20 to 110

C.

65

D.

60

IIFT 2016 16. In a certain village, 22% of the families own agricultural land, 18% own a mobile phone and 1600 families own both agricultural land and a mobile phone. If 68% of the families neither own agricultural land nor a mobile phone, then the total number of families living in the village is:

12. The number of students who opted for Finance and Strategy as area of specialization is A.

43

B.

46

C.

44

A.

20000

D.

40

B.

10000

C.

8000

D.

5000

13. Average salary of students who opted for all three specializations is A.

15

IIFT 2015

B.

18

C.

14

D.

12

17. The business consulting division of TCS has overseas operations in 3 locations: Singapore, New York and London. The Company has 22 analysts covering Singapore, 28 covering New York and 24 covering London. 6 analysts cover Singapore and New York but not London, 4 analysts cover Singapore and London but not New York, and 8 analysts cover New York and London but not Singapore. If TCS has a total of 42 business analysts covering at least one of the three locations: Singapore, New York and London, then the number of analysts covering New York alone is

14. If for students opting for only one area of specialization the ratio of average salaries of students opting for Marketing, Finance and Strategy are in the ratio of 1 : 2 : 3 then the average salary of students with Finance only as specialization is A.

9.6

B.

9

C.

10

D.

10.2

448

A.

14

B.

28

Set Theory and Venn Diagrams C.

5

D.

7

opted for C; 25 visitors opted for both A and C, 22 opted for both A and B, while no visitor opted for both B and C. 40 visitors did not opt for ride A and B, or both. How many visited with the entry pass on that day?

IIFT 2014 18. In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark scored by students majoring in Mathematics is 45 and that of students majoring in Biology is 60. However, the combined average mark in English, of students of these two majors, is 50. What is the maximum possible number of students who major ONLY in Physics?

A.

102

B.

115

C.

130

D.

135

E.

150 XAT 2016

20. 70% of the students who joined XLRI last year play football, 75% play cricket, 80% play basketball and 85% play carom. The minimum percentage of students who play all four games is:

A.

30

B.

25

A.

5%

C.

20

B.

10%

D.

15

C.

15%

E.

None of the above

D.

20%

E.

None of these

XAT 2017

XAT 2013

19. In an amusement park along with the entry pass a visitor gets two of the three available rides (A, B and C) free. On a particular day 77 opted for ride A, 55 opted for B and 50

ANSWER KEY 1.

C

2.

A

9.

D

10. A

11. A

12. C

17. D

18. D

19. E

20. B

3.

D

4.

A

5.

B

13. B

449

6.

A

14. A

7.

D

15. D

8.

*

16. A

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS

2.

Explanation: Use options. If we know that there are 20 volunteers in FR, then y + 16 = 20 and so y = 4. We can now find the exact number of volunteers involved in various projects. Hence correct option is [A]. Information given in [B] and [C] are already known to us from the given data.

Common explanation for questions 1 to 4: Let number of volunteers involved in all 3 projects be x. Then, number of volunteers involved in ER alone = 2x. Number of volunteers involved in TR alone = 2x – 1. Also, 10 volunteers are involved in TR project along with at least one more project. Since total 17 volunteers are involved in TR project, we can say that 17 – 10 = 7 volunteers are involved in TR project only. Therefore,

3.

Answer: D Explanation: As per the conditions given in the question, revised Venn diagram is drawn below. Volunteers in FR = 14 + y; those in TR = 16; those in ER = 21 – y. None of the statements in [A], [B] or [C] will necessarily follow.

2x – 1 = 7 or x = 4. Refer to the following Venn diagram.

Since number of volunteers involved in FR alone is equal to the number of volunteers having additional involvement in ER project, FR alone = z + 4. Also, total number of volunteers for TR is 17, and if number of volunteers for TR and FR alone is y, then those for TR and ER only = 6 – y.

4.

Answer: A Explanation: As per the conditions given, 5 = 8 – y  y = 3. Let number of volunteers in FR only, TR only and ER only become p each. Then, number of volunteers in FR = p + 5 + 0 + (3 + 1) = p + 9. Those in ER = p + 5 + 0 + 5 = p + 10; those in TR = p + 5 + 0 + (3 + 1) = p + 9. So, ER had the maximum number of volunteers.

Sum of all the elements = 37. Therefore, (z + 4) + z + 8 + 17 = 37 or z = 4. Total number of volunteers for FR = z + 4 + z + 4 + y = y + 16. Total number of volunteers for TR = 17

5.

Total number of volunteers for ER = z + 8 + 4 + 6 – y = 22 – y. 1.

Answer: A

Answer: B Explanation: On the basis of the information given, draw the following Venn diagram.

Answer: C

Explanation: If FR has maximum number of volunteers, then y + 16 > 22 – y  y > 3. So, minimum value of y = 4.

450

Set Theory and Venn Diagrams number of people having none of the options = 25 – 23 = 2. 8.

Answer: Incorrect question Explanation: Let us draw the diagram first.

x + 48 + 18 + 2 = 100 – 24 = 76  x = 8. Number of people who have read exactly two consecutive issues = (people reading Jul + Aug) + (people reading Aug + Sept) = 7 + 2 = 9. We will not count those reading Jul + Sept, as they are not consecutive issues of the magazine. 6.

Answer: A Explanation: Tn = {n, n + 1, n + 2, n + 3, n + 4} T1 = {1, 2, 3, 4, 5}, T2 = {2, 3, 4, 5, 6}, T3 = {3, 4, 5, 6, 7}, T4 = {4, 5, 6, 7, 8}, T5 = {5, 6, 7, 8, 9}, T6 = {6, 7, 8, 9, 10} So, from n = 1 to n = 6, there are 5 sets having a multiple of 6. Out of 96 sets, number 5 of sets having a multiple of 6 = th of 96 6 = 80.

7.

Since nobody plays Volleyball and Tennis, w = z = t = y = 0. Also, 20 play Basketball, Badminton and Volleyball, so x = 20. Only Basketball and Badminton = q = 25. Only Basketball and Tennis = p = 30. Basketball and Tennis = p + v + w + z = 55 or 30 + v + 0 + 0 = 55 or v = 25. Finally, Basketball and Volleyball = w + z + x + r = 60 or 0 + 0 + 20 + r = 60 or r = 40. So, Basketball and Badminton = v + z + x + q = 25 + 0 + 20 + 25 = 70.

Answer: D Explanation: On the basis of information provided, we can draw the following Venn diagram.

Total number of people having at least one of the options = 15 + 5 + 1 + 2 = 23. Therefore,

451

Quantitative Aptitude Simplified for CAT

So, Strategy only = 2x; Marketing only = 2x – 2 and Finance only = 2x + 2. Based on given information, we can draw the following Venn diagram:

Note: If we add all the elements of Basketball, we get 30 + 25 + 25 + 20 + 40 + a, which is more than 100. So, there was this error in the question. 9.

Answer: D Explanation: Total number of even values of n 1300  100  1  601 . = 2 Number which are even multiples of 7, that is numbers which are multiples of 14 are: 112, 126, …, 1288. Number of such numbers 1288  112  1  85 . = 14 Number which are even multiples of 9, that is numbers which are multiples of 18 are: 108, 126, …, 1296. Number of such numbers 1296  108 =  1  67 . 18 Numbers which are even multiples of 7 as well as 9, that is which are multiples of 126 are: 126, 252, …, 1260. Number of such 1260  126 numbers =  1  10 . 126 Now, number of numbers which are even multiples of 7 or 9 = n(14p  18q) = 85 + 67 – 10 = 142. Number of even numbers which are not divisible by 7 nor of 9 = 601 – 142 = 459.

As per the information given, (2x + 2y) – (2x – 2 + y) = 22  y = 20 So, 2y + y + z = 92  z = 92 – 60 = 32. Also, 6x + 92 + x = 120  x = 4. Now, revised Venn diagram is drawn below:

10. Answer: A Explanation: Let x students failed in all three subjects. Number of people failing in at least one subject = 100 – 1 = 99. So, 99 = 36 + 43 + 50 – (26 + 3x) + x  x = 2.

11. Answer: A Explanation: Number of students who opted for Finance = 10 + 40 + 4 + 20 = 74.

Common explanation for questions 11 to 14:

12. Answer: C

Let number of students taking all 3 specializations = x. Then, number of students taking exactly 1 specialization = 6x. Out of those taking exactly 2 areas, (Finance + Strategy) = 2(Marketing + Finance). Out of those taking exactly 1 area, (Strategy) = (Marketing) + 2 = (Finance) – 2. So, (Strategy) + (Marketing) + (Finance) = (Strategy) + [(Strategy) – 2] + [(Strategy) + 2] = 6x or (Strategy) = 2x.

Explanation: Number of students who opted for Finance and Strategy = 40 + 4 = 44. 13. Answer: B Explanation: Number of students who opted for exactly one area of specialization = 24. Number of students who opted for exactly two areas of specialization = 92.

452

Set Theory and Venn Diagrams 17. Answer: D

Number of students who opted for all three areas of specialization = 4. Let the average salary of those who opted for all three specializations is p. Now, using information given in the table, we get 24  10  92  15  4 p  14.1  p = 18 lacs. 120

Explanation: From the information given and using the formula of 3 Venn diagrams, we get 42 = 22 + 28 + 24 – (6 + x + 4 + x + 8 + x) + x, where x = number of analysts covering all the 3 locations.  x = 7. Hence number of analysts covering New York alone = 28 – 6 – 8 – 7 = 7.

14. Answer: A

18. Answer: D

Explanation: Let average salary of students taking only Marketing, only Finance and only Strategy be k, 2k and 3k. Then, 6  k  10  2k  8  3k  10 . 24 Therefore, k = 4.8. Therefore, average of students with Finance only as specialization = 2k = Rs.9.6 lacs.

Explanation: Let number of students majoring in Maths only, Physics only and Biology only be x, y and z. We can draw the following Venn diagram.

15. Answer: D Explanation: At least 80 students study neither Spanish nor Mandarin. Total number of students = 290. Therefore, maximum number of students who study at least one language = 290 – 80 = 210. Minimum number of students who study both languages = 100 + 120 – 210 = 10. Therefore, maximum number of students who study Spanish but not Mandarin = 120 – 10 = 110. Similarly, maximum number of students who study both languages = smaller value of 100 and 120 = 100. Therefore, minimum number of students who study Spanish but not Mandarin = 120 – 100 = 20 Hence, the range could be any number from 20 to 110.

Let M and B be the total scores of the students majoring in Mathematics and Biology respectively. Then, M = 45(x + 6); B = 60(z + 15); M + B = 50(x + z + 21). Therefore, 45(x + 6) + 60(z + 15) = 50x + 50z + 1050  5x – 10z = 120  x = 2z + 24. To maximize y, z = 0 (note that x cannot be 0). So, x = 24. Also, total number of students = 60. So, x + y + z + 21 = 60  24 + 0 + x = 39  15.

16. Answer: A

19. Answer: E

Explanation: Let total number of families in the village be x. Then, as per the question x – 0.68x = 0.22x + 0.18x – 1600  x = 20000.

Explanation: Let the Venn diagram be as shown in the figure.

453

Quantitative Aptitude Simplified for CAT

Total number of people opting for at least one ride = 77 + 55 + 50 – (22 + 25 + 0) + 0 = 135. Total number of people who visited with the entry pass = 135 + 15 = 150. 20. Answer: B Explanation: Students not playing Football = 30%; not playing Cricket = 25%; not playing Basketball = 20%; not playing Carrom = 15%. Hence, maximum number of students not playing at least one of the games = 30 + 25 + 20 + 15 = 90%. Minimum Percentage of students playing all four games = 100 – 90 = 10%.

40 did not opt for A and B, or both. So, 40 opted for only C or none of the rides. If c opted for only C, then 40 – c opted for none of the rides. From the diagram, we can say that c = 50 – 25 – 0 – 0 = 25. So, those opting for none of the rides = 40 – 25 = 15.

454

Permutation and Combination

Chapt er 14

Permutation and Combination INTRODUCTION Permutation and Combination is all about “counting”. There are various ways to count. The most basic way to count is “direct counting”. But, direct counting may not be always possible or easy. In light of this, some principals or rules are developed to simplify the counting process. We will study some of these rules in the following pages. It has been found that many students face difficulty in dealing with this chapter. This is because they end up mugging up the formulae without learning the underlying principles. So, at the time of applying those formulae, students try to recall the relevant formula but are unable to do so. Moreover, if a new situation is presented in a particular question, establishing its link with the relevant principle becomes all the more difficult. While writing this chapter, this problem is addressed in a style that promotes concept based learning instead of rote learning.

Fundamental Principle of Counting The first and foremost principle of counting is “The fundamental principal of counting”. There are basically two rules regarding this.

Product rule According to this rule, if an event can happen in ‘m’ ways and subsequent to the occurrence of this event, another event can happen in ‘n’ ways, then both the events together can happen in ‘m  n’ ways. For example, if there are 3 ways to go from Delhi to Agra and 4 ways to go from Agra to Jhansi, then there are 3  4, that is, 12 ways to go from Delhi to Jhansi via Agra. In this case, the first event is ‘to go from Delhi to Agra’ and the second event is ‘to go from Agra to Jhansi’. Therefore, when we want to go from Delhi to Jhansi via Agra, both the events are happening together and hence the product rule will operate. In fact, this rule can be extended to more than 2 events. So, if the third event can happen in ‘p’ ways, fourth in ‘q’ ways, and so on, all these events together can happen in ‘m  n  p  q ....’ ways.

455

Quantitative Aptitude Simplified for CAT

Therefore, in the above example, if we have to go further from Jhansi to Nagpur for which if there are 5 ways, then total number of ways of going from Delhi to Nagpur via Agra and Jhansi is 3  4  5, that is, 60 ways. It is important to understand here that for the product rule to be applicable, the two events should be independent of each other. That is, if an event can happen in ‘m’ ways and subsequent to the occurrence of this event, another event can happen in ‘n’ ways, the occurrence of second event and the number of ways it can occur should not be affected by the occurrence of the first event and the number of ways it can occur. For example, we can easily understand that the number of ways of going from Agra to Jhansi does not depend upon the number of ways of going from Delhi to Agra, and so on.

Sum rule According to this rule, if an event can happen in ‘m’ ways and another event can happen in ‘n’ ways, then either of the events can happen in ‘m + n’ ways. For example, if there are 3 ways to go from Delhi to Agra and 4 ways to go from Agra to Jhansi, then the number of ways of moving out of Agra city is 3 + 4, that is, 7 ways. Now, you can move out of the city Agra by going either to Delhi (which will require 3 ways) or to Jhansi (which will require 4 ways) and therefore total number of ways is 7. In this case, the first event is ‘to go from Agra to Delhi’ and the second event is ‘to go from Agra to Jhansi’. Therefore, when we want to move out of city Agra, either of the events are happening and hence the sum rule will operate. In fact, this rule can be extended to more than 2 events. So, if the third event can happen in ‘p’ ways, fourth in ‘q’ ways, and so on, then either of these events can happen in ‘m + n + p + q + ...’ ways. If Rajesh has 8 kinds of goggles and 5 kinds of caps, the number of ways in which he can wear goggles or a cap is 8 + 5 = 13 ways. Example 1 A man goes to a shop to buy a chair and a table. There are 11 different chairs and 13 different tables. In how many different ways can he buy a chair and a table? Solution He can choose any one of the 11 chairs in 11 ways and any one of the 13 tables in 13 ways. So, he can choose a chair and table in 11  13 = 143 ways. Here, the first event is choosing a chair and the second event is choosing a table. Since he has to choose a chair and a table, both the events are happening together. Example 2 In the previous example, in how many different ways can he buy a chair or a table? Solution In this case, he has to buy either a chair or a table. Therefore, addition rule will apply and hence the number of ways is 11 + 13 = 24 ways. In any case, out of 11 + 13 = 24 objects (chairs + tables), he has to choose any one of them. Therefore, the required number of ways is 24. Example 3 How many two digit natural numbers are there in total? How many of them have all digits distinct?

456

Permutation and Combination Solution In a 2-digit number, there are two places: ten’s place and unit’s place, and we have to place digits from 0, 1, 2, …, 9 in these places. However, the digit 0 cannot come in ten’s place. So, ten’s place can be filled in 9 ways, that is, ten’s place can be filled by digits 1, 2, 3, …, or 9. Unit’s place can be filled in 10 ways, as the digit 0 can also be placed there. So, total number of ways = 9 × 10 = 90, which is the total number of digits available. If digits are to be distinct: In this case, ten’s place can be filled in 9 ways only (1, 2, 3, …, 9), and unit’s place can be filled in 9 ways instead of 10 ways (because the digit which would have come at ten’s place would not appear in unit’s place and so number of options available for unit’s place will be 9. Total number of 2 digit numbers with distinct digits will be 9 × 9 = 81. Example 4 How many 4 digit numbers are there in each of the following cases? (i)

All digits are distinct

(ii)

All digits are same

(iii) At least one of the digits repeats Solution (i)

If all digits are distinct, then 1000’s place can be filled in 9 ways; 100’s place can be filled in 9 ways (the digit which appears in 1000’s place won’t come in 100’s place) 10’s place can be filled in 8 ways Unit’s place can be filled in 7 ways So, total number of ways = 9 × 9 × 8 × 7 = 4536 ways. ___ ___

___

___

9 × 9 × 8 × 7 (ii)

= 4536 ways

If all digits are same, then 1000’s place cannot be 0. So, it can be filled in 9 ways; Rest of the places will necessarily have to be filled with the same digit. So, each of these will have 1 way each. Therefore, total number of ways = 9 × 1 × 1 × 1 = 9 ways.

(iii) We can approach this question by thinking that there will be many cases, that is: (a) Two digits are same and rest are different; (b) Three digits are same and rest are different; (c) All four digits are same. There will be multiple possibilities within each case. We also need to take care that in case digit zero repeats, then leftmost place cannot be filled with zero. Therefore, such an approach should be avoided. The best approach is to find total number of cases where there is no repetition of digits and subtract this from the total number of 4-digit numbers. Total number of numbers = 9000; Total number of numbers without repetition = 9 × 9 × 8 × 7 = 4536 ways. Therefore, total number of numbers where at least one of the digits repeats = 9000 – 4536 = 4464.

457

Quantitative Aptitude Simplified for CAT

Remember that in most of the questions where the word ‘at least’ is mentioned, the best approach would be to find total number of cases and subtract those cases where ‘none’ happens. For example, in this example, ‘At least one of the digits repeats’ = ‘total number of 4-digit numbers’ – ‘no digit repeats’.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to apply the Fundamental Principle of Counting in basic problems.

Example 5 If 4-digit numbers are formed using the digits 1, 2, 3, 4, 5, 6 and 7 only, then (i)

how many numbers can be formed?

(ii)

how many numbers can be formed which are divisible by 5?

(iii) how many numbers can be formed which are divisible by 4? Solution (i)

Since nothing is mentioned regarding repetition of digits, we will assume that digits can be repeated. So, ___ ___ ___ ___ 7 × 7 × 7 × 7

(ii)

= 74 = 2401.

If the number is divisible by 5, then the unit’s digit will be 5. Rest of the places can be filled in 7 ways each. So, total number of ways = 7 × 7 × 7 × 1 = 73.

(iii) For divisibility by 4, we need to look at the rightmost two digits. If unit’s digit is 2 or 6, then ten’s place digit will be all odd digits, 1, 3, 5, 7. If unit’s digit is 4, then ten’s place digit will be all even digits, 2, 4, 6. These are 11 cases. The rest of the places can be filled in 7 ways each. Therefore, total number of ways = 7 × 7 × 11 = 539. Example 6 In the previous example, how will you answer all the parts if digits are not repeated? Solution (i)

In case of no repetition, ___ ___

___ ___

7 × 6 × 5 × 4 (ii)

= 840.

If the number is divisible by 5, then the unit’s digit will be 5. The left most place can be filled in 6 ways because digit 5 has already been used in the unit’s place and digits cannot repeat. Similarly, 100’s place can be filled in 5 ways and 10’s place in 4 ways. So, ___

___ ___ ___ .

6 × 5 × 4 × 1

= 120

Therefore, total number of numbers divisible by 5 = 120. (iii) For divisibility by 4, we need to look at the rightmost two digits. If unit’s digit is 2 or 6, then ten’s place digit will be all odd digits, 1, 3, 5, 7.

458

Permutation and Combination If unit’s digit is 4, then ten’s place digit will be all even digits, 2, 4, 6. But, if 4 appears in ten’s place, that would be repetition of digits and so won’t be allowed. So, there are 10 cases. For rest of the places, ___ ___ ___ ___ 5 × 4 × 10

Therefore, total number of ways = 5 × 4 × 10 = 200. Example 7 How many 5-digit numbers are formed using the digits 1, 2, 3, 4 and 5 only where no digit repeats? Solution The unit’s place can be filled in 5 ways, ten’s place in 4 ways, 100’s place in 3 ways, and so on. Therefore, total number of ways = 5 × 4 × 3 × 2 × 1 = 5! = 120. This is akin to arranging 5 objects in 5 different positions, which is done in 5! ways. So, we can now say that number of ways of arranging r different objects in r different locations is given by r!.

Permutation and Combination Combination First, let’s understand what is combination. Let there be 5 friends: A, B, C, D and E who want to go to a restaurant, but have money with which only 3 friends can eat out. So, only 3 of these 5 friends can go to restaurant, and it can be any 3 of these 5 friends. In how many ways can we select 3 of these 5 friends who can then go to restaurant? Instead of physically counting which all friends will be selected to go to restaurant, we use the expression 5C3 to denote the number of ways of selecting 3 out of 5 friends. The value of 5C3 is

5! = 10. So, there are 10 ways in which 3 of the 5 friends can be selected. 3! (5  3)!

In general, nCr denotes the number of ways of selecting r different objects from n different objects. The value of n! nCr is given by . r !(n  r )! Please note that if, for example, we select A, B and D, it does not matter in which order we made the selection, that is whether we select A first, then B and then D or whether we select D first, then A and then B, it is counted only once. Therefore, nCr counts only selection and not arrangement of those selected. Also note that if we are selecting r objects from n objects, then it automatically means that we are rejecting n – r objects. So, number of ways of selecting r objects is same as number of ways of rejecting n – r objects. In other words, nCr = nCn–r. One can prove this by using the formula of nCr. Having understood Combinations, note that example 6, part (i) can also be solved by selecting 4 different digits from 7 different digits, which is 7C4 ways. Then, the selected 4 digits will be arranged amongst themselves in 4! ways. So, total number of ways = 7C4 × 4! = 840. Example 8 In how many ways can we draw 7 cards from a standard deck of 52 cards?

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Quantitative Aptitude Simplified for CAT

Solution The situation is equivalent to selecting 7 different objects from 52 different objects. The required number of ways is given by 52C7. Example 9 How many 7 letter words can be constructed using 26 alphabets if each word contains exactly 3 vowels? (An alphabet can be used more than once) Solution 3 places for the vowels can be chosen in 7C3 ways. For each of the three places there are 5 vowels. So, vowels can be placed in 5 × 5 × 5 = 53 ways. Remaining 4 places are filled by 21 consonants. Therefore, consonants can be arranged in 21 × 21 × 21 × 21 = 214 ways. Therefore, total number of words with exactly 3 vowels is 7C3  53  214. Example 10 A committee of 3 men and 5 women is to be formed from 7 men and 10 women. Find the number of ways in which this can be done if (i)

a particular woman is always included.

(ii)

a particular woman is always excluded.

Solution (i)

If a particular woman is always included, then we have to choose 4 women from the remaining 9 which can be done in 9C4 = 126 ways. Men can be chosen in 7C3 ways = 35 ways. So, total number of ways = 126 × 35 = 4410 ways.

(ii)

If a particular woman is always excluded, then 5 women will be chosen from remaining 9 women. So, required total number of ways of forming the committee = 9C5 × 7C3 = 4410 ways.

Example 11 In the previous example, in how many ways can the committee of 5 persons be formed so that at least one woman is chosen? Solution Number of ways of choosing at least one woman = Total number of ways – number of ways in which no woman is chosen. Therefore, required number of ways = 17C5 – 7C5. Recall that choosing at least one woman means ‘total’ – ‘none’.

Permutation Now, let’s understand what is Permutations. Let there be 5 alphabets: a, b, c, d and e. From these 5 alphabets, we need to form words using any 3 of these alphabets, irrespective of whether the words are meaningful or not. For this, we need to first select 3 alphabets out of 5, which is done in 5C3 ways. After having selected 3 alphabets, we need to arrange them in all possible orders. This effectively means that we have to allocate the 3 alphabets to 3 different positions. As seen already, number of ways of arranging 3 alphabets = 3! = 6.

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Permutation and Combination So, number of ways in which the 3 letter words can be formed = 5C3 × 3! = 10 × 6 = 60 ways. Note that 5C3 × 3! is also written as 5P3, where P refers to Permutation. So, permutation includes selection of different objects followed by arrangement of those selected objects. In general, nPr is the number of ways in which r different objects are to be chosen followed by their arrangement. nPr

= nCr × r! =

n! n!  r!  . r !(n  r )! (n  r )!

Ideally, one should avoid using nPr directly. Instead, one should prefer to use nCr × r!, that is make the selection of objects followed by arrangement of the selected objects, for ease of understanding. Example 12 Find the number of ways of picking a 1st, 2nd, and 3rd place winner in an IPL team from a group of 10 participating teams. Solution We need to select 3 teams from 10 teams, which is done in 10C3 ways. Which one of the three will get 1st place, which 2nd and which 3rd, means we need to arrange the 3 selected teams, which is done in 3! = 6 ways. So, the required number of ways = 10C3 × 3! ways. Alternatively, We can say that the first position can be chosen from 10 teams contestants in 10 ways, second position in 9 ways and third position in 8 ways. So, total number of ways = 10 × 9 × 8, which is same as 10C3 × 3!. Example 13 Find 6P4 and 25C22 Solution 6P4

=

25C22

6! 6!  = 6 × 5 × 4 × 3 = 360. (6  4)! 2! =

25! 25!  = 2300. 22!(25  22)! 22!3!

Example 14 How many five letter words (with distinct letters) can be formed from the letters of the word ‘MANGOES’? Solution In this, we have to select 5 letters from 7 letters of the word ‘MANGOES’ and arrange those letters to form all possible words. So, required number of ways = 7C5 × 5! = 2520. Example 15 How many numbers divisible by 2 and between 30000 to 90000 can be formed from the digits 3, 4, 5, 6, 7, 8, 9 with none of the digits being repeated in any number? Solution The number should be divisible by 2  4, 6, or 8 should be in unit’s place. This can be done in 3 ways with the digits given.

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Quantitative Aptitude Simplified for CAT

The number is between 30000 and 90000. Therefore, ten thousands’ place can be filled by 3, 4, 5, 6, 7 or 8 that is in 6 ways. But this may lead to repetition of digits. If 4 appears in unit’s place, then ten thousand’s place can be filled in 5 ways (3, 5, 6, 7, 8). Similarly, if 6 appears in unit’s place, then ten thousand’s place can be filled in 5 ways (3, 4, 5, 7, 8), and so on. After filling two places, the remaining three places can be filled by remaining 5 numbers in 5C3 × 3! = 60 ways. Therefore, total number of numbers that can be formed with the given digits and satisfying the given conditions = 3  5  60 = 900.

Applications of Fundamental Principle of Counting So far, we have discussed the basic concepts of ‘Fundamental Principle of Counting’, along with basic understanding of the terms ‘Permutation’ and ‘Combination’. Now, we will see varied applications of all of these.

Distribution of objects Let there be 5 balls to be distributed into 3 boxes. In how many ways can this be done? While we solve such question, the obvious question arises whether the balls (as well as boxes) are to be treated distinct from each other, or not. Let us take the case when the balls as well as boxes are distinct.

When objects are distinct There are two ways to attempt the problem. Approach 1: We can argue that first box can receive the balls in 5 ways as there are 5 balls in all. The second box can also receive the balls in 5 ways and third box can also receive in 5 ways. So, total number of ways = 5  5  5 = 53 = 125 ways. Approach 2: The other way of looking at the problem could be: first ball can go to any of the boxes in 3 ways as there are 3 boxes in all. The second ball can also go in 3 ways, and so on with each of the ball. So, total number of ways of distributing the balls to boxes = 3  3  3  3  3 = 35 = 243. The first approach is incorrect. We should analyze the reason for the former approach being incorrect. In the first approach, the first box can receive balls in 5 ways. Once one of the 5 balls goes into the first box, we are left with 4 balls and so the second box is left with only 4 choices, and not 5 choices to choose from. So, the second box cannot receive the balls in 5 ways. There is another fundamental problem in this approach. By saying that the first box can receive balls in 5 ways, we are allowing only 1 ball in 1 box, whereas we can have more than 1 ball in a particular box. So, the number of balls which go into second box depends upon the number of balls which go into first box. When the number of ways of performing an event is dependent upon the number of ways of performing a previous event, then the fundamental principle of counting cannot be applied. In the second approach, the events are independent, that is the number of ways in which second ball can go to any of the boxes does not depend upon the number of ways in which the first ball goes to any of the boxes. Therefore, we can apply the fundamental principle of counting. Therefore, in such cases when the things (here, balls) to be distributed are distinct, we should be looking at the situation from the point of view of balls, that is the thing to be distributed. We can conclude that whenever we have to distribute r distinct objects into n distinct containers, the number of ways = n  n  n  …  n (n appears r times) = nr. Note that in the above case, there are cases where multiple balls are in a single box and hence some of the boxes may be empty. Also note that the above case is valid only in case the balls are distinct.

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Permutation and Combination What is the number of ways of distributing 5 distinct balls to 3 distinct boxes such that no box is empty? We should realize that there are 3 boxes and so either of the three cases exists: 

All the boxes have at least one ball, that is no box is empty;



Exactly one box is empty;



Exactly two boxes are empty

All the three boxes cannot be empty! So, from 35 ways, we should subtract those ways where at least one of the boxes is empty. Required number of ways = 35 – (two boxes are empty) – (one box is empty). When two boxes are empty, exactly one box is filled with all the 5 balls, and this is possible in 3 ways: put all the balls in first box, or second box or third box. When one box is empty, we need to select the box which is to be kept empty, and which is done in 3C1 = 3 ways. At the same time, the rest of the two boxes, say A and B must have at least one ball. Since there are 5 balls, first ball can go in 2 ways, second also in 2 ways, and so on. Total number of ways = 25. In these 25 ways, there is one way in which all the balls go in box A leaving B empty and there is another way in which all the balls go in B leaving A empty. So, required number of ways in which exactly one box is empty = 3C1  (25 – 2). So, number of ways of distributing the balls to boxes such that no box is empty = 35 – 3 – 3C1  (25 – 2). What is the number of ways of distributing 5 distinct balls to 3 distinct boxes such that every box has exactly one ball? In this case, only 3 balls are used as there are only 3 boxes. We need to choose 3 balls out of 5 balls, which will get distributed in the boxes. The number of ways of choosing 3 balls = 5C3. The selected balls can be distributed to the 3 boxes in 3! ways. So, total number of ways = 5C3  3! = 60 ways. The same question can also be solved in this manner: first box can receive balls in 5 ways, second box in 4 ways and third box in 3 ways. So, total number of ways = 5  4  3 = 60 ways.

EXPERT SPEAK Scan this QR Code to watch a video that explains Distribution of dissimilar objects under various situations.

When objects to be distributed are similar In the above cases, we have considered all the objects to be distributed as distinct from each other. What will happen when all the objects become similar to each other? In such cases, we cannot follow the above mentioned approach. In how many ways can we distribute 7 similar balls in 4 different boxes? Here, two cases arise 

Some of the boxes may be empty



No box is empty

In case some of the boxes may be empty, the number of ways is given by n+r–1Cr–1, where n is the number of objects to be distributed and r is the number of boxes, where the objects will get distributed. So, required number of ways of distributing 7 similar balls into 4 distinct boxes = 7+4–1C4–1 = 10C3 = 120.

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Note that in this case, some boxes may be empty. This also includes cases where no box is empty. In case we want to find out number of ways of distributing these balls so that no box is empty, then we use the formula n–1Cr–1 = 7–1C4–1 = 6C3 = 20. Therefore, number of ways of distributing 7 similar balls into 4 distinct boxes such that at least one box is empty = 10C3 – 6C3 = 120 – 20 = 100. Example 16 Find the number of ways of distributing 15 similar balls into 5 distinct boxes such that exactly three boxes are filled. Solution First of all, we need to select the three boxes which are to be filled. The number of ways to select the two boxes = 5C3 = 10. In these 3 boxes, balls should be distributed so that none of these three boxes remains empty, the number of ways for which is = n–1Cr–1 = 15–1C3–1 = 14C2 = 91. Therefore, total number of ways = 10  91 = 910. Example 17 Find number of ways of distributing 15 similar balls into 5 distinct boxes such that first box contains at least one ball, second box contains at least 2 balls, third box contains at least 3 balls, and rest of the boxes may contain any number of balls. Solution From 15 similar balls, 1 ball can be chosen in 1 way (not 15C1 ways). Put this ball in the first box. Now, choose 2 balls from remaining 14 balls, for which number of ways is again 1. Put these two balls in second box. Similarly, choose 3 balls from remaining 12 balls and put in third box, which is done in 1 way. Now, remaining 9 balls can go any of the boxes. Number of ways for this = n+r–1Cr–1 = 9+5–1C5–1 = 13C4.

Number of whole number, and natural number, solutions of equations Let us learn to find the number of whole number solutions of the equation: v + w + x + y + z = 15. Also find the number of natural number solutions of the same equation. Some of the whole number solutions will be (0, 1, 2, 3, 9), (1, 3, 2, 5, 4), (0, 0, 0, 3, 12), and so on. Realize that this is same as distributing 15 similar balls into 5 different boxes such that boxes may be empty. So, number of whole number solutions = n+r–1Cr–1, where n = value of the constant = 15 and r = number of variables = 3. Number of whole number solutions = 15+5–1C5–1 = 19C4. In case of natural number solutions, none of the variables can be 0. This is same as distributing 15 similar balls such that no box is empty. So, number of natural number solutions = n–1Cr–1 = 15–1C5–1 = 14C4. Example 18 Find the number of solutions of the equation w + x + y + z = 12, where w ≥ 2, x ≥ 1, y, z ≥ 0. Solution Distribute 2 balls to box w, 1 ball to box x. Now, number of ways of distributing 12 – 3 = 9 balls in the 4 boxes is same as number of whole number solutions = 9+4–1C4–1 = 12C3. Example 19 Find the number of terms in the expansion of (a + b + c + d)20.

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Permutation and Combination Solution Number of terms in the expansion is given by the formula n+r–1Cr–1 where n = value of power = 20 and r = number of variables = 4. Therefore, number of terms = 23C3. When each object is distributed to exactly one container When each object is distributed to exactly one container, then the number of objects and containers becomes same. Let’s understand through the following example. Example 20 Let there be 5 men, each owning 1 TV of different brand. (i)

In how many ways can the TVs be distributed to the men, irrespective of whether the TV goes to the right owner or not?

(ii)

In how many ways can the TVs be distributed to the men, such that each TV goes to the right owner only?

(iii) In how many ways can the TV be distributed to the men, such that at least one of the TVs goes to wrong owner? Solution (i)

This situation is similar to the one where we have to arrange 5 objects in a straight line fashion, and so it is given by 5! = 120.

(ii)

First TV can be given to the right owner in only 1 way. Similarly, second TV can be given to the right owner in 1 way, and so on. So, total number of ways = 1 × 1 × 1 × 1 × 1 = 1 way only.

(iii) The question effectively means = (Total number of ways) – (all the TVs go to respective right owners) = 5! – 1 = 119. Example 21 In the previous example, (i)

In how many ways can the TVs be distributed to the men, such that each TV goes to wrong owner only, that is no TV goes to its respective right owner?

(ii)

In how many ways can the TVs be distributed to the men, such that exactly two TVs go to the respective right owner?

(iii) In how many ways can the TVs be distributed to the men, such that at least one TV goes to the right owner? Solution (i)

This question is done by using the formula of derangements, that is 1 1 1 1 n!  1     ...  (1) n  where n is the number of objects all of which go to wrong 1! 2! 3! n!   locations. Here, all the 5 TVs are expected to go to wrong owners. So, required number of ways = 1 1 1 1 1 5!  1       = 44 ways. 1! 2! 3! 4! 5!  

(ii)

Select those two TVs first, which will go to the right owner. This is done in 5C2 = 10 ways. The number of ways to give these 2 TVs to right owners = 1. Each of the rest of the 3 TVs go to wrong owners in 1 1 1 3!  1     = 2 ways. So, total number of ways = 5C2 × 1 × 2 = 20 ways. 1! 2! 3!  

(iii) Required number of ways = (Total number of ways) – (Number of ways in which all the TVs go to wrong owners) = 5! – 44 = 120 – 44 = 76 ways.

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Arrangement of distinct objects with conditions Let there be 7 boys and 4 girls to be seated in a straight line fashion. The total number of ways of doing this is 11! because we have 10 objects each of which is distinct and hence it is a case of arrangement of those 11 objects. Example 22 Let there be 6 boys and 4 girls to be seated in a straight line fashion. (i)

In how many ways can they be seated so that all the girls sit together?

(ii)

In how many ways can they be seated so that no two girls sit together?

(iii) In how many ways can they be seated so that girls sit alternately? Solution (i)

Consider all the girls to be one object or one entity. Then, along with 6 boys, there are 7 entities, which can be arranged in 7! ways. Further, within each case of 7!, the 4 girls can arrange among themselves in 4! ways. Therefore, required number of ways = 7! × 4!.

(ii)

In cases where no two girls sit together, we need to make the boys be seated. All the boys can be seated in 6! ways. Around these boys, we have 7 vacant seats, where the 4 girls need to be seated. We first choose 4 vacant positions from 7 available positions in 7C4 ways. Now, girls will arrange among themselves in 4! ways. Hence total number of ways = 6! × 7C4 × 4!. Alternatively, after boys are seated in 6! ways, we need to seat 4 girls in 7 vacant positions. First girl will have 7 options to be seated, second will have 6 options, third will have 5 and fourth will have 4 positions. So, number of ways for all to be seated = 6! × 7 × 6 × 5 × 4, which is same as 6! × 7C4 × 4!

(iii) For girls to be seated alternately, boys have to seated first, for which number of ways is 6!, as shown below: a

B1

b

B2

c

B3

d

B4

e

B5

f

B6 g

Also shown are vacant positions a, b, c, d, e, f and g. If girls sit alternately, they will sit in ‘abcd’ or ‘bcde’ or ‘cdef’ or ‘defg’, which means 4 positions. In each position, the girls will arrange among themselves in 4! ways. So, required number of ways = 6! × 4 × 4!.

Arrangement of objects, some of which are similar Let there be a word MATHS, which consists of 5 letters, all of which are distinct. The number of words which can be formed using all the letters of this word = number of arrangements of 5 distinct objects = 5! = 120. Let there be another word MANAGE, which consists of 6 letters, 2 of which are similar. The number of words which can be formed using all the letters of this word will not be 6! = 720, because 2 of the letters: A, A, are similar and hence their arrangement 2! needs to be discounted by dividing 6! ways by 2!. So, effective number of different words =

6! = 360. 2!

In general, when there are n objects, p of them are of one kind, another q of them are of one kind, another r of them are of one kind, and so on, the total number of arrangements of all the n objects is given by:

n! . p!q!r !... Example 23 In how many ways can 4 coins of Rs.2, 3 coins of Rs.5 and 5 coins of Rs.10 be stacked one above the other?

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Permutation and Combination Solution It is same as arranging 12 coins, 4 of one type, 3 of another type and 5 of another type. So, number of ways =

12! . 4!3!5!

Example 24 In how many ways can the letters of the word ‘MANAGEMENT’ be arranged such that (i)

the two A’s never appear together?

(ii)

all the vowels come together?

Solution (i)

The situation of no two A’s appear together is similar to the situation of no two girls sit together (as discussed in example 21). So, first we will arrange rest of the letters. 8! Number of ways to arrange rest of the letters (M N G E M E N T) = . 2!2!2! The two A’s can be placed in 9 vacant positions in 9C2 ways. So, required total number of ways 8! = × 9C2. 2!2!2!

(ii)

Treat all the vowels as one entity. So, there are 6 + 1 = 7 entities, which can be arranged in 7! 4! . The 4 vowels A A E E can be arranged among themselves in ways. So, total number of = 2!2! 2!2! 7! 4! ways =  . 2!2! 2!2!

Example 25 In how many ways can 15 identical silver spoons and 12 identical golden spoons be arranged in a row so that no 2 golden spoon are together? Solution First arrange 15 silver spoons in a row. This can be done in 1 way (since they are identical). Now there are 16 places for the 12 identical golden spoons. The golden spoons can be placed in with golden spoons. So, required number of ways = 16C12.

16C12

ways

Selection of objects when any number of them can be selected Let there be 7 distinct objects. In how many ways can the objects be selected? Here, it is not clear as to how many objects can be selected at a time. In such cases, we will take all possible cases including “none” as well as “all” being selected. The number of ways of selecting no object = 7C0; The number of ways of selecting one object = 7C1; The number of ways of selecting two objects = 7C2; The number of ways of selecting three objects = 7C3; The number of ways of selecting four objects = 7C4; The number of ways of selecting five objects = 7C5; The number of ways of selecting six objects = 7C6;

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Quantitative Aptitude Simplified for CAT

The number of ways of selecting seven objects = 7C7. So, total number of ways of making the selection = 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7. From binomial theorem, we know that the above expression is also equal to 27. We can understand the answer of 27 logically also. Assume that there is a box in which we can any number of objects from a box having 7 distinct objects given. The first object may or may not be there. So, the number of ways to choose the first object is 2. Similarly, the number of ways to choose the second object is also 2, and so on. Using fundamental principle of counting, the total number of ways = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27. Example 26 (i)

Let there be a set A = {1, 2, 3, 4, 5}. How many subsets are there in all?

(ii)

How many subsets are there where the elements 2, 3 and 5 are surely there?

(iii) How many subsets are there where the elements 2, 3 and 5 are not there? Solution (i)

This is similar to making a set any number of elements from the given 5 elements. So, the element 1 may be there in the set or may not be there. So, the number of ways to choose the element 1 is 2. Similarly, the number of ways to choose 2 is also 2, and so on. So, total number of ways = 2 × 2 × 2 × 2 × 2 = 25. As we have seen in the chapter: Set Theory, that the number of subsets is given by the standard formula 2n, where n is the number of elements of the given set.

(ii)

The number of ways to choose element 2 is 1, that is just choose it! Similarly, the number of ways to choose 3 or 5 is 1 each. So, total number of ways = 2 × 1 × 1 × 2 × 1 = 22 = 4.

(iii) The number of ways to not choose element 2 is 1, that is just choose it! Similarly, the number of ways to not choose 3 or 5 is 1 each. So, total number of ways = 2 × 1 × 1 × 2 × 1 = 22 = 4. Example 27 Let there be 5 distinct apples, 4 distinct oranges and 3 distinct bananas. (i)

In how many ways can the fruits be selected?

(ii)

In how many ways can the fruits be selected so that at least one fruit is selected?

(iii) In how many ways can the fruits be selected so that at least one fruit of each type is selected? (iv) In how many ways can the fruits be selected so that at least one apple is selected? Solution (i)

Since there are 5 + 4 + 3 = 12 fruits, the number of ways = 212.

(ii)

Since 212 includes one case where none of the fruits is selected, the required number of ways = 212 – 1.

(iii) Number of ways to select at least one apple = 25 – 1. Number of ways to select at least one orange = 24 – 1. Number of ways to select at least one banana = 23 – 1. So, required number of ways = (25 – 1)(24 – 1)(23 – 1). (iv) To select at least one apple, number of ways = 25 – 1. The rest of the fruits can be selected in 24 × 23 ways. So, total number of ways = (25 – 1)(24 × 23) = (25 – 1)27. So far, we have considered cases where the objects being selected are distinct objects. What will happen if the objects being selected are similar? Let there be 7 objects, and all of them are similar. Whether we select first object or second one makes no difference. The difference will lie in selecting different number of objects. So, one may select no object or 1

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Permutation and Combination object or 2 objects or 3 objects, and so on up to selecting all the objects. So, number of ways to make the selection = 8. Example 28 Let there be 5 similar apples, 4 similar oranges and 3 similar bananas. (i)

In how many ways can the fruits be selected?

(ii)

In how many ways can the fruits be selected so that at least one fruit is selected?

(iii) In how many ways can the fruits be selected so that at least one fruit of each type is selected? (iv) In how many ways can the fruits be selected so that at least one apple is selected? Solution (i)

Number of ways of choosing apples = 6; that of choosing oranges = 5; that of choosing bananas = 4. So, required number of ways = 6 × 5 × 4 = 120.

(ii)

Required number of ways = 120 – 1 = 119.

(iii) Number of ways to select at least one apple = 5. Number of ways to select at least one orange = 4. Number of ways to select at least one banana = 3. So, required number of ways = 5 × 4 × 3 = 60. (iv) Number of ways of choosing at least one apple = 5. Number of ways of choosing orange = 4 + 1 = 5 and number of ways of choosing banana = 3 + 1 = 4. So, required number of ways = 5 × 5 × 4 = 100.

Circular Arrangements Let there be 4 persons: A, B, C and D who are required to be seated around a circular table, that is who are required to be arranged in a circular fashion. Imagine that there are 4 chairs around the given circular table. The first person A can be seated on any of the four chairs in only 1 way, and not 4 ways. This is so because all the chairs are similar and are not numbered or colored differently to become distinguishable. Once A sits on any of the chairs, the rest of the three persons can be seated with respect to A in 3! ways = 6 ways. For greater clarity, refer to the following figures:

The 2 arrangements shown are not different from each other. In both the diagrams, to the right of A, we have D, to the left we have B and opposite to A we have C. So, a different sitting position of A does not tantamount to a different arrangement! In general, n persons can be arranged in a circular manner in (n – 1)! ways. Example 29 (i)

In how many ways can 6 Men and 4 Women sit around a circular table?

(ii)

In how many ways can 6 Men and 4 Women sit around a circular table so that all the women always sit together?

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Quantitative Aptitude Simplified for CAT

(iii) In how many ways can 6 Men and 4 Women sit around a circular table so that no two women sit together? (iv) In how many ways can 6 Men and 4 Women sit around a circular table so that women sit alternately? Solution (i)

We have total 10 people who can be seated around a circular table in 9! ways.

(ii)

All the women can be treated as a single entity. So, total number of entities = 6 + 1 = 7. So, number of ways of arranging 7 entities = 6!. Further, in each of these 6! arrangements, the women can arrange amongst themselves in 4! ways. Therefore, total number of ways = 6! × 4!.

(iii) First, men would be seated in 5! ways. Now, there will be 6 vacant positions and 4 women can be seated in these positions in 6 × 5 × 4 × 3 = 360. Therefore, total number of ways = 5! × 360. (iv) Again, men would sit first in 5! ways. Let 6 vacant positions are numbered as A, B, C, D, E and F, and women are named as P, Q, R and S. Now, women will sit in A B C D positions or B C D E positions, and so on, till F A B C. These are 6 different positions. So, total number of ways = 5! × 6 × 4!. Note that when garlands are formed using different flowers or necklace is formed using beads, then we notice a slight difference in the number of ways compared to circular arrangement. Refer to the following example. Example 30 How many different necklace can be made using 5 different beads? Solution Garland is circular and hence number of different garlands that can be made appears to be 4!. However, in such cases, we need to divide 4! by 2. To understand this, refer to the following figure.

The 4! arrangements counts these two figures as distinct, but they are mirror images of each other. Note that in case of garlands, if we look at the first figure from the other side, we will get its mirror image arrangement, and hence should not be treated as a different arrangement. That is why we should do half of 4!. So, required 5! . number of arrangements = 2 In general, the number of ways in which n different beads can form a necklace = ways in which n flowers of different colors can form a garland =

(n  1)! . Similarly, number of 2

(n  1)! . 2

Applications to Geometrical Figures Here, we will study application of the principles of permutation and combination to geometrical shapes.

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Permutation and Combination Example 31 Let there be 15 points in 2D space all of which are non-collinear. We can join any two of them to form a line. (i)

How many such lines can be made in all?

(ii)

How many triangles can be made in all?

Solution (i)

To form a line, we need 2 points. Each selection of 2 points leads to each line. So, number of lines that can be made = number of ways in which 2 points can be selected = 15C2.

(ii)

To form a triangle, we need 3 points. So, required number of triangles = 15C3.

Example 32 In the previous example, if 6 of the 15 points are collinear, then (i)

how many lines can be formed?

(ii)

how many triangles can be formed?

Solution (i)

We will begin by assuming that all the points are non-collinear and hence number of lines possible = 15C2. Out of these, 6 points are collinear and hence they cannot form 6C2 lines. However, these 6 points can be joined to form a single line. So, required number of lines = 15C2 – 6C2 + 1.

(ii)

Similarly, the number of triangles = 15C3 – 6C3. The collinear points can never form a triangle.

Example 33 20 points, no 4 of which are coplanar, are given in space. How many tetrahedrons can they form? Solution Since no 4 points are coplanar, every set of 4 points determine a unique tetrahedron  Number of tetrahedrons = 20C4. Example 34 What is the number of diagonals in a decagon? Solution A decagon can be formed by joining 10 non-collinear points. Total number of lines that can be formed by joining 10 non-collinear points = 10C2. Out of these, 10 form the sides, and the rest are diagonals. So, number of diagonals = 10C2 – 10 = 35. In general, in a polygon of n sides, total number of diagonals = nC2 – n =

n(n  1) n(n  3) n . 2 2

In a polygon of n sides, from 1 vertex, we can draw (n – 3) diagonals. Since there are n vertices, total number of diagonals = n(n – 3). But, diagonal from vertex A to some vertex D is same as that from D to A. So, number n(n  3) of different diagonals = . 2 Example 35 In the above mentioned decagon, (i) how many triangles can be formed whose vertices are the vertices of the polygon? (ii) how many of these triangles do not share any side with the polygon?

471

Quantitative Aptitude Simplified for CAT

Solution (i)

Number of triangles = number of ways of choosing 3 vertices from 10 vertices = 10C3.

(ii)

Total number of triangles = 10C3 = 120. Out of these, some triangles share 1 side with the decagon, some share 2 sides with the decagon and some do not share any side with the decagon. There won’t be any triangle sharing all the three sides with the decagon.

Let us count number of triangles sharing 1 side with the decagon. If one of the sides, say AB, is shared, then the third vertex can be chosen in 6 ways. Since the shared side can be chosen in 10 ways, total number of triangles sharing 1 side with the decagon = 10 × 6 = 60. To count number of triangles sharing 2 sides, we observe that if the vertices are A, B, C, D, …, J, then various triangles possible are: ABC, BCD, CDE, …, JAB. These are 10 triangles. Therefore, number of triangles sharing 2 sides with the decagon = 10. So, total number of triangles which do not share any side with the decagon = 120 – 60 – 10 = 50.

Group Formation Let there be 12 students in a class and we have to form 3 groups of 4 students each. We will first choose 4 students from 12 students in 12C4 ways. We will then choose another set of 4 students from remaining 8 students in 8C4 ways. Finally, we will choose last set of 4 students from leftover 4 students in 4C4 ways. Total number of ways = 12C4 × 8C4 × 4C4. If the groups are ABCD, EFGH and PQRS, then, in 12C4 × 8C4 × 4C4, we have counted the order of forming the groups. For example, in one of the cases, 12C4 forms the group ABCD, 8C4 forms the group EFGH and 4C4 forms the group PQRS. In another case, which is taken as a different case, 12C4 forms the group EFGH, 8C4 forms the group PQRS and 4C4 forms the group ABCD. But, in the end, the groups formed are these only: ABCD, EFGH and PQRS. Since the order in which groups are formed is not to be counted, we need to discount this superfluous counting by dividing 12C4 × 8C4 × 4C4 by 3!. 12

Therefore, number of ways of forming 3 groups of 4 persons each =

C 4  8 C 4  4 C4 . 3!

Note that division by 3! is required if group size is same. If the group size is not same, then we would not divide by 3! or r! in general. Example 36 In a pack of 52 playing cards, in how many ways can we form 4 groups of 13 cards each? Solution 52

Required number of ways =

C13  39 C13  26 C13  13 C13 4!

Example 37 In the previous example, in how many ways can the 4 groups of 13 cards each be distributed to 4 players? Solution Distribution of groups of cards can be done in 4! ways. So, required number of ways 52

=

C13  39 C13  26 C13  13 C13 × 4! = 52C13 × 39C13 × 26C13 × 13C13. 4!

472

Permutation and Combination Example 38 From a group of 12 students, in how many ways can we form 2 groups having 5 and 7 students? Solution Groups size is different. So, required number of ways = 12C5 × 7C7. Example 39 From a group of 12 students, in how many ways can we form 5 groups, where 3 groups have 2 students each and the 2 groups have 3 students each? Solution 12

Number of ways =

C 2  10 C2  8 C2  6 C 3  3 C 3 . 3!2!

473

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is A.

200

B.

216

C.

235

D.

256

assigned one task. In how many ways can the assignment be done? A.

5.

Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S? A.

228

B.

216

C.

294

D.

192

5

B.

10

C.

9

D.

15

D.

360

E.

716

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos? A.

17

B.

16

C.

18

D.

15

E.

19

Let S be the set of all pairs (i, j) where 1 i < j n, and n 4. Any two distinct members of S are called "friends" if they have one constituent of the pairs in common and "enemies" otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1, 2) and (2, 3) are also friends, but (1, 4) and (2, 3) are enemies. CAT 2007 6.

For general n, how many enemies will each member of S have? A. B. C.

CAT 2005 4.

192

Directions for questions 6 and 7: Read the given information and answer the questions that follow:

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose? A.

180

C.

CAT 2007

CAT 2005 3.

B.

CAT 2006

CAT 2005 2.

144

D.

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be

E.

474

n–3 1 2 (n  3n  2) 2 2n – 7 1 2 (n  5n  6) 2 1 2 (n  7n  14) 2

Permutation and Combination 7.

For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members?

Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

1 2 (n  5n  8) 2

A.

60

B.

75

B.

2n – 6

C.

45

C.

1 n(n – 3) 2

D.

90

E.

72

A.

D. E. 8.

9.

n–3 1 2 (n  7n  16) 2

CAT 2008 10. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is

In a tournament, there are n teams T1, T2, ...,Tn, with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common: T1 and T2, T2 and T3, T3 and T4, …., Tn–1 and Tn, and Tn and T1. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together? A.

n(k – 1)

B.

k(n – 1)

C.

n(k – 2)

D.

k(n – 2)

E.

(n – 1)(k – 1)

A.

1170

B.

630

C.

792

D.

1200

E.

936 CAT 2008

11. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed? CAT 2007

A.

499

B.

500

Directions for questions 9 and 10: Read the given information and answer the questions that follow:

C.

375

D.

376

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

E.

501

A

CAT 2008 12. What is the number of distinct terms in the expansion of (a + b + c)20?

C

D P

A.

231

B.

253

C.

242

D.

210

E.

228 CAT 2008

Directions for questions 13 and 14: The question is followed by two statements A and B. Indicate your responses based on the following directives:

B

475

Quantitative Aptitude Simplified for CAT

14. If the number of players, say n, in the first round was between 65 and 128, then what is the exact value of n?

Mark A] if question can be answered from A alone but not from B alone. Mark B] if question can be answered from B alone but not from A alone. Mark C] if question can be answered from A alone as well as from B alone.

A.

Exactly one player received a bye in the entire tournament.

B.

One player received a bye while moving on to the fourth round from third round.

Mark D] if question can be answered from A and B together, but not from any of them alone.

CAT 2008 15. Suppose n is an integer such that the sum of the digits of n is 2, and 1010 < n < 1011. The number of different values for n is

Mark E] if question cannot be answered even from A and B together. In a single elimination tournament, any player is eliminated with a single loss. The tournament is played in multiple rounds subject to the following rules:

A.

11

B.

10

C.

9

(a) If the number of players, say n, in any round is even, then the players are grouped n pairs. in to 2

D.

8 CAT 2004

16. In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

The players in each pair play a match against each other and the winner moves on to the next round. (b) If the number of players, say n, in any round is odd, then one of them is given a bye, that is, he automatically moves on to the next round. The remaining (n − 1) players are (n  1) grouped into pairs. The players in 2 each pair play a match against each other and the winner moves on to the next round. No player gets more than one bye in the entire tournament.

B

A

n players move on 2 to the next round while if n is odd, then (n  1) players move on to the next round. 2 The process is continued till the final round, which obviously is played between two players. The winner in the final round is the champion of the tournament.

Thus, if n is even, then

The entry list for the tournament consists of 83 players.

B.

The champion received one bye.

15

B.

56

C.

120

D.

336 CAT 2004

17. A new flag is to be designed with six vertical stripes using some or all of the colors yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same color is

13. What is the number of matches played by the champion? A.

A.

A.

12 × 81

B.

16 × 192

C.

20 × 125

D.

24 × 216

CAT 2008 CAT 2004

476

Permutation and Combination 18. There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such that every two towns are connected with three direct lines if they belong to the same zone, and with only one direct line otherwise. How many direct telephone lines are required? A.

72

B.

90

C.

96

D.

144

B.

78

C.

71

D.

69

d)

If the second letter is n, then the third letter is e or u

e)

If the second letter is p, then the third letter is same as the first letter.

21. How many strings of letters can possibly be formed using the above rules?

19. An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, … , 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion when read upside down for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise? 80

If the second letter is m, then the third letter is any vowel, which is different from the first letter.

CAT 2003 (R)

CAT 2003 (R)

A.

c)

A.

40

B.

45

C.

30

D.

35

22. How many strings of letters can possibly be formed using the above rules such that the third letter of the string is e? A.

8

B.

9

C.

10

D.

11

23 There are two parallel lines with 100 points on each. How many triangles can be formed using these 200 points? CAT 2003 (R)

A.

10000

20. Using only 2, 5, 10, 25, and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 68 paise, and Re 1.01 to the persons?

B.

200C3

C.

990000

D.

None of these

A.

19

B.

20

C.

17

D.

18

CAT 2012 24. How many words can be formed from the word TRIGONOMETRIC so that no two vowels are together? CAT 2003 (R)

Directions for questions 21 and 22: Answer the questions on the basis of the information given below: A string of three English letters is formed as per the following rules: a)

The first letter is any vowel.

b)

The second letter is m, n or p.

A.

8!9! (2! ) 4 4!

B.

8!9! (2! )2 4!

C.

8! 5! (2! ) 4

D.

8!5! (2! )4 4!

CAT 2013

477

Quantitative Aptitude Simplified for CAT

25. Find the number of triangles which can be formed from 8 non-collinear points such that no more than 1 vertex is common between any of those triangles? A.

56

B.

24

C.

7

D.

15

30. There are two boxes, each containing 100 tokens numbered 1 to 100. One token each is taken from each of the boxes. In how many cases the number appearing on token taken from first box is more than that from second box?

CAT 2013 26. In how many ways can 5 different books be distributed to A, B and C such that A and B each must get at least one book? A.

120

B.

180

C.

160

D.

150

A.

4950

B.

9900

C.

10000

D.

4500 CAT 2016

31. A playschool contains 4 boys and y girls. On every Wednesday during winter, five students, of which at least three are boys, go to Zoological Garden, a different group being sent every week. At the Zoological Garden, each boy in the group is given a ball. If the total number of balls distributed is 368, then the value of y is

CAT 2014 27. In how many ways can one pay Rs.100 using denominations of Rs.2 and Rs.5 only?

A.

5

B.

6

A.

11

C.

7

B.

10

D.

8

C.

9

D.

20

IIFT 2016 32. Which of the following regarding arrangement of ‘RIYADH’ is/are true:

CAT 2015 28. How many arrangements of the word AMDABAD are there such that at least once A is followed by D? A.

420

B.

210

C.

378

D.

380

29. There are 4 points A, B, C and D on a circle. In how many ways can the 4 arcs AB, BC, CD and DA be colored by 3 colors so that no two consecutive arcs are colored by same color? 24

B.

60

C.

120

D.

None of these

i.

Two vowels can be arranged together in 120 ways

ii.

Vowels do not occur together in 240 ways

Which of the above statements are true?

CAT 2016

A.

statements the word

A.

Statement (i) only

B.

Statement (ii) only

C.

Both statements (i) and (ii)

D.

None of the above IIFT 2016

33. A reputed paint company plans to award prizes to its top three salespersons, with the highest prize going to the top salesperson, the next highest prize to the next salesperson and a smaller prize to the thirdranking salesperson. If the company has 15 salespersons, how many different

CAT 2016

478

Permutation and Combination arrangements of winners are possible (Assume there are no ties)?

having at least one of their digits repeated is:

A.

1728

A.

98185600

B.

2730

B.

97428800

C.

3856

C.

100000000

D.

1320

D.

None of the above

IIFT 2016

IIFT 2014

34. In an MBA entrance examination, a minimum is to be secured in each of the 6 sections to qualify the cut-offs. In how many ways can a candidate fail to secure the cutoffs? A.

60

B.

61

C.

62

D.

63

38. In an Engineering College in Pune, 8 males and 7 females have appeared for Student Cultural Committee selection process. 3 males and 4 females are to be selected. The total number of ways in which the committee can be formed, given that Mr. Raj is not to be included in the committee if Ms. Rani is selected, is: A.

IIFT 2016 35. During the essay writing stage of MBA admission process in a reputed B-School, each group consists of 10 students. In one such group, two students are batch mates from the same IIT department. Assuming that the students are sitting in a row, the number of ways in which the students can sit so that the two batch mates are not sitting next to each other, is:

1960

B.

2840

C.

1540

D.

None of the above IIFT 2014

39. Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels? A.

700

B.

504000

A.

3540340

C.

3528000

B.

2874590

D.

7056000

C.

2903040

D.

None of the above

IIFT 2013 40. In a sports meet for senior citizens organized by the Rotary Club in Kolkata, 9 married couples participated in Table Tennis mixed double event. The number of ways in which the mixed double team can be made, so that no husband and wife play in the same set, is

IIFT 2015 36. In the board meeting of a FMCG Company, everybody present in the meeting shakes hand with everybody else. If the total number of handshakes is 78, the number of members who attended the board meeting is: A.

7

B.

9

C.

11

D.

13

A.

1512

B.

1240

C.

960

D.

640 IIFT 2013

41. A student is required to answer 6 out of 10 questions in an examination. The questions are divided into two groups, each containing 5 questions. She is not allowed to attempt

IIFT 2015 37. The total number of eight-digit landline telephone numbers that can be formed

479

Quantitative Aptitude Simplified for CAT

more than 4 questions from each group. The number of different ways in which the student can choose the 6 questions is A.

100

B.

160

C.

200

D.

280

C.

32

D.

84

41

E.

None of the above.

A.

7 14

D.

44. Sara has just joined Facebook. She has 5 friends. Each of her five friends has twentyfive friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?

42. Eight points lie on the circumference of a circle. The difference between the number of triangles and the number of quadrilaterals that can be formed by connecting these points is B.

21

XAT 2017

IIFT 2012

A.

C.

≥ 105

B.

≤ 123

C.

< 125

D.

≥ 100 and ≤ 125

E.

≥ 105 and ≤ 123 XAT 2013

IIFT 2012

45. How many whole numbers between 100 and 800 contain the digit 2?

43. An institute has 5 departments and each department has 50 students. If students are picked up randomly from all 5 departments to form a committee, what should be the minimum number of students in the committee so that at least one department should have representation of minimum 5 students? A.

11

B.

15

A.

200

B.

214

C.

220

D.

240

E.

248 XAT 2013

ANSWER KEY 1.

A

2.

B

3.

C

4.

A

5.

9.

D

10. A

11. D

12. A

13. D

14. D

15. A

16. B

17. A

18. B

19. C

20. A

21. D

22. C

23. C

24. A

25. C

26. B

27. A

28. A

29. D

30. A

31. D

32. D

33. B

34. D

35. C

36. D

37. A

38. C

39. C

40. A

41. C

42. B

43. C

44. B

45. B

480

C

6.

D

7.

D

8.

A

Permutation and Combination

ANSWERS AND EXPLANATIONS 1.

There are 3 Frenchmen (F1, F2 and F3) and 3 Englishmen (E1, E2 and E3). Let’s assume that E3 knows French. We can arrange these 6 men in the following way. E1 E2 E3 F1 F2 F3 The message will be conveyed from E1 to E2, E2 to E3 and so on till it reaches F3 which means five calls would be made. In the last call, that is from F2 to F3, F2 will tell his secret to F3 and in the same call F3 will tell his secret to F2 too. F2 will then call up F1 to convey F3’s secret, F1 will call E3 and so on till it reaches E1 which means four calls. Therefore, total of 9 calls would be made.

Answer: A Explanation: Let number of girls = x; let number of boys = y. There are 45 game in which both the players were girls. Therefore, xC2 = 45 x! x ( x  1)   45  x = 10. 2!( x  2)! 2 There are 190 games in which both the players were boys. Therefore, yC2 = 190  y(y – 1) = 380  y = 20. The total number of games in which one player was a boy and the other was a girl = 10 × 20 = 200.

2.

4.

Explanation: Task 2 can be done in 2 ways (either person 3 or 4). Task 1 can be done in 3 ways (either “3 or 4”, or 5, or 6) The remaining 4 tasks can be assigned to remaining 4 persons in 4! = 24 ways. Total number of ways = 2  3  24 = 144.

Answer: B Explanation: Let the 5-digit number be a b c d e Here, a, c and e are odd positions. If 5 occurs at e, then either a is filled with odd digit or c is filled with odd digit. If a is filled with odd digit, this is done in 2 ways. In that case, c cannot have odd digit. So, c must be filled with even digit, which is done in 2 ways. Now, remaining positions b and d can be filled in 2 and 1 ways respectively. So, total number of ways = 2 × 2 × 2 × 1 = 8. Similarly, if c were filled with odd digit, then number of ways is again 8. So, 5 appears in unit’s digit in 16 ways. Similarly, 1 also appears in unit’s place in 16 ways, and 3 appears in unit’s place in 16 ways. When 2 appears at e, then a and c must have odd digits, which can be filled in 3C2 × 2! = 6 ways. Then, remaining positions can be filled in 2 × 1 ways. So, total number of ways = 6 × 2 = 12 ways. Similarly, 4 appears at unit’s place in 12 ways. Sum of digits at unit’s place = 16(5 + 3 + 1) + 12(2 + 4) = 216.

3.

Answer: A

5.

Answer: C Explanation: Let x, y and z be the number of 1 Miso, 10 Misos and 50 Misos coins. To pay a bill of 107 Misos, the ordered triplets (x, y, z) can be obtained as: When z = 0, then (x, y) can be (107, 0), (97, 1), (87, 2), and so on till (7, 10). These are 11 ways. When z = 1, then (x, y) can be (57, 0), (47, 1), and so on till (7, 5). These are 6 ways. When z = 2, then (x, y) can be (7, 0). This is only 1 way. Total number of ways = 11 + 6 + 1 = 18 ways.

6.

Answer: D Explanation: Let x be the number of enemies of any one pair. When n = 4, x = 1; when n = 5, x = 3; when n = 6, x = 6. When we put the values of n, we get the corresponding values of x. We see from the

Answer: C Explanation:

481

Quantitative Aptitude Simplified for CAT

options that only

1 2 (n – 5n + 6) satisfies the 2

permutations of the word H H V V will lead to each way of going from A to E. Number of 4! permutation of H H V V = = 6 ways. 2!2! The number of ways of going from E to F (by shortest path) = 1 (diagonal path). The number of ways of going from F to B 6! = = 15. 4!2! Total number of ways of going from A to B by shortest possible paths = 6 × 1 × 15 = 90.

given condition. Alternatively, Enemies of every pair are the pairs formed with all the numbers except the two in the member itself. Therefore, if there are n elements then each member has n–2C2 1 = (n2  5n  6) enemies. 2 7.

Answer: D

10. Answer: A

Explanation: The element which is common to these two friends will also have to be available in other members for them to be friends with these two members. Suppose the two friends are (a, b) and (a, c). The other member must have element ‘a’ and one more element from remaining n – 3 elements. So, number of members of S who will be common friends of both these members will be n – 3. 8.

Explanation: The number of ways of going from A to B (as found in the previous question) = 90. The number of ways of going from B to the top right corner of D region = 6C1 = 6. The number of ways of going from top right corner of D region to C = 2. There is 1 more way of going from B to C, that is move vertically till the end and then left till C. So, the number of ways of going from B to C = 6  2 + 1 = 13. Required total number of ways = 90  13 = 1170.

Answer: A Explanation: Since there are n teams each having k players, there will be nk players. But in each pair of teams, one player is common. So, in n pairs of teams, n players will be common. So, total number of players participating in the tournament = nk – n = n(k – 1).

9.

11. Answer: D Explanation: All integers must be 4 digit numbers. The left most digit space can be filled in 3 ways (1, 2 or 3). The rest of the three spaces each can be filled in 5 ways (0, 1, 2, 3, 4). Moreover, 4000 is also part of the list. So, total number of ways = (3  5  5  5) + 1 = 376.

Answer: D Explanation: Refer to the diagram below:

A

C

12. Answer: A E

D

Explanation: Use the formula: n+r–1Cr–1, where n is the power of the multinomial and r is the number of terms in the multinomial. Therefore, 20+3–1C3–1 = 22C2 = 231.

P F

13. Answer: D B First, Neelam goes from A to E, then from E to F, then from F to B. When she goes from A to E by shortest path, she always moves two horizontal steps and two vertical steps, that is H H V V. Each of the

Explanation: Using statement A, 1 of the players gets a bye and 41 more players go to round 2. After second round, 21 players go to third round, 11 to fourth round, 6 to fifth round, 3 to sixth round and 2 to 7th round. In each round, a

482

Permutation and Combination player plays only once. Since champion goes till 7th round, he must have played 7 matches assuming he does not get any ‘bye’, about which no information is given. Using statement B, we cannot answer the question. Using both the statements, we know that the champion gets only one ‘bye’ and hence both the statements together are sufficient to solve the question.

particular zone = 3 × 3 = 9. Since there are 4 zones, total number of direct lines between cities of same zone = 9 × 4 = 36. Each city of zone A is connected by 1 direct line to each city of rest of the zones. So, number of lines = 9. Since zone A has 3 cities, total number of lines connecting cities of zone A to cities of rest of the zones = 9 × 3 = 27. Each city of zone B is connected by 1 direct line to each city of rest of the zones. We have already counted the number of connections of zone B with zone A. So, number of direct lines between cities of zone B to cities of zone C and D = 6 × 3 = 18. Similarly, number of direct lines between cities of zone C to cities of zone D = 3 × 3 = 9. Total number of direct lines = 36 + 27 + 18 + 9 = 90.

14. Answer: D Explanation: Using statement A, n can be 124 in round 1, or 127 in round 1 and so on. So, A alone is not sufficient. Using statement B, n can be 124 as well as 122. Combining the two statements, we get a unique answer, that is n = 124. Hence, both the statements together are sufficient to solve the question.

19. Answer: C Explanation: Total number of codes possible = 9 × 9 = 81. Digits causing confusion are: 0, 1, 6, 8, 9. Number of codes which can be formed using these digits = 4 × 4 = 16. Out of these, 10, 60, 80 and 90 won’t cause confusion, because these codes when seen upside down would become invalid codes. Further, 69 and 96 won’t cause confusion, because 69 when seen upside down would still be read as 69 and same goes for 96. So, number of codes which can cause confusion is 16 – 4 – 2 = 10. So, number of codes which won’t cause confusion = 81 – 10 = 71.

15. Answer: A Explanation: Here n is going to be an eleven-digit number. If first digit is 1, then another 1 can be fixed at 10 places. So there are 10 ways, and rest of the digits will be 0. If first digit is 2, there is only one number and so there is only 1 way for that. Therefore, total number of ways = 11. 16. Answer: B Explanation: In every path, the person moves 5 steps horizontally and 3 steps vertically. So, total number of ways = 5+3C3 = 8C3 = 56.

20. Answer: A Explanation: Since 50 + 10 + 10 + 2 + 2 + 2 + 2 = 78 paise, so number of coins required is 7. Similarly, 50 + 10 + 5 + 2 + 2 = 68 paise. So, number of coins required is 5. Also, 50 + 25 + 10 + 10 + 2 + 2 + 2 + 2 = 101 paise. So, number of coins required is 7. Hence total number of coins required = 7 + 5 + 7 = 19 coins.

17. Answer: A Explanation: The first strip can be colored in 4 ways, second strip in 3 ways, third one in 3 ways and so on. Thus, total no of ways = 4 × 3 × 3 × 3 × 3 × 3 = 12 × 81. 18. Answer: B Explanation: Let zones be A, B, C and D. Each zone has 3 cities. Number of direct lines within any

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Quantitative Aptitude Simplified for CAT

21. Answer: D

Alternatively, Each vertex will appear only thrice. For example, vertex A can form group with B, C, D, E, F, G and H in only 3 ways without repeating appearance of any of these vertices. So, A will form groups like ABC, ADE, AFG. Similarly, B will also form groups in 3 ways. We can neglect H. If we write all these groups together and mix the alphabets, then each alphabet appears 3 times. So, number of groups possible 21 7. = 3

Explanation: If second letter is m, then number of strings possible = 5 × 4 = 20. If second letter is n, then number of strings possible = 5 × 2 = 10. If second letter is p, then number of strings possible = 5 × 1 = 5. So, total number of ways = 35. 22. Answer: C Explanation: If third letter is e, then following cases arise: Second letter is m. Then, number of ways = 4 ×1=4 Second letter is n. Then, number of ways = 5 × 1 = 5. Second letter is p. Then, number of ways = 1 × 1 = 1. So, total number of ways = 10.

26. Answer: B Explanation: Number of books which C may get can be 0, 1, 2 or 3. If C gets 4 or 5 books, then at least one of A and B would not get any book. When C gets 0 books, then all the 5 books go to A and B. Now, 5 books can be distributed to A and B in 25 ways. But this includes the case when all the books go to A and none goes to B, and also the case when all the books go to B and none goes to A. So, number of ways = 25 – 2. If C gets 1 book, then he can receive the book in 5C1 ways. The remaining 4 books can go to A and B both in 24 – 2 ways. Proceeding likewise, total number of ways = (25 – 2) + 5C1(24 – 2) + 5C2(23 – 2) + 5C3(22 – 2) = 30 + 70 + 60 + 20 = 180.

23. Answer: C Explanation: To form a triangle, we choose 2 points from first line and 1 point from second line, OR we choose 1 point from first line and 2 points from second line. So, number of triangles = 100C2 × 100C1 + 100C1 × 100C2 = 990000. 24. Answer: A Explanation: First, we will arrange consonants. Consonants are: T R G N M T R C. Number of ways to 8! . Between arrange all the consonants = 2!2! these consonants, there are 9 vacant positions (including corner ones), where vowels I O O E I can be placed and arranged 5! in 9C5 × . Hence, total number of ways 2!2! 8!9! = . (2! ) 4 4!

27. Answer: A Explanation: We want number of solutions of 2x + 5y = 100. The solution sets are: (0, 20), (5, 18), (10, 16), …, (50, 0). These are 11 solutions. 28. Answer: D Explanation: Total number of words that can be formed 7! = = 420. 3!2! To find number of words in which all the D’s come before all the A’s, we will first place them as D D A A A, and now place M and B in the 6 vacant spots. M can be placed in these spots in 6 ways. Once M goes to any 1 place, the number of vacant spots now becomes 7.

25. Answer: C Explanation: Let the 8 non-collinear points be A, B, C, D, E, F, G and H. Then, as per desired condition, the triangles possible are: ABC, ADE, AFG, BDF, BEG, CDG, CEF.

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Permutation and Combination Now, B can go to these vacant spots in 7 ways. Therefore, total number of ways in which all the D’s come before all the A’s = 6 × 7 = 42. Therefore, required number of ways = 420 – 42 = 378.

Also, 4 boys and 1 girl go y number of times and therefore, number of balls received = 4y. Total number of balls distributed = 3[2y(y – 1)] + 4y = 6y2 – 2y = 368  6y2 – 2y = 368  3y2 – y – 184 = 0. Therefore, y = 8.

29. Answer: D

32. Answer: D

Explanation: Let the colors be P, Q and R. Arc AB can be colored by any of the 3 colors. The adjacent arcs are BC and DA. Here, two cases arise. Case 1: If arcs BC and DA are of same color, then those colors can be any of the two remaining colors (the one not used in arc AB). In that case, arc CD can be colored in 2 ways: the color which is used in arc AB or the one not used in arc BC or DA. Total number of ways = 3 × 2 × 2 = 12. Case 2: If arcs BC and DA are of different colors, then these arcs can be colored in 2 ways. In that case, arc CD can be colored in only 1 way: the color used in arc AB, which is same as the color not used in the arcs BC and DA. Total number of ways = 3 × 2 = 6. Hence, total number of ways = 12 + 6 = 18.

Explanation: (i)

Two vowels can be arranged together in RIYADH in 5! × 2! = 240 ways. Hence,

statement (i) is not true. (ii)

Total arrangements where vowels are not together

= Total arrangements possible – arrangements where vowels are together = 6! – 240 = 720 – 240 = 480. Therefore, statement (ii) is also not true. 33. Answer: B Explanation: First select the top three salesmen out of 15 and then arrange them as first, second and third. Therefore, total number of ways = 15C3 × 3! = 2730.

30. Answer: A 34. Answer: D Explanation: If token taken from second box is numbered 1, then tokens that can be taken from first box can be 2, 3, 4, …, 100. These are 99 cases. If token taken from second box is numbered 2, then tokens that can be taken from first box can be 3, 4, …, 100. These are 98 cases. Proceeding likewise, total number 99  100 of ways = 99 + 98 + 97 + … + 1 = 2 = 4950.

Explanation: In section 1, he may clear the cutoff or not clear the cutoff. So, this can be done in 2 ways. In section 2, he may clear the cutoff or not clear the cutoff. So, this can be done in 2 ways. Likewise, in each section, he may or may not clear the cutoff which will happen in 2 ways. So, total number of ways = 2 × 2 × 2 × 2 × 2 × 2 = 26 = 64. Out of these 64 ways, there is one way in which he will clear the cutoff in each of the sections = 1. So, number of ways in which he fails to secure the cutoff = 64 – 1 = 63.

31. Answer: D Explanation: Since the group has to have at least three boys, number of ways in which a group of five can be formed = (4C3 × yC2) + (4C4 × yC1) 1 = [4 × (y)(y – 1)] + [(1)(y)] = 2y(y – 1) + y. 2 So, 3 boys and 2 girls go 2y(y – 1) number of times and therefore, number of balls received = 3[2y(y – 1)].

35. Answer: C Explanation: Number of ways in which other 8 students can sit = 8!

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Quantitative Aptitude Simplified for CAT

grouped in 2 ways. Therefore, total number of mixed double teams = 9C2 ×7C2 × 2 = 1512.

The two students who are batch mates can be seated in 9 × 8 = 72 ways. So, total number of ways = 2903040.

41. Answer: C

36. Answer: D

Explanation: The number of questions in the two sections that can be selected is (4, 2), (3, 3) and (2, 4). Therefore, total number of ways = 5C4 × 5C2 + 5C3 × 5C3 + 5C2 × 5C4 = 50 + 100 + 50 = 200.

Explanation: If number of members is n, then total number of handshakes = nC2 = 78. So, n = 13. 37. Answer: A Explanation: The total number of 8-digit landline telephone numbers that can be formed having at least one of their digits repeated = ‘The total number of 8-digit landline numbers’ – ‘The number of 8-digit landline numbers in which no digit is repeated’. The total number of 8-digit landline numbers = 108 The number of 8-digit landline numbers in 10! = 1814400 which no digit is repeated = 2 Therefore, number of required landline numbers = 98185600.

42. Answer: B Explanation: The number of triangles formed using the 8 points = 8C3 = 56 The number of quadrilaterals formed using the 8 points = 8C4 = 70. The difference = 14. 43. Answer: C Explanation: In the worst case scenario, we may end up choosing 4 students from each department by selecting 20 students (4 from each department). The 21st student chosen will be the 5th student from at least one of the departments.

38. Answer: C Explanation: Selecting 3 males from 8 and 4 females from 7 can be done in 8C3 × 7C4 = 56 × 35 = 1960 Raj and Rani together cannot be in the committee. Selection of both Raj and Rani can happen in 7C2 × 6C3 = 21 × 20 = 420 Required number of ways = 1960 – 420 = 1540.

44. Answer: B Explanation: For Maximum invites: Out of 25 of each of the 5 friends of Sarah, one is Sarah herself. That leaves 24 distinct friends for each of the 5 friends. Hence, 24 × 5 = 120 friends have to be invited. Also Sarah’s 5 friends are also to be invited. Hence, 120 + 5 = 125 friends have to be invited. However, 2 of them know each other and so we have to reduce 2 people from the list as we have already included them in our earlier calculations. Hence, a total of 125 – 2 = 123 people have to be invited. For Minimum invites: Let us consider the case when each friend of Sarah is also friend with other friends of Sarah. That leaves 20 friends for each of Sarah’s friends. Let those 20 friends be common for all 5 of them.

39. Answer: C Explanation: Number of ways of selecting 4 consonants and 3 vowels = 8C4 × 5C3 = 70 × 10 = 700. Arrangement of 7 letters will be done in 7! ways. So, total number of words that can be formed = 700 × 7! = 3528000. 40. Answer: A Explanation: Two men can be selected in 9C2 ways. After selecting two men, two women can be selected in 7C2 ways from (9 – 2 = 7) women, so that no husband and wife play in the same set. Also, these selected 4 people can be

486

Permutation and Combination So total number of people invited = 20 + 5 = 25. Only option 2 satisfies both the criteria.

all. So, 100’s place can be filled in 6 ways (1 to 7 except 2), 10’s place in 9 ways and unit’s place also in 9 ways. Total number of ways = 6 × 9 × 9 = 486. Including the number 800, total number of numbers not containing 2 is 487. Remaining numbers = 701 – 487 = 214.

45. Answer: B Explanation: From 100 to 800, total number of numbers = 800 – 99 = 701. Out of these, we will find the number of numbers which do not contain 2 at

.

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Quantitative Aptitude Simplified for CAT

Chapt er 15

Probability BASIC CONCEPTS Probability is defined as the chance of the occurrence of an event. In other words, it is the likelihood of an event to occur or not to occur. Probability is one of the most important mathematical concepts that you use or come across in your day-to-day life. When we toss a coin, the possible outcomes are {Head, Tail}. But the actual outcome is unknown. Similarly, when we roll a dice, the outcomes possible are {1, 2, 3, 4, 5, 6}. But the actual outcome is unknown. The set of all possible outcomes in a random experiment is known as Sample Space. A random experiment is one whose outcome is unknown and hence random. When one coin is tossed, sample space = {H, T}. Number of elements in sample space = 2 = 21. When two coins are tossed, sample space = {HH, HT, TH, TT}. Number of elements in sample space = 4 = 22. When three coins are tossed, sample space = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}. Number of elements in sample space = 8 = 23. In general, when n coins are tossed, number of elements in sample space = 2n. We can easily understand the logic by applying the fundamental principle of counting. For example, when 5 coins are tossed, then the first coin can show Head or Tail. So, two outcomes are possible. The second coin also can show Head or Tail, and so on. So, ___ ___

___

___

___

2 × 2 × 2 × 2 × 2

= 25

Similarly, for n coins, sample space = 2 × 2 × 2 × 2 × 2… (n times) = 2n. When one dice is rolled, sample space = {1, 2, 3, 4, 5, 6}. Number of elements in sample space = 6 = 61. When two dice are rolled, sample space = {(1, 1), (1, 2), (1, 3), … (2, 1), (2, 2), (2, 3), …, (6, 4), (6, 5), (6, 6)}

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Probability Number of elements in sample space = 36 = 62. In general, when n dice are rolled, number of elements in sample space = 6 × 6 × 6 × … (n times) = 6n. Any subset of the sample space is called an event. The sample space S of an experiment serves as the universal set for all possible outcomes of the experiment and an event E of the experiment is a set of all possible outcomes favorable to event E. Listed below are some types of events. For example, when a dice is rolled, Sample space = {1, 2, 3, 4, 5, 6}. If an event is described as: the outcome is more than 4, then the set of outcomes favorable to the occurrence of the event = {5, 6}. Probability =

Number of outcomes favorable to the event . Sample Space

In the current example, number of elements in sample space is 6 whereas number of elements in the event is 2 1 2. Hence probability of occurrence of a number more than 4 in the roll of a dice =  . 6 3 When two coins are tossed, the probability of occurrence of both Heads =

1 . 4

Total probability of occurrence of all the elementary events in a sample space is always 1. An event which will never occur is called an impossible event. Therefore, probability of an impossible event = 0. An event which will definitely occur is called a sure event. Probability of a sure event = 1. Two or more events that do not share any favorable outcomes are called mutually exclusive events. In other words, if occurrence of one precludes the occurrence of the other, the events are called mutually exclusive events. For example, when a dice is rolled, two events A and B are described as: A = number is less than 3; B = number is more than 5. There is no element which is common to A and B. So, A and B are mutually exclusive events. Two or more events are said to be exhaustive events if the union of all the elements of those events results in Sample Space. For example, when a dice is rolled, A = number is odd number; B = number is even number. Their union is the sample space. So, A and B are exhaustive events. Also note that these are mutually exclusive events as well. If we roll a dice, then two events A and B may be Mutually Exclusive, but not Exhaustive: A = Occurrence of even number; B = occurrence of odd prime number. Exhaustive, but not Mutually Exclusive: A = Occurrence of number < 4; B = occurrence of number > 2. Mutually Exclusive as well as Exhaustive: A = Occurrence of even number; B = occurrence of odd number. Neither Mutually Exclusive nor Exhaustive: A = Occurrence of even number; B = occurrence of prime number.

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Quantitative Aptitude Simplified for CAT

Two events A and B are said to be equally likely, if they have equal likelihood of occurrence. For example, when an unbiased coin is tossed, both head and tail have an equal possibility of appearing on the visible face of the coin and hence, these are equally likely events. The complement of an event E in S is the set of those sample points in S that are not present in E. The complement of event E is denoted by E or Ec. If E is the event of occurrence of odd number on the roll of a dice, then the complement of this event will be occurrence of even number on the same roll of dice. Therefore, if probability that Manoj will pass an exam is 0.7, then the probability that he will fail = 1 – 0.7 = 0.3. Two events are said to be independent if the occurrence of one event does not affect the occurrence of the second event. For example, if the first child of a couple is a boy, there is no effect on the chances of the second child being a boy and thus the two events are independent events. Example 1 If three coins are tossed, find the probability of occurrence of (i)

all the coins showing Head

(ii)

None of the coins showing Head

(iii) Exactly 2 Head (iv) At least 2 Head Solution (i)

(ii)

Number of elements in Sample Space = 8. Favorable case is HHH. So, number of favorable cases = 1. 1 Hence, required probability = . 8 1 7 Probability of none showing Head = complement of all showing Head = 1 –  . 8 8

(iii) Events showing two Head = {HHT, HTH, THH}. Number of cases = 3. 3 Probability of exactly two Head = . 8 (iv) Events showing at least 2 Head = {HHT, HTH, THH, HHH}. Number of cases = 4. 4 1 Probability of exactly two Head =  . 8 2 Example 2 If seven coins are tossed, find the probability of occurrence of (i)

all the coins showing Head

(ii)

None of the coins showing Head

(iii) Exactly 3 Head (iv) At least 2 Head Solution (i)

Number of elements in Sample Space = 2n = 27. Favorable case is HHHHHHH. So, number of favorable cases = 1. 1 1 . Hence, required probability = 7  128 2

490

Probability

(ii)

Probability of none showing Head = complement of all showing Head = 1 –

1 127 .  27 128

(iii) Number of cases showing 3 Head = all permutations of HHHTTTT. Number of cases = Required probability =

7!  35 . 3!4!

35 . 128

(iv) Number of cases showing at least 2 Head = 1 – [P(0 Head) + P(1 Head)] 1 1 Now, P(0 Head) = P(All Tail) = 7  . 128 2 Number of cases showing 1 Head = all permutations of HTTTTTT = Therefore, P(1 Head) =

7! 7 6!

7 . 27

 1  7   120  15 Therefore, required probability = 1 –  .   128 128  128 16 Example 3 Let two dice be rolled. (i)

Find the probability that at least one of the numbers is 3.

(ii)

Find the probability that the sum of the numbers appearing is a prime number.

Solution (i)

Sample Space = 36. Now, 3 appears at least once in the following cases: (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), and also (1, 3), (2, 3), (4, 3), (5, 3), (6, 3). 11 . Number of favorable cases = 11. So, required probability = 36 Alternatively, To count the number of cases where 3 appears at least once, we better count the number of cases where 3 does not appear at all. Number of ways in which 3 does not appear on the first dice is 5. Same is true for second dice. So, 3 does not appear in both the dice = 5 × 5 = 25 cases. So, 36 – 25 11 = 11 cases are the ones where 3 appears at least once. Hence required probability = . 36 (ii)

Sum of numbers appearing should be prime number. The prime numbers possible are: 2, 3, 5, 7, 11. For sum to be 2, favorable cases = {(1, 1)}. For sum to be 3, favorable cases = {(1, 2), (2, 1)}. For sum to be 5, favorable cases = {(1, 4), (2, 3), (3, 2), (4, 1)}. For sum to be 7, favorable cases = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}. For sum to be 11, favorable cases = {(5, 6), (6, 5)}. Total number of favorable cases = 15. 15 5 So, required probability = .  36 12

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Quantitative Aptitude Simplified for CAT

Example 4 4 dice are rolled together. Find the probability that the product of the numbers appearing ends with 5. Solution Sample Space = 64. For the product of numbers to end in 5, all the numbers appearing must be odd numbers and the number 5 must appear at least once anywhere. Odd numbers on a dice can be 1, 3 or 5. Number of cases in which all numbers are odd = 3 × 3 × 3 = 33. However, in this, some cases are there where 5 does not appear at all. We will remove those cases. Number of cases where 5 does not appear at all = number of cases where 1 or 3 only appears = 2 × 2 × 2 × 2 = 24. Therefore, number of cases where 5 appears at least once = 34 – 24 = 81 – 16 = 65. The required probability =

65 . 64

Example 5 Using the digits 1, 2, 3, …, 9, all possible 9-digit numbers are formed without repetition of digits. Find the probability that (i)

The number is divisible by 5.

(ii)

The number is divisible by 4.

(iii) The number is a prime number. Solution (i)

Sample Space = 9!. For the number to be divisible by 5, unit’s digit must be 5. The rest of the places 8!1 1 can be filled with remaining 8 digits in 8! ways. Therefore, required probability =  . 9! 9

(ii)

For the number to be divisible by 4, the number formed by the last two digits should be divisible by 4. For that, unit’s digit must be even number: 2, 4, 6 or 8. When unit’s digit is 2 or 6, ten’s place digit should be 1, 3, 5, 7, 9 (all odd numbers); When unit’s digit is 4 or 8, ten’s place digit should be 2, 4, 6, 8 (all odd numbers). However, since repetition is not allowed, the last two digits cannot be 44 or 88. Hence total number of cases = 16. Rest of the 7 places can be filled by the remaining 7 digits in 7! ways. 7!16 2 Required probability =  . 9! 9

Alternatively, To count the number of cases divisible by 4, we can also say that the last two digits should be 12, 16, 20, 24, … 96. These are 22 numbers. However, remove the numbers 20, 40, 60 and 80, and also 44 and 88. We are 7!16 2 left with 16 numbers. So, required probability =  . 9! 9 (iii) All the 9! permutations of the digits will always have all the digits 1, 2, 3, 4, …, 9 whose sum is 45, which is always divisible by 3 as well as 9. So, all the 9! numbers are divisible by 3 as well as 9 and hence no number is a prime number. So, required probability = 0.

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Probability Example 6 Find the probability that in a throw of two dice, the sum of the numbers appearing on their faces is odd. Solution For the sum of the numbers on the dice to be odd, the first number should be odd and the second number should be even or the first number should be even and the second should be odd. 1

1

1 1

1

1

1

 P(A) =           .  2 2  2 2 4 4 2 Alternatively, Half of the numbers will have their sum as odd and the other half as even. So, required probability =

1 . 2

Example 7 A committee of 4 persons is to be formed out of 6 men and 5 women. Find the probability that the committee contains exactly 3 women. Solution Sample Space = number of ways of selecting 4 persons from 6 + 5 = 11 persons = 11C4 = 330 To choose exactly 3 women, we need to also choose 1 man. Number of ways for that = 6C1 × 5C3 = 6 × 10 = 60. Therefore, required probability =

60 2 .  330 11

Example 8 Two squares are chosen at random on a chessboard. What is the probability that they are not from the same row or column? Solution For each square chosen, there are 7 squares in the same row and another 7 squares in the same column. So, there are 14 squares which must not be chosen. So, we need to choose one of the remaining 49 squares. So, required probability =

64 49 7   . 64 63 9

Addition Law of Probability As per Addition Law of Probability, P(A  B) = P(A) + P(B) – P(A  B) where P(A  B) = Probability of occurrence of event A or event B; P(A) = probability of occurrence of event A; P(B) = probability of occurrence of event B; P(A  B) = = Probability of occurrence of event A and event B. If A and B are mutually exclusive events, P(A  B) = P(A) + P(B) [Since P(A  B) = 0]

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Quantitative Aptitude Simplified for CAT

Example 9 From a pack of 52 cards, one card is drawn at random. Find the probability of occurrence of Red card or King card. Solution Let A = event of occurrence of Red card; B = event of occurrence of King card. P(A) =

26 4 2 , P(B) = , P(A  B) = 52 52 52

Here, we want probability of event A or event B. So, we will apply law of addition of probability. P(A  B) = P(A) + P(B) – P(A  B)  P(A  B) =

26 4 2 28 7 .     52 52 52 52 13

Alternatively, We know that there are total 26 Red cards which includes 2 King cards. There are also 2 Black King cards. So, there are total 26 + 2 = 28 cards favorable to the desired event. So, required probability =

28 7 .  52 13

Example 10 From a pack of 52 cards, two cards are drawn at random. Find the probability of occurrence of both cards as Red card or King card. Solution Let A = event of occurrence of both cards as Red cards; B = event of occurrence of both cards as King cards. 26

P(A) =

C2 , P(B) = 52 C2

4

C2 , P(A  B) = 52 C2

2

C2 C2

52

Here, we want probability of event A or event B. So, we will apply law of addition of probability. P(A  B) = P(A) + P(B) – P(A  B) 26

 P(A  B) =

C 2  4 C 2  2 C 2 26  25  4  3  2  1 52 .   52 52  51 221 C2

Example 11 There are 20 tables and 15 chairs in a showroom. Of these, 3 tables and 2 chairs are defective. From the lot of 35 pieces of furniture, 2 pieces are chosen at random. Find the probability that both are tables or both are non-defective. Solution Event A = both are tables; Event B = both are non-defective. Total number of pieces which are non-defective = 35 – 5 = 30. 20

P(A) =

35

C2 , P(B) = C2

30 35

C2 , P(A  B) = probability that both are tables as well as non-defective = C2

Therefore,

494

17 35

C2 . C2

Probability P(A  B) = P(A) + P(B) – P(A  B) 20

 P(A  B) =

C2  30 C 2  17 C2 . 35 C2

Product Rule for Independent Events If two events A and B are independent events, then P(A  B) = P(A) × P(B), and vice versa, that is, If P(A  B) = P(A) × P(B), then the two events A and B are independent events. Refer to the following examples. Example 12 If probability of A solving a question in a test is 0.6 and probability of B solving the same question in the same test is 0.7, what is the probability that (i)

Both A and B solve the problem

(ii)

The problem is solved

(iii) Exactly one of them solve the problem (iv) Both of them do not solve the problem Solution Since the event of solving of question by A is independent of event of solving of question by B, we can apply product rule. (i)

Probability of both A and B solving the problem = P(A  B) = P(A) × P(B) = 0.6 × 0.7 = 0.42.

(ii)

Probability that the problem is solved = probability that at least one of them solves the problem = P(A solves, B does not solve) + P(A does not solve, B solves) + P(both A and B solve) = 0.6 × 0.3 + 0.4 × 0.7 + 0.6 × 0.7 = 0.88.

(iii) Probability that exactly one of them solves the problem = P(A) × P(B’) + P(B) × P(A’) = 0.6 × 0.3 + 0.7 × 0.4 = 0.18 + 0.28 = 0.46. (iv) Probability that both of them do not solve the problem = P(A’) × P(B’) = 0.4 × 0.3 = 0.12 Alternatively, P(A  B) = P(A) + P(B) – P(A  B) = 0.6 + 0.7 – 0.42 = 0.88. Alternatively, Probability that at least one of them solves the problem = 1 – P(none of them solves the problem) = 1 – (0.4)(0.3) = 1 – 0.12 = 0.88. Example 13 If probability of A solving a question in a test is 0.6, probability of B solving the same question in the same test is 0.7 and probability of C solving is 0.9, what is the probability that the problem is solved? Solution Required probability = 1 – P(none of them solves the problem) = 1 – (0.4)(0.3)(0.1) = 1 – 0.012 = 0.988. Example 14 2 2 3 for A, for B and for C. If all of them fire 5 3 5 independently at the same target, calculate the probability of exactly one of them hitting the target.

In a shooting competition the probability of hitting the target is

495

Quantitative Aptitude Simplified for CAT

Solution P(A) =

2 3 , P(A) = 5 5

P(B) =

2 1 , P(B) = 3 3

P(C) =

3 2 , P(C) = . 5 5

Probability of exactly one of them hitting the target = Probability that A hits the target and B and C do not + Probability that B hits the target and A and C do not + Probability that C hits the target and A and B do not = P(A  B  C) + P(A  B  C) + P(A  B  C) =  2  1  2    3  2  2    3  1  3   4  12  9  25  1  5 3 5   5 3 5   5 3 5  75 75 75 75 3 Example 15 What is the probability of getting a total of 9 exactly once in 3 throws of a pair of dice? Solution A total of 9 can occur with a pair of dice in 4 ways: (3,6) (4,5) (5,4) (6,3)  In a throw of a pair of dice, the probability of getting a total of 9 =  The probability of not getting a total of 9 = 1 –

4 1  66 9

1 8  9 9

1 8 Probability for exactly one throw (out of three throws) of a pair of dice having the total 9 =       9  9

2

Since there are 3 different ways in which one throw of the given pair of dice has a total of 9 and the other two throws don’t, probability of throwing a total of 9 exactly once in 3 throws 2

64 1 8 = 3 ×      = . 9 9 243    

Example 16 Two friends A and B appeared for a job interview where the number of vacancies is unlimited. The probability of selection of A is

1 2 and that of B is . Find the probability of both of them getting selected. 5 7

Solution Probability that both of them get selected = P(A  B) = P(A) × P(B) (because both the events are independent events). Therefore, required probability =

1 2 2 .   5 7 35

496

Probability

Binomial Probability Let there be a random experiment which is performed n times, where the outcome of each experiment is either a success or failure. Out of n trials, we want to find the probability of getting exactly r successes. In such a case, the probability of getting r successes is given by P(r successes) = nCr × pr × qn–r, where p = probability of success in one trial; q = probability of failure in one trial = 1 – p; n = total number of trials; r = number of successes desired. Example 17 A dice is rolled 5 times. Find the probability of getting 6 exactly 2 times. Solution Here, getting a six is defined as success. So, probability of success, p = = 1

1 and so probability of failure 6

1 5  . Also, n = 5, r = 2. Success happens 2 times and failure happens 3 times. 6 6 2

3

1 5 Therefore, required probability = 5C2 ×      .  6  6

Example 18 In a shooting competition, probability of a shooter hitting bulls-eye is

2 . He shoots at the target 8 times. Find 5

the probability that he hits at least once. Solution Probability of success =

2 3 , probability of failure = . Total number of trials = 8. Number of successes 5 5

required, r ≥ 1. So, 0

8

2 3 P(r ≥ 1) = 1 – P(r = 0) = 1 – 8C0 ×      .  5  5

Example 19 If the probability that a student will not complete schooling is

1 , what is the probability that out of 4 students, 3

at least 3 will complete schooling? Solution 3

3

Probability that exactly 3 will complete and 1 will not = 4C3  2   1  3

2 Probability that all 4 will complete =   3 



4

=

3

= 4C1

32 1  2 .    = 3 3 81  

16 32 16 48 16 . Hence, p =    . 81 81 81 81 27

497

Quantitative Aptitude Simplified for CAT

Example 20 To open a lock, a key is taken out at random from a collection of n keys. If the lock is not opened with the key, it is put back into the collection and another key is tried. If the lock can open with only one key in the collection, find the probability that the lock opens in n trials. Solution Probability that the lock opens in a trial is

1 as there is exactly one key (in the given collection of n keys), n

which can open the lock.

1  Therefore, probability that the lock does not open in a particular trial is  1   . n  So, P(lock opens in one of the n trials) = 1 – P(lock does not open in any of the n trials) n

1 = 1 – [P(the lock does not open in a particular trial)]n = 1   1   . n 

Conditional Probability In case of some events, the probability that an event A takes place depends on whether another event B does or does not take place. For example, suppose two cards are drawn from a pack of cards one after the other without replacement. Let A = Event that the first card is a spade. B = Event that the second card is a spade. C = Event that the first card is not a spade. The sample space of the event B consists of 51 sample points as the first card is not replaced. The probability of occurrence of B when A has taken place or probability of B under A is denoted by P(B/A). If A occurs, then there are only 12 spade cards left.  P(B/A) =

4 12 = . 51 17

If C occurs then there are 13 spade cards left,  P(B/C) =

13 51

The conditional probability of occurrence of event B given that event A has occurred, is P(B/A) =

P(A  B) , where P(A)  0 P( A)

Similarly, P(A/B) =

P( A  B) , where P(B)  0 P(B)

We can solve the above example by the use of formula as well. P(B/A) =

P(A  B) . P( A)

Now, P(A) =

13 ; P(A  B) = Probability of both cards being spade = 52

498

13 52

C2 . C2

Probability

Therefore, required probability =

  

13

C2   C2  12 4   . 13 51 17 52

52

Example 21 Let two coins be rolled. Two events are described below: A = At least one of the numbers is 3. B = the sum of the numbers appearing is a prime number. Find P(A/B) and P(B/A). Solution Refer to example 3. P(A) =

11 15 , P(B) = . Also note that number of cases common to A and B are: {(3, 2), (2, 3), (3, 4), (4, 3)}. 36 36

Therefore, P(A  B) =

4 . 36

 4    4 P ( A  B)  36    Now, P(A/B) = . P (B)  15  15    36   4    4 P ( A  B)  36    . P(B/A) = 11 P ( A)   11    36 

Example 22 A box contains 6 white balls and 3 black balls and another box contains 4 white balls and 5 black balls. Find the probability that a ball selected from one of the boxes is a white ball? Solution Let A be the event that the first box is selected, B be the event that the second box is selected and C be the event that a white ball is selected. P(A) =

1 ; 2

P(C/A) =

P(B) = 6 ; 9

1 2

P(C/B) =

4 9

The probability of selecting the first box and choosing a white ball 1 6 6 = P(A  C) = P(A) × P(C/A) =        2   9  18

The probability of selecting the second box and choosing a white ball 1 4 4 = P(B  C) = P(B) × P(C/B) =        2   9  18

499

Quantitative Aptitude Simplified for CAT

Now, the events (A  C) and (B  C) are mutually exclusive events. Required probability = P(A  C)  (B  C) =

6 4 10 5    18 18 18 9

Example 23 Two coins are tossed together. Find the probability that the both the coins show head, given that at least one of them shows head. Solution Let event A = at least one of the coins shows head; B = both the coins show head. 1   P ( A  B)  4  1   . We want to find P(B/A). Therefore, required probability = P ( A) 3  3   4  

Baye’s Theorem Let’s gain some basic understanding of Baye’s Theorem, which is actually an application of Conditional Probability. Consider the following example. Example 24 In a locality, 40% of the residents are men and the rest are women. If 25% of the men and 10% of the women are employed, find the probability that (i)

a person selected at random is an employed person.

(ii)

an employed person selected at random is a man.

Solution (i)

Percentage of women in the locality = 60%. Now, if total number of persons is x, then number of employed persons = 25% of 40% of x + 10% of 60% of x = 0.25 × 0.4x + 0.1 × 0.6x. Therefore, probability of the person selected at random to be an employed person 0.25  0.4 x  0.1  0.6 x = = 0.25 × 0.4 + 0.1 × 0.6. x

(ii)

From the employed persons, that is 0.25 × 0.4x + 0.1 × 0.6x, number of employed men = 0.25 × 0.4x. 0.25  0 .4 5 Therefore, required probability =  . 0 .25  0.4  0 .1  0 .6 8

Part (ii) above is an example of Baye’s theorem.

Geometrical Probability Here, we will apply the theory of probability to cases which involve coordinate system of axes or geometrical shapes. Following examples will clarify the concept. Example 25 Let 2 < x < 5. Find the probability that x is positive if (i)

x is a real number.

(ii)

x is an integer.

500

Probability Solution (i)

(ii)

If x is a real number, then sample space will be the length of the line (on number line) from 2 to 5, which is 7. Of this, x is positive if 0 < x < 5, that is, length of the line = 5. Therefore, required 5 probability = . 7 4 2 If x is an integer, then x = {1, 0, 1, 2, 3, 4}. Required probability =  . 6 3

Example 26 Let 2 < x < 5 and 3 < y < 4. Find the probability that, (i)

xy > 0.

(ii)

x + y > 0.

(iii) xy > 0, given that x + y < 0. Solution (i)

For xy > 0, either x > 0 and y > 0, or x < 0 and y < 0. So, required probability 5 4 2 3 26 =     . 7 7 7 7 49

(ii)

To solve this, we need to draw the inequalities on coordinate axes as shown below.

A E

B

4

5

-2 D

-3

F

C

Sample space is the area enclosed by ABCD = 7 × 7 = 49. We want x + y > 0  y > x. Now, line EF is y = x. Coordinates of E and F are (2, 2) and (3, 3). The favorable region = area of ABCFE = area (ABCD) – area (DEF) 1 1 = 49 –  ED  DF  49   5  5 = 49 – 12.5 = 36.5. 2 2 36 .5 73 Therefore, required probability = .  49 98 (iii) Area common to xy > 0 and x + y < 0 is area in the 3rd quadrant in the above diagram. So, common 6 12 area = 3 × 2 = 6. Sample Space = area of DEF = 12.5. Therefore, required probability = .  12 .5 25

501

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

The student mess committee of a reputed Engineering College has n members. Let A be the event that the Committee has students of both sexes and let B be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events A and B are independent is A.

2

B.

3

C.

4

D.

None of the above

4.

12 1024

B.

11 1024

C.

11 256

D.

12 256

D.

0.7

In the MBA Programme of a B – School, there are two sections A and B. students in Section A and

A. B. C. D.

th

th

of the

of the

23 72 11 36 5 12 17 36 IIFT 2014

5.

Suppose there are 4 bags. Bag 1 contains 1 black and a2 – 6a + 9 red balls, bag 2 contains 3 black and a2 – 6a + 7 red balls, bag 3 contains 5 black and a2 – 6a + 5 red balls and bag 4 contains 7 black and a2 – 6a + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is A.

16 a  6 a  10

B.

20 a  6 a  10

C.

1 16

D.

None of these

IIFT 2015 3.

4 9

1 4

students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is:

The internal evaluation for Economics course in an Engineering programme is based on the score of four quizzes. Rahul has secured 70, 90 and 80 in the first three quizzes. The fourth quiz has ten True-False type questions, each carrying 10 marks. What is the probability that Rahul’s average internal marks for the Economics course is more than 80, given that he decides to guess randomly on the final quiz? A.

0.4

IIFT 2015

IIFT 2016 2.

C.

In a reputed engineering college in Delhi, students are evaluated based on trimesters. The probability that an Engineering student fails in the first trimester is 0.08. If he does not fail in the first trimester, the probability that he is promoted to the second year is 0.87. The probability that the student will complete the first year in the Engineering College is approximately:

2

2

IIFT 2013 6.

A.

0.8

Two trains P and Q are scheduled to reach New Delhi railway station at 10.00 AM. The probability that train P and train Q will be

B.

0.6

late is

502

7 9

and

11 27

respectively. The

Probability probability that train Q will be late, given that train P is late, is

A.

8 . Then the 9

B.

probability that neither train will be late on a particular day is A. B. C. D.

C.

40 81 41 81 77 81 77 243

D.

IIFT 2012 9.

IIFT 2013 7.

A survey was conducted to test relative aptitudes in quantitative and logical reasoning of MBA applicants. It is perceived (prior to the survey) that 80 percent of MBA applicants are extremely good in logical reasoning, while only 20 percent are extremely good in quantitative aptitude. Further, it is believed that those with strong quantitative knowledge are also sound in data interpretation, with conditional probability as high as 0.87. However, some MBA applicants who are extremely good in logical reasoning can be also good in data interpretation, with conditional probability 0.15. An applicant surveyed is found to be strong in data interpretation. The probability that the applicant is also strong in quantitative aptitude is A.

0.4

B.

0.6

C.

0.8

D.

0.9

The answer sheets of 5 engineering students can be checked by any one of 9 professors. What is the probability that all the 5 answer sheets are checked by exactly 2 professors? A. B. C. D.

20 2187 40 2187 40 729 None of the above IIFT 2012

10. The probability that in a household LPG will last 60 days or more is 0.8 and that it will last at most 90 days is 0.6. The probability that the LPG will last 60 to 90 days is A.

0.40

B.

0.50

C.

0.75

D.

None of the above IIFT 2012

11. A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first? IIFT 2013

8.

6 55 12 73 14 55 7 50

A.

Ashish is studying late into the night and is hungry. He opens his mother’s snack cupboard without switching on the lights, knowing that his mother has kept 10 packets of chips and biscuits in the cupboard. He pulls out 3 packets from the cupboard, and all of them turn out to be chips. What is the probability that the snack cupboard contains 1 packet of biscuits and 9 packets of chips?

B. C. D. E.

5 36 8 36 15 36 21 36 None of the above XAT 2017

503

Quantitative Aptitude Simplified for CAT

12. Ramesh plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively. Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

of both colors). A marble is randomly drawn from the first bag followed by another randomly drawn from the second bag, the probability of both being red is

the probability of both marbles being blue? A. B. C.

A.

0.004

B.

0.006

D.

C.

0.216

E.

D.

0.994

E.

0.996

multiple of 1023 is:

a2 , then ‘b – a’ would b2

a is b

be: 8 15

C.

21

D.

23

E.

45

None of these above

15. Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let ‘a’ denote the probability that at least one of the cards drawn is a king, and ‘b’ denote the probability of not drawing a king. The ratio

13. The probability that a randomly chosen positive divisor of 1029 is an integral

B.

1 16 2 16 3 16 4 16

XAT 2014 XAT 2015

A.

5 . What is 16

A.

≥ 0.25 and ≤ 0.5

B.

≥ 0.5 and ≤ 0.75

C.

≥ 0.75 and ≤ 1.0

D.

≥ 1.0 and ≤ 1.25

E.

≥ 1.25

XAT 2014

XAT 2013

14. Aditya has a total of 18 red and blue marbles in two bags (each bag has marbles

ANSWER KEY 1.

B

2.

B

9.

B

10. A

3.

A

11. C

4.

D

12. E

5.

D

13. D

504

6.

B

14. C

7.

B

15. E

8.

C

Probability

ANSWERS AND EXPLANATIONS 1.

Probability of exactly one girl = P(Girl from A and Boy from B) + P(Girl from B and Boy from A) 1 5 4 3 17 =     . 4 9 9 4 36

Answer: B Explanation: P(A) = probability of the event that both males and females are present in the committee 1 2 3 n  1 (n  1)n n  1    ...    . 2n 2 n n n n P(B) = probability that at most one female is 1 present in the committee = . n P(A  B) = probability of only one female in 1 the committee = n For independent events, P(A  B) = P(A) × P(B) 1 n1 1 n1    1  n = 3. So,  n 2 n 2

2.

5.

Explanation: Each bag has a2 – 6a + 10 balls. Probability 1 of selecting a particular bag = . 4 Probability that the selected ball is black 1 3 5    2  2  2  1  a  6a  10 a  6a  10 a  6 a  10   7  4  2   a  6 a  10 

1 16 4  ,  2   2 4  a  6a  10  a  6a  10 whose maximum value is 1 when a2 – 6a + 10 = 4 (which is true for real value of a).

=

Answer: B Explanation: Average marks of the first three quizzes = 80. So, for Rahul to have average internal marks more than 80, he has to score more than 80 marks in the last quiz. This is possible if he attempts 10 questions or 9 questions correctly. Number of ways this can be done = 1 + 10C9 = 1 + 10 = 11. Total number of ways the quiz can be solved = 210 = 1024. 11 Therefore, required probability = . 1024

3.

6.

Answer: B Explanation: Let P be the event that train P is late and Q be the event that train Q is late. P (Q  P ) Conditional Probability, P(Q/P) = P (P ) 8 7 56 .  P(Q  P) = P(Q/P) × P(P) =   9 9 81 Now, P(Q  P) = P(Q) + P(P) – P(Q  P) 11 7 56 40    = . Therefore, probability 27 9 81 81 that neither train will be late = 1 – P(Q  P) 40 41  =1– . 81 81

Answer: A Explanation: We assume that if the student fails in first trimester, he will fail in first year. Therefore, the probability that the student will complete the first year in the Engineering college = P(he passes first trimester) × P(he is promoted to second year, given that he passed in first trimester) = 0.92 × 0.87.

4.

Answer: D

7.

Answer: B Explanation: Let there be 100 MBA applicants who were surveyed. 80 of them are good in logical reasoning and 20 are good in quantitative aptitude. Number of applicants who are good in quantitative aptitude as well as data interpretation = 0.87 × 20. Number of applicants who are good in logical reasoning as well as data interpretation

Answer: D Explanation:

505

Quantitative Aptitude Simplified for CAT

Therefore, number of ways for answer sheets to be checked by exactly 2 professors = 9C2 × (25 – 2) = 36 × 30. 36  30 40 Required probability =  . 5 2187 9

= 0.15 × 80. Probability that the applicant is good in quantitative aptitude, given that he is found to be good in data interpretation 0.87  20 29   0.6 . = 0.87  20  0.15  80 49 8.

10. Answer: A

Answer: C

Explanation: Probability that the LPG will last ≥ 60 days = 0.8. So, probability that the LPG will last < 60 days = 1 – 0.8 = 0.2. Also, probability that the LPG will last ≤ 90 days = 0.6. Therefore, probability that the LPG will last ≥ 60 days and ≤ 90 days = (Probability that a LPG will last ≤ 90 days) – (Probability that a LPG will last < 60 days) = 0.6 – 0.2 = 0.4.

Explanation: There are at least three packets of chips in the cupboard. There are 10 packets in all. Let number of packets of chips be x and that of biscuits be y. Then, (x, y) ≡ (3, 7) or (4, 6) or (5, 5) or (6, 4) or (7, 3) or (8, 2) or (9, 1) or (10, 0). The number of ways in which three packets of chips can be drawn = 3C3 + 4C3 + 5C3 + 6C3 + 7C3 + 8C3 + 9C3 + 10C3 = 330. The number of ways in which three packets of chips can be drawn when there are 9 packets of chips = 9C3 = 84. 84 14  . Therefore, required probability = 330 55 9.

11. Answer: C Explanation: Sample space = 6 × 6 = 36. Total number of cases in which both the numbers are distinct = 6 × 5 = 30. Out of 30 cases, in half of the cases, first number will be less than second number and in the other half of the cases, the first number will be more than second number. Therefore, there are 15 cases in which first number is less than the second number. 15 So, required probability = . 36

Answer: B Explanation: Answer sheet of first engineering student can be checked by any of the professors in 9 ways. Answer sheet of second engineering student can be checked by any of the professors in 9 ways. Likewise, answer sheet of fifth engineering student can be checked by any of the professors in 9 ways. Therefore, total number of ways in which the answer sheets can be checked by the professors = 9 × 9 × 9 × 9 × 9 = 95. If the answer sheet goes to exactly 2 professors, then we will first choose those 2 professors, which can be done in 9C2 ways. Now, the answers sheets can be checked by them in 25. But, in 25 ways, there in 1 way in which all are checked by first professor and 1 way in which all are checked by second professor. These 2 ways need to be subtracted from 25 ways.

12. Answer: E Explanation: Required Probability = 1 – P(receiving no gift) = 1 – (0.4 × 0.2 × 0.1 × 0.5) = 1 – 0.004 = 0.996. 13. Answer: D Explanation: Number of factors of 1029 or 229 × 529 = (29 + 1)(29 + 1) = 30 × 30. Now, 1029 = 1023 × k, where k is integer. We need to find all factors of k, thereby ensuring 1023 × k is an integral multiple of 1023. Since k = 106 = 26 × 56, number of factors of k = (6 + 1)(6 + 1) = 7 × 7.

506

Probability

Therefore, probability =

7 7 72  2 a 30  30 30

When (m, n, k) = (2, 16, 2), then R1 + B1 = 2  R1 = 1, B1 = 1. Also, since R1 × R2 = 5k = 10, so R2 = 10 and hence B2 = 6. Therefore, probability of both marbles being 16 3  blue = . 2  16 16 When (m, n, k) = (8, 10, 5), then R1 + B1 = 8. Also, since R1 × R2 = 5k = 25, R1 = R2 = 5 each. Therefore, B1 = 3 and B2 = 5. Therefore, probability of both marbles being 53 3  . blue = 8  10 16

= 7, b = 30. So, b – a = 30 – 7 = 23. 14. Answer: C Explanation: Let number of red marbles in the first and second be R1 an R2, and number of blue marbles in the two bags be B1 and B2. Probability of both marbles drawn being red R1  R 2 5 =  (R1  B1 )  (R 2  B2) 16 So, R1 × R2 = 5k and (R1 + B1)(R2 + B2) = 16k. Also, (R1 + B1) + (R2 + B2) = 18 (total number of marbles). Let R1 + B1 = m and R2 + B2 = n. So, m + n = 18 (m, n ≥ 2, as both the colors are available in each bag) and m × n = 16k. (m, n, k) could be (2, 16, 2), (8, 10, 5).

15. Answer: E Explanation: a 2 C1  4 C1  2 C 2 9 The ratio = 1.5.   4 b 6 C2 While finding the ratio of a to b, the sample space (being same) will get cancelled.

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Chapter 16

Plane Geometry INTRODUCTION The word Geometry literally means measurement of earth (‘Geo’ means ‘earth’ and ‘metry’ means ‘measurement’) and this is how the mathematical principles of geometry evolved. In this chapter, we will discuss the basic principles of geometry which will be found helpful for the entrance exams. The treatment of topic is simple and lucid, so that the students develop interest in the subject. The logical proofs of most of the results are given here to show you the application of basic principles involved in those results. Note that (a)

Angle at a point on a straight line is always 180°.

(b)

Angle around a geometrical point is 360°.

Lines and Angles A geometrical point is a dimensionless quantity, that is, it has no thickness whatsoever. A straight line is composed of infinite geometrical points in an array. A straight line only has length and no breadth. A straight line has two ends, both of which can be extended infinitely. A ‘line’ would henceforth mean ‘straight line’. If two line segments have a common point of origin, then they enclose an angle. If the angle enclosed is less than 90°, then it is called acute angle. If the angle enclosed is equal to 90°, then it is called right angle. If the angle enclosed is more than 90° and less than 180°, then it is called obtuse angle. If the angle enclosed is more than 180°, then it is called reflex angle. Let us say we have a system of two lines. Three cases arise: (1) The lines are parallel to each other with a certain (non-zero) gap between them (fig 1). (2) The lines are parallel to each other with no gap between them (fig 2). This means the lines are coincident. This means that a line is parallel to itself. (3) The lines intersect each other in a point (fig 3).

508

Plane Geometry In case the two lines intersect, vertically opposite angles are equal. Therefore,  = and  = .

Fig 1

Fig 2

Fig 3

In case of non-coincident parallel lines, if there is a third line which intersects these parallel lines, then such a line is called transversal line. If p and q are the parallel lines and r the transversal line, then (p, r) and (q, r) form two systems of intersecting lines. Since vertically opposite angles are equal, a = c = f = g and b = d = e = h (refer to fig 4).

Fig 4

Fig 5

Basic Proportionality Theorem (BPT) If we have a system of three parallel lines (p, q and r) and two transversals x and y. The two transversals intersect the parallel lines in the points A, B, C, D, E and F (fig 5). Basic Proportionality Theorem, (also known as Thale’s Theorem), says “the lengths of the transversals intercepted between the parallel lines are AB DE  , irrespective of the gap between the parallel lines”. proportional, that is, BC EF Converse of this is also true, that is, if the lengths of the transversals intercepted between three lines are proportional, then the three lines are parallel.

Triangles A unique line passes through two distinct points. If a single line passes through three distinct points, then such points are called collinear points. If the three points are not collinear, then three lines are needed to connect them together and the shape so generated is a triangle. We can also say that a triangle is a three sided closed polygon. Note that the sum of the interior angles in a triangle is 180°. Two points are worth noting in triangles: (i)

Sum of two sides of a triangle is more than the third side. Alternatively, difference of two sides of the triangle is less than the third side. If this condition is not met, then the triangle cannot be constructed. For example, if the sides are 3, 4 and 8, then sum of two sides is 3 + 4 = 7 which is less than 8, and hence with these sides, a triangle cannot be constructed.

(ii)

In a triangle ABC, if we extend the side BC to P so that ACP is an exterior angle, then this exterior angle is equal to the sum of opposite interior angles, that is, ACP = ABC + BAC

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Quantitative Aptitude Simplified for CAT

Example 1 Prove that the sum of the angles in a triangle is always 180°. Solution Draw a triangle ABC and a line PAQ parallel to BC passing through A. Then, PAB + BAC + CAQ = 180°. But, PAB = ABC and CAQ = ACB. Therefore, PAB + BAC + CAQ = ABC + BAC + ACB = 180°.

Example 2 Show that in a triangle ABC, if D and E are mid-points of AB and AC, then DE is half of BC. Solution Since D and E are mid-points of the given sides, therefore

AD AE 1   . DB EC 1

Hence DE is parallel to BC (according to converse of BPT). Further, ABC and ADE are similar triangles, because A = A, B = D,C = E. Therefore,

AB BC  or DE = BC. AD DE

This is known as mid-point theorem. Example 3 A triangle is impossible if the ratio of its sides is A.

4:5:3

B.

4 : 7 : 12

C.

7:5:8

D.

1:1:1

Solution The triangle whose sum of any two sides is less than or equal to the third side is an impossible triangle. Since 4 + 7 < 12, this triangle is not possible. Hence correct option is [B]. Example 4 In a ABC, the true statement is A.

AC2 = AB2 + BC2

B.

AC = AB + BC

C.

AC > AB + BC

D.

AC < AB + BC

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Plane Geometry Solution Sum of any two sides is more than the third side. Hence correct option is [D].

Types of triangles There are two ways to classify triangles. Classifying triangles on the basis of the sides: Scalene triangles are those triangles whose all three sides are of unequal lengths. Isosceles triangles are those whose any two sides are equal and the third is not equal to the other two. Equilateral triangles are those whose all the three sides are of same length. Area of an equilateral triangle =

1 1 3 3 2 (Base) (Height)   a  a a , where a is the side. 2 2 2 4

1 1  a Area of isosceles triangle = (a) (height)  a b2    2 2  2

2



1 a 4b2  a2 . 4

When a = b, this reduces to the area of an equilateral triangle. Area of all the triangles having same base and lying between same pair of parallel lines is same.

Classification of triangles Classifying triangles on the basis of angles: Acute triangles are those whose all the three angles are acute. Right triangles are those whose one of the angles is right angle. Obtuse triangles are those whose one of the angles is obtuse. In a right angled triangle ABC (as drawn below), Pythagoras’ theorem is applicable, according to which AB2 + BC2 = AC2. The natural numbers which follow this formula are known as Pythagorean Triplets. So, (3, 4, 5), (5, 12, 13) are some of the common triplets. 32 + 42 = 52; 52 + 122 = 132. More such triplets are: (7, 24, 25), (8, 15, 17) and so on. All multiples of triplets are also triplets. So, if (3, 4, 5) is a triplet, then (6, 8, 10) is also a triplet.

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Also, length of altitude BD from B on AC can be found using the formula for area of triangle. Now, Area of ABC = BD =

AB  BC 1 1  BC  AB   AC  BD  BD = . So, if the sides of the triangle are (3, 4, 5), then 2 2 AC

34 = 2.4. 5

There is a standard way of drawing a triangle. When the vertices of the triangle are named as A, B and C, then the sides are named as ‘a’, ‘b’ and ‘c’ such that side of length ‘a’ is opposite to vertex A, side of length ‘b’ is opposite to vertex B and same for ‘c’. This standard will be followed throughout this chapter unless mentioned otherwise.

Let us drop an altitude from vertex A to the side BC, meeting the side at D.

By elementary principles of trigonometry, we know that AD = b sin C or c sin B. Therefore, b sin C = c sin B or

b c  . sinB sin C

Similarly, for the altitude from vertex B, meeting AC in E, BE = a sin C or c sin A. Therefore,

a c  . sinA sin C

a b c   , which is known as sine rule. This rule helps us understand sinA sin b sin C that the sides of a triangle are proportional to sines of the angles. Further,

Combining the above results,

cos A =

b2  c2  a2 ; 2bc

cos B =

c2  a2  b2 ; 2ca

cos C =

a2  b2  c2 . 2ac

This is known as cosine rule. Observing the pattern, one should be able to remember the results. If angle A is 90°, then cos A = 0 and hence b2 + c2 = a2 which is Pythagoras’ Theorem.

Similarity of triangles Two triangles are “similar” if they have same shape but may have different sizes. Therefore, similar triangles are those whose corresponding angles are equal and corresponding sides are proportional. In triangles ABC and DEF (shown below), A = D; B =E; C = F and

AB BC AC   DE EF DF

and therefore, ABC and DEF are similar. This is written as ABC ~ DEF.

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Plane Geometry

It is extremely important to understand that there must be correct correspondence. So, if ABC ~ DEF, then A corresponds to D, B to E and C to F. That means the order is important. If, on the other hand, if A corresponds to D, B to F and C to E, then we would say that ABC ~ DFE and not ABC ~ DEF. This is also true of congruence of triangles. There are certain rules using any of them we can prove that two given triangles are similar. In the following rules, S corresponds to Side and A corresponds to Angle. For example, SAS means Side Angle Side.

AAA Rule According to this rule, if all the three angles in a triangle are correspondingly equal to the angles in the other triangle, then the two triangles are similar, that is, in ABC and DEF, A = D, B = E and C = F ABC ~ DEF.

AA Rule According to this rule, if any two angles in a triangle are correspondingly equal to any two angles in the other triangle, then the two triangles are similar, that is, in ABC and DEF, A = D and B = E ABC ~ DEF. This is because if two angles are correspondingly equal, then the third has to be equal as the sum of the angles in a triangle is 180°.

SAS Rule According to this rule, if two sides are proportional to the corresponding sides in the other triangle and the included angle (that is, angle between the sides) are correspondingly equal, then the two triangles are similar. AB BC  and ABC = DEF, then that is, in ABC and DEF, DE EF ABC ~ DEF. An important point to be noted is that the angle should be the included angle and not any other angle. If the angle is not an included angle, then the triangles may not be similar. That is why the rule is ‘SAS’ and not ‘SSA’. This is shown diagrammatically in the following figures.

AB BC  and ACB = DFE (non-included angles are equal). But the triangles are not DE EF similar as ABC DEF.

Here, we can see that

SSS Rule According to this rule, all the three sides of one triangle are correspondingly proportional to all the three sides of the other triangle, that is, in ABC and DEF,

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Quantitative Aptitude Simplified for CAT

ABC ~ DEF. If two triangles are similar, then

s1 h1 m1 R1 r1      s2 h2 m2 R2 r2

A1 , A2

where s1, h1, m1, R1, r1 and A1 refers to side, height, median, circumradius, inradius and area of one triangle and s1, h1, m1, R1, r1 and A1 refers to the same things for the other triangle. We will understand the meaning of terms like median, circumradius and inradius later.

Congruence of triangles Two triangles are congruent when they have same shape and size. Obviously, if two triangles are congruent, then one can be completely superimposed on the other. For two triangles to be congruent, all the corresponding sides must be equal. Obviously, when the sides are correspondingly equal, the angles automatically become correspondingly equal, that is, if ABC and DEF are congruent, then AB = DE, BC = EF and AC = DF. This is written as ABC DEF.

Like rules of similarity, there are certain rules and by using any of them we can prove that two given triangles are congruent.

ASA Rule According to this rule, if two angles and one of the sides of a triangle are equal to the corresponding two angles and side of the other triangle, then the two triangles are congruent. It is important to note here that the side need not be enclosed between the angles. This is so because two angles being correspondingly equal means the third angle also becomes equal to the corresponding third angle of the other triangle. Therefore, a corollary to the given rule is AAS rule (Angle Angle Side rule).

SAS Rule According to this rule, if two sides and the included angle in one triangle are equal to the corresponding sides and angle in the other triangle, then the two triangles are congruent. that is, in ABC and DEF, AB = DE; BC = EF and ABC = DEF, then ABC ~ DEF. Once again, like in rule of similarity, the angle should be an included angle and not any other angle.

SSS Rule According to this rule, if all the three sides of one triangle are correspondingly equal to all the three sides of the other triangle, then the two triangles are congruent, that is, in ABC and DEF, AB = DE, BC = EF and AC = DF ABC ~ DEF.

RHS Rule Here, RHS means Right angle Hypotenuse Side. According to this rule, if hypotenuse and one side of one triangle are correspondingly equal to the hypotenuse and the one side of the other triangle, then the two triangles are congruent, that is, in ABC and DEF, given that B = E = 90°, then if AC = DF and AB = DE (or BC = EF), then the triangles are congruent.

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Plane Geometry Example 5 ABC is right angled at A and AD is perpendicular to the hypotenuse. Find A.

AB2 AC

B.

AB AC

C.

AB2 AD

D.

AB AD

BD . DA

Solution Triangles ADB and CDA are similar triangles. Therefore,

AD BD AB   . Hence correct option is [B]. CD DA AC

Example 6 If AD is altitude to the hypotenuse of a right triangle ABC, then (i)

ABC and CAD are similar

(ii)

ADB and CDA are similar

(iii) ADB and CAB are similar A.

(i) and (ii)

B.

(i) and (iii)

C.

(ii) and (iii)

D.

All the three

Solution The three triangles ADB,CDA and CAB are similar. Hence correct option is [C]. Note that because of the similarity of these triangles, we get the following results. Let AB = a, AC = b, BD = m, CD = n and AD = p (refer to the following figure).

Since ADB and CDA are similar,

AD DB AB p m a   or   . We obtain CD DA CA n p b

p2 = mn. Similarly, ADB and CAB are similar,

AD DB AB p m a   or   . CA AB CB b a mn

We obtain a2 = m(m + n). Similarly, from the similarity of CDA and CAB, we obtain b2 = n(m + n). These three are important results and must be remembered.

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Quantitative Aptitude Simplified for CAT

Example 7 ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. DEF is similar to ABC. If EF = 4 cm, then perimeter of DEF is A.

7.5 cm

B.

15 cm

C.

22.5 cm

D.

30 cm

Solution Since the ratio of EF to BC is 2 : 1, we find that DEF has each corresponding dimension double to that of ABC. Therefore, perimeter of DEF = 2(3 + 2 + 2.5) = 15 cm. Example 8 In the figure, AB, CD and XY are parallel. If AB = 3 cm and CD = 5 cm, find XY. A.

2 cm

B.

1.25 cm

C.

1.75 cm

D.

1.875 cm CAT 2003 (C)

Solution Since AB is parallel to XY, ABC ~ XYC. Let XY = m. Then,

AB AC 3 AC m  AC   or . Therefore, XC = . XY XC m XC 3

Similarly for the ‘similar’ triangles DAC and YAX, AX =

5 AC  . Therefore, m AX

m  AC . Adding these two equations, we get 5

15 1 1 AC = XC + AX = m  AC    or m = = 1.875 cm. 8  3 5 Hence correct option is [D]. Infinite lines can pass through a given point. Those lines may not be in one plane. Such lines are collectively called concurrent lines and the point is called point of concurrency. Every triangle has a unique set of 4 points: centroid (denoted by G), orthocentre (denoted by O), circumcentre (denoted by S) and incentre (denoted by ).

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Plane Geometry

Centroid In a triangle, if any vertex is joined to the mid-point of the opposite vertex, then the line so obtained is referred to as median. The three medians (corresponding to three vertices in a triangle) are always concurrent where the point of concurrency is called centroid, denoted by G. Centroid of a triangle always divides the medians in the ratio of 2 : 1. For example, in ABC, if AD is the median, and G the centroid, then AG : GD = 2 : 1. Further, every median bisects the area of the triangle, that is, if AD is the median of ABC, then Ar (ABD) = Ar (ACD).

Reason: If BC = a, then BD = DC =

a . 2

If the height of the triangle ABC is h, then area (ABD) = Hence we can say that Ar (ABD) = Ar (ACD) =

1 a 1   h   [Ar (ABC)] 2 2 2

1  [Ar (ABC)] 2

Now, GD is the median of the triangle GBC and hence Ar (GBD) = Ar (GCD). Similarly, Ar (GAF) = Ar (GBF) and Ar (GAE) = Ar (GCE). Since GD is

1 1 rd of AD (centroid divides the median in the ratio of 2 : 1), Ar (GBC) = Ar (ABC) 3 3

All this implies that Ar(GAB) = Ar(GBC) = Ar(GCA) =

1 Ar(ABC). 3

Moreover, areas of all the six triangles generated are also equal, and equal to

Apollonius’ Theorem Let the lengths of the medians AD, BE and CF be l, m and n (respectively). Then, according to Apollonius’ Theorem, 2l2 +

1 2 a = b 2 + c 2; 2

2m2 +

1 2 b = c2 + a2; 2

2n2 +

There are three ways to prove the theorem. (i)

Using cosine rule.

(ii)

Using a property regarding parallelograms.

(iii) Using Pythagoras’ Theorem.

Proof using cosine rule Refer to the figure above.

517

1 2 c = a2 + b2. 2

1 Ar (ABC). 6

Quantitative Aptitude Simplified for CAT

Applying cosine rule to ABC, we get cos B =

a2  c 2  b2 2ac 2

 a 2 2   c   2 Applying cosine rule to ABD, we get cos B =  a 2   c  2 2

Equating the two,

a2  c 2  b2 2ac

 a2 + c2 – b2 = 2 

 a 2 2   c   2 =  a 2   c  2

1 a2 + 2c2 – 2l2  2l2 + a2 = b2 + c2. 4 2

Proof using a property regarding parallelograms Extend the median AD so that AD = DP (refer to the diagram below). Complete the parallelogram. Now, there is a result in a parallelogram that says “sum of the squares of the sides = sum of the squares of the diagonals” Applying this result in the new parallelogram, we get AP2 + BC2 = AC2 + CP2 + PB2 + BA2, or (2l)2 + a2 = 2(b2 + c2)  2l2 + a2 = b2 + c2. Hence the result.

The students are advised to prove Apollonius’ Theorem by Pythagoras’ Theorem, themselves. Example 9 In ABC, the mid-points of the sides BC, CA and AB are D, E and F respectively. If area of ABC = 4 cm2, what is the area of DEF? Solution Since D is the midpoint of BC, AD is the median and hence area of ABD = Further, F is the midpoint of AB. So, area of FBD =

1 (Area of ABC). 2

1 1 (Area of ABD) = (Area of ABC). 2 4

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Plane Geometry

Similarly, area of each of the 4 small triangles is each other. Hence area of DEF =

1 (Area of ABC), and so area of these triangles are equal to 4

1 (Area of ABC) = 1 cm2. 4

Example 10 In ABC, AB is divided into 5 equal parts, BC into 7 equal parts and AC into 4 equal parts. What is the area of PQR (as shown in the diagram), if AB is an equilateral triangle of side 5 cm? Solution Refer to the following diagram. AB is divided by 4 points (shown) into 5 equal parts; BC is divided by 6 points into 7 equal parts; CA is divided by 3 points into 4 equal parts.

5 5 ar(ABC) = k. Since P divides AB in the ratio of 3 : 2, PQ 7 7 2 5 divides the area of ABQ in the same ratio. So, ar(PBQ) = × k. 5 7

Let area of ABC = k. If we draw AQ, then ar(ABQ) =

3 1 2 1 2 1 k  area of RQC = × k. Also, area of ABR = k  area of APR = k. So, 7 2 7 2 5 2 3 1 2 5 1 2  3 2 1 51  21  30  ar(APR) + ar(BPQ) + ar(RQC) = × k + × k + × k    k k . Therefore, k 5 2 5 7 2 7  10 7 7  70  70  Similarly, area of AQC =

area of PQR =

19 3 3 19 95 3 k . Since area of triangle ABC, k =  52   cm2.  52 , area of PQR = 70 4 4 70 56

Orthocentre A perpendicular dropped from any vertex to the opposite side, is called altitude. The three altitudes in a triangle are always concurrent and the point of concurrency is called orthocentre, denoted by O. Two results regarding orthocentre are worth mentioning. A.

In ABC, if O is the orthocentre, then BOC = 180° –A. In any quadrilateral, sum of all the four angles is always 360°. This is so because when a diagonal is drawn, we get two triangles and each triangle contributes 180°.

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Quantitative Aptitude Simplified for CAT

Fig 1

Fig 2

Now, in the quadrilateral AFOE (fig 1 above), F = E = 90°. Therefore, FOE + A = 180°. But, FOE = BOC (vertically opposite angles). Therefore, BOC + A = 180° or BOC = 180° – A. B.

Of the four points A, B, C and O, each of the points is the orthocentre for the triangle formed by the other three points. Therefore, for OBC, A is the orthocentre, and likewise (refer to fig 2 above). This also shows that orthocentre may lie outside the triangular region. In general, (i)

orthocentre will be outside the triangle if it is an obtuse triangle.

(ii)

orthocentre will be inside the triangle if it is an acute triangle.

(iii)

orthocentre will be on the triangle if it is a right triangle. To be more precise, orthocentre for right angled triangle is the right angle vertex. In ABC, right angled at B (fig below), if altitudes are dropped from A and C, they will meet in B and hence the vertex B is the orthocentre.

Circumcentre Locus of a point which is equidistant from two given points is called perpendicular bisector. So, if the two given points are A and B, then locus of P is such that PA is always equal to PB. Obviously, perpendicular bisector of AB passes through the mid-point of the line joining A and B and is perpendicular to it. The three perpendicular bisectors are concurrent and the point of concurrency is called circumcentre, denoted by S. The three dotted lines shown are perpendicular bisectors of the sides of the triangle.

Since S is on the perpendicular bisector of A and B, SA = SB. Similarly, S being on the perpendicular bisector of B and C, SB = SC. Therefore, SA = SB = SC and hence with S as the centre, a circle can be drawn passing through the points A, B and C. Such a circle is called circumcircle and the radius of such a circle is called circumradius (denoted by R).

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Plane Geometry

If three distinct points are collinear, then no circle can pass through all of them. If the three points are noncollinear, then a unique circle passes through them. This is only because three points which are non-collinear form a triangle and every triangle has a unique circumcentre and hence a unique circumcircle. In figure above, diameter AC’ is drawn and C’ is joined with B. Now, ACB = AC’B = C. Moreover, ABC = AB c c   2R . Therefore, sine rule becomes: 90 (as AC’ is the diameter). Therefore, sin C   AC 2R sin C a b c    2R . sinA sinB sinC This is a standard result.

Refer to figure above. (i)

circumcentre will be outside the triangle if it is an obtuse triangle, ABC.

(ii)

circumcentre will be inside the triangle if it is an acute triangle, APQ.

(iii) circumcentre will be on the triangle if it is a right triangle, ADE. To be more precise, circumcentre for right angled triangle is the mid-point of the hypotenuse. In ADE, right angled at A, if a circumcircle is drawn, then DE becomes the diameter (angle in a semi-circle is right angle). Therefore, midpoint of DE (diameter or hypotenuse) is the circumcentre. For a right triangle ABC, median to the hypotenuse is half of the hypotenuse. Proof: Let ABC be right angled at B. Let D be the mid-point of the hypotenuse AC. We have seen above that D will be the circumcentre. BD will be the median to the hypotenuse, which is also the radius of the circle. Therefore, BD = Radius of the circle =

1 1 1 of the diameter =  AC = (hypotenuse) 2 2 2

For a right triangle ABC, the line joining orthocentre and circumcentre is the median to the hypotenuse and also the circumradius. This is now easy to understand as the right angle vertex is the orthocentre and the mid-point of the hypotenuse is the circumcentre.

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Quantitative Aptitude Simplified for CAT

Incentre If a line bisects the angle, then the line is called as angle bisector. The three angle bisectors in a ABC are concurrent and the point of concurrency is called incentre, denoted by . The dotten lines shown are the angle bisectors.

From , drop perpendiculars on the sides BC and AC meeting in D and F. Now, DB FB, because (i)

side B is common,

(ii)

BD = BF (because B is angle bisector), and

(iii) DB = FB = 90° (by construction) Therefore, D = F. Similarly, D = E. Hence, with  as the centre and D or E or F as the radius, a circle (called incircle) can be drawn which touches all the three sides of the triangle. The radius of the circle is called inradius, denoted by r. Important Result: If  is the incentre of ABC, then BC = 90 +

1 A. 2

Proof: In ABC, A + B + C = 180°. Also, in BC, BC + BC + CB = 180° But, BC =

1 1 ABC and CB = ACB 2 2

BC +

1 (B + C) = 180° 2

BC +

1 1 (180 – A) = 180 BC = 90 + A. 2 2

Note: (i)

In an equilateral triangle, all the four points, that is centroid, orthocentre, circumcentre and incentre are coincident, that is, same.

(ii)

In an isosceles triangle, all the four points, that is centroid, orthocentre, circumcentre and incentre are collinear.

(iii) Centroid of any triangle divides the line joining orthocentre and circumcentre in the ratio of 2 : 1, that is, if G, O ands S are centroid, orthocentre and circumcentre, then OG : GS = 2 : 1. This also implies that the three points are always collinear irrespective of the nature of triangle.

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Plane Geometry Example 11 In triangle PQR, P = 70°. The internal bisectors of angles Q and R meet at S. The value of QSR will be A.

95°

B.

160°

C.

135°

D.

125°

Solution S is obviously the incentre. Therefore, QSR = 90 +

1 P = 125°.Hence correct option is [D]. 2

Example 12 In the ABC, the bisectors of the exterior angles B and C meet at O. If BAC = 65° and ABC = 75°, find BOC. A.

57.5

B.

50

C.

60

D.

none of these

Solution O is obviously the excentre.

ABC + ACB + BAC = 180°. Similarly, Therefore, BOC =

180  B 180  C  + BOC = 180. 2 2

1 1 1 (B + C) = (180 – A) = 90° – A = 90 – 32.5 2 2 2

= 57.5°. Hence correct option is [A]. Example 13 With each side of a triangle as diameter, three circles are drawn. The point of intersection of the three common chords is A.

centroid

B.

orthocenter

C.

circumcenter

D.

incenter

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Quantitative Aptitude Simplified for CAT

Solution Refer to the diagram below:

In the diagram shown, we have drawn two circles for simplicity. Since angle in a semi-circle is 90°, ADB = 90° and similarly, ADC = 90°. Therefore, AD is altitude from A on BC. But, AD is the common chord. Similarly, for other common chords. Therefore, all the common chords are nothing but altitudes. Therefore, point of intersection of common chords is orthocenter (point of intersection of altitudes). Example 14 An equilateral triangle is inscribed in a circle of radius 10 cm and a square is inscribed in this triangle. What is the side of the square? Solution Center of the circle is the circumcenter of the equilateral triangle, which divides the height of the triangle in the ratio of 2 : 1. If radius of the circle is 10 cm, then height of the equilateral triangle = 15 cm. Let side of the square be x. Also, APQ is also equilateral triangle, whose side will be x (because PQ = x). Height of APQ =

3 3 x . So, AD = x + x, which is 15. Therefore, x = 2 2

30 cm. 3 2

Angle Bisector Theorem According to the theorem, if AN is the angle bisector of BAC in ABC, then

In ABN, using sine rule,

Also, in ACN,

AB BN  , where ANB = . sinα sin(A/2)

AC NC AC NC    . sin(180-) sin(A/2) sinα sin(A/2)

524

AB BN  . AC NC

Plane Geometry

AB BN  . AC NC

Therefore,

The angle bisector theorem also applies if AN is the external angle bisector instead of internal angle bisector. In the following diagram, AN is the external angle bisector of BAC. Therefore, AN bisects the angle into two equal angles, each is . BAC = 180 – 2 BAN = 180 – . Let ANC = . Applying sine rule in ABN,

Similarly, in ACN,

Therefore,

BA sin sin   . BN sin(180-) sin

AC sin  . NC sin

AB AC AB BN    . BN NC AC NC

In ABC, let AD, BE and CF be angle bisectors intersecting in the point (incentre). Then, A : D = (b + c) : a, where a, b and c are the sides of the triangle as per the standard. This is an important result of angle bisector theorem. Proof: Since AD is the angle bisector of BAC,

Adding 1, we get

AB BD  . AC CD

AB  BC BD  DC bc a bc b      . AC CD b DC a DC

In ACD, C is the angle bisector. Therefore,

Combining, we get ,

CA AI b AI    . CD ID DC ID

b  c AI  that is, A : D = (b + c) : a. a ID

Example 15 A triangle ABC is given and AD is an angle bisector, meeting BC at D. A. Triangle ABD may be equilateral triangle B. Triangle ABD can be isosceles triangle C. Triangle ABD can be a right triangle D. Exactly two of the above three options are correct. Solution: D Option [A]: If ABD is equilateral triangle, then ABD = 60, BAD = 60. Since AD is angle bisector, BAC = 120  ACB = 0, which is not possible. Hence [A] is incorrect option. Triangle ABD can be isosceles triangle. Hence [B] is correct option. Triangle ABD can also be right triangle (in case ABC is isosceles triangle). Hence [C] is also correct option.

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Quantitative Aptitude Simplified for CAT

Hence correct option is [D].

Area of triangle Area of ABC = =

1 1 1 1 abc  base  ht  ab sin C  = bc sin A = ca sin B = = rs 2 2 2 2 4R

s( s  a)( s  b)( s  c) , where

a, b and c are the sides of the triangle, A, B and C are the angles of the triangle, s = semi-perimeter =

a b c , R = circumradius and r = inradius. 2

Example 16 Prove that area of a triangle is given by (i)

abc 4R

(ii)

rs

Solution (i)

Area = =

(ii)

1 1  c  ab (sin C) = ab   2 2  2R 

(because

a b c    2R ) sinA sinB sinC

abc (by sine rule) 4R

If  is the incentre, then Ar (ABC) = Ar (BC) + Ar (CA) + Ar (AB) 1 Ar (BC) = ar, where a is the base and r is the ht. 2

1 1 br and Ar (AB) = cr. 2 2 1 Therefore, Ar (ABC) = (a + b + c)r = rs. 2

Similarly, Ar (CA) =

Example 17 The radius of the incircle in the diagram will be A.

1.8 cm

B.

2 cm

C.

2.5 cm

D.

3.6 cm

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Plane Geometry

Solution Area of triangle = rs or r =

Area . Area = 24 and s = 12. Therefore, r = 2 cm. Hence correct option is [B]. s

EXPERT SPEAK Scan this QR Code to watch a video that explains how to use properties of triangles to problems that have actually appeared in the CAT.

Example 18 The base of a triangular field is three times altitude. If the cost of cultivating the field at Rs. 24 per sq metre is Rs. 324, find its base and height. Solution Let x be the altitude. Then base = 3x. Therefore, area =

24 

1 3 x2 . Therefore, (3 x )( x )  2 2

3 x2 = 324 or x = 3. Therefore, altitude = 3 and base = 9. 2

Example 19 One side other than the hypotenuse of a right-angled isosceles triangle is 4 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is A.

8 cm

B.

4 2 cm

C.

4 cm

D.

2 2 cm

Solution The hypotenuse will be 4 2 . If length of perpendicular is x, then area of triangle =

1 1  4 4  x   4  4 or x = 2 2 cm. 2 2

Example 20 ABC is an equilateral triangle. BE is perpendicular to CA meeting CA at E. Then AB2 + BC2 + CA2 is equal to A.

2 BE2

B.

3BE2

C.

4BE2

D.

6BE2

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Quantitative Aptitude Simplified for CAT

Solution 3 a, where a is the side of the triangle. Therefore, 2 AB2 + BC2 + CA2 = 3a2 = 4BE2. Hence correct option is [C].

BE = altitude of the triangle =

Example 21 There is a river and two persons A and B are at a certain distance away from the bank of the river (obviously on the same bank!). They are located as shown in the diagram below. What is the shortest way for A to go to the river and reach B?

Solution For the shortest distance, the path should be as drawn below.

A should look at the image of B in water, which is at B’, follow the line of sight from A along B’. Upon reaching the bank of the river, he should then move towards B. This will be the shortest path. This is so because the bank of the river acts as the perpendicular bisector of BB’. And so, P is equidistant from B and B’. So, AP + PB = AP + PB’ = AB’ is the distance which A would have to travel. Since AB’ is the straight line path and straight line is the shortest distance between any two points, we can say that A – P – B is the shortest route. Example 22 Of all the triangles with the same base length and equal areas, the one which has the least perimeter is A.

equilateral

B.

right angled

C.

isosceles

D.

none of these

Solution Let ABC be the triangle. Let the base we are talking of in the question be BC. Perimeter = AB + BC + CA. Since BC is constant for all the triangles, we want to minimize the value of AB + AC. When area is constant, the value of AB + AC will be minimized when they are equal in value. Therefore, the triangle must be isosceles. Hence correct option is [C]. Example 23 Three sides and one of the heights of a triangle are four consecutive integers. Find the smallest integer.

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Plane Geometry A.

10

B.

11

C.

12

D.

such a triangle is impossible

Solution Check options. Let the triangle be ABC with AD as the altitude. We observe that AD = 12, AC = 13, BC = 14 and AB = 15. Note that DC = 5 and DB = 9 (using Pythagoras’ Theorem). Example 24 A line AB (of length 10 cm) and a point P are such that APB is always a right angle. How many such points P are possible if area of APB is (i)

20 cm2

(ii)

25 cm2

(iii) 30 cm2 Solution Since APB must be always 90, we can draw a circle with AB as diameter and P on this circle. The area of APB is maximum when P is on the perpendicular bisector of AB. Area of such a triangle is equal to 1  10  5  25 cm2. At any other point, the area will be less than this. Therefore, under any circumstances, the 2 area cannot exceed 25 cm2. Therefore, in case of (c) when area needs to be 30 cm2, no such triangle is possible and hence no point P can exist. When area needs to be 25 cm2, there are only two points P possible: one above AB and other below AB. For the case of area 20 cm2, there will be 4 points possible: two above AB and two below AB.

EXPERT SPEAK Scan this QR Code to watch a video that is a continuation of the previous video on Triangles.

Quadrilaterals Four sided closed figures are called quadrilaterals. In a quadrilateral ABCD, if we draw diagonal AC, then the quadrilateral breaks into two triangles, whose sum of angles is 180° each. Therefore, sum of the angles in a quadrilateral is always 360°. Let ABCD be a quadrilateral (Fig 1 below). Join A with C and let length of AC be d. Area of the quadrilateral will be the sum of the areas of the two triangles ABC and ADC (refer to figure below). Area of ABC =

1 1 dp. Area of ADC = dq. 2 2

Therefore, area of the quadrilateral ABCD = =

1 (p + q)d 2

1 (sum of perpendiculars from the opposite vertices on the diagonal) (length of that diagonal). 2

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Quantitative Aptitude Simplified for CAT

Fig 1

Fig 2

In the same quadrilateral, join B with D (refer to fig 2 above) and let length of BD be g. Let the two diagonals intersect at O. Let OB = g1 and OD = g2. Then, p = g1 sin, where  is the angle enclosed by the two diagonals. Similarly, q = g2 sin. Therefore, area of the quadrilateral =

1 1 1 1 1 (p + q)d = (g1sin + g2sin)d = (g1 + g2)(d)(sin) = gd sin = (AC)(BD)sin 2 2 2 2 2

=

1 (product of diagonals)(sine of the angle between the diagonals) 2

This is an important formula for calculating the area of any quadrilateral. Note that the quadrilateral obtained by joining (in order) the mid-points of any quadrilateral is always a parallelogram whose area is half of the area of the quadrilateral. In the diagram below, ABCD is any quadrilateral and PQRS is a parallelogram.

Logic: P and Q are the mid-points of AB and BC respectively. Then, PQ is parallel to AC and half of AC (by mid-point theorem for triangle ABC). Similarly, for triangle ADC, RS is parallel to AC and half of it. Therefore, PQ is parallel to RS and is of same length as RS. Similarly, PS is parallel and equal to QR. Hence PQRS is a parallelogram. As regards the area, let AC = d1 and BD = d2. and let the angle between AC and BD be . Then PQ =

d1 d and PS = 2 and angle between PQ and PS is . 2 2

1 1 1  1  Area of the parallelogram = (PQ)(PS) sin =  d1   d2  sin d1d2 sin = (Area of ABCD) 4 2 2  2  Hence, area of the parallelogram PQRS is half of the area of the quadrilateral ABCD. Alternatively, Ar (BPQ) =

1 1 Ar (BAC) and Ar (DSR) = Ar (DAC). 4 4

Therefore, Ar (BPQ) + Ar (DSR) = Similarly, Ar (APS) =

1 1 [Ar (BAC) + Ar (DAC)] = Ar (ABCD) 4 4

1 1 Ar (ADB) and Ar (CRQ) = Ar (CDB). 4 4

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(I)

Plane Geometry

Therefore, Ar (APS) + Ar (CRQ) =

1 1 [Ar (ADB) + Ar (CDB)] = Ar (ABCD) 4 4

Adding (I) and (II), Ar (BPQ) + Ar (DSR) + Ar (APS) + Ar (CRQ) = Hence Ar (PQRS) =

(II)

1 Ar (ABCD) 2

1 Ar (ABCD). 2

We have seen earlier that through three non-collinear points, a unique circle passes. Through four noncollinear points, a circle may or may not pass. If a circle passes, then the four points are called concyclic points. Moreover, the quadrilateral obtained by joining those 4 points is called cyclic quadrilateral. Therefore, those quadrilaterals are cyclic which can be inscribed in a circle. In the cyclic quadrilateral ABCD (diagram below), opposite angles are supplementary, that is, sum of the opposite angles is always 180°.

This is so because  = 2ADC and  = 2ABC. Adding them, we get 2(ADC + ABC) =  + = 360°. Therefore, ADC + ABC = 180°. If the four sides of a cyclic quadrilateral are a, b, c and d, then area = ( s  a)( s  b)( s  c)( s  d ) . Let a quadrilateral be such that a circle can be inscribed in it (refer to the figure below). Mark the points of contact as K, L, M and N. Then, AL = AM (pair of tangents from A). Let us call this ‘p1’. Similarly, BM = BN = p4; CN = CK = p3 and DK = DL = p2. AB + CD = AM + MB + DK + CK = p1 + p2 + p3 + p4. AD + BC = AL + LD + BN + CN = p1 + p2 + p3 + p4. Hence, AB + CD = AD + BC. This is a standard result.

If this quadrilateral were cyclic, then semi-perimeter, s =

AB  BC  CD  DA = p1 + p2 + p3 + p4. 2

If AB = a, BC = b, CD = c and DA = d, then s = a + c = b + d Therefore, s – a = c, s – b = d, s – c = a and s – d = b.

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Quantitative Aptitude Simplified for CAT

Area of cyclic quadrilateral =

(s  a)( s  b)(s  c )( s  d)  abcd .

Common types of quadrilaterals are square, rectangle, rhombus, parallelogram, trapezium and kite.

Square In a square, the following observations hold: (i)

All the four sides are equal; let us call them ‘a’.

(ii)

All the four interior angles are right angles.

(iii) Diagonals are equal (let us call them‘d’) and bisect each other at right angles. (iv) Area = a2 = (v)

1 2 d. 2

Let P be any point inside the square (ref fig 2 below). Then, PA2 + PC2 = PB2 + PD2.

Fig 1

Fig 2

Rectangle In a rectangle, the following observations hold: (i)

Opposite sides are equal, but adjacent ones are not. So, length is ‘a’ and breadth (or width) ‘b’.

(ii)

All the four interior angles are right angles.

(iii) Diagonals are equal (let us call them ‘d’) and bisect each other, but not at right angles. (iv) Area = ab = (v)

1 2 d sin, where is the angle between the diagonals. 2

Let P be any point inside the rectangle (ref fig 2 below). Then, PA2 + PC2 = PB2 + PD2.

Fig 1

Fig 2

Rhombus In a rhombus (refer fig 1 below), following observations hold: (i)

All the four sides are equal, let us call them ‘a’.

(ii)

The interior angles are not right angles. Angles at the opposite vertices are equal.

(iii) Diagonals are unequal (let us call them, ‘d1’ and ‘d2’) and bisect each other at right angles. (iv) Area = a2 sin  =

1 d1d2, where  = angle between adjacent sides. 2

532

Plane Geometry

Fig 1

Fig 2

Parallelogram In a parallelogram (refer fig 2 above), following observations hold: (i)

Opposite sides are parallel and equal, but adjacent ones are not. So, length is ‘a’ and breadth (or width) ‘b’.

(ii)

Interior angles are not right angles. Angles at the opposite vertices are equal.

(iii) Diagonals are unequal (let us call them, ‘d1’ and ‘d2’) and bisect each other, but not at right angles.

1 d1d2sin, where is the angle between adjacent sides and is the angle 2 between the diagonals.

(iv) Area = ab sin  =

Property 1 The angle bisectors of the four angles of the parallelogram enclose a rectangle. Refer to the diagram below. AP is bisector of DAB and BP is bisector of CBA. Now DAB + CBA = 180°. Therefore, PAB + PBA = 90°. Therefore, APB = 90°. Similarly, Q, R and S are also right angles. This is sufficient condition for PQRS to be a rectangle.

Property 2 Sum of squares of sides of a parallelogram is equal to sum of squares of diagonals of the parallelogram. Let ABCD be a parallelogram.

Let AB = a and BC = b. Let AC = p and BD = q. Using cosine rule in ABC, cos B =

In BCD, cos C =

a2  b2  p2 ; 2ab

a2  b2  q2 . 2ab

But cos C = cos(180 – B) = –cosB.

533

Quantitative Aptitude Simplified for CAT

Therefore,

a2  b2  p2 a2  b2  q2  or 2(a2 + b2) = p2 + q2. Hence the result. 2ab 2ab

Property 3 Area of a parallelogram on the same base between a pair of parallel lines is same.

Trapezium In a trapezium, following observations hold: (i)

Only one pair of opposite sides is parallel and the other pair of opposite sides is called oblique sides.

(ii)

If ABCD is a trapezium, where AB || CD, then A and D are supplementary angles. Similarly, B and C are also supplementary.

(iii) The diagonals are unequal and do not bisect each other, nor is the angle between them a right angle. (iv) Area =

1 1 (a + b)h = d1d2sin, 2 2

where a and b are parallel sides, h = distance between parallel sides, d1 and d2 are diagonals and is the angle between the diagonals. In the trapezium ABCD, join AC. Now, Area of the trapezium = area of ABC + Area of ADC = (v)

1 1 1 ah + bh = (a + b)h. 2 2 2

If P is the intersection of the diagonals, then PAB ~ PCD.

(vi) Ar(ADC) = Ar(BCD) (area of triangles on the same base and between same parallel lines). Therefore, Ar(PAD) = Ar(PBC) (upon subtracting the area of PDC from the areas of ADC and BCD).

Property 1 Draw a line parallel to the parallel sides of a trapezium. Let it meet the non-parallel sides in P and Q (refer to the diagram). Then, AP : DP = BQ : CQ. The proportionality holds true because of Basic Proportionality Theorem. If AP : DP = m : n, then the length of PQ is given by x =

mb  na . mn

If the line PQ passes through the point of intersection of the diagonals, then PQ =

2ab = Harmonic mean of parallel sides. ab

534

Plane Geometry Let ABCD be a trapezium and AC and BD are the diagonals with R as the point of intersection. Let PQ be the line parallel to the parallel lines through R (refer to the figure below).

In CDA andRPA, since RP is parallel to CD, we can say that the two triangles are similar. Therefore,

AP AD  . PR DC

Similarly, BAD and RPD are similar. Therefore, Adding the two equations,



PD AD  . PR AB

AP PD AD AD    PR PR DC AB

AD ab 1 1  AD     PR  . PR a b a b  

Proceeding similarly on the oblique side BC, we get RQ 

Therefore, PR + RQ = PQ =

ab . ab

2ab = Harmonic mean of the parallel sides. a b

In such a case, the ratio in which P divides the oblique side AD is same as the ratio of parallel sides, that is, AP : PD = m : n = a : b.

Property 2 A trapezium inscribed in a circle is called isosceles trapezium, whose oblique sides are equal. Alternatively, if oblique sides are equal, then the trapezium can be inscribed in a circle. KITE In a kite, the following observations hold: (i)

Two adjacent sides are of same length and the other pair of adjacent sides are also of same length. In figure 1 below, AB = AD and BC = CD.

Fig 1 (ii)

Fig 2

Interior angles are not necessarily right angles. One pair of opposite angles are equal. In figure 1, ABC = ADC, but BAD BDC. This result follows from (i) above.

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Quantitative Aptitude Simplified for CAT

(iii) The two diagonals are unequal and intersect each other at right angles, where one of the diagonals is bisected by the other. In the figure, BD is bisected by AC, but not vice versa. (iv) Area = Ar (ABD) + Ar (CBD) =

1 2 1 a sinβ  b2 sin  2 2

Area = Ar (ABC) + Ar (ADC) = 2 Ar (ABC) = ab (sin ) Area =

(figure 2) (figure 2)

1 d1 d2 (where d1 and d2 are diagonals, which intersect at right angle). 2

Example 25 A trapezium is cut out of a rectangle whose length is 30 cm and breadth is 12 cm. If the parallel sides lie on the longer sides of the rectangle and bear to one another the ratio 1 : 4, and if the ratio of area of the trapezium to the area of the rectangle is 1 : 2, then find the lengths of the parallel sides of the trapezium. Solution Let the parallel sides be of lengths a and 4a (as per the given ratio). Then, area of the trapezium =

1 1 (a + 4a)(12) = (30)(12) or a = 6. Therefore, parallel sides are 6 and 24. 2 2

Example 26 Find the area of trapezium whose sides are 4, 8, 5 and 6 where 4 and 8 are parallel sides. Solution Refer to the figure below. Through A, draw a line parallel to BC meeting CD in E. Now, ABCE becomes a parallelogram and hence CE = 4. Therefore, DE = 4 and AE = 5. Area of ADE = =

s( s  a)( s  b)( s  c) . Now, s = 7.5. So, ar (ADE)

7.5(7.5  6)(7.5  5)(7.5  4)  7.5  1.5  2.5  3.5 .

1 ar(ABCE), because both the shapes have same base and same height. Therefore, area of 2  75   15   25   35  3  5  5  5  3 trapezium = 3 Ar(ADE) = 3   7  11.25 7         100  10   10   10   10  Ar(ADE) =

Alternatively, Extend DA and CB to meet in point P. Since AB = half of CD and AB || CD, A and B are the midpoints of PD and PC respectively. Therefore, PA = AD = 6 and PB = BC = 5. Now, we also know that area of PAB is one-fourth of the area of PDC, or one-third the area of ABCD. Therefore, ar(ABCD) = 3 Ar(PAB). PAB has sides 4, 5 and 6 and hence its area is same as that of ADE found in above method.

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Plane Geometry

Example 27 A point P on the side AB of the parallelogram ABCD divides AB in the ratio 2 : 5. If the area of triangle APD is 20 units, find the area of the triangle PDC. Solution Let ht of the parallelogram = h. Further, let AP = 2. Then PB = 5 and hence AB = 7. Therefore, area of APD =

1 (2)(h) = 20 units. Therefore, h = 20 units. 2

Therefore, area of PDC =

1 (7)(20) = 70 units. 2

Example 28 There is a rectangular piece of paper ABCD (length = 8 cm and width = 4 cm) which is folded such that opposite corners A and C are joined, thereby creating a crease in the paper (as shown in the figure below). Find the length of the crease.

Solution Refer to the figure below.

When paper is folded, it happens about Q and so we can say that CQ = QA. If DQ = x, then QC = 8 – x and so AQ = 8 – x. Applying Pythagoras’ Theorem, we get 42 + x2 = (8 – x)2  x = 3 = DQ. Similarly, PB = x = 3 cm. So, QR = 8 – 3 – 3 = 2 cm.

537

Quantitative Aptitude Simplified for CAT

Therefore, length of crease, PQ =

42  22  2 5 cm.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to apply properties of Quadrilaterals to various shapes such as Trapeziums and Rectangles.

Example 29 Find the area of a rhombus one side of which measures 20 cm and one diagonal 24 cm. Solution The length of semi-diagonal is 12 cm. Therefore, the other semi-diagonal is 1 1 d1d2 = (24)(32) = 384 units. of the rhombus = 2 2

202  122  16 . Therefore, area

Example 30 Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm broad at the rate of Rs. 20/m. Solution Let the length of the carpet be x metre. Then, area of carpet = area of the room. Therefore, 0.75x = 13 × 9 or x = 156 m. Cost of the carpet = 156 × 20 = Rs.3120. Example 31 A rectangular grassy plot is 112 m by 78 m. It has a gravel path 2.5 m wide round it on the inside. Find the area of the path and the cost of constructing it at Rs. 2 / m2. Solution Area of the total rectangular plot = 112  78 = 8736 m2. Length and breadth of the inner rectangle is 109.5 m and 75.5 m. Total area of inner rectangle = 8267.25 m2. Therefore, total area of the gravel path = 8736 – 8267.25 = 468.75. Cost of construction = 2 × 468.75 = Rs 937.5. Example 32 The length of the diagonal of rhombus is 80% of the length of the other diagonal. Then, the area of the rhombus is how many times the square of the length of the longer diagonal? A.

4 5

B.

2 5

C.

3 4

D.

1 4

538

Plane Geometry Solution If length of the longer diagonal is d, then length of the other diagonal = 0.8d. Area of the rhombus =

1 d1 d2 = 0.4d2. Area of the square of length d = d2. 2

Therefore, area of rhombus =

4 2 or times the area of the square. 10 5

Hence correct option is [B]. Example 33 ABCD is a trapezium in which AB || CD and AB = 2CD. If its diagonals intersect each other at O, then ratio of areas of triangles AOB and COD is A.

1:2

B.

2:1

C.

1:4

D.

4:1

Solution The triangles AOB and COD are similar triangles.

COD = AOB. Also, OCD = OAB and ODC = OBA. Therefore, triangles are similar. Therefore,

A1 s1 2 22   . So, the required ratio is 4 : 1. A2 s22 12

Example 34 The cross section of a canal is trapezium in shape. If the canal is 10 m wide at the top and 6 m wide at bottom and the area of cross section is 640 m2, the depth of canal is A.

40 m

B.

80 m

C.

160 m

D.

384 m

Solution Let d be the depth of the canal. Then area of the canal =

1 (10+6)d= 640 or d = 80 m. 2

Example 35 In a quadrilateral ABCD, if ABC = 104, ADC = 128 and AB = BC = 2 cm, then find BD.

539

Quantitative Aptitude Simplified for CAT

Solution The given data is satisfied only when there exists a circle such that the vertices A, C and D lie on a circle and B is the center of the circle (as shown in the diagram below).

ABC = 104. The outer angle ABC = 360 – 104 = 256, whose half is the ADC, that is, 128. Therefore, B must be the center of the circle. Therefore, AB, BC and BD are all the radii of the circle. Therefore, BD = 2 cm. Example 36 The four sides of a garden are known to be 20, 16, 12 and 10 rods. It is also known that it has the greatest possible area for those sides. Find the area. Solution The garden will have the greatest area only if the quadrilateral is cyclic. The area of the cyclic quadrilateral is given by (s  a)( s  b)( s  c)(s  d)  (29  20)(29  16)(29  12)(29  10) = 37791 .

540

Plane Geometry

PRACTICE EXERCISE 1.

Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is

10

B.

12

C.

14

D.

16 CAT 2005

4.

An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?

A.

4

B.

5

A.

75

C.

6

B.

90

D.

7

C.

120

D.

135

E.

150

CAT 2005 2.

A.

In ABC shown below, BC = 12 cm, DB = 9 cm, CD = 6 cm and BCD = BAC. What is the ratio of the perimeter of ADC to that of BDC? A.

7 9

B.

8 9

C.

6 9

D.

5 9

CAT 2006 Directions for question 5: Each question is followed by two statements A and B. Indicate your responses based on the following directives: Mark A. if the question can be answered using A alone but not using B alone. Mark B. if the question can be answered using B alone but not using A alone. Mark C. if the question can be answered by A and B together, but not by either A or B alone. Mark D. if the question cannot be answered even after using A and B together. 5.

Rahim plans to draw a square JKLM with a point O on the side JK but is not successful. Why is Rahim unable to draw the square? A. The length of OM is twice that of OL. B. The length of OM is 4 cm. CAT 2007

CAT 2005 3.

6.

A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is

541

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC? A.

17.05

B.

27.85

Quantitative Aptitude Simplified for CAT

C.

22.45

C.

0.6

D.

32.25

D.

0.45

E.

26.25

CAT 2004 CAT 2008

7.

8.

10. A rectangular sheet of paper, when halved by folding it at the mid-point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist? A.

5

B.

21

C.

10

D.

15

A.

4 2

E.

14

B.

2 2

CAT 2008

C.

2

Consider a square ABCD with midpoints E, F, G, H of AB, BC, CD and DA respectively. Let L denote the line passing through F and H. Consider points P and Q, on L and inside ABCD, such that the angles APD and BQC both equal 120°. What is the ratio of the area of ABQCDP to the remaining area inside ABCD?

D.

A.

4 2 3

B.

2 3

C.

10  3 3 9

D.

1

E.

None of these CAT 2004

Directions for questions 11 to 13: Answer the questions on the basis of the information given below: Consider three circular parks of equal size with centers at A1, A2, and A3 respectively. The parks touch each other at the edge as shown in the figure (not drawn to scale). There are three paths formed by the triangles A1A2A3, B1B2B3, and C1C2C3, as shown. Three sprinters A, B and C begin running from points A1, B1 and C1 respectively. Each sprinter traverses her respective triangular path clockwise and returns to her starting point.

1 3

2 3 1 CAT 2008

9.

A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son's head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 m, 1.8 m and 0.9 m respectively, and the father is standing 2.1 m away from the post, then how far (in m) is the son standing from his father? A.

0.9

B.

0.75

CAT 2003 (R) 11. Let the radius of each circular park be r, and the distances to be traversed by the sprinters A, B and C be a, b and c, respectively. Which of the following is true?

542

A.

b–a=c–b=3 3r

B.

b–a=c–b=

3r

Plane Geometry

a c = 2(1 + 2

C.

b=

D.

c = 2b – a = (2 +

15. In the figure (not drawn to scale) given below, if AD = CD = BC, and BCE = 96°, how much is DBC?

3 )r 3)

12. Sprinter A traverses distances A1A2, A2A3, and A3A1 at average speeds of 20, 30 and 15 respectively. B traverses her entire path at a uniform speed of (10 3 + 20). C traverses distances C1C2, C2C3 and C3C1 40 40 at ( 3 + 1), ( 3 + 1) and 120, 3 3 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint? A.

B1, C1

B.

B3, C3

C.

B1, C3

D.

B1, Somewhere between C3 and C1

A.

32

B.

84

C.

64

D.

cannot be determined CAT 2003 (R)

16. Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side a in the following steps: the original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is

13. Sprinters A, B and C traverse their respective paths at uniform speeds of u, v and w respectively. It is known that u2 : v2 : w2 is equal to Area A : Area B : Area C, where Area A, Area B and Area C are the areas of triangles A1A2A3, B1B2B3 and C1C2C3 respectively. Where would A and C be when B reaches point B3? A.

A2, C3

B.

A3, C3

C.

A3, C2

D.

Somewhere between A2 and Somewhere between C3 and C1.

A3,

14. In the figure (not drawn to scale) given below, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD is parallel CP. In ARC, ARC = 90°, and in PQS, PSQ = 90°. The length of QS is 6. What is ratio AP : PD?

A. B.

1:1

2 :1

C.

[n(4 – P)] : [4n – P]

D.

(4n – P) : n(4 – P) CAT 2003 (R)

A. B. C. D.

17. Let S1 be a square of side a. Another square S2 is formed by joining the mid-points of the sides of S1. The same process is applied to S2 to form yet another square S3, and so on. If AI, A2, A3, be the areas and P1, P2, P3……… be the perimeters of the S1, S2 S3 ……… respectively, then the ratio of P1  P2  P3  ... = A1  A2  A3  ...

10 : 3 2:1 7:3 8:3 CAT 2003 (R)

543

Quantitative Aptitude Simplified for CAT



2 1 2 A.



1 2



B.

1 4

C.

3 4

D.

1

a



2 2 2 C.

A.

a 2 2 2

B.





a



2 12 2 D.

CAT 2012



21. In a rectangle ABCD, there is a triangle AEF, where E lies on BC and F on CD. It is given that AD is 20, EF is 17, EC is 15 and the line joining the mid-points of AF and EF is of length 6.5. Find the area of AFE.

a CAT 2003 (R)

18. A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle? A.

16.665

B.

16.565

C.

15.465

D.

14.365

A.

115

B.

120

C.

110

D.

130 CAT 2013

22. Diagonals of rhombus are in ratio of 4 : 5. If area is 12, find area of square of side which is equal to the longer diagonal. A.

36

B.

30

CAT 2003 (R)

C.

25

19. In the following diagram, find the ratio of area of OAB to ODC.

D.

35 CAT 2014

23. 3 sides of a triangle are in AP. If area of the triangle is 6, find the longest side. A.

8

B.

4

A.

1:4

C.

5

B.

1:2

D.

6

C.

2 : 16

D.

0.5 : 8

CAT 2014 24. Sides of equilateral triangle are 2x  y, 3x  4y  2, x + 2y + 2. Find perimeter.

CAT 2012 20. ABCD is a square of side 2 units. Find the area of shaded region.

A.

5x + 1

B.

5x + 2

C.

x + 2y – 2

D.

3x – 2y + 4 CAT 2014

544

Plane Geometry 25. 2 poles of length 6 m and 10 m are 7 m apart. The top of each pole is joined to the base of opposite pole. Find the height of the intersection. A.

4m

B.

3.5 m

C.

3.25 m

D.

3.75 m

B.

7

C.

25 2

D.

9

B.

3:4

C.

2:5

D.

1:2

A.

15

B.

20

C.

30

D.

40 CAT 2016

31. In ABC, AB = 8 and BC = 10 and AD and CE are medians intersecting at right angles. Find the length of AC.

27. In a square ABCD, P and Q are mid-points of AB and BC. Find the ratio of area of DPQ to that of square? 5:8

3:8

A = 120 and area of ABC = 75 3 , then what is the length of BC?

CAT 2014

A.

D.

30. In an isosceles triangle ABC, with AB = AC,

26. In ABC, D is a point on AB and E on DC. Now, Ar(ADE) = 4, Ar(AEC) = 6 and Ar(BDE) = 5. Find the area of BEC.

15 2

3:2

CAT 2015

CAT 2014

A.

C.

48 ) : 1

B.

(5 +

48 ) : 1

C.

2:1

D.

3:2

B.

3 3 :16

C.

6.54

D.

7.87

A.

12

B.

18

C.

20

D.

Can't be determined

33. In a cyclic quadrilateral, longer diagonal bisects the smaller one at right angle. If the ratio of area of quadrilateral to the circle is 4 : 3, what is the ratio of shorter diagonal to the radius of the circle?

29. In an equilateral triangle ABC, AD, DE and EF are altitudes on BC, AC and DC resp., find the ratio of EF to AB.

3:4

5.73

CAT 2016

CAT 2015

A.

B.

32. A triangle ABC is such that there is a point F on AB with FB = 6 cm and E is midpoint of AC. If CF and BE intersect at Q and BQ = QE, then the length of AB is.

28. If area of a rectangle is equal to that of a square whereas perimeter of rectangle is twice that of the square, find the ratio of longer side to that of shorter side. (7 +

4.67

CAT 2016

CAT 2015

A.

A.

A.

3:2

B.

2:3

C.

4:3

D.

4 : 3 CAT 2016

545

Quantitative Aptitude Simplified for CAT

34. A ladder 7.6 m long is standing against a wall and the distance between the foot of wall and the base of the ladder is 6.4 m. If the top of the ladder now slips by 1.2 m, then the foot of the ladder shifts by approximately:

C.

510 m2

D.

630 m2 IIFT 2013

38. The perimeter of a right-angled triangle measures 234 m and the hypotenuse measures 97 m. Then the other two sides of the triangle are measured as

A.

0.4 m

B.

0.6 m

C.

0.8 m

A.

100 m and 37 m

D.

1.2 m

B.

72 m and 65 m

C.

80 m and 57 m

D.

None of the above

IIFT 2015 35. A ladder just reaches a window that is 8 m high above the ground on one side of the street. Keeping one end of the ladder at the same place, the ladder is moved to the other side of the street so as to reach a 12 m high window. If the ladder is 13 m long, what is the width of the street? A.

14.6 m

B.

15.8 m

C.

15.2 m

D.

15.5 m

IIFT 2012 39. AB, CD and EF are three parallel lines, in that order. Let d1 and d2 be the distances from CD to AB and EF respectively. d1 and d2 are integers, where d1 : d2 = 2 : 1. P is a point on AB, Q and S are points on CD and R is a point on EF. If the area of the quadrilateral PQRS is 30 square units, what is the value of QR when value of SR is the least?

IIFT 2014 36. There is a triangular building (ABC) located in the heart of Jaipur, the Pink City. The length of the one wall in east (BC) direction is 397 feet. If the length of south wall (AB) is perfect cube, the length of southwest wall (AC) is a power of three, and the length of wall in southwest (AC) is thrice the length of side AB, determine the perimeter of this triangular building. A.

3209 feet

B.

3213 feet

C.

3773 feet

D.

3313 feet

B.

400 m2

B.

10 units

C.

Slightly greater than 10 units

D.

Slightly less than 20 units

E.

Slightly greater than 20 units

40. ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1 : 3. The ratio of the triangle APS and rectangle ABCD is

37. In an engineering college there is a rectangular garden of dimensions 34 m by 21 m. Two mutually perpendicular walking corridors of 4 m width have been made in the central part and flowers have been grown in the rest of the garden. The area under the flowers is 320 m2

Slightly less than 10 units

XAT 2017

IIFT 2013

A.

A.

A.

36 128

B.

39 128

C.

44 128

D.

48 128

E.

64 128

XAT 2017

546

Plane Geometry 41. In the figure below, AB = AC = CD. If ADB = 20, what is the value of BAD?

A.

686 3

B.

1000

C.

961 2

A.

40

B.

60

C.

70

D.

650 3

D.

120

E.

None of the above

E.

140

XAT 2016 45. Study the figure below and answer the question:

XAT 2016 42. Akhtar plans to cover a rectangular floor of dimensions 9.5 m and 11.5 m using tiles. Two types of square shaped tiles are available in the market. A tile with side 1 meter costs Rs.100 and a tile with side 0.5 m costs Rs.30. The tiles can be cut if required. What will be the minimum cost of covering the entire floor with tiles? A.

10930

B.

10900

C.

11000

D.

10950

E.

10430

XAT 2016 Four persons walk from Point A to Point D following different routes. The one following ABCD takes 70 minutes. Another person takes 45 minutes following ABD. The third person takes 30 minutes following route ACD. The last person takes 65 minutes following route ACBD. If all were to walk at the same speed, how long will it take to go from point B to point C? A. B. C. D. E.

XAT 2016 43. A square piece of paper is folded three times along its diagonal to get an isosceles triangle whose equal sides are 10 cm. What is the area of the unfolder original piece of paper? A.

400 cm2

B.

800 cm2

C.

800 2 cm2

D.

1600 cm2

E.

Insufficient data to answer

10 min. 20 min. 30 min. 40 min. Cannot be answered as the angles are unknown.

46. ABCD is a quadrilateral such that AD = 9 cm, BC = 13 cm and DAB = BCD = 90. P and Q are two points on AB and CD respectively, such that DQ : BP = 1 : 2 and DQ is an integer. How many values can DQ take, for which the maximum possible area of the quadrilateral PBQD is 150 sq.cm? A. B. C. D. E.

XAT 2016 44. The difference between the area of the circumscribed circle and the area of the inscribed circle of an equilateral triangle is 2156 sq. cm. What is the area of the equilateral triangle?

14 12 10 9 8 XAT 2016

47. The figure below has been obtained by folding a rectangle. The total area of the

547

Quantitative Aptitude Simplified for CAT

figure (as visible) is 144 m2. Had the rectangle not been folded, the current overlapping part would have been a square. What would have been the total area of the original unfolded rectangle? A.

128 m2

B.

154

m2

C.

162 m2

D.

172 m2

E.

None of the above

B.

> 12 and < 14

C.

> 14 and < 16

D.

> 16 and < 18

E.

Cannot be determined XAT 2014

50. There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the ladder is at point A. After the first window is fixed, the foot of the ladder is pushed backwards to point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is

XAT 2015

48. In the diagram below, CD = BF = 10 units and CED = BAF = 30. What would be the area of triangle AED? (Note: the diagram below may not be proportional to scale).

A.

10 and < 12

548

Plane Geometry other of length ‘b’, where, b > a. A possible

B.

> 10 and ≤ 10.5

b value of , is: a

C.

> 10.5 and ≤ 11

D.

> 11 and ≤ 11.5

E.

> 11.5

A.

≥ 5 and < 8

B.

≥ 8 and < 11

C.

≥ 11 and < 14

D.

≥ 14 and < 17

E.

> 17

XAT 2014 54. In a square PQRS, A and B are two points on PS and SR such that PA = 2AS, and RB = 2BS. If PQ = 6, the area of the triangle ABQ is

XAT 2014 53. Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. The diagonal AC intersects the line segments PR and QR at M and N respectively. What is the area of the quadrilateral PQNM?

A.

6

B.

8

C.

10

D.

12

E.

14

> 9.5 and ≤ 10

A.

XAT 2013

ANSWER KEY 1.

C

2.

A

3.

B

4.

E

5.

A

6.

E

7.

C

8.

E

9.

D

10.

B

11.

A

12.

C

13.

B

14.

C

15.

C

16.

A

17.

C

18.

D

19.

D

20.

D

21.

C

22.

B

23.

C

24.

B

25.

D

26.

A

27.

A

28.

A

29.

D

30.

C

31.

B

32.

B

33.

C

34.

C

35.

C

36.

D

37.

C

38.

B

39.

E

40.

A

41.

D

42.

A

43.

A

44.

A

45.

C

46.

A

47.

C

48.

D

49.

B

50.

E

51.

E

52.

D

53.

D

54.

C

549

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS 1.

4.

Answer: E Explanation: Since BPC is equilateral and ABCD is a square, AB = BC = CD = DA = BP = PC.

Answer: C Explanation:

Now, ABP = 90 – 60 = 30. Since BP 1 = BA, BAP = BPA = × 150 = 75. 2 Therefore, APD = 360 – (75 + 75 + 60) = 150. 2.

Answer: A

5.

Explanation: BCD ~ BAC (because BCD = BAC and B is common). BC BD BC 2 122   BA    16 Therefore, BA BC BD 9 BC CD 12 6     AC  8 . Moreover, BA AC 16 AC Ratio of perimeter of ADC to that of BDC = (AD + DC + CA) : (BD + DC + CB) = (7 + 6 + 8) : (9 + 6 + 12) = 21 : 27 = 7 : 9. 3.

Answer: A Explanation: The maximum length of OM = x 2 where x is the side the square. Using statement A, if OM is twice that of OL, then OM is more than the allowed length. So, statement A alone is sufficient to solve the question. Statement B alone doesn’t explain why he is unable to draw the square. So, statement B alone is not sufficient.

6.

Answer: B

Answer: E Explanation: AB = 17.5 cm, AC = 9 cm. Let BC = x and circumradius of the triangle = R. Then, abc (17.5)(9)( x ) Area, A =  . Area is also 4R 4R 1 equal to ( x)(3) . Therefore, 2 (17.5)(9)  3  R = 26.25 cm. 2R

Explanation: Use options. Let number of tiles along width be x. If number of tiles along longer edge is 10, then number of white tiles = 8 × 2 + 2x = 2x + 16. Number of red tiles = 8 × (x – 2). Therefore, 2x + 16 = 8x – 16 or 6x = 32, which is not possible. So, [A] is incorrect option. If number of tiles along longer edge is 12, then number of white tiles = 10 × 2 + 2x = 2x + 20. Number of red tiles = 10 × (x – 2). Therefore, 2x + 20 = 10x – 20 or x = 5, which is possible.

7.

Answer: C Explanation: The range of possible values of x is: (15 – 8) < x < (15 + 8), i.e. 7 < x < 23. But x should be so chosen that the triangle is obtuse angled. The triangle will be right triangle if x = 17 (if x is the hypotenuse)

550

Plane Geometry Or x = 225  64  161  12.69 (if 15 is the hypotenuse). So, x can be {8, 9, 10, 11, 12, 18, 19, 20, 21, 22}, which are 10 values possible.

2x : 2 = 2 : x  x = 2 . Therefore, area of smaller rectangle = 2 2 . Common explanation for questions 11 to 13:

8.

Answer: E

From the given figure, we can calculate the values of each side of the given triangles. 1 C1C2B2 = × 60 = 30. Also, C2A2P 2 = 60. Using Pythagoras’ Theorem, we get, C1C2

Explanation:

= 2r + 2r 3 ; B1B2 = 2r + r 3 and A1A2 = 2r. Let side of the square be 10. Then, BF = 5 and 5 . QF = 3 1  5  Therefore, Area of BQC = (10)   . 2  3 Area of the two triangles (BQC + QPD) 50 50 = . So, area of ABQCDP = (10)2 – . 3 3 Required ratio = 2 3  1 . 9.

Answer: D Explanation: The shadow of head of father and son coincide so diagram will be like this:

11. Answer: A Explanation: From the calculation given above, a = 6r, b = 6r + 3r 3 , c = 6r + 6r 3 . 12. Answer: C Explanation: Let value of r = 30 units. Then time taken by A to cover distance A1A2, A2A3 and A3A1 = 3, 2 and 4 units of time respectively. Total time = 9 units. Similarly, for B, time taken to cover 6r + 3r 3 or 90(2 + 3 ) units of distance at the

If the shadow of their heads coincide at point D, then ACH and ABI will be the same. In 4.2 5.1 ABI and ACH,   x = 0.45. 2.1 2.1  x

speed of 10 3 + 20, would be 9 units. Similarly, C would have covered 2/3rd of its total distance in 9 units of time. Hence B would have reached B1 and C would have reached C3 while A has covered her sprint.

10. Answer: B Explanation: Let the longer side of bigger rectangle is 2x. So the ratio of longer side and smaller side is 2x : 2. The dimensions of the smaller rectangle are 2 and x. So, ratio = 2 : x. As per the question,

13. Answer: B Explanation: Hence, the time when B reaches B3 A would be at A3 and C would be at C3.

551

Quantitative Aptitude Simplified for CAT

14. Answer: C

Therefore, P1 + P2 + P3 + … = a a 1   4  a   ...   4 a 2  1  2   1  2 . 

Explanation: ACB ~ PQB. Therefore AP : PB = CQ : QB = 4 : 3  AP = 4k, PB = 3k Also CPB ~ QDB. Therefore, CQ : QB = PD : DB = 4 : 3. So, D divides PB in the ratio of 4 : 3. 4 12k Therefore, PD = (3k )  7 7 12k So, AP : PD = 4k : = 7 : 3. 7



4a 2

2 1 A1 + A2 + A3 + … a2 a2 1   ...  a2  2a2 . = a2  1 2 4  1 2  

15. Answer: C Therefore,

Explanation: Let DAC = x. Then, ADC = x. Therefore, DCB = 2x and DBC = 2x. So, CDB = 180 – 4x. ADB = 180 – 4x + x = 180 – 3x. Now, (180 – 3x) + 96 = 180 or x = 32. So, DBC = 2x = 64.

P1  P2  P3  ... 4a 2  2 A1  A2  A3  ... 2a 2 1





2 2 1 2 

a

  22  2 

.

a

18. Answer: D Explanation:

16. Answer: A Explanation: First process: We can assume that 4 equal squares are cut. If side of the original square is 4 cm, then side of each square = 2 cm = diameter of each circle. Therefore, area of scrap = 42 – 4 × (1)2 = 16 – 4. Second process: Area of scrap = 42 – (2)2 = 16 – 4. Therefore, ratio = 1 : 1.

If hypotenuse reduces by 35%, it becomes 65% of original length. Likewise, base and height of the triangle also becomes 65% of original lengths. Therefore, new area = (0.65)2 × 34 = 14.365.

17. Answer: C

19. Answer: D

Explanation:

Explanation: Length of DC = 2 + 2 + 4 = 8. So, AB : DC = 2 : 8 = 1 : 4. Therefore, ratio of area of OAB to ODC = 1 : 16. 20. Answer: D Explanation: Let O be the centre of the square. Area of shaded region inside the square AQOP 1 = half of area of OPQ = th of area of AQOP. 4 Therefore, area of shaded region 1 1 = th of area of square ABCD =  4  1 . 4 4

Side of first square = a; side of second square a a = ; side of third square = ; and so on. 2 2

552

Plane Geometry 21. Answer: C Explanation:

The distance between the poles is redundant. Since AB is parallel to XY, ABC ~ XYC. Let XY = m. AB AC 6 AC   Then, or . Therefore, XC XY XC m XC m  AC = . 6 Similarly for the ‘similar’ triangles DAC and 10 AC YAX,  . Therefore, m AX m  AC AX = . Adding these two equations, we 10 get 1 1  AC = XC + AX = m  AC    or m  6 10  15 =  3.75 m. 4

P and Q are midpoints of AF and EF. Since PQ = 6.5, AE = 13. Also, BE = 5. Therefore, AB = 12. Also, EF = 17, EC = 15, so FC = 8. Since DC = AB = 12, DF = 4. Area of AFE = Area of ABCD – (Area of ADF + Area of EFC + Area of ABE) = 12 × 20 – 1 1 1   (4)(20)  (8)(15)  (5)(12)  = 110. 2 2 2  22. Answer: B Explanation: Diagonals of rhombus are 4k and 5k. So, area 1 1 =  d1  d2   4 k  5k = 12. So, k2 = 1.2 2 2 Area of square = 25k2 = 25 × 1.2 = 30.

26. Answer: A 23. Answer: C

Explanation:

Explanation: Let us assume that the triangle is right angled triangle. Since sides are in AP, and using options, we can easily imagine that the sides of the triangle can be 3, 4 and 5. The area of this triangle is also 6. Therefore, longest side of the triangle is 5. 24. Answer: B Since ratio of area of ADE to AEC is 4 : 6 = 2 : 3, and both the triangles have same height (considering base is DC and altitude is dropped from A on DC), we can say that DE : EC = 2 : 3. Therefore, if area of BDE = 5, area 3 15 of BEC =  5  . 2 2

Explanation: 2x – y = 3x – 4y – 2 = x + 2y + 2  x – 3y = 2. Perimeter = 6x – 3y = 6x – x + 2 = 5x + 2. 25. Answer: D Explanation:

553

Quantitative Aptitude Simplified for CAT

27. Answer: A

x 3 1 x 3 . Therefore, EF : AB   2 2 4 x 3 = : 2x  3 : 8 . 4

= Explanation:

30. Answer: C Explanation:

1 × p2 × sin 120 2 = 75 3 . Therefore, p = 10 3 . Using cosine rule, If AB = p, then area =

Let the side of the square be 2. Then, PB = BQ = 1. Area of APD + BPQ + DCQ 1 1 1 =  2  1   1  1   2  1  2.5 . 2 2 2 Therefore, required ratio = 2.5 : 22 = 5 : 8.

BC  

2

2

 10 3    10 3   2  10 3  10 3  cos120 2 2  10 3    10 3   10 3  10 3   30

28. Answer: A Alternatively, If length of BC = 2x, then BD = x, where is AD is altitude on BC. Also, DAB = 60. x x  So, AD = . Area of ABC tan60 3

Explanation: Let length and breadth of rectangle be a and b, and side of square be x. Then, ab = x2; 2(a + b) = 2(4x) = 8x. Therefore, 2

ab a2  b2  2ab ab =    16  4   a2 + b2 – 14ab = 0

1  x  x2 (2 x)   75 3 .   2 3  3 Therefore, x2 = 225  x = 15  BC = 30. =

2

a a     14    1  0  b  b  k2 – 14k + 1 = 0  k



31. Answer: B Explanation:

14  14 2  4(1)(1) 14  192  . 2(1) 2

 7  48 Since a > b, k must be more than 1 and so ratio of longer to shorter side = 7 + 48 .

Let AD = p and CE = q. Also let AC = x.

29. Answer: D

2

2

 2q   p 2 Now, in CGD,     5  3  3  4q2 + p2 = 225,

Explanation:

2

2

 q  2p  2 In AGE,       4  q2 + 4p2 3 3     = 144. Adding these equations, we get 5q2 + 5p2 = 369. In AGC, AC2 =

Let AB = 2x. Then, DC = x x 3  DE = DC sin 60 = . 2 Now, EDF = 30. So, EF = ED sin 30

2

2

4(q2  p2 )  2q   2p       9  3  3 4  369 4  41 164     32.8 95 5 5 Therefore, AC  5.73.

554

Plane Geometry 32. Answer: B

Therefore, DE = 7.62  2.92  49.35  7 m. So, AD = 7 – 6.4 = 0.8 m.

Explanation:

35. Answer: C Explanation: Distance of the foot of ladder from one end (where height of window is 8 m) = 132  82  105 m. Distance of the foot of ladder from the other end (where height of window is 12 m) = 5 m. Therefore, width of the street = 5 + 105  15.2 m.

Draw EG parallel to CF. Now, AEG ~ ACF. So, AG : GF = 1 : 1, or AG = FG. Also, BQF ~ BEG. Since Q is midpoint of BE, F is midpoint of BG. So, BF = FG. Therefore, BF = FG = GA = 6. So, AB = 18 cm.

36. Answer: D

33. Answer: C

Explanation: BC = 397. Let AB = a3 and AC = 3n. Also, AC = 3 × AB Therefore, 3n = 3 × a3  3n–1 = a3. Now, perimeter, p = BC + AC + AB = 397 + 3n + 3n–1  (p – 397) = 3n–1 × (3 + 1) = 3n–1 × 4. Thus the LHS of the above equation should be a multiple of 3 and 4. For that p should be 3k + 1 type. The only option is 3313.

Explanation: If longer diagonal bisects the smaller one at right angle, the longer diagonal is perpendicular bisector of smaller one and so the longer diagonal passes through the center of the circle and in also the diameter of the circle. Area of the quadrilateral 1 1 = d1 d2  (2r )(d2 ) , where d1 2 2 = longer diagonal, d2 = smaller diagonal, r = radius of the circle. Ratio of area of quadrilateral to the circle 1 (2r )(d2 ) 4 d 4   2  . = 2 2 3  r 3 r

37. Answer: C Explanation: The following diagram displays the garden along with the paths,

34. Answer: C Explanation:

Length and breadth of each small rectangle = 15 and 8.5 m. Therefore, area under the flowers = 4 × (15 × 8.5) = 510 m2.

Original position of ladder = AB = 7.6 m; AE = 6.4 m. Therefore, BE = 7.62  6.42  16.8  4.1. If ladder slips by 1.2 m, the new position of ladder is CD. Now, CE = BE – BC = 4.1 – 1.2 = 2.9 m.

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Quantitative Aptitude Simplified for CAT

38. Answer: B

Therefore, area of APS = 20 – 1.25 – 1.75 – 8 = 9.

Explanation: Use options. Perimeter = 234 = 97 + a + b  a + b = 137. This is true in all the three options. Since hypotenuse is the longest side, [A] option is ruled out. Checking [B], 722 + 652 = 9409 = 972.

Therefore, required ratio =

9 . 32

41. Answer: D Explanation: AC = CD and ADB = 20°  CAD = 20°  ACB = (20 + 20)° = 40°. So, AC = AB  ABC = 40°  BAD = 180 – 40 – 20 = 120°.

39. Answer: E Explanation:

42. Answer: A Explanation: Consider the diagram below:

Area of PQS = 2 × Area of QRS (because base QS is same and height of PQS is twice the height of QRS. 1 Therefore, area of QRS = 10 =  QS  h . 2 SR is least when SR = h, and least integer value of h is 1. So, QS = 20 and hence value of QR will be more than 20.

Akhtar uses 9 × 11 tiles to cover area shown by A. Consider area shown by B and C. Akhtar will cut tiles with side 1 meter to make two pieces of the dimensions 1 × 0.5. So he will 20 have to use = 10 tiles. Now to cover D, 2 he will use a tile with side 0.5 meters. Thus, the minimum cost of covering entire floor with tiles = (99 + 10) × 100 + 1 × 30 = 10930.

40. Answer: A Explanation:

43. Answer: A Explanation: If the side of the isosceles triangle is 10 cm then we get the side of the square as 20 cm on unfolding it. Hence, the area of square = 400 cm2. Then, area of rectangle = 32. Since RS : SQ = 1 : 3, RT : TD 1 1 = 1 : 3. Therefore, RT =  2  = PV. 4 2 Area of ABVT = 8 × 2.5 = 20. 1 Also, TS =  4  1 . So, SV = 8 – 1 = 7. 4 1 Now, area of ATS =  1  2.5  1.25 ; area 2 1 of PVS =  7  0.5  1.75 ; area of ABP 2 1 = 82  8 ; 2

44. Answer: A Explanation: Let r and R be the inradius and circumradius. Then, R = 2r. As per the question, difference in the area of the two circles = 2156. So, R2 – r2 = 2156  4r2 – r2 = 2156 686  3r2 = 2156  r = . 3 Side of the triangle = R 3 = 2r 3

556

Plane Geometry 47. Answer: C

686  2 686 . 3 Therefore, area of the triangle 2 3  2 686   686 3 . =   4

= 2 3

Explanation: Had the rectangle not been folded, the overlapping part would have been a square of side 6. While unfolding, the increase in area 1 = Area of the triangle =  6  6  18 2 Area of the given figure = 144 square meters Therefore, area of unfolded rectangle = 144 + 18 = 162 m2.

45. Answer: C Explanation: Let the distances be as shown in the diagram.

48. Answer: D Explanation: ECD = 60 = BCF  BFC = 30. Also, CED = BAF = 30. Therefore, AFB = 60. So, AFC = 90.

Since speeds of all the persons is same, distance is proportional to time. So, a + b + e = 70; a + c = 45; d + e = 30; d + b + c = 65 Adding first and last equation, we get (a + b + e) + (d + b + c) = 70 + 65  (a + 2b + c + d + e) = 135. Adding second and third equation, we get (a + c) + (d + e) = 45 + 30 = 75. Subtracting, we get 2b = 60 or b = 30. 46. Answer: A

ED = CD cot 30 = 10 3 ; CD = 10. 10 BC = 10 tan 30 = . Also, AB = 10 tan 3

Explanation:

60 = 10 3 . Therefore, length of AD = 10 3 +

=

30  10  10 3



1  2k  9 ; Area of DQB 2





49. Answer: B

1  k  13 . 2

Therefore, area of PBQD =

+ 10

3

40  10 3  . 3 3 Area of AED  40  10 3  1 10 3  =   50 3  4 . 2 3  

=

Area of DPB =

10

Explanation: Let QS = x. We get the following figure

31 k ≤ 150 2

300 . 31 So, the values of k can be 1, 2, 3, …, 9.

k≤

557



Quantitative Aptitude Simplified for CAT

AC. So, we need all the three statement to find the length of AC. 52. Answer: D Explanation:

In any triangle, the sum of any two sides must be greater than the third side. Similarly, the difference between any two sides must be smaller than the third side. Hence, In ∆QRS, 17 – 5 < x < 17 + 5 or 12 < x < 22. Similarly, in ∆QPS, 9 – 5 < x < 9 + 5 or 4 < x < 14. Common range is 12 < x < 14.

8  2 2 . So,

Side of the squares are 3 and a + b = 3 or b = 3 – a.



So, a2 + (3 – a)2 = 2 2

50. Answer: E



2

8

 2a2 – 6a + 1 = 0  a 6  36  8 6  2 7 3  7 . =   4 4 2 3 7 Since a < b, a = and so b 2 3 7 3 7 . = 3  2 2 Therefore,

Explanation: The following figure represents the length and the position of the ladder,







3 7 3 7 b 3 7 16  6 7     83 7 2 a 3 7 2 3 7  15.94.

53. Answer: D

AC = BD = 30 m (length of ladder). 26 3 = 0.866  . CE = 26 m. So, sin 2 = 30 2 Therefore, 2 = 60 and so  = 30. So, CEA ~ BED. Also, BD = AC = 30 m. So, BE = CE = 26. 26 Further, EA = EC cot 60 = . Therefore, 3 26 AB = 26 –  10.98 m. 3

Explanation: Let the breadth be 3x and the length be y. 3xy = 90 xy = 30

51. Answer: E Explanation: Unless we know which of the vertices is the right angle, we cannot find the length of AC. From statement III, we get to know which vertex is the right angle. Besides knowing which vertex is the right angle, if we know only AE, then we would not know the length of AB or BC and hence would not know the length of

V is midpoint of WR. PW x PW || EV  EV =  = FV 2 4 Therefore, EF = ∆MPA ~ ∆MEV.

558

x . 2

Plane Geometry So, ratio of heights of ∆MPA to ∆MEV = AP : EV = 4 : 1. If height of ∆MPA = 4k, then height of ∆MEV y = k and so 0.5y = 4k + k = 5k or k = . 10 y Therefore, height of ∆MEV = . 10 Area of MEV 1 x y xy 30    0.375 . =   2 4 10 80 80 Similarly, ∆VFN ~ ∆CRN. Therefore, ratio of heights of ∆VFN to that of x ∆CRN = VF : CR = : 1.5 x = 1 : 6. 4 If height of ∆VFN = m, then height of ∆CRN = 6m. y Therefore, 0.5y = m + 6m = 7m  m = 14 = height of ∆VFN. Therefore, area of VFN 1 x y xy 30 =      0.27 . 2 4 14 112 112 Also, area of PQFE



1 x  y 1.5 xy 1.5  30 45    11.25  x    2 2 2 4 4 4

A(PQNM) = A(PQFE) – A(∆MEV) + A(∆VFN) = 11.25 – 0.375 + 0.27 ≈ 11.145. 54. Answer: C Explanation:

1 22  2 . 2 Area of PQA = Area of RQB 1 =  4  6  12 . 2 Therefore, area of ABQ = (6)2 – 12 – 12 – 2 = 36 – 26 = 10. Area of ASB =

559

Quantitative Aptitude Simplified for CAT

Chapter 17

Polygons and Circles BASIC CONCEPTS Polygons Polygons are n sided closed figures. They may be concave or convex.

Difference between convex and concave polygons CONVEX POLYGON

CONCAVE POLYGON

A line joining any two vertices lies completely within the polygon.

There exist at least two vertices such that the line joining them lies partly outside the polygon.

Each interior angle must be less than 180°.

At least one of the interior angles must be more than 180°.

Extend any side of the polygon on both sides. All the other vertices lie on the same side of the polygon.

Extend any side of the polygon on both sides. At least one of the vertices does not lie on the same side as the other vertices.

We will restrict ourselves to convex polygons and that too regular polygons. Regular convex polygons are those polygons whose (i)

all sides are equal and

(ii)

all angles are equal.

If all sides are equal, then all the angles may not be equal. For example, in a rhombus all sides are equal, but not angles.

560

Polygons and Circles In the figure below, pentagon ABCDE is such that all sides are equal but all angles are not equal, as BCDE forms a square and ABE is an equilateral triangle.

Moreover, if angles are equal, then sides need not necessarily be equal. For example, in a rectangle all angles are equal, but all sides are not equal. So, to prove a polygon to be a regular polygon, we have to prove that all sides are equal and also all angles are equal.

Interior and exterior angles Sum of all interior angles in a polygon (regular or general) = (n – 2) × 180. Let us say we have a polygon of n sides. From any vertex, say A, draw all the possible diagonals. Observe that the polygon gets divided into (n – 2) triangles for n sided polygon (refer to the diagram below).

The sum of the angles in a triangle is 180°. Therefore, sum of all the angles in all these triangles is given by (n – 2)180°, which is also written as (2n – 4) right angles. This is the sum of all the interior angles in a polygon of n sides. Note that this result is true for any convex polygon, regular or not. Sum of all exterior angles in a polygon (regular or general) = 360.

Extend all the sides of the polygon as shown in the diagram above. Interior angles are a, b, c, d, … Therefore, corresponding exterior angles = (180 – a), (18 – b), (180 – c), … Sum of exterior angles = (180 – a) + (180 – b) + (180 – c) + … = 180n – (a + b + c + d + …) = 180n – (Sum of interior angles) = 180n – (n – 2) × 180 = 360. Hence the result. Note that the result is true for regular as well as general polygons. Also note that sum of exterior angles does not depend upon the number of sides of the polygon.

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Quantitative Aptitude Simplified for CAT

1 th of the sum of all the n angles. Therefore, if ‘’ is the sum of all the interior angles and ‘i’, each interior angle (for regular polygons), then

In case of a regular polygon, all angles are equal. Therefore, each interior angle is

= (n – 2)180°  i =

(n  2)180o n

Let each exterior angle be ‘e’ and sum of all the exterior angles be E. E = 360°  e =

360o . n

Area of regular polygons To find the area of a regular polygon of n sides, let each side of the polygon be of length ‘a’ and center of the polygon be O (refer to figure below). Join O with two consecutive vertices A and B. Triangle OAB is an isosceles triangle. Drop perpendicular from O on the side AB meeting it in P. Now, AP = PB =

Also, AOB =

a . 2

360o 180 o . Therefore, AOP,  = . n n

If h is the ht of the triangle OAB, then (by elementary trigonometry), AP AP  180   180   tan   tan   tan     OP h  n   n  a 180o h = cot . 2 n

Therefore, area of OAB =

1 a2 180o . ah  cot 2 4 n

Area of the regular polygon = n (Area of OAB) =

na2 180 o . cot 4 n

Number of diagonals Let us join all the vertices of a polygon with each other. All the lines that get drawn are called diagonals of the polygon. There are two ways to count the number of diagonals:

562

Polygons and Circles Method 1: Let there be n distinct points in a 2D plane such that no three of them are collinear. They can be joined to each other in nC2 ways, which is the number of lines connecting n non-collinear points. Out of these, n lines will be the sides of the polygon. The remaining lines will be the diagonals of the polygon. Therefore, number of diagonals of the polygon is given by nC2 – n =

n(n  3) . 2

Method 2: From any vertex, draw all possible diagonals. This vertex cannot join with itself, nor with the two adjacent vertices to form a diagonal. So, this vertex can join with the rest of the (n – 3) vertices to form diagonals. The same can be done from every other vertex. Therefore, total number of diagonals that can be drawn = n(n – 3). But, vertex A joining with vertex D is same as vertex D joining with vertex A. Therefore, only half of n(n – 3) is the number of distinct diagonals. Hence, number of diagonals of the polygon is given by

n(n  3) . 2

It is important to note that the number of diagonals does not depend upon whether the polygon is regular or not. Example 1 Each interior angle of a regular polygon is 150°. Find the number of sides of the polygon. A.

12

B.

10

C.

13

D.

14

Solution Each interior angle =

180(n  2) = 150. Therefore, number of sides, n = 12. Hence correct option is [A]. n

Alternatively, If each interior angle is 150, then each exterior angle = 30, which is equal to

360 . So, n = 12. n

Example 2 For a convex polygon to be a regular polygon, A.

‘all the angles are equal’ is a necessary and sufficient condition

B.

‘all the sides are equal’ is a necessary and sufficient condition

C.

‘all the angles and sides are equal’ is a necessary and sufficient condition

D.

none of the above conditions is sufficient

Solution From the discussion already mentioned earlier, we need to prove ‘all sides are equal’ as well as ‘all angles are equal’, to prove a polygon to be regular one. Hence correct option is [C].

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Quantitative Aptitude Simplified for CAT

Example 3 A polygon has 27 diagonals. The number of sides of the polygon is A.

9

B.

10

C.

11

D.

12

Solution Number of diagonals =

n(n  3) = 27 or n = 9. Hence correct option is [A]. 2

Example 4 The sum of all exterior angles of a convex polygon of n sides is A.

4 right angles

B.

2n right angles

C.

(2n – 4)90°

D.

n/2 right angles

Solution We know that the sum of all interior angles is always 360° irrespective of the number of sides. Hence correct option is [A]. Example 5 The ratio of the number of sides of two regular polygons is 1 : 2 and the ratio of their interior angles is 2 : 3. The number of sides of these polygons (respectively) is A.

4, 8

B.

5, 10

C.

7, 10

D.

8, 4

Solution Use the options. Since ratio of number of sides is 1 : 2, either [A] or [B] is correct option. If the number of sides are 4 and 8, then interior angles are 90° and 135. Therefore, ratio of interior angles = 2 : 3. Hence correct option is [A]. Example 6 Four triangles from the four corners of a square piece of paper (of side 5 cm) are cut such that the remaining piece of paper is a regular octagon. Find the side of the regular octagon. CAT 2003 (C) Solution If x is the length of one of the sides of the triangles being cut out, then hypotenuse of the triangle = side of the regular octagon = x 2 . The remaining length of the side after two triangles on one side are cut = 5 2x. Since 5 it is a regular octagon, we have x 2 = 5 – 2x or x = cm. 2 2

564

Polygons and Circles

Therefore, side of the octagon = x 2 

5 2 2 2



5 2 1

5





2 1 .

Example 7 All the sides of a hexagon are produced on both sides to meet so as to form a star shaped figure. The sum of the angles at the vertices of the star is A.

180

B.

720

C.

360

D.

540

Solution Let the hexagon be ABCDEF. Let the sides be produced to meet in the points P, Q, R, S, T, W (as shown in the figure below). We observe that PTR and WQS are triangles, each of whose sum of interior angles is 180. So, sum of all the angles at the vertices of the star = 360.

Example 8 If a regular octagon is inscribed in a circle of radius r, what is the perimeter of the octagon? Solution Let a be the side of the octagon. The value of AOB = 45°, where A and B are any two consecutive vertices of the octagon. Therefore, using cosine rule,

1  1    a2 = r2 + r2 – 2r2 cos 45° or a = r  2  1   . Therefore, perimeter = 8r  2  1    6.12r. 2 2  

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Quantitative Aptitude Simplified for CAT

EXPERT SPEAK Scan this QR Code to watch a video that explains how to apply properties of Polygons to various problems actually asked in the previous years' CAT papers.

Circles Note that as number of sides of the polygon approaches infinity (thereby the length of the side approaches zero), the polygon approaches a circle. Thus, a circle is a limiting case of a polygon. A circle is defined as the locus of all the points in 2D plane whose distance from a fixed point is constant. This fixed point is called the center of the circle and the constant distance is called the radius. In the following figure, O is the center and OP is the radius. Therefore, a circle is completely understood in terms of its radius and location of the center.

Chord of a circle is the line joining any two points on the circumference of the circle. Diameter is the longest chord and which passes through the center of the circle.

Arc, sector and segment Any two points on the circle divide the circumference into two arcs. The smaller one is called minor arc (arc AnB) and the bigger one is called major arc (arc AmB). The angle  subtended by minor arc AnB at the center is called degree measure of arc AnB and the angle  subtended by major arc AmB at the center is called degree measure of arc AmB. If angle subtended at the center by an arc is 360°, then the arc is none other than circumference, whose length is equal to 2r.    If angle subtended at the center by an arc is , then length of the arc is    2r .  360 o     Therefore, length of minor arc AnB =    2r .  360 o     Similarly, length of major arc AmB =    2r ,  360 o 

where r is the radius of the circle. The region enclosed by the two radii and the arc is called sector. The shaded region shown (in the figure below) is the sector.

566

Polygons and Circles

If angle subtended at the center by an arc is 360°, then the area of the sector is none other than the area of the circle, r2.  α  If angle subtended at the center by an arc is , then area of the sector is   πr2 . o  360   α  2 Therefore, area of sector whose angle at the center is , is given by    πr .  360 o 

If the points A and B are joined, they form a chord. This chord divides the area of the circle into two areas. The smaller area (or region) is called minor segment and the other one is called major segment.

To find the area of the minor segment (shown shaded), we should subtract the area of triangle OAB from area of the sector OAB. 1  α  Area of minor segment = Ar (sector OAB) – Ar (OAB) =   πr2  r2 sin  . o 2  360 

Area of major segment = Area of circle – Area of minor segment  α   1 α  1 2  = πr2     πr2  r2 sin    πr2  1    r sin  . o  2 360o  2    360  

Properties related to circles (i)

Two chords of same length in a circle are equidistant from the center of the circle.

(ii)

Perpendicular bisector of the chords of the circle passes through the center of the circle. That is precisely the reason that the perpendicular bisectors of the three sides of any triangle ABC is the circumcenter! In case of a quadrilateral ABCD, if all the four perpendicular bisectors are concurrent, then the quadrilateral is obviously a cyclic one, where each side of the quadrilateral acts as the chord of the circumcircle!

(iii) Let P be any point in the major arc AB. Then, the degree measure of the minor arc AB of the circle is twice the angle subtended by it at P. In other words, AOB = 2APB.

567

Quantitative Aptitude Simplified for CAT

This means that angle in the same segment are always same. (iv) Of all the rectangles inscribed in a given circle, square has the maximum area. (v)

Of all the triangles inscribed in a given circle, equilateral triangle has the maximum area.

Example 9 The difference between the circumference and the radius of a circle is 37 cm. The area of the circle (in cm2) is A.

148

B.

111

C.

154

D.

259

Solution 2r – r = 37. Let  =

22 . Then r = 7 cm. Therefore, area of the circle = r2 = 154 cm2. 7

Hence correct option is [C]. Example 10 The radius of front and back wheels of a road roller are in the ratio of 2 : 3. When the road roller has travelled 4 km, what is the ratio of the number of revolutions made by the front and back wheels. Solution Let the radius of the front and back wheels be 2 m and 3 m. Perimeters are 4 m and 6 m respectively. When the road roller has travelled 4 km, the front wheel has made Similarly, the back wheel has made

4000 number of revolutions. 4

4000 . 6

The ratio of the number of revolutions = 3 : 2. We can solve this question like this also: lcm of 2 and 3 is 6. Therefore, when the distance travelled is 6 units, front wheel makes 3 revolutions and back wheel makes 2 revolutions. Therefore, ratio of the number of revolutions = 2 : 3, no matter what the actual distance is! Example 11 A circle has two parallel chords of lengths 6 cm and 8 cm. If the chords are 1 cm apart and the center is on the same side of the chords, the diameter of the circle will be A.

5 cm

B.

6 cm

C.

10 cm

D.

12 cm

568

Polygons and Circles Solution According to the diagram (as shown below),

r2  32  r2  42  1 or r = 5. Therefore, diameter = 10 cm.

Alternatively, Use options. If chords are 6 and 8 cm long, then diameter has to be more than 8 cm. Either [C] or [D] is correct. If diameter is 10 cm, then radius = 5, which satisfies the data of the question. Hence correct option is [C]. Example 12 Inside a square (side 4 cm) is drawn a circle such that the circle does not touch any side of the square at all. If the two diagonals of the square are drawn such that they do not cross the circular portion, we find that the five regions so generated have same area. Find the length of the diagonal from the circumference of the circle to the corresponding vertex of the square. Solution Area of the circle =

Therefore, r2 =

4 2 16 1 th the area of the square = .  5 5 5

16 4 16 or r = . The length of the semi-diagonal = 2 2. 5 5 2

Hence, required length of the diagonal = 2 2 

16  1.82 . 5

Example 13 A very large rope is tied around the equator of earth and we are left with 1 m extra length of rope. When the rope is tied end to end uniformly around the equator, the rope forms a circle with radius bigger than the radius of earth. Find the height of the rope above ground level. Solution Let radius of the earth be R. Then, 2R + 1 = 2(R + h), where h is the height of the rope above ground level (refer to the diagram given below).

569

Quantitative Aptitude Simplified for CAT

Therefore, h =

1  0.16 m. 2

Note that whatever be the radius of the earth, h will remain 0.16 m. Therefore, if we were left with 1 m extra length even on Jupiter, the answer will still be the same. Example 14 In a very large grazing area there is a cemented platform (with three walls around it) in the shape of an equilateral triangle of side 10 cm. A cow is tethered to one vertex of the triangle by a rope of length 15 cm. What is the total area that can be grazed by the cow? Solution Let the triangle be ABC where the cow is tethered to the vertex A. The grazed area is as shown in the diagram below (shaded portion).

The area of the portion of the bigger circle =

300 2    15   589 cm2 . 360

The total area of the portion of the smaller circles = 2 

120 2    5   52.36 cm2 . 360

Therefore, total area = 641.36 cm2. Example 15 In a circular pond, a fish starts from a point on the edge of the pond, swims 143 m due north, to reach another point on the edge, turns east and swims 24 m to reach yet another point on the edge. What is the diameter of the pond? Solution Since the two lines of paths are perpendicular, the line joining the beginning and the end of the paths is the diameter. Therefore, diameter =

1432  242  145 m.

Example 16 The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km/hr?

570

Polygons and Circles Solution The circumference of the wheel = 2r = 2 

22  70 = 440 cm = 4.4 m. 7

The speed required = 66000 m/hr = 1100 m/min. Therefore, number of revolutions required =

1100 rev/min 4.4

= 250 rev/min. Example 17 Inside a semi-circle of diameter AB (= 12 cm) and center T, two semi-circles with diameters TA and TB are drawn. A complete circle is drawn touching all the three semi-circles. Find the radius of this complete circle. CAT 2003 (C) Solution Length of diameter of the biggest circle = 12 cm. Therefore, radius of each of the semi-circles = 3 cm. Let radius of the complete circle be r. Let the center of the two semi-circles be P and Q and that of the complete circle be O. Then, in triangle OPT (where T is the center of the biggest circle), OT = 6 – r. OP = 3 + r and PT = 3. Therefore, (3 + r)2 = 32 + (6 – r)2 or r = 2 cm.

Example 18 The 8” pizza sells for $4.99 at my favorite pizza store. The store claims they have a great deal on the large 14” pizza, which is specially priced at $ 13.75. What is the per cent discount the store is offering? Solution The small pizza has a diameter of 8” and therefore an area proportional to 82. The large pizza has a diameter of 14” and therefore an area proportional to 142. If the pizzas are priced in accordance with their areas, then 2

 14  Price of large pizza/Price of small pizza =   = 3.06 or  8  Price of large pizza = 3.06 × 4.99 = $15.27. 13.75   Thus, discount on large pizza =  1    100  10% . 15.27  

Example 19 A circle is inscribed in a square as shown below. The dimensions of the rectangle drawn is as shown. Find the radius of the circle.

571

Quantitative Aptitude Simplified for CAT

CAT 2003 (C) Solution The diagram is redrawn as below:

Therefore, (r – 2)2 + (r – 4)2 = r2 or r = 10 units. Example 20 The area of the largest triangle that can be inscribed in the outermost circle is

3 3 cm2. Find the total 4

circumference of all five circles (given that the centers of all the circles are collinear). A.

2

B.

3

C.

4

D.

Cannot be determined

Solution Area of the triangle =

3 3 . Therefore, side of the triangle = 4

3 . So, radius of the circle, R = 1 cm.

Let the radius of the smaller circles be a, b, c and d. Then, 2(a + b + c + d) = 2R, where R is the radius of the biggest circle. So, (a + b + c + d) = R = 1. Total circumference of all the circles = 2(a + b + c + d + R) = 2(1 + 1) = 4. Hence correct option is [C].

572

Polygons and Circles Example 21 A curve is composed by connecting 6 quarter circumferences with different sizes as shown in the figure. Length of the side of the small square is 2 cm. Find the length of the curve. A.

42

B.

84

C.

21

D.

7

Solution Radius of the innermost circle = 2 cm; radius of the next circle = 4 cm; that of next one = 6 cm and so on. So, length of the curve =

1 × 2 × (2 + 4 + 6 + 8 + 10 + 12) = 21. 4

Hence correct option is [C]. Example 22 In a square of side 10 cm, with each vertex as center, 4 quarter circles are drawn, which are tangent to each other at the center of the square. Find the shaded area. A.

50– 100

B.

100( – 1)

C.

75 – 100

D.

25

Solution Area of 4 quarters circles = area of one circle. Radius of each quarter circle = half of diagonal =

1 10 2  5 2 . 2





573

Quantitative Aptitude Simplified for CAT

When we add 4 quarter circles, the shaded area gets counted twice and unshaded area gets counted only once. So, we end up counting shaded area and unshaded area extra once, that is area of the whole square extra. Therefore,



Area of shaded region = Area of circle – Area of square =  5 2

2  102 = 50 – 100.

Hence correct option is [A].

EXPERT SPEAK Scan this QR Code to watch a video that explains the applications of Area of sectors and segments, as also Pythagoras' Theorem to problems based on Circles.

Secants and tangents to a circle Tangent is a line which touches the circle at one point only, whereas a secant is a line which cuts the circle in two distinct points. We can say that a tangent is a limiting case of a secant. (i)

If a point P lies on the circle, then only one tangent (through P) can be drawn to the circle. It is an important property that tangent through P and the radius to P are always perpendicular.

(ii)

If the point P lies outside the circle, then 2 tangents are possible. Both these tangents are of equal length. The angle bisector of the pair of tangents passes through the center of the circle.

(iii) Finally, if P lies inside the circle, then no tangent through P is possible.

Property If from a point P inside the circle (refer to fig 1), 2 chords AB and CD are drawn, then PA  PB = PC  PD If from a point P outside the circle (refer to fig 2), 2 secants PAB and PCD are drawn, then PA  PB = PC  PD. If from a point P outside the circle (refer to fig 3), 1 secant PAB and 1 tangent PT is drawn, then PA  PB = PT2.

Fig 1

Fig 2

Fig 3

Refer to figure1. In BPD and CPA, BPD = CPA (vertically opposite angles are equal) BDP = CAP (as BDC = CAB, angle in the same segment are equal) Therefore, BPD ~ CPA 

PB PC   PA  PB = PC  PD. PD PA

Refer to figure 2.

574

Polygons and Circles The quadrilateral ABDC is cyclic. Therefore, BAC + BDC = 180°. But, BAC + PAC = 180°. Therefore, BDC = PAC. Similarly, ABD = PCA. Therefore, PBD ~ PCA 

PB PC  or PA  PB = PC  PD. PD PA

Refer to figure 3. In PAT and PTB, APT = TPB PTA = PBT (alternate segment theorem, refer to the next section) Therefore, PAT ~ PTB 

PA PT  or PA  PB = PT2. PT PB

Alternate Segment Theorem Draw a circle and line PQ tangent to the circle meeting the circle at T. From T, draw two chords TA and TB. Then, according to the theorem, BAT = BTQ and ABT = ATP (Fig. 1).

Fig 1

Fig 2

Refer to figure 2. Let ABT = . Therefore, angle at the center O, AOT = 2. Since AOT is isosceles triangle, OAT = OTA = 90° – . OT is radius and PTQ is tangent. Therefore, OT is perpendicular to PTQ. OTP = 90° ATP =  = ABT. Hence the result. Similarly, we can prove that BAT = BTQ. Alternatively, From T, draw diameter meeting the circle in C. Now, TCA = TBA = . CAT = 90° (angle in a semi-circle). Therefore, ACT = 90 – ACP = 90 – (90 –  =  

Example 23

575

Fig 3

Quantitative Aptitude Simplified for CAT

Two chords AB and CD in a circle intersect in a point P inside a circle such that AP = 8 cm, BP = 2 cm. CP = 5r and DP = r. Find the radius of the circle. Solution Using the result that PA × PB = PC × PD, we obtain (5r)(r) = (8)(2) or r =

16  1.789 . 5

But this is not possible as the diameter would be 3.578 which is less than the length of the chords AB (10 cm). Such a case is therefore impossible. Example 24 In the figure, the length of TR will be A.

6 cm

B.

4.5 cm

C.

5 cm

D.

8 cm

Solution Triangle TPR is similar to triangle QTR. This is so because TPR = QTR (Alternate segment theorem) PTR = PTQ + QTR = PTQ + TPR = TQR. Therefore, by AA rule of similarity, the said triangles are similar. Therefore,

TP PR TR 4 TR   , or   TR = 8 cm. QT TR QR 3 6

System of two circles and common tangents The present discussion will be on two circles. Common tangent is a line simultaneously tangent to two curves. The number of common tangents to the two circles varies depending upon the position of the two circles with respect to each other. In case the two circles are completely outside each other, there will be four common tangents (i)

two are direct common tangents, and

(ii)

two indirect (or transverse) common tangents.

P, Q, R, S, A, B, C and D are the points of contact of the common tangents to the two circles. Therefore, PQ and RS are direct common tangents, whereas AB and CD are indirect common tangents to the two circles.

576

Polygons and Circles

The indirect common tangents intersect in T1 and direct common tangents, when extended, meet in T2. C1 and C2 are the centers of the circles.

Properties 1.

C1, C2, T1 and T2 are always collinear.

2.

T1 divides the line joining the centers C1 and C2 internally in the ratio of their radii, that is, C1 T1 r1  , where r1 and r2 are the radii of the two circles. C2 T1 r2

3.

T2 divides the line joining the centers C1 and C2 externally in the ratio of their radii, that is,

C1 T1 r1  C2 T1 r2

Lengths of Common Tangents

Fig 1

Fig 2

In figure 1, AB is the direct common tangent of length l. E and C are the centers of the two circles. AE = r1 and BC = r2. And AB is perpendicular to the two radii. Now, through C, draw a line parallel to AB and let it meet AE at D. Then, ABCD is a rectangle. In the right triangle DCE, DC = l, DE = r1 – r2 and EC = d. Applying Pythagoras’ Theorem, d2 = l2 + (r1 – r2)2  l = The length of the direct common tangents, l =

2

d2   r 1 r 2  . 2

d2   r 1 r 2  .

Similarly, in figure 2, AB is the indirect common tangent of length m. E and C are the centers of the two circles. AE = r1 and BC = r2. And AB is perpendicular to the two radii. Now, through C, draw a line parallel to AB and let it meet AE (extended) at D. Then, ABCD is a rectangle.

577

Quantitative Aptitude Simplified for CAT

In the right triangle DCE, DC = m, DE = r1 + r2 and EC = d. Applying Pythagoras’ Theorem, d2 = m2 + (r1 + r2)2  m = The length of indirect common tangents, m =

2

d2   r 1 r 2  . 2

d2   r 1 r 2  .

From the above formulae, we can say that l > m. With respect to the system of two circles, five cases arise: Case 1: When the two circles are completely outside each other. In such a case, d > r1 + r2, where d = distance between the centers of the circles; r1 and r2 are the radii of the two circles. As d > r1 + r2, l and m are both real and non-zero. Hence, there exist four common tangents, all non-zero.

Case 2: When the two circles are touching each other externally. In such a case, d = r1 + r2. 2

As d = r1 + r2, m = d2   r 1 r 2   0 . Therefore, indirect common tangent is of length zero. Hence, there exist three common tangents: two direct and one indirect.

Let two circles touch each other externally and let’s draw a common tangent which when produced meet in P. We can keep on drawing circles each of which touches the common tangents and also the previous circle (refer to the figure below). The radii of these circles are in geometric progression. So, r1 r2 r3    ... r2 r3 r4

Case 3: When the two circles are intersecting each other at two distinct points. In such a case, |r1 – r2| < d < r1 + r2. Here, d must be more than |r1 – r2|, because when d becomes equal to |r1 – r2|, then the circles touch each other internally. Since d < r1 + r2, m is imaginary, and since d > |r1 – r2|, the length of direct common tangent is still not zero. Hence, there will be two common tangents, both direct.

578

Polygons and Circles

Case 4: When the two circles are touching each other internally. In such a case, d = |r1 – r2|. Since d = |r1 – r2|, length of direct common tangent =

2

d2   r 1 r 2   0 . Moreover, d being obviously less

than r1 + r2, indirect common tangent is imaginary. Hence, there will only one common tangent, which is direct common tangent.

Case 5: When one of the circles is inside the other without touching. In such a case, d < |r1 – r2|. The two circles need not be concentric. Since d < |r1 – r2|, the length of direct as well as indirect common tangent is imaginary and hence non-existent (by formula). No common tangent can be drawn.

Example 25 Two circles of different radii touch each other externally at the point of contact P, through which a common tangent is drawn. A bigger circle touches both these circles internally, and cuts the common tangent at the points A and B. Let AP = BP = 5 cm. Find the area outside the two circles (touching externally) but inside the biggest circle. Assume that the centers of the three circles are collinear.

579

Quantitative Aptitude Simplified for CAT

Solution Let the radius of the two circles be r1 and r2. Draw a line CD passing through the centers of the two circles and also P. If R is the radius of the biggest circle, then R = r1 + r2. Also, using the result that PA × PB = PC × PD, we obtain (2r1)(2r2) = 52 or r1r2 =

52 . 4

Required area = [R2 – r12 – r22] = [(r1 + r2)2 – r12 – r22] = (2r1r2) = 12.5 cm2. Example 26 Two circles with centers A and B intersect at P and Q. Then, A.

AB is not the perpendicular bisector of PQ

B.

PQ is the perpendicular bisector of AB.

C.

The figure formed by joining the mid-points of AP, PB, BQ and QA is a rectangle.

D.

area of APQ = area of PBQ.

Solution AB is the perpendicular bisector of PQ. Therefore, [A] is the wrong statement. PQ is not the perpendicular bisector of AB unless the two circles are of same radius. Therefore, [B] is also not the correct statement.

The mid-point of AP and AQ will be parallel to PQ and half of it. Similarly, for other mid-points. For it to be a rectangle, the four sides must be perpendicular, which is the case because the line AB and PQ themselves are perpendicular to each other. Area of the given triangles APQ and PBQ will be equal only if the two circles are of same radii. Hence correct option is [C]. Example 27 The distance between the centers of two circles with radii 9 cm and 4 cm is 13 cm. The length of direct common tangent between them is A.

10 cm

B.

12 cm

C.

11 cm

D.

10.5 cm

580

Polygons and Circles Solution 2

132   9  4   12 cm.

Length of direct common tangent, l = Example 28

The ratio of radii of two concentric circles is 2 : 1. From a point P on the outer circle, a pair of tangents to the inner circle is drawn meeting the inner circle at A and B. The centroid of triangle PAB A.

lies inside the inner circle

B.

lies on the inner circle

C.

lies outside the inner circle

D.

depends upon the radii

Solution

Let OA = r and OP = 2r. So, PA =

(2r)2  r2  r 3 = PB. Let OPA = . Now, sin  =

 = 30 and so APB = 60. PD = length of altitude = If G is the centroid, then PG =

OA r 1   . Therefore, OP 2r 2

3 3 r 3  r . 2 2

2 2 3  PD   r  r . Therefore, centroid of PAB lies on the inner circle. 3 3 2

Hence correct option is [B]. Example 29 Three circles, all of radius 5 cm, intersect each other such that the area common to all the three circles is nil. Find the area of the equilateral triangle formed by joining the centers of the circles. A.

100 3

B.

75 3 4

C.

74 3 4

D.

98 3 4

Solution Refer to the fig drawn. Radius = OA = 5 cm. Therefore, AB = 5 3 . Therefore, area of ABC =

3 5 3 4





2



75 3 . Hence correct option is [B]. 4

581

Quantitative Aptitude Simplified for CAT

Example 30 Two circles with radii 9 cm and 4 cm touch each other externally. Find the radius of a circle which touches these two circles as well as a common tangent to the two circles. Solution PQ is common tangent to the circles of radius 9 and r, QR is the common tangent to the circles of radius 4 and r, PR is the common tangent to the circles of radius 9 and 4. So, PQ =

(9  r)2  (9  r)2  6 r

QR =

(4  r)2  (4  r)2  4 r

PR =

(9  4)2  (9  4)2  12

Now, PQ + QR = PR  6 r  4 r  12  r = 1.44 cm.

.

EXPERT SPEAK Scan this QR Code to watch a video that will help you understand Tangent Secant Theorem better, and also problems based on Common Tangents.

Example 31 A circle of unit radius touches the coordinate axes. Another circle is such that it touches the given circle and also the coordinate axes. How many such circles are possible? Find the radii of all. CAT 2004 Solution All the circles possible are drawn below. There are 4 circles possible: 2 in the first quadrant (if the given circle is in the first quadrant), 1 in second and 1 in fourth quadrant. The ones in 2nd and 4th quadrants are of same

582

Polygons and Circles radius (that is, 1 units). Let us find the radius of the circles in the first quadrant. Let the radius of the circle (with center P) and which is bigger than the unit radius be r. Then, OP = r + 1 + 2 . Let the perpendicular from P on the x-axis is Q. Then PQ = OQ = r. Using Pythagoras’ theorem, (r + 1 + = r2 + r2 or 2r2  r =

2 1

2 )2

. Note that the ratio of radii of two such consecutive circles is always going to be

2 1

2  1 : 2  1 . Therefore, radius of the smallest circle is

583

2 1 2 1

.

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellants kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

Two identical circles intersect so that their centers, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is A.

 4

B.

 1 2

C.

 5

D.

2 1

A.

3 2

B.

1+

C.

4 3

D.

4.5 CAT 2005

CAT 2005 2.

5.

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm? A.

1 or 7

B.

2 or 14

C.

3 or 21

D.

4 or 28

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1 : 2, and DF is perpendicular to N such that NL : LM = 1 : 2. The length of DH in cm is

CAT 2005 3.

A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point? A.

3.88%

B.

4.22%

C.

4.44%

D.

4.72%

A. B.

C.

D.

2

2 1



2

3

2 1



2

2

2 1



3 CAT 2005

6. CAT 2005

4.

2 2 1

Four points A, B, C, and D lie on a straight line in the X-Y plane, such that AB = BC =

584

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

Polygons and Circles

C.

  2r  2  3  r 1  5 

D.

2r  3

A. B.

2r 1  3

9. CAT 2005 Directions for questions 7 and 8: Answer the questions on the basis of the information given below:

E.

 2 4

A semi-circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semi-circle (in sq. cm) will be 32 50 40.5 81 indeterminable CAT 2006

10. Two circles with centres P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the AQP in degrees?

CAT 2006

8.

 2 2

A. B. C. D. E.

A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminium of width 2 units, as shown below. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

7.

D.

A. B. C. D. E.

The proportion of the sheet area that remains after punching is:

Between 0 and 90 Between 0 and 30 Between 0 and 60 Between 0 and 75 Between 0 and 45

A.

 2 8

B.

6 8

C.

4 4

D.

 2 4

A.

 3  3 4

E.

14  3  6

B.

2 3  3 2

C.

4 3  3 2

CAT 2007 11. Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the other. What is the area (in sq cm) of the intersecting region?

Find the area of the part of the circle (round punch) falling outside the square sheet. A.

 4

D.

4 3  3 2

B.

 1 2

E.

C.

 1 4

2 3  3 2

CAT 2008

585

Quantitative Aptitude Simplified for CAT

Directions for questions 12 to 14: Answer the questions on the basis of the information given below:

C.

9

D.

None of the above CAT 2004

In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.

16. In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If CBE = 65°, then what is the value of DEC?

CAT 2004

A.

35

B.

55

C.

45

D.

25 CAT 2004

12. What is the ratio of the length of PQ to that of QO? A.

1:4

B.

1:3

C.

3:8

D.

3:4

17. Let C be a circle with centre P0 and AB be a diameter of C. Suppose P1 is the mid point of the line segment P0 B, P2 is the mid point of the line segment P1B and so on. Let C1, C2, C3, ...be circles with diameters P0 P1, P1 P2 , P2 P3 ...respectively. Suppose the circles C1, C2, C3, are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is

13. What is the radius of the circle II? A.

2 cm

B.

3 cm

C.

4 cm

D.

5 cm

A.

8:9

B.

9 : 10

C.

10 : 11

D.

11 : 12

14. The length of SO is CAT 2004 A.

8 3 cm

B.

10 3 cm

C.

12 3 cm

D.

14 3 cm

18. On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?

15. N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N? A.

5

B.

7

A.

7.5

B.

7

C.

7.75

D.

None of the above

CAT 2004

586

Polygons and Circles CAT 2003 (R)

19. In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with center at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is  : 3 . The line segment DE intersects at E such that ODC = ADE. What is the ratio AE : AD?

A.

1:

3

B.

1:

2

C.

1:2 3

D.

1:2

22. Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDE? A.

1 3

B.

1 2

C.

2 3

D.

5 6

CAT 2003 (R) 23. In the following figure, BC is diameter, BAP = 25, PD = QE. Find PQE. CAT 2003 (R)

20. The length of the circumference of a circle equals the perimeter of a triangle of equal sides, and also the perimeter of a square. The areas covered by the circle and square are c, t, and s, respectively. Then, A.

s>t>c

A.

100

B.

c>t>s

B.

110

C.

c>s>t

C.

105

s>c>t

D.

120

D.

CAT 2012

CAT 2003 (R)

24. Two circles have same radius 10 cm. Length of common chord is 10 2 cm. Find the length of line segment PQ.

21. In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ATC = 30° and ACT = 50°, then the angle BOA is

A.

10 2 2

A.

100

B.

150

C.

80

D.

Cannot be determined

B.

10 1 2

C.

587

10 2 1 2

Quantitative Aptitude Simplified for CAT

20

D.

2 2

CAT 2012 25. Three concentric semi-circles are drawn as shown below. The radii of the circles are r, 2r and 3r. PQ and PR are tangents to smallest and second smallest circles. T is the centre of all the circles. Find the ratio of area of PQTR to the shaded region.

A.

20 cm2

B.

35 cm2

C.

52 cm2

D.

75 cm2 CAT 2013

28. Let PQRSTU be a regular hexagon. The ratio of the area of the triangle PRT to that of the hexagon PQRSTU is

2 A. B.

C. D.



2 5



3 2 2 5 3



2 5

A.

0.3

B.

0.5

C.

1

D.

None of the above IIFT 2016



29. If in the figure below, XYZ = 90° and the length of the arc XZ = 10π, then the area of the sector XYZ is

3 4 2 2 5 3

CAT 2013 26. There are two circles of radii 10 and 8 cm and distance between their centers is 15. Find the approximate length of the common chord. A.

11.5 cm

B.

9.8 cm

C.

13.2 cm

D.

14.5 cm

A.

10

B.

25

C.

100

D.

None of the above IIFT 2015

CAT 2013

30. ABCDEF is a regular hexagon and PQR is an equilateral triangle of side a. The area of the shaded portion is X and CD : PQ : : 2 : 1. Find the area of the circle circumscribing the hexagon in terms of X.

27. In ABC, A = 90. The diameter of the semi-circle lies on the side BC, and the semi-circle touches the sides AB and AC. The centre of the circle divides BC in the ratio of 3 : 4. Also, AB + AC = 28 cm. Find the approximate area of the shaded region.

588

Polygons and Circles

A. B.

A.

243 3

23 3

B.

486 3

42

C.

729 3

D.

4374 3

E.

None of these

16

X X

5 3

C.

2

X

3 3

D.

XAT 2016

2 3X

34. In the figure below, two circular curves create 60 and 90 angles with their respective centres. If the length of the bottom curve Y is 10, find the length of the other curve.

IIFT 2014 31. In a circular field, there is a rectangular tank of length 130 m and breadth 110 m. If the area of the land portion of the field is 20350 m2 then the radius of the field is A.

85 m

B.

95 m

C.

105 m

D.

115 m

A.

2

B. IIFT 2012 C.

32. AB is a chord of a circle. The length of AB is 24 cm. P is the midpoint of AB. Perpendiculars from P on either side of the chord meets the circle at M and N respectively. If PM < PN and PM = 8 cm. then what will be the length of PN? A.

17 cm

B.

18 cm

C.

19 cm

D.

20 cm

E.

21 cm

15

20 2 3 60 2

D.

20 3

E.

15 XAT 2016

35. A circular road is constructed outside a square field. The perimeter of the square field is 200 ft. If the width of the road is 7 2 ft and cost of construction is Rs.100 per sq. ft. Find the lowest possible cost to construct 50% of the total road.

XAT 2017 33. ABC and XYZ are equilateral triangles of 54 cm sides. All smaller triangles like ANM, OCP, QPX and so on are also equilateral triangles. Find the area of the shape MNOPQRM.

A.

Rs.70400

B.

Rs.125400

C.

Rs.140800

D.

Rs.235400

E.

None of the above XAT 2015

589

Quantitative Aptitude Simplified for CAT

36. The center of a circle inside a triangle is at a distance of 625 cm from each of the vertices of the triangle. If the diameter of the circle is 350 cm. and the circle is touching only two sides of the triangle, find the area of the triangle.

hits the edge of the pool. It then turns east and swims for 400 feet before hitting the edge again. What is the area of the pool? A.

62500

B.

125000

240000

C.

250000

B.

387072

D.

500000

C.

480000

E.

Cannot be answered from the given data

D.

506447

E.

None of the above

A.

XAT 2013 39. The radius of a circle with center O is 50 cm. A and C two points on the circle, and B is a point inside the circle. The length of AB is 6 cm, and the length of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance OB.

XAT 2015 37. There are two circles C1 and C2 of radii 3 and 8 units respectively. The common internal tangent, T, touches the circles at points P1 and P2 respectively. The line joining the centers of the circles intersects T at X. The distance of X from the center of the smaller circle is 5 units. What is the length of the line segment P1P2? A.

≤ 13

B.

> 13 and ≤ 14

C.

> 14 and ≤ 15

D.

> 15 and ≤ 16

E.

> 16 XAT 2014

38. At the center of a city’s municipal park there is a large circular pool. A fish is released in the water at the edge of the pool. The fish swims north for 300 feet before it hits the edge of the pool. It then turns east and swims for 400 feet before it

A.

26

B.

25

C.

24

D.

23

E.

22 XAT 2013

ANSWER KEY 1. B

2. D

3. D

4. B

5. B

6. A

7. B

8. D

9. B

10. C

11. E

12. B

13. B

14. C

15. B

16. D

17. D

18. B

19. A

20. B

21. A

22. B

23. B

24. C

25. A

26. B

27. A

28. B

29. C

30. A

31. C

32. B

33. B

34. A

35. B

36. B

37. C

38. A

39. A

590

Polygons and Circles

ANSWERS AND EXPLANATIONS 1.

The circumference of circle = 2r. Total Distance of B = 4 r = 12.56r If they have to return to their starting point at the same time, then speed ratio = 12.56 : 12. 0.56 Required Percentage =  100  4.72% . 12

Answer: B Explanation: 4.

Answer: B Explanation:

Area of common portion AOBO = 2 × Area of  90  1 segment AOB = 2 .      = 1  2 2  360  2.

Path traveled by ant = AP + PQ + QD  Here, AP = length of quarter arc = (1)2 2   = ; PQ = 1; QD = . 2 2 Hence, AP + PQ + QD =  + 1

Answer: D Explanation: 5.

OM =

OB2  MB2  202  122  16

OL =

OD2  LD2  202  162  12

Explanation: MN = AB = 3 cm AE : EB = 1 : 2  AE = 1 cm, EB = 2 cm. 3 1 HL = EO = EB – OB = 2   . 2 2 LN : LM = 1 : 2  LN = 1 cm, ML = 2 cm. 3 1 LO = ML – MO = 2   . 2 2 In DOL, OD2 = OL2 + DL2 (1.5)2 = (0.5)2 + (DL)2  DL = 2 .

Therefore, ML = 16 – 12 = 4 or 16 + 12 = 28. 3.

Answer: B

Therefore, DH =

Answer: D Explanation:

6.

2

1 2 2 1 .  2 2

Answer: A Explanation: PQS is right angled triangle.

If radius of circle is r. Length or Rectangle is 4r. Breadth is 2r. The perimeter of rectangle is 12r.

591

Quantitative Aptitude Simplified for CAT

circle. Then, AP = AQ = QP AQP is equilateral and hence AQP will be 60.

PQ = r 3 . Since QPS = 30, 1 1 QS = PS = (2r) = r. 2 2 Therefore, perimeter of PQSR = 2r + 2r 3



11. Answer: E Explanation:



= 2r 1  3 . 7.

Answer: B Explanation: The area common to the circle and square = area of right triangle (whose right angle vertex is P) + area of the semi-circle. The right triangle is isosceles triangle (because diameter from P is along the diagonal of the square). Length of perpendicular side of the right 2  2. triangle = 2 2 1 Area of the triangle =  2  1 . 2 1  Area of the semi-circle = (1)2 =  2 2 Proportion of the sheet area that remains  2    2   1  2   6       . after punching = 2 8 2

Let the two circles intersect at P and Q and the centers of the circles be A and B. So, AP = AB = BP = 1 cm and hence APB is equilateral and similarly AQB is also equilateral. So, PAQ = 120. Area of the intersecting region = Sum of the areas of the two segments 1 3  120  2 = 2    (1)(1) sin120    . 360 2 3 2  

 

8.

12. Answer: B Explanation: Since OP : OQ = ratio of radii of circles I to II = 4 : 3, therefore, PQ : QO = 1 : 3.

Answer: D

13. Answer: B

Explanation:

Explanation: Since PO = 28 and ratio of PQ : QO = 1 : 3, PQ = 7 cm. Also, ratio of radii of the circles = 4 : 3. Therefore, radius of circle I = 4 cm and that of II = 3 cm.

 2   Required area = (1)2 –  1   = . 2 2 

9.

Answer: B Explanation: ACD ~ DCB 

AC CD  DC CB



2 6 .  6 CB

14. Answer: C

Therefore, BC = 18. Therefore, diameter = 20 and hence radius of circle = 10 cm. Area of 1 semi-circle = (10)2 = 50. 2

Explanation: OQ = OP – PQ = 28 – 7 = 21 cm. QS = radius of II = 3 cm. Therefore, SO =

10. Answer: C

212  32  432  12 3

15. Answer: B

Explanation: The smallest AQP will be 0 when the two circles touch each other externally. The angle will be maximum when circle with center P passes through the center Q of the other

Explanation: Given number of pairs = number of diagonals n(n  3) in a polygon of n sides = = 14. 2 So, n = 7.

592

Polygons and Circles 16. Answer: D

length and breadth of rectangle = a and b. Area of the circle = x2 Ratio of area =  : 3 = x2 : ab. If x2 = 1, then ab = 3 . By Pythagorean theorem, a2 + b2 = 4x2 = 4 3 We get, a2 + 2 = 4  a2 = 3 or 1  a = 1 or a 3 . Accordingly, b = 3 or 1. Therefore, longer side is 3 and shorter side = 1. Since ODC = ADE, EAD ~ BCD. Therefore, AE : AD = CB : CD = 1 : 3 .

Explanation: CBE = 65  CAE = 65  ACE = 25 (as AEC = 90, angle in a semicircle). Therefore, CED = 25 (alternate interior angle). 17. Answer: D Explanation: Let the diameter AB be 2r. So P0B = r, P1B r r and so on. = , and P2B = 2 4 r r Diameter of C1 = ; diameter of C2 = and 2 4 so on. So total shaded region will be  r 2  r 2  r 2            ...   8  16    4   .   r 2  1  r 2     1  12 16  1   4 Area of original circle = r2. Therefore, required ratio = 11 : 12.

20. Answer: B Explanation: 2r = 3x = 4y (where r = radius of circle, x = side of triangle and y = side of square). 1 1 1 Therefore, r : x : y = : :  6 : 4  : 3 2 3 4 Therefore, c : t : s = r2 :

3 2 2 x :y 4

3 (4 )2 : 92 4 So, 36 : 4 3 : 9. Therefore, c > t > s.

= 36 :

21. Answer: A 18. Answer: B Explanation: CAB = 50 + 30 = 80. By alternate segment theorem, ABC = ACT = 50. Therefore, ACB = 180 – 50 – 80 = 50. So, AOB = 2 × 50 = 100.

Explanation:

22. Answer: B

AC

=

2

2

8  2  2 15 cm.

altitude CP from C on AB =

Therefore, BP =

Explanation: Join O with A, C and E. We can easily observe that Ar(AOC) = Ar(ABC), and so on. So, Ar(ACE) is half of area of the hexagon.

Therefore,

2  2 15 15  . 8 2

 15  22     2 

2

23. Answer: B

1   AQ 2

Explanation: Since PD = QE, we can say that PQDE is isosceles trapezium  PQ || DE. From this, we can also say that AP = AQ. Therefore, CAQ = BAP = 25. Also, BAC = 90. Therefore, PAQ = 90 – 25 – 25 = 40. So, AQP = APQ = 70  PQE = 110.

1 1 1 . Therefore, CD = PQ = 8 – – 2 2 2 = 7 cm.

=

19. Answer: A Explanation: Let the diameter of the circle = 2x. Also,

593

Quantitative Aptitude Simplified for CAT

24. Answer: C

 s(s  a)(s  b)(s  c)

Explanation: Let the centers of the two circles be A and B and let the intersection points be C and D. The two circles have same radius. Therefore, AC = CB = BD = DA = 10 cm. Also, length of common chord is 10 2 cm, ACBD is a square. Therefore, AB = 10 2 cm.

 16.5(16.5  10)(16.5  8)(16.5  15)  16.5(6.5)(8.5)(1.5) . 1  1367.4375  (15)(h) 2 Length of common chord = 2h = 9.86 cm.

27. Answer: A Explanation: Let center of the circle be O. Let the circle touch AB and AC at P and Q. Join OP and OQ. We know that OP = OQ and AP = AQ. Also, A = 90. OQ is parallel to AB. So, BAC ~ OQC. Since BO : OC = 3 : 4, AQ : QC = 3 : 4. Similarly, BP : PA = 3 : 4. But AP = AQ. So, the ratio of AQ : QC = 3 : 4 can also be written as 12 : 16 and the ratio AP : PB = 4 : 3 can also be written as 12 : 9. So, ratio of BP : PA : AQ : QC = 9 : 12 : 12 : 16. We are given that 12 48 AB + AC = 28. So, PA =  28  49 7 = OP = OQ. AB = 12 and BC = 16. Therefore area of shaded region = Area of triangle – Area of semi-circle

Now, AP = AB – BP = 10 2  10 = BQ. Therefore, PQ = 10 2  2AP



= 10 2  2 10 2  10







= 20  10 2  10 2  2 

10 2 1 2

.

25. Answer: A Explanation: Area of shaded region 1 3 r 2 =  (2r)2  r 2  . 2 2 Area of PQTR = Area of PQT + Area of PRT. In PQT, QT = r and PT = 3r. Therefore, PQ



=

2



1 1  48   12  16      2 2  7  = 96 – 73.86 = 22.14 cm2  20 cm2.

=

28. Answer: B

(3r)2  r2  2 2r .

Explanation:

Similarly, in PRT, RT = 2r and PT = 3r. Therefore, PR =

(3r)2  (2r)2  5r .

Therefore, area of PQTR 1 1 = (r) 2 2r  (2r) 5r  2 2





   2  5  r2 . 2  5  r2 2  2  5   Required ratio =   3 r 2     2 

3

Join O with P, R and T. We can easily observe that Ar(POR) = Ar(PQR), and so on. So, Ar(PRT) is half of area of the hexagon.

26. Answer: B

29. Answer: C

Explanation: Let the centers of the circles be A and B and the intersection points be P and Q.

Explanation: Y is the centre of the circle. Length of arc XZ 90 =  2 r  10   r = 20 cm. 360 So, Area of sector XYZ

Then, area of ABP

594

Polygons and Circles

=

60  2r  10  r = 30 + AQ. 360 1 Now, APQ = BPQ = (360 – 90) = 135. 2

90 1  r 2    (20) 2  100 . 360 4

30. Answer: A Explanation: PQ = a  CD = 2a. Area of shaded region = Area of hexagon ABCDEF – area of triangle PQR 3 3 3 2 23 3 2 = (2a)2  a  a = X  a2 2 4 4 4X = 23 3 Area of circle = r2 = (2a)2 = 4a2  4X  16 = 4  X.    23 3  23 3

Also, AQB = 60, so AQP = 30. AQ AP  AP = 15 2 .  sin135 sin 30 Therefore, length of arc X 90 15 =  2 15 2  . 360 2





35. Answer: B Explanation:

31. Answer: C Explanation: As per the question, area of land portion = πr2 – (130 × 110) = 20350  r2 = 34650  r = 105 m. Side of the square = 50 ft  diagonal of the square = 50 2 ft. Note that cost will be lowest when the inner circle touches the corners of the square. Therefore, radius of inner circle = 25 2 ft. Radius of outer circle = 25 2 + 7 2 = 32 2 ft. Now, 50% of the area of the road 2 2 1 =   32 2  25 2  = 1254 ft2. 2   Therefore, lowest possible cost = 1254 × 100 = Rs.125400.

32. Answer: B Explanation: PA = PB = 12 cm. Now, PA × PB = PM × PN  12 × 12 = 8 × PN  PN = 18. 33. Answer: B Explanation: YZ is divided into 3 equal parts. So, MN 54 =  18 cm. Side of the hexagon 3 MNOPQRM is 18 cm. Therefore, area of the 3 3 hexagon = (18)2  486 3 . 2



 



36. Answer: B Explanation:

34. Answer: A Explanation:

OA = OB = OC = 625 cm = radius of the circumcircle, R. Radius of the smaller circle = 175 cm = OP = OQ. Since OP  AB and OQ  AC, BP

P and Q are the centers of the circles X and Y respectively. APB = 90 and AQB = 60. If radius of the circle Y is r, then AQ = r. Given that length of the arc Y is 10, we get

595

Quantitative Aptitude Simplified for CAT

AP1X ~ BP2X. AP BP 3 8 32 So, 1  2    P2 X  . P1 X P2 X 4 P2 X 3

= OB2  OP2  6252  1752  600 = AP = AQ = QC. Therefore, AB = AC = 1200 cm. Since ABC is isosceles, AR  BC. Now, ARC ~ AQO. So, AR RC AC AR RC 1200      . AQ QO AO 600 175 625 Therefore, AR = 1152 cm and RC = 336 cm  BC = 672 cm 1 × 672 × 1152 Therefore, area of ABC = 2 = 387072 cm2.

Therefore, P1P2 = 4 +

32 44 .  3 3

38. Answer: A Explanation: If radius of the pool is r, then diameter = 2r. (2r)2 = 3002 + 4002 = 250000  2r = 500  r = 250. Therefore, area of the pool = r2 = (250)2 = 625000.

37. Answer: C

39. Answer: A

Explanation:

Explanation: In any triangle, the difference of any 2 sides < 3rd side. Consider ∆OBC. OC = 50 cm  7.07, BC = 2. So, OB > 7.07 – 2 = 5.07  OB2 > (5.07)2  OB2 > 25.

596

Mensuration

Chapter 18

Mensuration INTRODUCTION Mensuration deals with the study of finding the area and perimeter of 2-dimensional shapes and surface area and volume of 3-dimensional shapes. Common two-dimensional shapes are (i)

Triangles: scalene, isosceles, equilateral, right, acute and obtuse;

(ii)

Quadrilaterals: square, rectangle, rhombus, parallelogram, trapezium, kite, cyclic quadrilaterals;

(iii) Polygons: pentagon, hexagon, and so on. (iv) Circles Common three-dimensional shapes are cube, cuboid, right circular cone, right circular cylinder, sphere, tetrahedron, frustums, and so on. In the previous chapter, we have already discussed the area and perimeter of all 2D shapes. Here, we will discuss volume and surface area of 3D shapes only.

Cube A Cube is a 3D shaped figure with each face being a square. The edge is obtained when 2 faces intersect each other. In the following figure, faces ABEF and EFGH intersect at the edge EF. We can count the number of edges to be 12. The Vertex is obtained when 3 faces intersect each other. In the following figure, faces ABEF, EFGH and BCGF intersect each other at the vertex F. We can count the number of vertices to be 8. Therefore, a cube has 6 faces, 8 vertices and 12 edges.

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Quantitative Aptitude Simplified for CAT

Since each face is a square and area of each square is a2 (length of each side of the square = a = length of edge of cube), the surface area of the whole cube = 6a2, as there are 6 squares. Note that ‘surface area’ is taken to mean ‘total surface area’. Lateral surface area of any 3D shape is that portion of the surface area which is on the ‘sides’. In case of cube, the surfaces ABFE, BCGF, CDHG and ADHE constitute lateral surface area. Therefore, lateral surface area of cube = 4a2, as there are 4 squares. Volume of the cube is a3. In the above figure, AG is the diagonal (or body diagonal) of the cube. Length of body diagonal is the largest distance between any two points on the surface of the cube. The length of AG can be found by using Pythagoras’ Theorem in ACG. So, AB = a and BC = a  AC = a 2 . Further, CG = a. So, AG =

AC2  CG2 

a 2 

2

 a2  a 3 .

Cuboid A Cuboid is a 3D shaped figure with each face being a rectangle. Some of the faces can be squares. A cuboid also has 6 faces, 8 vertices and 12 edges.

The length, breadth and height of the cuboid are l, b and h. Each face is a rectangle. Area of ABCD = lb = Area of EFGH. Area of ADHE = bh = Area of BCGF. Area of ABFE = lh = Area of DCGH. Therefore, surface area of the whole cuboid = 2(lb + bh + hl). In case of cuboid, the surfaces ABFE, BCGF, CDHG and ADHE constitute lateral surface area. Therefore, lateral surface area of cuboid = 2lh + 2bh = 2h(l + b). Volume of the cuboid = l × b × h. In the above figure, AG is the diagonal (or body diagonal) of the cuboid.

l2  b2 . Further, CG = h.

AB = l and BC = b  AC = So, AG =

AC2  CG2 



l2  b2



2

 h2  l2  b2  h2 .

598

Mensuration

Sphere and Hemi-Sphere In case of sphere, Total surface area = 4r2 = Curved/Lateral surface area. 4 3 r . 3

Volume =

In case of hemi-sphere, two cases arise: (i)

Hemi-sphere is solid;

(ii)

Hemi-sphere is hollow (with thickness)

In case (i), Curved surface area = 2r2, and Total surface area = Curved surface area + Area of circle = 2r2 + r2 = 3r2.

In case (ii), we have hemisphere with thickness. So, it will have inner and outer radius, r1 and r2. Thickness of the hemisphere = r2 – r1.

Fig 1

Fig 2

In the above figure, the side view and top view of the hollow hemi-sphere with thickness are shown. Capacity of the hemispherical bowl = volume of inner hemisphere = Volume of the material used in making the hemi-sphere =

2 r1 3 . 3

2  r23  r1 3 3





Curved Surface area of the bowl = 2(r22 + r12). Total surface area of the bowl = Curved surface area + area of the circular disc (as seen in figure 2 above) = 2(r22 + r12) + (r22 – r12).

Right Circular Cylinder In a right circular (solid) cylinder, Volume = r2h, Curved Surface Area = 2rh, Total Surface Area = Curved Surface Area + Area of the two circles at the ends of the cylinder

599

Quantitative Aptitude Simplified for CAT

= 2rh + 2(r2) = 2r(r + h).

Right Circular Cone In a right circular (solid) cone, Volume =

1 2 r h, 3

Curved Surface Area = rl, where l is called the slant height. Total Surface Area = Curved Surface Area + Area of the circle at the end of the cone = rl + r2 = r(r + l). Note that r, h and l form a right angle triangle and hence Pythagoras’ Theorem is used to find the slant height. So, l2 = r2 + h2 or l =

r 2  h2 .

The surface area and volume of the above described solids are summarized in the table below:

Solid

Lateral/Curved

Total Surface Area

Volume

4a2

6a2

a3

Cuboid

2h(l + b)

2(lb + bh + lh)

lbh

Sphere

4r2

4r2

4 3 r 3

(Solid) Hemi-Sphere

2r2

3r2

2 3 r 3

Right Circular Cylinder

2rh

2r(r + h)

r2h

rl

r(r + l)

1 2 r h 3

Cube

Right Circular Cone

Surface Area

Example 1 A conical tent is to accommodate 11 persons. Each person must have 4 m2 of the space on the ground and 20 m3 of air to breathe. Find the height of the cone. Solution Base area must be 11  4 = 44 m2 = r2  r2 = 14.

600

Mensuration

Volume required = 11  20 = 220 m3. Therefore,

1 2 r h = 220  h = 220  h = 15 m. 3

Example 2 A cylindrical pencil is sharpened to produce a perfect cone at one end with no overall loss of length. If the diameter of the pencil is 1 cm and the length of the conical portion is 2 cm, find the volume of the shavings. Solution Radius of the pencil = 0.5 cm. Volume of the cylinder portion of the pencil which got converted to the conical shape = r2h = (0.5)2(2) = 0.5 cm3. Volume of the conical shape =

1 × volume of the corresponding cylinder. 3

Therefore, required volume of the shavings =

2   0.5  cm3. 3 3

Example 3 A cuboid of length l, breadth b and height h is given. If l increases by 10%, b by 20% and h by 15%, what is the percentage increase in volume? What is the percentage increase in surface area? Solution Original Volume = lbh. Final volume = (1.1)l (1.2)b  (1.15)h = (1.518) lbh. Therefore, percentage increase in volume = 51.8%. Old surface area = 2(lb + bh + hl). New surface area = 2(1.32lb + 1.38bh + 1.265hl). Note that the overall percentage increase cannot be found out! Example 4 Find the length of the longest pole that can be placed in a room 5 m long, 4 m broad and 3 m high. Solution The longest pole will be the length of the diagonal that runs across the volume of the room. Therefore, longest pole =

52  42  32  5 2 m.

Example 5 The length of the longest pole that can be placed on the floor of a room is 10 m and the length of the longest pole that can be placed in the room is 10 2 m. The height of the room is A. 6 m B. 7.5 m C. 8 m D. 10 m Solution



Let h be the height of the room. Then, h2 + 102 = 10 2



2

. Therefore, h = 10 m.

Example 6 The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 5 5 cm. Its surface area is

601

Quantitative Aptitude Simplified for CAT

A.

361 cm2

B.

125 cm2

C.

236 cm2

D.

486 cm2

Solution l + b + h = 19 and

l2  b2  h2  5 5 or l2 + b2 + h2 = 125. We know that

(l + b + h)2 = 192 = 361  l2 + b2 + h2 + 2(lb + bh + hl) = 361. Therefore, 125 + 2(lb + bh + hl) = 361  2(lb + bh + hl) = 361 – 125 = 236 cm2. Therefore, surface area = 236 cm2. Hence correct option is [C]. Example 7 The area of four walls of a room is 77 m2. The perimeter of the floor is 22 m. The height of the room is A.

3.5 m

B.

5.4 m

C.

6.77 m

D.

7.7 m

Solution Let h be the height of the room. Then, area of the 4 walls = 2lh + 2bh = h[2(l + b)] = 22h which is given to be 77 m2. Therefore, h = 3.4 m. Example 8 A wall 8 m long, 6 m high and 22.5 cm thick is made up of bricks each measuring 25 cm  11.25 cm  6 cm. Find the number of bricks require to build the wall. Solution 8 m is a multiple of 25 cm, 6 m is a multiple of 6 cm and 22.5 cm is a multiple of 11.25 cm. Therefore, we can divide the volume of the wall by the volume of each brick to get the number of bricks. Therefore, number of 800  600  22.5 bricks required = = 6400 bricks. 25  11.25  6 Example 9 A well with 14 m inside diameter is dug 8 m deep. The earth taken out of it is evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment. Solution Volume of mud taken out = ×72 × 8 = 1232 m3. Since width of the embankment is 21 m, the area of the embankment = (282 – 72) = (282 – 72) = 2310 m2. Let the height of the embankment be h. Then, volume of the embankment = 2310h which must be equal to 1232 8 the volume of the mud taken out. Therefore, h = m.  2310 15 Example 10 A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.

602

Mensuration Solution The wire is cylindrical in shape. Let the length of the wire (height of the cylinder) be h. Then,

4    9 3    0.22  h or h = 24300 cm or 243 m. 3

Example 11 A river 2 m deep and 45 m wide is running at the rate of 3 km/hr. The amount of water that runs into the sea per minute, is A.

4500 m3

B.

27000 m3

C.

3000 m3

D.

2700 m3

Solution Speed of the river = 3 km/hr = 50 m/min. The volume of water that flows per minute = 2  45  50 = 4500 m3/min. Example 12 A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into water until it is completely immersed. The water level in the vessel will rise by A.

4.5 cm

B.

2.25 cm

C.

4/9 cm

D.

2/9 cm

Solution The volume of water displaced will be same as the volume of the sphere immersed. Volume of the sphere =

4    3 3  36  . 3

If h is the height by which the water level rises, then (4)2h = 36 or h = 2.25 cm. Example 13 What is maximum number of spheres of radius 10 cm that can surround a sphere S of radius 10 cm, each of the surrounding spheres touching S? Solution In one plane, maximum 6 spheres can surround the sphere S. Further, above this structure of spheres, we have 6 interstitial spaces in which 3 more spheres can be placed in alternate spaces. Similarly, below this structure of spheres, we can place 3 more sphere in alternate spaces. So, a total of 12 spheres can be placed around S, each of which touches S. Example 14 From a circular piece of paper, a sector of 120 angle is cut out. The two ends the remaining portion are joined end to end to form a cone. Find the semi-vertical angle of the cone.

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Quantitative Aptitude Simplified for CAT

Solution If we remove a sector of 120, then the piece of paper left with be of sectoral angle 240. The two ends are A and B, which when joined will form a cone (as shown below). The center of the circle, O will become the vertex of the cone. Note that slant height of the cone will be r. The major arc of the circular piece of paper will be same as the 240 2 circumference of the base of the cone. If radius of the cone is R, then  2r = 2R or R =  r . 360 3 Refer to the figure (of cone). In the right triangle, sin  =

AP R 2 2   . Therefore,  = sin–1 . OA r 3 3

Example 15 With each corner of a cube as the center of the sphere, 8 spheres are drawn whose radii are half of the edge of the cube. Find the ratio of the volume of cube occupied by spherical parts to that of the cube. A.

1:2

B.

: 6

C.

: 3

D.

:4

Solution At each corner, spherical part inside the cube will be one-eighth of the volume of sphere. Volumes of eight such spherical parts = volume of one full sphere =

So, the required ratio of volumes =

a3 3



6a

4  a   3  2

3



a3 , where a = edge of the cube. 6

 . 6

Hence correct option is [B]. Example 16 A right angled triangle (3, 4, 5) is rotated about hypotenuse to generate a diamond. Find the volume. Solution The diamond is a double cone with common base as shown in the figure below. The axis of rotation is the hypotenuse. The upper cone has height h, and so lower cone has height 5 – h. Radius of the cone = altitude from B or B’ on AC = B’P = r, say. Then, area of ABC =

1 1 12 ×r×5= ×3×4r= . 2 2 5 2

Volume of the double cone =

2

2

1  12  1  12  1  12  48  .  h    (5  h)     5  3  5  3  5  3  5  5

604

Mensuration

Example 17 A rectangular piece of paper measuring 4  8 is revolved (perpendicular to the plane of the rectangle) through 270° about an axis parallel to the longer side and 5 units away from the center of the rectangle. Find the total surface area of the shape generated. A.

180 + 64

B.

100 + 64

C.

180 – 64

D.

180

Solution The following diagrams clarify the final shape obtained. The total surface area =

3  2(7)(8)  2(3)(8)  2(72  32 )   2(4  8) = 180 + 64. 4

Example 18 An insect crawls along a cylindrical surface starting from a point A on the bottom and reaches the top by taking a spiral path. It is known that the insect takes exactly 5 complete spirals and reaches a point B exactly 4 vertically above the point A. If the diameter and height of the cylinder is m and 15 m, find the total distance  travelled by the insect. Solution If we take a rectangular piece of paper and fold along length or breadth, we get hollow cylinder. If the rectangular piece of paper had a line drawn along its diagonal, then this line will appear as a spiral path along cylindrical surface.

605

Quantitative Aptitude Simplified for CAT

If we divide the height of cylinder into 5 equal parts, we get 5 cylinders, each having height of 3 m and having 1 complete spiral. This spiral will be the length of the diagonal when the cylindrical surface is opened to form a rectangular piece of paper (refer to the figure below. Here A is the point from where the insect started and A’ is the point of completion of 1 spiral). Length of this rectangular piece of paper = circumference of the circular base of cylinder  2 = 2r = 2   = 4 m. 

Therefore, length of one spiral = length of the diagonal =

32  42  5 m.

Length of 5 complete spirals = 25 m.

EXPERT SPEAK Scan this QR Code to watch a video that explains how to visualize 3D shapes and what kind of skills are required in solving problems based on Mensuration.

Example 19 A sphere is inscribed in a right circular cone whose ratio of radius to the height is 1 : 3. Find the ratio of the radii of the cone to the sphere. Solution Let the vertex of the cone be A, center of the circular base be B, the two ends of the base diameter be C and D, and the center of the sphere be O. Let the radii of the cone and sphere be R and r and height of the cone be h. In ACD, CD = 2R, and the circular boundary of the sphere acts as the incircle for ACD. Area of (isosceles) ACD =

1 1 1 (CD)(AB)  (2R)h  (2R)(3R) . 2 2 2

Area is also equal to rs, where r is inradius of the circle and s is the semi-perimeter. h2  R2  (3R)2  R2  R 10 . Therefore, s =

AC =





So, r R  R 10 

2R  R 10  R 10  R  R 10 . 2

1 R 1  10 (2R)(3R)   . 2 r 3

606

Mensuration

Example 20 If surface area of sphere A is 300% higher than that of sphere B, then volume of B is how much % less than that of A? Solution 300   2 2 2 2 Surface area of Sphere A, SA = SB  1    4SB  4rA = 4(4rB)  rA = 4rB  rA = 2rB 100  

Volume of sphere A, VA =

4 4 4  rA 3   (2rB )3  8  rB 3   8VB . 3 3 3 

 8 1  If VB = 1, then VA = 8. Required percentage =    100 = 87.5%.  8 

Example 21 If h, c and V respectively are the height, curved surface and volume of a cone, prove that 3Vh3 – c2h2 + 9V2 = 0. Solution Curved surface area of cone, c = rl (where l is the slant height); Volume of cone, V = 1  1  3Vh3 – c2h2 + 9V2 = 3   r 2h  h3 – (rl)2h2 + 9  r2h  3  3 

1 2 r h  Now, 3

2

= 2r2h4 – 2r2l2h2 + 2r4h2

= 2r2h2(h2 – l2 + r2). Since l2 = r2 + h2, we get 2r2h2(h2 – l2 + r2) = 2r2h2 × 0 = 0. Hence proved. Example 22 Which of the following is a true statement? A.

Of all the rectangles having same perimeter, square has the minimum area

B.

Of all the rectangles having same area, square has the maximum perimeter

C.

Of all the triangles that can be circumscribed around a given circle, equilateral has the maximum area

D.

Of all the 3D shapes having same volume, sphere has the minimum surface area

Solution Standard result. Hence correct option is [D].

607

Quantitative Aptitude Simplified for CAT

Most of the 3D shapes mentioned above can be categorized as either Pyramid or Prism. So, let us now discuss ‘Pyramids’ and ‘Prisms’ and the difference between them. At the base of a prism or pyramid, we have a polygon of n sides.

Pyramids If the side walls are inclined such that they converge at a geometrical point above the base, then such shapes are called Pyramids. Also note that the side walls are inclined to the base and they converge at a point, and so side walls are always triangle shaped. You must have heard of Pyramids in Egypt. Since the base polygon of such pyramids is a square, they are called square pyramids. The pyramids are named after the base shape. So, we can have triangular pyramid (also known as tetrahedron), pentagonal pyramid, circular pyramid (also known as right circular cone), and so on. A regular pyramid is one in which the perpendicular dropped from the vertex falls exactly in the center of the base shape. This will happen if the lengths of inclined edges are equal. A regular tetrahedron is one in which all the six edges are of same length, that is, each of the four triangles is an equilateral triangle. Therefore, every face of the tetrahedron acts as the base.

Square Pyramid

Triangular Pyramid

Cone

In the above figures, the dotted vertical line is the perpendicular dropped from the vertex to the base.

Prisms If the side walls are vertical such that the top shape and size is same as the base shape and size, then such shapes are called Prisms. Most common example of Prism is “Triangular Prism”. If you observe the prism, you will find that the base is triangular in shape and the side walls are perpendicular to the base so that the other end is another triangle of same shape and size. Prisms are also named after the base shape. So, we can have square prism (whose base is square), rectangular prism (whose base is rectangle), circular prism (also known as right circular cylinder), pentagonal prism, and so on.

Cuboid

Cube

608

Triangular Prism

Mensuration

Cylinder Pentagonal

Prism

A regular prism is one in which the perpendicular from the center of the base passes through the center of the same shape on the other end and is perpendicular to it. Therefore, in case of right circular cylinder, there are two circular bases and the line joining their centers is perpendicular to both the circles. Therefore, we note two features that differentiate pyramids from prisms. Pyramids

Prisms

The side walls are inclined to the base.

The side walls are perpendicular to the base.

The side walls are triangular in shape.

The side walls are rectangular in shape.

Sphere or hemi-sphere is neither prism nor pyramid. The following table lists the formulae for regular pyramids or prisms only. Let Base Area = A, Base Perimeter = p, Height of the figure = h, Slant height = l. Pyramids

Prisms

Volume

1  A h 3

Ah

Curved/Lateral Surface Area

1 pl 2

ph

1    p l + A 2 

(p  h) + 2A

Total Surface Area

It is important to understand the curved/lateral surface area of the pyramids. Let us take the example of square based pyramid (refer to figure below).

Since we are talking about regular solids, so OA = OB = OC = OD. From O, draw a line along the surface OBC meeting BC in P, the mid-point of BC. Is the slant height OB or OP? Lateral surface area = Ar (OAB) + Ar (OBC) + Ar (OCD) + Ar (ODA) = 4  Ar (OBC). ABCD is a square of side ‘a’. Since OBC is isosceles, OP is perpendicular to BC.

609

Quantitative Aptitude Simplified for CAT

Then, area of OBC =

1 1 (BC)(OP) = (a)(OP). 2 2

Therefore, lateral surface area = 4  Ar(OBC) =

1 1 (4a)(OP) = (Base Perimeter)(Slant Height). 2 2

Therefore, slant height is OP and not OB.

Tetrahedron Surface area and volume of regular tetrahedron In the figure below, all the four triangles are equilateral triangles of side ‘a’. Lateral Surface Area = Ar (OAB) + Ar (OBC) + Ar (OCA) = 3  Ar (OAB) = 3 

3 2 3 3 2 a  a . 4 4

Total surface area = Ar (OAB) + Ar (OBC) + Ar (OCA) + Ar (ABC) = 4  Ar (OAB) =4

3 2 a  3a2 . 4

Volume =

1 3 2 1 (Base Area)(Height) =  a   h, where h = OG. 3 4 3

OG is the perpendicular from O to the ABC. It is very important to realize that the point G is equidistant from A, B and C, i.e., it is the circumcenter of ABC.But, this being an equilateral triangle, circumcenter is same as centroid. Now, AD is the median and G divides it in the ratio of 2 : 1. Therefore, AG =

In the right triangle OGA, right angled at G, OA2 = OG2 + GA2  a2 = h2 +

Therefore, volume of the tetrahedron =

2 2 3 a .  AD   a 3 3 2 3

2 a2 or h = a . 3 3

1  3 2   2  a3 2 a   a  .    3  3  4 12   

Important Drop perpendicular from each of the vertices of the regular tetrahedron on the opposite surface. These four lines are called medians of the tetrahedron. These medians are concurrent and meet at a point called centroid, G of the tetrahedron. If one of the vertices is A and the corresponding foot of perpendicular is P, then AG : GP = 3 : 1.

610

Mensuration

Frustum When a pyramid is cut by a plane parallel to the base plane, we get a frustum and a smaller pyramid. Observe the following series of diagrams, in which we show how to obtain a frustum.

Note that in the frustum, the top and bottom surface have same shape but different sizes. Lamp shades and pedestals are examples of square frustum, whereas bucket and vase are examples of conical frustum.

Surface Area and Volume of a Frustum Surface areas and volume of a frustum depends upon the type of frustum. Let us calculate the same for conical frustums. In general, whenever we want to find the surface area or volume of frustum, first of all we need to complete the corresponding pyramid and use similarity of triangles. This is elaborated below. Complete the pyramid as shown in the figure.

Let the radii of the top and bottom surface of the frustum be r1 and r2 and the height of the frustum be h. Let AB be x. Then, ABC ~ ADE. Therefore,

hr1 hr2 AB AD x x h     xr2 = xr1 + hr1  x  x+h  . BC DE r1 r2 r2  r1 r2  r1

Slant height of the frustum, CE is given by, CE2 = (BD)2 + (DE – BC)2 or l2 = h2 + (r2 – r1)2. Lateral Surface Area of the conical frustum = Lateral Surface Area of larger cone – Lateral Surface Area of smaller cone = r2(AE) – r1(AC) = r2

(x  h)2  r22 – r1 x2  r12

611

Quantitative Aptitude Simplified for CAT

= r2





2

2 2  r22  r1 2 h2   r2  r1   hr2   hr1  2 2    r2  r1    r1  r2  r1  r2  r1   r2  r1 

= (r1 + r2)l.

Hence, total surface area = (r1 + r2)l + r12 + r22. Volume =

=

hr2 hr1  1 2 1  r2  h  x   r1 2 x   r22   r1 2   3 3 3 r2  r1 r2  r1 

πh  r23  r13  1 2 2 2    πh r1  r2  r1  r1r2 3  r2  r1  3





.

Although the formulae for surface area and volume of frustum of right circular cone are given here, the students should not memorise these results, as the surface area and volume would vary as the shape of the frustum changes. For example, the same formulae would not be valid if it were a square frustum or pentagonal frustum. The students must learn how to arrive at these results and then use that technique for solving questions. Example 23 Find the volume of the square frustum whose dimensions are 4 cm and 8 cm and whose depth is 10 cm. Solution Complete the square pyramid whose total height be h. Then, by similarity of triangles, =

h  10 2  or h = 20. h 4

Therefore, volume of frustum = volume of complete pyramid – volume of smaller pyramid = [(64  20) – (16  10)] =

1 1120 1 (1280 – 160) =  373 . 3 3 3

Euler’s Theorem According to this theorem, for any flat surfaced 3-dimensional solid, (Number of Faces) + (Number of Vertices) = (Number of Edges) + 2, or F + V = E + 2. Applying this for cube, F = 6, V = 8, E = 12. Therefore, 6 + 8 = 12 + 2. Applying this for square based pyramid, F = 5, V = 5, E = 8. Therefore, 5 + 5 = 8 + 2. Applying this for Pentagonal Prism, F = 7, V = 10, E = 15. Therefore, 7 + 10 = 15 + 2. Example 24 A model of Egyptian Pyramid in an exhibition is found to have a height of 3 inches and is made of wood.

612

Mensuration 1 th that of the given 4 pyramid, what would be the new height of the pyramid, keeping the weight of the pyramid constant.

If another similar pyramid is to be built (model only) using material whose density is

Solution If the mass is to be the same, then density is inversely proportional to volume. Also, the volumes are directly proportional to the cubes of the heights for objects that are geometrically similar. Therefore, the heights are inversely proportional to the cube roots of the densities. 1

 Density of miniature  3 Thus, Height of model = Height of miniature ×    Density of Model 

 Height of model = 3  (41/3) = 4.76 inches. Example 25 From a regular tetrahedron whose all 6 edges are of length 10 cm, a triangular frustum is created by a cutting plane parallel to one of the bases and passing through the centroid of the tetrahedron. What is the volume of the frustum? Solution Centroid of tetrahedron divides the height in the ratio of 3 : 1. Volume of the tetrahedron =

a3 2 , where a = length of the edge. 12

Height of the smaller tetrahedron =

3 th of the height of the whole tetrahedron. 4

 3 Therefore, volume of the smaller tetrahedron =    4 =

3



27 th of volume of the whole tetrahedron 64

27  a3 2  27  63 2       . 64  12  64  12 

 2   37 333 2 27   2  63   1   63    Therefore, volume of the triangular frustum =  .      64 64   12 32  12  

Example 26 What is the volume of a regular tetrahedron whose height is 5 cm? Solution The height of the tetrahedron, h = a

Therefore, volume =

2  3 5  12  2 

2 3 = 5. So, a = 5 . 3 2

3



125 3 . 8

Example 27 Find volume of largest (regular) tetrahedron that can be inscribed in a sphere of radius 9 cm.

613

Quantitative Aptitude Simplified for CAT

Solution Radius = 9 cm. Radius is at the centroid of the tetrahedron and so is equal to tetrahedron. So, height of the tetrahedron = 12 cm  side of tetrahedron = 12

2  3   12 Volume of tetrahedron =  12  2 

3 th of the height of The 4

3 . 2

3

 216 3 .

Example 28 A cube of 10 cm dimension is given. The mid-points of three edges (which converge to a vertex O) are joined forming the triangle ABC and a cutting plane cuts along this plane ABC. Find the volume of the shape cut out. Solution The shape cut out will be a tetrahedron whose base edges (AB or BC or AC) will be 5 2 cm and inclined edges (AO or BO or CO) = 5 cm each.





Here, AD = 5 2 

Therefore, OG =

3 5 6 25 6 5 6   AG =  .   3  2  3 2 2

5 6 52     3 

2

5 1

Therefore, volume of tetrahedron =

6 5 .  9 3

1 3 5 2  3 4





2

 5 125  .  6 3 

Example 29 A piece of paper of the shape of an equilateral triangle ABC of side 10 cm, is given. The three corners are folded to create 4 congruent equilateral triangles. The three vertices of the original triangle are joined together. Find the volume of the shape obtained. A.

500 2 12

B.

500 2 3

C.

125 2 3

D.

125 2 12

Solution The shape formed will be tetrahedron whose each edge will be 5 cm.

614

Mensuration

Volume of tetrahedron =

2 3 2 125 2 a   53  . 12 12 12

Hence correct option is [D]. Example 30 Find the volume of a swimming pool whose length is 100 m, width 30 m, depth on shallow side is 4 ft and that on deeper side is 15ft, assuming that the base of the pool is continuously sloping (3 feet = 1 m) Solution The shape of the swimming pool is trapezoidal prism, whose base is the side wall the pool along the length (which is trapezium shaped). Area of trapezium =

1  4  14  2    100  300 m . 2 3 

So, volume of pool = (Base Area) (Height) = 300 × 30 = 9000 m3.

EXPERT SPEAK Scan this QR Code to watch a video that is a continuation of the previous video on Mensuration and explains how to visualize 3D shapes and what kind of skills are required to solve problems based on Mensuration.

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Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

the radius of the circle circumscribing that triangle will be

Consider a right circular cone of base radius 4 cm and height 10 cm. A cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone. Find the largest possible total surface area (in sq. cm) of the cylinder. A. B. C. D. E.

CAT 2004

100  3 80 3 120 7 130  9 110 7

A.

equal to the side of the cube

B. C. D. CAT 2008

3 times the side of the cube

1 3

times the side of the cube

impossible to find from the given information

Directions for questions 4 to 6: Answer the questions on the basis of the information given below:

Directions for question 2: Question below is followed by two statements A and B. Indicate your responses based on the following directives: Mark A. if the question can be answered using A alone but not using B alone. Mark B. if the question can be answered using B alone but not using A alone. Mark C. if the question can be answered by A and B together, but not by either A or B alone. Mark D. if the question cannot be answered even after using A and B together.

Consider a cylinder of height h cm and radius r 2 cm as shown in the figure (not drawn to =  scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string's length is the minimum length required to wind n turns). CAT 2003 (R)

2.

4.

ABC Corporation is required to maintain at least 400 Kilolitres of water at all times in its factory, in order to meet safety and regulatory requirements. ABC is considering the suitability of a spherical tank with uniform wall thickness for the purpose. The outer diameter of the tank is 10 meters. Is the tank capacity adequate to meet ABC requirements? A: The inner diameter of the tank is at least 8 meters. B: The tank weighs 30,000 kg when empty, and is made of a material with density of 3 gm/cc. CAT 2007

3.

What is the vertical spacing (in cm) between two consecutive turns?

A.

h n

B.

h n

C.

If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then

D.

616

h n2

Cannot be determined

Mensuration 5.

The same string, when wound on the exterior four walls of a cube of side n cm, starting at point C and ending at point D, can give exactly one turn. The length of the string, in cm, is

9.

Sum of all the edges of a cuboid is 84 cm and its surface area is 341 cm2. If a sphere is circumscribed around the cuboid, what is the diameter of the sphere? A.

10 cm

B.

20 cm

C.

5 cm

D.

Cannot be determined CAT 2016

6.

A.

2 n

B.

n 17

C.

n

D.

n 13

10. A right circular cylinder has a height of 15 and a radius of 7. A rectangular solid with a height of 12 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. The volume of the liquid is

In the setup of the previous two questions, how is h related to n? A.

h= n 2

B.

h = n 17

C.

h=n

D.

h = n 13

C.

49(5 – 24)

D.

49(15 – 8)

A.

110

B.

220

C.

440

D.

660

12. A right circular cylinder has a radius of 6 and a height of 24. A rectangular solid with a square base and a height of 20, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. If water is then poured into the cylinder such that it reaches the rim, the volume of water is

3 4 1 2

Swimming pool has length of water surface 60 m, width 25 m and depth at shallow and deeper ends are 1.6 m and 4.4 m. Find the volume of water. A. B. C. D.

180( – 5)

IIFT 2015

CAT 2003 (R) 8.

B.

11. If a right circular cylinder of height 14 is inscribed in a sphere of radius 8, then the volume of the cylinder is

A square tin sheet of side 12 inches is converted into a box with open top in the following steps: The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four comers of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box? A. B. C. D.

147(5 – 8)

IIFT 2016 CAT 2003 (R)

7.

A.

m3

5000 4500 m3 4000 m3 3500 m3

A.

288( – 5)

B.

288(2 – 3)

C.

288(3 – 5)

D.

None of the above IIFT 2014

13. Your friend’s cap is in the shape of a right circular cone of base radius 14 cm and

CAT 2014

617

Quantitative Aptitude Simplified for CAT

height 26.5 cm. The approximate area of the sheet required to make 7 such caps is

A.

9%

B.

16%

A.

6750 sq cm

C.

25%

B.

7280 sq cm

D.

50%

C.

8860 sq cm

E.

None of the above

D.

9240 sq cm

XAT 2015 IIFT 2013

17. Diameter of the base of a water – filled inverted right circular cone is 26 cm. A cylindrical pipe, 5 mm in radius, is attached to the surface of the cone at a point. The perpendicular distance between the point and the base (the top) is 15 cm. The distance from the edge of the base to the point is 17 cm, along the surface. If water flows at the rate of 10 meters per minute through the pipe, how much time would elapse before water stops coming out of the pipe?

14. A hemispherical bowl is filled with hot water to the brim. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If diameter of the bowl is the same as that of the vessel, the volume of the hot water in the cylindrical vessel is A.

60% of the cylindrical vessel

B.

80% of the cylindrical vessel

C.

100% of the cylindrical vessel

D.

None of the above IIFT 2012

15. The Volume of a pyramid with a square base is 200 cm3. The height of the pyramid is 13 cm. What will be the length of the slant edges (i.e. the distance between the apex and any other vertex), rounded to the nearest integer? A.

12 cm

B.

13 cm

C.

14 cm

D.

15 cm

E.

16 cm

A.

< 4.5 minutes

B.

≥ 4.5 minutes but < 4.8 minutes

C.

≥ 4.8 minutes but < 5 minutes

D.

≥ 5 minutes but < 5.2 minutes

E.

≥ 5.2 minutes XAT 2014

18. A rectangular swimming pool is 48 m long and 20 m wide. The shallow edge of the pool is 1 m deep. For every 2.6 m that one walks up the inclined base of the swimming pool, one gains an elevation of 1 m. What is the volume of water (in cubic meters), in the swimming pool? Assume that the pool is filled up to the brim.

XAT 2017

A.

528

16. A solid metal cylinder of 10 cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3 : 4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

B.

960

C.

6790

D.

10560

E.

12960 XAT 2014

ANSWER KEY 1. A

2. B

3. A

4. A

5. B

6. C

7. D

8. B

9. A

10. A

11. D

12. C

13. D

14. C

15. C

16. D

17. D

18. D

618

Mensuration

ANSWERS AND EXPLANATIONS 1.

4.

Explanation: Considering negligible width of string, vertical spacing of between two consecutive turns h should be . n

Answer: A Explanation: Let r and h be the radius and height of the 10 10  h 20  5r  , or h = . cylinder. Then, 4 r 2 Total surface area of the cylinder,  3r2   20  5r  S = 2r2 + 2r  .  = 2  10r  2  2    Differentiating S with respect to r, dS we get = 2(10 – 3r) = 0, dr 10 5 or r = and hence h = . 3 3 Therefore, maximum total surface area of the 10  10 5  100 cylinder, S = 2  .     3  3 3 3

2.

5.

Answer: B Explanation: When we open the lateral faces of the cube, we get the following:

Therefore, length of the string = CD = 6.

Answer: B

n2  (4n)2  n 17 .

Answer: C Explanation: In the first case when the string is wound around cylinder, circumference of the base circle = 2r = 4 cm. To find length of the string wound around cylinder, we open the cylinder to get a rectangular piece of paper. Therefore, length of each winding the string

Explanation: Using statement A, inner radius is at least 4 meter. So, minimum volume possible 4 =  (4)3  268.1 , which does not meet the 3 safety standards. Statement A is not sufficient as the volume may be less than 400 kilolitres or more than 400 kilolitres. Using statement B, volume of material used 30,000,000 = 10,000,000 cc = 10 m3. = 3 Volume of the tank using outer radius (5 m) 4 =   5 3 = 523.6 m3. 3 Therefore, capacity of the tank = 523.6 – 10, which is more than 400 m3. Therefore, second statement alone is sufficient. 3.

Answer: A

2

 h 42    .  n So, length of the complete string =

2

 h = n 42     h2  16n2 .  n In the second case, when the string is wound around cube, length of the string = n 17 . Therefore, h2  16n2  n 17  h2 + 16n2 = 17n2  h = n.

Answer: A

7.

Explanation: If the side of a triangle is x, then length of diagonal = x 3 = length of the side of the triangle. So, circumradius of this equilateral triangle  1  = x 3  x.  3

Answer: D Explanation: Volume of the box = (12 – 2x)2x = 144x + 4x3 – 48x2. To maximize volume, we will differentiate the function. We get, 144 + 12x2 – 96x = 0  x2 – 8x + 12 = 0  x = 6 or 2. Therefore, x = 2. Hence correct option is [D].

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Quantitative Aptitude Simplified for CAT

Alternatively, use options.

8.

A.

x = 3  volume = 6 × 6 × 3 = 108.

B.

x = 4  volume = 4 × 4 × 4 = 64.

C.

x = 1  volume = 10 × 10 × 1 = 100.

D.

x = 2  volume = 8 × 8 × 2 = 128.

Answer: B Explanation: The shape of the swimming pool is trapezoidal prism, whose base is the side wall the pool along the length (which is trapezium shaped). Area of trapezium 1 = (1.6  4.4)  60  180 m2. 2 So, volume of pool = (Base Area)(Height) = 180 × 25 = 4500 m3.

9.

Diameter of the cylinder = 2r. Diameter of sphere = 16 and height of cylinder = 14. Using Pythagoras’ Theorem, (2r)2 + 142 = 162 or r2 = 15. Therefore, volume of cylinder = r2h 22 =  15  14 = 660. 7 12. Answer: C Explanation: Diagonal of the square base of the rectangular solid = diameter of the cylinder = 12. Therefore, side of the square base 12 6 2. = 2 Volume of liquid = volume of cylinder – volume of rectangular solid

Answer: A Explanation: 4(a + b + c) = 84  a + b + c = 21. Surface area = 2(ab + bc + ca) = 341. Now, (a + b + c)2 = 212  a2 + b2 + c2 + 2(ab + bc + ca) = 441  a2 + b2 + c2 + 341 = 441  a2 + b2 + c2 = 100. Therefore, length of diagonal



= a2  b2  c2  10 = diameter of the circumscribed sphere.

= (6)2 × 24 – 6 2

2



 20 = 288(3 – 5).

13. Answer: D 10. Answer: A

Explanation:

Explanation: Diagonal of the square base of the rectangular solid = diameter of the cylinder = 14. Therefore, side of the square base 14 = 7 2. 2 Volume of liquid = volume of cylinder – volume of rectangular solid



= (7)2 × 15 – 7 2



2

Slant height = 142  26.52  29.97  30 cm. Hence area of sheet required to make 7 such caps = 7 × rl = 22 × 14 × 30 = 9240 sq cm. 14. Answer: C Explanation: Let radius of the spherical bowl be r. Then, radius of the vessel = r. Height of the vessel r = . 1.5 2 Volume of hemi-spherical bowl = r 3 and 3  r  volume of cylindrical vessel = r2   .  1.5 

 12 = 147(5 – 8).

11. Answer: D Explanation:

620

Mensuration Since both the volumes are same, volume of hit water in the vessel is same that in bowl. 15. Answer: C Explanation: Volume of a square base pyramid 1 1 = × Base area × Height = a2h 3 3

1 2 a (13)  a = 3

600 = 6.8 cm. 13 The side of the base square = 6.8 cm  length of the diagonal = 6.8 2 cm. Therefore, length of half of diagonal = 3.4 2 cm. Measure of the slant height

 200 =

=

 3.4 2 

2

Radius of the cone = 13 cm. In ABC, AC = 17 and BC = 15. Since (8, 15, 17) is a Pythagorean triplet, AB = 8. So, FB = 13 – 8 = 5 cm. Now, DAF ~ DCE. So, DE EC h 5 75 h= .    DF FA h  15 13 8 Volume of frustum = Volume of bigger cone – Volume of smaller cone 1  75   2 75     132   15   5  3  8 8    1   132  195  52  75   38  = 1295. Volume rate of flow of water through the pipe = (0.5)2(1000) cm3/min. Therefore, time taken (in min) 1295 = = 5.18 min. (0.5)2 1000

 132  192.12  14 cm.

16. Answer: D Explanation: Flat surface area of the cylinder = 2 ×  × 72 = 98 cm2 Volume of the cylinder =  × 72 ×10 = 490  cm3 3 Volume of cone A = × 490  = 210  cm3 7 1 = × (flat surface area of cone A) × 10 3 Flat surface area of cone A = 63  cm2 4 Volume of cone B = × 490  = 280  cm3 7 =

18. Answer: D Explanation: Since the elevation increases by 1 m for every 2.6 m walk along the incline, the horizontal

1 × (flat surface area of cone B) × 10 3

distance = 2.62  12  2.4 m. So, for every 2.4 m walk along the length of the pool, the depth increases by 1 m. For a horizontal walk of 48 m, depth increases by 20 m. So, depth of the deeper end of the pool = 20 + 1 = 21 m. 1  Volume of water =  (1  21)48  (20) 2  = 10560 cubic metre.

Flat surface area of cone B = 84  cm2 Total flat surface area of cones = (63 + 84)  = 147  cm2 Percentage change in the flat surface area 147  98 =  100  50% . 98 17. Answer: D Explanation:

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Chapter 19

Coordinate Geometry and Trigonometry COORDINATE GEOMETRY Coordinate geometry is the one of the less important topics asked in MBA entrance exams and mostly easy and direct questions are asked from this topic. Therefore, the focus here is only on elementary applications. In coordinate geometry, we have a coordinate plane obtained by drawing two lines perpendicular to each other. The horizontal line is called x-axis, and the vertical line is called y-axis. By convention, the right direction of xaxis is the positive direction and the left direction is negative. Similarly, north direction and south direction of yaxis are respectively the positive and negative directions. The point of intersection of x-axis and y-axis is called origin, denoted by O. These two axes divide the entire region into four areas, called quadrants as shown below. Quadrants are denoted as I, II, III and IV.

If we locate a point P in the first quadrant, then the distance of P from y-axis is called abscissa and that from xaxis is called ordinate. So, if abscissa of P is ‘a’ and ordinate of P is ‘b’, then the point P is represented as P (a, b) as shown above. (a, b) are called coordinates of the point P. Note that the coordinates (a, b) are different from (b, a). Therefore, (a, b) is called an ordered pair.

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Coordinate Geometry and Trigonometry

Distance Formula Let there be any two points A and B, whose coordinates are (x1, y1) and (x2, y2) resp. Then, the distance between them is given by d=

(x1  x 2 )2  (y1  y 2 )2

This is known as distance formula. This is obtained by using Pythagoras’ Theorem (AP2 + BP2 = AB2), as shown in the diagram below.

Example 1  1 1  What is the length of side of the equilateral triangle with vertex (1, 1) and circumcenter  ,  ? 3  3

Solution Length of the side of equilateral triangle is equal to 3R , where R is circumradius. 2

R=

1  1    1   1  3 3  

2



8 8 2 2 . . Therefore, side = 3  3 3

Example 2 Let there be a circle which passes through (1, 2), (0, 5) and (6, 7). Diameter of the circle will be (approximately) A. 12.50 B. 4.5 C. 10.5 D. 7.1 Solution Let the coordinates of the center be (a, b). Its distance from each of the given points is constant. So, (a –1)2 + (b – 2)2 = (a – 0)2 + (b – 5)2, and (a – 1)2 + (b – 2)2 = (a – 6)2 + (b – 7)2.  1 + 4 – 2a – 4b = 25 – 10b and 1 + 4 – 2a – 4b = 36 + 49 – 12a – 14b  –a + 3b = 10 and a + b = 8  4b = 18 or b = 4.5 and so a = 8 – b = 3.5. Therefore, radius of the circle =

 3.5  1 2   4.5  2 2

 12.5 .

Hence correct option is [A].

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Example 3 Find a point on the x-axis which is equidistant from the points (7, 6) and (–3, 4). Solution Let the point on x-axis be (a, 0). Therefore, (a – 7)2 + (0 – 6)2 = (a + 3)2 + (0 – 4)2  49 – 14a + 36 = 9 + 6a + 16  40 + 20 = 20a  a = 3. So, the point on x-axis is (3, 0). SLOPE The line joining A and B is inclined to x-axis at an angle, say . Then tan  is called the slope of the line. By perpendicular . Therefore, elementary trigonometry, we know that tan  = base tanθ 

BP y 2  y1  AP x 2  x1

Slope is generally denoted by m.

Section Formula If a point P (x, y) lies on the line joining A and B and if it is given that the point P divides the line segment AB in the ratio m : n, then we can find the coordinates (x, y) of point P as  mx 2  nx1 my 2  ny1   mn , mn   

This is known as section formula.  x  x 2 y1  y 2  If P is mid-point of AB, then  1 , 2 2  

If the coordinates of point P are given, then we can find the ratio in which P divides the line joining A and B, provided P lies on the line AB. If this ratio comes out to be positive, then P lies between A and B, whereas if this ratio comes out to be negative, then P lies outside the line segment between A and B but lies on the line AB. This will be clearer from the example below. Example 4 Find the ratio in which the x-axis divides the line joining the points (2, 3) and (4, –7). Also find the ratio for yaxis. Also find the coordinates of the point of the line joining these points with the x-axis and y-axis.

624

Coordinate Geometry and Trigonometry Solution Let the coordinates of the point of intersection of the line joining the given points with x-axis be (a, 0). Let the required ratio be  : 1. Then, using the y-coordinates, 0=

3  1  ( 7)   or  = 3 : 7, which is the required ratio. 1

Since the ratio is positive, the point (a, 0) lies between the given points. To find coordinates of this point, a =

3  4  7  2 26 . Therefore, coordinates of the point on x-axis are  37 10

 26  , 0 .   10  Let the coordinates of the point of intersection of the line with y-axis be (0, b). 2 1  4   or  = –1 : 2. Since the ratio is negative, the point (0, b) lies outside the line segment 1

Then, 0 =

between the given points. To find the coordinates of this point, b =

(1)  (7)  2  3 13  . Therefore, coordinates of the point on x-axis 1  2 1

are (0, 13). By plotting these points, we observe the confirmation of the above fact (as shown below).

Example 5 Find the points of trisection of the line joining the points (0, 3) and (6, –3). Solution The points of trisection will divide the line joining the given points in the ratio 2 : 1 and 1 : 2. When it divides in  0  1  6  2 3  1  (3)  2  the ratio 2 : 1, then the point, say P, is  ,   (4,  1). When it divides in the ratio 1 : 12 12    0  2  6  1 3  2  (3)  1  2, then the point, say Q, is  ,   (2, 1). 12 12   Example 6 Let A (2, 5) and B (–3, 2) be two points. (i)

Find the slope of the line joining them.

(ii)

What is the angle between the line joining them and y-axis? Is the angle between the line joining the points and the x-axis acute or obtuse?

(iii) Find the coordinates of a point P such that A is the mid-point of PB. In which quadrant does P lie?

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Quantitative Aptitude Simplified for CAT

Solution 25 3  . 3  2 5

(i)

Slope =

(ii)

If  is the angle which the line makes with x-axis, then tan  = slope = Then angle with y-axis = 90 –  = 90 – tan–1

3 . 5

3 . 5

The angle of the line joining the given points with x-axis is acute because slope is positive. (iii) Let coordinates of P be (a, b). Then, using section formula 2=

a  (3) 2 b or a = 7. Similarly, 5 = or b = 8. 2 2

Therefore, coordinate of P  (7, 8). Hence the point P lies in first quadrant. Example 7 Girish starts from origin and moves 4 units towards east, then 5 units towards north and then 30° south-east till he reaches a point which is exactly on the east side of the place from where he started the journey. (i)

What are the coordinates of Girish now?

(ii)

What would have been the coordinates if he had moved 30° south-east in the last step?

(iii) Also find the total distance travelled by Girish in both the cases. Solution His movement is shown in the figure below:

5

 5  . Therefore, coordinates of C   4  , 0 . 3 3  

(i)

Distance AC = 5 tan 30 =

(ii)

If he had moved 30 south-east, then he would have followed the path BD (shown in the figure above by dotted line). Now, AD = 5 cot 30 = 5 3 . Therefore, coordinates of C  4  5 3, 0 .



(iii) Total distance travelled by Girish in first case = 4 + 5 + BC = 9 + 5 sec 30 = 9  Total distance in second case = 4 + 5 + 5 cosec 30 = 9 + 5 × 2 = 19. Example 8 Find the point which is three-fourth of the way from (3, 1) to (–2, 5). Solution The required point divides the line joining points in the ratio 3 : 1.

626



10 3

.

Coordinate Geometry and Trigonometry

 3(2)  1(3) 3(5)  1(1)   3  , , 4 . Therefore, coordinates of the point     3 1 31   4   Example 9 Let P be a point of trisection of A (2, 3) and B (4, –5). If Q is the mid-point of A and B, then (i)

find the distance between P and Q

(ii)

find the ratio AP : PQ : QB

Solution (i)

Method 1: First of all, find the coordinates of P and Q and then use distance formula. Since P divides AB in the ratio of 2 : 1 or 1 : 2, Coordinates of P can be

 2  1  4  2 3  1  (5)  2   10 7   2  2  4  1 3  2  (5)  1   8 1  either  , ,  or  ,     , . 21 2 1 21 2 1    3 3     3 3  24 35 Coordinates of Q are  ,   (3, 1) . 2   2 One must realize that distance PQ will be same irrespective of which coordinates of P are chosen. 2

Using distance formula, distance PQ =

 10   7   3   1   3   3 

Method 2: Since Q is the mid-point of AB, we can say that AQ = Similarly, AP = = (ii)

1 6

2



1 16 17   . 9 9 3

1 AB. 2

1 1 AB. Therefore, PQ = AQ – AP = AB 3 6

 4  2 2   5  3 2



1 1 17 4  64  (2) 17  . 6 6 3

If P is a point closer to A than to B, then AP =

1 1 1 AB. PQ = AB and QB = AB. 2 3 6

Therefore, AP : PQ : QB = 2 : 1 : 3. Alternatively, If AB = 6, then AP = 2 and AQ = 3 and so PQ = 1. Therefore, AP : PQ : QB = 2 : 1 : 3.

Centroid and Incentre of a Triangle If the coordinates of the three vertices of ABC are A (x1, y1), B (x2, y2) and C (x3, y3), then the coordinates of  x  x 2  cx 3 x1  x 2  cx 3  centroid G, is given by  1 ,  3 3    ax  bx 2  cx 3 ay1  by 2  cy 3  and the coordinates of incentre , is given by  1 ,  a b c a b c  

where a, b and c are the length of the sides of the triangle as shown below. We can also use the concept that the incentre is a point equidistant from the sides AB, BC and AC of the triangle, to find the coordinates of incentre.

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Quantitative Aptitude Simplified for CAT

Example 10 Find the circumcenter and orthocentre of the triangle whose vertices are (0, 0), (3,

3 ),(0, 2 3 ).

Solution The triangle is equilateral as all the three sides are of equal length. Therefore, centroid of the triangle is same as circumcentre as well as orthocentre.  030 0 3 2 3  , Centroid     1, 3 3 3  





Example 11 Find the incentre of the triangle formed by (0, 0), (5, 12) and (16, 12). A.

(7, 9)

B.

(9, 7)

C.

(–9, 7)

D.

(–7, 9)

Solution Since the triangle is in the first quadrant, the incentre must be in the first quadrant only. Therefore, options [C] and [D] are impossible. Method 1: The three sides are a = 11, b = 20 and c = 13. Hence, incentre (using the formula is given by

 11  0  20  5  13  16 11  0  20  12  13  12   ,  , i.e., (7, 9). 11  20  13 11  20  13   Hence correct option is [A]. Method 2: Use options. The incentre should be equidistant from the three sides of the triangle.

Area of a Triangle If the coordinates of the vertices of a triangle are given, then area of the triangle is given by x1 1 x2 2 x3

y1

1

y2

1 =

y3

1

1 [(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3)]. 2

Collinearity We can check for collinearity of three points A, B and C using any of the following conditions. 1.

AB + BC = AC, or

2.

Slope of AB = Slope of BC = Slope of AC, or

3.

Area of ABC = 0, or

4.

Equation of the line joining any two of the points passes through the third point.

Example 12 Find the area of a triangle whose vertices are (a, a), (a + 1, a + 1) and (a + 2, a).

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Coordinate Geometry and Trigonometry Solution Area of triangle =

x1 1 x2 2 x3

y1

1

y2

1 

y3

1

a a 1 1 a1 a1 1 2 a2 a 1

=

1 [a(a + 1 – a) – a(a + 1 – (a +2)) + 1(a(a + 1) – (a + 1)(a + 2)] 2

=

1 [a + a – 2(a + 1)] = –1. 2

Since area cannot be negative, area is taken as 1. Example 13 A (6, 3), B (–3, 5), C (4, –2) and D (x, 3x) are four points. If ratio of areas of DBC and ABC is 1 : 2, then find the value of x. Solution Using the formula of area of triangle, we get Area of DBC =

x 3x 1 6 3 1 1 1 3 5 1 = 14x – 7 and that of ABC = 3 5 1 2 2 4 2 1 4 2 1

49 49 11 . Since ratio of areas is 1 : 2 we get 14x – 7 = or x = . 2 4 8

Example 14 Prove that the coordinates of the vertices of an equilateral triangle cannot all be rational. Solution Let A (x1, y1), B (x2, y2) and C (x3, y3), be the coordinates of the vertices of an equilateral triangle and let the numbers x1, y1, x2, y2, x3, y3 be all rational. Then, using the determinant formula for area of a triangle, the area 1 of this triangle will be [(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3)] which is a rational number, because x’s and 2 y’s are all rational numbers. Also, if the triangle is equilateral and if ‘a’ is the side, then a2 = (x1 – x2)2 + (y1 – y2)2 will also be a rational number.

3 2 3 2 a . Since a2 is a rational number, the area a must be an irrational 4 4 number because of presence of 3 . But, earlier we have said that the area is a rational number. This is a contradiction. Hence, all the coordinates of the vertices of an equilateral triangle cannot be rational.

Area of an equilateral triangle =

Example 15 x1 If x 2 x3

y1

1

a1

h1

1

y2

1  a2

b2

1,

y3

1

b3

1

a3

then the two triangles with vertices (x1, y1), (x2, y2), (x3, y3) and (a1, b1), (a2, b2), (a3, b3) A. must be similar

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B. C. D.

must be congruent must be equilateral and congruent None of these

Solution Equality of areas of triangles does not conclusively prove that the triangles are similar or congruent. Hence correct option is [D]. Example 16 If the points A (3, 4), B (7, 7) and C (a, b) are collinear and AC = 10, then what is the value of (a, b)? Solution AC2 = (a – 3)2 + (b – 4)2 = 100 Also,

b7 3 (slope of BC = slope of AB).  a7 4

So, 4b – 28 = 3a – 21 or b =

 3a  7   4 (a – 3)2 +  4  

(i) (ii)

1 (3a + 7). Putting this in (i), we get 4

2

= 100  25(a2 – 6a + 9) = 1600  a2 – 6a – 55 = 0  a = 11 or –5.

Therefore, (a, b)  (11, 10), (–5, –2). Example 17 If the vertices P, Q and R of a triangle PQR are rational points, then which of the following points of the triangle PQR is (are) always rational point(s)? A.

Centroid

B.

Incentre

C.

Circumcentre

D.

Orthocenter

Solution Only a centroid will satisfy this condition, due to the very nature of the formula of the coordinates of centroid. In case of other points, radical sign “ ” may occur. Hence correct option is [A]. Example 18 Find the area of the triangle whose vertices are the point of intersection of the lines 2x + y = 12, x – 4y + 7 = 0 and 3x + 7y = 3 with the line 4x + 3y = 12. Solution Since all the three points lie on the line 4x + 3y = 12, the three points are collinear and so the area is 0. Example 19 Amit, Rahul and Sunil are standing at points (2, 5), (5, 10) and (10, 30) respectively. If they decide to meet at the centroid of the triangle formed by the points representing their positions, who will arrive first, if they walk at the speeds which are in the ratio of 4 : 2 : 5 respectively, by taking the shortest route available to them?

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Coordinate Geometry and Trigonometry Solution

 17  The coordinates of the centroid   , 15  .  3  Distances of Amit, Rahul and Sunil from the centroid are 2

1021  11  ;    100  3  3  Times taken are

2

 2    25   3

229 and 3

2

 13     225   3 

1021 /16 63.81 229 / 4 57.25  ;  and 3 3 3 3

2194 respectively. 3

2194 / 25 87.76  . 3 3

Therefore, least time is taken by Rahul. Example 20 A triangle is formed by 3 lines whose equations are: y = x; y = 2x; x + y = 6. Find the area of the triangle formed by joining the mid-points of the sides of the original triangle. Solution 1 th of the area 4 of the original triangle. To find area of original triangle, we need to find coordinates of the vertices.

The area of the triangle formed by joining the midpoints of the sides of the original triangle is

y = x and y = 2x intersect in (0, 0). y = x and x + y = 6 intersect in (3, 3). y = 2x and x + y = 6 intersect in (2, 4). 0 0 1 1 1 Therefore, area of the triangle = 3 3 1   12  6   3 . 2 2 2 4 1

Therefore, area of the triangle formed by joining the midpoints of the original triangle =

3 . 4

Equations of Straight Line An equation of straight line is always a linear (or first degree polynomial) equation in the variables x and/or y. Therefore, equations like 2x + 5y = 7, x = 7, 2y = 17 are all equations of straight lines. From the knowledge of plane geometry, we understand that a line is uniquely defined if (i)

the two distinct points through which it passes are known, or

(ii)

a point through which it passes, and the slope of the line are known.

Point-point form of equation of line If the coordinates of two distinct points are (x1, y1) and (x2, y2), then the equation of the line joining these points is given by: y – y1 =

y 2  y1 (x – x1) x 2  x1

This is known as point-point form of the equation of line.

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Point-slope form of equation of line Since =

y 2  y1 = m is the slope of the line, the equation of the line is also given by x 2  x1

y – y1 = m(x – x1) This is known as point-slope form of the equation of line.

Slope-intercept form of equation of line The slope-intercept form of line is given by y = mx + c, where m is the slope of the line and c is the y-intercept. By y-intercept, we mean the distance between the origin and the point where the line intersects y-axis. Therefore, if the equation of line is y = 4x + 7, then we can say that the slope of the line is 4 and the line cuts (positive side of) y-axis at a distance of 7 units from origin. This form of equation of line is useful in identifying the slope of the line. For example, if the equation of line is given as 2x + 5y = 17, then the equation can be rewritten as y= 

2 17 2  slope of the line =  . x 5 5 5

Observe that since slope is negative, the line makes an obtuse angle with the positive direction of x-axis.

General form of equation of line The general form of equation of a line is given by ax + by + c = 0 The slope of this line is 

a a and y-intercept is  . b b

Intercepts form of equation of line The intercepts-form of the equation of line is given by

x y  1. a b

where (a, 0) and (0, b) are the x-intercept and y-intercept of the line.

Important points (i)

Equation of x-axis is y = 0 and that of y-axis is x = 0.

(ii)

Slope of x-axis is 0 and that of y-axis is .

Example 21 A line passes through (1, 5) and (–2, 3). (i)

What is the equation of the line in the general form, slope-intercept form, and intercepts form.

(ii)

What are the x and y intercepts of the line

(iii) What is the length of the line between the axes (iv) What is the slope of the line (v)

What is the area of the triangle enclosed between the line and the coordinate axes.

Solution (i)

Using point-point form, the equation of line is

 35  y–5=   (x – 1)  3y – 15 = 2x – 2  2x – 3y + 13 = 0.  2  1 

632

Coordinate Geometry and Trigonometry Therefore, general form of equation of line is: 2x – 3y + 13 = 0. Slope intercept form of equation of line is: y = Intercepts form of equation of line is: (ii)

2 13 . x 3 3

x y  1. (13 / 2) (13 / 3)

We can read the x-intercept and y-intercept from the intercepts form of equation of line, that is, x13 13 intercept =  and y-intercept = . 2 3 2

(iii) Length of the line between the axes = (iv) Slope of the line = (v)

 13   13       2   3 

2



13 13 . 6

2 . 3

Area of the required triangle =

1  13   13  169  .     2  2   3  12

Example 22 A line passes through a point P(3, 4) and makes an angle of 135° with the positive side of x-axis. Find the equation of the line. Solution Slope of the line = tan 135° = –1. Since a point and the slope is known, we would use point-slope form to get the equation of the line. y – y1 = m(x – x1) or y – 4 = –1(x – 3) or x + y = 7. Example 23 The points (1, 3) and (5, 1) are the opposite vertices of a rectangle. The other two vertices lie on the line y = 2x + c. Find the value of c. Solution

 1  5 3 1  The mid-point of (1, 3) and (5, 1) lies on the line y = 2x + c. Mid-point   ,   (3, 2). 2   2 Substituting these coordinates in the equation of the line, we get: 2 = 2(3) + c or c = –4. Example 24 Find the equation of the hour hand at 4 o’clock. Solution The hour hand makes and angle of –30° with the positive direction of x-axis. It passes through origin. Therefore, slope is m = tan (–30°) = 

y = mx or y = 

1 3

1 3

. Therefore, equation of line is as below:

x or x + y 3 = 0.

Example 25 |x| + |y| = 1, represents

633

Quantitative Aptitude Simplified for CAT

A.

a circle

B.

rectangle

C.

square

D.

triangle

Solution The given equation can be written as: x + y = 1 (x, y > 0) x – y = 1 (x > 0, y < 0) y – x = 1 (x < 0, y > 0) –x – y = 1 (x, y < 0) When we draw these lines, the shape is a square. Hence correct option is [C].

Parallelogram Let A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4) be the four vertices of a parallelogram. Since diagonals bisect each other, coordinates of mid-point of AC will be same as that of BD. Therefore,

x1  x 3 x 2  x 4 y  y 3 y2  y 4  and 1  . 2 2 2 2

Therefore, x1 + x3 = x2 + x4 and y1 + y3 = y2 + y4 is an important result useful to find the coordinates of one of the vertices if the coordinates of the other three vertices are given.

Area of a parallelogram Area of a parallelogram can be found out if only three vertices are given, as we know that the area of a parallelogram is twice the area of the triangle. Therefore, if the three coordinates are A (x1, y1), B (x2, y2) and C (x3, y3), then area of parallelogram is given by x1

y1

1

x2

y2

1

x3

y3

1

irrespective of where the 4th vertex lies.

634

Coordinate Geometry and Trigonometry Example 26 If the coordinates of the three vertices of a parallelogram (in order) are (1, 4), (3, 7) and (9, 16), find the coordinates of the fourth vertex. Solution x4 + x2 = x1 + x3  x4 = 1 + 9 – 3 = 7; y4 + y2 = y1 + y3  y4 = 4 + 16 – 7 = 13. Hence coordinates of the fourth vertex is (7, 13).

Angle between Two Lines Let  be the angle between two lines y = m1x + c1 and y = m2x + c2. The slopes of the lines are m1 and m2. m1  m2 Then, tan  = 1  m1m2

Condition for lines to be parallel If lines are parallel, = 0 and hence m1 = m2.

Condition for lines to be perpendicular If the lines are perpendicular, then  = 90°, which means that the denominator in the above formula must be zero. We obtain m1m2 + 1 = 0 or m1m2 = –1. Therefore, equation of a line parallel to ax + by + c = 0 is ax + by + k1 = 0, and equation of a line perpendicular to ax + by + c = 0 is bx – ay + k2 = 0. Example 27 The equations of the sides of a triangle are x + y – 5 = 0; x = y + 1 and y – 1 = 0. The coordinates of the circumcenter are A.

(3, 1)

B.

(1, 2)

C.

(2, –2)

D.

(1, –2)

Solution The lines x + y – 5 = 0; x = y + 1 are perpendicular. Therefore, it is a right triangle. So, the coordinates of the circumcenter will be mid-point of the hypotenuse. Since the coordinates of the end points of the hypotenuse are (4, 1) and (2, 1), the coordinates of the mid-point will be (3, 1). Hence correct option is [A].

635

Quantitative Aptitude Simplified for CAT

Alternatively, y = 1 is the equation of hypotenuse. Only (3, 1) passes through this line. Example 28 If the lines 2x + ky = 1 and 3y – x = 3 are perpendicular, then find the value of k. Solution

 2   1  2 Since the lines are perpendicular, m1m2 = –1. Therefore,      1 or k = . k 3 3     Example 29 For the equation y = 3x + 2, which of the following statements is false? A.

Equation of a line through (1, 2) and parallel to the given line is y = 3x – 1.

B.

Equation of a line through (1, 2) and perpendicular to the given line is x + 3y = 7.

C.

Equation of a line through (2, –1) and perpendicular to the given line is x + 3y + 1 = 0.

D.

at least one of the above statements is false.

Solution The first three options [A], [B] and [C] are correct. Therefore, [D] is definitely a false statement. Example 30 Find the coordinates of the foot of the perpendicular drawn from the point (1, 2) to the line 3x + y + 1 = 0. A.

(1, –4)

B.

(–1, 2)

C.

 4 7  ,   5 5

D.

 4 17   ,  5 5 

Solution SHORTCUT All the four options satisfy the given line. For the point to be the foot of perpendicular, the given line must be perpendicular to the line joining the foot of the perpendicular and the given point. Slope of the given line is –3.  4 7 1 Therefore, slope of its perpendicular must be . This is true for the case of   ,  only. 3  5 5

Concurrent Lines If three lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0, are concurrent, then a1

b1

c1

a2

b2

c2  0

a3

b3

c3

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Coordinate Geometry and Trigonometry

Distance between a Point and a Line Shortest distance between a point (x1, y1) and a line ax + by + c = 0 is given by, d =

ax1  by1  c a2  b2

.

Distance between Two Parallel Lines Distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by, d =

Example 31 Three lines 3x – y = 2, 5x + ay = 3 and 2x + y = 3 are concurrent. Find a. Solution 3 1

2

5

a

3 or 3(–3a + 3) + 1(–15 + 6) – 2(5 – 2a) = 0 or –5a – 10 = 0 or a = –2.

2

1

3

Example 32 Find the shortest distance between (3, 1) and 3x + 2y + 9 = 0. Solution Shortest distance, d =

3(3)  2(1)  9 2

3 2

2



20 . 13

Example 33 Find the shortest distance between x – 6y + 4 = 0 and 4x = 24y – 6.

637

c1  c2 a2  b2

.

Quantitative Aptitude Simplified for CAT

Solution Equation 4x = 24y – 6 can be rewritten as x – 6y + 1.5 = 0. Therefore, shortest distance, d =

4  1.5 2

1 6

2



2.5 . 37

EXPERT SPEAK Scan this QR Code to watch a video that explains the application of basic formulae of Coordinate Geometry including the kinds of problems asked in the past.

Position of a Point with Respect to a Line Let there be a point A (x1, y1) and a line whose equation is ax + by + c = 0. A line divides the coordinate region in two parts, one is called ‘negative’ and the other ‘positive’ with respect to the line. Accordingly, this point A could have three positions with respect to the line: (i)

in the negative region

(ii)

in the positive region

(iii) on the line To identify which region does the point lie in, put the coordinates of the point in the expression ax + by + c. The value of the expression will be either negative, or positive, or zero, depending upon whether the point lies on one side, or the other side or on the line. If there are two points A (x1, y1) and B (x2, y2) and the given line is ax + by + c = 0. Then, both the points are either on the same side, or on the opposite side of the line. This can be easily known by putting the coordinates of A and B in the expression ax + by + c. If the sign of ax1 + by1 + c is same as that of ax2 + by2 + c, then both the points are on the same side of the line, whereas if the expressions have opposite signs, then the points are on the opposite side of the line.

If the points are on the opposite side of the line, then the line divides the line joining the two points in a certain ratio internally. Similarly, if the points are on the same side of the line, then the line divides the line joining the two points in a certain ratio externally. In any case, this ratio is given by 

ax1  by1  c . ax 2  by 2  c

Directions for following examples: There are two points (2, 1) and (5, 7) and a line 4x + 3y = 12. Example 34 Which of the two points lies on the same side of the line as the origin lies?

638

Coordinate Geometry and Trigonometry Solution For origin, 4x + 3y – 12 = 0 + 0 – 12 = –12; For (2, 1), 4x + 3y – 12 = 8 + 3 – 12 = –1; For (5, 7), 4x + 3y – 12 = 20 + 21 – 12 = 11. Therefore, (2, 1) lies on the same side of the given line as origin lies. Example 35 Find the ratio in which the line divides the line joining the two points? Solution Required ratio = 

4(2)  3(1)  12 1 1   . 4(5)  3(7)  12 11 11

Example 36 Does (3, –1) lie between the lines 2x + y = 6 and 4x + 2y + 1 = 0, or outside these lines? Solution 2(3) + (–1) – 6 = –1; 4(3) + 2(–1) + 1 = 11. Also check for origin. So, 2(0) + 0 – 6 = –6. So, (3, –1) lies on same side of 2x + y = 6 as origin lies. Checking for origin for the other line, we get 4(0) + 2(0) + 1 = 1. So, (3, –1) lies on the same side of 4x + 2y + 1 = 0 as origin lies. Therefore, (3, –1) lies between the given lines. Example 37 A triangle is enclosed by a line 4x + 3y = 12 in the first quadrant. Find the distance between orthocentre and circumcenter of this triangle. Solution The triangle is right angled triangle. So, orthocenter is the right angle vertex (0, 0) and circumcenter is midpoint of hypotenuse. The x-intercept is (3, 0) and y-intercept is (0, 4). Therefore, midpoint of hypotenuse is (1.5, 2). Required distance =

1.5 2  22  2.5 .

Example 38 Consider a triangle ABC with the vertices at (0, 0), (0, 10) and (10, 0). How many points are there inside this triangle whose coordinates are integral coordinates (excluding all the points on the boundary)? Solution Let us consider a 4th vertex whose coordinates are (10, 10). If we join all the 4 vertices, we get a square. In this square, on the x-axis, there are points (0, 0), (1, 0), and so on till (10, 0) and that makes 11 points, each of whose coordinates are integral. This goes on till (10, 10) and hence total number of integral coordinates in this square (including the boundary) is 11  11 = 121. On the boundary, we have 11 + 11 + 9 + 9 = 40 points. Excluding the boundary, we have 121 – 40 = 81 points.

639

Quantitative Aptitude Simplified for CAT

Along the line joining (10, 0) and (0, 10), the number of points is 9 (excluding the end points). The number of points without these points = 81 – 9 = 72. Half of these number of points is the required number of points = 36. Alternatively, The equation of line joining (10, 0) and (0, 10) is: x + y = 10. The points along the bottom row are: (1, 1), (2, 1), (3, 1), …, (8, 1), which are 8 points. The points along the next row are: (1, 2), (2, 2), (3, 2), …, (7, 2), which are 7 points. This process goes on. So, total number of points = 8 + 7 + 6 + … + 1 =

n(n  1) 8  9  = 36. 2 2

TRIGONOMETRY Trigonometry is an integral part of geometry. Some results of trigonometry like sine rule, cosine rule, area of triangle (in the form of trigonometric function), have been discussed in the earlier chapters. In the current chapter, we will discuss a few more trigonometric results followed by ‘Heights and Distances’, which is an important application of the results of trigonometry, dealing with measurement of heights of buildings, towers, tree and so on.

Basic Results We know that every right triangle has ‘base’, ‘perpendicular’ and ‘hypotenuse’, as shown in the diagram.

We also know that sin  =

Perpendicular AB Base BC  = and cos  =  . Hypotenuse AC Hypotenuse AC 2

2

AB2  BC2 AC2  AB   BC     2  1 , which is a standard result. Therefore, sin2+ cos2 =      AC   AC  AC2 AC Dividing both sides of the result by cos2, we get sin2  2

1 

cos 

1 cos 2 

or tan2 + 1 = sec2, which is also a standard result.

If we divide sin2 + cos2 = 1 by sin2, we get 1

cos 2  sin2 



1 sin2 

or 1 + cot2 = cosec2, which is again a standard result.

So, we have following standard results (i)

sin2 + cos2 = 1

(ii)

1 + tan2 = sec2

(iii) 1 + cot2 = cosec2

640

Coordinate Geometry and Trigonometry Further, sin (90 – ) = cos ; cos (90 – ) = sin ; tan (90 – ) = cot ; cot (90 – ) = tan ; sec (90 – ) = cosec ; cosec (90 – ) = sec . In fact results pertaining to ‘tan’, ‘cot’, ‘cosec’ and ‘sec’ can be easily obtained by results pertaining to ‘sin’ and ‘cos’. Therefore, following results are given only for ‘sin’ and ‘cos’. sin (90 – ) = cos 

cos (90 – ) = sin 

sin (90 + ) = cos 

cos (90 + ) = –sin 

sin (180 – ) = sin 

cos (180 – ) = –cos 

sin (180 + ) = –sin 

cos (180 + ) = –cos 

sin (270 – ) = –cos 

cos (270 – ) = –sin

sin (270 + ) = –cos

cos (270 + ) = sin

sin (360 – ) = –sin 

cos (360 – ) = cos 

sin (360 + ) = sin 

cos (360 + ) = cos

There is no need to remember all these results. For example, sin (180 – ) = sin [90 + (90 – )] = cos [(90 – )] = sin . Likewise, every result can be reduced to sin (90 ± ) or cos (90 ± ). Thus only these four results need to be remembered. Following results are also important. (i)

sin2 = 2 sincos

(ii)

cos2 = cos2 – sin2 = 2cos2 – 1 = 1 – 2sin2

(iii) tan 2 =

2 tan  1  tan2 

.

Trigonometric Functions of Common Angles Common angles dealt with in trigonometry are 0°, 30°, 45°, 60° and 90°. Angles more than 90° can be reduced (using trigonometry) to less than or equal to 90°. Let us see the values of trigonometric functions of these angles.



0

30

45

60

90

sin

0

1 2

1

3 2

1

cos

1

3 2

1

1 2

0

2

2

641

Quantitative Aptitude Simplified for CAT

Values of other trigonometric functions of these angles, like tan , cot, and so on can be found using the above sin30 1 /2 1   table. For example, tan 30° = . cos 30 3 /2 3 The graphs of sin  and cos  are drawn below. The knowledge of graphs is useful in many instances.

From the graph, we can see that cos 180° = –1 and sin 180° = 0, whereas cos 270° = 0. Example 39 Find the value of sin8x + 4sin6xcos2x + 6sin4xcos4x + 4sin2xcos6x + cos8x. Solution Let sin2x = a and cos2x = b. Then, sin8x + 4sin6xcos2x + 6sin4xcos4x + 4sin2xcos6x + cos8x = a4 + 4a3b + 6a2b2 + 4ab3 + b4 = (a + b)4 = (sin2x + cos2x)4 = 1. Example 40 If sin  + sin 2= 1, then prove that cos 2 + cos 4 = 1. Solution cos2  + cos4  = cos2 (1 + cos2 ) = (1  sin2 )[1 + (1  sin2 )] = (sin )(1 + sin ) = sin  + sin 2= 1. Example 41 If sin  + cosec= 2, then find the value of sin8 + cosec8. Solution sin  + cosec= 2  sin  +

1 = 2  sin  = 1. Therefore, sin8 + cosec8 = 1 + 1 = 2. sin 

Example 42 ABC is a right angled triangle, right angled at C. D is the midpoint of BC. DAC =  and BAC = . Prove that

tan  1  . tan  2

Solution We know that tan  

CD CB 2CD  and tan   . AC AC AC

642

Coordinate Geometry and Trigonometry

tan  CD 1   tan  2CD 2

Example 43 Show that: cos  sin  –

sin  cos(90  ) cos  cos  sin(90  ) sin   0 sec(90  ) cos ec(90  )

Solution cos  sin  –

sin  cos(90  ) cos  cos  sin(90  ) sin   sec(90  ) cos ec(90  )

= cos  sin  –

sin  sin  cos  cos  cos  sin   cos ec  sec 

= cos  sin  – (sin3 cos  + sin cos3 ) = cos  sin  – (sin cos )(sin2  + cos2 ) = cos  sin  – (sin cos ) = 0 Example 44 Find the value of  (0° <  < 90°), if 2cos2  =

1 . 2

Solution 2cos2  = cos  =

1 1 1  cos2 =  cos  =  . Since 0° <  < 90°, cos  is positive and hence 2 4 2

1  or 60. = 2 3

Example 45 If sin 15 = 0.26, find sin 75. Solution sin 75 = cos 15. Now, sin2 + cos2= 1 sin2 15 + cos2 15 = 1 cos215 =

1  sin2 15  1  (0.26)2  1  0.0676  0.9324  0.9656  0.97 .

Example 46 Evaluate (i)

cosec(65° + ) – sec(25° – ) – tan(55° – ) + cot(35° + )

643

Quantitative Aptitude Simplified for CAT

sin80o

(ii)

cos10

o

cos 35o

(iii)

sin55

o

 sin59o sec 31 o



tan27o tan63o sin30

o

 3 tan2 60o

(iv) (tan 1°)(tan 2°)(tan 3°).......(tan 89°) 1 24

(v)

sin230°cos245° + 4tan230° + sin290° – 2cos290° +

(vi)

sin 39 + 2 (tan11 tan 31 tan 45 tan 59 tan 79) – 3(sin221 + sin269) cos 51

Solution (i)

cosec(65° + ) – sec(25° – ) – tan(55° – ) + cot(35° + ) = sec[90° – (65° + )] – sec(25° – ) – tan(55° – ) + tan[90° – (35° + )] = sec(25° – ) – sec(25° – ) – tan(55° – ) + tan(55° – ) = 0. sin80o

(ii)

cos10

o

cos 35o

(iii)

sin55o

 sin59o sec 31 o 



tan27o tan63o sin 30o

cos10  cos 31 sec 31 = 1 + 1 = 2. cos10

 3 tan2 60o 

sin55o sin55o



cot 63o tan63o

= 1 + 2 – 3(3) = –6. Example 47 If tan 2 =

2 tan  1  tan2 

, then find the value of tan 15°.

Solution Let  = 15°. Then, tan 2 = tan 30° =



1 3



2 tan  1  tan2 

2 tan  1  tan2 

 1 – tan2 = 2 3 tan 

 tan2 + 2 3 tan   tan =

2 3  12  4 2 3  16    3 2  2 3 2 2

Note that positive value is taken because  is in first quadrant. Example 48 If tan  =

4 , show that 3

1  sin  1  . 1  sin  3

Solution Refer to the figure below. Since tan  =

4 4 , sin  = . So, 3 5

644

sin30o

 3 tan2 60o

Coordinate Geometry and Trigonometry 4 1 1  sin  5   4 1  sin  1 5

1 5 1. 9 3 5

Example 49 If the angle of elevation of a cloud from a point h metres above a lake is  and the angle of depression of its 2hsec  reflection in lake is , prove that the distance of the cloud from the point of observation is . tan   tan  Solution B is the point h meters above lake, C is the cloud, and D is its reflection in lake water. In BEC, let BE = x. Then, CE = x tan . In BED, DE = x tan . Also, CE + h = DE – h or h =

x

1 x (DE – CE) =  tan   tan   2 2

2h . tan   tan 

Now, BC = BE sec  = x sec  =

2hsec  . tan   tan 

Example 50 A tower in a city is 750 m high and a multi-storeyed hotel at the city centre is 50 m high. The angle of elevation of the top of the tower from the top of the hotel is 30°. A building, h metre high, is situated on the straight road connecting the tower with the city center at a distance of 1 km from the tower. Find the value of h if the top of the hotel, the top of the building and the top of the tower are in straight line. Also find the distance of the tower from the city center.

645

Quantitative Aptitude Simplified for CAT

Solution tan 30 =

750  50 , where x is the distance between the hotel and building. 1000  x

Or 1000 + x = 700 3 or x = 700 3 – 1000 Also, tan 30 =

h  50 x 700 3  1000 750 3  1000 h=  50   50  = 172.65 m. x 3 3 3

So, distance of the tower from the city center = 1000 + x = 700 3 .

Example 51 The angle of elevation of the top of a tower from two points P and Q at distances of ‘a’ and ‘b’ m respectively, from the base and the same straight line with it, are complementary. Prove that the height of the tower is ab metres. Solution h = a tan  = b tan (90 – ) Therefore, h2 = (a tan )[b tan (90 – )] = ab. So, h =

ab metres.

Example 52 From the foot of a hill, the angle of elevation of the top of a tower is 45°. After walking 2 km upwards along the slope of the hill, which is inclined at 30°, the same is found to be 60°. Find the height of the hill along with the tower. Solution AE is the tower and CE is the hill. Now, tan 45 =

AB = 1. So, AB = BC = x. BC

DC = 2 km. So, FD = 2 sin 30 = 1 km and FC = 2 cos 30 =

646

3 km.

Coordinate Geometry and Trigonometry So, GD = BF = x 

3 . Also, AG = x – BG = x – DF = x – 1.

In AGD, tan 0 

AG x 1   x 3  3  x 1  x  GD x  3

Therefore, height of the hill = BE = x tan 30 =

2 3 1



2 3 1

1 3

2 



. 3 1 2 3

 1

1 3

km.

Also, height of the tower, AE = x – BE =

2

 1   1   3 1  3

 1  1 2 3 1  1   km.   3 3 3 3 





Example 53 Two poles of equal heights are standing opposite to each other on either side of a road, which is 100 m wide. From a points between them on the road, the angles of elevation of their tops are 30° and 60°. Find the position of the point and also the heights of the poles. Solution BP = h cot 60 =

So, h =

 1   4  ; DP = h cot 30 = h 3 . So BD = h   3  h  = 100. 3  3   3

h

100 3  25 3 . 4

Distance of P from B =

h 3

= 25 m. So, P is 25 m from B and 75 m from D.

647

Quantitative Aptitude Simplified for CAT

Example 54 There is a ladder resting against a wall such that when its base is pulled outwards by x, the top of the ladder drops by y and the angle of inclination of the ladder changes from 45° to 30°. Find the ratio of x : y. Solution When the angle of inclination is 45°, the point of contact with the wall and the ground is A and B, and when the angle of inclination is 30°, then the respective contact points are C and D. Then, BD = x and AC = y. Let the length of the ladder be m and the place where wall meets the ground be O. m m = OB and OC = m sin 30° = . Then, OA = 2 2 Therefore, y = OA – OC =

m 2

Similarly, OD = m cos 30° = x:y=





3 2 :



m . 2

m 3 m 3 m and therefore, x = OD – OB =  . 2 2 2



2 1 .

Example 55 Find the side of the square field in the centre of which is standing a pole of length 10 m such that the side of the square subtends an angle of 60° at the top of the pole. Solution OA = OB  OAB is isosceles triangle. So, OAB = OBA = 60 and therefore OAB is an equilateral triangle. Let the side of the square field be x. Then, OA = OB = x. Also, PB = half of the diagonal of square =

1 x x 2  . 2 2





In OPB, OB2 = OP2 + PB2  OP =

 x  x2     2

2



x 2

= 10

 x = 10 2 m.

648

Coordinate Geometry and Trigonometry Example 56 An aeroplane when 3000 m high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation point are 60° and 45° respectively. How many metres is one plane higher than the other? Solution Let plane Q be h metres above plane P. Refer to the figure below. x = PB cot 45 = PB. Also, x = 3000 cot 60 =

3000 3

 1000 3 .

Therefore, h = BQ – BP = 3000 – 1000 3 = 1000 3





3  1  1268 m.

Example 57 A ladder is resting against a wall making an angle of 45° with the ground. If the base of the ladder is slipped outwards by a distance 5 m, it makes an angle of 30° with the ground. Find the height through which the ladder slipped. Solution BD = 5 m. Let AC = x.

OB  5

In COD, OC = OD tan 30 = (OB + 5) tan 30 =

3

Now, AO = OB  x + OC = OB  x = OB – OC = OB –

.

OB  5 3

Also, AB = CD and AB = OB 2 = CD. 2

 OB  5  2 OC2 + OD2 = CD2     (OB  5)  OB 2 3  



2



 (OB + 5)2 + 3(OB + 5)2 = 6(OB)2  2(OB + 5)2 = 3(OB)2  2(OB2 + 25 + 10OB) = 3(OB)2  OB2 – 20 OB – 50 = 0  OB =

20  400  4(50)  10  5 6 2

649

.

Quantitative Aptitude Simplified for CAT

Therefore, x = OB –

OB  5 3

 1  5 = 10  5 6  1   3 3  





Example 58 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. The angle subtended by the flagstaff at a point on the plane is  – . From the same point on the ground, the angle of htan elevation of the top of the tower is . Prove that the height of the tower is . tan-tan Solution BC = x tan . Also, h + BC = x tan . So, h + x tan  = x tan  h = x tan  – x tan  x =

h tan-tan htan tan-tan

 height of tower, BC = x tan  =

Example 59 A car is speeding at 36 km/hr towards a building. The driver observes that the angle of elevation changed from 30° to 45° in 5 sec. In how much time will another car travelling at 72 km/hr reach the base of the building from the place where the first car started? Solution Speed of the car = 36 km/hr = 10 m/s. So, distance AB = 5 × 10 = 50 m. Also, BC = 10t. Now, h = 10t (tan 45) = 10t, and h = (10t + 50) tan 30 =

10t  50 3

.

650

Coordinate Geometry and Trigonometry

So, 10t =

10t  50 3

50

t= 10



3 1



 

5 3 1



.

So, time taken by the car whose speed is 72 km/hr will be half of the time taken by the car whose speed is 5

36 km/hr. Therefore, time taken by the faster car = 2



3 1



.

EXPERT SPEAK Scan this QR Code to watch a video that explains the application of basic formulae of Trigonometry, Heights and Distances.

651

Quantitative Aptitude Simplified for CAT

PRACTICE EXERCISE 1.

A. B. C. D.

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower? A. B. C. D.

 6 7 8

5 3 1

CAT 2014 4.



3 2

30 150 45 60



 3  2

3 1

A boat is being rowed away in still water, from a 210 m high cliff at the speed of 3 km/hr. What is the approximate time taken for the angle of depression of the cliff at the boat to change from 60 to 45? A.

5 min

B.

4 min

C.

1 min

D.

2 min

CAT 2003 (R) 2.

IIFT 2014

Two straight roads R1 and R2 diverge from a point A at an angle of 120°. Ram starts walking from point A at a uniform speed of 3 km/hr. Shyam starts walking at the same time from A along R2 at a uniform speed of 2 km/hr. They continue walking for 4 hours along their respective roads and reach points B and C on R1 and R2 respectively. There is a straight line path connecting B and C. Then Ram returns to point A after walking along the line segments BC and CA. Shyam also returns to A after walking along line segments CB and BA. Their speeds remain unchanged. The time interval (in hours) between Ram's and Shyam's return to the point A is: A.

10 19  26 3

B.

2 19  10 3

C.

19  26 3

D.

19  10 3

5.

A.

30 m

B.

18 m

C.

40 m

D.

20 m IIFT 2012

6.

If 5 ≤ x ≤ 15, then the value of sin 30 + cos x – sin x will be: A.

Between –1 and –0.5 inclusive

B.

Between –0.5 and 0 inclusive

C.

Between 0 and 0.5 exclusive

D.

between 0.5 and 1 inclusive

E.

None of the above XAT 2017

7.

3.

There are two buildings, one on each bank of a river, opposite to each other. From the top of one building – 60 m high, the angles of depression of the top and the foot of the other building are 30° and 60° respectively. What is the height of the other building?

In ABC, 3 cos A + 4 sin B = 6 and 4 cos B + 3 sin A = 1. Find ACB which is acute angle.

652

A person standing on the ground at point A saw an object at point B on the ground at a distance of 600 meters. The object started flying towards him at an angle of 30 with the ground. The person saw the object for the second time at point C flying at 30 angle with him. At point C, the object

Coordinate Geometry and Trigonometry changed direction and continued flying upwards. The person saw the object for the third time when the object was directly above him. The object was flying at a constant speed of 10 kmph. XAT 2016

B.

16 25

C.

5 6

D.

24 25

E.

A single solution is not possible XAT 2015

9. Find the angle at which the object was flying after the person saw it for the second time. You may use additional statement(s) if required. Statement I: After changing direction the object took 3 more minutes than it had taken before. Statement II: After changing direction the object travelled an additional 200 3 m. Which of the following is the correct option? A.

Statement I alone is sufficient to find the angle but statement II is not. (2) Statement II alone is sufficient to find the angle but statement I is not.

B.

Statement I and Statement II are consistent with each other.

C.

Statement I and Statement II are inconsistent with each other.

D. 8.

A person is standing at a distance of 1800 meters facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at 30 elevation from his eye-level. Immediately, the person starts walking towards the tower. At 5.10 pm., the person noticed that the tip of the minute hand made an angle of 60 with respect to his eye-level. Using three-dimensional vision, find the speed at which the person is walking. The length of m the minute hand is 200 3 ( 3 = 1.732). A. B. C. D. E.

7.2 km/hour 7.5 km/hour 7.8 km/hour 8.4 km/hour None of the above XAT 2015

10. Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

Neither Statement I nor Statement II is sufficient to find the angle.

The parallel sides of a trapezium ABCD are in the ratio of 4 : 5. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD (as shown below). ABCD has a perimeter equal to 1120 meters and PBCD has a perimeter equal to 1000 meters. Find sin(ABC), given 2DAB = BCD.

A. B. C. D.

780 800 820 741 CAT 2005

11. Find the distance between the lines x + y = 4 and x + y = 2.

A.

4 5

A.

6 2

B.

4 2

C. D.

3 2 3 CAT 2014

653

Quantitative Aptitude Simplified for CAT

12. Two lines are 2x + 3y = 6 and 2x + 3y = 12. Another line is such that it is parallel to the given lines and equidistant from them. Find the distance of the new line from origin. A.

B.

15 and 40

C.

17 and 30

D.

17 and 40

E.

Multiple solutions are possible

13 XAT 2015

5

B.

16. Point of intersection of the diagonals of a square is origin and the coordinate axes are drawn along the diagonals. If the side is of length a, then one which is not the vertex of the square is

13 8

C.

13 9

D.

13

A.

(a 2 , 0)

CAT 2016

B.

(0,

13. Suppose the two sides of a square are along the straight lines 6x – 8y = 15 and 4y – 3x = 2. Then the area of the square is

C.

(

D.

(

A.

2.52

B.

3.61

C.

4.33

D.

None of the above

14. Circle C1 has a radius of 3 units. The line segment PQ is the only diameter of the circle which is parallel to the X axis. P and Q are points on curves given by the equations y = ax and y = 2ax respectively, where a < 1. The value of a is:

B. C. D. E.

1 6

2 3

1 3

6

1 6 None of the above

, 0)

a 2

, 0)

A.

45

B.

65

C.

75

D.

90

A.

45 miles

B.

109 miles

C.

136 miles

D.

181 miles

19. If the mid-points of the sides of a triangle are (2, 1), (0, –3) and (4, 5), then the coordinates of the centroid of the triangle are

XAT 2014 15. Two diagonals of a parallelogram intersect each other at coordinates (17.5, 23.5). Two adjacent points of the parallelogram are (5.5, 7.5) and (13.5, 16). Find the lengths of the diagonals. A.

2

18. Sanjay stays 6 miles south of Lalbaug, which is 10 miles east of Dibrugarh. Krishna stays 3 miles north of Dibrugarh. How far does Sanjay stay from Krishna?

1 6

a

17. Pair of tangents are drawn from a point P(16, 7) to the circle whose center has coordinates C(1, 2) and whose radius is 5 units. If Q and R are the points of contact of the tangents with the circle, then find the area of the quadrilateral PQCR.

IIFT 2016

A.

a ) 2

A. B. C. D.

15 and 30

654

(2, 1) (0, 2) (1, 3) Cannot be determined

Coordinate Geometry and Trigonometry

26. If sin  = 20. The points (a, b + c), (b, a + c), (c, a + b) are A. B. C. D.

(tan  + sec )2.

vertices of an equilateral triangle collinear points concyclic vertices of a right triangle

21. The points (–1, 1) and (1, –1) are symmetrical about the line A.

y+x=0

B.

y=x

C.

x+y=1

D.

x–y=1

 13   ,0   5 

B.

 5   ,0   13 

C.

(–7, 0)

D.

 1    ,0   5 

A.

1

B.

2

C.

3

D.

4

27. A ladder 8 m long is rested against a building such that it touches a point 8 m below the top of that building. At the foot of the ladder, the angle of elevation of the top of the building is 60°. Find the height of the building.

22. A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). The coordinates of the point A are A.

3 , find the value of 5

A.

10 m

B.

12 m

C.

14 m

D.

16 m

28. Two flagstaffs of equal lengths are mounted on either side of a horizontal plank of wood which in turn is mounted at its middle on a pole as shown in the figure. The angles of elevation of the two flagstaffs from a point A are 30° and 60° respectively. If the height of the top of the flagstaffs from the ground is 10 m, then find the height of the flagstaffs.

23. If sec x – tan x = 4, find the value of sec x + tan x.

A.

10/3 m

B.

13/3 m

A.

1/4

C.

17/3 m

B.

1/2

D.

20/3 m

C.

1

D.

0

29. A vertically straight tree, 15 m high, is broken by wind and the tree is bent in such a way that its top just touches the ground and makes an angle of 60° with the ground, whereas the other end (which is stuck on the ground) is inclined such that angle at the point of breakage is 90°. Find the area of the triangle enclosed by the tree and the ground.

24. If ABCD is a cyclic quadrilateral, then find the value of cosA + cosB + cosC + cosD. A.

0

B.

1

C.

-1

D.

1/2

25. If cos  = A.

0

B.

1

C.

-1

D.

1/2



225 3 2  3

1 2 sec  , find the value of 2 1  tan2 

A.

4



225 3 2  3 B.

655



4



Quantitative Aptitude Simplified for CAT



225 3 2 3  3 C.



50 A.

4



225 3 2 3  3 D.

B.

3 1



3 50



 

3 1



3

4



50 2 3  1

30. From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 45° respectively. Find the height of the tower.

C.

3



50 2 3  1 D.

 

3

ANSWER KEY 1. A 9.

2. B

3.

A

4.

D

5.

C

6.

E

7.

D

8.

A

D

10. A

11. C

12. D

13. B

14. A

15. D

16. A

17. C

18. D

19. A

20. B

21. B

22. A

23. A

24. A

25. B

26. D

27. B

28. D

29. A

30. A

656

Coordinate Geometry and Trigonometry

ANSWERS AND EXPLANATIONS 1.

3.

Explanation: Given equations are: 3 cos A + 4 sin B = 6 and 4 cos B + 3 sin A = 1. Squaring each equation, we get 9 cos2 A + 16 sin2 B + 24 cos A sin B = 36, 16 cos2 B + 9 sin2 A + 24 cos B sin A = 1 Adding, we get 9 (cos2 A + sin2 A) + 16 (cos2 B + sin2 A) + 24 (cos A sin B + cos B sin A) = 37  25 + 24 (cos A sin B + cos B sin A) = 37 1  (cos A sin B + cos B sin A) = 2 1  A + B = 30 or 150.  sin (A + B) = 2 Therefore, ACB = 150 or 30. Since ACB is acute, ACB = 30.

Answer: A Explanation:

Let speed of the car be v and the time taken from new position to reach the base be t. Then, h = vt tan 60 = (10v + vt) tan 45  t = 10 tan45  (tan60)  1

10 3 1

10 



3 1

3 1

 5



3 1

 4.

2.

Answer: A

Answer: D Explanation:

Answer: B Explanation:

AB = 3 × 4 = 12 km; AC = 2 × 4 = 8 km. 8 2  122  BC2 Using cosine rule, cos 120 = 2(8)(12) 2 2  –96 = 64 + 144 – BC  BC = 304  BC = 304 . Therefore, time interval between the Ram’s and Shyam’s return

 12  

304  8 2

   12 

304  8

Let us say the time taken from B to A be t hours. Then, distance BA = 3t = 3000t m. 210 Now, BC = 210 cot 60 =  70 3 m. 3



t=



210  70 3 21  7 3  hours 3000 300

21  7 3 21  7 3  60  min 300 5 = 1.776 min.

2

=

1 1  20  304     2 3

  20  



Also, 210 = 70 3  3000t tan45

 304 

5.

6

Answer: C Explanation:

4 19  20 10  2 19   . 6 3

657

Quantitative Aptitude Simplified for CAT

Statement II: After changing directions, the object travels 200 3 m, which is same as before. This can happen only if the object stays on the course as before without changing any direction. Thus, we can clearly see that the two angles from the statements are inconsistent with each other. 8. Let BD = x. Then, x = 60 cot 60° = Then, EC =

60 3

60 3

m.

Explanation: Let DAB = θ. Then, BCD = 2θ. PBCD is a parallelogram. Therefore, DPB = 2θ. PBC = PDC = 180 – 2θ. ABP = θ (because BPD = 2θ = exterior angle to BAP). Therefore, PA = PB. So, ABC = θ + 180 – 2θ = 180 – θ. 1 BC : AD = 4 : 5. Also, BC = PD. So, AP = BC 4 = PB = CD. It is given that AB + BC + CD + DA = 1120. So, AB + BC + 0.25BC + 1.25BC = 1120  AB + 2.5BC = 1120. Also, BP + PD + DC + BC = 1000  0.25BC + BC + 0.25BC + BC = 1000  2.5BC = 1000  BC = 400. So, AB = 120 and PA = PB = 100. Now, applying sine rule in BAP, we get 100 120   120 sin  sin  sin(180  2) 3 = 100 sin 2 = 200 sin  cos   cos  = 5 4  sin  = . Therefore, sin (ABC) 5 4 = sin (180 – ) = sin  = . 5

m.

So, AE = x tan 30 =

60

tan 30 = 20 m. 3 Therefore, height of the building, DC = AB – AE = 60 – 20 = 40 m.

6.

Answer: E Explanation: sin 30 + cos x – sin x = 0.5 + cos x – sin x. In the range 5 < x < 15, cos x > sin x. So, cos x – sin x > 0. Therefore, value of the given expression is more than 0.5. Moreover, the value of the given expression will never be equal to 0.5 because cos x would never be equal to sin x for 5 < x < 15.

7.

Answer: A

Answer: D Explanation: From the given data, CAB = 30°; CBA = 30° and AB = 600 m Therefore, BC = AC = 200 3 m.

9.

Answer: D Explanation: The person is initially at P and PB = 1800 m. Therefore, BD = 1800 tan 30 = 600 3 m. At 5:10, the person reaches A (refer to second figure below) and the minute hand moves through 60. We also know that length of minute hand = 200 3 m. So, CD = CF = 200 3 m and FCE = 60.

Statement I: After changing the direction the object took 3 more minutes than it had taken before. In 3 minutes at the speed of 10 km/hr, it can travel 500 m, which is CD. So, we can find the angle.

658

Coordinate Geometry and Trigonometry

=

2(0)  3(0)  9 2

2 3

2



9 . 13

13. Answer: B Explanation: The two lines are 4y – 3x = 2 and 4y – 3x = −7.5 Side of the square = Distance between the Therefore, CE = 200 3 cos 60 = 100 3 m. So, DE = 100 3 m. This means FG = DB – DE = 600 3  100

two parallel lines =

2



9.5  1.9 5

Therefore, area of square = 1.92 = 3.61. 14. Answer: A Explanation: Points P and Q have the same Y coordinate but their X coordinates differ by 6 units (Since the diameter = 6 units) We have two possible cases. Case 1: y = ax = 2ax+6 1  2a6 = 1  a = 6 . 2 Case 2: y = 2ax = ax+6  a6 = 2  a = 6 2 . But it is given that a < 1. Therefore, case 2 is rejected.

Also, BG = EF = FC sin 60 = 200 3 sin 60 = 300 m. Therefore, AB = AG2  BG2  5002  3002  400 m. Distance travelled by the person in 10 minutes = 1800 – 400 = 1400 m = 1.4 km. 1.4  60 Speed = = 8.4 km/hr. 10 10. Answer: A Explanation: Number of points from (1, 1) to (39, 1) = 39 Number of points from (1, 2) to (38, 2) = 38 Similarly, the point in the topmost row is (1, 39) and so number of points = 1. Total points = 1 + 2 + 3 + … + 39 39(40) =  780 2

15. Answer: D Explanation: (13.5, 16) and (5.5, 7.5) are adjacent points of the parallelogram and (17.5, 23.5) is the point of intersection of two diagonals of the parallelogram. We know that diagonals of a parallelogram bisect each other. Therefore, length of one diagonal

11. Answer: C Explanation: Distance between the lines x + y – 4 = 0 and x + y + 2 = 0 is 12  12

2

4 3

3 = 500 3 m. In DFGA, FAG = 60. Therefore, AG = FG cot 60 = 500 3 cot 60 = 500 m.

4  2

2  ( 7.5)

2 2 = 2 (17.5  5.5)  (23.5  7.5) = 40 cm.

3 2.

Length of the other diagonal 2 2 = 2 (17.5  13.5)  (23.5  16) = 17 cm.

12. Answer: D

16. Answer: A

Explanation: Equation of line equidistant from the given lines and parallel to them is given by 6  12 2x + 3y = = 9. 2 Distance of this line from origin

Explanation: Since the side is of length a, the length of a semi-diagonal will be . Therefore, 2 coordinates of the vertex cannot be (a 2 , 0).

659

Quantitative Aptitude Simplified for CAT

17. Answer: C

21. Answer: B

Explanation:

Explanation: The best strategy for solving this question is to visualize the location of points and identify the equation of line. In this case, it will be y = x. Alternatively, for two given points to be symmetrical about a line, the mid-point of the line joining the given points should lie on the line. Also, the line joining the given points must be perpendicular to the required line. Both these properties are satisfied only for the line y = x.

Area of the quadrilateral PQCR = 2  area of triangle PQC. Length of tangent from P to Q =

PC 2  QC2  (152  52 )  52  15 .

1  15  5 . 2 Therefore, area of the quadrilateral 1 = 2   15  5 = 75 units. 2

Therefore, area of PQC =

18. Answer: D

22. Answer: A

Explanation: Let the coordinates of Sanjay be (0, 0). Then, coordinates of Lalbaug will be (0, 6). Coordinates of Dibrugarh will be (–10, 6) and coordinates of Krishna will be (–10, 9). Therefore, distance of Sanjay from Krishna =

Explanation: The abscissa must lie between 1 and 5. 23. Answer: A Explanation: sec2x – tan2x = 1  (sec x – tan x)(sec x + tan x) = 1  4(sec x + tan x) = 1 1  (sec x + tan x) = . 4

102  92  181 miles.

19. Answer: A Explanation: The centroid of the given triangle will be same the centroid of the triangle formed by the midpoints of the given sides. Therefore, centroid  20 4 1 35  ,  or (2, 1). 3 3  

24. Answer: A Explanation: In a cyclic quadrilateral, A + C = 180 = B + D. So, cos A + cos B + cos C + cos D = cos A + cos B + cos (180 – A) + cos (180 – B) = cos A + cos B – cos A – cos B = 0.

20. Answer: B

25. Answer: B

Explanation: Let us put a = 1, b = 2 and c = 3. Then the coordinates become A (1, 5), B (2, 4) and C (3, 3). The slope of AB = –1 and slope of BC = –1. Therefore, points are collinear. Alternatively, Using the formula of area of triangle, we get a bc 1

Explanation: 2 sec  2 sec  2   = 2 cos  = 1. 2 2 1  tan  sec  sec  26. Answer: D Explanation: 3 3 sin  =  tan  = (refer to the figure 4 5 5 4 below). Also, cos  = or sec  = 4 5

ab  c b  c 1

1 1 b a c 1  ab  c a c 1 2 2 c ab 1 ab  c ab 1

2

3 5 Therefore, (tan  + sec )2 =     4 .  5 4

1 bc 1



1 (a  b  c) 1 a  c 1 2 1 ab 1

= 0. Therefore, the points are collinear.

660

Coordinate Geometry and Trigonometry 29. Answer: A Explanation: AC = BC cosec 30 = 2BC = 2x. Now, using Pythagoras’ Theorem, x2 + (15 – x)2 = (2x)2 = 4x2  2x2 + 30x – 225 = 0  x 30  900  1800 30  30 3   4 4 15 3  15 = . Therefore, area of triangle 2

27 Answer: B Explanation: Let OC be x. Then the height of the building is x + 8. x8 So, OB = (x + 8) cot 60 = . 3 2

Now,

x2

+

 x8   =  3 



82

or

x2

+

x 2 64 16x   = 64 3 3 3  4x2 + 16x – 128 = 0  x2 + 4x – 32 = 0  x = 4. Therefore, the height of the building = 4 + 8 = 12 m.

1 x 2 3  15 3  15  (x)(2x) sin60     2 2 2  



2

3 2



225 3 2  3 225 3 42 3  8 4 . =





30. Answer: A Explanation: Let BD = x. Then, x = 50 cot 45° = 50 m. Then, EC = 50 m. 50 So, AE = x tan 30 = 50 tan 30 = . 3 Therefore, height of the tower, DC = AB – AE

28. Answer: D Explanation: Let the height of the flagstaff be h and the length of the plank be x. Then, 10 10 10 20 x   10 3   tan30 tan60 3 3.

= 50 –

h  tan30  h = x tan 30° x  20  1 20   .  3 3  3

Moreover,

661

50 3

50 =



3 1 3



m.

CAT 2017 Solved Paper Slot 1

SOLVED PAPERS  CAT 2017 QA and DILR Tests Slot 1  CAT 2017 QA and DILR Tests Slot 2  CAT 2018 QA and DILR Tests Slot 1  CAT 2018 QA and DILR Tests Slot 2

P-1

Quantitative Aptitude Simplified for CAT

P-2

CAT 2017 Solved Paper Slot 1

CAT 2017 Solved Paper Slot 1 Quantitative Aptitude (QA) 1.

Arun’s present age in years is 40% of Barun’s. In another few years, Arun’s age will be half of Barun’s age. By what percentage will Barun’s age increase during this period?

2.

A person can complete a job in 120 days. He works alone on Day 1. On day 2, he is joined by another person who can also complete the job in exactly 120 days. On day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

3.

4.

5.

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi’s savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs. 59500, then Ravi’s total monthly savings (in Rs) is

6.

If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures she makes a profit of 20%? (A)

Give a discount of 5% on retail price

(B) Give a discount of 5% on retail price (C) Increase the retail price by 2% (D) Sell at retail price

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and lightest weighs 53 kg. What is the maximum possible number of people in the group?

7.

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before train’s departure. The distance (in km) from his home to the railway station is

A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is (A)

6: 2

(B)

7 :2

(C)

2 5 :3

(D) 3 : 2

P-3

Quantitative Aptitude Simplified for CAT

8.

Suppose C1, C2, C3, C4 and C5 are five companies. The profits made by C1, C2 and C3 are in the ratio 9 : 10 : 8 while the profits made by C2, C4 and C5 are in the ratio 18 : 19 : 20. If C5 has made a profit of Rs. 19 crore more than C1, then the total profit (in Rs.) made by all the companies is

destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

(A)

438 crore

(C) 24%

(B) 435 crore

(D) 28%

(A)

(C) 348 crore

13. If a and b are integers of opposite signs such that (a + 3)2 : b2 = 9 : 1 and (a – 1)2 : (b – 1)2 = 4 : 1, then the ratio a2 : b2 is

(D) 345 crore 9.

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is (A)

(A)

(C) 1 : 4 (D) 25 : 4

35

14. A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

(C) 60 (D) 65 10. A stall sells popcorn and chips in packets of three sizes: large, super and jumbo. The numbers of large, super and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

(A)

9.5

(B) 10 (C) 4.5 (D) 6

1:1

15. The area of the closed region bounded by the equation |x| + |y| = 2 in the twodimensional plane is

(B) 8 : 7 (C) 4 : 3 (D) 6 : 5

(A)

4

(B) 4

11. In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is (A)

9:4

(B) 81 : 4

(B) 50

(A)

30%

(B) 25%

(C) 8 (D) 2 16. From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq. ft, of the remaining portion of triangle ABC is

6%

(A)

(B) 8% (C) 10% (D) 12% 12. If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are

P-4

225 3

(B)

500 3

(C)

275 3

(D)

250 3

CAT 2017 Solved Paper Slot 1 17. Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semicircle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm, then the area, in sq. cm, of the region enclosed by BPC and BQC is (A)

23. The value of log0.008

9 – 18

(C) 9

(A)

1 3

(B)

2 3

(C)

5 6

(D)

7 6

(D) 9

10

(B) 50 (C) 60 19. A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 cm3. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is

3

(B)

9 4

(C)

4 9

(D)

1 3

(A)

101

(B) 99 (C) 87 (D) 105

20. Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is

26.* For how many integers n, will the inequality (n – 5)(n – 10) – 3(n – 2)  0 be satisfied? 27. If f1(x) = x2 + 11x + n and f2(x) = x, then the largest positive integer n for which the equation f1(x) = f2(x) has two distinct real roots, is __________ .

21. Suppose log3x = log12y = a, where x, y are positive numbers. If G is the geometric mean of x and y, then log6G is equal to

28. If a, b, c and d are integers such that a + b + c + d = 30, then the minimum possible value of (a – b)2 + (a – c)2 + (a – d)2 is __________ .

a

(B) 2a a 2

29. Let AB, CD, EF, GH and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K and O so as to form a triangle?

(D) a 22. If x + 1 = x2 and x > 0, then 2x4 is (A)

(A)

25. The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y and z are positive integers such that x  40, y  12 and z  12 is

(D) 20

(C)

81 – 7 is

24. If 92x–1 – 81x–1 = 1944, then x is

18. A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to

(A)

3

equal to

(B) 18

(A)

5  log

6+4 5

(B) 3 + 5 5 (C) 5 + 3 5 (D) 7 + 3 5

P-5

Quantitative Aptitude Simplified for CAT

Data Interpretation and Logical Reasoning (DILR)

1  30. The shortest distance of the point  , 1  2  from the curve y = |x – 1| + |x + 1| is

(A)

Directions for questions 35–38: Read the information given below and answer the questions that follow.

1

(B) 0 (C)

2

(D)

3 2

Healthy Bites is a fast food joint serving three items: burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare up to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However, fries cannot be prepared in anticipation of future orders.

31. If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is (A)

2:3

(B) 3 : 2 (C) 3 : 4 (D) 4 : 3 32. In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers? (A)

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however, the order is considered to be completely served only when all the items of that order are served.

16

(B) 20 (C) 14

The table below gives the orders of three clients and the times at which they placed their orders:

(D) 15 33. If f(x) =

5x  2 and g(x) = x2 – 2x – 1, then 3x  5

the value of g(f(f(3))) is

Client No.

(A)

2

1

10: 00

(B)

1 3

1 burger, 3 portions of fries, 1 order of ice cream

2

10:05

2 portions of fries, 1 order of ice cream

3

10:07

1 burger, 1 portion of fries

Time

(C) 6 (D)

2 3

35. Assume that only one client's order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.

34. Let a1, a2, a3, …, an be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + a3 + … + a3n = 1830, then what is the smallest positive integer m such that m(a1 + a2 + a3 + … + an) > 1830? (A)

Order

At what time is the order placed by Client 1 completely served?

8

(B) 9

(A)

(C) 10

10:17

(B) 10:10

(D) 11

(C) 10:15 (D) 10:20

P-6

CAT 2017 Solved Paper Slot 1 36. Assume that only one client's order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.

The kids surveyed were further divided into two groups based on whether their mothers dropped out of school before completing primary education or not. The table below gives the number of kids in different types of schools for mothers who dropped out of school before completing primary education:

At what time is the order placed by Client 3 completely served? (A)

10:35

(B) 10:22 (C) 10:25 (D) 10:17 37. Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

G

P

O

Total

NE

4200

500

300

5000

W

4200

1900

1200

7300

S

5100

300

300

5700

Total

13500

2700

1800

18000

It is also known that: 1.

In S, 60% of the surveyed kids were in G. Moreover, in S, all surveyed kids whose mothers had completed primary education were in school.

2.

In NE, among the O kids, 50% had mothers who had dropped out before completing primary education.

3.

The number of kids in G in NE was the same as the number of kids in G in W.

At what time is the order placed by Client 2 completely served? (A)

10:10

(B) 10:12 (C) 10:15 (D) 10:17 38. Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

39. What percentage of kids from S were studying in P? (A)

Also assume that the fourth client came in only at 10:35. Between 10:00 and 10:30, for how many minutes is exactly one of the employees idle? (A)

37%

(B) 6% (C) 79% (D) 56%

7

40. Among the kids in W whose mothers had completed primary education, how many were not in school?

(B) 10 (C) 15 (D) 23

(A)

Directions for questions 39–42: Read the information given below and answer the questions that follow.

300

(B) 1200 (C) 1050 (D) 1500

A study to look at the early learning of rural kids was carried out in a number of villages spanning three states, chosen from the North East (NE), the West (W) and the South (S). 50 four-year old kids each were sampled from each of the 150 villages from NE, 250 villages from W and 200 villages from S. It was found that of the 30000 surveyed kids 55% studied in primary schools run by government (G), 37% in private schools (P) while the remaining 8% did not go to school (O).

41. In a follow up survey of the same kids two years later, it was found that all the kids were now in school. Of the kids who were not in school earlier, in one region, 25% were in G now, whereas the rest were enrolled in P; in the second region, all such kids were in G now; while in the third region, 50% of such kids had now joined G while the rest had joined P. As a result, in all three regions put together, 50% of the kids who were earlier out of school had joined G. It

P-7

Quantitative Aptitude Simplified for CAT

was also seen that no surveyed kid had changed schools.

1.

No one is below the 80th percentile in all 3 sections.

What number of the surveyed kids now were in G in W?

2.

150 are at or above the 80th percentile in exactly two sections.

(A)

3.

The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.

4.

Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.

6000

(B) 5250 (C) 6750 (D) 6300 42. In a follow up survey of the same kids two years later, it was found that all the kids were now in school. Of the kids who were not in school earlier, in one region, 25% were in G now, whereas the rest were enrolled in P; in the second region, all such kids were in G now; while in the third region, 50% of such kids had now joined G while the rest had joined P. As a result, in all three regions put together, 50% of the kids who were earlier out of school had joined G. It was also seen that no surveyed kid had changed schools.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

What percentage of the surveyed kids in S, whose mothers had dropped out before completing primary education, were in G now? (A)

43. What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?

94.7%

(B) 89.5% (C) 93.4% (D) Cannot be determined from the given information

(A)

3 or 10

(B) 10

Directions for questions 43–46: Read the information given below and answer the questions that follow.

(C) 5 (D) 7 or 10 44. If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET? 45. If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?

Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET: P-8

CAT 2017 Solved Paper Slot 1 46. If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more than 100, how many candidates had to sit for the separate test for BIE? (A)

3.

47. What is Amda's score in F? 48. What is Zooma's score in S? 49. Benga and Delma, two countries categorized as happy, are tied with the same total score. What is the maximum score they can have?

299

(B) 310 (C) 321 (D) 330

(A) (B) (C) (D)

Directions for questions 47–50: Read the information given below and answer the questions that follow. Simple Happiness index (SHI) of a country is computed on the basis of three parameters: social support (S), freedom to life choices (F) and corruption perception (C) . Each of these three parameters is measured on a scale of 0 to 8 (integers only). A country is then categorized based on the total score obtained by summing the scores of all the three parameters, as shown in the following table: Total Score Category

0-4 Very Unhappy

All the 3 countries, which are categorised as happy, have the highest score in exactly one parameter.

5-8

9-13

14-19

20-24

Unhappy

Neutral

Happy

Very

14 15 16 17

50. If Benga scores 16 and Delma scores 15, then what is the maximum number of countries with a score of 13? (A)

0

(B) 1 (C) 2 (D) 3 Directions for questions 51–54: Read the information given below and answer the questions that follow. There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular- skilled employees (RE). During the next five months, the division has to complete five projects every month. Out of the 25 projects, 5 projects are "challenging", while the remaining ones are "standard". Each of the challenging projects has to be completed in different months. Every month, five teams – T1, T2, T3, T4 and T5, work on one project each. T1, T2, T3, T4 and T5 are allotted the challenging project in the first, second, third, fourth and fifth month, respectively. The team assigned the challenging project has one more employee than the rest.

Happy

Following diagram depicts the frequency distribution of the scores in S, F and C of 10 countries – Amda, Benga, Calla, Delma, Eppa, Varsa, Wanna, Xanda, Yanga and Zooma;

In the first month, T1 has one more SE than T2, T2 has one more SE than T3, T3 has one more SE than T4, and T4 has one more SE than T5. Between two successive months, the composition of the teams changes as follows:

Further, the following are known: 1. Amda and Calls jointly have the lowest total score, 7, with identical scores in all the three parameters. 2. Zooma has a total score of 17.

a.

P-9

The team allotted the challenging project, gets two SE from the team which was allotted the challenging project in the previous month.

Quantitative Aptitude Simplified for CAT

In exchange, one RE is shifted from the former team to the latter team. b.

Directions for questions 55–58: Read the information given below and answer the questions that follow.

After the above exchange, if T1 has any SE and T5 has any RE, then one SE is shifted from T1 to T5, and one RE is shifted from T 5 to T1. Also, if T2 has any SE and T4 has any RE, then one SE is shifted from T2 to T4, and one RE is shifted from T4 to T2.

In a square layout of size 5m x 5m, 25 equal sized square platforms of different heights are built. The heights (in metres) of individual platforms are as shown below:

Each standard project has a total of 100 credit points, while each challenging project has 200 credit points. The credit points are equally shared between the employees included in that team. 51. The number of times in which the composition of team T2 and the number of times in which composition of team T4 remained unchanged in two successive months are: (A) (B) (C) (D)

(2,1) (1, 0) (0, 0) (1,1)

2

4

3

9

5

3

2

8

7

8

4

6

5

3

9

5

1

2

1

7

6

3

9

(i)

A and B are in the same row or column

(ii)

A is at a lower height than B

(iii) If there is/are any individuals(s) between A and B, such individual(s) must be at a height lower than that of A.

(0, 2)

(B) (0, 3)

Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.

(C) (1, 2) (D) (1, 3) 53. Which of the following CANNOT be the total credit points earned by any employee from the projects? (A)

1

Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:

52. The number of SE in T1 and T5 for the projects in the third month are, respectively: (A)

6

140

Rows in the layout are numbered from top to bottom and columns are numbered from left to right.

(B) 150 (C) 170 (D) 200

55. How many individuals in this layout can be reached by just one individual?

54. One of the employees named Aneek scored 185 points. Which of the following CANNOT be true?

(A)

(A)

Aneek worked only in teams T1, T2, T3, and T4 (B) Aneek worked only in teams T1, T2, T4, and T5 (C) Aneek worked only in teams T2, T3, T4, and T5 (D) Aneek worked only in teams T1, T3, T4, and T5

3

(B) 5 (C) 7 (D) 8 56. Which of the following is true for any individual at a platform of height 1 m in this layout? (A)

P-10

They can be reached by all the individuals in their own row and column

CAT 2017 Solved Paper Slot 1 60. Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:

(B) They can be reached by at least 4 individuals (C) They can be reached by at least one individual (D) They cannot be reached by anyone 57. We can find two individuals who cannot be reached by anyone in (A) (B) (C) (D)

the last row the fourth row the fourth column the middle column

(A)

58. Which of the following statements is true about this layout?

(B) 120 (C) 96

(A)

Each row has an individual who can be reached by 5 or more individuals (B) Each row has an individual who cannot be reached by anyone (C) Each row has at least two individuals who can be reached by an equal number of individuals (D) All individuals at the height of 9 m can be reached by at least 5 individuals

(D) 60 61. Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following: 1. Both cities are in G1

Directions for questions 59–62: Read the information given below and answer the questions that follow.

2. Between A and any city in G2 3. Between B and any city in G3

A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.

4. Between C and any city in G4 Then, the minimum number of direct flights that satisfies the underlying principle of the airline is __________ . 62. Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:

The underlying principle that they are working on is the following: Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day. 59. If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is: (A) (B) (C) (D)

54

1.

Both cities are in G1

2.

Between A and any city in G2

3.

Between B and any city in G3

4.

Between C and any city in G4

However, due to operational difficulties at A, it was later decided that the only flights that would operate at A would be those to and from B. Cities in G2 would have to be assigned to G3 or to G4.

45 90 180 135

P-11

Quantitative Aptitude Simplified for CAT

65. A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M- NB. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.

What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose? Directions for questions 63–66: Read the information given below and answer the questions that follow. Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N). The roads from A to M, and from N to B, are both short and narrow. In each case, one car takes 6 minutes to cover the distance, and each additional car increases the travel time per car by 3 minutes because of congestion. (For example, if only two cars drive from A to M, each car takes 9 minutes.) On the road from A to N, one car takes 20 minutes, and each additional car increases the travel time per car by 1 minute. On the road from M to B, one car takes 20 minutes, and each additional car increases the travel time per car by 0.9 minute.

How many cars would the police department order to take the A-M-N-B route so that it is not possible for any car to reduce its travel time by not following the order while the other cars follow the order? (Assume that the police department would never order all the cars to take the same route.) 66. A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M- NB. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.

The police department orders each car to take a particular route in such a manner that it is not possible for any car to reduce its travel time by not following the order, while the other cars are following the order. 63. How many cars would be asked to take the route A-N-B, that is Akala-Nanur-Bakala route, by the police department?

If all the cars follow the police order, what is the minimum travel time (in minutes) from A to B? (Assume that the police department would never order all the cars to take the same route.)

64. If all the cars follow the police order, what is the difference in travel time (in minutes) between a car which takes the route A-N- B and a car that takes the route A-M-B? (A)

1

(A)

26

(B) 0.1

(B) 32

(C) 0.2

(C) 29.9

(D) 0.9

(D) 30

P-12

CAT 2017 Solved Paper Slot 1

ANSWERS AND EXPLANATIONS Quantitative Aptitude (QA) 1.

5.

Explanation: Let Ravi’s monthly savings be Rs. 100. Then, money invested in fixed deposits = Rs. 50 Money invested in stocks = 30% of Rs. 50 = Rs. 15 Money going into savings bank account = 50 – 15 = Rs. 35 So, total money deposited by him in the bank = 50 + 35 = Rs. 85

Ans. 20% Explanation: Let Barun’s present age be x years. Then, Arun’s present age = 0.4x After ‘y’ years, ages become (x + y) and (0.4x + y) years. As per the question, 0.4x + y =

1 (x + y) 2

 0.1x = 0.5y 

y=

1 x 5

So, total monthly savings = 59500 ×

Since Barun’s age has increased by ‘y’ years, the required percentage increase in Barun’s age = 2.

6.

y 1  100   100 = 20% x 5

4.

So, retail price =

102 = 120  1  15    100  

To gain 20%, the seller should sell at retail price. 7.

Ans. B Explanation: Let v and u be the speed of motor boat and that of stream. Also, let distance from home to office be x. Then, total travel time

n(n  1) = 120 2

=

Ans. 11 Explanation: For maximum number of persons, we should carry people of minimum weights in the lift. Quotient when 630 is divided by 53 is 11. So, the maximum number of persons is 11.

x x 2vx   v  u v  u v 2  u2

When the speed of motor boat gets doubled, then total travel time =

x x 4vx ,   2v  u 2v  u 4v 2  u2

which is 25% of the earlier travel time. So, 4vx 4v 2  u2

Ans. 20 Explanation: Let the distance from his home to the railway station be x. Then,

 0.25 

2vx v 2  u2

 2(v2 – u2) = 0.25(4v2 – u2) = v2 – 0.25u2  2(v2 – u2) = v2 – 0.25u2  v2 = 1.75u2  v:u=

x x 10  (10) 1    12 15 60 3



Ans. D Explanation: Let cost price be Rs. 100. Then, selling price = Rs. 102, which is after 15% discount on retail price.

 n(n + 1) = 240  n = 15 days 3.

100 85

= Rs. 70000

Ans. 15 Explanation: Let the work required to be done is 120 units, which each person alone can do in 120 days. So, Work done by each person in 1 day = 1 unit. Now, work done on day 1 = 1 Work done on day 2 = 1 + 1 = 2 Work done on day 3 = 1 + 1 + 1 = 3; and so on. If work is completed in n days, then 1 + 2 + 3 + 4 + … + n = 120 

Ans. 70000

x = 20 km

P-13

7 : 2.

Quantitative Aptitude Simplified for CAT

8.

9.

Ans. A

12. Ans. D

Explanation: Ratio of profits of C1, C2 and C3 = 9 : 10 : 8 = 81 : 90 : 72 Ratio of profits of C2, C4 and C5 = 18 : 19 : 20 = 90 : 95 : 100 So, ratio of profits of all the 5 companies = 81 : 90 : 72 : 95 : 100 Difference of profits of C5 and C1 = Rs. 19 crore So, 100k – 81k = 19 or k = 1 So, total profit by all the companies = 81k + 90k + 72k + 95k + 100k = 438k = Rs. 438 crore

Explanation: Let cost price (CP) be Rs. 100 per toy. Then, selling price (SP) = 120 And, so, marked price (MP)

Ans. D

So, discount that should be given on printed price

=

Total cost of toys = 60 × 100 = Rs. 6000 Profit per toy = 120 – 100 = Rs. 20 So, total profit = 60 × 20 = Rs. 1200 Since 10 toys are destroyed in fire, the remaining 50 are to be sold for Rs. 7200 to maintain the same profit of Rs. 1200. So, SP per toy =

Explanation: Let the number of girls and the number of boys appearing for an admission test be 200 and 100. Then, number of girls getting admission = 30% of 200 = 60 Number of boys getting admission = 45% of 100 = 45 Total students not getting admission = 200 + 100 – 60 – 45 = 195 Required percentage =

120 = Rs. 200 1  0.4

=

7200 = 144 50

200  144  100 = 28% 200

13. Ans. D Explanation: If (a + 3)2 : b2 = 9 : 1, then

a3 3 3 or  .  1 b 1

So, a + 3 = 3b or a + 3 = –3b Let a + 3 = 3b. Then a and b will have same sign, and so it is ruled out. So, a + 3 = –3b or a = –3b – 3. (a – 1)2 : (b – 1)2 = 4 : 1  (–3b – 3 – 1)2(1) = 4(b – 1)2  9b2 + 24b + 16 = 4(b2 – 2b + 1)  9b2 + 24b + 16 = 4b2 – 8b + 4  5b2 + 32b + 12 = 0  (5b + 2)(b + 6) = 0

195  100 = 65% 300

10. Ans. A Explanation: 7x + 17x + 16x = 6y + 15y + 14y  40x = 35y  8x = 7y So, ratio of number of jumbo popcorn to jumbo chips packets = 16x : 14y = 8x : 7y = 1 : 1 (because 8y = 7y)



b= 

2 , –6. 5

But a and b are integers. So, b = –6 and a = 15 Therefore, a2 : b2 = 152 : 62 = 25 : 4

11. Ans. B Explanation: Let the price of good quality mangoes be pg. So, the price of the medium quality mangoes, pm, is given by:

14. Ans. A Explanation: Let the average score of boys in mid-semester be x. Then, average score of girls = x + 5 Total score of the entire class = 20x + 30(x + 5) = 50x + 150 In the final exam, average score of girls = x + 5 – 3 = x + 2.

1 pm = pg; 2

CP of all mangoes = 80pg + 40pm = 80pg + 20pg = 100pg Also, SP = 120(0.9pg) = 108pg So, profit percent = 8%

P-14

CAT 2017 Solved Paper Slot 1 17. Ans. B

Let the average score of boys be y. 30( x  2)  20 y 50 x  150 Then,  2 50 50

Explanation:

 30x + 20y + 60 = 50x + 150 + 100  20y – 20x – 190 = 0  y = x + 9.5 So, the average score of boys increased by 9.5. 15. Ans. C Explanation: The graph of the equation is drawn below.

BC = AB 2 = 6 2 = Diameter of arc BQC So, area of semicircle BQC =



1  3 2 2

2

= 9

Area of region between BC and BPC = Area of segment = Area of sector ABPC – Area of DABC It is a square of side AB = 2 2 Hence, area = 8

=

90 1  (6)2   6  6 = 9 – 18 360 2

Therefore, area of region enclosed by BPC and BQC

16. Ans. B Explanation: AD is the median. Since G is the centroid of ABC, AG : GD = 2 : 1.

= 9 – (9 – 18) = 18 18. Ans. B Explanation: If volumes of the 5 cubes are x3, x3, 8x3, 27x3 and 27x3, then the volume of the original cube = 64x3 and so the sides of the original and 5 new cubes are 4x, x, x, 2x, 3x and 3x. Total surface area of original cube = 6(4x)2 = 6 × 16x2

Therefore, Ar(GBC) =

Total surface area of the new cubes

1 Ar(ABC) 3

= 6[(x)2 + (x)2 + (2x)2 + (3x)2 + (3x)2] = 6 × 24x2

Now, Ar(ABC) =

s(s  a)(s  b)(s  c) , a  b  c 40  25  35 Where s =   50 2 2

Therefore, percentage by which the sum of the surface areas of these five cubes

So, Ar(ABC) =

exceeds the surface are of the original cube

50(50  40)(50  25)(50  35)

=

= 50  10  25  15  250 3 Therefore, area of remaining portion of ABC =

2 500 2 Ar(ABC) =  250 3  3 3 3

P-15

6  24 x2   6  16 x2   100 6  16 x2 

= 50%

Quantitative Aptitude Simplified for CAT

19. 6

22. Ans. D

Explanation:

Explanation: 2x4 = 2(x2)2 = 2(x + 1)2 = 2(x2 + 2x + 1) = 2(x + 1 + 2x + 1) = 6x + 4 Now, solution of the quadratic equation, x2 – x – 1 = 0, can be calculated as below: x = 1 14  1 5



2

2

Since x > 0, x = 1  5 2

Therefore, 6x + 4 = 3 + 3 5 + 4 = 7 + 3 5 23. Ans. C Explanation: BC = 3 cm Volume of cylinder = 9 cm3 Now, volume = r2h  9 = r2(3)

log0.008 5 + log =

 r = 3 cm = BQ Since PB = Radius of sphere = 2 cm,

 

PQ = 22  3

2

=

= 1 cm

RP = 2 cm So, RQ = 2 + 1 = 3 cm Required vertical distance = RQ + BC = 3 + 3 = 6 cm

=

 23  2  log 3  10  

+8–7

log 5 +1 1  6  log  5  1 5 1  6 6

Explanation: 92x–1 – 81x–1 = 1944  92x–1 – 92x–2 = 1944 Let 92x = k. Then,

15  20 = 12 km 25

k k = 1944  9 81 8k  = 1944  k = 19683 = 92x 81

So, time taken =

log 5

24. Ans. B

Explanation: Hypotenuse = 25 km So, length of perpendicular from A to the hypotenuse =

81 – 7

log 5 4 log 3 –7  2 log 0.008  1  log 3   2 

= 

20. 24

3

12  60 minutes = 24 minutes 30

21. Ans. D



Explanation: x = 3a, y = 12a.

39 = 34x  x =

9 4

25. Ans. B a

a

6a

Now, G = xy  3  12 = Taking log on both sides, we get: log G = a log 6  log6G = a

Explanation: The equation can be written as: x = 25 + y + z. Minimum value of y and z is 1 each. So, minimum value of x is 27.

P-16

CAT 2017 Solved Paper Slot 1 For each value of y + z, there is a unique value of x. Now, y + z can be 2, 3, 4, …, 15 (because x  40).

29. 160 Explanation:

When y + z = 2, the number of solutions is 1, i.e. (1, 1);

= 11C3 =

When y + z = 3, the number of solutions is 2, i.e. (1, 2) and (2, 1)

Out of these, vertices A, O, B cannot form triangle. Similarly, the set of three points lying on the diagonal cannot form triangles.

Total number of triangles possible

When y + z = 4, the number of solutions is 3; and so on, till y + z = 13 whose number of solutions is 12.

11! = 165 3!8!

So, 5 triangles are not possible. Hence, total number of triangles = 160

When y + z = 14, the number of solutions will be 11, i.e., (2, 12), (3, 11), …, (12, 2).

30. Ans. A Explanation: For –1  x  1, y = 2, that is, the curve is a straight horizontal line.

When y + z = 15, the number of solutions will be 10, i.e., (3, 12), (4, 11), …, (12, 3). So, total number of solutions

1 So, the shortest distance of  , 1  from the 2

= (1 + 2 + 3 + … + 12) + (11 + 10)



given curve = 1

= 78 + 21 = 99

31. Ans. A

26. 11 Explanation:

Explanation:

The given inequality becomes

(a + 6d)2 = (a + 2d) (a + 16d)

n2 – 15n + 50 – 3n + 6  0



a2 + 12ad + 36d2 = a2 + 18ad + 32d2

 n2 – 18n + 56  0



4d2 = 6ad



Since d is positive, it cannot be zero and hence can be cancelled from both sides. We get

4  n  14

The integer solutions are: 4, 5, 6, …, 14, which are 11 in number.

2d = 3a  a : d = 2 : 3

27. 24

32. Ans. A

Explanation:

Explanation: Number of ways of distribution of 7 identical erasers among 4 kids so that each gets at least one eraser is given by the formula n–1Cr–1, where n is number of objects (erasers) and r is number of kids.

f1(x) = f2(x) 

x2 + 11x + n = x

 x2 + 10x + n = 0 For distinct real roots, b2 – 4ac > 0, that is, (10)2 – 4(1)(n) > 0  n < 25

So, number of ways = 7–1C4–1 = 6C3 = 20

Therefore, the greatest integer value of n is 24.

However, this includes (4, 1, 1, 1), (1, 4, 1, 1), (1, 1, 4, 1) and (1, 1, 1, 4).

28. 2

So, required number of ways = 20 – 4 = 16

Explanation: For minimum possible value of (a – b)2 + (a – c)2 + (a – d)2, difference of a with b, c and d should be minimum, that is, a should be as close to b, c and d as possible. So, a = b = 7, c = d = 8, is one of the possible values. So, the minimum value of (a – b)2 + (a – c)2 + (a – d)2 = 2

33. Ans. A Explanation: f(3) =

5(3)  2 17  3(3)  5 4

17  5   2 17  93 4    f(f(3)) = f     3 17   5 31  4  3    4 

g(f(f(3))) = g(3) = 9 – 6 – 1 = 2

  P‐17 

Quantitative Aptitude Simplified for CAT 

37. Ans. A

34. Ans. B Explanation: a = 3, d = 7 – 3 = 4. Now, a1 + a2 + a3 + … + a3n =

Explanation: For the order of client 1, one of the employees, say Anish, will be busy on burger till 10:10, while Bani would work on ice cream till 10:02, whereas automatic fryer will make fries in 5 minutes. The automatic fryer would work on the order of client 2 at 10:05 (at the time of placing of order) for next 5 minutes. So, order of fries of client 2 would be completely served at 10:10. Bani would start work on ice cream for client 1 at 10:05 (at the time of placing of order) for next 2 minutes, and would be done by 10:07. So, the order of client 2 would be completely served by 10:10.

3n 3n (12n + 2) = 1830 2(3)  (3n  1)4   2 2

 n(6n + 1) = 610  n = 10 Now, a1 + a2 + a3 + … + an =

n n 2(3)  (n  1)4   (4n + 2) 2 2

= 5 × 42 = 210 So, 210m > 1830 or m > 8.7… Therefore, smallest positive integer value of m = 9

Data Interpretation and Logical Reasoning (DILR)

38. Ans. B

35. Ans. B

Explanation: If Anish prepares burger for client 1, then he is busy till 10:10. Meanwhile, Bani starts preparing ice cream at 10:00 till 10:02, and remains idle till 10:05, when client 2 places order. Then, Bani starts preparing ice cream from 10:05 to 10:07. Then the order of client 3 comes at 10:07. While Anish is busy preparing burger for client 1, Bani would start preparing burger which will take 10 minutes and would be completed by 10:17. From 10:10 to 10:17, Anish would be idle. After 10:17, both the employees would be idle. Therefore, exactly one of the employees is idle for 3 (only Bani) + 7 (only Anish) = 10 minutes. After 10:17, since both the employees are idle, that would not be counted.

Explanation: Client 1 has ordered 1 burger, 3 portions of fries and 1 order of ice cream, at 10:00. One of the employees will work on burger and take 10 minutes. During the same time, second employee will work on the order of ice cream and take 2 minutes. Also, fries won’t need any employee and so would be done in 5 minutes. The time taken for ice cream and fries will be done along with the time taken for burger and hence would not be counted separately. Therefore, the order of client 1 would be completed by 10:10. 36. Ans. C Explanation: The order of client 1 would be completely served by 10:10 and then work on order of client 2 would be started. Fries take 5 minutes and ice cream takes 2 minutes. So, overall time taken would be 5 minutes. The order of client 2 would be served by 10:15. Since client 3 has ordered burger, the order would be served in next 10 minutes and hence would be done by 10:25.

39. Ans. A Explanation: We can make the following table mentioning number of kids for both categories: whose mothers have dropped out of primary education, and whose mothers have not dropped out of primary education.

  P‐18 

CAT 2017 Solved Paper Slot 1 G

P

O

Total

Dropped out

Completed

Dropped out

Completed

Dropped out

Completed

NE

4200

1050

500

1150

300

300

7500

W

4200

1050

1900

3850

1200

300

12500

S

5100

900

300

3400

300

0

10000

Total

13500

3000

2700

8400

1800

600

30000

Total number of kids from S = 50  200 = 10000 Now, 60% of these are in G. So, number of kids from S in G = 60% of 10000 = 6000 Of the remaining 4000, 300 are those who are in O and whose mothers have dropped out of primary education (given in the table). So, remaining 3700 are studying in P. So, required percentage = 37%

= 5100 + 300 + 300 = 5700 Number of kids from these who are in G = 5100 + 300 = 5400 Required percentage =

5400  100 = 94.7% 5700

Common explanation for questions 43–46: We know that 200 students scored above 90th percentile overall in CET. We can draw Venn diagram representing number of persons scoring 80th percentile or above in the three sections. From the information given, Only Physics = Only Chemistry = Only Maths = a Also, y = 0. And p + q + r = 150 So, 3a + x = 200 – 150 = 50 Number of candidates below 80th percentile in Physics = 2a + r Number of candidates below 80th percentile in Chemistry = 2a + q Number of candidates below 80th percentile in Maths = 2a + p

40. Ans. A Explanation: As we can see from the table, the number of kids in W whose mothers had completed primary education and who are not in school is 300. 41. Ans. A Explanation: In W, there are total 1500 kids who were not going to school. Since total number of kids who were not going to school was 2400, the number of kids from this 2400 who are in G now is 1200. Naturally, all the 1500 kids from W cannot be the ones who shift to G. If 50% of these join G, then 750 joined G. It is given that 25% of kids of a particular region not going to school join G. If that region is NE, then 25% of 600 = 150 are now in G. Also, the 300 kids from S who were not going to school are all now in G. That makes a total of 750 + 150 + 300 = 1200 Therefore, number of surveyed kids now in G in W = 4200 + 1050 + 750 = 6000

It is also given that (2a + r) : (2a + q) : (2a + p) = 4 : 2 : 1. Therefore, (2a + r) + (2a + q) + (2a + p) = (4 + 2 + 1) k = 7k

42. Ans. A Explanation: Total surveyed kids in S, whose mothers had dropped out before primary education

P-19

Quantitative Aptitude Simplified for CAT

 6a + 150 = 7k  6a + 147 + 3 = 7k  6a + 3 = Multiple of 7 Therefore, a = 3, 10, 17, … 3a + x = 50. So, the maximum value of a can be 16. So, a = 3 or 10

= a + p + q + x = 3 + (150 – 90) + 41 = 104 If a = 10, then k = 30 So, 2a + r = 4k = 120  r = 100 So, number of candidates who are at or above 80th percentile in P = a + p + q + x = 10 + (150 – 100) + 20 = 80 But number of candidates at or above 80th percentile in P is more than 100. So, number of candidates will be 104. Therefore, number of candidates who had to sit for the separate test for BIE = (400 – 104) + a = 296 + 3 = 299

43. Ans. A Explanation: Amongst those who have secured above 90th percentile, the candidates who have secured 80th or above percentile in only P are not eligible for AET, and so are required to appear in a separate test to be conducted by BIE. The number of such candidates is a, whose value is either 3 or 10.

Common explanation for questions 47–50: The given data can be tabulated as below: Scores

44. 60

S

F

C

1

2

1

2

1

3

0

Explanation: If x is a multiple of 5, a can be 10 only. So, x = 20. Also, (2a + r) : (2a + q) : (2a + p) = 4 : 2 : 1  (20 + r) : (20 + q) : (20 + p) = 4 : 2 : 1  20 + r = 4k, 20 + q = 2k and 20 + p = k 210 Since 6a + 150 = 7k, k = = 30 7 So, 20 + r = 4k = 120 or r = 100 Similarly, 20 + q = 2k = 60 or q = 40 Also, p = 30 – 20 = 10 The number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET = q + x = 40 + 20 = 60

3

3

2

4

4

3

1

1

5

2

3

6

1

7

1

1

Total

10

10

1

10

A and C had a total score of 7, with identical scores in all these parameters. So, it can only be 1, 2 and 4 or 3, 3, 1. Also, Zooma has a score of 17, and all three countries in the happy category had the highest score in exactly one parameter, it can only have a 7 in F, 6 in S and 4 in C as a score of 7 in S and 6 in C would be the scores of the other two countries and it cannot have a 7, 7 and 5 as there is no country which scored a 5 in C.

45. 170 Explanation: Refer to the previous solution. So, x = 20. For getting shortlisted for the AET for AIE, one should be at or above the 90th percentile overall and at or above 80th percentile in at least two subjects. So, number of such candidates = p + q + r + x = 150 + 20 = 170

47. Ans. 1

46. Ans. A Explanation: We know that 6a + 150 = 7k If a = 3, then k = 24 So, 2a + r = 4k = 96  r = 90 So, number of candidates who are at or above 80th percentile in P

Explanation: A and C have a score of 1 in either S or F or C. But there is no score of 1 in S and there is only 1 country with a score of 1 in C. So, the score of 1 is possible only in F. So, Amda’s score in F is 1.

P-20

CAT 2017 Solved Paper Slot 1 48. Ans. 6

Month 1

Explanation: As explained above, the score of Zooma in S is 6. 49. Ans. B

T1

T2

T3

T4

T5

No. of SE

4

3

2

1

0

No. of RE

1

1

2

3

4

After the said changes in team structure in subsequent months, the distribution of employees in the months 2 to 5 is as below:

Explanation: On the basis of score of Zooma, we can identify the scores of Benga and Dalma. The possible scores of Benga and Dalma are: Benga: S (5), C (6) and F (5). Total = 5 + 6 + 5 = 16. Dalma: S (7), C (3) and F (5). Total = 7 + 3 + 5 = 15. But they had same score. So, it can only be 15.

T1

T2

T3

T4

T5

No. of SE

1

4

2

2

1

No. of RE

3

1

2

2

3

No. of SE

0

1

4

3

2

No. of RE

4

3

1

1

2

No. of SE

0

1

2

5

2

No. of RE

4

3

2

0

2

Month 2

Month 3

50. Ans. B Explanation: If Benga scored 16 and Dalma scored 15, then the maximum possible values remaining are as below:

Month 4

Score

S

F

C

Month 5

3

3

2

3

No. of SE

0

0

2

4

4

4

3

1

0

No. of RE

4

4

2

0

1

5

1

1

0 51. Ans. B

The maximum score of another country can be S – 5, F – 5 and C – 3. Only one country can have a score of 13 as no other country can have a score of 5 in any parameter.

Explanation: The composition of T2 did not change from 3rd to 4th month, whereas the composition of T4 always changed.

Common explanation for questions 51–54: The challenging project is allotted to T1 in the first month, to T2 in the next month, and so on. Also, the team which is given challenging project has one more employee than the rest. Since there are a total of 21 employees, the team working on challenging project will have 5 employees, whereas others will have 4 employees each. Also, in the first month, T1 has 1 more SE than T2, T2 has 1 more SE than T3, and so on. Since there are total of 10 SEs, the distribution between them will be: 4 in T1, 3 in T2, 2 in T3, 1 in T4 and 0 in T5. So, the distribution of employees in the first month will be as given below:

52. Ans. A Explanation: As we can see from the above table, the number of SE in T1 and T5 for the third month are 0 and 2 respectively. 53. Ans. B Explanation: The number of challenging projects and standard projects in which an employee could have worked can be (5, 0), (4, 1), (3, 2), (2, 3), (1, 4) or (0, 5). The credit points in case of challenging project are 200, which are divided equally among the 5 employees.

P-21

Quantitative Aptitude Simplified for CAT

55. Ans. C

So, credit points earned by an employee in 200 case of challenging project = = 40 5 The same in case of standard project 100 = = 25 4 So, total points that can be earned by any employee can be: 5  40 + 0  25 = 200; Each decrease of challenging projects results in loss of 40 points, and each increase of standard project results in gain of 25 points. So, there is a net loss of 40 – 25, that is, 15 points. So, points earned in subsequent cases are: 185, 170, 155, 140 and 125. So, a score of 150 points is not possible for any employee.

Explanation: The persons who can be approached by only one individual are shown below as highlighted. 6

1

2

4

3

9

5

3

2

8

7

8

4

6

5

3

9

5

1

2

1

7

6

3

9

56. Ans. D Explanation: Since they are of the smallest heights, they cannot be reached by anyone, as per the conditions given in the data (condition II).

54. Ans. D

57. Ans. C

Explanation: Since Aneek secured 185 credits, he worked in four challenging projects and one standard project. Option A: The sequence in which Aneek could have worked in the five months could be – T1 in challenging project, T2 in challenging project, T3 in challenging project, T4 in challenging project) and T4 in standard project. So, this is a possible case. Option B: Aneek could have worked in T1 in challenging project, T2 in challenging project, T4 in standard project, T4 in challenging project and T5 in challenging project. This is also possible. Option C: Aneek could have worked in T2 in standard project, T2 in challenging project, T3 in challenging project, T4 in challenging project and T5 in challenging project. This is also possible. Option D: Aneek could have worked in T1 in challenging project in the first month. He can work in T1 or T5 in the second month. In either case, he cannot work in T3 without working in T2 first. If we assume that he worked in T3 in the first month, he could not have worked in four teams in the five months. Similarly, we can rule out the other possibilities for this option.

Explanation: In the fourth column, there are two heights: 2 and 1, which cannot be reached by anyone as per the conditions given. 58. Ans. C Explanation: First row does not have any individual who can be reached by 5 or more individuals. So, option A is not correct. Third row has all the individuals who are reachable. So, option B is also not correct. In row 2, individual at height 9 can be reached by only 3 persons. So, option D is also not correct. 59. Ans. C Explanation: If every city is to be connected to every city by a direct flight, then number of routes = 10C2 = 45 On any route, there has to be 4 flights: morning to and fro, and evening to and fro. So, minimum number of direct flights that can be scheduled = 4  45 = 180

P-22

CAT 2017 Solved Paper Slot 1 64. Ans. B

60. Ans. C

Explanation: The difference in time taken would be only due to time difference between M-B and A-N routes, which is 1 – 0.9 = 0.1 minute.

Explanation: Each of the three hubs will have a direct flight with each of the remaining 7 cities. So, there will be 21 routes. Also, each hub will connect with each other in 3 routes. So, there are total 24 routes and each route will have 4 flights: morning and evening ‘to and fro’. So, there will be a total of 4  24 = 96 flights to be scheduled to meet the conditions given.

65. Ans. 2 Explanation: Case I: There is no car on M-N route. Then, 2 cars go on A-M-B route and 2 on A-NB route. Travel time for the car on A-M-B route = 9 + 20.9 = 29.9 minutes A-N-B route = 21 + 9 = 30 minutes If one car on route A-M-B breaks the rule and follows A-M-N-B route, the travel time for that car will be 9 + 7 + 9 = 25 minutes So, this case is not possible, as following this route, it is possible for a car to reduce travel time be not following the police order. Case II: There is one car on M-N route. Then, there are further two possibilities: (a) 3 cars go on A-M-B route with one car later on following M-N route, or (b) 2 cars go on A-M-B route with one car later on following M-N route. In case of (a), time taken for car following AM-B route = 12 + 20.9 = 32.9 If one of the cars following A-M-B route, breaks the rule and follows M-N route, then time taken by that car = 12 + (7 + 1) + 12 = 32 Similar is the problem in case of (b) option. Therefore, two cars must be directed through M-N such that any car breaking the police order cannot reduce the travel time.

61. Ans. 40 Explanation: The number of routes between the cities of G1 = 3C2 = 3 The number of routes between A and any city in G2 = 1  3 = 3 The number of routes between B and any city in G3 = 1  2 = 2 The number of routes between C and any city in G4 = 1  2 = 2 Total number of routes = 10 Total number of flights = 4  10 = 40 62. Ans. 4 Explanation: The direct flights between A and any city in G2 would now not be available. However, the cities in G2 would now connect with G3 or G4. So, this does not lead to any reduction in flights. The reduction would be for the flights between A and C, which are now not available. There are a total of 4 flights between A and C, which would now not be available.

66. Ans. B

63. Ans. 2

Explanation: The minimum time would be taken when 3 cars go on A-M route, and later 2 cars divert to M-N route, and one car goes along A-N-B route. Time taken for the car on A-M-B route = 12 + 20 = 32 minutes Time taken for the car on A-M-N-B route = 12 + 8 + 12 = 32 minutes Time taken for the car on A-N-B route = 20 + 12 = 32 minutes

Explanation: There are four cars and the time taken through each route is almost the same. So, two cars should go via A-M-B route and the other two through A-N-B route. In case three cars are directed to go through any of the routes, one of the three cars can break the police order and reduce its travel time, which should not be possible.

P-23

Quantitative Aptitude Simplified for CAT

CAT 2017 Solved Paper Slot 2 4.

Quantitative Aptitude (QA) 1.

The numbers 1, 2, …, 9 are arranged in a 3 x 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value. If the top left and the top right entries of the grid are 6 and 2 respectively, then the bottom middle entry is __________ .

2.

3.

(A) (B) (C) (D)

In a 10 km race, A, B and C, each running at uniform speed, get the gold, silver and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

5.

Bottle 1 contains a mixture of milk and water in 7 : 2 ratio and Bottle 2 contains a mixture of milk and water in 9 : 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3 : 1 ratio? (A) (B) (C) (D)

Arun drove from home to his hostel at 60 miles per hour. While returning home, he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then total distance travelled by him is

Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in domestic market. If the remaining 8840 shirts are left for exports, then the number of shirts produced in the factory is (A) (B) (C) (D)

27 : 14 27 : 13 27 : 16 27 : 18

6.

P-24

55 miles 60 miles 65 miles 70 miles

13600 13000 13400 14000

The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of

CAT 2017 Solved Paper Slot 2 the original 22, then the average height, in inches, of the remaining 20 toddlers is

7.

8.

9.

11. If a, b and c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio of 2 : 1, then which one of the following is a possible value of (a + b + c)?

(A) 30 (B) 28 (C) 32 (D) 26 The manufacturer of a table sells it to a wholesale dealer at a profit of 10%. The wholesale dealer sells the table to a retailer at a profit of 30%. Finally, the retailer sells it to a customer at a profit of 50%. If the customer pays Rs. 4290 for the table, then its manufacturing cost (in Rs.) is

(A)

(B) 205 (C) 207 (D) 210 12. A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B?

(A) 1500 (B) 2000 (C) 2500 (D) 3000 A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed, then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open, then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open, then in how many hours the full tank becomes half-full?

(A)

364

(B) 378 (C) 380 (D) 388 13. Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs. 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal’s share, in rupees, is

(A) 20 (B) 30 (C) 40 (D) 45 Mayank buys some candies for Rs. 15 a dozen and an equal number of different candies for Rs. 12 a dozen. He sells all for Rs. 16.5 a dozen and makes a profit of Rs. 150. How many dozens of candies did he buy altogether?

(A)

100

(B) 200 (C) 300 (D) 400

(A) 50 (B) 30 (C) 25 (D) 45 10. In a village, the production of food grains increased by 40% and the per capita production of food grains increased by 27% during a certain period. The percentage by which the population of the village increased during the same period is nearest to (A)

201

14. Consider three mixtures – the first having water and liquid A in the ratio 1 : 2, the second having water and liquid B in the ratio 1 : 3, and the third having water and liquid C in the ratio 1 : 4. These three mixtures of A, B and C, respectively, are further mixed in the proportion 4 : 3 : 2. Then the resulting mixture has (A)

The same amount of water and liquid B

(B) The same amount of liquids B and C

16

(C) More water than liquid B

(B) 13

(D) More water than liquid A

(C) 10 (D) 7

P-25

Quantitative Aptitude Simplified for CAT

15. Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq. cm) of a square with AC as one side is

(B) 1785 (C) 1875 (D) 1877

(A) 3 2 (B) 3 (C) 4 (D)

22. If x is a real number such that log35 = log5 (2 + x), then which of the following is true? (A) (B) (C) (D)

3

16. The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is (A) (B) (C) (D)

23. Let f(x) = x2 and g(x) = 2x, for all real x. Then the value of f(f(g(x)) + g(f(x))) at x = 1 is (A) (B) (C) (D)

1300 1340 1480 1520

(A) (B) (C) (D)

–5 –6 –7 –8

25. If 9 (A) (B)

18. ABCD is a quadrilateral inscribed in a circle with centre O. If COD = 120° and BAC = 30°, then the value of BCD (in degrees) is __________ .

(C) (D)

19. If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is __________ .



2 1



x

1 2

– 22x–2 = 4x – 32x–3, then x is

3 2 2 5 3 4 4 9

27. Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

cm, then

the area, in sq. cm, of the triangle ABC is __________ .

If the sum of the numbers in the new sequence is 450, then a5 is __________ .

21. If the product of three consecutive positive integers is 15600, then the sum of the squares of these integers is (A)

1 2 3 4

26. If log (2a × 3b × 5(C) is the arithmetic mean of log (22 × 33 × 5), log (26 × 3 × 57) and log (2 × 32 × 54), then a equals __________ .

20. Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC and CA is 4

16 18 36 40

24. The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x – (a + 5) = 0 is

17. The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has equation y = 3x + c, then c is (A) (B) (C) (D)

0 9. Now, pairs of (a, b) possible are (10, 90), (12, 36) and (18, 18).

 32x–3 = 22x–3, which is possible only when power is zero, That is, 2x – 3 = 0  x =

450 = 90, 5

which is also the middle number of these 5 numbers. Therefore, the 5th number in the new sequence = 94, which is given to be equal to 2a3. So, a3 = 47. Therefore, a5 = 51

Explanation: Given equation can be written as: 

1 [log (29 × 36 × 512)] = log (23 × 32 × 54) 3

As per the question, log (2a × 3b × 5c) = log (23 × 32 × 54) So, a = 3

25. Ans. A

(9x)

1 [log (22 × 33 × 5)(26 × 3 × 57) 3

3 satisfies the equation. 2

3 is the solution of the equation. 2

P-36

CAT 2017 Solved Paper Slot 2  p + q + r = 2, where p, q, r  0 Number of solutions of the equation = n+r–1Cr–1 = 2+3–1C3–1 = 4C2 = 6

Or, x =

So, no solution exists for x >

7 . 2

5 7  x  , equation becomes 2 2

For

30. Ans. 50.

2x – 5 + 7 – 2x = 2 Or, 2 = 2.

Explanation: One of the given digits is not used and the sum of the remaining digits must be a multiple of 3 because the required number is divisible by 6. So, the digits to be used can be: (2, 3, 4, 6) or (0, 2, 3, 4) or (0, 2, 4, 6). Using (2, 3, 4, 6), numbers formed should end with even number. So, unit’s digit can be 2 or 4 or 6 and the other 3 digits can be arranged in 3! ways. So, total number of cases = 3 × 3! = 18 Using (0, 2, 3, 4), the unit’s digit can be 0 or 2 or 4. If unit’s digit is 0, then other 3 digits can be arranged in 3! ways. So, total number of cases = 3! × 1 = 6 If unit’s digit is 2 or 4, then leftmost digit can be filled in 2 ways, and middle two places in 2 × 1 ways. So, number of cases = 2 × 2 × 2 × 1 = 8 Using (0, 2, 4, 6), all numbers formed will be even numbers. Leftmost place can be filled in 3 ways, next one in 3 ways, next 2 ways and unit’s place in 1 way. So, total number of ways = 3 × 3 × 2 × 1 = 18 Total number of ways = 18 + 6 + 8 + 18 = 50

5 7 x is a feasible solution range. 2 2 5 For x < , equation becomes 2

So,

5 – 2x + 7 – 2x = 2 Or, x =

5 2

5 . 2 5 7 and x = satisfies the Also, note that x = 2 2

So, no solution exists for x


7 2

7 , equation becomes 2

Therefore, a1 + a2 + a3 + … + a100 =

2x – 5 + 2x – 7 = 2

P-37

1 1 1 1 + + +…+ 25 299  302 58 8  11

Quantitative Aptitude Simplified for CAT

For Thin Crust pizza,



1 1 1 1 1 1 1  1 1  1 1 1             ...    3  2 5  3  5 8  3  8 11  3  299 302 



1 3

1   1 1  1 1  1 1   1   2  5    5  8    8  11   ...   299  302          



1 1 1  1  300  50 25      3  2 302  3  2  302  302 151

Total number ordered by party 1 = 0.6  120 = 72 Total number ordered by party 2 = 0.55  120 = 66 So, the number ordered by party 3 = 300 – 72 – 66 = 162

For Normal Cheese pizza, Total number ordered = 0.52  800 = 416 Total number ordered by party 3 = 0.65  560 = 364 Total number ordered by party 2 = 0.3  120 = 36 So, the number ordered by party 1 = 416 – 364 – 36 = 16 If party 1 ordered 72 thin crust, then it ordered 120 – 72 = 48 deep dish If party 2 ordered 66 thin crust, then it ordered 120 – 66 = 54 deep dish

Data Interpretation and Logical Reasoning (DILR) Common explanation for questions 35–38: The number of pizzas ordered by party 3 = 70% of 800 = 560 Number of pizzas ordered by each of party 1 and 2 = 120 Total number of thin crust pizzas ordered = 0.375  800 = 300 So, total number of deep dish pizzas ordered = 500

We can now develop the following table for the number of pizzas ordered. Thin Crust

Deep Dish

Total

Normal Cheese

Extra Cheese

Normal Cheese

Extra Cheese

Party 1

X

72 – X

16 – X

32 + X

120

Party 2

Y

66 – Y

36 – Y

18 + Y

120

Party 3

Z

162 – Z

364 – Z

34 + Z

560

Total

300

500

800

= 0.65  560 = 364 The number ordered by party 2 = 0.3  120 = 36 So, the number ordered by party 1 = 416 – 364 – 36 = 16

35. Ans. B Explanation: Total number of thin crust pizzas ordered = 0.375  800 = 300 Of these, the number ordered by party 1 = 0.6  120 = 72 The number ordered by party 2 = 0.55  120 = 66 So, the number ordered by party 3 = 300 – 72 – 66 = 162

37. Ans. B Explanation: T-NC = 50% of NC = 50% of 36 = 18 In the above table, Y = 18 So, required difference between T-EC and D-EC for party 2 = (66 – Y) – (18 + Y) = 48 – 2Y = 48 – 2×18 = 12

36. Ans. C Explanation: Total number of NC ordered = 0.52  800 = 416 Of these, the number ordered by party 3

P-38

CAT 2017 Solved Paper Slot 2 38. Ans. A Explanation: Cost of T-EC = Rs. 500 So, cost of D-EC = Rs. 550 Cost of T-NC = Cost of D-NC = D-EC =

3 of price of 5

3  550 = Rs. 330 5

E5

5

30

E6

7

3

35

2

9

21 101

E7

4

16

30

5

5

41

Total

15

76

78

20

43

60

It is also given that before the change, E1 = E4 + 6. From the table above, we see there were 31 students in E1 before the change. So, 31 – 6 = 25 students must have been there in E4 before the said change. For this, both the blank cells in the row headed by E4 must be 1 each.

Since 25% of NC delivered to party 1 were of Deep Dish type, so 16 – X = 25% of 16 = 4 Or, X = 12 Total bill of party 1 = 330  16 + 500  (72 – 12) + 550  (32 + 12) = Rs. 59480

Common explanation for questions 39–42: The table given in the data is redrawn below.

E1

E2

E3

E4

E5

E6

Total

E1

9

5

10

1

4

2

31

E2

0

34

8

0

2

2

46

E3

2

6

25

2

35

1

3

2

4

25

E1

E2

E3

E4

E5

E6

Total

E4

9

5

10

1

4

2

31

E5

5

34

8

2

2

46

E6

7

3

6

25

2

35

E7

4

16

30

E4

3

2

4

23

Total

16

76

78

E5

5

35

E6

7

3

It is given that before the change process, E4 had 2 more students than E6. Since E4 has 25 students, E6 will have 23 students, for which both the blank cells in the row pertaining to E6 will be 1 each.

E1 E2 E3

2

14 30 2

9

21 101

E7

4

16

30

5

5

41

Total

15

76

78

20

43

60

It is given that the number of students in E2 increased by 30 after the change process. After the change, the number of students in E2 = 76. The sum of elements in the row headed by E2 is 46. This sum can be 46 or 47 or 48, depending upon the values taken by blank cells. If this sum is more than 46, then number of students in E2 after change would be more than 76, which is not possible. So, number of students shifting from E2 is exactly equal to 46. So, the blank cells in the row pertaining to E2 have 0 values each. E1

E2

E3

E4

E5

E6

Total

E1

9

5

10

1

4

2

31

E2

0

34

8

0

2

2

46

E3

2

6

25

2

35

3

2

4

23

E4

14

14

1 30

35

2

9

21

5

5

41

101

20

44

60

E1

E2

E3

E4

E5

E6

Total

E1

9

5

10

1

4

2

31

E2

0

34

8

0

2

2

46

E3

2

6

25

2

35

E4

1

3

2

4

25

E5

14

5

1 30

35

E6

1

7

3

1

2

9

23

E7

4

16

30

5

5

41

101

Total

17

76

78

21

44

60

It is given that after the reshuffle, the number of students in E4 is 3 more than those in E1. Currently, the gap between the number of students = 21 – 17 = 4. To decrease the gap, number of

P-39

Quantitative Aptitude Simplified for CAT

students shifting from E5 to E1 should be 1 and remaining blank cells in E4 column should be 0. E1

E2

E3

E4

E5

E6

Total

E1

9

5

10

1

4

2

31

E2

0

34

8

0

2

2

46

E3

2

6

25

0

2

35

E4

1

3

2

14

1

4

25

E5

1

5

0

30

E6

1

7

3

1

2

9

23

E7

4

16

30

5

5

41

101

Total

18

76

78

21

44

60

=

61  23  100 = 165% 23

The next highest percentage change is in E3 given by: =

79  36  100 = 119.4% 36

42. Ans. A Explanation: The students of E1 after the change is less than 20. So, all the students who shifted from E1 will shift back. Since 5 students shifted from E1 to E2. If they come back to E1, then E2 will have 76 – 5 = 71 students. Similarly, we can do for other electives. The revised table is:

36

Since there are a total of 300 students and the total students in the last column add up to 297. So, the remaining 3 blank cells will surely have value of 1 each. The table finally looks as shown below.

E1

E2

E3

E4

E5

E6

Total

E1

31

0

0

0

0

0

31

E2

0

34

8

0

2

2

46

E3

2

6

25

0

1

2

36

E1

E2

E3

E4

E5

E6

Total

E4

1

3

2

14

1

4

25

E1

9

5

10

1

4

2

31

E5

1

5

1

0

30

1

38

E2

0

34

8

0

2

2

46

E6

1

7

3

1

2

9

23

E3

2

6

25

0

1

2

36

E7

4

16

30

5

5

41

101

E4

1

3

2

14

1

4

25

Total

40

71

69

20

41

59

300

E5

1

5

1

0

30

1

38

E6

1

7

3

1

2

9

23

E7

4

16

30

5

5

41

101

Total

18

76

79

21

45

61

300

The descending order of the electives is: E2, E3, E6, E5, E1, E4. 43. Ans. C Explanation: Since gold coins are also distributed equally, let each gets x gold coins. Total value of all the assets = 70 + 50 + 3  30 + 3x = 210 + 3x So, each one gets 70 + x. One of the persons will definitely get house of Rs. 50 lakhs. Then, that person won’t get any of the three flats, because in that case, the assets with her will be 50 + 30 + x, which is not possible. So, the person who gets house of Rs. 50 lakhs will get Rs. 20 lakhs from bank deposits. So, there will be one person who gets 2 flats, and so she will get Rs. 10 lakhs from bank deposits. Finally, the third one will

39. Ans. C Explanation: The electives which had a decrease in enrolments after the change process are E1 and E4. 40. Ans. D Explanation: The correct sequence is: 18, 76, 79, 21, 45, 61. 41. Ans. D Explanation: The percentage change in E6

P-40

CAT 2017 Solved Paper Slot 2 get 1 flat and Rs. 40 lakhs from bank deposits. So, Person 1 = 1 house + Rs. 20 lakhs + gold Person 2 = 2 flats + Rs. 10 lakhs + gold Person 3 = 1 flat + Rs. 40 lakhs + gold Since Neeta received the least amount in bank deposits and Geeta received the highest amount in bank deposits, Seeta received Rs. 20 lakhs in bank deposits.

So, Geeta got the remaining Rs. 20 lakhs in bank deposits.

Common explanation for questions 47–50: We can collate the given data in the form of table as shown below:

44. Ans. 2

Light Repair

Moderate Repair

Extensive Repair

1

NO

YES

NO

2

Explanation: Referring to previous solution, since Neeta received the least amount in bank deposits, she would have received Rs. 10 lakhs in bank deposits and hence she received 2 flats.

3

NO NO

4

45. Ans. B Explanation: Let the number of gold coins distributed among Neeta, Seeta and Geeta were 2x, 3x and 4x respectively. Total gold coins = 9x Seeta got

Dorm

YES

NO

NO

5

NO

YES

NO

6

YES

NO

NO

7

NO

NO

YES

8

YES

NO

NO

9

NO

YES

NO

10

NO

47. Ans. D

2 1  rd of total assets, and also 6 3

Explanation: From the above table, we see that Dorm 1 needs moderate repair. So, option A is not correct. Dorm 5 needs moderate repair and hence would cost Rs. 3 crores or Rs. 4 crores. So, option B is also not correct. Dorm 7 needs extensive repair. So, option C is also not correct. Dorm 10 may need light repair or extensive repair. So, option D may not be necessarily true.

3 1  rd of gold coins. 9 3 1 So, she got rd of assets other than gold 3

also. Assets other than gold = Rs. 210 lakhs So, she got Rs. 70 lakh assets + Gold. She cannot get 3 flats. So, she got a house and also Rs. 20 lakhs in bank deposits. Since Geeta didn’t get Rs. 30 lakhs, the one who got Rs. 30 lakhs in bank deposits must be Neeta. Also, she cannot get all three flats, as her assets are less than that of Seeta. So, Geeta got all three flats. So, Neeta got Rs. 30 lakhs + 2x gold coins and Seeta got Rs. 70 lakhs + 3x gold coins. Ratio of their assets = 1 : 2. So, (30 + 2x) : (70 + 3x) = 1 : 2 Or, x = 10 So, total number of gold coins = 9x = 90

48. 19 crores Explanation: Total cost of odd numbered dorms = Cost of 4 dorms in moderate category + Cost of dorm 7 (which is the costliest repair) Cost of dorms in moderate category = 3 dorms requiring 3 crores each + 1 dorm requiring 4 crores each = 9 + 4 = Rs. 13 Cost of repairing dorm 7 = Rs. 6 crores Total cost = Rs. 19 crores

46. Ans. 20 Explanation: Since Neeta got Rs. 30 lakhs in bank deposits and Seeta got Rs. 20 lakhs in bank deposits.

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Quantitative Aptitude Simplified for CAT

49. Ans. 3

50. Ans. D

Explanation: If any dorm of women requires light repair, then the maximum cost of that dorm can be Rs. 2 crores. In that case, rest of the 3 dorms require a cost of Rs. 18 crores, and so average cost is Rs. 6 crores. But we do not have 3 dorms requiring Rs. 6 crores cost. So, none of the women’s dorms requires light repairs. If none of the women’s dorms required repair cost of Rs. 3 crores, then 4 dorms would require cost of 4 + 5 + 6 + 6 = Rs. 21 crores, but we need total cost to be Rs. 20 crores. So, at least one dorm requiring cost of Rs. 3 crores must be there. If there are two dorms requiring Rs. 3 crores cost, then remaining 2 dorms would require cost of Rs. 14 crores, averaging Rs. 7 crores, which is impossible. So, the only possibility is 1 dorm requiring cost of Rs. 3 crores. So, the rest of the 3 dorms require cost of Rs. 17. If we include cost of Rs. 4 crores, then the remaining 2 dorms require cost of Rs. 13, which is again not possible. So, the only possibility is: 3 + 5 + 6 + 6 = Rs. 20 crores. So, all the dorms requiring extensive repairs belong to women. Now, there are 3 dorms needing extensive repairs, of which dorm 7 is already known to be requiring extensive repair. The dorms whose status is not known are 2, 4 and 10. Of these, two would require extensive repairs, because total number of dorms requiring extensive repairs is 3. Dorms 2 and 4 both cannot be the ones which require extensive repairs, because only one of the dorms from 1 to 5 is a women’s dorm. So, dorm 10 has to be women’s dorm. For other dorms given in other options, we are not sure of whether they belong to women’s or not. Since one of the dorms from 1 to 5 require extensive repairs and that belongs to women’s, dorms 1, 3 and 5 (needing moderate repairs) cannot belong to women (because only one dorm from 1 to 5 is a women’s dorm). So, dorm 9 needing moderate repairs is a women’s dorm, and the woman’s dorm requiring moderate repair would cost Rs. 3 crores. So, dorm 9 requires repairs costing Rs. 3 crores.

Explanation: From previous solution, we can say that dorm 10 is a women’s dorm. Common explanation for questions 51–54: From the information given, we can draw the following table. Cup Number

Location of Tea

Rank

1 2

Even number; Minimum Rating

3

2x (not highest)

4 5

x

6

Himachal

Tea from Ooty gets the highest rating. Only two cups got the ratings in even numbers. So, one of them has to be cup 3. If x = 1, then rating of cup 3 is 2, which is not possible because rating of cup 2 is lowest even number. So, x can be 3 or 5. If x = 5, then rating of cup 3 is 10, which is also not possible as it cannot be the highest. So, x has to be 3, and so rating of cup 3 is 6. Also, since rating of cup 2 is the minimum, it has to be less than the rating of cup 5, and so its rating is 2. Cup Number

Location of Tea

Rank

1

Assam > Wayanad > Munnar Since lowest rating is 2, so no tea got rating of 1. Also, out of 6 teas, 2 teas have got even rating, the rest of the 4 teas have got odd

P-42

CAT 2017 Solved Paper Slot 2 ratings starting from 3. So, the odd numbered ratings are 3, 5, 7 and 9.

The only option that may be true is B. In that case, table will be as shown below.

51. Ans. 7 Explanation: The second highest rating was 7. 52. Ans. 4 Explanation: Cup 1 cannot be of Ooty, as its rating is less than 6. So, cup 4 is the only possibility from Ooty. Cup

Location of Tea

Rank

1

5

2

2

3

6

4

Ooty

5 6

Cup

Location of Tea

Rank

1

Darjeeling

5

2

Munnar

2

3

Assam

6

4

Ooty

9

5

Wayanad

3

6

Himachal

7

55. Ans. C Explanation: The pieces are placed as shown below:

9 3

Himachal

c5

g5

7 a3

53. Ans. B

g3 c2

Explanation: Tea from Munnar did not get minimum rating, and we already know that Ooty > Assam > Wayanad > Munnar, then the ratings of Assam, Wayanad and Munnar will be 6, 5 and 3 respectively. So, ratings of Wayanad = 5

g1 Queen is at c5. It can attack pieces at a3, c2, g1 and g5. So, 4 pieces are under attack. 56. Ans. D Explanation:

54. Ans. B

h8 d7

Explanation: If cups containing teas from Wayanad and Ooty had consecutive numbers, then it can be cup 3 or 5. But if it is cup 3, then its rating will be 6, which is not possible, because rating of Assam is higher than that of Wayanad. So, Wayanad must be in cup 5. Now, cup 5 cannot be from Assam. Option A is not correct. Also, tea from Wayanad has rating 3. Option C is also not correct. Since rating of Munnar is less than that of Wayanad, the rating of Munnar will be 2, and so tea from Darjeeling cannot be minimum. Option D is also incorrect.

h7

b4 a3 a1 If queen is at f8, then it can attack h8 and b4. So, 2 pieces are under attack. If queen is at a7, then it can attack a3 and d7. So, 2 pieces are under attack. If queen is at c1, then it can attack a1 and a3. So, 2 pieces are under attack.

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Quantitative Aptitude Simplified for CAT

If queen is at d3, then it can attack h7, a3 and d7. So, 3 pieces are under attack. So, the maximum number of pieces are under attack when the queen is at d3.

= 7 (in the row) + 7 (in the column) + 7 (one diagonal) + 6 (another diagonal) = 27 So, number of squares not under attack = 64 – 27 – 1 = 36

57. Ans. C

Common explanation for questions 59–62:

Explanation:

Based on the information given, row numbers 1 to 20 have extra charges except for middle seat (remember for rows 1, 12 and 13, it charges Rs. 1000 extra for any of the seats).

h8 d7

h7

J, A and B must be in aisle seat so that the sum of charges will be Rs. 4600, and in row numbers 10, 11 and 12 as they paid different amounts.

b4

Row No.

a3

a1 Queen cannot be at any position in column a, b, d or h, in the first place, so as to meet the condition in the question. In case of column c, the queen will always be able to attack in any row number. In case of column e, only e2 is the position where the queen cannot attack any of the pieces. Similarly, checking for other positions, we find that there are only 4 positions: e2, f2, g2 and g5, from where the queen cannot attack any of the pieces.

A

C

D

10

J

M

11

A

400  1 = 400

12

B

1000  1 = 1000

E

F

Extra Charges 500  2 = 1000

K/G

1000  2 = 2000

20

P

200  1 = 200

21

T

No extra charge

13

58. Ans. C

B

G/K

G and K can interchange their positions and also can be in row 1 or 12. Moreover, they need not be in window seat. They may be in aisle seat.

Explanation: In the following, queen is at d5.

59. Ans. A Explanation: Manik was sitting in row 10, as can be seen in the table above.

d5

60. Ans. C Explanation: Jayanta paid Rs. 500 extra for his choice of seat. 61. Ans. D Explanation: Gargi paid Rs. 1000 extra for her choice of seat.

Now, the shaded positions are under attack. The number of squares which are under attack

P-44

CAT 2017 Solved Paper Slot 2 65. Ans. 15

62. Ans. D Explanation: Tapesh sat on seat number 21, which had no extra charges.

Explanation: If the original scan is TIMTRL, then any two will be chosen and their positions will be interchanged in 6C2  1 = 15 ways (we have not counted those cases where the scans were chosen but their positions were not interchanged). This includes choosing T and T and their positions interchanged, but that does not lead to a new case. So, total number of interchanged positions allowed = 14 There will also be 1 case where there is no interchange at all, which is also obviously allowed. So, total number of cases = 14 + 1 = 15

63. Ans. 11 Explanation: Exactly one scan cannot be out of order. There are two cases possible: either all scans are in correct order, or exactly two are interchanged. If all scans are in correct order, number of ways = 1 If exactly two are interchanged, then these two can be chosen from five in 5C2 = 10 ways. Total number of ways = 11 64. Ans. C Explanation: Let the original scan be TIMRL. The number of cases if it is the original scan =1 To find number of cases in case it is not original scan, then TI, IM, MR and RL each can interchange in 1 way each. So, that makes 4 ways. Further, TI and MR can interchange together, which is 1 way. Also, IM and RL can also interchange, which is 1 way. Finally, TI and RL can interchange, which is 1 way. Total number of ways = 8

66. Ans. C Explanation: The original scan sequence is LRLTIM. One of the sequences is the case of no change. It a pair of letters flips, then there is shifting of one place only, and all such shifts are allowed. So, the shifts allowed are: LR, RL, LT, TI, IM, (LR, LT), (LR, TI), (LR, IM), (RL, TI), (RL, IM), (LT, IM), (LR, LT, IM). So, a total of 13 scan sequences are possible.

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Quantitative Aptitude Simplified for CAT

CAT 2018 Solved Paper Slot 1 Quantitative Aptitude (QA) 1.

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio (A) (B) (C) (D)

2.

4p If log1281 = p, then 3   is equal to 4p (A)

log68

(B) log616 (C) log28 (D) log416

17 : 25 19 : 24 18 : 25 21 : 25

5.

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

6.

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be (A) (B) (C) (D)

3.

4.

93 96 23 26

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

(A) (B) (C) (D)

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25 26 27 28

CAT 2018 Solved Paper Slot 1 7.

If x is a positive quantity such that 2 x  3log5 2 , then x is equal to (A)

12. Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

log58

3 5 5 (C) 1 + log3 3 (D) log59

(B) 1 + log5

8.

(A)

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be (A)

(B) (C) (D)

13. John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs. __________ .

192 3

(B) 164 3 (C) 248 3 (D) 188 3 9.

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is (A) (B) (C) (D)

14. In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

16 20 22 26

10. The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is (A)

7 33 6 19 1 4 1 5

(A)

13

(B)

11

(C)

12

(D)

14

15. Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

3

(B) 4

(A)

(C) 5

(B) 36

(D) 6

(C) 40

32

(D) 45

11. If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals__________ .

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Quantitative Aptitude Simplified for CAT

16. Given that x2018y2017 = the value of x2 + y3 is (A)

31 4

(B)

33 4

(C)

35 4

(D)

37 4

20. How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

1 and x2016y2019 = 8, 2

21. Let f(x) = min{2x2, 52 − 5x}, where x is any positive real number. Then the maximum possible value of f(x) is __________ . 22. The number of integers x such that 0.25 < 2x < 200, and 2x +2 is perfectly divisible by either 3 or 4, is __________ . 23. In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

17. In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct? (A)

N  200

(B) 201  N  242

1

(C) 243  N  252

(A)

  2   3 3 

(B)

  2   4

(C)

  2   4 3 

(D)

  2   6

(D) N  253

1

18. A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts? (A)

1

1

24. Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is __________ .

96

(B) 86 (C) 98 (D) 84 19. Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD? (A)

25. If u2 + (u − 2v − 1)2 = −4v(u + v), then what is the value of u + 3v?

24, 10

(A)

0

(B)

1 4

(C)

1 2

(D)



(B) 24, 12 (C) 25, 9 (D) 25, 10

P-48

1 4

CAT 2018 Solved Paper Slot 1 26. If log2(5 + log3 a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to (A)

31. When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

32

(B) 40 (C) 59 (D) 67 27. In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is (A)

(A)

(B) 18 (C) 20 (D) 22

24 3

32. Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

(B) 18 3 (C) 32 3 (D) 12 3 28. Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is (A)

5 2

(B)

1 6

(C)

3 2

(D)

3 6

16

(A)

4:9

(B) 2 : 5 (C) 3 : 8 (D) 1 : 3 33. Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is __________ . 34. While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is __________ .

29. Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Data Interpretation and Logical Reasoning (DILR)

30. A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft The tip of the cone is cut off with a plane which is parallel to the base and 9 ft. from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is __________ .

Directions for questions 35–38: Read the information given below and answer the questions that follow. Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low. 1.

P-49

Contamination levels at three pumps among P1 – P5 were recorded as high.

Quantitative Aptitude Simplified for CAT

2.

P6 was the only pump among P1 – P10 where the contamination level was recorded as low.

3.

P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.

4.

High contamination levels were not recorded at any of the pumps P16 – P20.

5.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

(B) Contamination levels at P10 and P14 were recorded as the same. (C) Contamination levels at P13 and P17 were recorded as the same. (D) Contamination level at P14 was recorded to be higher than that at P15. Directions for questions 39–42: Read the information given below and answer the questions that follow. An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

35. Which of the following MUST be true? (A)

The contamination level at P10 was recorded as high.

(B) The contamination level at P12 was recorded as high. (C) The contamination level at P13 was recorded as low.

39. In how many different ways can the ATM serve a customer who gives 500 rupee notes as her preference?

(D) The contamination level at P20 was recorded as medium.

40. If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences?

36. What best can be said about the number of pumps at which the contamination levels were recorded as medium? (A)

More than 4

(B) At most 9 (C) Exactly 8

41. What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?

(D) At least 8 37. If the contamination level at P11 was recorded as low, then which of the following MUST be true? (A)

The contamination level at P18 was recorded as low.

(A) (B) (C) (D)

(B) The contamination level at P12 was recorded as high. (C) The contamination level at P14 was recorded as medium.

42. What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?

(D) The contamination level at P15 was recorded as medium. 38. If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE? (A)

13 12 16 10

(A) (B) (C) (D)

Contamination levels at P11 and P16 were recorded as the same.

P-50

750 800 900 1400

CAT 2018 Solved Paper Slot 1 (C) Q3

Directions for questions 43–46: Read the information given below and answer the questions that follow.

(D) Q4 45. During which quarter was the percentage decrease in sales from the previous quarter’s sales the highest?

The multi-layered pie-chart below shows the sales of LED television sets for a big retail electronics outlet during 2016 and 2017. The outer layer shows the monthly sales during this period, with each label showing the month followed by sales figure of that month. For some months, the sales figures are not given in the chart. The middle-layer shows quarter-wise aggregate sales figures (in some cases, aggregate quarter-wise sales numbers are not given next to the quarter). The innermost layer shows annual sales. It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December) of that year.

(A)

Q2 of 2016

(B) Q1 of 2017 (C) Q2 of 2017 (D) Q4 of 2017 46. During which month was the percentage increase in sales from the previous month’s sales the highest? (A)

March of 2016

(B) October of 2016 (C) March of 2017 (D) October of 2017 Directions for questions 47–50: Read the information given below and answer the questions that follow. 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites: 1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. 2. The number of satellites serving all three of B, C, and S is 100. 3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.

43. What is the percentage increase in sales in December 2017 as compared to the sales in December 2016? (A)

The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

22.22

(B) 28.57 (C) 38.46

47. What best can be said about the number of satellites serving C?

(D) 50.00 44. In which quarter of 2017 was the percentage increase in sales from the same quarter of 2016 the highest? (A)

(A)

Cannot be more than 800

(B) Must be at least 100 (C) Must be between 450 and 725

Q1

(D) Must be between 400 and 800

(B) Q2

P-51

Quantitative Aptitude Simplified for CAT

48. What is the minimum possible number of satellites serving B exclusively? (A)

100

(B) 200 (C) 250

5.

Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.

6.

Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.

7.

Daisy minors in Operations.

51. Who are the students from the institute Z?

(D) 500

(A)

49. If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S? (A)

(B) Bandita and Chitra (C) Chitra and Daisy

Exactly 475

(D) Adriana and Daisy

(B) At most 475

52. Which subject does Deb minor in?

(C) At least 475

(A)

(D) No conclusion is possible based on the given information

Finance

(B) Marketing (C) Operations

50. If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE? (A)

Adriana and Bandita

(D) Cannot be determined uniquely from the given information 53. Which subject does Amit major in?

All 1600 satellites serve B or C or S

(B) The number of satellites serving B is more than 1000

(A)

(C) The number of satellites serving C cannot be uniquely determined

(C) Operations

(B) Marketing (D) Cannot be determined uniquely from the given information

(D) The number of satellites serving B exclusively is exactly 250

54. If Chitra majors in Finance, which subject does Bandita major in?

Directions for questions 51–54: Read the information given below and answer the questions that follow.

(A)

Three students are from X, three are from Y, and the remaining two students, both female, are from Z.

2.

Both the male students from Y minor in Finance, while the female student from Y majors in Operations.

3.

Only one male student majors in Operations, while three female students minor in Marketing.

4.

One female and two male students major in Finance.

Finance

(B) Marketing

Adriana, Bandita, Chitra, and Daisy are four female students, and Amit, Barun, Chetan, and Deb are four male students. Each of them studies in one of three institutes - X, Y, and Z. Each student majors in one subject among Marketing, Operations, and Finance, and minors in a different one among these three subjects. The following facts are known about the eight students: 1.

Finance

(C) Operations (D) Cannot be determined uniquely from the given information Directions for questions 55–58: Read the information given below and answer the questions that follow. Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees: 1.

P-52

The numbers of bureaucrats in the research and teaching committees are equal, while the

CAT 2018 Solved Paper Slot 1 number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. 2.

3.

they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the first or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.

59. What is the minimum number of different numerals needed to fill a 3 × 3 square matrix? 60. What is the minimum number of different numerals needed to fill a 5 × 5 square matrix?

60% of the politicians are in the administration committee, and 20% are in the teaching committee.

61. Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5 × 5 matrix?

55. Based on the given information, which of the following statements MUST be FALSE? (A)

The size of the research committee is less than the size of the administration committee

(A)

(B) In the teaching committee the number of educationalists is equal to the number of politicians

9

(B) 16 (C) 25 (D) 4

(C) The size of the research committee is less than the size of the teaching committee

62. Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5 × 5 square matrix?

(D) In the administration committee the number of bureaucrats is equal to the number of educationalists 56. What is the number of bureaucrats in the administration committee?

(A)

57. What is the number of educationalists in the research committee?

(C) 9

(B) 16 (D) 4

58. Which of the following CANNOT be determined uniquely based on the given information? (A)

25

Directions for questions 63–66: Read the information given below and answer the questions that follow.

The size of the teaching committee

A company administers a written test comprising of three sections of 20 marks each – Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited.

(B) The size of the research committee (C) The total number of educationalists in the three committees (D) The total number of bureaucrats in the three committees Directions for questions 59–62: Read the information given below and answer the questions that follow. You are given an n × n square matrix to be filled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if

P-53

Quantitative Aptitude Simplified for CAT

Ten candidates appeared for the written test. Their marks in the test are given in the table below. Some marks in the table are missing, but the following facts are known:

63. Which of the following statements MUST be true? 1.

Jatin’s composite score was more than that of Danish.

1.

2.

Indu scored less than Chetna in DI.

3.

Jatin scored more than Indu in GA.

No two candidates had the same composite score.

2.

Ajay was the unique highest scorer in WE.

3.

Among the four recruited, Geeta had the lowest composite score.

4.

Indu was recruited.

5.

Danish, Harini, and Indu had scored the same marks in GA.

6.

(A)

(B) Both 2 and 3 (C) Only 1 (D) Only 2 64. Which of the following statements MUST be FALSE?

Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s. Candidate

Ajay

(A)

WE

8

Bala

(C) Bala’s composite score was less than that of Ester

GA 16

9

11 12

Chetna

19

4

Danish

8

15

Ester

12

18

16

Falak

15

7

10

Geeta

14

Harini

5

(D) Chetna scored more than Bala in DI 65. If all the candidates except Ajay and Danish had different marks in DI, and Bala's composite score was less than Chetna's composite score, then what is the maximum marks that Bala could have scored in DI? 66. If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE?

6

Indu

8

Jatin

16

Harini’s composite score was less than that of Falak

(B) Bala scored same as Jatin in DI

Marks out of 20 DI

Both 1 and 2

14

P-54

CAT 2018 Solved Paper Slot 1

ANSWERS AND EXPLANATIONS Quantitative Aptitude (QA) 1.

3.

Ans. B

Explanation:

Explanation:

Let each filling pipe fill an empty tank in x hours and each draining pipe empty a filled tank in y hours. Then, as per the question

Let the cost price of A and B be Rs x and Rs y per kg respectively. Then, 3x + 2y =

6 5 1 5 6 1 and     x y 6 x y 60

40 (3 + 2) 1.1

And, 40 (2 + 3) 1.05 The equations become 3.3x + 2.2y = 200 And, 2.1x + 3.15y = 200 Equating Eqs. (1) and (2), we get 3.3x + 2.2y = 2.1x + 3.15y  1.2x = 0.95y  x : y = 19 : 24 Therefore, required ratio is 19 : 24.



2x + 3y =

2.

Ans. 10 hours

36 30 25 30 1   1 and   x y 12 x y

36 25 1 11  x = 12  1  x x 12 12 Therefore, y = 15

 (1)

If the time taken for 1 draining and 2 filling pipes is t, then

(2)

2 1 1    t = 10 hours 12 15 t

4.

Ans. A Explanation:

 4  log12 81  4p 3   3  4  log 81  4p  12 

Ans. A Explanation:

 4  4log12 3   3   4  4log12 3 

Let x be the number of students who like both. If everyone likes at least one of the two, then 200 = 105 + 134 – x  x = 39 Therefore, the number of students who like only burger = 134 – 39 = 95 If all who like pizza are part of those who like burger, then x = 105 and so number of students who like only burger = 134 – 105 = 29 Therefore, the number of students who like only burger lies between 29 and 95, inclusive of both. From the options, 93 is a possible number of such students.

 1  log12 3  = 3   1  log12 3   log12  log 3  = 3   log12  log 3   log 4   3   log 36   2log 2   3   2log 6  = log623 = log68 5.

Ans. 60 Explanation: Let the number of tests taken the CAT aspirant be n and the average of all n tests be x.

P-55

Quantitative Aptitude Simplified for CAT

According to question, nx – 20 × 10 = (n – 10)(x + 1) And, nx – 30 × 10 = (n – 10)(x – 1) Eq. (1) simplifies to nx – 200 = nx + n – 10x – 10  10x – n = 190 Eq. (2) simplifies to nx – 300 = nx – n – 10x + 10 10x + n = 310 Subtracting Eq. (3) from Eq. (4), we get 2n = 120  n = 60 6.

8. (1)

Explanation:

(2)

The area of T1 = Area of T2 =

(3)

(4)

=

3 3 3 2 (24)2 + (12)2 + (6) + … 4 4 4

3 3 (24)2 (24)2 3(24)2 4 4    192 3 = 1 3 3 1 4 4 9.

38(n  30)  30 y x= n For maximum value of x, y should be minimum and so y can be 51 only. So,

Ans. B Explanation: Let the cost of paint B be x. Then, cost of paint A = x + 8 Cost price of the mixture

38(n  30)  30  51 n 38n  390 390 = 38  x n n The larger the value of n, the larger the value of x. Since the largest value of n is 39, the largest value of x is 38 – 10 = 28.

x=

=

264 = 240 10   1    100  

If amounts of paint A and B are n and 10 – n respectively, then n(x + 8) + (10 – n)x = 240 8n + 10x = 240 For maximum value of x, n should be minimum and we know that n cannot be less than 5, else amount of B will exceed that of A. Therefore, when n = 5, 10x = 240 – 40  10x = 200  x = 20

Ans. B Explanation: Given that 2x  3log5 2 Taking log on both sides, we get log 2x = log  3log5 2  log 2x = (log52)(log 3) (log 2)(log 3) log 5

10. Ans. C Explanation:

 x = log53  x = 1 + log53 – 1  x = 1 + log53 – log55  x = 1 + log5

3 (12)2 4

3 2 (6) , and so on. 4 So, the areas of the triangles form an infinite 1 GP with common ratio . Therefore, 4 Sum of area of all the triangles

Ans. D

 x log 2 =

3 (24)2 4

Area of T3 =

Explanation: Let the average age of people below 51 years and those aged 51 and above be x and y respectively. Also, let n be the number of people below the age of 51 years. According to question, nx + 30y = 38(n + 30)

7.

Ans. A

Let Narayan reaches B in t hours. Then, Partha reaches B in t + 4 hours. Also, Partha reaches mid-point of A and B two hours before Narayan reaches B, that is, 2 hours before t hours. So, time taken for

3 5

P-56

CAT 2018 Solved Paper Slot 1 13. Ans. Rs. 121,000

Partha to reach the mid-point of A and B is t – 2 hours. If Partha takes t – 2 hours to travel half the distance, she will take 2(t – 2) hours to reach B. So, 2(t – 2) = t + 4  t = 8 hours Therefore, time taken for Partha to reach B = t + 4 = 12 hours So, speed of Partha =

Explanation: Let each instalment be Rs. x. Then, Future value of loan borrowed = Future value of amount paid in instalments  210000(1.1)2 = x(1.1) + x  210000(1.1)2 = 2.1x  x = Rs. 121,000

60 = 5 km/hr 12

14. Ans. A Explanation:

11. Ans. 54

Let the radius of the circle be r.

Explanation: Suppose f(9) = a and f(10) = b. Then, f(11) = a + b f(12) = a + 2b f(13) = 2a + 3b f(14) = 3a + 5b f(15) = 5a + 8b According to question, f(11) = 91  a + b = 91 (1) And, f(15) = 617  5a + 8b = 617 (2) Multiplying Eq. (1) by 5, we get 5a + 5b = 455 (3) Subtracting Eq. (3) from Eq. (2), we get 8b – 5b = 617 – 455  b = 54 So, f(10) = 54

From the figure, we have RQ = 1 cm, BR = 2 cm and DQ = 3 cm Applying Pythagoras’ theorem in PRB and PQD, we get

r 2  32

PQ =

And, PR = r 2  22 . According to question, PR – PQ = 1 

r 2  22 –

r 2  32 = 1

Now, using options, we observe that r = satisfies the above equation.

13

15. Ans. A

12. Ans. A Explanation:

Explanation:

Let Raju and Lalitha originally had marbles 4x and 9x respectively. If Lalitha gave n number of marbles to Raju, then as per question

Let the efficiencies of one human and one robot per day be h and r respectively. As per the question, we get (15h + 5r) × 30 = (5h + 15r) × 60  5h = 25r or h = 5r Let t be the time taken by 15 humans working together to finish the work. Then, (15h + 5r) × 30 = 15h × t  (15h + h) × 30 = 15h × t  t = 32 days

4x  n 5  9x  n 6

 24x + 6n = 45x – 5n 21x = 11n Required fraction =

n 21 7   9 x 9(11) 33

P-57

Quantitative Aptitude Simplified for CAT

18. Ans. A

16. Ans. B Explanation:

Explanation:

Given that

Let the wholesaler bought peanuts at Rs. p per kg. So, he bought walnut at Rs. 3p per kg. Sales revenues from peanuts = 8 × 1.1p = 8.8p Sales revenue from walnuts = 16 × 1.2 × 3p = 57.6p Total cost for the shopkeeper = 8.8p + 57.6p = 66.4p Selling price of the mixture for the shopkeeper = Rs. 166 per kg So, cost price of the mixture for the shopkeeper

1 and x2016y2019 = 8 2 Taking logarithm on both sides, we get

x2018y2017 =

1 2 And, 2016 log x + 2019 log y = log 8

2018 log x + 2017 log y = log

Let log x = p and log y = q. Above equations becomes 2018p + 2017q = –log 2 (1) And, 2016p + 2019q = 3 log 2 (2) Solving Eqs. (1) and (2), we get 3log 2  2016 p  2018p + 2017   = –log 2 2019    2019 × 2018p + 2017 × 3 log 2 – 2017 × 2016p = –2019 log 2  (2019 × 2018 – 2017 × 2016)p = –2019 log 2 – 2017 × 3 log 2  8070p = –8070 log 2  p = – log 2

 p = log

=

Therefore, 66.4p = 132.8 × [(8 – 3) + (16 – 5)]  p = 32 So, the wholesaler bought the walnuts at 3p or Rs. 96 per kg.

1 2

19. Ans. A

1 x= 2 Therefore,

Explanation: The length, breadth of rectangle and diameter of the circle forms a right triangle and so Pythagoras’ theorem can be applied. Using options, we observe that 242 + 102 = 262 is true. So, the possible length and breadth of the rectangle are 24 and 10 respectively.

1 x2018y2017 = 2 2018

1 1  y 2017     2 2    y2017 = 22018–1  y2017 = 22017  y = 2 Therefore, x 2 + y3 =

166 = Rs. 132.8 25   1    100  

20. Ans. 502

1 1 33  23   8  4 4 4

Explanation: To form two digit numbers, we need to select 2 digits from given 9 digits, which can be done in 9C2 ways. We cannot arrange these two digits chosen as the digits have to appear in the ascending order. So, number of two digit numbers possible is 9C2. Similarly, number of 3 digit numbers possible = 9C3, and so on.

17. Ans. C Explanation: Marks obtained by the candidate = 36 According to question, 68  36 = 0.45N  1   100   So, N = 250

P-58

CAT 2018 Solved Paper Slot 1 Total number of numbers = 9C2 + 9C3 + 9C4 + … + 9C9 = 2(9C2 + 9C3 + 9C4) + 9C8 + 9C9 = 2(36 + 84 + 126) + 9 + 1 = 502

If length of OC = x, then Area of COD =

 3 2 . x , which is equal to 12 4

So,

 3 2 x = 12 4

21. Ans. 32

1

  2 x=   3 3 

Explanation: The graph of g(x) = 2x2 and h(x) = 52 – 5x are drawn as shown in the figure.

24. Ans. 12 Explanation: Let d be the distance AP, so distance BP = 3d. Also, let the speed of car 1 be 2x and that of car 2 be x. d 2x 3d 6d Time taken by car 2 to reach P =  x 2x Time difference is 1 hour. So,

Time taken by car 1 to reach P =

The graph of f(x) = The parabolic curve below the line segment PQ = Straight line part (y = 52 – 5x), otherwise Since x is positive real number, the maximum value of f(x) will be the ordinate of Q. Now, 2x2 = 52 – 5x  2x2 + 5x – 52 = 0  2x2 + 13x – 8x – 52 = 0  x(2x + 13) – 4(2x + 13) = 0  x = 4, –13 The x-coordinate of Q is 4. So, its ordinate = 2x2 = 32

6d d  1 2x 2x 5d  1 2x d 1   hours = 12 minutes 2x 5

Therefore, time taken by car 1 in reaching P from A is 12 minutes.

22. Ans. 5

25. Ans. D

Explanation: If 0.25 < 2x < 200, then x can be –2, –1, 0, 1, 2, 3, 4, 5, 6, 7. For 2x + 2 to be divisible by 3 or 4, x can be 0, 1, 2, 4 and 6. So, there are 5 integer values of x possible.

Explanation: Given equation is: u2 + (u − 2v − 1)2 = −4v(u + v)  2u2 + 4v2 + 1 – 4uv – 2u + 4v + 4uv + 4v2 =0  2u2 – 2u + (8v2 + 4v + 1) = 0  4u2 – 4u + (16v2 + 8v + 1) + 1 = 0  4u2 – 4u + (4v + 1)2 + 1 = 0  (2u – 1)2 + (4v + 1)2 = 0

23. Ans. A Explanation: R = Area of sector OAB =

60    (1)2  360 6

Therefore, area of OCD =

1 1 , v – 2 4 Therefore,

 u



12 Since OC = OD and angle between OC and OD is 60, COD is equilateral.

u + 3v =

P-59

1 1  1  3     2 4 4  

Quantitative Aptitude Simplified for CAT

26. Ans. C

x z  25    144    904 . z x

Explanation:

x  k , then given equation becomes z 144 25k + = 904 k  25k2 – 904k + 144 = 0

If

log2(5 + log3 a) = 3  5 + log3 a = 23  log3 a = 8 – 5  a = 33 = 27 log5(4a + 12 + log2 b) = 3  4a + 12 + log2 b = 53  4(27) + 12 + log2 b = 125  log2 b = 125 – 108 – 12  log2 b = 5  b = 25 = 32 Therefore, a + b = 27 + 32 = 59

k= =

904  896 2(25)

= 36 or Since

27. Ans. C

4 25

4 x 1, k = 25 z

Therefore, z 25 = r2  x 4 5 r= 2

Explanation: Area of parallelogram = Base × Height Area of parallelogram = CD × AP  72 = 9 × AP  AP = 8 cm In APD, AP = 8, AD = 16 So, PD =

904  9042  4(25)(144) 2(25)

29. Ans. 15 Explanation:

16  8  8 3 2

2

Let the speeds of trains T and S be 4v and 3v respectively. Also, let distance between X and Y be 5d. Then, XZ = 3d and YZ = 2d. As per the question,

1 Area of APD =  PD  AD 2 1   8 3  8  32 3 2

3d 2d  1 4v 3v

28. Ans. A Explanation:

d d  1  = 12 12v v 5d 5    12 = 15 hours 4v 4



Given that 5x, 16y, and 12z are in an arithmetic progression, so 16y – 5x = 12z – 16y  32y = 5x + 12z

30. Ans. 198

5 x  12z y= 32

Explanation:

Since x, y and z are in GP, so y2 = xz

Height of the smaller cone = 12 – 9 = 3 ft 1 th of the 4 original cone, volume of smaller cone will be

Since height of smaller cone is

2

 5 x  12z    = xz  32   25x2 + 144z2 + 120xz = 1024xz  25x2 + 144z2 = 904xz

3

1 1  th of the volume of the original    64 4 cone. Volume of the original cone

P-60

CAT 2018 Solved Paper Slot 1  8x2 + 8y2 = 5(x2 + y2 + 2xy)  3x2 + 3y2 = 10xy

1 2 1 22 2 r h =   4  12 3 3 7 Therefore, volume of the frustum

=

x y  3   + 3   = 10 y x  

63 1 22 2 =    4  12 = 198 ft3 64 3 7

3 = 10 k  3k2 – 10k + 3 = 0

 3k +

31. Ans. C Explanation:

 (3k – 1)(k – 3) = 0

Let A alone does the work in 4x days. Then, time taken by B = 5x days Initially, A works for 2x days (when half the job is done). As per the question,

 k = 3,

1 3

Since EB < CG, so x < y and so k = 3  EB : CG = 1 : 3 33. Ans. 52

1  4 4  95    2  4 x 5 x  100 36 9   20 x 20 Therefore, x = 4

Explanation: Refer to the following Venn diagram:

Time taken for B alone to finish the entire job = 5x = 20 days 32. Ans. D Explanation: Refer to the following figure:

Since no one studies only P and everyone studies at least one of the three subjects, so H  E = 74 If x students study H, then number of students studying E is also x. So, 74 = x + x – (20 + 10)  x = 52 34. Ans. 40

If AE = x and EB = y, then EF =

Explanation:

x2  y 2

Let the other two numbers be x and y. Then, 73xy – 37xy = 720  xy = 20 For minimum possible value of x2 + y2, x = y

Area (EFGH) = x2 + y2, and Area (ABCD) = (x + y)2 Now, Area (EFGH) = 62.5% of Area (ABCD) So, Area (EFGH) =  x 2 + y2 =

5 × Area (ABCD) 8

and so x2 = 20 or x = 20 So, minimum possible value of x2 + y2 = x2 + x2 = 2x2 = 2 × 20 = 40

5 × (x + y)2 8

P-61

Quantitative Aptitude Simplified for CAT

Data Interpretation and Logical Reasoning (DILR) Common explanation for questions 35–38: The contamination levels from P1 to P10 are as following: P1

P2

P3

P4

P5

P6

High

Medium

High

Medium

High

Low

P7

P8

P9

P10

P6 is the only pump with contamination level recorded as low. Moreover, 3 pumps from P1 – P5 recorded as high and they cannot be consecutive to comply with condition 3. So, it can be P1, P3 and P5 only recording as high and so P2 and P4 have to be medium. Now, P7 and P8 can be recorded either high or medium. Also note that none from P7 – P10 can be recorded as low. So, two cases are possible: Case I P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

H

M

H

M

H

L

H

H

M

H

P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

H

M

H

M

H

L

M

M

H

M

L: Low M: Medium H: High Case II

For condition 4, two cases arise: Case A P16

P17

P18

P19

P20

L

M

L

M

L

P16

P17

P18

P19

P20

M

L

M

L

M

Case B

The possible number of contaminations – low, medium and high – are given below: L = 3, H = 6, M = 11, OR L = 4, H = 8, M = 8 Note that 5 or more L’s are not possible. If L = 5, H would be 10 and so M would be 5, in which case it won’t be possible to fill P11 to P15 keeping all conditions intact. Combining all the above, the following three cases are possible: Case 1: P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

P11

P12

P13

P14

P15

P16

P17

P18

P19

P20

H

M

H

M

H

L

H

H

M

H

M

H

M

H

M

L

M

L

M

L

P-62

CAT 2018 Solved Paper Slot 1 Case 2: P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

P11

P12

P13

P14

P15

P16

P17

P18

P19

P20

H

M

H

M

H

L

H

H

M

H

L

M

H

M

H

M

L

M

L

M

P1

P2

P3

P4

P5

P6

P7

P8

P9

P10

P11

P12

P13

P14

P15

P16

P17

P18

P19

P20

H

M

H

M

H

L

H

H

M

H

M

L

H

M

H

M

L

M

L

M

Case 3:

35. Ans. A

40. Ans. 6

Explanation:

Explanation:

From the given three cases possible, P10 is always recorded as high, whereas other options are not always true.

There are a total of 50 notes of 500 rupee denomination. To maximize number of customers, we need to give minimum number of Rs. 500 notes. As per the solution of previous question, minimum number of Rs. 500 notes that can be given to any given customer who has chosen Rs. 500 note as his preference, is 8.

36. Ans. C Explanation: As we can see from the three possible cases, number of pumps at which contamination levels were recorded as medium is always 8.

50  6.25 8 That is, required number of customers can be maximum 6.

So,

37. Ans. C Explanation:

41. Ans. B

From case 3, it is clear that the contamination level at P14 was recorded as medium when the contamination level at P11 was low.

Explanation: To maximize number of customers, we need to minimize number of Rs. 500 notes. If we do not use any note of Rs. 500, then 25 notes of Rs. 200 would be needed, which defies the condition of maximum 20 notes per withdrawal. Similarly, we cannot have 1 note of Rs. 500 dispensed, as that would mean minimum 22 notes of Rs. 200. If we use 2 notes of Rs. 500, then we would need 20 notes of Rs. 200, which again defies the condition. If we use 3 notes of Rs. 500, then we would need 17 notes of Rs. 200 and 1 note of Rs. 100, which again defies the condition. With 4 notes of Rs. 500, we can have 15 notes of Rs. 200, which satisfies the condition.

38. Ans. A Explanation: From case 1, it is clear that the contamination levels at both P11 and P16 were recorded as medium, when contamination level at P15 was medium. 39. Ans. 7 Explanation: Number of notes of Rs. 500, Rs. 200 and Rs. 100 can be: (10, 0, 0), (9, 0, 5), (9, 1, 3), (9, 2, 1), (8, 3, 4), (8, 4, 2), (8, 5, 0) So, only 7 ways are possible.

P-63

Quantitative Aptitude Simplified for CAT

So, every customer can be given the combination of (4, 15, 0), and so maximum 50 = 12.5 or 12 customers can be served. 4

figures for May and June will be 50 and 60 respectively. Similarly, sales figures of October, November and December of 2016 are also in AP. So, the sales figures of November and December are 120 and 140 respectively. The sales figure of August 2017 = 220 – 60 – 70 = 90 The sales figure of December 2017 = 500 – 150 – 170 = 180 These figures are placed in the above table (in bold letters), including missing quarterly figures.

42. Ans. C Explanation: If total number of notes to be dispensed is the smallest possible, then we should aim to give maximum notes of highest denomination. The customer who is giving Rs. 500 as preference can be given (10, 0, 0). Since there are 50 such customers, we would need 500 notes of Rs. 500 denomination. Those who are preferring Rs. 100 denomination, can be given (8, 0, 10) requiring minimum total number of notes. This means 8 × 50 = 400 notes of Rs. 500 denomination. So, a total of 900 notes of Rs. 500 denomination are required.

43. Ans. B Explanation: Required percentage increase =

44. Ans. A Explanation:

Common explanation for questions 43–46: Data from pie chart can be tabulated as following:

The percentage increase in Q1 is the highest, which is =

Month

2016

2017

January

80

120

February

60

100

March

100

160

April

40

60

May

50

75

June

60

65

July

75

60

August

120

90

September

55

70

October

100

150

November

120

170

December

140

180

Quarters

2016

180  140  100 = 28.57% 140

2017

380  240  100 = 58.33% 240

45. Ans. C Q1

Q2

240

150

Explanation:

380

Only two quarters have experienced decrease, Q2 of 2016 over Q1 of 2016 and Q2 of 2017 over Q1 of 2017. Percentage changes in Q2 of 2016 over Q1 of 2016

200

150  240  100 = –37.5% 240 Percentage changes in Q2 of 2017 over Q1 of 2017

=

Q3

Q4

250

360

220

380  200  100 = –47.4% 380

500

46. Ans. D Explanation:

Since sales figures of April, May and June of 2016 are in AP and total of that quarter is 150, with first month sales as 40, the sales

Only in October 2017 sales has more than doubled than in September 2017, whereas in other months, the sales have increased by

P-64

CAT 2018 Solved Paper Slot 1 less than two times. Therefore, the percentage increase in sales is the highest for the month of October 2017 over previous month.

So, the minimum possible number satellites serving B exclusively = 250

47. Ans. C

of

49. Ans. B

Explanation:

Explanation:

Refer to the following Venn diagram:

Since 2x = 1450 – 5z, if z  100, the maximum value of x =

1450  5 z 1450  500 = 475  2 2

50. Ans. C Explanation: Number of satellites serving at least two among B, C and S = 2[x – (0.3y + 100 + z)] + 100 + z = 1200  2x – 0.6y – z – 100 = 1200  2x – 0.6y – z = 1300 (1) Also, y – 5z = 250, and 2x = 1450 – 5z. Substituting the values of x and y in Eq. (1), we get 1450 – 5z – 0.6(250 + 5z) – z = 1300 z=0 Therefore, number of satellites serving B = 2x = 1450, which is more than 1000. Number of satellites serving exclusively is y whose value is = 250 + 5z = 250 So, options (A), (B) and (D) are definitely true, and so option (C) must be FALSE.

Number of satellites serving B and C but not S = x – (0.3y + 100 + z) Number of satellites serving B and S but not C = x – (0.3y + 100 + z) Number of satellites serving B = 2x = y + 100 + 2[x – (0.3y + 100 + z)] = 2x + 0.4y – 2z – 100  0.4y – 2z = 100  y – 5z = 250 Also, 2x + 2(0.3y) + 2z = 1600  2x + 0.6(5z + 250) + 2z = 1600  2x = 1450 – 5z 1450 = 725 When z = 0, x = 2 Also, x  100 + z + 0.3y  725 – 2.5z  100 + z + 0.3y  3.5z  625 – 0.3y  3.5z  625 – 0.3(5z + 250)  5z  550 So, maximum value of 5z is 550, in which case minimum value of x 1450  5 z 1450  550 = = 450  2 2

Common explanation for questions 51–54: Adriana and Deb cannot be from Z, similarly, Daisy and Amit cannot be from Z, because none of the males are from Z. So, Bandita and Chitra are from Z. Also, since female student from Y majors in operations, that female cannot be Daisy as she is doing minor in Operations. So, Adriana is from Y doing major in Operations, and Daisy is from X. Accordingly, Amit is from X and Deb is from Y. Since both the male students from Y minor in Finance, which are Barun and Deb, so Amit and Chetan are the male students doing

48. Ans. C Explanation: Minimum possible value of y occurs for minimum value of z, which can be 0. When z = 0, y = 250 + 5z = 250

P-65

Quantitative Aptitude Simplified for CAT

Common explanation for questions 55–58:

major in Finance. If Daisy is doing minor in Operations, and three females are doing minor in Marketing, so Adriana, Bandita and Chitra are doing minor in Marketing. The above results can be developed in the form of the following table:

Given information can be tabulated as following:

Bureaucrats Name of student

Male/ Female

Institute

Adriana

Female

Bandita

Major

Minor

Y

Operations

Marketing

Female

Z

Ops/Fin

Marketing

Chitra

Female

Z

Ops/Fin

Marketing

Daisy

Female

X

Fin/Mtkg

Operations

Amit

Male

X

Finance

Ops/Mktg

Barun

Male

Y

Operations

Finance

Chetan

Male

X

Finance

Ops/Mktg

Deb

Male

Y

Mktg

Finance

Educationalists Politicians

Research Committee

Teaching Committee

Administration Committee

0.75x

0.75x

x

a

a–d

a+d

0.2p

0.2p

0.6p

As per the question, (0.75x + 0.75x + x) + (a + a – d + a + d) + p = 24  2.5x + 3a + p = 24. Since number of bureaucrats in Research Committee is 0.75x, the value of x must be a multiple of 4. If x = 4, then total bureaucrats = 2.5x = 10 If x = 8, then total bureaucrats = 2.5x = 20 Also, p must be a multiple of 5, otherwise 0.2p will be a fraction. So, x cannot be 8 because that would mean number of people will be 20 + 5 + educationalists, whereas total number of persons is 24 only. Therefore, x = 4 If p = 5, then total number of educationalists = 24 – 10 – 5 = 9 = 3a or a = 3 Note that p = 10 is not possible, as that would mean number of educationalists is 4, which cannot be equal to 3a. So, x = 4, p = 5 and number of educationalists = 9 and so a = 3 With a = 3, d can be 1 or 2.

51. Ans. B Explanation: From the above table, we can say that the students from Z are Bandita and Chitra. 52. Ans. A Explanation: Deb minors in Finance, as can be seen from the table above. 53. Ans. A Explanation:

When d = 1, table becomes

Amit majors in Finance, as can be seen from the table above.

Research Committee

Teaching Committee

Administration Committee

Bureaucrats

3

3

4

Explanation:

Educationalists

3

2

4

As per the statement 4 given in the information provided, only one female majors in Finance. So, if Chitra majors in Finance, then Bandita cannot major in Finance and so she majors in Operations.

Politicians

1

1

3

Research Committee

Teaching Committee

Administration Committee

Bureaucrats

3

3

4

Educationalists

3

1

5

Politicians

1

1

3

54. Ans. C

When d = 2, table comes

P-66

CAT 2018 Solved Paper Slot 1 55. Ans. C

So, minimum 4 different numerals are needed to fill a 3 × 3 matrix.

Explanation:

60. Ans. 4

Option (A) is definitely true. Options (B) and (D) may or may not be true, so they are not DEFINITELY FALSE. Option (C) is definitely false.

Explanation: Here also, only 4 different numerals are required to fill the 5 × 5 matrix.

56. Ans. 4

1

2

1

2

1

Explanation:

4

3

4

3

4

As we can see from the above two possible tables, the number of bureaucrats in Administration committee is 4.

1

2

1

2

1

4

3

4

3

4

1

2

1

2

1

57. Ans. 3

Note that 5 × 5 matrix can be taken as multiple 3 × 3 matrices super-imposed on each other. That’s why only 4 different numerals are needed.

Explanation: The number of educationalists in Research Committee is 3, even in both the possible tables.

61. Ans. D

58. Ans. A

Explanation: If you are allowed to make one mistake in a 5 × 5 matrix, then also there will be at least one matrix of 3 × 3 within this 5 × 5 matrix where we would still need 4 different numerals. So, the answer here is again 4.

Explanation: The size of teaching committee cannot be uniquely determined, as there are two possible values as given in above two tables: 5 or 6. Note that the size of administration committee also cannot be determined uniquely.

62. Ans. C Explanation: In this case, if the particular cell is not from the boundary, it will have 8 adjacent cells. So, that particular cell and those 8 adjacent cells must have different numerals. Therefore, minimum 9 different numerals are required to comply with the conditions given in the question.

59. Ans. 4 Explanation: If top left corner is filled with some number, say 1, then it will have three adjacent cells each of which must have a different number, to comply with the conditions given. 1

2

4

3

Common explanation for questions 63–66: Reworking on the table, on the basis of the information given, we get the following:

Now, the remaining cells can be filled using these digits only without defying the condition given. This is shown below. 1

2

1

4

3

4

1

2

1

P-67

Marks out of 20 DI

WE

GA

Composite Score

Ajay

8

A

16

32 + A

Bala

B

9

11

2B + 20

Chetna

19

4

12

54

Danish

8

15

C

31 + C

Candidate

Quantitative Aptitude Simplified for CAT

Candidate

Marks out of 20

The revised table is as following:

Composite Score

DI

WE

GA

Ester

12

18

16

58

Falak

15

7

10

47

Geeta

14

D

6

34+D

Harini

5

E

C

10 + E + C

Indu

F

8

C

2F + 8 + C

Jatin

G

16

14

2G + 30

Candidate

As per the information, only four were recruited and Geeta had the lowest composite score amongst those four. Indu was also recruited. So, Indu’s composite score was more than Geeta’s composite score. Ajay’s score in WE was unique highest. So, A = 19 or 20 Jatin scored 100% in one section, and that section can be only DI. So, G = 20 Jatin’s composite score = 70 Indu also scored 100% in one section. So, either F = 20 or C = 20 Since composite score of Jatin is 10 more than that of Indu, score of Indu = 60. If F = 20, then C would be 12, in which case she would not qualify and hence would not get recruited. So, C = 20, and so F = 16 Since Geeta got recruited, she must have scored more than 14 in WE. Geeta cannot score 20 in WE as Ajay has unique highest score in WE. So, even if Geeta scored 19 in WE, her composite score would be 53, which is fourth highest (since she is recruited amongst four). Chetna does not qualify for recruitment because her score in WE is less than 70% marks in WE and GA. Score of Ester in all sections is more than 70% and his composite score is definitely more than that of Geeta, so Ester is recruited. If A = 19, then Ajay and Danish would have same composite score, which is not allowed. So, A = 20

Marks out of 20

Composite Score

DI

WE

GA

Ajay

8

20

16

52

Bala

B

9

11

2B + 20

Chetna

19

4

12

54

Danish

8

15

20

51

Ester

12

18

16

58

Falak

15

7

10

47

Geeta

14

19

6

53

Harini

5

E

20

30 + E

Indu

16

8

20

60

Jatin

20

16

14

70

Since Ajay qualifies and is not recruited, the score of Geeta must be more than 52 and hence D must be more than 18. Since D cannot be 20, the value of D = 19, and so Geeta’s composite score = 34 + 19 = 53. 63. Ans. A Explanation: From the final table, we can say that statements 1 and 2 are definitely true. 64. Ans. B Explanation: Option (B) is definitely false, because if option (B) were true, then Bala’s composite score would be 60, matching with the score of Indu. 65. Ans. 13 Explanation: As per the information, 2B + 20 < 54 or B < 17, so B’s maximum value can be 13. 66. Ans. 14 Explanation: The maximum score of Harini in WE can be 14. If her score in WE were 17, then her composite score would match with the score of Falak.

P-68

CAT 2018 Solved Paper Slot 2

CAT 2018 Solved Paper Slot 2 Quantitative Aptitude (QA) 1.

and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

If p3 = q4 = r5 = s6, then the value of logs(pqr) is equal to (A)

47 10

(A)

(B)

24 5

(C) 1 : 4

(C)

16 5

(B) 2 : 7 (D) 2 : 9 4.

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

5.

If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

(D) 1 2.

The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is (A)

3:2

(B) 4 : 3 (C) 5 : 4 (D) 8 : 5 3.

1:2

(A)

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B

16

(B) –32 (C) 48 (D) 64

P-69

Quantitative Aptitude Simplified for CAT

6.

7.

On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is

12.



If A = {62n – 35n – 1: n = 1, 2, 3, ...} and B = {35(n – 1): n = 1, 2, 3, ...}, then which of the following is true? (A)

1 1 1 1    log 2 100 log 4 100 log 5 100 log 10 100 1 1 1 =?   log 20 100 log 25 100 log 50 100

(A) –4 (B) 0 (C) 10

Neither every member of A is in B nor every member of B is in A

(D)

(B) Every member of B is in A.

13. The value of the sum 7  11 + 11  15 + 15  19 + ...+ 95  99 is

(C) At least one member of A is not in B (D) Every member of A is in B and at least one member of B is not in A 8.

9.

(A) 80707 (B) 80730 (C) 80751

If N and x are positive integers such that NN = 2160 and N2 + 2N is an integral multiple of 2x, then the largest possible x is __________ .

(D) 80773 14. A parallelogram ABCD has area 48 sq cm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is (A)

(A) (B) (C) (D)

16. There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

(B) 3 : 10 (C) 1 : 3 (D) 2 : 5 10. Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is __________ . 11. The smallest integer n for which holds, is closest to (A)

>

s6 s6 s≠6 5s7

15. The smallest integer n such that n3 – 11n2 + 32n – 28 > 0 is __________ .

1:4

4n

1 2

(A) (B) (C) (D)

220 : 149 229 : 141 239 : 161 251 : 163

17. From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is

1719

39

(A)

(B) 37

88 + 12

(B) 86 + 8

(C) 35

(C) 82 + 24

(D) 33

(D) 80 + 16

P-70

CAT 2018 Solved Paper Slot 2 18. Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his efficiency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work? (A) (B) (C) (D)

retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y. 23. In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is __________ .

11 12 13.5 14.5

19. The area of a rectangle and the square of its perimeter are in the ratio 1 ∶25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio (A) (B) (C) (D)

1:3 1:4 3:8 2:9

24. On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is __________ .

20. A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is (A)

25. Let t1, t2,… be real numbers such that t1 + t2 + … + tn = 2n2 + 9n + 13, for every positive integer n ≥ 2. If tk = 103, then k equals __________ .

5 3

(B) 6 2 (C) 2π (D) 8

26. How many two-digit numbers, with a nonzero digit in the unit’s place, are there which are more than thrice the number formed by interchanging the positions of its digits?

21. A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along? (A) (B) (C) (D)

(A)

6

(B) 5 (C) 7 (D) 8 27. The arithmetic mean of x, y and z is 80, and xy that of x, y, z, u and v is 75, where u = 2 yz and v = . If x ≥ z, then the minimum 2 possible value of x is __________ .

4:12 pm 4:24 pm 4:36 pm 4:48 pm

28. Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

22. Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest

P-71

Quantitative Aptitude Simplified for CAT

(A)

34. A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0, 0) is

20

(B) 23 (C) 45 (D) 48

(A)

29. Let f(x) = max{5x, 52 – 2x2}, where x is any positive real number. Then the minimum possible value of f(x) is __________ .

(B) 2 2 units (C) 4 2 units (D) 8 units

30. If a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is __________ .

Data Interpretation and Logical Reasoning (DILR)

31. For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1, 2, 3, 4}, Q = {2, 3, 5, 6}, R = {1, 3, 7, 8, 9}, S = {2, 4, 9, 10}, then the number of elements in (PΔQ)Δ(RΔS) is (A)

Directions for questions 35–38: Read the information given below and answer the questions that follow. According to a coding scheme the sentence Peacock is designated as the national bird of India is coded as 5688999 35 1135556678 56 458 13666689 1334 79 13366.This coding scheme has the following rules:

6

(B) 7 (C) 8 (D) 9 32. A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is (A)

50%

(B) 55%

1.

The scheme is case-insensitive (does not distinguish between upper case and lower case letters).

2.

Each letter has a unique code which is a single digit from among 1, 2, 3, …, 9.

3.

The digit 9 codes two letters, and every other digit codes three letters.

4.

The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.

Answer these questions on the basis of this information.

(C) 48% (D) 52%

35. What best can be concluded about the code for the letter L?

33. A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now (A)

4 units

(A)

1

(B) 6 (C) 8 (D) 1 or 8

20.5

36. What best can be concluded about the code for the letter B?

(B) 25.4 (C) 30.3

(A)

(D) 35.2

1

(B) 3 (C) 3 or 4 (D) 1 or 3 or 4

P-72

CAT 2018 Solved Paper Slot 2 37. For how many digits can the complete list of letters associated with that digit be identified? (A)

the number of Old visitors buying Gold tickets was ________ . 41. If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number of Middle-aged visitors buying Gold tickets was ________ .

0

(B) 1 (C) 2 (D) 3

42. Which of the following statements MUST be FALSE?

38. Which set of letters CANNOT be coded with the same digit? (A)

(A)

X, Y, Z

(B) S, U, V

(B) The numbers of Old and Middle-aged visitors buying Economy tickets were equal

(C) I, B, M (D) S, E, Z

(C) The numbers of Gold and Platinum tickets bought by Young visitors were equal

Directions for questions 39–42: Read the information given below and answer the questions that follow.

(D) The numbers of Old and Middle-aged visitors buying Platinum tickets were equal

Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day: 1.

140 tickets were sold.

2.

The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.

3.

Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.

4.

The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

Directions for questions 43–46: Read the information given below and answer the questions that follow. An agency entrusted to accredit colleges looks at four parameters: faculty quality (F), reputation (R), placement quality (P), and infrastructure (I). The four parameters are used to arrive at an overall score, which the agency uses to give an accreditation to the colleges. In each parameter, there are five possible letter grades given, each carrying certain points: A (50 points), B (40 points), C (30 points), D (20 points), and F (0 points). The overall score for a college is the weighted sum of the points scored in the four parameters. The weights of the parameters are 0.1, 0.2, 0.3 and 0.4 in some order, but the order is not disclosed. Accreditation is awarded based on the following scheme:

39. If the number of Old visitors buying Platinum tickets was equal to the number of Middleaged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold? (A)

The numbers of Middle-aged and Young visitors buying Gold tickets were equal

32

(B) 34 (C) 38 (D) 36 40. If the number of Old visitors buying Platinum tickets was equal to the number of Middleaged visitors buying Economy tickets, then

P-73

Range

Accreditation

Overall score  45

AAA

35  Overall score < 45

BAA

25  Overall score < 35

BBA

15  Overall score < 25

BBB

Overall score < 15

Junk

Quantitative Aptitude Simplified for CAT

Eight colleges apply for accreditation, and receive the following grades in the four parameters (F, R, P, and I): F

R

P

I

A-one

A

A

A

B

Best Ed

B

C

D

D

Cosmopolitan

B

D

D

C

Dominance

D

D

B

C

Education Aid

A

A

B

A

Fancy

A

A

B

B

Global

C

F

D

D

High Q

C

D

D

B

from the corresponding product, while its centre represents the Product popularity and Market potential scores of the product (out of 20). The shadings of some of the boxes have got erased.

It is further known that in terms of overall scores: 1.

High Q is better than Best Ed;

2.

Best Ed is better than Cosmopolitan; and

3.

Education Aid is better than A-one.

The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:

43. What is the weight of the faculty quality parameter? (A)

0.1

(B) 0.2 Promising

Blockbuster

Doubtful

No-hope

>10

>10

≤10

≤10

the

Product popularit y score

45. What is the highest overall score among the eight colleges?

Market potential score

>10

≤10

>10

≤10

46. How many colleges have overall scores between 31 and 40, both inclusive?

The following facts are known:

(C) 0.3 (D) 0.4 44. How many colleges accreditation of AAA?

(A)

receive

1.

Alfa and Bravo had the same number of products in the Blockbuster category.

2.

Charlie had more products than Bravo but fewer products than Alfa in the No-hope category.

3.

Each company had an equal number of products in the Promising category.

4.

Charlie did not have any product in the Doubtful category, while Alfa had one product more than Bravo in this category.

5.

Bravo had a higher revenue than Alfa from products in the Doubtful category.

0

(B) 1 (C) 2 (D) 3 Directions for questions 47–50: Read the information given below and answer the questions that follow. Each of the 23 boxes in the picture below represents a product manufactured by one of the following three companies: Alfa, Bravo and Charlie. The area of a box is proportional to the revenue

P-74

CAT 2018 Solved Paper Slot 2 6.

Charlie had a higher revenue than Bravo from products in the Blockbuster category.

7.

Bravo and Charlie had the same revenue from products in the No-hope category.

8.

Alfa and Charlie had the same total revenue considering all products.

Directions for questions 51–54: Read the information given below and answer the questions that follow. Fun Sports (FS) provides training in three sports – Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known: 1. The number of students enrolled only in L is double the number of students enrolled in all the three sports. 2. There are a total of 17 students enrolled in G. 3. The number of students enrolled only in G is one less than the number of students enrolled only in L. 4. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L. 5. The maximum student enrollment is in L. 6. Ten students enrolled in G are also enrolled in at least one more sport.

47. Considering all companies' products, which product category had the highest revenue? (A)

Promising

(B) Doubtful (C) Blockbuster (D) No-hope 48. Which of the following is the correct sequence of numbers of products Bravo had in No-hope, Doubtful, Promising and Blockbuster categories respectively? (A)

2, 3, 1, 2

(B) 1, 3, 1, 3 (C) 1, 3, 1, 2 (D) 3, 3, 1, 2

51. What is the minimum number of students enrolled in both G and L but not in K?

49. Which of the following statements is NOT correct? (A)

52. If the numbers of students enrolled in K and L are in the ratio 19 : 22, then what is the number of students enrolled in L?

Alfa's revenue from Blockbuster products was the same as Charlie's revenue from Promising products

(A)

(B) Bravo and Charlie had the same revenues from No-hope products

(C) 22 (D) 18

(C) The total revenue from No-hope products was less than the total revenue from Doubtful products

53. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?

(D) Bravo's revenue from Blockbuster products was greater than Alfa's revenue from Doubtful products 50. If the smallest box on the grid is equivalent to revenue of Rs.1 crore, then what approximately was the total revenue of Bravo in Rs. crore? (A)

17

(B) 19

54. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and L?

24

(B) 30 (C) 34 (D) 40

(A)

P-75

7

Quantitative Aptitude Simplified for CAT

(B) 5

(C) Cxqi

(C) 8

(D) Dipq

(D) 6

58. The complete list of brands whose profits went up in 2017 from 2016 is:

Directions for questions 55–58: Read the information given below and answer the questions that follow.

(A)

There are only four brands of entry level smartphones called Azra, Bysi, Cxqi, and Dipq in a country. Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below: Market Share %

Unit Selling Price (Rs.)

Profitability

Azra

40

15,000

10

Bysi

25

20,000

30

Cxqi

15

30,000

40

Dipq

20

25,000

30

Brand

(C) Cxqi, Azra, Dipq (D) Azra, Bysi, Dipq Directions for questions 59–62: Read the information given below and answer the questions that follow. The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

In 2017, sales volume of entry level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 as they were in 2016.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.

55. The brand that had the highest revenue in 2016 is: (A)

The following facts are known about the outlet on a particular day: 1. The amount of L used by the outlet to buy C equals the amount of L it received by selling C. 2. The amounts of L used by the outlet to buy A and B are in the ratio 5:3. 3. The amounts of L the outlet received from the sales of A and B are in the ratio 5:9. 4. The outlet received 88000 units of L by selling A during the day. 5. The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C. 6. The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

Azra

(B) Bysi (C) Cxqi (D) Dipq 56. The brand that had the highest profit in 2016 is: (A)

Azra

(B) Bysi (C) Cxqi (D) Dipq 57. The brand that had the highest profit in 2017 is: (A)

Azra, Bysi, Cxqi

(B) Bysi, Cxqi, Dipq

Azra

(B) Bysi

P-76

CAT 2018 Solved Paper Slot 2 61. What was the base exchange rate of currency B with respect to currency L on that day?

59. How many units of currency A did the outlet buy on that day? 60. How many units of currency C did the outlet sell on that day? (A)

62. What was the buying exchange rate of currency C with respect to currency L on that day?

3000

(B) 6000

(A)

(C) 19000

0.95

(B) 1.10

(D) 22000

(C) 1.9 (D) 2.2

Directions for questions 63–66: Read the information given below and answer the questions that follow. Seven candidates, Akil, Balaram, Chitra, Divya, Erina, Fatima, and Ganeshan, were invited to interview for a position. Candidates were required to reach the venue before 8 am. Immediately upon arrival, they were sent to one of three interview rooms: 101, 102, and 103. The following venue log shows the arrival times for these candidates. Some of the names have not been recorded in the log and have been marked as ‘?’. Time Person

7:10 am

7:15 am

7:25 am

7:30 am

7:40 am

7:45 am

Akil, ?

?

?

Chitra

Fatima

?

Additionally here are some statements from the candidates: Balaram: I was the third person to enter Room 101. Chitra: I was the last person to enter the room I was allotted to. Erina: I was the only person in the room I was allotted to. Fatima: Three people including Akil were already in the room that I was allotted to when I entered it. Ganeshan: I was one among the two candidates allotted to Room 102. 63. What best can be said about the room to which Divya was allotted? (A)

65. When did Erina reach the venue? (A)

Definitely Room 101

(B) 7:15 am

(B) Definitely Room 102

(C) 7:25 am

(C) Definitely Room 103

(D) 7:45 am

(D) Either Room 101 or Room 102

66. If Ganeshan entered the venue before Divya, when did Balaram enter the venue?

64. Who else was in Room 102 when Ganeshan entered? (A)

7:10 am

(A)

Akil

7:10 am

(B) 7:15 am

(B) Divya

(C) 7:25 am

(C) Chitra

(D) 7:45 am

(D) No one

P-77

Quantitative Aptitude Simplified for CAT

ANSWERS AND EXPLANATIONS Quantitative Aptitude (QA) 1.

(8A + 18B)  60 = (7A + 27B)  t becomes (12A + 18B)  60 = (10.5B + 27B)  t  t = 48 minutes

Ans. A Explanation: p = s2, q = s6/4 = s3/2, r = s6/5. So, logs(pqr) = logs(s2.s3/2.s6/5) = logs(s2+3/2+6/5) =

2.

5.

Explanation: xy  xy 2  (x + y)2  4xy  x2 + y2  2xy  2xy  97  xy  48.5 Therefore, product cannot be equal to 64.

20  15  12 47  10 10

Ans. B Explanation: 11k  x 47  14k  x 56  616k + 56x = 658k + 47x  42k = 9x  14k = 3x Ratio of Bimal’s new score to that of his original score = (14k + x) : 14k = (3x + x) : 3x = 4 : 3

3.

6.

Ans. 24 Explanation: Refer to the following figure:

Ans. C Explanation: Ratio of speeds of car 1 and 3 = Ratio of distances travelled = AQ : BQ = 2 : 1 Ratio of speeds of car 2 and 3 = Ratio of distances travelled = AP : BP = 1 : 2 Therefore, ratio of speeds of car 2 to car 1=1:4

4.

Ans. D

CPB = 90 (because angle in a semicircle is right angle). So, CP is altitude on AB. Similarly, BQ is altitude on AC. Now, 1 1 AB  PC  AC  BQ 2 2  AB  PC  AC  BQ

Area of ABC =

Ans. 48 minutes Explanation: Let A and B be the quantity of water coming from pipes A and B per minute respectively. As per the question, (10A + 45B)  30 = (8A + 18B)  60 = (7A + 27B)  t, where t is the required time in the question. Now, (10A + 45B)  30 = (8A + 18B)  60  10A + 45B = 16A + 36B  6A = 9B  A = 1.5B Using this,

 30  20 = 25  BQ  BQ = 7.

600 = 24 cm 25

Ans. D Explanation: A = {0, 1225, 46550, …}, B = {0, 35, 70, 105, …} So, B contains all the multiples of 35 including 0. In A, the expression 62n – 35n – 1 is divisible by 5 as well as 7 and so every element of A is

P-78

CAT 2018 Solved Paper Slot 2

8.

divisible by 35, including 0, but A doesn’t contain all multiples of 35. So, every member of B is not necessarily in A.

= 0.5 + 1 + 2 = 3.5 hours

Ans. 10

car 1. Car 2 travels first 50 km in 0.5 hours, then next 50 km in 1 hour. When car 1 reach B, time taken is 3.5 hours, of which 0.2 + 0.5 + 1 = 1.7 hours is already spent by car 2. In the remaining time of 3.5 – 1.7 = 1.8 hours, distance travelled by car 2 = 25  1.8 = 45 km So, distance between car 2 and B when car 1 reaches B = 50 – 45 = 5 km

Car 2 starts

Explanation: NN = 2160  NN = 3232  N = 32 N2 + 2N = 322 + 232 = 210 + 232 = 210(1 + 222) If this is to be an integral multiple of 2x, then the maximum value of x is 10. 9.

Ans. C

11. Ans. A

Explanation: Let the strengths of salt solutions A, B and C be x%, y% and z% respectively. Then, x + 2y + 3z = 6  20 = 120 (1) And, 3x + 2y + z = 6  30 = 180(2) Multuplying Eq. (1) by 3, we get 3x + 6y + 9z = 360 (3) Subtracting Eq. (2) from Eq. (3), we get 4y + 8z = 180  y + 2z = 45  y = 45 – 2z Now, from Eq. (1), we get x = 120 – 2(45 – 2z) – 3z = 30 + z

Explanation: 1719 > 1619  1719 > 438 So, if 4n is to be more than 1719, n has to be more than 38. The only option possible is (A). 12. Ans. D Explanation: 1 1 1 1 1 1 1       log2 100 log 4 100 log 5 100 log10 100 log 20 100 log 25 100 log 50 100

= log1002 – log1004 + log1005 – log10010 + log10020 – log10025 + log10050  2 5 20  = log100     50  4 10 25  

 2y + 7z  Strength of D =  %  9 



2(45  2z)  7z 9



90  3z 30  z  9 3

1 1  log100 10   log10 10  2 2

13. Ans. A Explanation: The nth term of the series, tn = (4n + 3)(4n + 7) = 16n2 + 40n + 21 So, sum of n terms of the series = (16n2 + 40n + 21) = 16n2 + 40n + 21

Strength of A = 30 + z Required ratio =

30  z : (30  z) = 1 : 3 3

10. Ans. 5

16n(n  1)(2n  1) 40n(n  1)   21n 6 2 If the number of terms in the given series is n, then

=

Explanation: Time taken for car 1 to reach B =

20 = 0.2 hours after start of 100

50 50 50   100 50 25

95 = 7 + (n – 1)(4)

P-79

Quantitative Aptitude Simplified for CAT

 n = 23 So, required sum of the series 

And, radius = 12 cm Therefore, AB = 24 cm,

16  23(23  1)(2  23  1) 40  23(23  1)   21  23 6 2

768 = 32 cm 24 Hence, perimeter of leftover portion

And, so BC =

= 69184 + 11040 + 483 = 80707

= 2  32 + 24 + (12) = 88 + 12

14. Ans. B Explanation: Area of parallelogram = Base  Height  48 = 8h  h = 6 Now, AD = s, which is either equal to 6 or more than 6, because s will act as hypotenuse of the triangle whose altitude is 6 cm.

18. Ans. C Explanation: Let Ramesh and Ganesh can complete a work in x and y days, working alone. Then, 1 1 1   x y 16

15. Ans. 8

Also,

Explanation: n – 2 is a factor of n3 – 11n2 + 32n – 28. So, n3 – 11n2 + 32n – 28 = (n – 2)(n2 – 9n + 14) = (n – 2)(n – 7)(n – 2) = (n – 2)2(n – 7) So, (n – 2)2(n – 7) > 0  n > 7 So, the smallest integer n is 8.

10  7 7  10 (0.7)  1    x y x y    14 17     1 y   x 1 1  3  14      1 x y y 

16. Ans. C

 y = 24 days, x = 48 If Ganesh had worked alone after Ramesh got sick, then

Explanation: Let the ratio of A to B in drum 2 be x : y. Then, x 18 3  4 13 25 xy  7 y 3  4 7 25 xy 54  25  21  25

14 3  1 16 y

7 t  1 16 24  t = 13.5 days

19. Ans. B

4x x  y 13  4y 7 xy

Explanation: If the length and the breadth of the rectangle is x and y, then xy : [2(x + y)]2 = 1 : 25  25xy = 4(x2 + y2 + 2xy)  4x2 + 4y2 = 17xy

 7  54(x + y) + 7  100x = 13  21(x + y) + 13  100y  378x + 378y + 700x = 273x + 273y + 1300y  805x = 1195y  x : y = 239 : 161

 4k +

x 4 = 17, where k  y k

So, 4k2 – 17k + 4 = 0  4k(k – 4) – 1(k – 4) = 0

17. Ans. A

1 4 So, ratio of shorter to longer side is 1 : 4.

Explanation: If the area of semicircle = 72, then Area of circle = 144

 k  4,

P-80

CAT 2018 Solved Paper Slot 2 So, A alone can complete a work in 6 hours which B can do in 4 hours. This means that initially the tank gets emptied at 2 pm. If A and B work together, then time taken by them to complete the work

Alternatively, Use options, If ratio is 1 : 3, then length and breadth can be taken as 3 and 1. So, area = 3 and perimeter = 8 and ratio of area to square of perimeter becomes 3 : 64, which is not 1 : 25. If ratio is 1 : 4, then length and breadth can be taken as 4 and 1. So, area = 4 and perimeter = 10 and ratio of area to square of perimeter becomes 4 : 100, which is same as 1 : 25.

46 = 2.4 hours 46 = 2 hours, 24 minutes Therefore, time when they will complete the work = 4:24 pm

=

22. Ans. 4000 Explanation: Net interest earned by Gopal after returning Ankit’s dues = Interest earned by Gopal – Interest paid by Gopal to Ankit = 0.1(X + Y) – 0.08X – Y = 0.02X + 0.1Y Net interest accrued to Ankit = 0.08X Therefore, 0.02X + 0.1Y = 0.08X  0.06X = 0.1Y  5Y = 3X In the second situation, net interest retained by Gopal = 0.1(X + 2Y) – 0.08X – 2Y = 0.02X + 0.2Y. Now, (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150  Y = 1500 Therefore, X = 2500 And, so X + Y = 4000

20. Ans. A Explanation: Let the center of the circle be O and chord be AB. If AOB = 60, and we know that OA = OB, then A = B = 60. So, OAB is equilateral triangle and so OA = OB = 5 cm Let the length of the chord which subtends an angle of 120 be x. Then, x=

52  52  2  5  5  cos120  5 3

21. Ans. B Explanation: On Monday A alone completed the work by 8 pm and on Tuesday B alone completed the work by 6 pm. This means that B takes 2 hours less than A to complete the work. So, if A does a work in x hours, B does the same work in x – 2 hours. So, ratio of work by A to B = (x – 2) : x Moreover, on Wednesday, A worked till 5 pm and then B alone worked for 2 hours to complete the work. Comparing with Monday, A had to work from beginning till 8 pm, that is, A worked from beginning till 5 pm and then from 5 pm to 8 pm. So, the work which is done by A from 5 pm to 8 pm is same as the work done by B from 5 pm to 7 pm. This is means the work which A takes 3 hours to complete is done in 2 hours by B. So, ratio of work done by A to B = 2 : 3, which is same as (x – 2) : x. We get, 2 : 3 = (x – 2) : x x=6

23. Ans. 1098 Explanation: If n is the number of girls at junior level, then nC2 = 153 or n = 18 So, number of boys at junior level = 43 – 18 = 25 If m is the number of boys at senior level, then mC2 = 276 or m = 24 So, number of girls at senior level = 51 – 24 = 27

P-81

Quantitative Aptitude Simplified for CAT

So, total number matches a boy plays against a girl = 18  25 + 24  27 = 1098

So, x + z = 240 – 30 = 210 Since x  z, the minimum value of x is more than or equal to half of 210, that is, 105. Therefore, the minimum value of x is 105.

24. Ans. 50 Explanation: If the two cars move in same direction, Relative speed = v – u Then,

28. Ans. B Explanation: a1  a2  a3  ...  a52 a2  a3  a4  ...  a52  1 52 51

350 =7 v u  v – u = 50 km/hr

 51(a1 + a2 + a3 + … + a52) = 52 (a2 + a3 + a4 + … + a52) – 52  51  51a1 – a2 – a3 – a4 – … – a51 – a52 = – 2652  51a1 – a2 – a3 – a4 – … – a51 – 100 = – 2652  51a1 – a2 – a3 – a4 – … – a51 = – 2552  51a1 = a2 + a3 + a4 + … + a51 – 2552 For largest value of a1, the values of a2 to a51 should be the largest keeping in mind that a1 < a2 < ... < a52 and that a52 = 100. So, a51 = 99, a50 = 98, …, a2 = 50

25. Ans. 24 Explanation: t1 + t2 = 2(2)2 + 9(2) + 13 = 39 t1 + t2 + t3 = 2(3)2 + 9(3) + 13 = 58  t3 = 58 – 39 = 19 t1 + t2 + t3 + t4 = 2(4)2 + 9(4) + 13 = 81  t4 = 81 – 58 = 23 t1 + t2 + t3 + t4 + t5 = 2(5)2 + 9(5) + 13 = 108  t5 = 108 – 81 = 27 We observe that t3, t4, t5, … follows an Arithmetic Series pattern. If we assume that t3 is the first term, then tk will be (k – 2)th term. So, tk = 103 = 19 + (k – 3)4  k = 24

 a1 =

a2  a3  a4  ...  a51  2552 = 23 51

29. Ans. 20 Explanation: Refer to the following figure:

26. Ans. A Explanation: The ten’s place of such a number must be 3 or more. Such numbers are: 51, 61, 71, 81, 91 and 92. 27. Ans. 105 Explanation: x + y + z = 3  80 = 240 x + y + z + u + v = 5  75 = 375

(1)

xy yz x+y+z+ = 375  2 2  3x + 4y + 3z = 2  375 = 750  (3x + 3y + 3x) + y = 750 Using Eq. (1) in Eq. (2), we get

(2)

The curve of f(x) is straight line portion to the left of A and to the right of B, and parabolic portion between the points A and B. For minimum value with x positive, we need to find the y-coordinate of B. So, 5x = 52 – 2x2 2x2 + 5x – 52 = 0  2x2 – 8x + 13x – 52 = 0

3(240) + y = 1125  y = 750 – 720 = 30

P-82

CAT 2018 Solved Paper Slot 2 33. Ans. D

 2x(x – 4) + 13(x – 4) = 0 13 2 When x = 4, f(x) = 5x = 20  x  4, 

Explanation:

So, the minimum value of f(x) = 20

700 10   = 1   (700  175)  100 

Final fraction of alcohol after 2 substitutions

30. Ans. 36



Explanation:

2

700 81 567 = .  875 100 875

2x2 − ax + 2 > 0

Therefore, required percentage of water

 Discriminant, a2 – 4(2)(2) < 0

567   = 1    100 = 35.2% 875  

 a2 < 16

34. Ans. A

So, the maximum integral value of a is 3. x2

− bx + 8 ≥ 0

Explanation: Let the perpendicular from A on BC be h. Then,

 Discriminant, b2 – 4(1)(8)  0  b2  32

1  8  h  32 2 h=8 So, point A will lie on the line x = –4. For A to have shortest distance with (0, 0), A must lie on x-axis. So, coordinates of A are (–4, 0) and hence its distance from (0, 0) is 4.

So, the minimum integral value of b is −5. So, maximum possible value of 2a – 6b = 2(3) – 6(−5) = 36 31. Ans. B Explanation: PQ = {1, 4, 5, 6} RS = {1, 2, 3, 4, 7, 8, 10}.

Data Interpretation and Logical Reasoning (DILR)

So, (PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10} The number of elements = 7 32. Ans. A

Common explanation for questions 35–38: The word of is coded as 79, which means 9 codes either o or f. The digit 9 also codes a letter from the word national. So, 9 codes o. Therefore, 7 codes f. In the word Peacock, 9 codes one more letter. Also, 9 appears thrice in this word and one 9 appears for o, so the rest of two 9’s appears for a single alphabet which appears twice, and the only letter is c. So, 9 also codes c. The code for is is 35. The code for as is 56. So, 5 codes s. Hence, 3 codes i and 6 codes a. In India, code for i is 3 and that for a is 6. So, code for n and d is 1 or 6. But code for bird is 1334, so the code for d is 1, and so the code for n is 6.

Explanation: Let the concentration of ethanol in solution S be p%. Then, concentration of resultant solution when 20% ethanol is mixed with S 20(1)  p(3) 4 If 4 litres of this is mixed with 4 litres of 20% solution, the concentration of resultant solution is,

=

20(1)  p(3)  4  20  4 4  31.25 44  p = 50%

e is coded as 5 from the word designated. The is coded as 458, of which code for e is 5.

P-83

Quantitative Aptitude Simplified for CAT

So, code for t is either 4 or 8. Since the word national does not have code 4, the code for t can be 8 only. So, h is coded as 4.

Platinum

g is the present in only one word – designated. Keeping in mind codes of the remaining letters of the word designated, we can say that code for g is 7. Similarly, the code for L is 1, using the word national.

Economy

38

Total

4Y

1

D, L

A

2Y

Young

I

4

H

5

S, E

6

A, N

7

F, G

8

T, P, K

9

O, C

85 – A

X

55

Y

140

 Y  20 So, revised table becomes

2 3

X

4Y  2Y  Y  140

The code for B and R can be either of 3 or 4. Letters

Old

A 2

Gold

Similarly, code for P and k is 8.

Code

Middle Aged

Young

Platinum

Gold

A 2 42 –

Economy Total

Middle Aged

Old 20 – 2X

A

X

85 – A

X

55

A 2

38 80

40

20

140

39. Ans. A Explanation: Number of middle aged visitors buying platinum tickets = 20 – 2X. So,

35. Ans. A Explanation: The code for letter L is 1.

A  2(20–2 X )  A 2 A   40– 4 X 2  A  80– 8 X This is a multiple of 8. The only possible option is 32.

36. Ans. C Explanation: The code for B is 3 or 4. 37. Ans. C Explanation: This can be done for digits 8 and 9 only. So, only 2 such digits are possible.

40. Ans. 3 Explanation: If number of middle-aged visitors buying economy tickets is 20 – 2X, then 38  (20–2 X )  X  55 X 3 Which is also the number of old visitors buying Gold tickets.

38. Ans. B Explanation: Since S and E are coded as 5, we can say that S, U and V cannot have same digit, because each digit codes only 3 letters except 9. Common explanation for questions 39–42: As per the information given, we can develop the following table:

41. Ans. 0 Explanation: If number of middle-ages visitors buying Gold tickets is Z, then

P-84

CAT 2018 Solved Paper Slot 2

 42–

So, i > p, or weight assigned to I is more than that assigned to P. High Q is better than Best Ed. If f and r are the weights assigned to F and R, then 30f + 20r + 20p + 40i > 40f + 30r + 20p + 20i  2i  f  r

A  Z  X  85– A 2

A  Z  X  43 2 Also, 

X  42– 

A 2

f r 2 Also, Best Ed is better than Cosmopolitan. So, 40f + 30r + 20p + 20i > 40f + 20r + 20p + 30i r>i So, r > i > p. If f > r, then i is less than f and r and so i f r . Also, since i is cannot be more than 2 more than average of f and r and i is less r, so i has to be more than f. So, r = 0.4, i = 0.3, f can be 0.2 or 0.1

A  X  42 2

i

So, the minimum value of

A  X is 43 and so 2

Z = 0. 42. Ans. B Explanation: Check each option. Option (A): If given statement is true, then A  A   42–    42–   X  85– A 2  2   84  X  85 , which can be true.

f  r 0.1  0.4   0.25 and i 2 2 is more than 0.25.

Option (B): If given statement is true, then 38  X  X  55, which doesn’t result in natural number value of X. So, option (B) must be FALSE.

If f = 0.1, then

f  r 0.2  0.4   0.3 and i is 2 2 not more than 0.3. So, f = 0.1 and so p = 0.2. The overall scores table is shown as following:

If f = 0.2, then

Common explanation for questions 43–46: The following table gives the value score instead of grades:

F (0.1)

R (0.4)

P (0.2)

I (0.3)

Overall Score

A-one

50

50

50

40

47

20

Best Ed

40

30

20

20

26

20

30

Cosmopolitan

40

20

20

30

25

20

40

30

Dominance

20

20

40

30

26

50

50

40

50

Education Aid

50

50

40

50

48

Fancy

50

50

40

40

Fancy

50

50

40

40

45

Global

30

0

20

20

Global

30

0

20

20

13

High Q

30

20

20

40

High Q

30

20

20

40

27

F

R

P

I

A-one

50

50

50

40

Best Ed

40

30

20

Cosmopolitan

40

20

Dominance

20

Education Aid

Since Education Aid is better than A-one, and above table shows that both have same values in F and R, then 40p + 50i > 50p + 40i, where p and i are the weights assigned to P and I respectively.

43. Ans. A Explanation: The weight of the faculty parameter is 0.1.

P-85

Quantitative Aptitude Simplified for CAT

44. Ans. 3

revenue is more than 6.5. Out of (1), (2) and (3), the number (3) has revenue more than 6.5. So, (3) is Bravo and so (1) and (2) are Alpha. From point 3, in Promising category (which has total number 3), Alpha = 1, Bravo = 1, Charlie = 1. Using point 8, total revenue of Alpha = total revenue of Charlie. Now, Total revenue of Alpha = 28 + unknown one from Promising category = 28 + x Total revenue of Charlie = 21 + unknown one from Promising category = 21 + y So, y – x = 7. Therefore, (8) is Alpha and (9) is Charlie and hence (10) is Bravo. The above can be summarised in the following table:

Explanation: As we can see from the table, only three colleges receive AAA accreditation. 45. Ans. 48 Explanation: The highest overall score is 48. 46. Ans. A Explanation: No college scores between 31 and 40, both inclusive. Common explanation for questions 47–50: From point 1, in Blockbuster category (which has total 7), number in Alpha = number in Bravo. Also, from point 6, Revenue from Charlie > Revenue from Bravo. There are already 2 in Alpha given, and so there must be minimum 2 in Bravo. If there were one more in Alpha, then there would be 3 in Bravo and that would leave only 1 in Charlie, in which case revenue from Charlie won’t be more than that from Bravo. So, Numbers in Charlie = 3, Bravo = 2, Alpha = 2. Also, (11) must be Charlie and (7) must be Bravo so that Revenue from Charlie > Revenue from Bravo. Total number in No-Hope category is 6. From point 2, Alpha is more than Charlie, which is more than Bravo, the only possible combination is Bravo = 1, Charlie = 2 and Alpha = 3. Using point 7, Revenue from Bravo = Revenue from Charlie. Since revenue of Bravo is 4, revenue of Charlie (from 2 products) must be 4. So, (4) must be Charlie and either of (5) or (6) can be Charlie and the other one Alpha. From point 4, in Doubtful category (which has total number 7), Alpha has one more than Bravo, with Charlie having 0. So, Alpha = 4, Bravo = 3. Also, using point # 5, revenue from Bravo > revenue from Alpha. We already know which two are Bravo in Doubtful category, and that their revenue total is 8. Since total revenue from Doubtful category is 29, total revenue from Bravo must be more than 14.5. For this, the third Bravo should appear where

Product Popularity Score

Market Potential Score

Area of the figure (Proportional Revenue)

Company

Categories

2

5

4

Alpha

No Hope

2

11

2

Bravo

Doubtful

2

16

1

Alpha

Doubtful

4

4

4

Bravo

No Hope

4

13.5

6

Alpha

Doubtful

(2)

4.5

18

9

Bravo

Doubtful

(3)

5

9

2

Alpha

No Hope

6

2

3

Charlie

No Hope

6

11

1

Alpha

Doubtful

8

6

1

C/A

No Hope

(5)

8

9

1

A/C

No Hope

(6)

8

12

6

Bravo

Doubtful

8

17

4

Alpha

Doubtful

11

2

4

Bravo

Blockbuster

11

6

2

Charlie

Blockbuster

11

18

2

Alpha

Promising

13

7.5

3

Alpha

Blockbuster

14

3

6

Bravo

Blockbuster

14.5

14.5

9

Charlie

Promising

16

9

6

Charlie

Blockbuster

17.5

11

3

Bravo

Promising

(10)

18

2

9

Charlie

Blockbuster

(11)

18

6

6

Alpha

Blockbuster

P-86

(1)

(4)

(7)

(8)

(9)

CAT 2018 Solved Paper Slot 2 47. Ans. C Explanation: Category-wise proportional revenues are: No Hope = 15 Doubtful = 29 Blockbuster = 36 Promising = 14 So, the product category ‘Blockbuster’ has the highest revenue.

Note that maximum student enrolment is in L. So, 8 + 5 + 4 + (6 – z) > 17 z 5 + 4 + z + 9  z < 2.5. So, the maximum value of z = 2

48. Ans. C Explanation: From the table, we can say the correct sequence is 1, 3, 1, 2. 49. Ans. D Explanation: Check all the options using table. We realize that revenue of Bravo from Blockbuster is 10, which is less than the revenue of Alpha from Doubtful category (12). So, option (D) is definitely incorrect.

51. Ans. 4 Explanation: We need to minimize ‘G and L but not K’, that is, 6 – z. Since maximum value of z is 2, the minimum value of 6 – 2 = 4.

50. Ans. C

52. Ans. C

Explanation: Total revenue of Bravo = 34 crores

Explanation: Number of students in K = 18 + z, and that in L = 23 – z. The ratio of students in K to L = 19 : 22, means z = 1. So, number of students in L = 23 – 1 = 22

Common explanation for questions 51–54: From the information given, we can develop the following Venn Diagram:

53. Ans. 2 Explanation: Since number of students in K went down by 1, there is a shift of 1 student from being in all 3 sports to ‘G and L only’. Of the remaining 3 students, if all 3 move to ‘K and L only’, then difference between L and G would be = [8 + (5 + 3) + 0 + (6 – z + 1)] – [7 + z + 0 + (6 – z + 1)], which should be 6  23 – z – 14 = 6  z = 3, which is not possible. So, one student must go to ‘G and K only’ and remaining 2 go to ‘K and L only’. Therefore, number of students in K and L only is 5 + 2 = 7.

Also, 17 = 2x – 1 + z + x + (10 – x – z) x=4 We also know that total number of students is 39. So, 39 = 17 + x + y + y + 2x y=5 The revised Venn diagram is:

P-87

Quantitative Aptitude Simplified for CAT

55. Ans. A

For difference between L and G to be 6, number of students in ‘G and K only’ must be 2. So, number of students in ‘G and K only’ = 2

Explanation: The brand that had the highest revenue in 2016 is Azra.

54. Ans. D

56. Ans. C Explanation: The brand that had the highest profit in 2016 is Cxqi.

Explanation: Earlier there were total 10 students who had at least one more sport apart from G. Now, 2 of them moved to ‘K and L only’. That leaves us with 8, of which 2 are in G and K only (as seen in the previous solution). So, number of students in ‘G and L only’ will be 6.

57. Ans. B Explanation: The brand that had the highest profit in 2017 is Bysi.

Common explanation for questions 55–58: If sales volume of the entire market in 2016 is N, then number of units sold of Azra, Bysi, Cxqi and Dipq are 0.4N, 0.25N, 0.15N and 0.2N. So, revenue of Azra in 2016 = 15000 × 0.4N = 6000N Profit = 10% of revenue = 600N Working similarly for other brands, we get the following table: Brand

Market Share %

Unit SP (Rs.)

Profitability

Revenue

Profit

Azra

40

15,000

10

6000N

600N

Bysi

25

20,000

30

5000N

1500N

Cxqi

15

30,000

40

4500N

1800N

Dipq

20

25,000

30

5000N

1500N

58. Ans. D Explanation: The brands whose profits increased in 2017 over 2016 are Azra, Bysi and Dipq. Common explanation for questions 59–62: 100 units of L = 1 unit of A 120 units of L = 1 unit of B 1 unit of L = 1 unit of C To buy 1 unit of A, the outlet uses 5% less, that is, 95 units of L. To buy 1 unit of B, the outlet uses 114 units of L. To buy 1 unit of C, the outlet uses 0.95 units of L. Tabulating the buying and selling rates, we get

The market size in 2017 becomes 1.4N. The new unit selling price of Cxqi is 0.6 × 30000 = Rs 18,000. New market share = 15 + 15 = 30% And that of Azra, Bysi and Dipq are 35%, 20% and 15% respectively. The following table shows figures for 2017: Market Share %

Unit SP (Rs.)

Profitability

Azra

35

15,000

10

5250N

735N

Bysi

20

20,000

30

4000N

1680N

Cxqi

30

18,000

20

5400N

1512N

Dipq

15

25,000

30

3750N

1575N

Brand

Revenue

A

B

C

Base Exchange rate

100k

120k

1k

Buying rate

95k

114k

0.95k

Selling rate

110k

132k

1.1k

Net Addition

3300 –

4800 –

51000 –

2500 =

4800 = 0

48000 =

800

Profit

No of currency units

3000

X

Y

Z

P

Q

R

bought No of currency units sold

Amount of L used to buy A = 95kX Amount of L used to buy B = 114kY Ratio = 95kX : 114kY = 5 : 3

P-88

CAT 2018 Solved Paper Slot 2 59. Ans. 1200

So, X : Y = 38 : 19 = 2 : 1. Amount of L received from sale of A = 110kP; to buy B = 132kQ. Ratio = 110kP : 132kQ = 5 : 9 So, P : Q = 2 : 3 Since the outlet received 88000 units of L by selling A, 110kP  88000

Explanation: As we can see from the table, the number of units of currency A bought by the outlet on that day is 1200. 60. Ans. C Explanation: As we can see from the table, the number of units of currency C sold by the outlet on that day is 19000.

800 P  k 1200 k The outlet started with 2500 units of A and ended with 3300 units of A, and we also 800 units of A, it is know that the outlet sold k 800   clear that the outlet bought  800   k   units of A, which is equal to X. 1200 Also, the outlet bought units of B, k which is equal to Y. Now, X : Y = 2 : 1 or X = 2Y 800 1200  800   2 k k 1600  800  k k 2 From statement 1, 0.95kZ = 1.1kR or Z : R = 22 : 19. Since difference is 3000, we can say that the outlet bought 22000 and sold 19000. The revised table is as following:

And so, Q 

A

B

C

200

240

2

Buying rate

190

228

1.9

Selling rate

220

264

2.2

Net Addition

3300 –

4800 –

51000 –

2500 =

4800 = 0

48000 =

Base

Exchange

61. Ans. 240 Explanation: As we can see from the table, the base exchange rate of currency B with respect to L is 240. 62. Ans. C Explanation: The buying exchange rate of currency C with respect to L is 1.9. Common explanation for questions 63–66: Room 102 had only 2 candidates allotted to, and Erina was allotted to the room which had no other candidate, and this cannot be 101, as room 101 had Balaram as the third person. So, 103 was allotted to Erina. Room 101 was allotted to Balaram, Akil, Fatima and one more person. Room 102 was allotted to Ganeshan and one more person. Room 103 was allotted to Erina. Fatima was the fourth one to enter 101. Since Chitra was the last person to enter the room she was allotted to, so it cannot be room 101, as Fatima enters after Chitra and Fatima is allotted room 101. So, Chitra was allotted room 102. The person entering at 7:45 am cannot be allotted room 101, as Fatima is the last person to enter room 101 (because there are a total of 4 person allotted room 101). So, the one entering at 7:45 am is allotted room 103, which is Erina. We can now say that Divya is allotted room 101.

rate

800 No

of

currency

3000

1200

600

22000

400

600

19000

units bought No. of currency units sold

P-89

Quantitative Aptitude Simplified for CAT

65. Ans. D

The table now looks as following: 7:10

7:10

7:15

7:25

7:30

7:40

7:45

am

am

am

am

am

am

am

Person

Akil

?

?

?

Chitra

Fatima

Erina

Room

101

102

101

103

Time

Explanation: We can see from the table, Erina reached the venue at 7:45 am. 66. Ans. C Explanation: If Ganeshan enters at 7:15 am, then Divya would enter at 7:25 am, which would mean that Balaram won’t be the entering room 101 as the third person. So, Ganeshan has to enter the venue at 7:10 am. Now, between Divya and Balaram, Divya comes before Balaram, to enable Balaram to be the third person to enter room 101. So, Divya enters at 7:15 and so Balaram enters at 7:25 am.

63. Ans. A Explanation: From the discussion, we see that Divya was allotted room 101. 64. Ans. D Explanation: Since Ganeshan entered room 102 and Chitra entered after Ganeshan, there was no one in the room when Ganeshan entered room 102.

P-90