ALL THAT MATTERS IN MATH - KG-VIII EXPLAINED EXAMPLES – Volume-I - PART II (A to Z mathematics Series) 9788195396214
A to Z of Mathematics - The Family Edition Did you know that 14 years of K–12 math curricula could be judiciously packed
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ALL THAT MATTERS IN
MATH The ONLY Way To Learn Math And Sharpen Habits of Thinking
Volume I – Part II KG-VIII Explained Examples
ALL THAT MATTERS IN MATH Volume I - Part II KG-VIII Explained Examples The Only Real Math Book
Copyright© Sandeep Srivastava, 2022 The moral right of the author has been asserted. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form, or by any means, without the prior permission in writing of the publisher, or be otherwise circulated in any form of binding or cover other than that in which it is published and without a similar condition, including this condition being imposed on the subsequent purchaser.
ISBN: 978-81-953962-1-4 Domain editor: Saloni Srivastava Contributory editor: Nikita Todarwal Editorial team: Nidhi Gandhi Cover design: Nikita Singh Publishing team: Dipti Chauhan (Lead), Manish Kumar, Yogesh Kumar, Tanu Gaur, Harmanpreet Singh, Manish Bhati Published in India in 2022 by Nextgen Books Private Limited, the publishing partner of IYCWorld Softinfrastructure Private Limited
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Our brain already knows all the math that will ever be known! Math education just needed a book of real mathematics.
PART II
A to Z Mathematics Series CONTENT
L
for Long Form Mathematical expressions that expand 1. Exponents and Numbers ............................................3 Writing and evaluating exponential form of numbers Working with zeroth and negative exponents Comparison of exponential numbers Addition of exponents of same base Multiplication of different bases with same exponents Division of same bases with different exponents Division of different bases with same exponents Interpreting power of a power Writing daily-life numbers as exponents Real-life application of exponents 2. Square of Numbers ...................................................24 Square numbers Numbers ending with digits 2, 3, 7, or 8 are non-perfect squares Zeros at end of a perfect square Square of even number is always an even number Square of odd number is always an odd number Square of numbers ending with 5 Difference between two consecutive squares Non-perfect squares between two consecutive square numbers Sum of first ‘n’ odd natural numbers Pythagorean triplet Patterns in square numbers 3. Cube of Numbers .....................................................31 Cubic numbers Cube of rational numbers To make perfect cube–by prime factorisation and multiplication To make perfect cube – by prime factorisation and division
Sum of ‘n’ consecutive cubes Difference between the squares of two consecutive triangular numbers 4. Logarithms and Antilogarithms ..............................36 Definition of logarithm Power rule Quotient rule Product rule Antilog of numbers Antilog law 5. Ratios ........................................................................40 Idea of ratio Simplest form of ratios Comparison of ratios Equivalent ratios – multiples of the given ratio 6. Rate – A Special Kind of Ratio ..................................50 Idea of rate Rate 7. Unitary Relationship ................................................54 Idea of unitary ratio Unitary ratio 8. Percentage ................................................................57 Idea of percentage How do we convert ratios to percentage? Difference between “what percentage of x is y?” and “what percentage is x of y?” Conversion of decimal into percentage Quantifying percentage Conversion of percentage into fraction or decimal form 9. Proportion ................................................................70 Idea of proportion Are the numbers in proportion? Finding the missing term
Continued proportion Mean proportion Direct proportion Inverse proportion 10. Arithmetic Progression .........................................82 Finding missing term in AP Characterising arithmetic progressions Using the general term of arithmetic progression How to assume terms in AP? Sum of n terms of an AP Applying the concept of arithmetic progression 11. Geometric Progression ..........................................94 Finding the term in geometric progression Finding the geometric progression Finding the sum of geometric progression Finding the sum to infinity of the geometric progression Applying the concept of geometric progression 12. Harmonic Progression ........................................101 Finding the term in harmonic progression 13. Algebraic Expressions .........................................103 Constants, variables/literals and expressions Creating algebraic expressions Finding values of algebraic expressions Addition and subtraction of algebraic expressions Multiplication of algebraic expressions Division of algebraic expressions 14. Algebraic Identities .............................................111 Identity : (a + b) (a – b) = a2 – b2 Identity : (a + b)2 = a2 + 2ab + b2 Identity : (a – b)2 = a2 – 2ab + b2 Identity : a3 + b3 = (a + b)3 – 3ab (a + b) Identity : a3 – b3 = (a – b)3 + 3ab (a – b) Identity : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Identity : a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) Factorising algebraic expressions 15. Divisibility Test ....................................................121 Utilising divisibility : By 2, 5, 10 Utilising divisibility : By 3, 6, 9 Utilising divisibility : By 4, 11
M
for Matter More about numbers, exploring factors, multiples 16. Factors and Multiples of Numbers ..........................125 Idea of factors Finding factors of numbers Idea of multiples Finding multiples of numbers 17. Prime and Composite Numbers ............................129 Defining prime and composite numbers Twin prime and co-prime numbers Prime triplets 18. Fundamental Law/Theorem of Arithmetic ...........132 Prime factorisation of numbers using division method Prime factorisation of numbers using factor tree method 19. Highest Common Factor (or Greatest Common Divisor) of Numbers ..............................................134 Finding HCF using factors Finding HCF using prime factorisation method Finding HCF using division method Finding HCF using Euclid’s division lemma and algorithm Application of HCF in real-life 20. Least Common Multiple of Numbers ....................140 Finding LCM by listing the multiples Finding LCM using prime factorisation method Finding LCM using division method Application of LCM in real-life 21. Applying Fundamental Theorem of Arithmetic ....144 Relationship between HCF and LCM 22. Square Root of a Number ......................................146 Finding square root of a number using repeated subtraction method Estimating square root of a number Finding square root of a number using prime factorisation method Finding square root of a number using long division method Finding square root of fractions Finding square root of decimal numbers
23. Cube Root of a Number ........................................153 Finding cube root of a number using prime factorisation method Estimating cube root of a number Finding cube root of a negative and rational number 24. The Order of Operation ........................................156 Using BODMAS rule to simplify the expressions
N
for Negative Numbers The numbers we created 25. Visualisation of Negative Numbers .....................161 Visualising situations involving negative numbers Visualising negative numbers on number lines 26. Operations on Negative Numbers .......................164 Addition of negative numbers Subtraction of negative numbers Multiplication of negative numbers Division of negative numbers
Zero property of multiplication of positive and negative numbers 34. Closure Property of Numbers ............................179 Closure property of addition of whole numbers and integers Closure property of subtraction of integers Closure property of multiplication of whole numbers and integers
O
for Organising (data) Numericalising geometry, exploring coordinate systems, graphs 35. Cartesian System: 1-Dimensional space ...............182 Representing situations using integers Plotting points in 1-dimensional space 36. Cartesian System: 2-Dimensional space ................183 Abscissa, ordinates and quadrants Plotting ordered pairs in quadrants
27. Commutative Property of Numbers ...................169
37. Graph – A Cartesian Plane ...................................185
Commutative law of addition Commutative law of multiplication
Dependent and independent variables in graph Plotting points (or ordered pairs) to make graphs Finding distance between two points Defining and using section formula Mid-point formula Centroid of a triangle - application of section formula Area of a triangle Visualising collinearity of three points
28. Associative Property of Numbers ......................172 Associative law of addition Associative law of multiplication 29. Distributive Property of Numbers .......................174 Distributive law of multiplication over addition Distributive law of multiplication over subtraction 30. Additive Identity or Zero Property .....................176 Additive identity law of positive and negative numbers 31. Multiplicative Identity Property .........................177 Multiplicative identity law of positive and negative numbers 32. Inverse Property ....................................................177 Additive inverse property of positive and negative numbers Multiplicative inverse property of positive and negative numbers 33. Zero Property of Multiplication ........................179
38. Application of Coordinate System – Statistical Graphs ..................................................196 Collecting and organising data Making pictograph of data Making bar graph from data Making histogram of data Making pie charts of data 39. Cumulative Frequency Distribution Tables and Curves ....................................................................209 Cumulative frequency table (ungrouped data) Cumulative frequency table (grouped data)
Introduction to Basics of Other Volumes 40. Comparing Quantities ..........................................214 Idea of profit and loss Idea of discount Idea of simple interest Idea of compound interest 41. Triangle and its Properties ...................................219 Idea of different parts of a triangle Classifying triangles on the basis of sides Classifying triangles on the basis of angles Idea of a median and altitude of a triangle Angle sum property of a triangle Exterior angle property of a triangle Triangle inequality property Pythagoras theorem 42. Congruency of Triangles ......................................225 Idea of congruency Idea of SSS congruency Idea of SAS congruency Idea of ASA congruency Idea of AAS congruency Idea of RHS congruency
Finding area between rectangles Finding area enclosed by two concentric circles Finding area of a trapezium Finding area of a polygon Finding volume of a cuboid Finding volume of a cube Finding volume of a cylinder Finding volume of a sphere Finding volume of a right circular cone Finding surface area of a cuboid Finding surface area of a cube Finding surface area of a cylinder 45. Visualising Solid Shapes .......................................254 Diagnosing solid shapes Drawing nets for building 3D shapes Euler’s relation for polyhedrons 46. Symmetry ..............................................................257 Idea of symmetry Identifying lines of symmetry Idea of reflection symmetry Idea of rotational symmetry
43. Understanding Quadrilaterals .............................230
47. Probability .............................................................261
Idea of quadrilaterals Idea of polygon’s interior angles Angle sum property of quadrilaterals Exterior angles of quadrilaterals/polygons Properties of parallelograms Properties of rectangles Properties of squares Properties of rhombuses Properties of kites and trapeziums
Idea of probability Probability of throwing of dice Probability of tossing coins
44. Mensuration .........................................................240 Finding perimeter and area of a triangle Finding perimeter and area of a rectangle Finding perimeter and area of a square Finding area of a parallelogram and rhombus Finding circumference and area of a circle
PART I
A
for Approximation The world of pre-number quantification 1. Approximate Number System Estimating quantities Visual estimation Visual isolation and estimation Estimating age
Estimating weight Estimating capacity
Forward or skip counting Backward counting
2. Subitising
13. Roman Numerals
Perceptual subitising Conceptual subitising Subitising as key to numeration Number patterns and number conversation Subitising with five frames Subitising with ten frames Subitising with domino tiles Part and whole in five and ten frames Splat subitising
Formation of Roman numerals
3. One to One Correspondence Comparison of quantities using one to one correspondence
B
for Bounty Learning to count 4. Ordinal Numerals Using ordinal numerals to count things Using ordinal numerals to know the position of things 5. Cardinal Numerals Using cardinal numerals to show quantity 6. Why do we count? Difference between ordinal and cardinal numerals 7. Unit is everything in Counting
C
for Calibration Learning to measure 14. Non-Standard Unit of Measurement Expressing quantities measurement
in
non-standard
units
of
15. Standard Unit of Measurement Standard unit of measurement for length, weight, volume Estimating quantities with respect to a standard unit 16. Conservation of Measurement Conservation of weight Conservation of volume 17. Unit of Measurement Measuring less than a unit Measurement using fractional units What will be counted, what measured? 18. Conversion of Units Conversion from one unit to another 19. Measuring Space
8. Make your own Number System
Exploring area and perimeter by counting number of squares Exploring volume
A number system – all about ‘0’ (starting number) and the ‘1’
20. Measuring Time
9. How to count?
Exploring calendar Reading time in clock Estimating time and time elapsed Creating timeline
Unit of counting
Counting of like things Counting of unlike things 10. What to count? Differentiate between like and unlike units Converting unlike units to like units 11. Counting Bigger Quantities Counting using tally marks 12. Types of Counting
D
for Decimal Numbers Quantities are expressed as numbers 21. Decimal Number System 0–9 Number system
IX
The idea of ‘10’ The idea of ‘0’ Reaffirming the concept of ‘0’ Digits, numerals and numbers
Base 2 number system 30. Ternary Number System Base 3 number system 31. Quinary Number System
22. Numbers as Packets
Base 5 number system
Numbers in packets Place value and face value
32. Octal Number System
23. Concept of Number Names
Base 8 number system
Unique names Combined (word) names Creating number names
33. Nonary Number System
24. Exploring Numbers
Base 12 number system
Base 9 number system 34. Dozenal Number System
Comparison of numbers Greatest and smallest number formation Expanded form of numbers Indian and International number system Odd and even numbers Prime and composite numbers
35. Hexadecimal Number System Base 16 number system 36. Octodecimal Number System Base 18 number system 37. Duotrigesimal Number System
25. Approximation of Numbers
Base 32 number system
F
Rounding off numbers 26. Implication of Numbers as Packets Carry – a property of numbers Borrow – a property of numbers Trailing zeros in multiplication Trailing zeros in division
for Fractional Quantities The real-world numbers 38. Fractional Quantity What is fractional quantity? The unit (1) has to be same for comparison
27. Graphically Representing Numbers Number line
39. What is Fraction?
28. Decimal Number Formation
Unit of the fraction Numerator and denominator in fraction Identifying equivalent fractions
Visualising numbers with a decimal point Expanded form of decimal numbers Representing decimal numbers through blocks Like and unlike decimal numbers Comparison of decimal numbers Addition of decimal numbers Subtraction of decimal numbers
40. Types of Fractions Like and unlike fractions Proper, improper and mixed fractions 41. Conversion of Mixed and Improper Fractions
E
Improper fraction to mixed fraction Mixed fraction to improper fraction
for Extended Number System Playing with numbers
42. Rational and Irrational Numbers Rational numbers
29. Binary Number System
X
Irrational numbers Rational numbers between two given rational numbers 43. Comparison and Operations on Rational Numbers Comparison of rational numbers or fractions Addition of rational numbers or fractions Subtraction of rational numbers or fractions Multiplication of rational numbers or fractions Division of rational numbers or fractions 44. Fraction and Decimal Numbers Fraction to decimal numbers Decimal numbers to fraction Relationship of 1 of measurement and 1 of fraction Unit fraction of decimal numbers 45. Real and Imaginary Numbers Real and imaginary numbers Complex numbers – combination of real and imaginary numbers 46. Integers Exploring integers 47. Representation of Real Numbers on Number Lines Representing different types of numbers on number line
G
for (Basic) Geometrical Ideas Visualising quantities 48. Point and Lines Defining a point Points at the interior, exterior or boundary of closed figures Collinear and non-collinear points Types of lines – lines, line segments and rays Intersecting, parallel and perpendicular lines 49. Angles Vertex and the angles Interior and exterior points in angles Types of angles – acute, obtuse, reflex, right, straight and complete angles Linear pair, adjacent and vertically opposite angles Complementary and supplementary angles Angles made by a transversal
50. Planes Defining a plane – parallel and perpendicular planes Differentiating open, closed and complex curves Polygons – closed figure made from line segments Circle – closed curve, equidistance from a point inside it 51. Construction Construction of a circle Construction of a line segment Construction of a perpendicular bisector of a line segment Construction of an angle using protractor Construction of an angle equal to a given angle Contraction of an angle bisector Construction of angles 30°, 60°, 90°, 120° using compass Construction of parallel lines Dividing a line segment into equal parts Construction of triangles Construction of quadrilaterals
H
for Higher The idea of sum, exploring addition 52. What to Add? Adding like things Adding unlike things 53. Counting Bigger Quantities – Addition Adding quantities by making groups Addition by counting objects in rows and columns Addition through forward skip counting Addition by using number line Addition with number blocks Addition with carry by using column addition method Addition with carry on carry using column addition method 54. Real-life Application of Addition Addition operation application
I
for Important Difference is important, exploring subtraction 55. Need for Subtraction Why we need subtraction?
XI
56. Subtraction – A Simple Introduction
66. Long Multiplication Method
Backward skip counting
Multiplication of two numbers Multiplication of multiple numbers Multiplication of decimal numbers
57. Components of Subtraction Expression Backward skip counting and subtraction expression Minuend, subtrahend and difference 58. Subtraction of Unlike Things Subtracting unlike things 59. Three Types of Subtraction Take away, comparison, part-of-a-whole 60. Visualising Subtraction Subtraction with number blocks Subtraction by using column subtraction method Subtraction with cascade borrow Subtracting multiple subtrahends 61. Real-World Application of Subtraction Application of subtraction operation
J
for Jump Rapid increase is common, exploring multiplication
67. Real-World Application of Multiplication Application of multiplication operation Addition and multiplication operation application
K
for Knife Making groups/packages is routine, exploring division 68. Repeated Subtraction and Division Expression Relation between repeated subtraction and division expression 69. Dividend and Divisor Exploring the nature of dividend and divisor 70. Quotient and Remainder Exploring the nature of quotient and remainder 71. Relationship Between Multiplication and Division
62. What is Multiplication?
Multiplication expression for division expression and vice versa
Faster addition of same quantities
72. Division by Subtraction
63. Components of Multiplication Expression
Division by repeated subtraction Faster way of division – bigger takeaway
What is a multiplier? What is a multiplicand? 64. Understanding the Difference between Expressions Difference between addition and multiplication expression Possible and impossible cases of multiplication
73. Long Division Method Division using long division method 74. Cascading Division
65. Multiplication Table
Solving division expressions from top to bottom
How to make multiplication table? Quantifying multiplication
75. Real-World Application of Division Division operation application
FOREWORD First things first This is the third book of the KG – VIII mathematics volume, the Reading Book is the first of the volume. This is a unique book of math in many ways, such as – 1. This is world’s first mathematics book for curricular study, arithmetic we study in schools is just a part of mathematics. Pertinently, the Draft Indian National Education Policy, 2019, stated that (true) mathematics has five dimensions:
MATHEMATICS COUNTING
ARITHMETIC
MATHEMATICS
REASONING
PROBLEM SOLVING
Hope you did register that one of the dimensions of mathematics is named “mathematics”. This just emphasizes the core of mathematics – mathematical thinking. ‘True mathematics’ sharpens critical, creative and logical thinking, and makes us better decision makers! This book is about all five of the above, up to a curricular level. 2. This is the world’s first book that broadly presents 10 Grades curricula (KG – VIII) in one volume; indeed, mathematics is rigidly hierarchical and logical, there is only that much needs to be learnt as a foundation. This book logically connects math concepts like never before. 3. This is the world’s first book of explained examples that is almost an independent book; of course, the Reading Book Volume I takes logical and critical thinking to a very different depth for each concept. This is the most complete book on foundations of math. 4. This is world’s first book that converts the entire curricula into practical, visualisable, and touchable domain of knowledge. And the result – ALL parents will be equally good in supporting math education. This book will kill the need for (wasteful, hurtful) tuition forever. 5. This is the world’s first book to fill all the gaps in school math education and ensure excellence in school math exams, including the traditionally difficult ‘word problems’. For example, the books explains why 1 is 16 ( ); to be right, it’s not 16 anyway. 2 This book is the ONLY way to master ‘school math’ with utmost confidence.
8 1 2
Back to square one Pertinently, the current global pandemic has created an unprecedented 2-year disruption in the formal education system. Unfortunately, the system can’t act to acknowledge the gaps, for example, a child in
Grade I in 2019-20 session will be offered Grade IV books in 2022-23, and taught accordingly. And let’s know that this disruption is not one-time event, and our lives and the world has changed forever.
Parents have to step in The only guarantee of continuity in children’s lives is family, and parents must take direct charge of academics, especially math (it’s anyway a struggle for the whole of humanity in good times). Moreover, math is the language of life and business because non-human intelligence is possible only for those situations that can be mathematised/logicalised. NO ONE WILL STRUGGLE IN MATH NOW.
Mathematics as a language Mathematics as a language is fundamentally different from ‘school math’, and it’s impossible for any human to be struggling in mathematics – it’s the easiest language ever invented – and it’s purely logical (learnable without any dedicated resources and experiments, or even a community of learners). Mathematics is NOT arithmetical methods, devoid of all logic!
Our books are magical Excelling in mathematics is now only a function of intent and effort at the family and community level.
Born equal Every child is born with the same cognitive and other capabilities. The quality of their education determines their future quality of life and contribution of their society. Math is the best domain to live that equality.
Friendship with math pays For school-going children, there are literally no model philomath adults. Fear of math is probably the most common social affliction. A happy association with math is almost a pre-requisite for a happy childhood and a happy school-life. A disproportionate amount of time, attention, and money is invested in supplementing and complementing math education at school, and this affects performance in other curricular and co-curricular subjects and interests. Struggling in math has a cascading effect on overall learning. Thus, math must be specially attended to.
How does every parent matter? Surprising as it may be, many of the important educational developments of children are such that the determinants of their parents’ ‘life and professional conditions’, for instance, their education, profession, economic or social status, and cultural environment, don’t matter. Therefore, every parent matters.
How is each parent unique? Dr Spock famously said “avoid doling out love (to children) only when earned”; indeed, as parents we must create conditions that help children easily ‘earn love’ if they must. He pointed out that not only are children unique but so are their circumstances. Parents are the reason why the circumstances are unique for every child. The uniqueness comes from the varied degree of commitment to parenting, being a role model learner (not necessarily academics), learning together, etc.
This is the world’s best primary-school math education @ a pizza’s cost. Saloni Srivastava XIV
The Best Way To Use This Book! This is a unique book of examples, there are many firsts for a math book to make the book a self-learning resource. For instance, every example with multiple steps in solution carries ‘never-seen-before’ basic logical explanation for each step. We have collated a set of instructions for the best use of the book and a careful reading of the same would ensure the necessary gains in mathematical thinking.
Detailed Index with 15 Chapters, 122 Lessons and 350 Learning Outcomes in Volume I (Part I & Part II)
M
for matter
M for Matter
More about numbers, exploring factors, multiples 16. Factors and Multiples of Numbers Idea of factors1 Example 1 Define factors of a number by writing all parts of 20 using multiplication.
XV
Lesson
M
Learning outcome
Chapter
More about numbers, exploring factors, multiples 16. Factors and Multiples of Numbers ................................. Idea of factors Finding factors of numbers Idea of multiples Finding multiples of numbers Defining prime and composite numbers
A brief explanation of the learning outcomes is given wherever required.
Sum of ‘n’ consecutive cubes26 Sum of ‘n’ consecutive perfect cubes = Square of the sum of ‘n’ consecutive numbers Example 1 Add: 13 + 23 + 33 + 43 + 53 + 63 + 73 Solution Procedure
Explanation
1 + 2 + 3 + 4 + 5 + 6 +7 = (1 + 2 + 3 + 4 + 5 + 6 + 7)2 3
3
3
3
3
3
Given ∵ 13 + 23 + 33 + … + n3 = (1 + 2 + 3 +…. + n)2 Here, n = 7. Since, 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. Since, 28 × 28 = 784.
3
= (28)2 = 784 Hence, 784 is the required sum of the given series.
Difference between the squares of two consecutive triangular numbers27 Difference between squares of two consecutive triangular numbers
Cube of difference of two triangular numbers
=
Example 1 What is the value of 102 – 62? Solution Procedure
Explanation
10 – 6 = (10 – 6) = 4 = 64 2
2
3
Since, difference between the squares of two consecutive triangular numbers equal to the cube of difference of two triangular numbers. Here, 6 and 10 are consecutive triangular numbers.
3
A triangular number is the count of objects arranged in an equilateral triangle. Pictorially,
1
3
6
10
Therefore, 32 – 12 = (3 – 1)3 = 23 and 62 – 32 = (6 – 3)3 = 33 Example 2 Find the next triangular number in the series 45, 55, … 26 27
New learning outcome Refer P 28
Comments explain the concept used to solve the example. Every learning outcome has a reference number. In the footer, the reference number lists the page number on which this concept can be found in the KG-VIII Reading Book, All that matters in Math, Volume I.
XVI
L L for Long Form
Mathematical expressions that expand Revising the arithmetic used in this chapter (a) ‘of ’ is considered as one of the arithmetic operations which means ‘multiplication’. 3 of 4 = 3 × 4 3 × 4 = 3 of 4 multiply
(b)
4 1 = 4 ÷ 1 3 3
4 (c) 3 = 4 5 3 2
4 ×
=
3 = 12 1
inverse multiply
÷
5 2
4 2 8 × = 3 5 15
=
inverse
(d) 3 + 2 4 5 3 2 and are unlike fractions. 4 5 For addition, first make them like fractions by taking the LCM of denominators. Procedure LCM of 4 and 5 is 20.
3 2 + 4 5 3×5 2×4 = + 4×5 5×4
∴
Explanation 2 4, 5 2 2, 5 5 1, 5 1, 1 LCM of 4 and 5 = 2 × 2 × 5 = 20 Finding equivalent fractions with same denominator (like fractions).
1
2 A to Z of Mathematics
Adding like fractions to get the sum.
15 8 + 20 20 15 + 8 = 20 23 = 20 =
In the same way, subtraction operation is performed. (e)
n
x=x
1 n 1
1
1
1
∴ 3 5 = 53 ; 2 5 = 5 = 52 ; 5 5 = 55 ; 4 5 = 5 4 (f) Multiplication of integers • (–) × (–) = (+) ⇒ (–3) × (–3) = 9 • (+) × (–) = (–) ⇒ 3 × (–3) = –9 • (–) × (+) = (–) ⇒ (–3) × 3 = –9 • (+) × (+) = (+) ⇒ 3 × 3 = 9 (g) (–x)n = (–1)n (x)n = –(x)n if n is an odd number. = xn if n is an even number. For example, (–3)2 = (3)2 = 9 [Exponent 2 is an even number.] (–4)3 = –(4)3 = –64 [Exponent 3 is an odd number.] (h) Solving equations using cross multiplication method. x+3
3 x+2 4 x+3 3 Thus, (Using cross multiplication method) = x+2 4 ⇒ 4 (x + 3) = 3(x + 2) [Multiply 4 with (x + 3) and 3 with (x + 2)] ⇒ 4x + (4 × 3) = 3x + (3 × 2) (Each term inside the bracket is multiplied.) ⇒ 4x + 12 = 3x + 6 ⇒ 4x – 3x = 6 – 12 ⇒ x = –6 (i) Exponents and their rules =
−−11
33 = 33 = 1 1 −−11
11 33 11 11 = = = = = 11÷÷ = 11×× = = 11 33 33 33 1 3 ) 3 (= = 13 11 3 ) of = 3 inverse (= −1
−1
−1
−1
−1
−1
1 −1 3 3 −1 ) −1 = (= 3 −1 −1
1 1 =1 ÷ =1 × 3 = 3 1 3 inverse of 3 1 3 3inverse 3 (= ) of=
(3 ) −1
1 = 3
3
=
−1
−1
−1
3
Mathematical expressions that expand 3
1. Exponents and Numbers Writing and evaluating exponential form of numbers1 Example 1 Simplify: (a) 34
(d)
2 (c) (2a)2 (3b)
(b) (–12)3
4 3
2
Solution Procedure (a) 3 =3×3×3×3 = 81 (b) (–12)3 = (–12) × (–12) × (–12) = [(–12) × (–12)] × (–12) = (144) × (–12) 4
= –1728
(2a ) 2 (c) (3b) 2 2a × 2a = 3b × 3b (2 × 2)×(a × a) = = (3 × 3)×(b × b) 4a 2 = 2 9b 4 (d) 3
2
Given Writing 34 in expanded form. Given Writing (–12)3 in expanded form. Multiplying the first two numbers. (−) × (−) = (+) and (–12) × (–12) = 144 ( + ) × ( −) = ( −)
Multiplying the product of first two numbers to the third number. Given Writing the numerator and denominator in expanded form. Separating the constants and variables in the numerator and denominator.
Given
=
4×4 3×3
Write the numerator and denominator in expanded form.
=
16 9
Simplifying
Example 2 Simplify: (–3) × (–2)3
1
Explanation
Refer P 243
4 A to Z of Mathematics Solution Procedure
Explanation
(–3) × (–2)3
Given
= –3 × (–2 × –2 × –2)
Expanding (–2)3
= –3 × (–8) = 24
(–) × (–) × (–) = (–); (–2 × –2 × –2) = –8 (–) × (–) = (+); (–3) × (–8) = 24
When the exponent of a negative number is odd, the expansion forms a negative number. When the exponent of a negative number is even, the expansion is a positive number. (–x)a = (–1)a (x)a = –xa (If ‘a’ is an odd exponent) a =x (If ‘a’ is an even exponent) Example 3 –2 –2 Express a × a × a × × in exponential form. 3 3 Solution Procedure
–2 –2 a × a × a × × 3 3
Explanation –2 a is multiplied by itself 3 times and is multiplied by itself 3 2 times.
2
–2 = a3 3 2
–2 Hence, a 3 is the simplified exponential form. 3 Example 4 Express 729 in exponential form with base 3. Solution Procedure
Explanation
Prime factors of 729 =3×3×3×3×3×3
By prime factorisation.
⇒ 729 = 36
3 is multiplied by itself 6 times.
Hence, 36 is the exponential form of 729 with base 3.
Mathematical expressions that expand 5
Working with zeroth and negative exponents2 Example 1 Simplify: (a) (4)–3
1 32 –1 (b) –3 ÷ 0 4 4
Solution
(c)
1 1 4
–3
× 3–5
Procedure (a) (4) =
1 43
a–n =
1 4 × 4 ×4 1 == 64
Simplifying –1
3 × 3 = 43 ÷ 1
Given
–1
= [64 ÷ 9]–1 64 −1 == 9 9 == 64 1 × 3–5 (c) –3 1 4
1 1 × 5 3 4 3 1 1 = × 4×4×4 3×3×3×3×3
=
1 1 × 64 243 1 = 64 × 243 =
2
Refer P 243
1 an
43 = 4 × 4 × 4
= =
1 32 (b) –3 ÷ 0 4 4
Explanation Given
–3
(d) 1 + 60 −1 4
1 1 = = a n and a0 = 1 −n 1 a an
43 = 4 × 4 × 4 = 64 and 3×3 = 9 .
a b
−1
=
1 b = a a b
Given
1 1 =a n and a − n = n −n a a
Writing the numbers in expanded form. 4 × 4 × 4 = 64 and 3 × 3 × 3 × 3 × 3 = 243 .
0
6 A to Z of Mathematics
= =
1 15552
64 × 243 = 15552
1 (d) 1 + 60 4 = (4 + 1)0 = (5)0 =1
Given
0
1 = a n and and a 00 ==11. –n a
a0 = 1
a) Anything to the zeroth power is ‘1’, i.e., 10 = 1. b) 00 is not defined. 1 –m –m c) a = m , where m is a positive integer and a is the multiplicative inverse of am. We will learn a about multiplicative inverse in ‘Chapter N’. Example 2 –1
–1 Simplify: 6 –1 + 3 + 50 2 Solution
Procedure –1
–1 3 –1 0 6 + + 5 2 –1
1 2 0 = + +5 6 3 –1
1+ (2× 2) 0 = +5 6 –1
5 = +1 6
=
6 +1 5
6 + (1× 5) 5 11 = 5
=
Explanation
Given ∵ a–n =
1 an
Adding the fraction inside the parenthesis by taking the LCM of 6 and 3 as 6. ∵ 50 = 1 1 an ∵ LCM of 5 and 1 is 5. ∵ a–n =
∵ 6 + (1 × 5) = 11
Example 3 By what number should (–8)–1 be multiplied so that the product would be equal to 10–1?
Mathematical expressions that expand 7
Solution Procedure
Explanation
Let (–8)−1 be multiplied by x to get 10−1.
Assumption
x × (−8)−1 = 10−1 ⇒ x = 10−1 ÷ (−8)−1
Product of two numbers = 10−1 (Given)
⇒x=
1 . an 1 1 1 −8 ∵ 10 ÷ −8 = 10 × 1
1 1 ÷ 10 –8
∵ a–n =
–8 10 –(4 × 2) ⇒x= 5×2 –4 ⇒x= 5 ⇒x=
Reducing –8 to its lowest equivalent 10 rational number.
Hence, the required number is
–4 . 5
Comparison of exponential numbers3 Example 1 Which is greater: 53 or 35? Solution Procedure
Explanation
53 = 5 × 5 × 5 = 125
Expanding 53.
35 = 3 × 3 × 3 × 3 × 3 = 243 Hence, 35 is greater than 53.
Expanding 35. ∵ 243 > 125
Example 2 Compare: Solution
1 × 2 –4 × 43 and 14 0 + 33 − 2 –1 16
Procedure 1 × 2−4 × 43 16 1 1 == × 4 × 43 16 2 1 1 = × ×4×4×4 = 16 2 × 2 × 2 × 2 1 1 × ×2×2×2×2×2×2 = 2 × 2 × 2× 2 2 × 2 × 2 × 2 3
New learning outcome
Explanation Given a −n =
1 an
Writing 24 and 43 in expanded form. ∵ Writing the numerator and denominator in the form of their prime factors.
8 A to Z of Mathematics
1 4 140 + 33 – 2–1
= =
....(i)
Given
1 = 1 + (3 × 3 × 3) − 2 1 = 1 + 27 − 2 1 = 28 – 2 28 × 2 1 == − 1× 2 2 56 1 == − 2 2
==
55 2
1 1 0 3 −1 33 aa3 =×=aa3× a and and 1; aa3 = 1; = a3 a3×a××a3a3 × a3aa–1 and = a −1 = a 0 = a a 3 × 3 × 3 = 27
LCM LCM of of 1 and 2 == 2
28 × 2 = 56 ....(ii)
55 1 > 2 4
On comparison,
Reducing the fraction to its lowest term.
56 1 56 − 1 55 − = = 2 2 2 2 From eq. (i) and eq. (ii), 55 1 = 27.5 and = 0.25 2 4
1 × 2−4 × 43 is smaller than(14 0 + 33 − 2−1 ). Thus, 16
Addition of exponents of same base4 2 4 8 Example 1 3 3 3 Simplify and express in exponential form: – × – × –
7
Solution
7
7
Procedure 2
4
3 3 3 – × – × – 7 7 7 3 = – 7
Explanation Given
8
∵ xm × xn = xm+n
2+ 4+ 8
14
Adding the exponents, 2 + 4 + 8 = 14
3 = – 7
14
3 Hence, – is the simplified exponential form. 7 x 2 Example 2 7 7 Simplify and express in exponential form: ×
2
4
Refer P 243
2
Mathematical expressions that expand 9
Solution Procedure x
7 7 × 2 2 = 7 2
Given
2
∵ xm × xn = xm+n
x+2
7 Hence, 2
Explanation
x+2
is the simplified exponential form.
Example 3 If
1 92
× 3m = ( 27 ) , then find the value of ‘m’. m
Solution Procedure 1 92
× 3m = ( 27 )
Explanation Given
m
1
⇒ (32 ) 2 × 3m = (33 )m ⇒ 31 × 3m = 33m
1
1
∵ 9 = 32, so 9 2 = (32 ) 2 and 27 = 33. ∵ (am)n = am × n 1
1
Resolving 2 in the power 9 2 of = (32 ) 2 .
∵ am × an = am + n
⇒ 31+m = 33m
Comparing the powers of both sides as the base is same on both sides.
⇒ 1 + m = 3m ⇒ 3m – m = 1 ⇒ 2m = 1 1 ⇒m= 2
Example 4 1 0.25 Find the value of 5 4 × (125 ) . Solution Procedure 1 4
5 × (125)0.25 = 50.25 × (53)0.25
Explanation Given
∵ 1 ÷ 4 = 0.25 and 125 = 5 × 5 × 5 = 53
10 A to Z of Mathematics
= 50.25 × 50.75
∵ (am)n = am x n Thus, 3 × 0.25 = 0.75
= 50.25 + 0.75
By using am × an = am+n ∵ 0.25 + 0.75 = 1.
= 51
∵ x1 = x
=5 Example 5 3 Find the value of x, when x 5 = 8. Solution Procedure 3 5
Given
x =8 3
5
5 so that one 3 side of the equation is without power form and calculation become easy, and obvious.
5
Raising both sides to the power
⇒ ⇒ (x 5 ) 3 = 8 3 3
⇒ ⇒ (x) 5
Explanation
×
5 3
∵ (am)n = am×n
5
= 83 5
⇒ x = (23 ) 3
3×
∵ 8 = 23 ∵ (am)n = am × n
5
⇒ ⇒ x = (2) 3 ⇒ x = 25 = 32
∵ 2 × 2 × 2 × 2 × 2 = 32
Example 6 Find the value of x such that 8x – 3 = 164 – x. Solution Procedure 8x – 3 = 164 – x ⇒ (23)x – 3 = (24)4 – x
⇒ 23(x – 3) = 24(4 – x) ⇒ 23x – 9 = 216 – 4x ⇒ 3x – 9 = 16 – 4x ⇒ 3x + 4x = 16 + 9 ⇒ 7x = 25 25 ⇒x= 7
Explanation Given ∵ 8 = 23 and 16 = 24. Thus, making the base of exponent same. ∵ (ab)c = ab × c Since the base is same on both sides, therefore equating the exponents and solving for x.
Mathematical expressions that expand 11
Multiplication of different bases with same exponents5 Example 1 (a) 24 × 44 × 54
(b) (–2)3 × (–3)3 × (–4)3
3 3 3 (d) ( 2) × ( 8 ) × ( 4)
(c) (0.5)2 × (0.4)2
Solution Procedure (a) 2 × 4 × 5 = (2 × 4 × 5)4 4
4
= 404 = 40 × 40 × 40 × 40 = 2560000 (b) (–2)3 × (–3)3 × (–4)3 = [(–2) × (–3) × (–4)]3 = [(6) × (–4)]3 = (–24)3 = (–24) × (–24) × (–24) = –13824 (c) (0.5)2 × (0.4)2
= (d)
Writing (–24)3 in expanded form. Multiplication of three negative numbers is always negative.
a n × bn = (a × b)n
2
Resolving the base 5 × 4 = 20 and 10 × 10 = 100
20 2 = 100 1 = 5
Given a n × bn × cn =(a × b× c)n (−) × (−) = (+) (+) × (−) = (−)
Given Write decimal in fractional form.
5 2 4 2 = × 10 10 4 5 = × 10 10
Explanation Given a n × bn × cn = (a × b × c)n 2 × 4 × 5 = 40 Writing 404 in expanded form.
4
2
Reducing the fraction to its lowest form. 20 1 × 20 1 = = 100 5 × 20 5
1 25
52 = 5 × 5 = 25
( 2) × ( 8) × ( 4 ) = ( 2 × 8 × 4) = ( 2 × 8 × 4) = ( 64 ) 3
3
3
3
Given
a n × bn × cn = (a × b × c)n
3
3
5
= (8)3
64 = 8 × 8 = 8
= 512
8 × 8 × 8 = 512
New learning outcome
12 A to Z of Mathematics Example 2 5 5 4 7 Simplify × and express the result in the exponential and standard form. 7 2 Solution Procedure 5
4 7 × 7 2 4 7 = × 7 2 28 = 14
5
Explanation Given ∵ (a)n × (b)n = (a × b)n
5
∵ 4 × 7 = 28 and 7 × 2 = 14.
5
∵
= 25
28 2 × 14 = =2 14 14
= 32 ∵ 25 = 2 × 2 × 2 × 2 × 2 = 32 Hence, 25 is the simplified exponential form and 32 is the simplified standard form. Example 3 37 Simplify 2 × 45 and express in the exponential form. 3 Solution Procedure 37 2 3
Explanation Given
5 × 4
= 37 – 2 × 45 = 35 × 45 = (3 × 4)5 = 125 Hence, 125 is the simplified exponential form.
xm m–n ∵ n = x ; x = 0 x n n ∵ (a) × (b) = (a × b)n
Example 4 What is the value of n that satisfies the equation: 2n.4n.8n.16 = 2–n.4–n.8-n Solution Procedure 2n.4n.8n.16 = 2–n.4–n.8–n ⇒ 2n(22)n(23)n(24) = 2–n(22)–n(23)–n
Explanation Given Simplifying to make base of all exponents same.
Mathematical expressions that expand 13
⇒ 2n(22n)(23n)(24) = 2–n(2–2n)(2–3n) ⇒ 2(n + 2n + 3n + 4) = 2[–n + (–2n) + (–3n)] ⇒ 6n + 4 = –6n ⇒12n + 4 = 0
4 1 ⇒n= – = – 12 3
⇒ ⇒n=–
∵ (ab)c = ab × c ∵ ab × ac = a(b + c) Since base is same on both sides, therefore equating the exponents and solving for n. Simplifying
1 3
1 Thus, the value⇒ of n = – . 3
Division of same bases with different exponents6 Example 1 Simplify: (–4)5 ÷ (–4)9 Solution Procedure
Explanation Given
(–4)5 ÷ (–4)9 (–4)5 = (–4)9 = (–4)5 – 9 = (–4)–4
1 ( –4)4 1 = 256
am = am – n an
∵ a –n =
=
1 an
∵ (–a)m = am if m is an even integer. Thus, (–4)4 = –4 × –4 × –4 × –4 = 256.
Example 2 Simplify (a8 ÷ a5) ÷ a3 and express it in exponential form. Solution Procedure (a8 ÷ a5) ÷ a3 8–5
a = a3 a3 = 3 a
6
Refer P 243
Explanation Given a8 am m–n ∵ n = a , i.e., 5 = a8 – 5 a a
Simplifying
14 A to Z of Mathematics
am a3 m–n = a , i.e., = a3 – 3 3 an a
= a3 – 3 = a0
∵ x0 = 1
=1 Hence, 1 is the simplified exponential form. Example 3 Evaluate:
532 + 528 528
Solution To solve such type of questions, take out the common exponential number in numerator and denominator and reduce it to its lowest equivalent. Procedure 32
Explanation
28
Given
5 +5 528 5(28+4 ) + 528 = 528 528 × 54 + 528 = 528 528 (54 + 1) = 528 = 54 + 1 = (5 × 5 × 5 × 5) + 1 = 625 + 1 = 626
32
5 can be written as 5(28 + 4). a ( m+n ) = a m × a n Taking 528 common. Reducing the expression. Write 54 in expanded form. 5 × 5 × 5 × 5 = 625
Example 4 54 × x10 y 5 Simplify and express in exponential form: 4 7 4
5 ×x y
Solution Procedure
Explanation
54 × x10 y 5 54 × x 7 y 4
Given
= 54 – 4 x10 – 7 y5 – 4 = 50 x3 y1
∵
= x3y Hence, x3y is the simplified exponential form.
xm m – n =x xn
∵ a0 = 1 and a1 = a
Mathematical expressions that expand 15
Example 5
3n + 3n + 1 Simplify: n + 1 3 – 3n Solution
Procedure
3n + 3n + 1 3n + 1 –3n =
Explanation Given
3n +(3n × 31 ) (3n ×3) –3n
(3n ×1)+(3n × 3) = n (3 ×3)–(3n ×1)
∵ am + n = am × an ∴ 3n + 1 = 3n × 31 We can write 3n as 3n × 1.
=
3n (1+3 ) 3n ( 3–1)
Taking 3n common.
==
3n × 4 3n × 2
Simplifying
=
3n × 2 2 3n × 2
Expressing ‘4’ in exponential form.
am m –n ∵ n =aam – n, simplifying the expression. a
= 3n – n × 22 – 1 = 30 × 21
=1×2 ∵ a0 = 1 =2 Hence, 2 is the simplified form of the given expression. Example 6
1 1 1 If 2x = 4y = 8z and xyz = 288, find the value of 2x + 4y + 8z . Solution
Procedure 2x = 4y = 8z ⇒ 2x = 22y = 23z x = 2y = 3z = k xyz = 288 k k ⇒ k = 288 2 3
Explanation Given Making the base of the exponents equal. Writing equation with base 2. ∵ 4 = 22 and 8 = 23 ....(i) Since the base is same, therefore equating the exponents and assuming to be equal to k. Given k ∵ x = k; 2y = k ⇒ y = 2 k And, 3z = k ⇒ z = 3
16 A to Z of Mathematics
⇒
∵ k3 = 288 × 6 = 1728 = 123
k3 = 288 6
⇒ k3 = 288 × 6 = 123 ⇒ k = 12 ∴ x = 12, y = 6, z = 4
....(ii) From eq. (i), x = k = 12; y =
1 1 1 + + 2x 4y 8z
k 12 k 12 = = 6; z = = =4 2 3 2 3
Given
1 1 1 + + 2(12) 4(6) 8(4) 1 1 1 = + + 24 24 32
=
From eq. (ii), x = 12, y = 6 and z = 4.
4+4+3 96 11 = 96
∵ LCM of 24 and 32 is 96.
=
Hence, the value of
1 1 1 11 + + is . 2x 4y 8z 96
Division of different bases with same exponents7 Example 1 1 −3 Evaluate: + 3 Solution
1 4
−3
1 −3 ÷ 2
Procedure 1 −3 + 3
1 4
−3
Given
1 −3 ÷ 2
= (33 + 43 ) ÷ (2)3
1 −n 1 = − n = a n a a
= (27 + 64) ÷ 8 = 91 ÷ 8 91 = 8
33 = 3 × 3 × 3 = 27; 43 = 4 × 4 × 4 = 64; 23 = 2 × 2 × 2 = 8
Hence, the value of the expression is
7
Explanation
New learning outcome
91 . 8
Mathematical expressions that expand 17
x Example 2 y x IfIfxxy y==yyxx,, then find find . y Solution
y y
Procedure x
Explanation 1 Raising the power of exponents by so that x has the y effective exponent of 1.
xy = y x ⇒ x = y y y
x
⇒ xy = yy x =y ⇒ y
x
x y
⇒ ⇒x=y
............(i)
x
Given
x y
a an ∵ = n b b
x y y =
x y
= ==
x x x
....(i)
n
x y x y
∵x=y
x y
[from eq.(i).]
xx yy
1
xx1
=x
x – 1 y
x y
xm = x m–n ; x ≠ 0 xn x
x – 1
Hence, y = x y
.
Example 3 Simplify (35 ×105 × 25)÷(57 × 65 ) and express it in the exponential form. Solution Procedure
(35 ×105 × 25)÷(57 × 65 ) 5 3355 ×10 5 × 25 ×10 × 25 = = 577 × 655 5 ×6 5 2 3355 × × (2 (2 × × 5) 5)5 × × (5) (5)2 = = 577 × (2 × 3)55 5 × (2 × 3)
Explanation Given
∵ 105 = (2 × 5)5, 25 = (5×5)=52, and 65 = (2 × 3)5.
18 A to Z of Mathematics ∵ (a × b)n = an × bn
35 × 25 × 55 × 52 = 57 × 25 × 35
=
∵ an × am = a(n + m) Using the law of exponents.
35 × 25 × 55 + 2 57 × 25 × 35
xm m–n ∵ n = x ; x =/ 0 x
= 35−5 × 25−5 × 55 + 2 − 7
= 30 × 20 × 50 ∵ x0 = 1 =1 Hence, 1 is simplified exponential form.
Interpreting power of a power8 Example 1
4
1 Simplify: (a) 3 2 Solution
(b) ( 2−3 ) 4
Procedure 4
1 (a) 3 2 14 = 3 4 (2 )
Explanation Given n
an a ∵ = n and (am)n = am × n. b b
=
1 212
∵ 14 = 1 and (23)4 = 23 × 4 = 212
=
1 4096
Multiplying 2 twelve times gives 4096.
1 is the simplified form of given expression. 4096 Given (b) ( 2−3 ) 4 1 1 4 a − n = n ∵ = 3 a 2 1 1 4 1 2 a = a2 ∵ = 3 2 4 1 n 1 = 1 ∵ = am 1 a m×n 3× 2 2
Hence,
8
Refer P 243
Mathematical expressions that expand 19
1
==
3
1 n 1 m = m×n ∵ a a
×4
22 1 == 3× 2 2 1 = 6 2
1 2×2×2×2×2×2 1 == 64 =
Writing the expanded form of 26.
∵ 2 × 2 × 2 × 2 × 2 × 2 = 64
Example 2 −2 2 −1 Simplify: 5 Solution
−1
Procedure −2 2 −1 5
−1 −2 = 5 −1 −2 = 5 −1 = 5 −1 = 5
= =
( –1)
4
54 1 625
Hence,
2× −1
−2
−2× −2
4
−1
Explanation Given
Opening the outermost bracket using the law of exponents, (am)n = am × n ∵ 2 × (−1) = −2 Opening the next bracket by using the law of exponents, (am)n = am×n. ∵ −2 × −2 = 4 n
n a a ∴ = n ∵ b b ∵ (–a)m = am, if m is an even integer. Thus, (–1)4 = 1 and (5)4 = 625.
1 is the simplified form of given expression. 625
20 A to Z of Mathematics Example 3
5
–3 3 –3 5 Evaluate: ÷ 4 4
3
Solution Procedure 5
–3 3 –3 5 ÷ 4 4 15
3
15
–3 –3 = ÷ 4 4 15 – 15
Given ∵ (xn)m = xn×m m
∵ x = xm – n n
= –3 4
= –3 4
Explanation
x
0
=1 Hence, 1 is the value of the given expression.
∵ x0 = 1
Example 4 (x m + n )2 . (x n + p )2 . (x p + m )2 Simplify: (x m . x n . x p )3 Solution Procedure (x m + n )2 . (x n + p )2 . (x p + m )2 (x m . x n . x p )3
x 2m + 2n . x 2n + 2 p . x 2p + 2m = x 3m . x 3n . x 3p
x 2m+2n + 2n+2p + 2p+2m = x 3m +3n +3p
Explanation Given ∵ (am)n = am×n ∵ am × an = am+n
x 4m + 4n +4p = 3m +3n + 3p x = x (4m + 4n +4p) – (3m + 3n + 3p) = x 4m + 4n + 4p – 3m – 3n – 3p
∵
am = a m-n n a
= x m+n+ p Hence, x m + n + p is the simplified form of the given expression.
Mathematical expressions that expand 21
Example 5 Solve for x: (81)–4 ÷ (729)2 – x = 94x Solution (81) ÷ (729) –4
Procedure = 94x
2–x
Explanation Given
⇒ {(92)–4 ÷ (93)2 – x} = 94x
Making the base of all terms same. As, 729 = 93 and 81 = 92. ∵ (am)n = amn
⇒ (9)–8 – (6 – 3x) = 94x
∵
(9)−8 ⇒ = 94x 6 − 3x (9)
⇒ –8 – 6 + 3x = 4x ⇒ –14 = 4x – 3x ⇒ x = –14 Hence, the value of the x is –14.
am = am − n n a
Equating the exponents of the equation since base are same on both sides of the equation.
Writing daily-life numbers as exponents9 Example 1 Write the number whose expanded form is 7 × 106 + 6 × 103 + 5 × 101 + 3 × 100. Solution Procedure Explanation 7 × 106 + 6 × 103 + 5 × 101 + 3 × 100 Given = (7 × 1000000) + (6 × 1000) + (5 × 10) + (3 × 1) Write 10n in expanded form and multiply them with = 7000000 + 6000 + 50 + 3 the multiplier. = 7006053 Hence, the required number is 7006053. Example 2 Express the number 70500000 in standard form. Solution Procedure 70500000 = 70500000.00
Explanation The number can be written as 70500000.00 by introducing a decimal. 7 ∴ 70500000.00 = 7.05 × 10 The decimal point is moved 7 places to the left. 7 Hence, 7.05 × 10 is the standard form of 70500000.
9
New learning outcome
22 A to Z of Mathematics
Standard form is a way of writing very large or very small numbers easily. If the number is more than 1, then move the decimal point to the left to get just one digit to the left of the decimal point and write the given number as the product of the number so obtained and 10n, where ‘n’ is the number of places the decimal point has been moved to the left. If the number is less than 1, then move the decimal point to the right to get just one digit to the left of the decimal point and write the given number as the product of the number so obtained and 10–n, where ‘n’ is the number of places the decimal point has been moved to the right. Example 3 Write the following in standard form. (a) 0.000589 (b) 438.930000 Solution Procedure (a) 0.000589 = 5.89 × 10-4 (b) 438.930000 = 4.38930000 × 102 = 4.3893 × 102
Explanation Given The decimal point is moved 4 places to the right. Given The decimal point is moved 2 places to the left. Remove the trailing zeros to the left of the decimal point from the decimal number as they do not change the number.
Example 4 Write 5.67853 × 105 in the usual form. Solution Procedure
Explanation
5.67853 × 10 = 567853 × 10–5 × 105
Given The decimal point is moved 5 places to the right, so that the number becomes a whole number. Multiply the given number by 10–5.
= 567853
xa × x b = xa + b
5
∴ 10–5 × 105 = 10–5 + 5 = 100 = 1
Hence, 567853 is the usual form of 5.67853 × 105. Usual form is a way of writing the exponential numbers in standard notation form. To write any exponential number in usual form, move the decimal point to the right to get all the digits to the left of the decimal point and write the given number as the product of the number so obtained and 10–n, where ‘n’ is the number of places the decimal point has been moved to the right.
Real-life application of exponents10 Example 1 One bag weighs 3.7 × 103 kg and another bag weighs 4.7 × 102 kg. What is the difference between the two weights of the bags? 10
New learning outcome
Mathematical expressions that expand 23
Solution Procedure Weight of 1 bag = 3.7 × 103 kg = 37 × 102 kg
Explanation Given The decimal point moved to 1 place right to make the power 103 equal to 102, the base and exponent of the other number. Given Taking 102 as common.
Weight of another bag = 4.7 × 102 kg Difference between the two weights = (37 – 4.7) × 102 kg = 32.3 × 102 kg Simplifying Hence, the difference between the weights of two bags is 32.3 × 102 kg.
Example 2 The size of a red blood cell is 0.000007 m and the size of a plant cell is 0.00001275 m. Compare these two. Solution Procedure Size of red blood cell = 0.000007 m = 7 × 10–6 m Size of plant cell = 0.00001275 m = 1.275 × 10–5 m ∵
Given The decimal point is moved 6 places to the right for standard form of exponential number. Given The decimal point is moved 5 places to the right for standard form of exponential number.
Size of red blood cell 7 × 10–6 = Size of plant cell 1.275 × 10–5
7 ×10−6+5 = 1.275
7 ×10−1 1.275 0.7 = 1.275
=
≈
Explanation
0.7 1 ≈ 1.3 2
xm n = x m– n x – 6 + 5 = –1 1 a 1 ∴7 × = 0.7 10
a–1 =
Half of 1.3 is 0.65, which is approximately equal to 0.7.
So, the size of a red blood cell is approximately half of the size of a plant cell.
24 A to Z of Mathematics
2. Square of Numbers Square numbers11 Example 1 Find the square of the following numbers. (a) (12)2
(b) (–4)2
3 2 (c) 8
2 (d) − 3
Solution Square of a negative or positive number is always positive. Procedure (a) (12)2 = 144
Explanation
(b) (–4) = 16
× 12 144= 16 12 (−4) × (=−4) (−4) × (−4) = 16
3 2 9 (c) = 64 8
3 3 99 × == 64 8 8 64
2 2 4 (d) − = 9 3
2 2 2 − 2 × −22 4 − × − = − = 3 3 3 × 3 33 9
2
2
Example 2 Check whether 225 is a perfect square. Solution Prime factorise the number and pair the similar factors to check if the number is a perfect square. Procedure 225 = 3 × 3 × 5 × 5 225 = 3 × 3 × 5 × 5
Explanation Prime factorising the number. Group the similar factors in pairs.
225 = 15 × 15 = 152 Thus, 225 is a perfect square number.
Check to see there is no factor unpaired.
Numbers ending with digits 2, 3, 7 or 8 are non-perfect squares12 Example 1 Check whether the given numbers are perfect squares. (a) 33 (b) 152652 (c) 39 (d) 5896457
(e) 48888
(f) 144
Solution Procedure (a) 33 is not a perfect square number. (b) 152652 is not a perfect square number. 11 12
Refer P 244 New learning outcome
Explanation Last digit of the number is 3. Last digit of the number is 2.
Mathematical expressions that expand 25
(c) 13 × 3 = 39 Thus, 39 is not a perfect square number. (d) 5896457 is not a perfect square number. (e) 48888 is not a perfect square number. (f) 144 = 2 × 2 × 2 × 2 × 3 × 3 = 12 × 12 = 122 Thus, 144 is a perfect square.
The last digit is 9, so 39 may be a square. Checking it by prime factorisation. Last digit of the number is 7. Last digit of the number is 8. The last digit is 4, so 144 may be a square. Checking it by prime factorisation.
Zeros at end of a perfect square13 Even number of zeros at the end of a perfect square is a perfect square. Example 1 Find out the perfect square numbers from the following. (a) 100000000 (b) 1044000 (c) 25000000 Solution Procedure Explanation (a) Since the count of zeros is even, therefore Counting the number of zeros at the end of the number. 100000000 is a perfect square number. There are 8 zeros. 100000000 = (10000)2 (b) Since the count of zeros is odd, therefore 1044000 is not a perfect square number. (c) Since the count of zeros is even, therefore 1000000 is a perfect square number. Since both 25 and 1000000 are perfect square numbers, therefore 25000000 is also a perfect square number. ⇒ 25000000 = (5000)2
Counting the zeros only at the end of the number. There are 3 zeros. Counting the number of zeros at the end of the number. There are 6 zeros. We know that 25 is a perfect square number as 25 = 5 × 5
Square of even number is always an even number14 Example 1 Check whether the square of 8468 is 71707023 or 71707024? Solution Since 8468 is an even number, therefore square of 8468 is also an even number. Thus, from options, 71707024 is the square of 8468. Example 2 What will be the unit digit of the square of 122? (a) 3 (b) 1
(c) 4
(d) 9
Solution We know that the square of an even number is always an even number. Therefore, unit digit of a square of an even number is always an even number. Thus, 4 is the unit digit of the square of 122.
13 14
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26 A to Z of Mathematics
Square of odd number is always an odd number15 Example 1 Check whether the square of 409 is 167281 or 167282? Solution Since 409 is an odd number, therefore square of 409 is also an odd number. Thus, from options, 167281 is the square of 409. Example 2 What will be the unit digit of the square of 187? (a) 2 (b) 9
(c) 4
(d) 0
Solution We know that the square of an odd number is always an odd number. Therefore, unit digit of a square of an odd number is always an odd number. Thus, 9 is the unit digit of the square of 187.
Square of numbers ending with 516 (n5)2 = n(n + 1) hundred + 25 where all digits of the number other than unit digit 5 is considered ‘n’. Example 1 Find out the square of 645. Solution Procedure (645) = [(64)(64 + 1) × 100] + 25 ⇒ (645)2 = [(64)(65) × 100] + 25 ⇒ (645)2 = (4160 × 100) + 25 ⇒ (645)2 = 416000 + 25 ⇒ (645)2 = 416025 Hence, the square of 645 is 416025. 2
Explanation We know that square of natural number that ends with 5 is (n5)2 = n(n+1) hundred + 25. Here, n5 is 645. And, n = 64.
Example 2 Find the square of 4.5. Solution
Procedure (4.5)2
Given Write decimal number in fractional form.
45 2 = 10
=
15 16
Explanation
[(4)(4 + 1) × 100 + 25] (10)2
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We know, (n5)2 = n (n + 1) hundred + 25 Here, n5 is 45. And, n = 4.
Mathematical expressions that expand 27
4 (4 + 1) × 100 = 4 × 5 × 100 = 2000
2000 + 25 100 2025 = 100 = 20.25 Hence, the square of 4.5 is 20.25.
=
Difference between two consecutive squares17 The difference between two consecutive squares is the sum of the two consecutive numbers. For natural number ‘n’, (n + 1)2 – n2 = (n + 1) + n Example 1 Find the difference between 282 and 272. Solution Procedure
Explanation
(28)2 – (27)2 = 28 + 27 = 55 Hence, the required difference is 55.
We know that a difference of squares of two consecutive natural numbers is equal to the sum of both consecutive numbers.
Example 2 Simplify:
(783)2 − (782)2 5
Solution Procedure
Explanation
(783) − (782)
Given
5 783 + 782 = 5 1565 = 5 = 313
We know that a difference of squares of two consecutive natural numbers is equal to the sum of both consecutive numbers. ∴ (783)2 − (782)2 = 783 + 782
2
2
Hence, the value of
(783)2 − (782)2 5
is 313.
Non-perfect squares between two consecutive square numbers18
Between two consecutive square numbers, n2 and (n + 1)2, there are ‘2n’ non-perfect square numbers. Example 1 Find out the number of non-perfect square numbers between 81 and 100. 17 18
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28 A to Z of Mathematics Solution Procedure 81 = 92 = n2 and 100 = 102 = (n + 1)2
Explanation Since both 81 and 100 are consecutive perfect square numbers of 9 and 10 respectively. n=9
Non-perfect square numbers between 81 and 100 = 2n = 2 × 9 = 18 Hence, there are 18 non-perfect square numbers between 81 and 100. Example 2 How many non-perfect square numbers lie between (998)2 and (999)2. Solution Procedure n = (998) ⇒ n = 998 2
2
Explanation 998 and 999 are consecutive numbers. ∴ Their squares are also consecutive squares.
Non-perfect square numbers between n = 998 2 2 (998) and (999) = 2n = 2 × 998 = 1996 Hence, there are 1996 non-perfect square numbers between (998)2 and (999)2.
Sum of first ‘n’ odd natural numbers19
The sum of first ‘n’ odd natural numbers is square of ‘n’ natural number. Example 1 Find out the sum of the following numbers. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 Solution Procedure Explanation 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 Sum of first 10 odd natural numbers (Given) = (n)2 Here, n = 10. = (10)2 = 100 Hence, 100 is the sum of the first 10 natural numbers. Example 2 Find the sum of all odd natural numbers that are less than 22. Solution Odd natural numbers that are less than 22 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21.
19
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Mathematical expressions that expand 29
Procedure Explanation 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 Sum of odd natural numbers that are less than 22. = (n)2 Here, n = 11 = (11)2 = 121 Hence, the sum of all odd natural numbers that are less than 22 is 121.
Pythagorean triplet20 For any natural number n greater than 1, Pythagorean triplet is 2n, n2 – 1 and n2 + 1. Example 1 Which of the following triplets are Pythagorean? (a) 10, 24, 26 (b) 20, 99, 101
(c) 6, 7, 8
⇒
⇒
Solution (a) 10, 24, 26 Let the smallest even number be 2n. Then, 2n = 10 10 =5 ⇒ n= 2 n2 – 1 = (5)2 – 1 = 25 – 1 = 24 n2 + 1 = (5)2 + 1 = 25 + 1 = 26 Hence, (10, 24, 26) is a Pythagorean triplet. (b) 20, 99, 101 Let the smallest number be 2n. Then, 2n = 20 20 = 10 ⇒n= 2 n2 – 1 = (10)2 – 1 = 100 – 1 = 99 n2 + 1 = (10)2 + 1 = 100 + 1 = 101 Hence, (20, 99, 101) is a Pythagorean triplet. (c) 6, 7, 8 Let the smallest number be 2n. Then, 2n = 6 6 ⇒ n= =3 2 2 n – 1 = (3)2 – 1 = 9 – 1 = 8 n2 + 1 = (3)2 + 1 = 9 + 1 = 10 Hence, (6, 7, 8) is not a Pythagorean triplet. (a) Pythagorean triplet is a set of three sides of a right angled triangle. (b) Pythagorean triplet is a set of either three even numbers or two odd and one even number. In both the cases, the smallest even number = 2n. (c) In Pythagorean triples, sum of squares of two smaller number is equal to the square of the largest number. 20
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30 A to Z of Mathematics Example 2 Find a Pythagorean triplet in which one number is 18. Solution We can get Pythagorean triplet by using general form 2n, n2 –1, n2 + 1. With trial and error method we try to put the 18 in the 3 terms and get a number such that n = natural number. Procedure
Explanation
Let us first take, n2 – 1 = 18. n2 – 1 = 18 2 ⇒ n = 18 + 1 = 19 From the above, we can see that 19 is not a perfect square number. So, when n2 – 1 = 18, we will not get n, which is an integer. Now, we will try to take n2 + 1 = 18. n2 + 1 = 18 ⇒ n2 = 18 – 1 = 17
Again we can see that when n2 + 1 = 18, 17 is not a perfect square and will not yeild n, which is an integer. ⇒ = 18 2n 2n = 18 Now, let us try with 2n = 18. 18 ⇒ =9 ⇒n= 2 Wen2get 9,2an can nbe2 +1 used to get=other ∴ – 1n==(9) – 1integer = 81 – which 1 = 80 and = 81+1 82 terms of Pythagorean triplet. 2 2 2 n – 1 = (9) – 1 = 81 – 1 = 80 and n + 1 = 81 + 1 = 82 Put value of n in n2 – 1 and n2 + 1. Hence, the required triplet is 18, 80, 82.
Patterns in square numbers21 Example 1 12 = 1 112 = 121 1112 = 12321 11112 = 1234321 111112 = 123454321 Observe the pattern and write the square of 1111111. Solution In the given pattern, the squares of the given numbers can be found by first writing the counting number up to the number of 1’s and then writing the reverse counting till 1. 12 = 1 112 = 121 1112 = 12321 11112 = 1234321 111112 = 123454321 1111112 = 12345654321 11111112 = 1234567654321 [∵ There are 7 ‘1’ in the given number.] Hence, the square of 1111111 is 1234567654321. To find the square of a big number is time-consuming and tedious. Thus, we find some patterns to calculate the square of a number without actual multiplication. 21
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Mathematical expressions that expand 31
Example 2 132 = 169 1032 = 10609 10032 = 1006009 100032 = 100060009 Observe the pattern and write the square of 1000003. Solution In the given pattern, the square of numbers which have 1 and 3 as first and last digit with zeros in between has the digits 1, 6, and 9 with same number of zeros as in the number, between 1 and 6, and 6 and 9. 132 = 169 2 103 = 10609 10032 = 1006009 100032 = 100060009 1000032 = 10000600009 10000032 = 1000006000009 [∵ There are 5 ‘0’ in the given number.] Hence, the square of 1000003 is 1000006000009.
3. Cube of Numbers Cubic numbers22 Example 1 Write the cube of the following numbers. (a) 5 (b) −9 Solution Procedure (a) Cube of 5 = (5)3 = 125 (b) Cube of −9 = (−9)3 = −729
Sign conventions followed in maths:
Explanation ∵ 53 = 5 × 5 × 5 ∵ (−9)3 = (−9) × (−9) × (−9)
( −) × ( + ) = ( −) ( −) × ( −) = ( + ) (+) × (−) = (−) (+) × (+) = (+) From the above sign conventions, we can conclude that cubes of negative numbers are always negative. Example 2 Show that –4096 is a perfect cube. Also, find the number whose cube is –4096.
22
Refer P 244
32 A to Z of Mathematics Solution Procedure
Explanation
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ⇒ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Prime factorisation of 4096. Grouping the same factors into triples.
So, 4096 is a perfect cube.
We can see that all prime factors are grouped into triples, and no factor left ungrouped.
∴ 4096 is a perfect cube of 16.
Taking one factor from each group, we find that 4096 is a perfect cube of 2 × 2 × 2 × 2 = 16 (−) × (−) = (+) ; (+) × (−) = (−)
∴ (−16) × (−16) × (−16) = −4096 Hence, −4096 is a perfect cube of −16.
Cube of rational numbers23
Example 1 Write the cube of the following numbers. 7 3 Solution
(b) −
(a)
2 11
Procedure
Explanation 3
3
7×7×7 7 = 3×3×3 3
7 343 (a) Cube of = 27 3 3
2 8 2 (b) Cube of − = − = − 11 11 1331
3
2 2 2 2 − = − × − × − 11 11 11 11
Example 2 Show that Solution
27 is a cube of a rational number. 125 Procedure
3× 3× 3 27 = 125 5 × 5 × 5
27 33 3 ⇒ = = 125 53 5 Hence,
23
3
27 3 is a cube of . 125 5
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Explanation Prime factors of 27 and 125. Grouping the prime factors into sets of three.
Mathematical expressions that expand 33
To make perfect cube – by prime factorisation and multiplication24 Example 1 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (a) 100 (b) 675 Solution Procedure (a) 100 = 2 × 2 × 5 × 5 100 is not a perfect cube. 100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000
Explanation Prime factors of 100. We cannot make sets of three same factor. Multiplied 100 by ‘2 × 5’, to get a perfect set of three same factor. Hence, 100 must be multiplied by 10 to get a perfect cube.
(b) 675 = 3 × 3 × 3 × 5 × 5
Prime factors of 675. Sets of three of factor ‘5’ cannot be made. 675 is not a perfect cube. 675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 Multiplying 675 by ‘5’ to get a perfect set of three same factors. Hence, 675 must be multiplied by 5 to get a perfect cube.
Example 2 Is 2700 a perfect cube? If not, find the smallest number by which 2700 must be multiplied to get a perfect cube. Solution Procedure 2700 = 3 × 3 × 3 × 2 × 2 × 5 × 5.
Explanation Prime factors of 2700.
2700 is not a perfect cube. Since, there is no triplet of factors 2 and 5. 2700 × 2× 5 = 3 × 3 × 3 × 2 × 2 × 2 × 5 × 5 × 5 = 27000 Multiplying 2700 by ‘2 × 5’, to get a perfect set of triples. Hence, 2700 must be multiplied by 10 to get a perfect cube.
To make perfect cube – by prime factorisation and division25 Example 1 Find the smallest number by which each of the following must be divided to obtain a perfect cube. (a) 1125 (b) 1372 Solution Procedure (a) 1125 = 3 × 3 × 5 × 5 × 5 1125 is not a perfect cube.
24 25
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Explanation Prime factors of 1125. There is no triplet of 3.
34 A to Z of Mathematics
In the factorisation of 1125, 3 appears only two times, therefore divide 1125 by 9 (3 × 3), to get a perfect set of three same factors.
1125 ÷ 9 = 5× 5× 5 = 125
Hence, 1125 must be divided by 9 to get a perfect square. Prime factors of 1372. (b) 1372 = 2 × 2 × 7 × 7 × 7 There is no triplet of 2. In the factorisation of 1372, 2 appears only two times, therefore divide 1372 by 4 (2 × 2), to get a perfect set of three same factors.
1372 is not a perfect cube.
1372 ÷ 4 = 7×7×7 = 343
Hence, 1372 must be divided by 4 to get a perfect square. Example 2 Find the smallest number by which 23625 must be divided to obtain a perfect cube. Solution Procedure
Explanation
23625 = 3 × 3 × 3 × 5 × 5 × 5 × 7 23625 ÷ 7 = 3 × 3 × 3 × 5 × 5 × 5
Prime factors of 23625. In the factorisation of 23625, 7 appears only one time. Therefore, divide 23625 by 7, to get a perfect set of three same factors.
Hence, 23625 must be divided by 7 to get a perfect cube.
Sum of ‘n’ consecutive cubes26 Sum of ‘n’ consecutive perfect cubes = Square of the sum of ‘n’ consecutive numbers Example 1 Add: 13 + 23 + 33 + 43 + 53 + 63 + 73 Solution Procedure 13 + 23 + 33 + 43 + 53 + 63 +73 = (1 + 2 + 3 + 4 + 5 + 6 + 7)2 = (28)2 = 784 Hence, 784 is the required sum of the given series. Example 2 Find the sum of cubes of first 25 natural numbers.
26
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Explanation Given ∵ 13 + 23 + 33 + … + n3 = (1 + 2 + 3 +…. + n)2 Here, n = 7. ∵ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 ∵ 28 × 28 = 784
Mathematical expressions that expand 35
Solution Procedure 1 + 2 + 3 + ………. + 253 = (1 + 2 + 3 + ……… + 25)2 3
3
Explanation Given 13 + 23 + 33 + ………. + n3 = (1 + 2 + 3 + ……… + n)2 Here, n = 25. Instead of adding the sum of first 25 numbers which is a tedious task, we use the fact that 1, 2, 3, ……., 25 forms an Arithmetic Progression.
3
n(n+1) 2 = 2 2 = 25 × 26 2
And, the sum of an AP for natural numbers is n(n+1) given by . 2 = [325] 2 Here, n = 25. = 105625 We will discuss about AP later in this Chapter. Hence, the sum of cubes of first 25 natural numbers is 105625.
Difference between the squares of two consecutive triangular numbers27 Difference between squares of two consecutive triangular numbers
Cube of difference of two triangular numbers
=
Example 1 What is the value of 102 – 62? Solution Procedure
Explanation
∵ 6 and 10 are two consecutive triangular numbers. Since, difference between the squares of two consecutive triangular numbers equal to the cube ∴ 102 – 62 = (10 – 6)3 = 43 = 64 of difference of two triangular numbers. A triangular number is a count of objects that can be arranged in an equilateral triangle. Pictorially,
1
3
6
Therefore, 32 – 12 = (3 – 1)3 = 23 and 62 – 32 = (6 – 3)3 = 33 Example 2 Find the next triangular number in the series 45, 55, … Solution The difference between two consecutive terms = 55 – 45 = 10 Now, add 1 to the difference obtained = 10 + 1 = 11 27
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10
36 A to Z of Mathematics
Therefore, the next term is 55 + 11 = 66. To get the next triangular number in series, add to the last triangular number 1 and the difference between the last two triangular numbers.
4. Logarithms and Antilogarithms Definition of logarithm28 If logx y = n, then y = xn. Example 1 Find x if log5(x – 7) = 1. Solution Procedure
Explanation
log5(x – 7) = 1
Given
∵ If logxy = n, then y = xn.
⇒ x – 7 = 51 ⇒x=5+7 ⇒ x = 12
A logarithm is the exponent or power to which a base must be raised to produce the given number.
Example 2 Find the value of y, if logxy = 100 and log2x = 10. Solution Solve log2 x = 10 to get the value of x, and substitute it in logxy = 100 to get the value of y. Procedure
Explanation Given
log2x = 10 ⇒ x = 210
logxy = 100 ⇒ y = x100
⇒ y = (210)100 ⇒ y = 21000
Power rule29
logb mn = n logbm, logaa = 1 and log101 = 0 Example 1 Find the value of log232.
28 29
Refer P 244 Refer P 249
....(i)
∵ If logxy = n, then y = xn. Given
∵ If logxy = n, then y = xn. ∵ x = 210 from eq. (i). ∵ (am)n = am×n
Mathematical expressions that expand 37
Solution Procedure
Explanation
Let n = log232
Assumption
⇒ n = log22
∵ logbmn = n logbm (Power rule)
∵ 32 = 2 × 2 × 2 × 2 × 2 = 25
5
⇒ n = 5 log22
∵ logaa = 1
⇒n=5
Example 2 Find the value of log1010000 + log101. Solution Procedure log1010000 + log101
Explanation Given
∵ 10000 = 104 and log101 = 0
= log10104 + 0
∵ logbmn = n logbm (Power rule)
= 4 log1010
∵ log1010 = 1
=4
Quotient rule30 loga
m = logam – logan n
Example 1 Find the value of (log1042 – log106) using log table. Solution Procedure
Explanation
log1042 – log106
Given
42 = log10 6
∵ logam – logan = loga
= log107
∵
m (Quotient rule) n
42 =7 6 From log table, log107 = 0.8450
= 0.8450 Example 2 Find the value of log5 2500 – 2 log52 Solution
Procedure log5 2500 – 2 log5 2 30
Refer P 249
Explanation Given
38 A to Z of Mathematics
= log5 2500 – log5 (2)2 = log5 2500 – log5 4
n loga m = loga mn
2500 = log5 4
m loga m – loga n = loga n
= log5 625
= log5 (5)4
625 = 5 × 5 × 5 × 5 = 54
= 4 log5 5
loga mn = n loga m
=4
logaa = 1
2500 625× 4 = = 625 4 1× 4
Product rule31
loga (m × n) = logam + logan Example 1 Find the value of log1050 + log102. Solution Procedure log1050 + log102
Explanation Given
∵ logam + logan = loga(m × n) [Product rule]
= log10(50 × 2)
∵ 50 × 2 = 100
= log10100
∵ 100 = 102
= log10102
∵ logamn = n logam
= 2 log1010
∵ log1010 = 1
=2
[Power rule]
Example 2 Solve for x, if log(x – 1) + log(x + 1) = log21 Solution Procedure
Explanation
log(x – 1) + log(x + 1) = log21
Given
⇒ log(x – 1) + log (x + 1) = 0
∵ logam + logan = loga(m × n) [Product rule]
⇒ log{(x – 1) (x + 1)} = 0
⇒ (x – 1) (x + 1) = antilog (0) ⇒ (x – 1) (x + 1) = 1 31
Refer P 249
∵ log21 = 0
Transposing log from left-hand side to righthand side, then it becomes an antilog. ∵ antilog (0) = 1
Mathematical expressions that expand 39 ∵ (a – b) (a + b) = a2 – b2
⇒ x2 – 1 = 1 ⇒ x2 = 2 ⇒x= ± 2
∵ ‘log’ of a negative number is not defined, therefore consider only positive value.
⇒x= 2
Antilog of numbers32 Example 1 Find the antilog of 1.2115 using the antilog table. Solution A log is a mathematical tool for compressing numbers that are too large or too small, and thus are difficult to handle. The compressed number can be converted back to its original form using the inverse operator antilog. Procedure Characteristic = 1 and mantissa = 2115
The value of column 1 of row .21 is 1626.
The mean difference is 2. ∴ Required number = 1626 + 2 = 1628 Now we must put the decimal place in the number. ∵ Characteristic = 1 ∴ There are 1 + 1 = 2 digits before the decimal place. Therefore, the antilog of 1.2115 is 16.28.
Explanation In 1.2115, characteristic is the whole number part, while mantissa is all the numbers after the decimal point. From mantissa part 2115, use 21 of 2115 to find the antilog table for the row .21 Use 1 of 2115, to find the column 1 for row of .21 Use 5 of 2115 to find the mean difference, in the mean difference column 5 for row .21 Add the value of column 1 and the mean difference. We add 1 to the characteristic part to get the number of places before the decimal point.
Antilog law33
If logax = b, then x = antilogab = ab Example 1 Find the value of x using logarithm table, when x = 4 195 . Solution Procedure x = 4 195
32 33
Refer P 247 Refer P 244
Explanation Given
40 A to Z of Mathematics
⇒ log x = log 4 195
Taking log on both sides 1
⇒ log x = log (195 ) 4 1 4
⇒ log x = log 195 ⇒ log x =
1 (2.2900) 4
⇒ log x = 0.5725
⇒ x = antilog (0.5725) ⇒ x = 3.7368
∵ log mn= n log m [Power rule] From log table, log 195 = 2.2900 2.2900 = 0.5725 4
Keeping x on the left-hand side of the equation and taking the log to the right-hand side will turn it into an antilog function. ∵ antilog (0.5725) = 3.7368 from antilog table.
5. Ratios
Idea of ratio34 Example 1 Which of the following is the appropriate statement with respect to the concept ‘ratio’? (a) Ratio can be scaled up or down by multiplying both antecedent and precedent by different numbers. (b) Ratios can represent quantities with the same or different units. (c) Ratios are dimensionless. (d) The positions of antecedent and consequent in a ratio can be interchanged. Solution (a) Incorrect statement as the ratio can be scaled up or down by multiplying both antecedent and precedent by the same number. (b) Incorrect statement. If the ratios represent quantities with different units then the units must be measurable. So, ratios can represent quantities with the same or different measurable units. For example, 1 kg: 1 fist full of grains is not a ratio because 1 fist full of grains is not a definite quantity. It depends on the size of one’s fist. (c) Correct statement as it is a ratio of definite measurable quantities irrespective of their units. (d) Incorrect statement, as the positions of antecedent and consequent in a ratio cannot be interchanged unless there is a change in the given statements. Example 2 Match the picture to the correct ratio.
34
Refer P 251
Mathematical expressions that expand 41
(a) Forks to spoons 1:2 (b) apples to oranges 2:1 (c) Dogs to cats
(d) boys to girls
3:1
3:2 (e) Cups to glasses 2:4 Example 3 Write a ratio as a fraction. (a) India games won to games lost (b) Australia games won to games lost (c) West Indies games lost to games played (d) England games lost to games won
Team India Australia West Indies England
Win 22 29 13 47
Loss 15 32 5 33
Solution Procedure (a) Ratio of India games won to games lost = 22:15 22 = 15 (b) Ratio of Australia games won to games lost = 29:32 29 = 32 (c) Ratio of West Indies games lost to games played = 5:18
Explanation ∵ Number of games India won = 22 Number of games India lost = 15 Writing ratio in fractional form. ∵ Number of games Australia won = 29 Number of games Australia lost = 32 Writing ratio in fractional form. ∵ Number of games West Indies lost = 5 Number of games West Indies played = 13 + 5 = 18
42 A to Z of Mathematics
5 18 (d) Ratio of England games lost to games won = 33:47 33 = 47 =
Writing ratio in fractional form. ∵ Number of games England lost = 33 Number of games England won= 47 Writing ratio in fractional form.
Simplest form of ratios35 Example 1 Write 320:160 in its simplest form. Solution Procedure
Explanation
For expressing the ratio 320:160 in its simplest form, divide the numbers with their highest common factor such that there is no scope for further division. We will learn more about HCF in ‘Chapter M’. Method 1 320:160 320 ÷ 160 = 160 ÷ 160
=
2 = 2 :1 1
Given HCF of 320 and 160 is 160. ∴ Dividing numerator and denominator by 160 to reduce the fraction to its simplest form. Simplest form (as no common factor other than 1 exists).
Hence, 2:1 is the simplest form of 320:160. Method 2 320:160 =
=
320 160
2×2×2×2×2×2×5 2×2×2×2×2×5
2 = 2:1 1 Hence, 2:1 is the simplest form of 320:160. =
Given Write ratio in fractional form. Prime factorising 320 and 160. Simplest form (as no common factor other than 1 exists).
HCF is the Highest Common Factor or Greatest Common Divisor of two or more numbers. It is the highest possible number that divides both the numbers. Example 2 Express the following in their simplest form of ratio. 35
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Mathematical expressions that expand 43
(a) A dozen to score
(b) 300 grams to 15 kg
(d) 20 minutes to 2 hours
(e) 150 cm to 2 m
(c) 400 paise to ₹5 (f) 3.2 kg to 4 kg
Solution Procedure (a) A dozen to score ∵ As dozen = 12 and score = 20, The ratio is 12:20. = =
Explanation Given Writing in ratio form.
12 20
Writing ratio in fractional form.
3 5
Reducing the fraction to its simplest equivalent fraction.
= 3:5 (b) 300 grams to 15 kg = 300 g:15 kg = 300 g:5 × 1000 g = 300:15000 300 = 15000 1 = 50 = 1:50 (c) 400 paise to ₹5 = 400 paise:₹5 = 400 paise:5 × 100 paise = 400:500 400 = 500 4 = 5 = 4:5 (d) 20 minutes to 2 hours = 20 minutes:2 hours = 20 minutes:2 × 60 minutes = 20:120 20 = 120 =
1 6
12 3 × 4 3 = = 20 5 × 4 5
Simplest form of ratio. Given Writing in ratio form. 1 kg = 1000 g Writing ratio in fractional form.
300 1× 300 1 = = 15000 50 × 300 50 Simplest form of ratio. Given Writing in ratio form. Converting rupee to paise. ₹1 = 100 paise Writing ratio in fractional form.
400 4 × 100 4 400 paise = = 500 5 × 100 5 500 paise
Simplest form of ratio. Given Writing in ratio form. Converting hour into minutes. 1 hour = 60 minutes Writing ratio in fractional form.
20 20 ×1 1 20 minutes = = 120 minutes 120 20 × 6 6
44 A to Z of Mathematics
= 1:6 (e) 150 cm to 2 m = 150 cm:2 m = 150 cm:2 × 100 cm = 150:200 150 = 200
50 cm 50 × 3 3 = == 00 cm 50 × 4 4
150 50 × 3 3 = = 200 50 × 4 4 Simplest form of ratio. Given Write in ratio form. Write ratio in simplest form. Removing the decimal and writing in simplest equivalent form.
= 3:4 (f) 3.2 kg to 4 kg = 3.2 kg:4 kg = 3.2:4 =
Simplest form of ratio. Given Writing in ratio form. Converting metre into centimetre. 1 m = 100 cm Writing ratio in fractional form.
4 5
3.2 32 4 × 8 4 = = = 4 40 5 × 8 5
= 4:5
Simplest form of ratio.
The first quantity of the ratio is called antecedent and the second quantity of the ratio is called precedent or consequent. Example 3 The number of boys and girls in a school are 480 and 384 respectively. Express the ratio of the number of boys to that of girls in the simplest form. Solution We have to express the ratio of the number of boys and girls, i.e., 480:384 in simplest form which means divide both 480 and 384 with the highest common factor such that there is no scope of further division. Procedure
Explanation
Number of boys = 480 Given Number of girls = 384 Given Expressing in ratio form. ∴ Ratio of number of boys to the number of girls = 480:384 To express the ratio in the simplest form, we will have to find the HCF of 480 and 384. It is 96. Writing the ratio in fractional form and dividing the 480 480 ÷ 96 ∴ = numerator as well as the denominator of the fraction 384 384 ÷ 96 with their HCF, i.e., 96. 5 The required simplest form of the ratio 480:384. = = 5:4 4
Mathematical expressions that expand 45
Example 4 In a year, Seema earns ₹150000 and saves ₹50000. Find the ratio of (a) money that Seema earns to the money she saves. (b) money that she saves to the money she spends. Solution We have to find different ratios of Seema’s earning to saving and Seema’s saving to expenditure. Seema’s earning and saving is already given, so we can calculate the expenditure. Use these values to calculate the asked ratios. Then, express the specified ratios in simplest form by dividing both numerator and denominator with the highest common factor. Procedure
Explanation
Money earned = ₹150000 Money saved = ₹50000 Money spent = ₹150000 – ₹50000 = ₹100000 ....(i) (a) Ratio of money earned to money saved 150000 = 50000 =
150000 ÷ 50000 3 = = 3:1 50000 ÷ 50000 1
(b) Ratio of money saved to money spent
50000 100000 50000 ÷ 50000 1 = = = 11:2 :2 100000 ÷ 50000 2
Given Given Money spent = Money earned – Money saved
Required ratio in fractional form.
Dividing numerator and denominator by their HCF i.e., 50000 to get the simplest form of the given ratio. From eq. (i), required ratio in fractional form.
=
Dividing numerator and denominator by their HCF i.e., 50000 to get the simplest form of the given ratio.
Example 5 The cost of a dozen pencils is ₹180 and the cost of 8 ball pens is ₹56. Find the ratio of the cost of a pencil to the cost of a ball pen.
Solution We have to find the ratio of the cost of 1 pencil to the cost of 1 ball pen. For this, first, calculate the cost of 1 pencil and 1 pen and then express the specified ratio in simplest form by dividing both numerator and denominator with their highest common factor. Procedure Cost of dozen pencils = ₹180 180 ⇒ Cost of 1 pencil = ₹ = ₹15 12 Cost of 8 ball pens = ₹56 56 ⇒ Cost of 1 ball pen = ₹ = ₹7 8
Explanation Given ....(i) 1 dozen = 12 Given ....(ii)
46 A to Z of Mathematics
∴ Required ratio =
15 = 15:7 7
From eq. (i) and eq. (ii), this is the simplest form of ratio.
Example 6 In a mixture of 60 litres, the ratio of milk and water is 2:1. If this ratio has to be changed to 1:2, then find the quantity of water to be further added. Solution Ratio of two quantities in the mixture is given. And, we have to find the changed quantity of things in the mixture if the ratio of things in the mixture is changed. Procedure
Explanation
The ratio of milk to water is 2:1.
Given
∴ Total parts of the mixture = 1 + 2 = 3
Water = 1 part and milk = 2 parts
Quantity of milk in the mixture 2 = 60 × litres = 40 litres 3 Quantity of water in the mixture = (60 – 40) litres = 20 litres
Total quantity of mixture = 60 litres 2 Ratio of milk to the mixture in fractional form = 3
Quantity of water
= Quantity of mixture – Quantity of milk Let the quantity of water to be added further Assuming x litre water is mixed to get the required ratio. be x litres. 40 Then, milk : water = 20 + x 1 40 Given new ratio of milk and water = 1:2 Now, = 2 20 + x By cross multiplication. ⇒ 20 + x = 40 × 2 ⇒ 20 + x = 80 ⇒ x = 60 Quantity of water to be added to change the ratio of milk to water to be 1:2 is 60 litres.
Equivalent ratios – multiples of the given ratio36 Example 1 Find three equivalent ratios of 8:12. What is the simplest form of these ratios? Solution Two ratios that have the same value or the simplest form are called equivalent ratios. Procedure We have 8:12 =
36
New learning outcome
8 12
Explanation Writing the given ratio in fractional form.
Mathematical expressions that expand 47
8 8÷4 2 = = = 2:3 12 12 ÷ 4 3
To obtain an equivalent ratio, divide the numerator and denominator of the given fraction by their common factors. To obtain an equivalent ratio, multiply or divide the numerator and denominator of the given fraction by the same number, i.e., 2, 3, 4 ...
8 8 × 2 16 = = = 16 : 24 12 12 × 2 24 8 8 × 3 24 = = = 24 : 36 12 12 × 3 36 Thus, 2 : 3, 16 : 24, and 24 : 36 are the equivalent ratios of 8 : 12. Here the simplest form of these ratios is 2 : 3. Example 2 Are the given ratios 4:9 and 10:18 equivalent? Solution Procedure 4:9 =
10 4 and 10:18 = 18 9
Explanation Writing the given ratio in fractional form.
First method The simplest form of the two fractional ratios will be equal if they are equivalent. Simplest form of
4 4 = 9 9
HCF of 4 and 9 is 1.
HCF of 10 and 18 is 2. 10 10 ÷ 2 5 = = Simplest form of 18 18 ÷ 2 9 As the simplest form of the two ratios are not equal, they are not equivalent ratios. Second method To find whether the given ratios are equivalent, use cross multiplication method. The product of cross multiplication will be equal if the two ratios are equivalent. 4 10 Assuming the two ratios are equivalent. = 9 18 Cross multiply the numerator of one ratio to the 4 10 = denominator of second ratio. 9 18
⇒ 4 × 18 = 10 × 9 72 ≠ 90 ∴ The two ratios are not equal. Hence, 4:9 and 10:18 are not equivalent ratios.
Example 3 Find the value of x to make equivalent ratios.
48 A to Z of Mathematics
(a)
x 6 = 12 72
(b)
x 9 = 5 15
(c)
24 12 = x 25
(d)
4 2 = x 7
Solution In a pair of equivalent fractional ratios, we can find the missing value of the incomplete fractional ratio by multiplying or dividing its given numerator or denominator by some number so it becomes equal to the given (complete) fractional ratio’s numerator or denominator respectively. Procedure
6 6 × 6 36 = = 12 12 × 6 72
(a)
36 6 = 12 72 ∴ x = 36 9 9÷3 3 = = (b) 15 5 15÷3 5
(c)
⇒
6 Therefore, finding equivalent ratio of that has 72 in 12 the denominator. Since incomplete fractional ratio has 5 in the denominator.
33 99 == 55 15 15 ∴x=3
9 Therefore, finding equivalent ratio of that has 5 in 15 the denominator.
12 12 × 2 24 = = 25 25 × 2 50
Since incomplete fractional ratio has 24 in the numerator. 12 Therefore, finding equivalent ratio of that has 24 in 25 the numerator.
24 12 = 50 25 ∴ x = 50 ⇒
(d)
Explanation Since incomplete fractional ratio has 72 in the denominator.
2 2×2 4 = = 7 7 × 2 14
2 = 4 7 14 ∴ x = 14 ⇒
Since incomplete fractional ratio has 4 in the numerator. 2 Therefore, finding equivalent ratio of that has 4 in the 7 numerator.
Comparison of ratios37 Example 1 Compare the ratios – 5:12 and 3:8. Solution For comparing ratios write them in fractional form, and compare them as we compare fractions. Use the LCM of denominators to convert fractional forms of ratio into like fractions, then compare numerators to get the largest ratio. The lowest common multiple (LCM) of two numbers is the lowest possible number that can be divisible by both numbers. We will learn more about LCM in ‘Chapter M’. Procedure 3 5 5:12 = and 3:8 = 8 12 37
Refer P 252
Explanation Writing the given ratios in fractional form.
Mathematical expressions that expand 49
5 5 × 2 10 3 3×3 9 = = and = = 12 12 × 2 24 8 8 × 3 24 10 9 > 24 24 5 3 i.e., > 12 8 Hence, 5:12 > 3:8.
∴
Using LCM to convert the two fractions into like fractions for comparison. LCM of 12 and 8 is 24. Comparing like fractions is easy, greater the numerator, greater would be the fraction, and thus the ratio. 10 is equivalent to 5 and 9 is equivalent to 3 . 24 12 24 8
Example 2 Write the following ratio in ascending order. 4:5, 3:10, and 5:8 Solution Procedure 4:5 =
5 4 3 , 3:10 = , 5:8 = 5 10 8
Explanation Writing the given ratios in fractional form.
4 4 × 8 32 = = 5 5 × 8 40 3 3 × 4 12 = = 10 10 × 4 40 5 5 × 5 25 = = 8 8 × 5 40
Using LCM to make fractional ratios become like fractions. LCM of 5, 10, and 8 is 40.
12 25 32 < < 40 40 40 3 5 4 ⇒ < < 10 8 5
Comparing numerators of the like fractions, greater the numerator, greater would be the fractional ratio.
∴
Hence, ⇒
3 5 4 < < . 10 8 5
Example 3 If a cricket team won 18 matches and lost 3 matches in the year 2019 and won 12 matches and lost 3 matches in the year 2020. In which year, the record of the team was better? Solution For comparing the record of a team, first, find the ratios of matches won to the total number of matches played by the team in the year 2019 and 2020. Then, compare both the ratios. Procedure In 2019, the ratio of matches won to the total number of matches = 18:21
Explanation Number of matches won in 2019 = 18 Total number of matches played by the team in 2019 = Won matches + lost matches = 18 + 3 = 21
50 A to Z of Mathematics
=
18 21
3×6 3×7 6 = 7
Writing the ratio in fractional form. Reducing the fraction into its simplest form.
=
....(i)
In 2020, the ratio of matches won to the total Number of matches won in 2020 = 12 number of matches = 12:15 Total number of matches played by the team in 2020 = Won matches + lost matches = 12 + 3 = 15 Writing the ratio in fractional form. 12 = 15 Reducing the fraction into its simplest form. 3×4 = 3×5 =
4 5
....(ii)
4 Now, compare 6 and . 5 7 6 6 × 5 30 = = 7 7 × 5 35 4 4 × 7 28 = = 5 5 × 7 35 30 28 > 35 35 6 4 ∴ > 7 5
Using LCM to make the fractional ratios become like fractions. LCM of 7 and 5 is 35.
Comparing numerators of the like fractions, greater the numerator, greater would be the fractional ratio.
Thus, the record of the team was better in the year 2019.
6. Rate - A Special Kind of Ratio Idea of rate38 Example 1 Which of the following is the appropriate statement with respect to the concept ‘rate’? (a) Rate is a ratio of two like units. (b) Rate is a special type of percentage. (c) Rate is a special ratio with dimensions. (d) Rates are used in pricing in grocery stores and departmental stores, measuring speed, or paying hourly wages and monthly fees. 38
Refer P 253
Mathematical expressions that expand 51
Solution (a) Incorrect statement as the rate is a ratio of two unlike units. (b) (Strictly speaking) Incorrect statement as the rate is a special type of ratio. (c) Correct statement as the rate is a ratio of measurements that have different units. (d) Correct statement Example 2 Identify the situations which describes rate and unit rate from the following: (a) 6 kilometres/hour (b) ₹10/10 paise (c) 5 litres/2 buckets (d) 3 marbles/rupee (e) 1 metre/1 centimetre (f) 30 pages/day (g) ₹500/10 packets (h) 12 chairs/a pair of tables Solution
Procedure (a) It is a rate. It is a unit rate. (b) It is not a rate as well as a unit rate. It is a simple ratio. (c) It is a rate.
(d) It is a rate. It is a unit rate. (e) It is not a rate as well as a unit rate. It is a simple ratio. (f) It is a rate. It is a unit rate. (g) It is a rate.
(h) It is a rate. It is a unit rate.
Explanation ∵ Rate is a ratio that has different units of measurement. Here, it is kilometre and hour. ∵ The precedent of the ratio is 1. ∵ The rate should have different units of measurement. Here, both the units are units of currency. ∵ Rate is a ratio that has different units of measurement. Here, it is litre and bucket. We consider a bucket to have a definite volume. ∵ Rate is a ratio that has different units of quantification. Here, it is marble and rupee. ∵ The precedent of the ratio is 1. ∵ A rate should have different units of measurement. Here, the two units are metre and centimetre. Both the units are units for measuring length. ∵ Rate is a ratio that has different units of quantification. Here, it is page and day. ∵ The precedent of the ratio is 1. ∵ Rate is a ratio that has different units of measurement. Here, it is rupee and packet. We consider a packet to have a definite measurement. ∵ Rate is a ratio that has different units of quantification. Here, it is chair and a pair of tables. ∵ Here, pair of tables is a unit. ∴ The precedent of the ratio is 1. (If we consider 2 tables instead of a pair of tables, then the unit is a table. And, the precedent of the ratio is 2. Therefore, it will not be a unit rate.)
52 A to Z of Mathematics Example 3 Find the unit rate for the situation. (a) 350 words in 7 minutes (c) 20 metres in 10 seconds Solution Procedure (a) Rate = 350 words/7 minutes Number of words in a minute 350 = words 7 ⇒ Unit rate = 50 words/minute (b) Rate = ₹60/6 bananas Price of one banana 60 = rupee/banana 6 ⇒ Unit rate = 10 rupee/banana (c) Rate = 20 metres/10 seconds Distance covered in 1 second 20 = metres/second 10 ⇒ Unit rate = 10 metres/second (d) Rate = 452 packets/4 cartons Number of packets per carton 452 = packets/carton 4 ⇒ Unit rate = 113 packets/carton
(b) ₹60 for 6 bananas (d) 452 packets in 4 cartons Explanation Given ∵ Unit rate = Number of words in a minute
Given ∵ Unit rate = Price of one banana
Given ∵ Unit rate = Distance covered in a second
Given ∵ Unit rate = Number of packets per carton
Rate39
Example 1 A train runs 200 km in 5 hours. How many kilometres does it run in 7 hours? Solution We have to find the distance travelled in 7 hours. For this, first, find the distance travelled by the train in an hour (speed). Then, multiply the distance travelled in 1 hour, by 7. Procedure Distance travelled in 5 hours = 200 km
39
Refer P 253
Explanation Given
Mathematical expressions that expand 53
⇒ Distance travelled in 1 hour (speed) 200 km = 5 hours
Speed =
= 40 km/hr .…(i) ∴ Distance travelled in 7 hours = 7 × 40 km = 280 km Hence, the train covers 280 km in 7 hours.
Distance travelled Time taken
And from eq. (i), Speed of train = 40 km/hr
Rate is a special ratio which has ‘1’ as precedent. Example 2 A baker can bake 150 pastries in 2 hours. How many pastries can he bake in 9 hours? What is the rate per hour at which he bakes the pastries? Solution We have to find the number of pastries which the baker can bake in 9 hours. For this, first find the rate at which the baker bakes the pastries, that is, number of pastries baked per hour and then multiply it by 9 hours. Procedure
Explanation
Number of pastries baked in 2 hours = 150 ⇒ Number of pastries baked in 1 hour =
150 2
⇒ Rate of baking pastries = 75 pastries/hour
Given Rate of baking pastries = Number of pastries baked per hour
∴ Number of pastries baked in 9 hours ∴ Number of pastries baked in ‘n’ hours = 9 × 75 pastries = Rate per hour × n = 675 pastries Here, n = 9 hours Hence, 675 pastries can be baked in 9 hours and rate of baking is 75 pastries per hour.
Example 3 Rehana writes about 14 mails in a week. How many mails will she write in 49 days? Solution We need to find the rate at which Rehana writes mails, and then find the number of mails she writes in 49 days. Procedure Number of mails Rehana writes in 1 week Given (7 days) = 14 1 week = 7 days 14 = 2 mails/day ∴ Rate of writing mails = 7 Or rate of writing mails per day = 2 mails/day
Explanation
54 A to Z of Mathematics ∴ Number of mails Rehana can write in 49 days ∴ Number of mails Rehana can write in ‘n’ days = 49 × 2 mails = Rate of writing mails × n = 98 mails Here, n = 49 days Hence, Rehana can write 98 mails in 49 days.
7. Unitary Relationship Idea of unitary ratio40 Example 1 Write the ratio and unit ratio of the following: (a) (b)
:
:
=?
=?
(c)
(d)
:
:
=?
=?
Solution We know that ratio is a comparison of two quantities. A unit ratio is an equivalent ratio where antecedent is a unit (1). Every ratio can be converted to a unit ratio. Ratio
Unit Ratio
(a)
:
= 1:4
:
= 1:4
:
= 2:6
:
= 1:3
:
= 3:9
:
= 1:3
(b)
(c)
40
Refer P 253
Mathematical expressions that expand 55
(d)
:
= 8:8
:
= 1:1
Example 2 Apply the concept of unit ratio to fill in the blanks. (a) 1 blue car out of 13 cars ; 5 blue cars out of ______cars. (b) 1 football out of 15 balls ; ________footballs out of 60 balls. (c) 1 building has 54 flats ; 7 buildings have _________ flats. (d) 1 child eats 2 pieces of cake ; __________ children will eat 20 pieces of cake. Solution Procedure (a) 1 blue car out of 13 cars Blue car : cars = 1:13 ⇒ Blue cars : cars = 1 × 5:13 × 5 ⇒ Blue cars : cars = 5:65 Hence, there are 5 blue cars out of 65 cars. (b) 1 football : 15 balls Football : balls = 1:15 ⇒ Footballs : balls = 1 × 4:15 × 4 ⇒ Footballs : balls = 4:60 Hence, there are 4 footballs out of 60 balls. (c) 1 building : 54 flats Building : flats = 1:54 ⇒ Buildings : flats = 1 × 7:54 × 7 ⇒ Buildings : flats = 7:378
Hence, 7 buildings have 378 flats. (d) 1 child eats 2 pieces of cake. Child : pieces of cake = 1:2 ⇒ Children : pieces of cake = 1 × 10:2 × 10 ⇒ Children : pieces of cake = 10:20
Hence, 10 children will eat 20 pieces of cake.
Explanation Given Writing in ratio form. For 5 blue cars, multiplying 1 blue car by 5. ∴ For equivalent ratio, multiplying 13 cars by 5 also. Given Writing in ratio form. For 60 balls, multiplying 15 balls by 4. ∴ For equivalent ratio, multiplying 1 football by 4 also. Given Writing in ratio form. For 7 buildings, multiplying 1 building by 7. ∴ For equivalent ratio, multiplying 54 flats by 7 also. Given Writing in ratio form. For 20 pieces of cake, multiplying 2 pieces of cake by 10. ∴ For equivalent ratio, multiplying 1 child by 10 also.
56 A to Z of Mathematics
Unitary ratio41 Example 1 In a computer class, there are 6 computers for every 24 students. How many computers will be required for 80 students? Solution Find the unitary ratio of student to computers. Then, multiply it by 80 to get number of computers needed by 80 students. Procedure
Explanation
For every 24 students, the number of computers required = 6
Given
Ratio of students to computers = 24:6 = 4:1
Reducing the ratio into simplest form. 24 6× 4 4 = = ∵ 6 6×1 1
∴ Unitary ratio of student to computers = 1:
1 Unitary ratio is an equivalent ratio with antecedent 1. 1 4 ∴ 4:1 = 1: 4
∴ For 80 students, By unitary relationship, 1 1 Number of computers required ∵ 80 × = 4 × 20 × = 20 1 4 4 = 80 × = 20 4 Therefore, 20 computers will be required for 80 students. Example 2 A worker is paid ₹280 for working 7 days. What should be paid to the worker for working 30 days?
Solution Find the unitary ratio of day to payment. Then multiply it by 30 to get the money that should be paid for working 30 days. Procedure Payment for 7 days = ₹280 Ratio of days to payment = 7:280 = 1:40 ∴ Unitary ratio of day to payment = 1:40
Explanation Given Reducing the ratio into simplest form. 7×1 1 : 280 = = ∵77:280 7× 40 40 Unitary ratio is an equivalent ratio with antecedent 1.
By unitary relationship. ∴ Payment for 30 days = 30 × ₹40 = ₹1200 Hence, ₹1200 should be paid to the worker for 30 days.
41
Refer P 253
Mathematical expressions that expand 57
Example 3 Aarya pays ₹6600 as rent for 3 months. How much does he have to pay for a whole year, if the rent per month remains the same? Solution We have to find the rent which has to be paid by Aarya for a year. For this, first, find the unitary ratio of month to rent. Then, multiply it by 12 months. Procedure Rent for 3 months = ₹6600 Ratio of months to rent = 3:6600 = 1:2200
Explanation Given 3×1 = 11:2200 : 2200 3 × 2200 For unitary ratio, antecedent should be 1. ∵ 3:6600 =
∴ Unitary ratio of month to rent = 1:2200 We know, 1 year = 12 months. By unitary relationship. ∴ Rent for 12 months = 12 × ₹2200 = ₹26400 Hence, Aarya will have to pay ₹26400 for the whole year.
8. Percentage
Idea of percentage42 Example 1 Which of the following is the appropriate statement with respect to the concept ‘applications of percentage’? (a) The most basic use of percentages is to compare two quantities, with the first quantity rebased to 100. (b) Extensive data can be evaluated in less time and with more accuracy using percentages. (c) Fractions cannot be compared using percentages. (d) In academics, percentages are used to evaluate the performance of students. Solution (a) Incorrect statement as the most basic use of percentages is to compare two quantities, with the second quantity rebased to 100. (b) Correct statement (c) Incorrect statement as fractions can be compared using percentages. (d) Correct statement Example 2 Calculate the percentage of the shaded region for each figure. (a) (b) (c)
42
Refer P 254
(d)
58 A to Z of Mathematics Solution Procedure (a) Ratio of shaded small squares to the total number of squares = 81:100 In percent form, we can write 81:100 as 81%. (b) Ratio of shaded small squares to the total number of squares = 28:100 In percent form, we can write 28:100 as 28%. (c) Ratio of shaded small squares to the total number of squares = 15:100 In percent form, we can write 15:100 as 15%. (d) Ratio of shaded small squares to the total number of squares = 44:100 In percent form, we can write 44:100 as 44%.
Explanation ∵ Number of shaded small squares = 81 Total number of squares = 100 ∵ 81:100 is read as 81 per (‘/’) 100 (‘cent’) = 81% ∵ Number of shaded small squares = 28 Total number of squares = 100 ∵ 28:100 is read as 28 per (‘/’) 100 (‘cent’) = 28% ∵ Number of shaded small squares = 15 Total number of squares = 100 ∵ 15:100 is read as 15 per (‘/’) 100 (‘cent’) = 15% ∵ Number of shaded small squares = 44 Total number of squares = 100
∵ 44:100 is read as 44 per (‘/’) 100 (‘cent’) = 44%
How do we convert ratios to percentage?43 Example 1 Express 12:20 in percent form. Solution Simply stated, we have to find the equivalent ratio of 12:20 in such a way that ‘20 is 100’, that is, the precedent becomes 100. And, in the process, we have to find the antecedent, ‘what will be 12, if 20 is 100’. Procedure ∵ 12:20 =
12 × 5 60 = = 60:100 20 × 5 100
Hence, 12:20 can be expressed as 60:100. In percent form, we can write 60:100 as 60%.
Explanation For percentage scale up the precedent 20 to 100. ∵ 100 ÷ 20 = 5, 5 is the multiplier for scaling the ratio. Multiply the antecedent by 5, so that the ratio remains equivalent. Equivalent ratio of 12 : 20 is 60 : 100, when precedent is 100. ∵ 60:100 is read as 60 per (‘/’) 100 (‘cent’) = 60%
Percent is special ratio which has 100 as precedent. Example 2 There are 50 people in the swimming club out of which 35 go for squad training as well. Calculate the number of people who go for squad training as a percentage of the number of swimming club members. Solution We have to express the ratio of people who go for squad training to the people in the swimming club in percentage form. So, scale up the antecedent and precedent of the ratio in a way that ‘precedent is 100’. 43
Refer P 254
Mathematical expressions that expand 59
And, in the process, we have to find ‘what will be the antecedent of the ratio, if the precedent is 100’. Procedure
Explanation
The ratio of people who went for squad training to the Total number of people in a swimming club people in the swimming club = 35:50 = 50 Total number of people who go for squad training = 35 Converting ratio into percentage
For percentage, scale up the precedent from 50 to 100. 100 ÷ 50 = 2, 2 is the multiplier for scaling the ratio. Multiply the antecedent and precedent by 2 so that the ratio remains equivalent.
Hence, 35:50 can be expressed as 70:100.
Equivalent ratio of 35:50 is 70:100 when precedent is 100.
35 × 2 70 = = 70:100 70 : 100 = 50 × 2 100
In percent form, we can write 70:100 as 70%.
∵ 70:100 is read as 70 per (‘/’) 100 (‘cent’) = 70%
Therefore, 70% of the swimming club members go for squad training as well. Example 3 In a fabric, cotton and synthetic fibres are in the ratio of 2:3. What is the percentage of cotton fibre in the fabric? Solution Cotton fibres are one part of a given kind of fabric. We have to express the amount of cotton fibres in the fabric as a percent. Procedure
Explanation
The given ratio of cotton and synthetic fibres = 2:3
It implies that for every 2 units of cotton fibres, there are 3 units of synthetic fibre. Let the unit of fibres be x (x could be 1 kg, 100 m, etc.), Assumption thus the amount of cotton fibre in a given fabric is 2x. Thus, total quantity of fibres in the given fabric ∵ Quantity of cotton fibre = 2x and = 2x + 3x = 5x Quantity of synthetic fibre = 3x The ratio of the cotton fibre to the total amount of fibre For percentage, scale the precedent from 5 to in the given fabric 100. = 2x:5x = 2:5 ∵ 100 ÷ 5 = 20, 20 is multiplier for scaling the Converting ratio into percentage ratio. Multiply antecedent and precedent by 2 × 20 20 so that the ratio remains equivalent. = 5 × 20 =
40 100
60 A to Z of Mathematics
Hence, 2x:5x can be expressed as 40:100. In percent form, we can write 40:100 as 40%.
Equivalent ratio of 2:5 is 40:100 when precedent is 100.
∵ 40:100 is read as 40 per (‘/’) 100 (‘cent’) = 40%
The ratio of cotton fibre in the given fabric expressed in percent is 40%.
Example 4 If ₹250 is to be divided amongst Ravi, Raju, and Roy such that Ravi gets two parts, Raju gets three parts, and Roy gets five parts. How much money will each get? Express them in percentages?
Solution In this given problem, ₹250 is to be divided between Ravi, Raju, and Roy in the ratio 2 parts : 3 parts : 5 parts, i.e., 2:3:5. Find the total number of parts money is divided, the amount received by each person, and then express the percentage of money each person will get. Procedure Total number of parts money needs to be divided = 2 + 3 + 5 = 10 (a) Ratio of money Ravi receives to the total money = 2:10 Therefore, 2 Money received by Ravi = of ₹250 10 2 = ₹ × 250 10
= ₹50 Hence, Ravi receives ₹50. ∵ The ratio of money Ravi receives to the total money is 2:10. ∴ Percent of total money Ravi receives 2 × 10 = 10 × 10 20 = 100 = 20:100 The percentage of money Ravi receives is 20%. (b) Ratio of money Raju receives to the total money = 3:10 Therefore, 3 Money Raju receives = of ₹250 10 3 = ₹ × ₹ 250 10 = ₹75 Hence, Raju receives ₹75.
Explanation ∵ The parts that the three boys will get can be written in terms of ratios as 2:3:5. ∵ No. of parts for Ravi = 2 Total number of parts = 10 Total money which is to be divided between Ravi, Raju, and Roy = ₹250
For percentage, scale up the precedent from 10 to 100. ∵ 100 ÷ 10 = 10, 10 is the multiplier for scaling the ratio. Multiply antecedent and precedent by 10, so that the ratio remains equivalent.
∵ In percent form, we can write 20:100 as 20%. ∵ No. of parts for Raju = 3 Total number of parts = 10 Total money which is to be divided between Ravi, Raju, and Roy = ₹250
Mathematical expressions that expand 61 ∵ The ratio of money Raju receives to the total money is 3:10. ∴ Percentage of total money Raju receives 3 × 10 = 3 × 10 = 10 × × 10 10 10 30 = 30 = 100 100 = 30:100 The percentage of money Raju receives is 30%. (c) Ratio of money Roy receives to the total money = 5:10 Therefore, 5 Money Roy receives = of ₹250 × 250 10 5 = ₹ × 250 10 = ₹125 Hence, Roy receives ₹125. ∵ The ratio of money Roy receives to the total money = 5:10. ∴ Percentage of total money Roy receives 5 × 10 = 10 × 10 50 = 100 = 50:100 The percentage of money Roy receives is 50%.
For percentage scale up the precedent from 10 to 100. ∵ 100 ÷ 10 = 10, 10 is the multiplier for scaling the ratio. Multiply antecedent and precedent by 10, so that the ratio remains equivalent.
∵ In percent form, we can write 30:100 as 30%. ∵ No. of parts for Roy = 5 Total number of parts = 10 Total money which is to be divided between Ravi, Raju, and Roy = ₹250
For percentage scale up the precedent from 10 to 100. ∵ 100 ÷ 10 = 10, 10 is the multiplier for scaling the ratio. Multiply antecedent and precedent by 10 so that the ratio remains equivalent.
∵ In percent form, we can write 50:100 as 50%.
Thus, Ravi gets 20%, Raju gets 30%, and Roy gets 50% of ₹250 which is ₹50, ₹75, and ₹125 respectively.
Example 5 What percent of the figure is shaded?
Solution In the above figure, there are three squares out of which one and a half of the squares are shaded. Procedure
Explanation
Total number of parts = 3 1 Total number of shaded parts = 1
Given
2
1
1 1× 2 +1 3 = = 2 2 2
Calculating the shaded part in improper fraction.
62 A to Z of Mathematics
3 3 parts of 3 parts is shaded. Therefore, ratio of the shaded part to 3 parts = : 3 2 2 For percentage scale up the precedent from 3 ∵ The ratio of shaded part to total squares is 3 : 3 . to 100. 2 ∴ Percentage of shaded parts 100 100 ∵ 100 ÷ 3 = , is the multiplier for 3 100 × 3 3 3 = 2 scaling the ratio. Multiply the antecedent 100 3× 3 100 and precedent by , so that the ratio 50 3 = remains equivalent. 100 = 50:100 Here,
Hence, in percent form, we can write 50:100 as 50%. Therefore, 50% of the figure is shaded.
∵ 50% is 50 per (‘/’) 100 (‘cent’).
Difference between “what percentage of x is y?” and “what percentage is x of y?”44 Example 1 Express x in percentage. (a) x of 96 = 48
(b) x of 9 = 3.6
Solution We need to find equivalent ratio that has 100 as precedent. Procedure (a) x of 96 = 48
⇒ x × 96 = 48 48 ⇒x= 96 1× 48 ⇒x= 2 × 48 1 ⇒x= 2
⇒ x = 1:2
Converting ratio into percentage 1 × 50 2 × 50 50 ⇒x= ⇒ 100 ⇒ x = 50:100 ⇒x= ⇒
Explanation Given
Reducing the fraction to its simplest form.
Writing fraction in ratio form. For percentage, scale up the precedent from 2 to 100. ∵ 100 ÷ 2 = 50, 50 is the multiplier for scaling up the ratio. Multiply antecedent and precedent by 50, so that the ratio remains equivalent.
Hence, 48:96 can be expressed as 50 : 100 and in percent form, we can write it as 50%. Therefore, 48 is 50% of 96 or x = 50% 44
Refer P 258
Mathematical expressions that expand 63
(b) x of 9 = 3.6 ⇒ x × 9 = 3.6 3.6 ⇒x= 9
Given
36 90 2 ×18 ⇒x= 5×18 2 ⇒x= 5 ⇒x=
Removing the decimal part. Reducing the fraction to its simplest form.
Write fraction in ratio form. ⇒ x = 2:5 Converting ratio into percentage For percentage, scale up the precedent from 5 2 × 20 to 100. ⇒ ⇒x= 5 × 20 ∵ 100 ÷ 5 = 20, 20 is the multiplier for scaling 40 up the ratio. Multiply the antecedent and ⇒ ⇒x= 100 precedent by 20, so that the ratio remains ⇒ x = 40:100 equivalent. Hence, 3.6:9 can be expressed as 40:100 and in percent form, we can write it as 40%. Therefore, 3.6 is 40% of 9 or x = 40%. Example 2 What percentage of ₹250 is ₹50? And what percentage is ₹250 of ₹50? Solution The question is divided into two parts.
Part I Here, the relevant ratio for the given situation is 50:250. For percentage, we scale down the ratio 50:250 in a way that the precedent ‘250 is 100’. And, in the process, we have to find ‘what will be 50, if 250 is 100’. 50:250 50 ∴ = 250 =
50 × 1 50 × 5
=
1 5
Procedure
Explanation Given Ratio in fractional form. Simplest form of ratio.
64 A to Z of Mathematics
Converting ratio into percentage 1× 20 = 5 × 20 20 = 100 = 20:100 Hence, 50:250 can be expressed as 20:100 and in percent form, we can write it as 20%.
For percentage, scale up the precedent from 5 to 100. ∵ 100 ÷ 5 =20, 20 is the multiplier for scaling up the ratio. Multiply the antecedent and precedent with 20, so that the ratio remains equivalent. ∵ 20% is 20 per (‘/’) 100 (‘cent’).
Therefore, ₹50 is 20% of ₹250. School arithmetic uses an abstract method for such conversions. It multiplies the given ratio with 100 to find the percentage, which is just a short-cut.
50 × 100 is ‘what percentage of 50 is 250’. Here, the relevant ratio is 50:250, and 250 50 × 100 = 20% ∵ 250 Therefore, ₹50 is 20% of ₹250.
Part II Here, the relevant ratio is 250:50. For percentage, we have to scale up the ratio 250:50 in a way that the precedent ‘50 is 100’. And, in the process, we would find ‘what will be 250, if 50 is 100’. Procedure
Explanation
250:50 Given 250 Ratio in fractional from. = 50 5× 50 Simplest form of ratio. = 1× 50 5 = 1 Converting ratio into percentage For percentage, scale up the precedent from 1 5 × 100 to 100. ∵ 100 ÷ 1 = 100, 1 × 100 100 is the multiplier for scaling up the ratio. 500 = Multiply the antecedent and precedent by 100 100, so that ratio remains equivalent. = 500:100 Hence, 250:50 can be expressed as 500:100 and in percent form, we can write it as 500%. Therefore, ₹250 is 500% of ₹50. School arithmetic uses an abstract method for such conversions. It multiplies the given ratio with 100 to find the percentage, which is just a short-cut.
Mathematical expressions that expand 65
Here, the relevant ratio is 250:50, and 250 × 100 is ‘what percentage of 250 is 50’. 50 250 × 100 = 500% ∵ 50 Therefore, ₹250 is 500% of ₹50.
Example 3 What percentage is 18 hours of 3 days?
Solution First, convert 18 hours and 3 days into same units. Then, find the equivalent ratio of the given quantity (18 hours : 3 days) in such a way that ‘precedent is 100’. And, in the process, we have to find ‘what will be the antecedent, if precedent is 100’. Procedure Numbers of hours in 3 days = (24 × 3) hours = 72 hours
Explanation 1 day = 24 hours
18 hours:3 days = 18 hours: 72 hours 18 = 72
Given
=
1 4
3 days = 3 × 24 hours = 72 hours Writing ratio in fractional form.
18:72 =
18 18 × 1 1 = = 72 18 × 4 4
= 1:4
Writing ratio in simplest form.
Converting ratio into percentage
For percentage, scale up the precedent from 4 to 100. ∵ 100 ÷ 4 = 25 25 is the multiplier for scaling up the ratio. Multiply the antecedent and precedent by 25, so that ratio remains equivalent.
=
1 × 25 25 = = 25 : 100 4 × 25 100
1 × 25 25 = 25 : 100 = 4 × 25 100
= 25:100
Hence, 1:4 can be expressed as 25:100. And, in percent form, we can write 25:100 as 25%. Hence, 18 hours is 25% of 3 days.
Conversion of decimal into percentage45 Example 1 Express the following decimal number as a percent. (a) 0.037 (b) 0.002
(c) 3.56
(d) 425.0
Solution To convert the decimal number into percentage, first convert the decimal number into fractional form, and write it in ratio form. Then, scale up or scale down the ratio in such a way that precedent is 100 to find the percentage. 45
Refer P 260
66 A to Z of Mathematics
Procedure (a) 0.037
Given
37 1000 = 37:1000 1 37 × 10 = 37 ÷10 1 = 1000 × 1000 ÷1010 =
=
Explanation
3.7 100
Decimal number in fractional form. Writing fraction in ratio form. For percentage, scale down the precedent from 1000 to 100. 1 ∵ 100 ÷ 1000 = 10 1 is the multiplier for scaling down the ratio. 10 1 Multiply the antecedent and precedent by , 10 so that the ratio remains equivalent.
Hence, 0.037 can be expressed as 3.7:100 and in percent form, we can write it as 3.7%. (b) 0.002
Given
2 1000 = 2:1000
Writing decimal number in fractional form.
=
=
=
2×
Writing fraction in ratio form.
1 10
1000 × 0.2 100
1 10
For percentage, scale down the precedent from 1000 to 100. 1 ∵ 100 ÷ 1000 = 10 1 is the multiplier for scaling down the ratio. 10 1 Multiply the antecedent and precedent by , 10 so that the ratio remains equivalent.
Hence, 0.002 can be expressed as 0.2:100 and in percent form, we can write it as 0.2%. (c) 3.56 =
356 100
= 356:100
Given Writing decimal number in fractional form. Writing fraction in ratio form, here precedent is 100.
Hence, 3.56 can be expressed as 356:100 and in percent form, we can write it as 356%. (d) 425.0
4250 10 = 4250:10 =
Given Write decimal number in fractional form by removing the decimal. Write fraction in ratio form.
Mathematical expressions that expand 67
For percentage, scale up the precedent from 10 to 100. ∵ 100 ÷ 10 = 10 10 is the multiplier for scaling up the ratio. Multiply the antecedent and precedent by 10, so that the ratio remains equivalent.
4250 ×10 10 ×10 42500 = 100 =
Hence, 425.0 can be expressed as 42500:100 and in percent form, we can write it as 42500%. Example 2 Match up decimal and percentage that are equivalent. Decimal 3.05 0.007 0.305 0.8 0.7 0.85
Percentage 80% 30.5% 70% 305% 85% 0.7%
Quantifying percentage46 Example 1 I have 50 marbles. 24% of them are red. How many red marbles do I have? Solution We have to find the number of red marbles which is equal to 24% of 50. Procedure Number of red marbles = 24% of 50 marbles 24 × 50 = 100 = 12 marbles Hence, there are 12 red marbles.
Explanation ∵ 24% is 24 per (‘/’) 100 (‘cent’). It is written as 24 .× 50 100
Example 2 8% of a sum of money is ₹480, find the sum.
Solution Let the required sum of money be ₹x. How do we see x? It is 8% of something (x) is ₹480. Procedure
8% of ₹x = ₹480 ⇒ 46
8 × x = 480 100
Refer P 256
Explanation Given
∵ 8% is 8 per (‘/’) 100 (‘cent’). It is written as 8 . × x 100
68 A to Z of Mathematics
⇒x=
480 ×100 = 6000 8
On cross multiplication.
Therefore, the required sum of money is ₹6000.
Example 3 What is 16% of 40% of 8?
Solution Let the required number be y, i.e., y is 16% of 40% of 8. But what is y? It is 16% of something, and that something is 40% of 8. Thus, to find y, we have to first find ‘something’, i.e., 40% of 8. Procedure Something = 40% of 8 40 = ×8 100 = 3.2 y = 16% of something 16 = × 3.2 100 = 0.512
Explanation ∵ 40% is 40 per (‘/’) 100 (‘cent’). It is written as
40 . 100
∵ 16% is 16 per (‘/’) 100 (‘cent’).
16 . 100 And, something = 3.2
It is written as
However, the two steps can be combined as shown below: Required number =
16 40 × × 8 = 0.512 100 100
∵ 16% of 40% of 8 =
16 40 × ×8 100 100
Hence, 16% of 40% of 8 is 0.512. Example 4 If 16% of 40% of a number is 8, then what is the number? Solution Let the required number be y, i.e., 16% of 40% of y is 8. But what is y? 8 is 16% of something, and that something is 40% of y. Thus, to find y, we have to first find ‘something’, i.e., 40% of y. Procedure Something = 40% of y 40 ×y 100 4 = y 10 =
16% of something = 8
Explanation ∵ 40% is 40 per (‘/’) 100 (‘cent’). 40 It is written as . 100
∵ Something =
4 y 10
Mathematical expressions that expand 69 ∵ 16% is 16 per (‘/’) 100 (‘cent’).
4 y=8 10 16 4 × y =8 ⇒ ⇒ 100 10 ⇒ ⇒ 16% of
It is written as
1000 64 1000 ⇒y= ⇒ 8 ⇒ y = 125 ⇒
16 . 100
Simplifying
⇒y =8× ⇒
However, the above steps can be combined as shown below: 16% of 40% of y = 8 16 40 × ×y =8 100 100
(Given)
Simplifying
(8 × 100 × 100) 16 × 40
⇒y=
⇒ y = 125
Hence, the required number is 125.
Conversion of percentage into fraction or decimal form47 Example 1 Express each of the following per cents as a decimal and fraction in the simplest form.
2 3
(b) 0.8%
(a) 6 %
(c) 45%
(d) 480%
Solution 2 3
Procedure
(a) 6 % 20 = % 3 20 1 1 × = = 3 100 15 1 = 0.066 15
47
Refer P 260
Explanation Converting mixed fraction to improper fraction.
20 20 % is per (‘/’) 100 (‘cent’). 3 3 20 1 20 × It is written as ÷ 100 = 3 100 3 Then, express the fraction in simplest form. Convert the fraction into decimal form. ∵
70 A to Z of Mathematics
(b) 0.8% 8 1 8 1× 8 1 × = = = = 10 100 1000 125 × 8 125 8 1 8 1× 8 1 × = = = 10 100 1000 125 × 8 125 8 = 0.008 1000
∵
8 8 % is per (‘/’) 100 (‘cent’). 10 10 8 1 8 × It is written as ÷ 100 = 10 100 10 Then, express the fraction in simplest form.
Convert the fraction into decimal form.
1 45 5×9 9 45 = = (c) ×45% == 100 100 5 × 20 20
1 45 5×9 9 45 × = = = 100 100 5 × 20 20 45 = 0.45 100 1 480 24 × 20 24 4 480 × = = =4 (d) 480% = 100 100 5 × 20 5 5
480 ×
Convert decimal number into a fraction.
1 480 24 × 20 24 4 = = = =4 100 100 5 × 20 5 5
480 = 4.8 100
∵ 45% is 45 per (‘/’) 100 (‘cent’).
45 . 100 Then, express the fraction in simplest form. It is written as
Convert the fraction into decimal form. ∵ 480% is 480 per (‘/’) 100 (‘cent’). 480 It is written as . 100 Then, express the fraction in the simplest form. But, it is an improper fraction. So, convert it to a mixed fraction. Convert the fraction into decimal form.
9. Proportion Idea of proportion48 Example 1 Explain that the two given figures (a) and (b) are in proportion using the concept of ratio.
1:3 (a)
2:6 (b)
Solution Here, we are comparing the size of two triangles and the size of two squares using a ratio. If these ratios are equivalent, then the two figures are in proportion. Since the equality of the two ratios is known as proportion.
48
Refer P 261
Mathematical expressions that expand 71
Procedure Ratio of size of triangles = 1:3
Explanation Given
Ratio of size of squares = 2:6
Given Writing the ratio in fractional form.
2 6
=
Reducing the fraction to its simplest form.
2×1 2×3 1 = 3 Ratio of size of squares = 1:3 1:3 = 2:6 =
Converting the fractional form into ratio. ∵ The ratio of the size of two triangles and the ratio of the size of two squares are equivalent. ∵ The two ratios are equivalent. ∴ They are in proportion.
⇒ 1:3 :: 2:6
Hence, the two figures are in proportion.
Example 2 Determine whether the x and y values in each table are proportional. (a) x 1 3 9 (b) x 2 1 3 4 5 y 2 6 18 10 y 8 4 12 20 Solution Procedure (a)
x
1
3
9
5
y
2
6
18
10
×3
∴
×5
×9
×3
Explanation Check the proportionality of x and y by checking whether the ratios of x and y are equivalent or not.
×9
1 3 9 5 = = = 2 6 18 10
Hence, x and y are in proportion.
×5
1 1× 3 1× 9 1× 5 = = = 2 2×3 2×9 2×5
∵ All ratios are equivalent.
72 A to Z of Mathematics
(b)
×2
Check the proportionality of x and y by checking whether the ratios of x and y are equivalent or not.
×4
×3
x
2
1
3
4
y
8
4
12
20
×3 ×2
×5 ×4
Hence, x and y are not in proportion.
2 1 3 4 = = ≠ 8 4 12 20 ∴ Ratios are not equivalent. ∵
Example 3 If 3 men take 1 second to pluck 3 flowers, then how much time would 1 man take to pluck all the three flowers? Solution
Procedure To pluck 3 flowers, 3 men take 1 second. ⇒ 1 man can pluck 1 flower in 1 second ⇒ 1 man plucks 3 flowers in 3 seconds.
Explanation Given Equation (i) From eq.(i)
From the above, we can say that more men can do a piece of work in less time.
Are the numbers in proportion?49 Example 1 Are 25, 15, 50, 30 in proportion? Solution We have to find that the given numbers are in proportion or not. For this, we know that proportion is a c simply a statement that two ratios are equal. It can be written in two ways: a:b :: c:d or = b d ⇒a×d=b×c Here, ‘a’ and ‘d’ are extreme terms and ‘b’ and ‘c’ are middle terms. Procedure
We have 25, 15, 50, 30. Product of extreme terms = 25×30 = 750 Product of middle or mean terms = 15×50 = 750 The four numbers 25, 15, 50, 30 are in proportion.
49
Refer P 261
Explanation
Given a, b, c, d are in proportion, if a×d = b×c. Since both the products are the same.
Mathematical expressions that expand 73
If four given numbers are to be checked for proportionality, then we cannot change the order of numbers. Example 2 A shopkeeper sold vanilla and strawberry ice creams. For every 11 ice creams sold, 4 of them were strawberry ice creams. Altogether 40 strawberry ice creams were sold. How many ice creams did the shopkeeper sell together? Solution We have to find the total number of ice creams sold by the shopkeeper. First find the number of vanilla ice creams sold by him and then find the total number of ice creams using proportionality. Procedure Number of strawberry ice creams sold for every 11 ice creams = 4 For every 11 ice creams sold, number of vanilla ice creams sold = 11 – 4 = 7 Total number of strawberry ice creams sold by the shopkeeper = 40 Let the total number of vanilla ice creams sold by the shopkeeper be x. Then, 4, 7, 40, and x are in proportion. 4 40 = 7 x ⇒ 4x = 40 × 7 40 × 7 ⇒x= 4 ⇒ x = 70 ⇒
Explanation Given ∵ For every 11 ice creams sold, 4 of them were strawberry ice creams. Given Assumption ∵ 4:7 :: 40:x
Hence, total number of ice creams sold by the shopkeeper = 40 + x = 40 + 70 = 110. Example 3 What number shall be added to each of the numbers 4, 7, 14, 22 to form the terms in a proportion? Solution Procedure Let the number to be added be x. Assumption Then, (4 + x), (7 + x), (14 + x), (22 + x) are in proportion.
4 + x 14 + x = 7 + x 22 + x ⇒ (4 + x) (22 + x) = (7 + x) (14 + x) ⇒ 88 + 4x + 22x + x2 = 98 + 7x + 14x + x2 ⇒ 88 + 26x + x2 = 98 + 21x + x2
Explanation
∵ (4 + x), (7 + x), (14 + x), (22 + x) are in proportion. By cross multiplication. Simplifying.
74 A to Z of Mathematics ⇒ 26x – 21x + x2 – x2 = 98 – 88 ⇒ 5x = 10 10 ⇒x= =2 5
Transposing 21x and x2 to the left-hand side of the equation and 88 to the right-hand side of the equation.
Hence, 2 is added to each of the numbers 4, 7, 14, 22 to form 6, 9, 16, 24 (new terms) which are in proportion. Example 4 If 2a − 3b = 0, find the value of (a−b):(a+b). Solution There are two ways in which we can find the value of this ratio. In the first method, replace the value of ‘b’ in the ratio with the value of ‘a’ using the relation 2a = 3b. In the second method, use componendo and dividendo rule as it is a quick way to perform calculations and reduce the number of expansions needed. Procedure 2a − 3b = 0 ⇒ 2a = 3b a 3 ⇒ ⇒ = b 2 a 3 ⇒ + 1= + 1 b 2
a + b 3+2 5 = = b 2 2 Again, a 3 ⇒ −1= −1 b 2
⇒
a −b 3−2 1 = = b 2 2 On dividing eq. (iii) by eq. (ii), we get a−b 1 = ⇒ a+b 5
⇒
Hence, the value of (a – b):(a + b) is 1:5.
Explanation Given Simplifying ....(i) By applying componendo, add 1 to both sides of the equation. ....(ii)
∵ LCM of b and 1 is b.
....(iii)
∵ LCM of 1 and b is b. LCM of 1 and 2 is 2.
LCM of 2 and 1 is 2. By applying dividendo, subtract 1 from both sides of the eq. (i).
∵
a−b b a+b b
=
1 2 5 2
⇒
a−b 1 = a+b 5
Mathematical expressions that expand 75
Proportion properties a c b d (a) If = , then = b d a c a c a b = , then = (b) If b d c d a+b (c) If a = c , then = b b d a –b (d) If a = c , then = b b d a+b (e) If a = c , then = a–b b d
[By invertendo property] [By alternendo property] c+d d c–d d c+d c–d
[By componendo property] [By dividendo property] [By componendo – dividendo property]
Finding the missing term50 Example 1 Find the value of x in the following proportionalities. 1 1 1 (a) 30:25 :: 42:x (b) : :: : x 9 5 18 Solution Procedure (a) 30:25 :: 42:x 30 42 ⇒ ⇒ 30= 42 ⇒ 25 = x 25 x ⇒ 30x = 42 × 25 ⇒ 30x = 42 × 25 ⇒ 42 × 25 ⇒ x = 42 × 25 ⇒ x = 30 ⇒ 30 ⇒ x = 35 ⇒ x = 35 ⇒
1 1 1 : :: : x 9 5 18 1 1 1 1 9 18 ⇒ = ⇒ ⇒ 1 9 = x18 1 x 5 5 1 1 1 ⇒ 1× x = 1× 1 ⇒ 9 × x =18 ×5 ⇒ 9 18 5 9 9 ⇒ x= ⇒ ⇒ x =18 × 5 18 × 5 1 ⇒ x= 1 ⇒ x =10 ⇒ 10
(b)
50
Refer P 261
Explanation Given ∵ 30, 25, 42, and x are in proportion. By cross multiplication.
Given
∵
1 1 1 , , and x are in proportion. 9 5 18
By cross multiplication.
76 A to Z of Mathematics Example 2 The first, second, and fourth terms of a proportion are 6, 10, and 25 respectively. Find its third term. Solution Procedure Explanation Let the third term be x. Then, 6, 10, x, 25 are in proportion. We know that, Product of extreme terms = Product of mean terms 6 × 25 = 10 × x Out of four terms, the extreme terms are 6 and 25, ⇒ 150 = 10x while the mean terms are 10 and x. ⇒ x=
150 10
⇒ x = 15 Therefore, the third term of the proportion is 15.
Continued proportion51 Example 1 If 3, x, 12 are in continued proportion, find the value of x. Solution Any three quantities are said to be in continued proportion if the ratio between the first and the second terms is equal to the ratio between the second and the third terms. a b If a, b, c are in continued proportion, then = . b c Procedure 3, x, 12 are in continued proportion. 3, x, x, 12 are in proportion.
Explanation
Given ∵ From the definition of continued proportion, a:b :: b:c. We know that, Product of extreme terms = Product of mean terms. ⇒ 3 × 12 = x × x ∵ Extreme terms = 3 and 12 ⇒ 36 = x2 Both mean terms = x ⇒ 62 = x2 ⇒x=6 Hence, the value of x is 6.
We consider only positive value of x.
Example 2 Three people P, Q, and R earned a profit of ₹100000 in a business. Their share of profit is as follows: P:Q = 3:5 and Q:R = 10:4. Find the share of each person.
Solution We have to find the share of each person in the profit. For this we should know the ratio of their profit. And, P:Q and Q:R are in continued proportion. So, first find P:Q:R and then find each person’s share.
51
New learning outcome
Mathematical expressions that expand 77
Procedure P:Q = 3:5 and Q:R = 10:4
Explanation Given Writing ratio in fractional form.
P 3 Q 10 = and = Q 5 R 4
To find the ratio P:Q:R, value of Q should be same in P 3× 2 6 = = P 3 Q 10 . For this, take LCM of 5 and 10. Q 5× 2 10 = and = Q 5 R 4 Q 10 = LCM of 5 and 10 = 10 R 4 Now required ratio of share of profit is P:Q:R = 6:10:4 Given Total profit earned by P, Q, and R = ₹100000 Sum of ratio = 6 + 10 + 4 = 20 ∵ P:Q:R = 6:10:4 6 6 Ratio of P’s share out of total profit = Share of P = ₹ × 100000 = ₹30000 20 20 10 10 Ratio of Q’s share out of total profit = Share of Q = ₹ × 100000 = ₹50000 20 20 4 4 Ratio of R’s share out of total profit = Share of R = ₹ × 100000 = ₹20000 20 20 Hence, the share of P, Q, and R in the profit is ₹30000, ₹50000, and ₹20000, respectively.
Mean proportion52
Example 1 Find the mean proportional between 9 and 25. Solution If a, b, c are in continued proportion, then b is the mean proportional between a and c. Thus, the mean proportional (b) =
ac .
Procedure Explanation Let x be the mean proportion between 9 and 25. Assumption If three numbers a, b, c are in proportion, then b is ⇒ x = 9 × 25 the mean proportional given by b = ac . ⇒ x = 225 ⇒ x = 15 × 15 ⇒ x = 15 Hence, the mean proportion between 9 and 25 is 15.
Example 2 Find two numbers whose mean proportional is 16 and their third proportional is 54.
52
New learning outcome
78 A to Z of Mathematics Solution Procedure Let two numbers be p and q. p:16 = 16:q p 16 = ⇒ 16 q ⇒ ⇒ pq =16 ×16 ⇒ ⇒ pq = 256 256 ⇒ ⇒ p= q
Explanation Assumption Mean proportional between p and q is 16. (Given)
....(i)
p:q = q:54 p q ⇒ ⇒ = q 54 ⇒ ⇒ p × 54 = q 2 256 × 54 = q 2 ⇒ ⇒ q ⇒ 256 × 54 = q 2 × q ⇒
Third proportional to p and q is 54. (Given)
⇒ ⇒ q = 3 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3 ⇒ ⇒ q = 2 × 2 × 2 × 3 = 24
Making triplets for cube root.
From eq (i), putting the value of p =
⇒qq23= 256 × 54 ⇒ 3 ⇒ ⇒ q = 256 × 54 p=
256 24
⇒ ⇒ p=
32 3
256 . q
Taking out one factor from each triplet. Putting q = 24 in eq. (i), p =
256 . q
Reducing fraction to its simplest form.
256 32 × 8 32 = = 24 3×8 3
Hence, the two numbers are 32 and 24. 3
Direct proportion53 Example 1 A car covers 432 km using 36 litres of petrol. How much distance would it cover using 25 litres of petrol? Solution Let the required distance be x km when 25 litres of petrol is used. Then, we have Quantity of petrol (in litres)
36
25
Distance (in km)
432
x
We know that less the quantity of petrol consumed, less is the distance covered, a case of direct proportion. 53
New learning outcome
Mathematical expressions that expand 79
∴ Ratio of the quantity of petrol = Ratio of distance covered Procedure
Explanation Since quantities are in direct proportion.
∴ 36 : 25 = 432 : x ⇒
Converting the ratio into fractional form.
36 432 = 25 x
By cross multiplication.
⇒ 36 × x = 432 × 25
432 × 25 36 12 × 36 × 25 ⇒ ⇒x= 36 ⇒ x = 12 × 25 ⇒ ⇒ x = 300 Therefore, the distance covered by the car using 25 litres of petrol is 300 km.
⇒ ⇒x =
Example 2 The amount of extension in an elastic spring varies directly with weight hanging on it. If weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm? Solution Let the required weight be x grams. Then, Weight (in grams)
150
x
Extension (in cm)
2.9
17.4
The amount of extension in the spring varies directly as the weight hanging on it, a case of direct proportion. ∴ Ratio of weight = Ratio of extension Procedure
∴ 150 : x = 2.9 : 17.4
Explanation Since quantities are in direct proportion.
150 2.9 Converting the ratio into fractional form. = x 17.4 ⇒ x × 2.9 = 150 × 17.4 By cross multiplication. 150 × 17.4 = 900 ⇒x= 2.9 Thus, 900 gm weight will produce an extension of 17.4 cm.
⇒
Inverse proportion54 Example 1 If 45 men can do a piece of work in 49 days, then in how many days will 35 men do it?
54
New learning outcome
80 A to Z of Mathematics Solution Let the required number of days be x. Then, we have Number of men
45
35
Number of days
49
x
Fewer men will take more days to finish the work, so it is a case of inverse proportion. ∴ Number of men is inversly proportional to number of days it takes to finish the work. Procedure
Explanation
∵ 45 men or 35 men are expected to do the same work. Since quantities are in inverse proportion.
∴ 45 : 35 = x : 49 ⇒
45 x = 35 49
Converting the ratio into fractional form.
⇒ 35 × x = 45 × 49 45 × 49 = 63 ⇒x= 35 Therefore, 35 men will complete the work in 63 days.
By cross multiplication.
Example 2 Shalu cycles to her school at an average speed of 12 km/hr. It takes her 20 minutes to reach the school. If she wants to reach her school in 15 minutes, what should her average speed be? Solution Let the required speed be x km/hr. Then, the given information can be put in a table as: Speed (in km/hr)
12
x
Time (in minutes)
20
15
It is obvious that more the speed less will be the time taken to cover the given distance. So, it is a case of inverse variation. ∴ Speed is inversely proportional to time taken to travel. Procedure As distance travelled is same, 12:x = 15:20 ⇒
12 15 = x 20
⇒ 12 × 20 = x × 15 ⇒x=
Explanation Since quantities are in inverse proportion. Converting the ratio into fractional form. By cross multiplication.
12 × 20 240 = = 16 15 15
Therefore, Shalu’s average speed should be 16 km/hr to reach the school in 15 minutes.
Mathematical expressions that expand 81
Example 3 A garrison of 500 men had provision for 27 days. After 3 days, a reinforcement of 300 men arrived. For how many more days will the remaining food now last? Solution Let the remaining food will last for x days. Since reinforcement comes after 3 days, therefore 500 men had provision left for (27 – 3) 24 days. Total number of men available after the reinforcement of 300 men came is 500 + 300 or 800. Men Number of days
500 24
800 x
Since more men will consume the food in less number of days and the less number of men will consume the food in more number of days. So, it is a case of inverse proportion. Procedure
As the amount of provision remains the same, ∴ 500 : 800 = x : 24 500 x = 800 24 ⇒ 800 × x = 24 × 500 500 × 24 ⇒x= 800 ⇒ x = 15 days
⇒
Explanation Ratio of number of men = Inverse ratio of number of days Converting the ratio in fractional form.
Hence, the remaining food will last for 15 days. Example 4 A speedboat, whose speed is 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. What is the speed of the stream in km/hr? Solution Let the speed of the stream be x km/hr. To find the speed of the stream, form the equation with the help of the relation of distance, speed, and time. But, in the case of a river, the direction of the flow of water and boat plays an important role. Procedure Speed of the stream = x km/hr Upstream speed of the boat = (15 – x) km/hr Downstream speed of the boat = (15 + x) km/hr
Explanation Assumption ....(i)
Speed of boat = 15 km/hr (given)
....(ii)
30 hr Time taken to go downstream = 15+ x
Distance = 30 km and from eq. (ii), downstream
30 Time taken to go upstream = hr 15 – x
Distance = 30 km and from eq.(i), upstream
speed = (15 + x) km/hr speed = (15 – x) km/hr
82 A to Z of Mathematics
Total time taken by boat = Time taken to go downstream + Time taken to go upstream ⇒ ⇒ ⇒ ⇒
30 30 1 + = 4 15 + x 15 – x 2 30 (15 − x) + 30(15 + x) 9 = (15 + x) (15 − x) 2
450 − 30x + 450 + 30x 9 = 225 − x 2 2 900 9 = ⇒ 225 − x 2 2 ⇒ 900 × 2 = 9 × (225 – x2)
⇒ ⇒
Total time taken by boat is 4 hrs 30 min (Given) 1 9 4 hr 30 min = 4 hrs = hrs 2 2 LCM of (15 + x) and (15 – x) is (15 + x) (15 – x).
(a – b) (a + b) = a2 – b2
⇒ 1800 = 2025 – 9x2 ⇒ 9x2 = 2025 – 1800 = 225 225 ⇒ x2 = 9 2 ⇒ x = 25 = 52 ⇒ x = 5
Hence, the speed of the stream is 5 km/hr. (a) If the boat is moving in the opposite direction of the flow of the stream (going upstream), then the effective speed of the boat slows down as it moving against the flow of the stream. Hence, the speed of the boat in still water – speed of the stream becomes the effective speed with which the boat moves. (b) If the boat is moving along the direction of the flow of the stream (going downstream), then the effective speed of the boat increases as it is moving with the flow of the stream. Hence, the speed of the boat in still water + speed of the stream becomes the effective speed with which the boat moves.
10. Arithmetic Progression Finding missing term in AP 55 Example 1 Find the missing number in the given AP: 7, 12, 17, ? , 27 Solution Procedure Difference = 12 – 7 = 5 12 = 7 + 5 17 = 12 + 5 ? = 17 + 5 = 22 ∴ The missing number in the given sequence is 22. 55
Refer P 266
Explanation Finding the difference between first two terms. Second term = First term + difference Third term = Second term + difference Fourth term = Third term + difference
Mathematical expressions that expand 83
Example 2 Find the missing term in the following AP. (b) 33 , 29 , 25 , y, 17
(a) 18, 10, 2, x, –14
2
Solution
2
2
2
Procedure
Explanation
(a) 18, 10, 2, x, –14 Difference = 10 – 18 = –8 –8
–8
Given Difference between two consecutive terms. Adding (–8) to each term to get the next term.
–8
–8
∴ 18,
10, 2, –6 , –14 Hence, the value of x = – 6 33 29 25 17 , , , y, (b) 2 2 2 2
Difference = –4 2
∴ 33 , 29 , 2 2
29 33 −4 − = 2 2 2 –4 2
–4 2
25 , 2
Difference between two consecutive terms.
–4 2
21 , 2
Hence, the value of y =
Given
Adding (
–4 ) to each term to get the next term. 2
17 2
21 2
Characterising arithmetic progressions56 Example 1 Which of the following sequences are in AP? For the sequences in AP, write the next two terms. (a) 2, 8, 14, 20,... (b) 1, –2, –5, –8,... (c) –2, 2, –2, 2, –2,... Solution Procedure (a) We have, a1 = 2, a2 = 8, a3 = 14, a4 = 20 a2 – a1 = 8 – 2 = 6 a3 – a2 = 14 – 8 = 6 a4 – a3 = 20 – 14 = 6
Explanation Finding the difference between two consecutive terms of the given sequence.
The given sequence is in AP with the common difference, d = 6 Therefore, the next two terms of the AP are 20 + 6 = 26 and 26 + 6 = 32. (b) We have, a1 = 1, a2 = –2, a3 = –5, a4 = –8 a2 – a1 = –2 –1 = –3 a3 – a2 = – 5 – (–2) = –5 + 2 = –3 a4 – a3 = –8 – (–5) = –8 + 5 = –3 56
Refer P 266
Finding the difference between two consecutive terms of the given sequence.
84 A to Z of Mathematics
The given sequence is in AP with the common difference, d = –3 Therefore, the next two terms of the AP are –8 + (–3) = –11 and –11 + (–3) = –14. (c) We have, a1 = –2, a2 = 2, a3 = –2, a4 = 2, a5 = –2 a2 – a1 = 2 – (–2) = 2 + 2 = 4 a3 – a2 = –2 – 2 = –4
Finding the difference between two consecutive terms of the given sequence.
As a2 – a1 is not equal to a3 – a2, therefore the given sequence is not in AP.
Using the general term of arithmetic progression57 Example 1 Find the tenth term of the AP: 4, 8, 12, . . . Solution Procedure a = 4, d = 8 – 4 = 4, and n = 10. We have an = a + (n – 1) d
Explanation Given General term formula of an AP.
So, a10 = 4 + (10 – 1) × 4 = 4 + 36 = 40 Therefore, the tenth term of the given AP is 40.
Putting the value of a, d, and n in a10 = a + (10 – 1)d to find the tenth term.
The general term (an) of an AP is the nth term with ‘a’ as first term and ‘d’ as common difference between any two consecutive terms. General term (an) = a + (n – 1) d Example 2 The sum of fifth and ninth terms of an AP is 72 and the sum of seventh and twelfth terms is 97. Find the AP. Solution We use the general term to find the value of a and d, thus the terms of AP. Procedure a5 + a9 = 72 and, a7 + a12 = 97 a5 + a9 = 72 ⇒ (a + 4d) + (a + 8d) = 72 ⇒ 2a + 12d = 72
a7 + a12 = 97 ⇒ (a + 6d) + (a + 11d) = 97 ⇒ 2a + 17d = 97
2a + 17d – (2a + 12d) = 97 – 72 ⇒ 5d = 25 ⇒d=5 57
Refer P 269
Explanation Given
an = a + (n – 1) d ....(i)
an = a + (n – 1) d ....(ii) Subtracting eq. (i) from eq. (ii), to get an equation with only one variable.
Mathematical expressions that expand 85
2a + 12 (5) = 72 ⇒ 2a + 60 = 72 ⇒ 2a = 12 ⇒a=6 Hence, the required AP is 6, 11, 16, 21, 26,…
Putting d = 5 in eq. (i), we get
a = 6 and d = 5
Example 3 How many two-digit numbers are divisible by 3? Solution Two digit numbers which are divisible by 3, have 3 as common difference between any two consecutive terms. Since the common difference is same, thus the sequence is an AP. The first two digit number divisible by 3 is 12 and last is 99. Thus, we can find the total number of terms in the sequence using the general term formula. Procedure 2-digit numbers are 10, 11, 12, .... 99. The list of two-digit numbers divisible by 3 is 12, 15, 18,... , 99. 12, 15, 18.., 99 is an AP. Here, a = 12 d = 15 – 12 = 3 an = 99
Explanation We already know. Two digit numbers which are multiples of 3. ∵ Common difference between any two consecutive terms is always 3 because the sequence is a multiple of 3. From the above series.
Putting the value of a, d, and an in the general term an = a + (n – 1) d formula to find n. ⇒ 99 = 12 + (n – 1) × 3 ⇒ 99 = 12 + 3n – 3 ⇒ 3n = 99 – 12 + 3 ⇒ 3n = 90 ⇒ n = 90 ÷ 3 = 30 Therefore, there are 30 two-digit numbers that are divisible by 3. Example 4 Which term of the AP 14, 19, 24, … is 299? Solution Here we are given ‘a’ of AP. We can find ‘d’ from the given sequence and can use the general term formula to find ‘n’ when nth term is given. Procedure Let 299 be the nth term of an AP. a = 14, d = 19 – 14 = 5
Explanation Given 14, 19, 24 …. is an AP.
86 A to Z of Mathematics
an = a + (n – 1)d
General term formula of an AP.
⇒ 299 = 14 + (n – 1)5 Putting the values of a = 14, d = 5, and an = 299 in the general term formula to find n. ⇒ 299 = 14 + 5n – 5 ⇒ 299 – 14 + 5 = 5n ⇒ 290 = 5n 290 ⇒n= = 58 5 Hence, 299 is the 58th term of the given sequence.
How to assume terms in AP?58 Example 1 The sum of three numbers in AP is –3, and their product is 8. Find the numbers. Solution Procedure Let the numbers be (a – d), a, (a + d). Sum of 3 numbers in AP = –3 ⇒ (a – d) + a + (a + d) = –3 ⇒ 3a = –3 ⇒ a = –1
Product of 3 numbers in AP = 8 ⇒ (a – d) (a) (a + d) = 8 ⇒ a (a2 – d2) = 8 ⇒ (–1) (1 – d2) = 8 ⇒ –1 + d2 = 8 ⇒ d2 = 9 ⇒ d = 3, –3 If d = 3, the numbers are –4, –1, 2. If d = –3, the numbers are 2, –1, –4.
Explanation This is a convenient way of assuming 3 consecutive terms in AP. Given ....(i) Given ∵ (a – d) (a + d) = a2 – d2 From eq.(i), a = –1. ∵ Square root of 9 is +3, –3. ∵ a – d = (–1) – (3) = –4; a = –1; a + d = –1 + 3 = 2 ∵ a – d = (–1) – (–3) = 2; a = –1; a + d = (–1) + (–3) = –4
Thus, the numbers are –4, –1, 2, or 2, –1, –4. (a) When we have to assume even number of terms in AP, it is convenient to take the two middle terms as (a – d) and (a + d) and common difference as ‘2d’. (b) When we have to assume odd number of terms in AP, it is convenient to take the middle term as ‘a’ and common difference as ‘d’. Example 2 Find four numbers in AP whose sum is 20 and the sum of whose squares is 120.
58
Refer P 271
Mathematical expressions that expand 87
Solution Procedure Let the consecutive terms in AP be (a – 3d), (a – d), (a + d), (a + 3d) Sum of 4 terms in AP = 20
Explanation This is the most convenient way of assuming 4 consecutive terms in AP.
⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 ⇒ 4a = 20 ⇒a=5 .... (i) Sum of the squares of 4 terms in AP = 120 ⇒ (a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 120
Adding all the four assumed terms and equating this to 20 (as given).
Given
Given Squaring all the assumed terms and equating them to 120 (as given). 2 2 2 2 2 2 2 ⇒ a + 9d – 6ad + a + d – 2ad + a + d + 2ad + a + On expanding the terms by using the formula: (a + b)² = a² + b² + 2ab 9d2 + 6ad = 120 2 2 (a – b)² = a² + b² – 2ab ⇒ 4a + 20d = 120 2 2 Dividing by 4 on both sides. ⇒ a + 5d = 30 ⇒ 25 + 5d2 = 30 ∵ a = 5, ∴ a2 = 25 [from eq. (i)] As square root of 1 = 1, –1 ⇒ 5d2 = 5 or d = 1, –1 If d = 1, then the numbers are 2, 4, 6, 8. ∵ a – 3d = 5 – 3 = 2; a – d = 5 – 1 = 4; a + d = 5 + 1 = 6; a + 3d = 5 + 3 = 8 If d = –1, then the numbers are 8, 6, 4, 2. ∵ a – 3d = 5 – 3(–1) = 8; a – d = 5 – (–1) = 6; a + d = 5 + (–1) = 4; a + 3d = 5 + 3 (–1) = 2 Thus, the four consecutive numbers in AP are 2, 4, 6, 8 or 8, 6, 4, 2. Their sum is 20 and sum of their squares is 120.
Sum of n terms of an AP59 Example 1 Ritu makes a 7 storied triangle starting with a base of 13 building blocks. Each storey has two blocks less than previous storey. How many blocks would be required to make such a triangle? Solution The number of bricks in the base 13, and the number of bricks in each subsequent storey is 2 less than in the preceding storey. So, the numbers of bricks in the stories form an arithmetic progression. Procedure Explanation First term of an AP is 13, common difference is –2, and the number of terms is 7. The total number of bricks in Ritu’s construction is equal to the sum of the seven terms of this arithmetic progression. Here, a = 13, d = –2 and n = 7. Given n Applying formula for finding the sum of an AP, Snn = [2a + (n − 1)d] where a, n, and d are known. 2
59
Refer P 270
88 A to Z of Mathematics
7 × [26 +(7 – 1)(–2)] 2 7 = × (26 – 12) 2 7 × 14 = = 49 2 =
Here, a = 13, n = 7, and d = –2.
Therefore, Ritu needs 49 blocks for the construction of a required triangle. n The sum of an AP can be easily found by using the formula, Sn = [2a + (n − 1)d] where a = first 2 term, n = number of terms, and d = common difference.
n We can also find the sum of an AP whose first and last terms are given, Sn = (a + l ) . 2 Here , l = last term, a = first term and n = number of terms. Example 2 The first term of an AP is 5, the last term is 45 and the sum of AP is 400. Find the number of terms and the common difference. Solution Use the sum formula for AP, here sum of AP, a and l are given. Procedure Let ‘d’ be the common difference and ‘n’ be the number of terms in an AP. First term, a = 5 Last term, l = 45 Sum of the AP, Sn = 400 ∵ Sn =
n (a + l) 2
n (5 + 45) 2 ⇒ 800 = 50n 800 ⇒n= = 16 50 Hence, there are 16 terms in the AP. ∵ l = a + (n – 1) d ⇒ 45 = 5 + (16 – 1) d ⇒ 45 = 5 + 15d ⇒ 15d = 40 8 ⇒d= 3
⇒ 400 =
Explanation Assumption Given
Formula of sum of an AP.
Putting the values of a, l, and Sn in the general formula of a sum of an AP to find n.
Putting the value of l, a, n in the general term formula of an AP to find d.
∵d=
40 5 × 8 8 = = 15 5 × 3 3
Hence, total number of terms in the given AP is 16 and the common difference is
8 . 3
Mathematical expressions that expand 89
Example 3 1 2 Find the number of terms in the AP: 20 + 19 + 18 + .............. whose sum is 300. Explain why do we 3 3 get two different number of terms as answer. Solution Procedure First term (a) = 20 Sum of n terms = 300 Common difference (d) = a2 – a1 1 = 19 – 20 3 58 = – 20 3 58 – 60 −2 –2 = = 3 33
n {2a + (n – 1) d} = Sn 2 ⇒
n −2 {2 × 20 + (n – 1)( )} = 300 3 2
⇒ n {2 × 20 + (n – 1)( ⇒ 40n –
−2 )} = 300 × 2 3
2 2 2 n + n = 600 3 3
120n – 2n2 + 2n = 600 3 ⇒ 120n – 2n2 + 2n = 1800 ⇒ 60n – n2 + n = 900 ⇒ n2 – 61n + 900 = 0 ⇒ n2 – 36n – 25n + 900 = 0
⇒ ⇒
Explanation Given Given 1 3 (19 × 3) + 1 58 = = 3 3
∵ a2 = 19
And, a = a1 = 20
General Sum formula of an AP.
Putting a = 20, d = formula of an AP.
−2 and sum = 300 in the 3
LCM of 1 and 3 is 3. Multiply 40n with 3 and then move denominator 3 of LHS to RHS. Divide LHS and RHS with 2. Solving by bringing all the terms on the LHS. Solving this equation using factoring of algebraic expressions. We’ll learn this in the end of this chapter.
⇒ n (n – 36) – 25 (n – 36) = 0 ⇒ (n – 25) (n – 36) = 0 Either of the factor is zero. ⇒ n = 25 or n = 36 So, sum of 25 terms = sum of 36 terms = 300. As sum of the series is 300. (Given) We know that the sum of 25 terms and 36 terms in the AP is 300. This is possible only when there are −2 some terms in AP which are negative. As d = , so there are also some negative terms in the AP. 3
90 A to Z of Mathematics Example 4 The sum of n, 2n, 3n terms of an AP is S1, S2, S3 respectively. Prove that S3 = 3(S2 – S1). Solution Procedure Explanation Let ‘a’ be the first term and ‘d’ be the common difference Assumption of the given AP. General formula for sum of n terms of an AP. n Sn = [2a + (n − 1)d] 2 S1 = Sum of n terms
n ⇒ S1 = [2a + (n – 1)d] 2
....(i)
S2 = Sum of 2n terms 2n ⇒ S2 = [2a + (2n – 1)d] 2
....(ii)
3n [2a + (3n – 1)d] 2
....(iii)
S2 – S1 =
n {2[2a + (2n – 1)d] – [2a + (n – 1)d]} 2 n = {4a + 4nd – 2d – 2a – nd + d} 2 n = {2a + 3nd – d} 2 n = [2a + (3n – 1)d] ....(iv) 2 3n ∴ 3(S2 – S1) = [2a + (3n – 1)d] 2
= S3
Using the sum formula of an AP.
S3 is the sum of 3n terms of an AP. (Given)
2n n [2a + (2n – 1)d] – [2a + (n – 1)d] 2 2
⇒ S2 – S1 =
Using the sum formula of an AP.
S2 is the sum of 2n terms of an AP. (Given)
and, S3 = Sum of 3n terms ⇒ S3 =
S1 is the sum of n terms of an AP. (Given)
Using the sum formula of an AP. Subtracting eq. (i) from eq. (ii). Solving further and arranging the terms in such a way that this equation can be composed to eq (iii).
From question and eq. (iv). Using eq. (iii).
Hence, S3 = 3(S2 – S1). Example 5 Find the number of terms that are needed to get Sn = 0 in the AP of 96, 93, 90, . . . . . . . . Solution From the given AP, we can find a and d. With Sn given, we can find n using the general formula for sum of AP.
Mathematical expressions that expand 91
Procedure
Explanation
a = 96, and Sn = 0
Given
d = 93 – 96 = –3
Calculating common difference. General formula for a sum of n terms of an AP.
n Sn = [2a + (n – 1)d] 2 n ⇒ [ 2 × 96 + (n – 1)(–3)] = 0 2
Putting the values of a, Sn, and d in the sum formula of an AP.
⇒ 192 – 3n + 3 = 0 ⇒ 195 – 3n = 0 ⇒ n = 65
Hence, 65 terms are needed to get the sum zero. Example 6 Find the sum of first 100 terms of the series 1 – 2 + 2 – 3 + 3 – 4 + 4 – 5 + ……… Solution We can visualise that the given AP series is an aggregate of two different AP series. We can find the sum of the two series separately and then add to get the sum of first 100 terms of the given AP. Procedure 1 – 2 + 2 – 3 + 3 – 4 + 4 – 5 + ……… . This series is composed of the series 1, 2, 3, ….and –2, –3, –4, –5 ……... Sum of 100 terms of given series = (1 + 2 + 3 + 4 + ……... to 50 terms) + [(–2) + (–3) + (–4) + ....to 50 terms] Sum of the first series = (1 + 2 + 3 + 4 + ... to 50 terms) 50
= [ 2(1) + (49)(1)] 2 = 25 (2 + 49) = 25 × 51 = 1275 Sum of the second series = [(–2) + (–3) + (–4) + ....to 50 terms] 50
....(i)
Explanation Given series Separating the two series. To get the sum of 100 terms of the given series, we add sum of 50 terms of the two series it is composed of. n Sn = [2a +(n − 1)d ] 2 And, n = 50, a = 1, d = 1.
n [2a +(n − 1)d ] 2 And, n = 50, a = –2, d = –1.
Sn =
= [ 2(–2) + (49)(–1)] 2 = 25 (–4 – 49) = 25 × (–53) = –1325 ....(ii) Sum of the given series = eq. (i) + eq. (ii) Hence, 1 – 2 + 2 – 3 + 3 – 3 + 4 – 5 + ... to 100 terms = 1275 + (–1325) = –50
92 A to Z of Mathematics
Applying the concept of arithmetic progression60 Example 1 Raman started working in a company at an annual salary of ₹50000. He will receive an annual increment of ₹2000 per year. After how many years will his salary be ₹80000?
Solution Raman received ₹50000 in the first year, ₹52000 in the second year, ₹54000 in the third year and so on, so the sequence will be 50000, 52000, 54000, 56000, … 80000. Here, the common difference is the annual increment. Therefore, the salaries form an AP. We need to find n, the number of years when Raman’s salary will be ₹80000. Procedure First term (a) = 50000, d = 2000, an = 80000
Explanation
From the series
an = a + (n – 1) d
General term formula of an AP.
Putting the values of a, d, n in the general term ⇒ 80000 = 50000 + (n – 1) 2000 formula of an AP. ⇒ 80000 – 50000 = (n – 1) 2000 ⇒ 30000 = (n – 1) 2000 30000 ⇒ = n −1 ⇒ 2000 ⇒ n – 1 = 15 ⇒ n = 15 + 1 = 16 Hence, after 16 years Raman’s annual salary will be ₹80000.
Example 2 On work getting delayed, a contracting company has to pay a penalty of ₹100 for the first day, ₹150 for the second, ₹200 for the third, and so on. Find out how much money the contractor has to pay if the work gets delayed by 30 days.
Solution If the work of the contractor gets delayed, then he will pay ₹100 on first day, ₹150 on second day, ₹200 on the third day, and so on. So, the sequence formed will be 100, 150, 200, 250. Here, the common difference between any two consecutive penalties is ₹50. Therefore, the penalties form an AP. We need to find the sum of 30 terms of the sequence. Procedure Explanation Here, a = 100 and n = 30. Given d = 150 – 100 Calculating common difference ∴ The amount of money that contractor has to pay if The sum Sn of n terms of an AP with the first term ‘a’ and common difference ‘d’ is work gets delayed by 30 days n 30 Sn = {2a + (n − 1) d} [2 × 100 + (30 – 1)50] = 2 2 = 15[200 + 1450] Simplifying = 15 × 1650 = 24750 Hence, the contractor has to pay ₹24750 as a penalty for the delay of work for 30 days.
60
New learning outcome
Mathematical expressions that expand 93
Example 3 A clock strikes once when its hour hand is at 1, twice when it is at 2, and so on. How many times does the clock strike in 24 hours? Solution A clock face has 1–12 numbers, hence in 24 hours, the hour hand goes through all the number twice. Hence, to find the number of times the clock strikes in 24 hours, we find the sum of times it will strikes in 12 hours and double it. Since the clock strikes once when its hour hand is at 1, twice when it is at 2 and so on, so the sequence of strikes from 1 hour to 12 hours is 1, 2, 3, 4, 5, 6,……. Here, the common difference between any two consecutive strikes is 1. Therefore, the strikes of clock form an AP. Procedure
Explanation
a = 1, n = 12 d=2–1 =1 Number of times the clock strikes in 12 hours =
From the series. Calculating common difference Applying the sum formula of n terms of an AP with the first term ‘a’ and common difference ‘d’.
n [2a + (n – 1)d] 2
Sn =
n {2a + (n – 1) d} 2
Putting a = 1, d = 2 – 1 = 1, n = 12 in the sum 12 [2(1) + (12 –1)(1)] formula of an AP. 2 = 6 (2 + 11) = 6 × 13 = 78 Simplifying Hence, the clock strikes 2 × 78 = 156 times in 24 hours. =
Example 4 A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find (a) the production in the first year. (b) the production in the tenth year. (c) the total production in the first 7 years. Solution Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in first, second, third, ..., years will form an AP sequence. Procedure
Explanation
Let us denote the number of TV sets manufactured in Assumption the nth year be an. an = a + (n – 1)d a3 = 600 ⇒ a3 = a + (3 – 1) d ⇒ 600 = a + 2d
General term formula of an AP. Where a = a1 = first term and d is the common difference. Given ....(i)
94 A to Z of Mathematics
a7 = 700 ⇒ a7 = a + (7 – 1) d ⇒ 700 = a + 6d ....(ii) (a + 6d) – (a + 2d) = 700 – 600 ⇒ 4d = 100 ⇒ d = 25 a + 2(25) = 600 ⇒ a = 600 – 50 = 550 (a) Production of TV sets in the first year is 550. (b) Production of TV sets in the tenth year, a10 = a + (10 – 1)d ⇒ a10= 550 + (9 × 25) = 550 + 225 = 775 Thus, 775 sets are produced in the tenth year. n (c) Sn = [2a +(n − 1)d)] 2 Total production of TV sets in first 7 years, 7 77 +=150 =−4375 S7 = = [1100 2 × 550 + (][71100 1)+×150 25 ] = 4375 2 22 7 = [1100 + 150 ] = 4375 2
Given
Subtracting eq. (i) from eq. (ii). This eliminates 'a', and only one variable 'd' is left. Putting d = 25 in eq. (i). a = 550 an = a + (n – 1) d
a = 550, n = 10, d = 25
Using the sum formula of an AP. Here, n = 7, a = 550 and d = 25.
Thus, the total production of TV sets in the first 7 years is 4375.
11. Geometric Progression Finding the term in geometric progression61 Example 1 1 1 Find the nth term of a GP if the first two terms are and . 2 4 Solution Procedure First term, (a) =
1 2
1 1 2 1 Common ratio, (r) = 4 = × = 1 4 1 2 2 th n-1 n term = ar
61
Refer P 271
Explanation Given Common ratio =
Second term First term
General term formula of a GP.
Mathematical expressions that expand 95
=
1 2
1 2
Putting the value of ‘a’ and ‘r’ in the formula.
n −1
n
1 1 1 = × × 2 2 2 =
−1
1 2n
am + n = am × an 1
−1
1−1
1 1 1 = 2 2 2
0
1 = =1 2
Hence, the nth term of the GP is 1 . 2n Geometric progression (GP) is a type of sequence where each succeeding term is found by multiplying each preceding term by a fixed number, which is called a common ratio. Example 2 1 1 1 Find the sixth term of the GP sequence: , , … 7 14 28 Solution To find any term in GP, find the common ratio between any two consecutive terms and the first term. Then, put the values of common ratio and first term in the formula of nth term. Procedure First term (a) =
Explanation Given
1 7
1
a 7 1 Common ratio (r) = 2 = 14 = = 1 a1 14 2
7
Tn = ar
General term formula of a GP.
n–1
1 1 ⇒ T6 = × 7 2
1 1 ⇒ T6 = × 7 2
6 -1
a=
5
1 1 ⇒ T6 = × 7 32
1 7×32 1 ⇒ T6 = 224
⇒ T6 =
Hence, the sixth term in the given GP sequence is
1 . 224
1 1 ,r= ,n=6 7 2
96 A to Z of Mathematics Example 3 Fifth term and third term of a GP sequence is 256 and 16 respectively. Find its eighth term. Solution Write both the given terms in the form of first term ‘a’ and common ratio ‘r’ Then, solve both the equations for the value of ‘a’ and ‘r’ and put them in the formula of nth term. Here, n = 8. Procedure n term = ar a5 = ar5–1 = ar4 = 256 a3 = ar3–1 = ar2 = 16 th
n–1
ar 4 256 = ar 2 16
Explanation General term formula of a GP. ....(i) Fifth term = 256 (Given) ....(ii) Third term = 16 (Given) Dividing eq. (i) by eq. (ii).
⇒ r2 = 16 ⇒r=4
Substituting the value of r in eq. (ii).
∴ a(4)2 = 16 ⇒a=1
an = arn–1
General term formula of a GP.
∴ a8 = ar8–1 ⇒ a8 = (1) (4)7 = 16384 Hence, the eight term in the GP sequence is 16384.
Putting the value of ‘a’ and ‘r’ in 8th term formula.
Finding the geometric progression62 Example 1 1 Write 4 terms of a GP sequence if a = 132 and common ratio, r = . 6 Solution Procedure an = ar
n–1
a, ar, ar2, ar3 a1 = a = 132
1 = 22 6 1 1 11 a3 = ar2 = 132 × × = 6 6 3 11 1 1 1 a4 = ar3 = 132 × × × = 6 6 6 18
a2 = ar = 132 ×
Explanation General term formula of a GP. Assuming 4 terms in GP sequence. 1 a1 = a = 132 and r = (Given) 6 Putting the values of ‘a’ and ‘r’ in the assuming four terms.
Therefore, the required terms of the GP sequence are 132, 22, 11 , and 11 . 3 18
62
Refer P 271
Mathematical expressions that expand 97
Example 2 Write the GP if first term is 2 and ninth term is 512. Solution Procedure First term, (a) = 2 an = arn–1
Explanation Given General term formula of a GP.
a9 = ar9-1 Ninth term = 512 (given) 8 ⇒ 512 = (2) × r 512 8 ⇒ =r 2 ⇒ r8 = 256 ⇒ r8 = 28 ⇒r=2 Now, find the second, third, fourth, fifth, sixth, seventh, and eighth terms of the GP. a1 = 2 First term = a (Given) a2 = 2 × 2 = 4 Second term = ar 2 a3 = 2 × (2) = 8 Third term = ar2 a4 = 2 × (2)3 = 16 Fourth term = ar3 a5 = 2 × (2)4 = 32 Fifth term = ar4 a6 = 2 × (2)5 = 64 Sixth term = ar5 a7 = 2 × (2)6 = 128 Seventh term = ar6 a8 = 2 × (2)7 = 256 Eighth term = ar7 a9 = 512 Ninth term = ar8 (Given) Hence, the GP is 2, 4, 8, 16, 32, 64, 128, 256, 512.
Finding the sum of geometric progression63 Example 1 Find the sum of the first 6 terms of the GP sequence: 4, 12, 36,…. Solution Find the value of first term ‘a’ and common ratio ‘r’ from the given GP. Then, put these values in the sum formula of GP. Procedure First term, a = 4
12 Common ratio, r = = 3 4 Number of terms, n = 6
Sn = 63
a(r n − 1) (r − 1)
Refer P 271
Explanation Given
General sum formula of a GP, when the value of ‘r’ is greater than 1.
98 A to Z of Mathematics
Putting the values of a, r, and n in the sum 4(36 − 1) formula of a GP. (3 − 1) 4(729 − 1) = 2 = 2 × 728 = 1456 Hence, the sum of the first six terms in GP sequence is 1456.
∴ S6 = Therefore,
Example 2 1 1 1 1 Find the sum of first 8 terms of the geometric series: 1 + + + + + ....... 4 8 16 2 Solution Procedure
Explanation
First term of the GP = a = 1 1 1 Common ratio = r = ÷ 1 = 2 2
Given
a(1 − r n ) Sn = 1−r
General sum formula of a GP, when the value of ‘r’ is smaller than 1.
8 1 1 1 – 2 ∴ S8 = 1 1– 2 1 1 1 – 256 ⇒ S8 = 1 2 256 – 1 1 ⇒ S8 = 256 1 2 255 ⇒ S8 = 256 1 2
Putting the values of a and r in the sum formula.
255 × 2 256 255 ⇒ S8 = 128
Here, n = 8.
⇒ S8 =
Therefore, the sum of 8 terms of the given GP is
255 . 128
Mathematical expressions that expand 99
Finding the sum to infinity of the geometric progression64 Example 1 −5 5 −5 5 Find the sum of terms in the infinite GP: , , , , ... 4 16 64 256 Solution Procedure First term, a1 of the given series = a = −
Explanation 5 4
5 1 5 4 Common ratio, (r) = 16 = × − = − 4 5 16 5 − 4
S∞ =
a 1− r
S∞ =
−
Given
r=
a2 a1
Formula for the sum of terms in the infinite GP, |r| < 1.
5 4
Putting the values of ‘a’ and ‘r’ in the sum formula of a GP.
1 1 − − 4 5 − 4 = 4 +1 4 5 − = 4 5 5 −4 == − 14 5 Hence, the sum of terms in the given infinite GP is –1 . 4
Applying the concept of geometric progression65 Example 1 In a particular year, a factory sells 500 kg of cement in the market. The quantity of cement that the factory sells increases by 20% every year. What is the total quantity of cement that the factory will sell in 7 years? Round off the answer to the nearest whole number. Solution Procedure Quantity of cement sold in first year (a) = 500 kg Increase in sale every year = 20%
64 65
New learning outcome New learning outcome
Explanation Given Given
100 A to Z of Mathematics
Quantity of cement sold in second year 120 = 600 kg 100 Quantity of cement sold in third year 120 = a3 = 600 × = 720 kg 100 The sequence formed will be 500, 600, 720, … It is a GP. = a2 = 500 ×
The number of years (n) = 7 and common ratio (r) = The total quantity of cement sold in 7 years 7
=
500 (1.2 – 1) 1.2 – 1
Required quantity = Quantity ×
(100 + Percentage ) 100
Each consecutive term increases by a ratio of
120 . 100
(Given)
120 = 1.2 100 Here, we calculate the total quantity of cement with the help of the sum formula of geometric progression, where r > 1 a(r n – 1) Sn = (r – 1)
500 (3.583 − 1) = 1.2 − 1
(1.2)7 = 3.583
500 × 2.583 0.2 = 6457.5 ≈ 6458 kg
3.583 – 1 = 2.583 and 1.2 –1 = 0.2
=
6457.5 is rounded off to the nearest whole number, i.e., 6458.
Hence, 6458 kg of cement is sold in 7 years. Example 2 A bob of a pendulum swing through an arc of 60 cm on its first swing. Each successive swing is 80% of the length of the previous swing. What is the total distance the bob travels before it comes to rest? Solution It is an example of an infinite GP. Use the formula of sum of an infinite GP to find the distance travelled by the bob before coming to rest. (Since we cannot count the number of swings of the pendulum, we consider it to be an infinite swings). Procedure The length of the arc, (a) = 60 cm 80 Common ratio, (r) = 100
Explanation Given Each successive swing is 80% of the length of the previous swing.
Mathematical expressions that expand 101
The total distance travelled by the bob before coming to rest 60 = 80 1− 100 =
a 1− r 80 Here, a = 60 and r = . 100 S∞ =
60 × 100 (100 − 80 )
60 × 100 20 = 300 cm =
Hence, the total distance travelled by the bob before coming to rest is 300 cm.
12. Harmonic Progression Finding the term in harmonic progression66 Example 1 Find the eighth term of the harmonic progression. 4 2 4 , , ,.... 19 9 17 Solution We know that HP is a reciprocal of AP terms. Thus, write the reciprocal of each term of the HP to form an AP. Now find the nth term of the AP using the formula, nth term = a + (n – 1)d. And, then the reciprocal of the nth term of the AP is the required answer, i.e., nth term of HP. Procedure HP is 4 , 2 , 4 ,.... 19 9 17 AP is 19 , 9 , 17 ,.... 4 2 4 In the AP, 19 First term, (a) = 4
Given Reciprocal of HP is AP. Given ....(i)
Common difference, (d) =
=
66
Refer P 276
Explanation
=
9 19 – 4 2
Common difference (d) = Term2 – Term1
19 9 , Term2 = 4 2 LCM of 2 and 4 is 4. Here, Term1 =
18 − 19 –1 4
4 ....(ii)
102 A to Z of Mathematics
an = a + (n – 1) d
General term formula of an AP.
⇒ a8= a + (8 – 1)d
Putting the values of ‘a’, ‘n’ and ‘d’ in the general term formula. 19 –1 From eq. (i) and eq. (ii), a = ,d= . 4 4 And, n = 8
19 + (8 – 1) 4 19 7 − ⇒ a8 = 4 4 12 ⇒ a8 = 4 ⇒ a8 = 3 ⇒ a8 =
−1 4
Hence, eighth term of HP =
1 3
HP is the reciprocal of AP.
Harmonic progression (HP) is a sequence of real numbers formed by taking the reciprocals of the terms of an arithmetic progression (AP). Example 2 If the sum of reciprocals of the first 13 terms of an HP series is 260, find the seventh term of the HP sequence. Solution Reciprocals of the first 13 terms of an HP will make an AP. And, we will use this fact to find the nth term of an HP by finding out the nth term of an AP. Procedure
Explanation
n 2a + ( n − 1) d 2 S13 = 260
General sum formula of an AP.
13 [2a + (13 – 1) d] = 260 2 2 ⇒ (2a + 12d) = 260 × 13 ⇒ 2 (a + 6d) = 40 ⇒ a + 6d = 20
Putting the values of n = 13 in the sum
Sn =
Sum of 13 terms of an AP = 260 (Given)
⇒
⇒ a7 = 20
Seventh term of an HP =
formula of AP.
....(i) By using general term formula of an AP, a7 = a + (7 – 1)d = a – 6d.
1 1 = a + 6d 20
So, the seventh term of the HP is
1 . 20
Term of an HP is reciprocal of the term of
an AP. And, form eq. (i).
Mathematical expressions that expand 103
13. Algebraic Expressions Constants, variables/literals and expressions67 Example 1 Separate constants and variables from the following terms: 2ax 8 −qm 3 −1 0, , y, , 74, 3x, − 4y, , pq, cd 3 19 4 7 2 Solution Terms 0 2ax 3 8 y 19 − qm 4 74 3x −4y 3 7 –1 pq 2 cd
Constants 0
Variables/ Literals No variables
2 and 3
a and x
8 and 19
y
−1 and 4
q and m
74
No variables
3 −4
x y
3 and 7
No variables
−1 and 2
p and q
No constants
c and d
Example 2 Write each of the following situations using literals, numbers, and basic operations. (a) The sum of x and y. (b) 3 more than a number x. (c) 100 less than literal p (d) Decrease m by n (e) 5 times a is subtracted from b. (f) 10 more than thrice a number y. (g) Quotient of x by 4 is added to y. (h) Quotient of x by y added to the product of x and y. Solution (a) x + y (b) x + 3 (c) p – 100 (d) m – n (e) b – 5a (f) 3y + 10 x (g) Quotient of x by 4 = 4 x Therefore, quotient of x by 4 added to y = + y. 4 x (h) Quotient of x by y = y And, the product of x and y = xy x Therefore, the sum of the quotient of x by y and product of x and y = + xy. y 67
Refer P 280
104 A to Z of Mathematics
Creating algebraic expressions68 Example 1 Identify the terms, coefficients, numerical coefficients, constant term, factors for 7mnp2 − 8mn + 6. Solution Procedure 7mnp − 8mn + 6 7mnp2 , −8mn, and 6 are three terms in the given expression. 2
Explanation Given Expression Terms in the given expression.
7 is the coefficient of mnp2; m is the coefficient of Coefficients in the term 7mnp2. 7np2; n is the coefficient of 7mp2; p2 is the coefficient of 7mn; 7m is the coefficient of np2; 7n is the coefficient of mp2; 7p2 is the coefficient of mn; mn is the coefficient of 7p2; np2 is the coefficient of 7m; mp2 is the coefficient of 7n; 7mn is the coefficient of p2; 7np2 is the coefficient of m; 7mp2 is the coefficient of n. –8 is the coefficient of mn, m is the coefficient of –8n, Coefficients in the term (−8mn). n is the coefficient of –8m, –8m is the coefficient of n; – 8n is the coefficient of m; mn is the coefficient of –8. In term 7mnp2, 7 is the numerical coefficient. Numerical coefficients in the term 7mnp2. In term (–8mn), (–8) is the numerical coefficient. 6 is the constant term. 7, m, n, p, and p are the factors of term 7mnp2. –8, m, and n are the factors of term (–8mn).
Numerical coefficients in the term (–8mn). Constant term in the expression 7mnp2 – 8mn + 6. Factors of the term 7mnp2. Factors of the term (–8mn).
Example 2 An electrician charges ₹45 per hour and spends ₹20 a day on gasoline. Write an algebraic expression to represent his earnings for one day. Solution
Procedure Explanation Given Charges per hour = ₹45 Expenditure on gasoline = ₹20/day (Fixed Charge) Given If the electrician works for x hr/day, his earning Assuming electrician works for x hours. = ₹45x ∴ Net earnings for one day Required expression = ₹45x − ₹20 = ₹(45x – 20) Hence, the expression for the net earning of the electrician of a day is ₹(45x – 20). 68
Refer P 280
Mathematical expressions that expand 105
Example 3 Mr. Roy had ₹y in his saving account before he was paid his weekly salary. His current saving balance is ₹4290. If Mr. Roy deposits all his earnings into the saving account, give an algebraic expression to represent his weekly earning. Solution
Procedure Total balance in the account before salary = ₹y Current balance after getting salary = ₹4290 x = ₹(4290 – y)
Explanation
Given Given Assuming x is the amount Mr. Roy earns each week. Hence, the required expression for Mr. Roy weekly earning is ‘₹(4290 – y)’. Example 4 John’s age is four less than twice Mary’s age. If Mary is now 18, how old will John be seven years from now? Solution Procedure
Explanation
Let Mary’s age be ‘x’ years and John’s age be ‘y’ years. Assumption y = 2x – 4 John’s present age, y = 2(18) – 4 = 36 – 4 = 32 years
....(i) ∵ John’s age is four less than twice Mary’s age. (Given) Putting the value of x in the eq. (i). Mary’s age, x = 18 years (Given)
Thus, John’s age after 7 years = (32 + 7) years = 39 years.
Finding values of algebraic expressions69 Example 1 If x = 1, y = −2, and z = 3, find the values of the following algebraic expressions. (a) x3 + y3 + z3 − 3xyz (b) 3xy4 – 15x2y + 4z Solution Procedure (a) x + y + z − 3xyz = (1)3 + (−2)3 + (3)3 – [3 × 1 × (−2) × 3] 3
3
3
= 1 + (−8) + 27 – [3 × (−6)] = 1 − 8 + 27 + 18 = 46 – 8 = 38 Hence, the value of x3 + y3 + z3 – 3xyz is 38. 69
New learning outcome
Explanation Given Substituting the given values of x, y, and z in the expression. As –2 × –2 × –2 = –8, 3 × 3 × 3 = 27 and 1 × −2 × 3 = –6 As, (+) × (–) = (–) and (–) × (–) = (+) By BODMAS rule, solving the brackets first.
106 A to Z of Mathematics
(b) 3xy4 – 15x2y + 4z = [3 × 1 × (–2)4] – [15 × (1)2 × (–2)] + (4 × 3) = [3 × 16] – [15 × –2] + (4 × 3) = 48 + 30 + 12 = 90 Hence, the value of 3xy4 – 15x2y + 4z is 90.
Given Substituting the given values of x, y, and z in the expression, x = 1, y = –2, and z = 3 (Given) –2 × –2 × –2 × –2 = 16 By the BODMAS rule, solving the brackets first.
BODMAS stands for ‘Brackets, Orders, Division, Multiplication, Addition, and Subtraction. So, BODMAS is an acronym that is used to remember the order of operations to be followed while solving expressions in mathematics. Example 2 Find of the sum of 13xy – 2xz + 5xy2 – 3x and 6xz – 3xy for x = 1, y = 2, and z = 3? Solution Procedure
Explanation
13xy – 2xz + 5xy2 – 3x and 6xz – 3xy
Given
(13xy – 2xz + 5xy2 – 3x) + (6xz – 3xy) = 13xy – 2xz + 5xy2 – 3x + 6xz – 3xy = 13xy – 3xy – 2xz + 6xz + 5xy2 – 3x = 10xy + 4xz + 5xy2 – 3x
Simplifying the expression by adding all the ‘like’ terms together.
= [10 (1 × 2)] + [4 (1 × 3)] + [5 (1) × (2)2] – 3(1)
Substituting the given values of x, y, and z in the expression, x = 1; y = 2; z = 3.
= [10 × 2] + [4 × 3] + [5 × 4] – 3
By BODMAS rule, solving the inner brackets first.
= 20 + 12 + 20 – 3 = 52 – 3 = 49
By BODMAS rule, solving the square brackets.
Hence, the sum of the two algebraic expressions is 49.
Addition and subtraction of algebraic expressions70 Example 1 Find the sum of 2x2 + x + 6 and 4x + x3 + 8. Solution Procedure 2x2 + x + 6 and x3 + 4x + 8
70
New learning outcome
Explanation Arranging the terms of the given expressions in descending powers of x.
Mathematical expressions that expand 107
2x 2 + x + 6 + x3 + 4x + 8 3 2 x + 2x + 5x +14
Writing all the addends in such a way that like terms come in same column. And, then adding them.
Example 2 Subtract 3x2 – 6x – 4 from 5 + x – 2x2. Solution Write the subtrahend below the minuend such that like terms come in same column. Change the sign of all the terms in the subtrahend to get additive inverse of the terms. Now add the like terms columns wise and write the same below the concerned term. Procedure –2x2 + x + 5 +3x2 (+) – 6x (+) –4 (–) –5x2 + 7x + 9
Explanation Arrange the terms of the given expressions in descending powers of x and change the sign of each term in the subtrahend and add columnwise.
Additive inverse of a number ‘x’ is the number (–x), that when added to a given number gives zero. Example 3 Subtract a2 – 3ab from 2a2 – 7ab. Solution Procedure (2a2 – 7ab) – (a2 – 3ab) = 2a2 – 7ab – a2 + 3ab = 2a2 – a2 – 7ab + 3ab = a2 – 4ab
Explanation Using brackets to group the minuend and subtrahend. Changing the sign of each term in the subtrahend to get additive inverse. Arranging ‘like’ terms.
Example 4 From the sum of 4x4 –3x3+ 6x2, 4x3 + 4x – 3 and −3x4 – 5x2 + 2x, subtract 5x4 – 7x3 – 3x + 4. Solution Procedure Explanation 2 3 4 2 The sum of 4x – 3x + 6x , 4x + 4x – 3 and −3x – 5x + 2x is given by (4x4 – 3x3 + 6x2) + (4x3 + 4x – 3) + (−3x4 – 5x2 + 2x) = 4x4 − 3x4 – 3x3 + 4x3 + 6x2 – 5x2 + 4x + 2x – 3 Arranging ‘like’ terms. 4 3 2 = x + x + x + 6x – 3 Simplifying the expression. 4 3 4 3 2 Now, we subtract 5x – 7x – 3x + 4 from x + x + x + 6x – 3. 4
3
108 A to Z of Mathematics
(x4 + x3 + x2 + 6x – 3) – (5x4 – 7x3 – 3x + 4) = x4 + x3 + x2 + 6x – 3 – 5x4 + 7x3 + 3x – 4 = (x4 – 5x4) + (x3 + 7x3) + x2 + 6x + 3x – 3 – 4 = –4x4 + 8x3 + x2 + 9x – 7
Changing the sign of each term of subtrahend to get additive inverse. Arranging ‘like’ terms. Simplifying
Multiplication of algebraic expressions71 Example 1 Find the product of 6xy and –3x2y3. Solution Procedure
= –18x1+2y1+3
Explanation Using brackets to separate the terms which are to be multiplied. Multiplying numerical coefficients and variables of both the terms separately. Using power rule, am × an = am+n
= –18x3y4
Required product
(6xy) × (–3x y ) 2 3
= {6 × (–3)} × {xy × x2y3}
Example 2 Multiply (3x + 5y) and (5x – 7y). Solution Procedure
Explanation
(3x + 5y) × (5x – 7y)
Using brackets to separate the terms that are to be multiplied.
= [3x × (5x – 7y] + [5y × (5x – 7y]
By using the distributive law of multiplication over addition, (a + b) × (c + d) = a × (c + d) + b × (c + d). (Properties of numbers have been dealt in ‘Chapter N’)
= [3x × 5x – 3x × 7y] + [5y × 5x – 5y × 7y]
Multiplying the binomial (5x – 7y) by each term of (3x + 5y).
= [(3 × 5) × (x × x)] – [(3 × 7) (x × y)] + [(5 × 5) × (y × x)] – [(5 × 7) × (y × y)] = 15x2 – 21xy + 25xy – 35y2 = 15x2 + 4xy – 35y2
Multiplying numerical coefficients and variables of terms separately.
Hence, the product of (3x + 5y) and (5x – 7y) is 15x2 + 4xy – 35y2.
71
New learning outcome
Mathematical expressions that expand 109
We can see that the multiplication of polynomial expressions is quite similar to the multiplication of numbers (e.g., multiplication of integers).
Division of algebraic expressions72 Example 1 Divide 8x2y3 by –2xy. Solution Procedure 2
3
2
8x y 8 x y = × × –2xy –2 x y
Explanation Dividing numerical coefficients and literals separately.
3
Using quotient law, xm ÷ xn = xm–n
8 2–1 3–1 2 = x y = –4xy –2
= –4xy2 Hence, the quotient of dividing 8x2y3 by –2xy is –4xy2. Example 2 Divide: 6x4 – 3x3 – 2x2 + 3x –1 by 2x – 1. Solution
3x3 – x + 1 2x – 1 6x4 – 3x3 – 2x2 + 3x – 1 6x4 (+) – 3x3 (–) – 2x2 + 3x – 1 – 2x2 + x
(+)
(–)
2x – 1 2x (+) –1 (–) 0 Hence, 3x3 – x + 1 is the required quotient. When dividing a polynomial by another polynomial, we use the long division method. Example 3 Divide [(z – y)2(x – y)2(x – z)2] by [x(y2 – z2) + y(z2 – x2) + z(x2 – y2)]. Solution For finding the quotient, we factorise the divisor into one term and reduce the expression into its simplest form. First find the factors of x(y2 – z2) + y(z2 – x2) + z(x2 – y2). Procedure x(y – z ) + y(z – x ) + z(x2 – y2) = xy2 – xz2 + yz2 – yx2 + zx2 – zy2 2
72
2
2
New learning outcome
2
Explanation Given Opening the brackets
110 A to Z of Mathematics
= x2z – x2y + xy2 – xz2 + yz2 – y2z = x2(z – y) – x(z2 – y2) + yz(z – y) = x2(z – y) – x[(z – y) (z + y)] + yz(z – y)
Arranging the terms in descending power of x.
a2 – b2 = (a – b) (a + b) Taking (z – y) common from all terms.
= (z – y) x 2 – x (z + y) + yz
Simplifying
2 = (z – y) x – zx – yx + yz 2 = (z – y) x – yx – zx + yz
= (z – y) [ x (x – y) – z (x – y)] = (z – y)(x – y)(x – z) Now,
(z – y ) ( x – y ) ( x – z) ( z – y )( x – y )( x – z ) 2
2
....(i) According to the question, dividing (z – y)2 (x – y)2 (x – z)2 by eq. (i)
2
Required quotient.
= ( z – y )( x – y )( x – z )
Example 4 If x3 + 5x2 + 10k leaves remainder –2x when divided by x2 + 2, then find the value of k. Solution Procedure x+5 2 x + 2 x3 + 5x2 + 10k + 2x x3
Explanation Dividing x + 5x + 10k by x2 + 2 by long division method. 3
2
(–)
(–)
5x + 10k – 2x + 10 + 5x2 2
(–)
(–)
10k – 2x – 10 Remainder = –2x + 10k – 10 From the above calculation. According to question, Remainder = –2x (Given) –2x + 10k – 10 = –2x ⇒ 10k – 10 = 0 10 ⇒k= =1 10 Hence, the value of k is 1, and the expression is x3 + 5x2 + 10. Example 5 When (x3 – 2x2 + px – q) is divided by (x2 – 2x – 3), the remainder is (x – 6). Find the value of p and q.
Mathematical expressions that expand 111
Solution Procedure x 2 x – 2x – 3 x3 – 2x2 + px – q x3 – 2x2 – 3x (–) (+)
Explanation Dividing (x – 2x + px – q) by (x2 – 2x – 3) by long division method. 3
2
(+)
(p + 3)x – q Remainder = (p + 3)x – q
From the above calculation.
According to the question, (p + 3)x – q = x – 6
Remainder = x – 6
(Given)
On comparing the coefficients of p and q on LHS Coefficient of x is (p + 3) in LHS and 1 in RHS. and RHS, we get Constant term is (–q) in LHS and (–6) in RHS. p+3=1 and –q = –6 ⇒ p = 1 – 3 = –2 and q=6 Hence, the value p = –2 and q = 6, and the dividend is x3 – 2x2 – 2x – 6.
14. Algebraic Identities Identity73 : (a + b) (a – b) = a2 – b2 Example 1 x 2 − 16 Simplify: x+4 Solution Simplify the expression of the numerator using the identity a2 − b2 = (a – b) (a + b) and then reduce the expression. Procedure x − 16 x+4 x2 − ( 4 ) = x+4 =
Explanation Given
2
16 = ( 4 )
2
( x – 4 )( x + 4 )
2
a2 – b2 = (a – b) (a + b) Here, a = x and b = 4
x+4 =x–4
Reducing the expression.
Example 2 Find the value of (68)2 – (32)2. Solution Procedure (68) – (32) 2
73
Refer P 283
2
Explanation Given
112 A to Z of Mathematics
= (68 + 32) (68 – 32) = 100 × 36 = 3600
∵ a2 – b2 = (a + b) (a – b) Here, a = 68 and b = 32.
Example 3 Use identity to evaluate: 102 × 98 Solution Procedure 102 × 98 = (100 + 2) (100 – 2) = (100)2 – (2)2 = 10000 – 4 = 9996
Explanation Given Write 102 and 98 in such a way that it matches with the identity. (a + b) (a – b) = a2 – b2
Identity74 : (a + b)2 = a2 + 2ab + b2 Example 1 Expand (x + 2y)2 Solution Procedure
Explanation
(x + 2y)
Given
= (x)2 + 2 (x)(2y) + (2y)2
(a + b)2 = a2 + 2ab + b2 Here, a = x, b = 2y. Expanded form.
2
= x2 + 4xy + 4y2 Example 2 Find the value of (36)2 + 1008 + (14)2. Solution Procedure (36)2 + 1008 + (14)2 = (36)2 + 2 × 36 × 14 + (14)2
= (36 + 14)2 = 50 × 50 = 2500 Example 3 Evaluate: (100.25)2 74
Refer P 283
Explanation Given Finding the factors of 1008 and writing them in such a way that it consists of 2, 36, and 14. (Factors have been dealt in detail in ‘Chapter M’). ∵ a2 + 2ab + b2 = (a + b)2 Here, a = 36 and b = 14.
Mathematical expressions that expand 113
Solution Procedure (100.25)2 = (100 + 0.25)2 = (100)2 + (2 × 100 × 0.25) + (0.25)2
25 ) + 0.0625 100 = 10000 + 50 + 0.0625 = 10050.0625 = 10000 + (2 × 100 ×
Explanation Given Write (100.25)2 in the form of (a + b)2. (a + b)2 = a2 + 2ab + b2 1002 = 10000 2 625 25 2 = = 0.0625 (0.25) = 100 10000
Example 4 If a + b = 15 and ab = 54, find the value of a2 + b2. Solution Procedure (a + b) = a + b + 2ab ⇒ (15)2 = a2 + b2 + 2(54) ⇒ 225 = a2 + b2 + 108 ⇒ a2 + b2 = 225 – 108 ⇒ a2 + b2 = 117 Hence, the value of a2 + b2 is 117. 2
2
2
Explanation Using the identity to find the value of a2 + b2. Putting the value of (a + b) and ab in the identity. Here, a + b = 15 and ab = 54.
Example 5 1 1 2 If x + = 6 , calculate x + 2 . x x Solution Procedure
x+
1 =6 x
Explanation Given
2
On squaring both sides.
1 2 ⇒ x + = (6) x 2
1 1 ⇒ x + + 2(x) = 36 x x 1 2 ⇒ x + 2 + 2 = 36 x 1 2 ⇒ x + 2 = 36 − 2 = 34 x 2
Hence, the value of x 2 +
1 is 34. x2
(a + b)2 = a2 + b2 + 2ab 1 Here, a = x and b = . x
114 A to Z of Mathematics
Identity75 : (a – b)2 = a2 – 2ab + b2 Example 1 Expand (3p – 2q)2. Solution Procedure
Explanation
(3p – 2q)2
Given
= (3p)2 – 2 (3p)(2q) + (2q)2
(a – b)2 = a2 – 2ab + b2 Here, a = 3p, b = 2q.
= 9p2 – 12pq + 4q2
Expanded form
Example 2 Find the value of (182)2 – 2 × 182 × 12 + (12)2. Solution Procedure (182) – 2 × 182 × 12 + (12)2 = (182 – 12)2 = 170 × 170 = 28900 2
Explanation Given ∵ a2 – 2ab + b2 = (a – b)2 Here, a = 182 and b = 12.
Example 3 Evaluate: (1093)2 Solution Procedure (1093) = (1100 – 7)2 = (1100)2 – 2 (1100) (7) + (7)2 = 1210000 – 15400 + 49 = 1194649 Hence, the value of (1093)2 is 1194649. 2
Explanation Given Write (1093)2 in the form of (a – b)2. (a + b)2 = a2 – 2ab + b2 Here, a = 1100 and b = 7.
Example 4 If a – b = 17 and ab = 60, find the value of a2 + b2. Solution Procedure (a – b) = a – 2ab + b2 (17)2 = a2 – 2 × 60 + b2 ⇒ 289 = a2 – 120 + b2 ⇒ a2 + b2 = 289 + 120 ⇒ a2 + b2 = 409 Hence, the value of a2 + b2 is 409. 2
75
Refer P 283
2
Explanation Using the identity to find the value of a2 + b2. Putting the value of (a – b) and ab in the identity. Here, a – b = 17 and ab = 60.
Mathematical expressions that expand 115
Identity76 : a3 + b3 = (a + b)3 – 3ab (a + b) Example 1 Expand (2x + 3)3 Solution Procedure (2x + 3)
3
= (2x)3 + (3)3 + 3 (2x) (3) (2x + 3) = 8x3 + 27 + 18x (2x + 3) = 8x3 + 27 + (18x) (2x) + (18x) (3) = 8x3 + 27 + 36x2 + 54x
Explanation Given (a + b)3 = a3+ b3 + 3ab (a + b) Here, a = 2x, b = 3. Multiplying each term in the brackets (2x + 3) with 18x. Expanded form
Example 2 Evaluate: (43)3 Solution Procedure (43) = (40 + 3)3 = (40)3 + (3)3 + 3(40)(3) (40 + 3) = 64000 + 27 + (360) × (43) = 64000 + 27 + 15480 = 79507 Hence, the value of (43)3 is 79507. 3
Explanation Given (a + b)3 = a3 + b3 + 3ab (a + b) Here, a = 40 and b = 3.
Example 3 If a3 + b3 = 91, a + b = 7, find the value of ab. Solution Procedure (a + b) = a + b + 3ab (a + b) (7)3 = 91 + 3ab (7) ⇒ 343 = 91 + 21ab ⇒ 21ab = 343 – 91 ⇒ 21ab = 252 ⇒ ab = 12 Hence, the value of ab is 12. 3
3
3
Example 4 1 1 If x − = 4 + 2 , find the value of x 3 − 3 . x x
76
Refer P 283
Explanation Using the identity to find the value of ab. Putting the value of (a + b) and (a3 + b3) in the identity. Here, a + b = 7 and a3 + b3 = 91.
116 A to Z of Mathematics Solution Procedure
x−
Explanation
1 = 4+ 2 x 3
Given
1 x – = 4 + 2 x
(
)
On cubing both sides.
3
3
(
−1 −1 −1 ⇒ x + + 3(x) x + = 4 + 2 ⇒ x x x 3 1 1 ⇒ x3 − 3 − 3 x − = 4 + 2 ⇒ x x 3
(
)
1 1 3 ⇒ ⇒ x − 3 − 3 x − = ( 4 ) + x x 3
((
)
3
( 2 ) + 3( 4 ) ( 2 ) ( 4 + 2 ) 3
))
11 ⇒ xx33 −– 33 – −3 34 4++ 22 == 64 64 + 2 2 + 48 ⇒ 48 22 ++24 24 xx 1 ⇒ ⇒ x 3 − 3 = 88 + 50 2 + 3 4 + 2 x 1 ⇒ x 3 − 3 = 88 + 50 2 + 12 + 3 2 ⇒ x 1 3 ⇒ x − 3 = 100 + 53 2 ⇒ x
(
)
(a + b)3 = a3 + b3 + 3ab (a + b) −1 Here, a = x and b = . x (a + b)3 = a3 + b3 + 3ab (a + b)
Here, a = 4 and b = Subtituting x −
1 =4+ x
2.
2 (Given)
1 Hence, the value of x 3 − 3 is 100 + 53 2 . x Example 5 If a + b = 2 and a2 + b2 = 8, find a3 + b3. Solution Procedure a + 2ab + b = (a + b) ⇒ 2ab = (a + b)2 – (a2 + b2) ⇒ 2ab = (2)2 – (8) ⇒ 2ab = 4 – 8 ⇒ 2ab = –4 ⇒ ab = –2 ∴ a3 + b3 = (a + b)3 – 3ab (a + b) 2
2
2
= (2)3 – 3(–2) (2) = 8 – 3(– 4) = 8 + 12 = 20
Explanation Using the identity to find the value of 2ab. Subtituting a + b = 2 and a2 + b2 = 8 (Given)
Use the identity to find the value of a3 + b3. Subtituting a + b = 2 and ab = –2.
Mathematical expressions that expand 117
Identity77 : a3 – b3 = (a – b)3 + 3ab (a – b) Example1 Expand (7m – 3n)3 Solution Procedure
Explanation
(7m – 3n)3
Given
= (7m)3 – (3n)3 – 3 (7m) (3n) (7m – 3n) = 343m3 – 27n3 – 63mn (7m – 3n) = 343m3 – 27n3 – (63mn)(7m) – (63mn)(–3n)
(a – b)3 = a3 – b3 – 3ab (a – b) Here, a = 7m, b = 3n. Multiplying each term in the bracket (7m – 3n) with 63mn. – (63mn) (–3n) = 189mn2 as (–) × (–) = (+)
= 343m3 – 27n3 – 441m2n + 189mn2 Example 2 Find the value of (35)3 – (5)3. Solution Procedure (35) – (5) = (35 – 5)3 + 3 × (35 × 5) (35 – 5) = (30)3 + 525 (30) = 27000 + 15750 = 42750 3
3
Explanation Given ∵ a3 – b3 = (a – b)3 + 3ab (a – b) Here, a = 35 and b = 5.
Example 3 If a – b = 7 and a3 – b3 = 973, then what is the value of ab? Solution Procedure a3 – b3 = (a – b)3 + 3ab (a – b) 973 = (7)3 + 3ab (7) ⇒ 973 = 343 + 21ab ⇒ 21ab = 973 – 343 630 ⇒ ab = 21 ⇒ ab = 30 Hence, the value of ab is 30.
77
Refer P 283
Explanation Using the identifty to find the value of ab. Putting the value of (a – b) and (a3 – b3) in the identity. Here, a – b = 7 and a3 – b3 = 973.
118 A to Z of Mathematics
Identity78 : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Example 1 Write (2p + 3q + 4r)2 in expanded form. Solution Procedure
Explanation
(2p + 3q + 4r) ⇒ (2p)2 + (3q)2 + (4r)2 + 2(2p)(3q) + 2(3q)(4r) + 2(4r)(2p) ⇒ 4p2 + 9q2 + 16r2 + 12pq + 24qr + 16pr
Given (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca Here, a = 2p, b = 3q, and c = 4r.
2
Example 2 If a + b + c = 13 and ab + bc + ca = 52, then find the value of a2 + b2 + c2. Solution Procedure (a + b + c) = a + b + c + 2(ab + bc + ca) 2
2
2
Explanation Using the identity to find the value of a2 + b2 + c2. Putting the value of (a + b + c) and (ab + bc + ca) in the identity. Here, a + b + c = 13 and ab + bc + ca = 52.
2
(13)2 = a2 + b2 + c2 + 2 × 52 ⇒ 169 = a2 + b2 +c2 + 104 ⇒ a2 + b2 + c2 = 169 – 104 ⇒ a2 + b2 + c2 = 65 Hence, the value of a2 + b2 + c2 is 65.
Example 3 Factorise the given expression: 9x2 + 49y2 + 25z2 – 42xy – 30xz + 70yz Solution Procedure 9x + 49y + 25z – 42xy – 30xz + 70yz = (–3x)2 + (7y)2 + (5z)2 + 2(–3x)(7y) + 2(–3x)(5z) + 2(7y)(5z) = (–3x + 7y + 5z)2 = (–3x + 7y + 5z)(–3x + 7y + 5z) 2
2
Explanation Given The expression resembles a2 + b2 + c2 + 2ab + 2bc + 2ca We can write a2 + b2 + c2 + 2ab + 2bc + 2ca as (a + b + c)2. Here, a = –3x, b = 7y, and c = 5z.
2
Identity79 : a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) Example 1 If p + q + r = 6 and pq + qr + rp = 11, find the value of p3 + q3 + r3 – 3pqr. Solution Procedure p+q+r=6 78 79
Refer P 283 Refer P 283
Explanation Given
Mathematical expressions that expand 119 On squaring both sides. ⇒ (p + q + r)2 = (6)2 ⇒ p2 + q2 + r2 + 2 (pq + qr + rp) = 36 (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) ⇒ p2 + q2 + r2 + 2 (11) = 36 (Given) pq + qr + rp = 11 2 2 2 ⇒ p + q + r = 36 – 22 = 14 .… (i)
p3 + q3 + r3 – 3pqr a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc 2 2 2 – ca) = (p + q + r) (p + q + r – pq – qr – rp) 2 2 2 = (p + q + r)[p + q + r – (pq + qr + rp)] = (6) (14 – 11) p + q + r = 6 and pq + qr + rp = 11 (Given) =6×3 p2 + q2 + r2 = 14 [From eq. (i)] = 18 Hence, the value of (p3 + q3 + r3 – 3pqr) is 18. Example 2 Find the value of a3 + b3 + c3 – 3abc, if a + b + c = 9 and a2 + b2 + c2 = 49. Solution Procedure (a + b + c) = a + b2 + c2 + 2ab + 2bc + 2ca ⇒ (9)2 = 49 + 2 (ab + bc + ca) ⇒ 81 = 49 + 2 (ab + bc +ca) 2
2
Explanation Finding the value of ab + bc + ca using the identity Putting the values of (a + b + c) and (a2 + b2 + c2) in the identity. Here, a + b + c = 9 and a2 + b2 + c2 = 49
⇒ ab + bc + ca = 81 – 49 = 32 = 16 ...(i) ....(i) 2 2 a3 + b3 + c3 – 3abc We know that, 2 2 2 = (a + b + c)( a + b + c – ab – bc – ca) a3 + b3 + c3 – 3abc = ( a + b + c) ( a2 + b2 + c2 – ab – bc – ca) = (9) [49 – (ab + bc + ca)] (given) a + b + c = 9 and a2 + b2 + c2 = 49 = 9 (49 – 16) From eq. (i), ab + bc + ca = 16 = 9 × 33 = 297 Hence, the value of a3 + b3 + c3 – 3abc is 297.
Factorising algebraic expressions80 Example 1 Factorise 2y + 6. Solution Procedure 2y = 2 × y ....(i) 6=2×3 ....(ii) So, we can factorise the whole expression: 2y + 6 = (2 × y) + (2 × 3)
80
New learning outcome
Explanation Both 2y and 6 have a common factor of 2. Adding eq. (i) and eq. (ii).
120 A to Z of Mathematics
⇒ 2y + 6 = 2(y + 3)
Taking the constant ‘2’ common from both of the grouped elements.
So, 2 and y + 3 are the two factors of 2y + 6. Factorising is the reverse process of expanding brackets. Example 2 Factorise the following algebraic expressions. (a) 16 (2x – 1)2 – 25z2 (b) 4a2 – 9b2 – 2a – 3b
(c) x2 + 4x – 5
(d) 11x2 + 33x – 110
(f) x2 + 13x + 30
(e) x2 − 7x + 10
Solution Procedure
Explanation
(a) 16(2x – 1)2 – 25z2 = {4(2x – 1)}2 – (5z)2
25z2 = 5z × 5z = (5z)2 and 16 = 42
= {4 (2x – 1) – 5z} {4 (2x – 1) + 5z}
Using identity: a2 – b2 = (a – b) (a + b)
= (8x – 4 – 5z) (8x – 4 + 5z)
Simplifying
= (8x – 5z – 4) (8x + 5z – 4) are the two factors of 16(2x – 1 )2 – 25z2. (b) 4a2 – 9b2 – 2a – 3b = {(2a)2 – (3b)2} – (2a + 3b) As 4a2 = 2a × 2a and 9b2 = 3b × 3b = (2a – 3b) (2a + 3b) – (2a + 3b) ∵ a2 – b2= (a – b) (a + b) = (2a + 3b) {(2a – 3b) – 1} Taking (2a + 3b) common. = (2a + 3b) (2a – 3b – 1) are the two factors of 4a2 – 9b2 – 2a – 3b. (c) x2 + 4x – 5
= x2 + 5x – x – 5 ∵ x2 + (a + b)x + ab = (x + a) (x + b) = x(x + 5) – 1(x + 5) = (x – 1)(x + 5) are the two factors of x2 + 4x – 5.
(d) 11x2 + 33x – 110 = 11(x2 + 3x – 10)
Taking 11 common from the expression.
= 11(x + 5x – 2x – 10) ∵ x2 + (a + b)x + ab = (x + a) (x + b) = 11[x(x + 5) – 2(x + 5)] = 11(x – 2) (x + 5) are the three factors of 11x2 + 33x – 110. 2
(e) x2 − 7x + 10
= x2 – 5x – 2x + 10 ∵ x2 + (a + b)x + ab = (x + a)(x + b) = x(x – 5) – 2( x – 5) = (x – 2)(x – 5) are the two factors of x2 – 7x + 10.
Mathematical expressions that expand 121
(f) x2 + 13x + 30
= x2 + 10x + 3x + 30 ∵ x2 + (a + b)x + ab = (x + a) (x + b) = x(x + 10) + 3(x + 10) = (x + 3) (x + 10) are the two factors of x2 + 13x + 30.
Example 3 343 Factorise the given expression: 27x3 – 63x2 + 49x – 27 Solution Procedure 27x 3 – 63x 2 + 49x –
Explanation
343 27
Given 2
7 7 7 = (3x)3 – (3)(3x)2 + (3)(3x) – 3 3 3 7 = 3x – 3
3
3
Resembles like a3 – 3a2b + 3ab2 – b3. ∵ a3 – 3a2b + 3ab2 – b3 = (a – b)3
15. Divisibility Test A convenient way of determininig whether a given number is divisible by a fixed divisor usually by examining its digits. Utilising divisibility81 : By 2, 5, 10 Example 1 Which of the following numbers are divisible by 2? (a) 4528 (b) 3259
(c) 15420
(d) 10520002
Solution Option (a), (c), and (d) are divisible by 2 as they end with even numbers. Example 2 Which of the following numbers are divisible by 5? (a) 2592 (b) 245
(c) 502090
(d) 5251
Solution Option (b), and (c) are divisible by 5 as they end with 5 and 0 respectively. Example 3 Which of the following numbers are divisible by 10? (a) 365 (b) 18560 (c) 256856 Solution Option (b) and (d) are divisible by 10 as both numbers end with 0.
81
New learning outcome
(d) 56852000
122 A to Z of Mathematics
(a) Divisibility by 2: Numbers ending with 0, 2, 4, 6, and 8 (even numbers) are divisible by 2. (b) Divisibility by 5: A number is divisible by 5 if it ends with 0 or 5. (c) Divisibility by 10: All numbers ending with 0 are divisible by 10.
Utilising divisibility82 : By 3, 6, 9 Example 1 Which of the following numbers are divisible by 3? (a) 24235 (b) 24867 (c) 10200120
(d) 32564
Solution Procedure (a) 24235 16 is not divisible by 3.
∴ 24235 is not divisible by 3. (b) 24867 27 is divisible by 3. ∴ 24867 is divisible by 3. (c) 10200120 6 is divisible by 3. ∴ 10200120 is divisible by 3. (d) 32564 20 is not divisible by 3. ∴ 32564 is not divisible by 3.
Explanation Sum of digits = 2 + 4 + 2 + 3 + 5 = 16
Sum of digits = 2 + 4 + 8 + 6 + 7 = 27
Sum of digits = 1 + 0 + 2 + 0 + 0 + 1 + 2 + 0 = 6
Sum of digits = 3 + 2 + 5 + 6 + 4 = 20
(a) Divisibility by 3: A number is divisible by 3 if the sum of all of its digits is divisible by 3. (b) Divisibility by 6: A number is divisible by 6 if it is divisible by 2 and 3. (c) Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Example 2 Which of the following numbers are divisible by 6? (a) 328 (b) 54235
(c) 24852
(d) 1020030
Solution For each number check the divisibility by both 2 and 3. Procedure Explanation The numbers divisible by 2 and 3 are divisible by 6. (a) 328 is divisible by 2. 328 ends with 8. 328 is not divisible by 3. Sum of digits = 3 + 2 + 8 = 13. 13 is not divisible by 3. Thus, 328 is not divisible by 6. 328 is divisible only by 2 and not both 2 and 3. (b) 54235 is not divisible by 2. 54235 ends with 5, so it is not divisible by 2. 82
New learning outcome
Mathematical expressions that expand 123
54235 is not divisible by 3. Thus, 54235 is not divisible by 6. (c) 24852 is divisible by 2. 24852 is divisible by 3. Thus, 24852 is divisible by 6. (d) 1020030 is divisible by 2. 1020030 is divisible by 3. Thus, 1020030 is divisible by 6.
Sum of digits = 5 + 4 + 2 + 3 + 5 = 19. 19 is not divisible by 3. 54235 is not divisible by both 2 and 3. 24852 ends with 2. Sum of digits = 2 + 4 + 8 + 5 + 2 = 21. 21 is divisible by 3. 24852 is divisible by both 2 and 3. 1020030 ends with 0. Sum of digits = 1 + 0 + 2 + 0 + 0 + 3 + 0 = 6. 6 is divisible by 3. 1020030 is divisible by both 2 and 3.
Example 3 Which of the following numbers are divisible by 9? (a) 1854 (b) 961
(c) 102509
(d) 99567
Solution Procedure Explanation (a) Sum of digits = 1 + 8 + 5 + 4 = 18 18 is divisible by 9. ∴ 1854 is divisible by 9. (b) Sum of digits = 9 + 6 + 1 = 16 16 is not divisible by 9. ∴ 961 is not divisible by 9. (c) Sum of digits = 1 + 0 + 2 + 5 + 0 + 9 = 17 17 is not divisible by 9. ∴ 102509 is not divisible by 9. (d) Sum of digits = 9 + 9 + 5 + 6 + 7 = 36 ∴ 99567 is divisible by 9.
36 is divisible by 9.
Utilising divisibility83 : By 4, 11 Example 1 Which of the following numbers are divisible by 4? (a) 1004 (b) 40458
(c) 255556
(d) 59900
Solution Procedure (a) 1004 is divisible by 4. (b) 40458 is not divisible by 4. (c) 255556 is divisible by 4. (d) 59900 is divisible by 4.
83
New learning outcome
Explanation Number formed by last 2 digits is 04 which is divisible by 4. The number formed by the last 2 digits is 58 which is not divisible by 4. The number formed by the last 2 digits is 56 which is divisible by 4. Any number that ends with 00 is divisible by 4.
124 A to Z of Mathematics
(a) Divisibility by 4: A number is divisible by 4 if the number formed by its last 2 digits is divisible by 4. (b) Divisibility by 11: A number is divisible by 11 if the difference of the sum of the digits at the odd places and the sum of the digits at the even places is a multiple of 11 or zero. Example 2 Is the number 7686311 divisible by 11? Solution Procedure Explanation The sum of its digits at odd places Sum of digits at odd places in 7686311. = 1 + 3 + 8 + 7 = 19 The sum of its digits at even places Sum of digits at even places in 7686311. = 1 + 6 + 6 = 13 The difference of the two sums Calculating the difference of the two sums. = 19 – 13 = 6, which is not divisible by 11. Therefore, 7686311 is not divisible by 11.
Example 3 Is the number 57981 divisible by 11? Solution
Procedure The sum of its digits at odd places = 1 + 9 + 5 = 15 The sum of its digits at even places = 8 + 7 = 15 Difference of the two sums = 15 – 15 = 0, which is divisible by 11. Therefore, 57981 is divisible by 11.
Explanation Sum of digits at odd places in 57981. Sum of digits at even places in 57981. Calculating the difference of the two sums.
M M for Matter
More about numbers, exploring factors, multiples 16. Factors and Multiples of Numbers Idea of factors1 Example 1 Define factors of a number by writing all parts of 20 using multiplication. Solution Let us see the ‘parts’ of 20 using multiplication (a) 2 × 10 = 20 (so, 2 and 10 are parts of 20 such that 2 parts of 10 each makes 20, and vice versa) (b) 4 × 5 = 20 (so, 4 and 5 are parts of 20 such that 4 parts of 5 each makes 20, and vice versa) (c) 20 × 1 = 20 (so, 1 and 20 are parts of 20 such that 20 parts of 1each make 20, and vice versa) Thus, 1, 2, 4, 5, 10, and 20 are the only possible parts of 20. There is no other way of making parts of 20. Hence,1, 2, 4, 5, 10, and 20 are the factors of 20.
Finding factors of numbers2 Example 1 Write all the factors of 144 and 68. Solution Procedure
Explanation 144 ÷ 1 = 144 ; 144 ÷ 2 = 72 ; 144 ÷ 3 = 48 ; All divisors and quotients are factors of the 144 ÷ 4 = 36 ; 144 ÷ 6 = 24 ; 144 ÷ 8 = 18 ; number. 144 ÷ 9 = 16 ; 144 ÷ 12 = 12
1 2
Refer P 288 Refer P 288
125
126 A to Z of Mathematics
6 × 24 = 144 8 × 18 = 144 9 × 16 =144 12 × 12 =144 1
2
3
4
6
8
9
12
16
18
24
36
48
72
144
4 × 36 =144 3 × 48 =144 2 × 72 =144 1 × 144 =144
We will stop here because both the factors (12) are repeated. Thus, the factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144. All divisors and quotients are factors of the 68 ÷ 1 = 68 ; 68 ÷ 2 = 34 ; 68 ÷ 4 = 17 number. 4 × 17 = 68 1
2
4
17
34
68
2 × 34 = 68 1 × 68 = 68
The factors of 68 are 1, 2, 4, 17, 34, and 68. A factor is a natural number that divides the given number without leaving any remainder. And, we can find the factors with the help of divisibility rules. Example 2 Maria was listing the factors of 147 below, but did not finish. Which factor is missing? 1, 3, 7, ?, 49, 147 Solution Procedure 147 ÷ 1 = 147 ; 147 ÷ 3 = 49 ; 147 ÷ 7 = 21
Explanation Finding factors using divisibility rules.
Factors of 147 are 1, 3, 7, 21, 49, 147. Thus, the missing factor of 147 is 21.
All the divisors and quotients are the factors. 21 is the also a factor of 147.
Example 3 Write all the factor pairs of 24. Solution Factor pairs of a number is a set of two factors such that on multiplying the pair, the product is the number itself.
More about numbers 127
Procedure 24 ÷ 1 = 24 ; 24 ÷ 2 = 12 ;
Explanation Finding factors using divisibility rules.
24 ÷ 3 = 8 ; 24 ÷ 4 = 6 Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Factor pairs of 24 are 1 × 24; 2 × 12; 3 × 8; 4 × 6.
Factor pairs of 24 that will give a product of 24.
Example 4 Which is the smallest and the greatest factor of any number? Solution The smallest factor of any number is 1 and the greatest factor of any number is the number itself. Example 5 Find two numbers whose difference is 3 and the product is 28. Solution We first list all the factors of 28, and then identify the factors pair with a difference of 3. Procedure All factors of 28 are 1, 2, 4, 7, 14, 28. Clearly, 7 – 4 = 3 and 7 × 4 = 28. Therefore, the required numbers are 7 and 4.
Explanation Finding factors using divisibility rules. 4 and 7 are factor pairs of 28.
Idea of multiples3
Example 1 Avyanna has a pack of cards numbered from 1 to 20. She picks three different number cards and all numbers are multiples of 6. Write what the numbers could be. Solution Procedure 20 cards numbered from 1 to 20.
1 , 2 , 3 , 4, 5, 6 , 7, 8, 9, 10 11, 12 , 13, 14, 15, 16, 17, 18 , 19, 20 Hence, the numbers on the cards are 6, 12, and 18.
Finding multiples of numbers4 Example 1 Write first five multiples of 5.
3 4
Refer P 294 Refer P 294
Explanation Given Writing numbers on the cards – 1 to 20 Drawing circles on the multiple of 6, 6 × 1 = 6 ; 6 × 2 = 12 ; 6 × 3 = 18
128 A to Z of Mathematics Solution Procedure Explanation To get the first five multiples of 5, we multiply 5 by 1, 2, The required multiples are: 5 × 1 = 5; 5 × 2 = 10; 5 × 3 = 15; 3, 4, and 5, respectively. 5 × 4 = 20; 5 × 5 = 25 Therefore, the first five multiples of 5 are 5, 10, 15, 20, and 25. Example 2 Identify the following numbers. (a) Fourth multiple of 6
(b) Second multiple of 3
(c) Tenth multiple of 10
(d) First multiple of 5
Solution Procedure
Explanation Finding fourth multiple of 6. Finding second multiple of 3. Finding tenth multiple of 10. Finding first multiple of 5.
(a) 4 × 6 = 24 (b) 2 × 3 = 6 (c) 10 × 10 = 100 (d) 1 × 5 = 5
Example 3 The values of A and B are given. Fill them in appropriate boxes to form a correct statement. S. No.
A
B
Statement
(a)
108
54
54 is a factor of
(b)
27
9
27 is a multiple of
9 .
(c)
70
35
35 is a factor of
70 .
(d)
1
5
5
(e)
10
100
108 .
is a multiple of
10 is a factor of
1 . 100 .
Example 4 Find the difference between the twenty-ninth multiple of 5 and eighteenth multiple of 3? Solution Procedure 29 × 5 = 145 18 × 3 = 54 The difference = 145 – 54 = 91 Hence, the required difference is 91.
Explanation Finding twenty-ninth multiple of 5. Finding eighteenth multiple of 3. Finding the difference between twenty-ninth multiple of 5 and eighteenth multiple of 3.
More about numbers 129 Example 5 Write the multiples of 4 which are greater than 24 and less than 38. Solution Procedure Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40….. Required multiples of 4 are 28, 32, and 36.
Explanation Listing multiples of 4.
Finding multiples of 4 which are greater than 24 and less than 38. Hence, the multiples of 4 which are greater than 24 and less than 38 are 28, 32, and 36. Example 6 Which is the largest 4-digit multiple of 12? Solution Largest 4-digit number is 9999. But, it is not necessary that is a multiple of 12. So, first divide 9999 by 12 to check whether it is a multiple of 12 or not. And, if the number is not divisible by 12, then subtract the remainder from 9999 to get the exact multiple of 12. Procedure Largest 4-digit number = 9999 12
Explanation We already know. Dividing 9999 by 12 to check whether it is divisible by 12 or not.
9999 833 – 96 39 –36 39 –36 3 The largest 4-digit number which is multiple ∵ 9999 is not exactly divisible by 12 as we get remainder 3. of 12 = 9999 – 3 = 9996 Hence, 9996 is the largest 4-digit multiple of 12.
17. Prime and Composite Numbers Defining prime and composite numbers5 Example 1 Identify prime and composite numbers: 5, 1, 16, 25, 19, 91, 41, 47 Solution Procedure 5 is a prime number. 1 is neither a prime nor a composite number.
5
Refer P 290
Explanation ∵ Factors of 5 are 1 and 5. ∵ Factors of 1 is only 1.
130 A to Z of Mathematics
16 is a composite number. 25 is a composite number. 19 is a prime number. 91 is a composite number. 41 is a prime number. 47 is a prime number.
∵ Factors of 16 are 1, 2, 4, 8, and 16. ∵ Factors of 25 are 1, 5, and 25. ∵ Factors of 19 are 1 and 19. ∵ Factors of 91 are 1, 7, 13, and 91. ∵ Factors of 41 are 1 and 41. ∵ Factors of 47 are 1 and 47.
Example 2 Write the factors of 108 and identify factors which are prime or composite amongst them. Solution Procedure Explanation Factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, Listing factors of 108. 54, 108. 1 is neither a prime nor a composite number. Identifying prime and composite numbers from them. Factors which are prime numbers are 2 and 3. Factors which are composite numbers are 4, 6, 9, 12, 18, 27, 36, 54, and 108. Example 3 Express 12 and 1012 as the sum of two prime numbers. Solution Procedure 12 = 5 + 7 1012 = 1009 + 3
Explanation Writing 12 as a sum of two prime numbers. Here, 5 and 7 are prime numbers. Writing 1012 as a sum of two prime numbers. Here, 1009 and 3 are prime numbers.
Example 4 Fill in the blanks. (a) The smallest composite number is 4 . (b) The smallest prime number is 2 . (c) 71 is the prime number nearest to 70. (d) All prime numbers are odd except 2 .
Twin prime and co-prime numbers6 Example 1 Write all the twin prime numbers between 1 and 20. Solution Two consecutive prime numbers with a difference of 2 are called twin prime numbers.
6
New learning outcome
More about numbers 131
Procedure
Explanation
The twin prime numbers between 1 and 20 are ∵ The difference between two consecutive prime (3, 5), (5, 7), (11, 13), (17, 19). numbers is 2. Example 2 Is (2, 3) an example of twin primes? Solution 2 and 3 are prime numbers, but there is no composite number between them and the difference between the two numbers is not equal to 2. Thus, (2, 3) is not an example of twin primes. Example 3 Check whether 28 and 57 are co-prime. Solution Two numbers having only 1 as a common factor are co-prime numbers. Procedure The factors of 28 are 1, 2, 4, 7, 14, and 28. The factors of 57 are 1, 3, 19, and 57. 28 and 57 are co-prime numbers.
Explanation Finding factors using divisibility rules. Finding factors using divisibility rules. ∵ 28 and 57 have only one common factor, i.e., 1.
Example 4 Put tick (✓) on the appropriate box. (a)
1
2
is co-prime with every number.
prime composite numbers are co-prime to each other. (b) Any two consecutive even numbers are always co-prime. (c) Any two (d) The difference between any twin prime is
1
2
.
Prime triplets7 Example 1 Check whether the given sets are prime triplets. (a) (11, 13, 17) (b) (5, 9, 11) Solution A set of three prime numbers which are prime triplets can be represented in the form of (n, n + 2, n + 6) or (n, n + 4, n + 6).
7
New learning outcome
132 A to Z of Mathematics
Procedure (a) In 11, 13, and 17, n = 11. Then, n + 2 = 11 + 2 =13 And, n + 6 = 11 + 6 = 17 ⇒ (n, n + 2, n + 6) = (11, 13, 17) Therefore, (11, 13, 17) is a prime triplet. (b) In 5, 9, and 11, n = 5. Then, n + 4 = 5 + 4 = 9 And, n + 6 = 5 + 6 = 11 ⇒ (n, n + 4, n + 6) = (5, 9, 11) Therefore, (5, 9, 11) is a prime triplet.
Explanation Prime triplets can be represented as n, n + 2, n + 6.
Prime triplets can be represented by n, n + 4, n + 6.
18. Fundamental Law/Theorem of Arithmetic Prime factorisation of numbers using division method8 Example 1 What are the prime factors of 100? Solution Procedure
Explanation 100 is divisible by 2, the smallest prime number. 50 is divisible by 2, the smallest prime number. 25 is divisible by 5, the next smallest prime number. 5 is divisible by 5, the next smallest prime number.
100 ÷ 2 = 50 ⇒ 50 ÷ 2 = 50 ⇒ 25 ÷ 5 = 5 ⇒ 5÷5 = 1 Thus, the prime factors of 100 are 2 × 2 × 5 × 5. 2 100 Finding factors using prime factorisation method. 2 50 5 25 5 5 1 Thus, the prime factors of 100 are 2 × 2 × 5 × 5.
For prime factorisation, check the number for divisibility with prime numbers, then use the prime numbers with which the number is divisible to divide the number, and get factors. Example 2 Find the smallest number having five different prime factors.
8
New learning outcome
More about numbers 133 Solution Procedure Explanation We know prime numbers are 2, 3, 5, 7, 11, 13, ∵ The smallest number having five different prime 17, 19, … factors will have five distinct smallest prime numbers Therefore, the prime factors of the required in increasing order. number are 2, 3, 5, 7, and 11. Hence, required smallest number = 2 × 3 × 5 × 7 × 11 = 2310 Example 3 Can we make 42 by multiplying only prime numbers? Solution Procedure Explanation 2 42 Finding factors using prime factorisation method. 3 21 7 7 1 Thus, 42 = 2 × 3 × 7 Hence, we can make 42 by multiplying prime factors, i.e., 2, 3, and 7.
Prime factorisation of numbers using factor tree method9 Example 1 Using the factor tree method, find all the prime factors of 1729 and arrange them in ascending order. Solution Procedure
Explanation Finding prime factors of 1729 using factor tree method.
1729 247
7 13
19
Therefore, the prime factors of 1729 in ascending order = 7 × 13 × 19. A factor tree is a diagram used to express the prime factors of a number. Each branch in the tree is split into factors (prime number in a circle and composite number in a rectangle). We will split the branches that have composite numbers (factors) until we get all the prime factors at the end. Example 2 Find the prime factors of the largest 4-digit number using the factor tree method.
9
Refer P 291
134 A to Z of Mathematics Solution Procedure The factor tree diagram of 9999 is as follows:
Explanation ∵ The largest 4-digit number is 9999.
9999 3333
3
1111
3 11
101
Therefore, the prime factors of 9999 = 3 × 3 × 11 × 101.
19. Highest Common Factor (or Greatest Common Divisor) of Numbers Finding HCF using factors10 Example 1 Find the HCF of 36 and 45. Solution The HCF of a given set of numbers is the biggest common factor of all the common factors. So find all the common factors of 36 and 45, and then the biggest common factor is the HCF. Procedure 36 ÷ 1 = 36 ; 36 ÷ 2 = 18 ; 36 ÷ 3 = 12
Explanation Finding factors of 36 using divisibility rules.
36 ÷ 4 = 9 ; 36 ÷ 6 = 6
∴ Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
All the divisors and quotients are the factors.
45 ÷ 1 = 45 ; 45 ÷ 3 = 15 ; 45 ÷ 5 = 9
Finding factors of 45 using divisibility rules.
∴ Factors of 45: 1, 3, 5, 9, 15, 45
All the divisors and quotients are the factors.
The highest common factor of 36 and 45 is 9.
∵ The common factors of 36 and 45 are 1, 3, and 9.
Thus, the HCF of 36 and 45 is 9. Example 2 What is the greatest common factor (GCF) of 90 and 60?
10
New learning outcome
More about numbers 135 Solution Procedure 90 ÷ 1 = 90 ; 90 ÷ 2 = 45 ; 90 ÷ 3 = 30 ;
Explanation Finding factors of 90 using divisibility rules.
90 ÷ 5 = 18 ; 90 ÷ 6 = 15 ; 90 ÷ 9 = 10
∴ Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. All the divisors and quotients are the factors. 60 ÷ 1 = 60 ; 60 ÷ 2 = 30 ; 60 ÷ 3 = 20 ;
Finding factors of 60 using divisibility rules.
60 ÷ 4 = 15 ; 60 ÷ 5 = 12 ; 60 ÷ 6 = 10
∴ Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. All the divisors and quotients are the factors. The greatest common factor of 90 and 60 is 30.
∵ The common factors of 90 and 60 are 1, 2, 3, 5, 6, 10, 15, and 30.
Thus, the GCF of 90 and 60 is 30.
Finding HCF using prime factorisation method11 Example 1 Find the HCF of 192 and 180 by using prime factorisation method. Solution Procedure The prime factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 The prime factors of 180 = 2 × 2 × 3 × 3 × 5
The highest common factor of 192 and 180 = 2 × 2 × 3 = 12 Therefore, the HCF of 192 and 180 is 12. Example 2 Find the HCF of 57 and 190.
11
Refer P 293
Explanation By prime factorisation method (using shortcut division) 2 192 2 180 2 96 2 90 2 48 3 45 2 24 3 15 2 12 5 5 2 6 1 3 3 1 ∵ The product of common prime factors of 192 and 180 (2, 2, and 3) give the HCF.
136 A to Z of Mathematics Solution Procedure The prime factors of 57 = 3 × 19 The prime factors of 190 = 2 × 5 × 19
The highest common factor of 57 and 190 is 19. Therefore, the HCF of 57 and 190 is 19.
Explanation By prime factorisation method (using shortcut division) 3 57 2 190 19 19 5 95 1 19 19 1 ∵ The only common factor is 19.
Example 3 Find the HCF of 24 and 96. Solution Procedure The prime factors of 24 = 2 × 2 × 2 × 3 The prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
The highest common factor of 24 and 96 is 23 × 3 = 8 × 3 = 24.
Explanation By prime factorisation method (using shortcut division) 2 96 2 24 2 48 2 12 2 24 2 6 2 12 3 3 2 6 1 3 3 1 ∵ The product of common factors of 24 and 96 (23 × 3) give the HCF.
Hence, the HCF of 24 and 96 is 24. Example 4 Find the HCF of 70, 105, and 175. Solution Procedure Explanation Let us first find the prime factors of 70, 105, and 175. Prime factors of 70 = 2 × 5 × 7 By prime factorisation method (using shortcut division) Prime factors of 105 = 3 × 5 × 7 2 70 3 105 5 175 Prime factors of 175 = 5 × 5 × 7 5 35 5 35 5 35 7 7 7 7 7 7 1 1 1
More about numbers 137
The highest common factor of 70, 105, and 175 = 5 × 7 = 35
∵ The product of common factors of 70, 105, and 175 (5 and 7) give the HCF.
Therefore, the HCF of 70, 105, and 175 is 35.
Finding HCF using division method12 Example 1 Find the HCF of 12 and 18 using division method. Solution Procedure 12
18 1 –12 6 12 2 –12 0
Explanation Treat the smaller number 12 as the divisor and the bigger number 18 as the dividend. Now, divide them. Next, the remainder 6 becomes the divisor and the divisor 12 becomes the dividend. Repeat this process till the remainder becomes zero. The last divisor is the HCF.
Hence, the HCF of 12 and 18 is 6. Example 2 Find the HCF of 16, 18, and 24 using division method. Solution When HCF of more than two numbers is to be found, find the HCF of any two numbers, and use the HCF of first two numbers as divisor for third, and proceed similarly. Procedure 16 18 1 –16 2 16 8 –16 0
Explanation Take up first two numbers, 16 and 18. Treat the smallest number 16 as the divisor and the bigger number 18 as the dividend. And, find the HCF.
Here, 2 is the HCF of 16 and 18. 2
24 12 –24 0
Here, the HCF of 2 and 24 is 2. Hence, the HCF of 16, 18, and 24 is 2. This division method discovers the HCF faster.
12
New learning outcome
Now, the HCF of the first two numbers – 2 becomes the divisor and the third number 24 becomes the dividend. Again, find the HCF.
138 A to Z of Mathematics
Finding HCF using Euclid’s division lemma and algorithm13 Example 1 Find the HCF of 55 and 215 and express it in form 55m + 215n using Euclid’s division algorithm. Solution To calculate the HCF of two positive integers a and b, we use Euclid’s division algorithm which is based on Euclid’s division lemma. According to Euclid’s division lemma, for any two positive integers, ‘c’ and ‘d’, the condition ‘c = dq + r’, where 0 ≤ r < d always holds true. Here, ‘q’ and ‘r’ are the quotient and remainder respectively. Procedure Applying division lemma to 55 and 215, we get 215 = (55 × 3) + 50
Explanation Given integers are 55 and 215 such that 215 > 55. ....(i) Divide c by d and apply Euclid’s division lemma to get the quotient and remainder, ‘q’ and ‘r’ respectively. c = dq + r, d > 0; c = 215 and d = 55.
Since the remainder 50 =/ 0, we apply the division lemma to the divisor 55 and the remainder 50. 55 = (50 × 1) + 5
....(ii) Here, r is not equal to 0. ∴ Apply Euclid’s division lemma to the divisor and the remainder. Here, c = 55 and d = 50. We consider the new divisor 50 and the new remainder 5 and apply division lemma. Here, r = 0, then divisor at this stage is the HCF of 50 = (5 × 10) + 0 ‘c’ and ‘d’. Here, c = 50, and d = 5. ∴ 5 is the HCF of 215 and 55. Now, we rearrange the above equations in such a way that we get an equation in the form 5 = 55m + 215n to get the value of m and n. 5 = 55 – (50 × 1) ⇒ 5 = 55 – [215 – (55 × 3)]
From eq. (ii) From eq. (i), 50 × 1 = 215 – (55 × 3) Opening the bracket on RHS. We rearrange the RHS as we are required to express it in form 55m + 215n.
⇒ m = 4 and n = –1
On comparing the two equations, 5 = (55 × 4) + [215 × (–1)] and 5 = 55m + 215n.
⇒ 5 = 55 – 215 + (55 × 3) ⇒ 5 = (55 × 4) + [215 × (– 1)]
Therefore, m = 4 and n = –1 in the equation 5 = 55m + 215n.
13
New learning outcome
More about numbers 139
Application of HCF in real-life14 Example 1 Reena has 24 red candies and Maya has 18 green candies. They want to arrange the candies in such a way that each row contains an equal number of candies and also each row should have only red candies or green candies. What is the greatest number of candies that can be arranged in each row? Solution To find the greatest number of candies that can be arranged in equal rows, we have to find the highest common factor of two numbers. Procedure Explanation Factors of 18 are 1, 2, 3, 6, 9, and 18. By using divisibility rule. Factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. By using divisibility rule. The HCF of 18 and 24 is 6. The common factors of 18 and 24 are 2, 3, and 6. So, the greatest number of candies that can be arranged in each row is 6. Example 2 Preeti has two pieces of cloth. One piece is 8 inches wide and the other piece is 36 inches wide. She wants to cut both pieces into strips of equal width that are as wide as possible. How wide should she cut the strips? Solution This is the direct application of HCF because we are cutting 8 and 36 inches wide pieces of cloth into equal width strips (common factor of 8 and 36) and we are looking for the widest possible strips (not any, but only the highest common factor of 8 and 36). Procedure Width of two pieces are 8 inches and 36 inches. 8
Explanation Given Finding the HCF of 8 and 36 using division method.
36 4 –32 4 8 2 –8 0 Thus, the HCF of 8 and 36 is 4. Hence, Preeti should cut 4 inches wide strips of both the material.
Example 3 A wine seller had three types of wine. 403 litres of first kind, 434 litres of second kind, and 465 litres of third kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing. Solution We need to find out the smallest number of casks of equal size in which the three types of wine can be packed without mixing them or any wine being left. In this, we first find the largest volume of casks which can be used for packing wine of 3 types – having volume – 403 L, 434 L, and 465 L. The volume of cask will be such that it would divide 403, 434 and 465 litres without leaving remainder. This is the HCF of 403, 434 and 465. 14
Refer P 299
140 A to Z of Mathematics
After finding the volume of the casks, we can find the number of casks, by dividing the three volumes of wines by the volume of casks and adding them together. Procedure
Explanation Finding the HCF of 403, 434, and 465 using division method.
31 465 15 403 434 1 –465 –403 0 31 403 13 –403 0 Hence, the volume of each cask must be 31 litres. ∵ HCF of 403, 434, and 465 is 31. Required number of casks Total volume of each wine ∵ Number of casks = Volume of 1 cask 403 434 465 = + + 31 31 31 = 13 + 14 + 15 = 42 Hence, the least possible number of casks of equal size required is 42.
20. Least Common Multiple of Numbers Finding LCM by listing the multiples15 Example 1 Find the LCM of 12 and 20 by listing the multiples. Solution The first common multiple of a set of numbers is the Lowest Common Multiple (LCM). So, here we list the multiples of 12 and 20 till we get the first common multiple. Procedure Explanation Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, For multiples of 12, multiply 12 by 1, 2, 3, 4, …. 108, 120, .... Multiples of 20 = 20, 40, 60, 80, 100, 120, .... For multiples of 20, multiply 20 by 1, 2, 3, 4, …. The least (smallest) common multiple of the two ∵ The common multiples of 12 and 20 are 60, 120, numbers is 60. and so on. Example 2 What is the LCM of two co-prime numbers? Solution Procedure The LCM of two co-prime is always their product.
15
Refer P 296
Explanation Since, co-prime numbers have only 1 as their common factor.
More about numbers 141
Finding LCM using prime factorisation method16 Example 1 Find the LCM of 112, 140, and 168 by prime factorisation method. Solution Write each number in the exponential form of prime factors. Then from all the prime factors of numbers, select distinct prime factors with highest exponents. The product of these numbers is LCM. Procedure Prime factors of 112 = 2 × 2 × 2 × 2 × 7 = 24 × 7 Prime factors of 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 Prime factors of 168 = 2 × 2 × 2 × 3 × 7 = 23 × 3 × 7
Explanation By using prime factorisation method, 2 112 2 140 2 168 2 56 2 70 2 84 2 28 5 35 2 42 2 14 7 7 3 21 7 7 1 7 7 1 1
The LCM of 112, 140 and 168 = 24 × 7 × 5 × 3 = 16 × 7 × 5 × 3 = 1680 Hence, the LCM of 112, 140, and 168 is 1680.
The highest exponent of prime factor base 2 is 24. The highest exponent of prime factor base 7 is 71. The distinct prime factors are 5 and 3.
Example 2 Find the LCM of 1165 and 4095 by prime factorisation method. Solution Procedure Prime factors of 1165 = 5 × 233 Prime factors of 4095 = 3 × 3 × 5 × 7 × 13 = 32 × 5 × 7 × 13
Explanation By using prime factorisation method,
The LCM of 1165 and 4095 = 32 × 5 × 7 × 13 × 233 = 954135
The highest exponent of prime factor base 3 is 32. The highest exponent of prime factor base 5 is 51. The highest exponent of prime factor base 7 is 71. The highest exponent of prime factor base 13 is 131. The highest exponent of prime factor base 233 is 2331.
Hence, the LCM of 1165 and 4095 is 954135.
Finding LCM using division method17 Example 1 Find the LCM of 40, 36, and 126.
16 17
Refer P 296 New learning outcome
5 1165 233 233 1
3 4095 3 1365 5 455 91 7 13 13 1
142 A to Z of Mathematics Solution The logic of finding LCM remains the same. This division method is just a faster way of getting the common multiples. Procedure
Explanation Finding LCM using shortcut division method. 2 40, 36, 126 Start by writing the numbers in a line separated by 2 20, 18, 63 commas. 2 10, 9, 63 Divide the numbers by the smallest prime number 3 5, 9, 63 which divides at least one of the numbers. The 3 5, 3, 21 quotient is written in the following line, again 5 5, 1, 7 separated by commas. The numbers not divisible 7 1, 1, 7 by a prime number are written as it is. Continue 1, 1, 1 dividing by prime numbers till the quotients in the last row is all 1s. LCM is the product of all the divisors. ∴ LCM of 40, 36, and 126 = 2 × 2 × 2 × 3 × 3 × 5 × 7 Multiplying all the prime factors to get the LCM. =8×9×5×7 = 2520 Hence, the LCM of 40, 36, and 126 is 2520. Example 2 Find the LCM of 26, 124, 369, and 1045. Solution Procedure 2 2 3 3 5 11 13 19 31 41
26, 13, 13, 13, 13, 13, 13, 1, 1, 1, 1,
124, 62, 31, 31, 31, 31, 31, 31, 31, 1, 1,
369, 369, 369, 123, 41, 41, 41, 41, 41, 41, 1,
1045 1045 1045 1045 1045 209 19 19 1 1 1
Explanation Finding LCM using shortcut division method.
Multiplying all the prime factors to get the LCM. ∵ LCM of 26, 124, 369, and 1045 = 2 × 2 × 3 × 3 × 5 × 11 × 13 × 19 × 31 × 41 = 621595260 Hence, the LCM of 26, 124, 369, and 1045 is 621595260.
More about numbers 143
Application of LCM in real-life18 Example 1 Eric exercises every 15 days and Ethan every 9 days. Eric and Ethan both exercised today. After how many days will they exercise together again? Solution This is the direct application of the Least Common Multiple. In this problem we have to figure out that when will they both exercise together next. So, we have to find the least common multiple of time. Procedure Explanation 3 15, 9 Finding LCM using shortcut division method. 3 5, 3 5 5, 1 1, 1 LCM of 15 and 9 is 3 × 3 × 5 = 45. Hence, they will exercise together again on the forty-fifth day, and after every forty-fifth day they will exercise together. Example 2 The traffic lights at three different road crossings change after every 48 sec, 72 sec, and 108 sec respectively. If they all change simultaneously at 9:30:00 hours, when will they again change simultaneously? Solution Figuring out the least time when the three traffic lights will change simultaneously is a direct application of finding Least Common Multiple (LCM). To find the smallest time interval, when all the lights will change simultaneously, we have to find the LCM of the interval at which the three traffic light change. 2 2 2 2 3 3 3
Procedure 48, 72, 108 24, 36, 54 12, 18, 27 6, 9, 27 3, 9, 27 1, 3, 9 1, 1, 3 1, 1, 1
LCM of 48, 72, and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds
Explanation Finding LCM using shortcut division method.
1 ∵ 1 sec = min = 432 minutes 60 60 = 7 minutes 12 seconds So, after every 7 min 12 sec, all the signals will change simultaneously. At 9:30:00 hours, if all the three signals change simultaneously, they will change again simultaneously after 7 min 12 sec. That is at 9:37:12 hours. 18
Refer P 299
144 A to Z of Mathematics Example 3 The number of soldiers in a regiment is a perfect square. Find the least number of soldiers in the regiment such that they can stand in rows of 15, 20, and 25. Solution In this problem, first find the smallest number which is exactly divisible by the given numbers 15, 20, and 25. That is, LCM of 15, 20, and 25. Then, we have to check to see whether the number is a perfect square. If not, then multiply the number with the smallest number such that it will become a perfect square. 2 2 3 5 5
Procedure 15, 20, 25 15, 10, 25 15, 5, 25 5, 5, 25 1, 1, 5 1, 1, 1
Explanation Finding LCM using shortcut division method.
LCM of 15, 20, and 25 = 2 × 2 × 3 × 5 × 5 = 300 ....(i) So, the number of soldiers required in a regiment such that they make a perfect square when they stand in rows of 15, 20, and 25 is 300. But, they have to form a perfect square. From eq. (i) The factors of 300 = 2 × 2 × 3 × 5 × 5 2 2 =2 ×3×5 From the factors we can see that on multiplying ∴ To form a perfect square, we need to multiply 300 × 3 such that 300 by 3, we can get a perfect square. 300 × 3 = 22 × 32 × 52 = 900 Hence, the smallest perfect square number of soldiers is 900, so they can stand in rows of 15, 20, and 25.
21. Applying Fundamental Theorem of Arithmetic Relationship between HCF and LCM19 Example 1 The HCF and LCM of the two numbers are 13 and 1989 respectively. If one of the numbers is 117, determine the other. Solution Procedure
Explanation
First number = 117 HCF of two numbers = 13 LCM of two numbers = 1989
Given
Let second number be x. (First number) × (Second number) = HCF × LCM
Assumption Using formula
19
Refer P 298
More about numbers 145
⇒x=
=
13 ×1989 117 13 × 9 × 221
Putting the given values in the formula.
13 × 9
= 221 Hence, the second number is 221. Example 2 The HCF of the two numbers is 16 and their product is 3072. Find their LCM. Solution Procedure HCF of two numbers = 16 Product of two numbers = 3072 Let LCM of two numbers be x. ∵ HCF × LCM = Product of two numbers 3072 ⇒x= 16 = 192 Hence, the LCM of the two numbers is 192.
Explanation Given Assumption Using formula Putting the given values in the formula.
Example 3 Given that HCF (450, 1350) = 450, LCM (450, 1350) = 135 × k, then what is the value of k? Solution Procedure HCF of 450 and 1350 = 450 LCM of 450 and 1350 = 135k Two numbers are 450 and 1350. ∵ HCF × LCM = Product of two numbers ⇒ 450 × 135k = 450 × 1350
Explanation Given
Using formula Putting the given values in the formula.
450 × 1350 450 × 135 ⇒ k = 10 Thus, the value of k is 10. ⇒k=
Example 4 HCF and LCM of two numbers are 6 and 60. If the two numbers are in the ratio of 2:5, then find the two numbers.
146 A to Z of Mathematics Solution Procedure HCF of two numbers = 6 LCM of two numbers = 60 Ratio of two numbers = 2:5 Let the two numbers be 2x and 5x. First number × second number = HCF × LCM ⇒ 2x × 5x = 6 × 60 ⇒ 10x2 = 360 360 ⇒ x2 = 10 ⇒ x2 = 36 ⇒ x2 = (6)2 ⇒x=6 ∴ 2x = 2 × 6 = 12 ∴ 5x = 5 × 6 = 30 Hence, the two numbers are 12 and 30.
Explanation Given ∵ Ratio of two numbers is 2:5.
Using formula Putting the given values in the formula.
∵ First number = 2x ∵ Second number = 5x
22. Square Root of a Number Finding square root of a number using repeated subtraction method20 Example 1 Find the square root of 100 by the repeated subtraction method. Solution The sum of first ‘n’ odd natural numbers is n2. We can find this by using the formula of sum of an arithmetic progression series. We can use this fact to find the square root of a number by repeated subtraction of odd numbers from that number till we get zero. To find the square root of 100 by the method of repeated subtraction, let us subtract the odd numbers 1, 3, 5, 7, 9, …… successively from the numbers 100 till we get zero (0). • 100 – 1 = 99 • 99 – 3 = 96 • 96 – 5 = 91 • 91 – 7 = 84 • 84 – 9 = 75 • 75 – 11 = 64 • 64 – 13 = 51 • 51 – 15 = 36 • 36 – 17 = 19 • 19 – 19 = 0 From 100, we have subtracted successive odd numbers starting from 1 and obtained 0 at the 10th step. ⇒ 100 = 10 Thus, the square root of 100 is 10.
20
Refer P 118
More about numbers 147
Estimating square root of a number21 Example 1 Without calculating the square root, find the number of digits in the square root of the following numbers. (a) 19600 (b) 221904 (c) 200000000 Solution To find the number of digits in the square root, make groups of two digits starting from the right side of the number and place bars. The number of bars gives the number of digits in the square root. Procedure
Explanation
Placing bars over 19600. (a) We have, 19600. Since there are three bars; therefore, the number of digits in its square root is 3. Placing bars over 221904. (b) We have, 221904. Since there are three bars; therefore, the number of digits in its square root is 3. Placing bars over 200000000. (c) We have, 200000000. Since there are five bars; therefore, the number of digits in its square root is 5. Example 2 Find the value of 274 to the nearest whole number. Solution To estimate the value of a square root of a number, list the perfect square numbers between which the given number lies. The estimated value will be the root of the nearest perfect square. Procedure We know that, 15 = 225; 162 = 256; 172 = 289.
Explanation
2
⇒ 16 < 274 < 17
So, 274 is approximately equals to 17.
∵ From above, we can say that value of 274 lies between 16 and 17. ∵ 274 is closer to 289 than 256.
Finding square root of a number using prime factorisation method22 Example 1 Find the value of
62500 .
Solution Procedure Given
62500
=
2 × 2× 5 × 5 × 5 × 5 × 5 × 5
=2×5×5 × 5 = 250 Hence, 21 22
Explanation
62500 is 250.
New learning outcome New learning outcome
Write Prime factors of 62500 and then make pairs of the same factors. Taking out one factor from each pair to get the square root.
148 A to Z of Mathematics Example 2 Find the value of
289 + 121 .
Solution Procedure
Explanation Given
289 + 121 = 17 × 17 + 11 × 11
Write prime factors of 289 and 121.
= 17 + 11 = 28
Taking out one factor from each pair to get the square root. 289 + 121 is 28.
Hence, the value of Example 3 Find the value of
(
108
) ÷ (3
)
432 .
Solution Procedure
Explanation Given
108 ÷ 3 432 =
=
Write prime factors of 108 and 432 and then make pairs of same factors.
2×2×3×3×3 3× 2×2×2×2×3×3×3
Taking out one factor from each pair and unpaired factor remaining underroot to get the square root of the number.
2×3× 3 3×2×2×3× 3
= 1 6 Hence, the value of
Reducing the expression.
(
108
) ÷ (3
)
432 is 1 . 6
Finding square root of a number using long division method23 Example 1 Evaluate: (a) 16384 (b) 10404
23
New learning outcome
More about numbers 149 Solution Procedure (a) 1 28
Explanation Finding the square root of 16384 by using the long division method.
1 1 6384 + 1 –1 22 63 +2 –4 4 248 1984 –1984 0 ∴ 16384 = 128 (b) 1 02
Finding the square root of 10404 by using the long division method.
1 1 04 04 + 1 –1 20 4 +0 –0 202 404 –404 0 ∴ 10404 = 102 Steps to find the square root using the long division method: (a) Starting with the digit at the unit's place, place bars over the digits of the number to make pairs of segments. In case of an odd count of digits in the number, there will be an unpaired digit in the segment at the end. (b) Start from the left-most segment. The largest number whose square is equal to or just less than the first segment is taken as a divisor and a digit in the quotient so that the product is the square. (c) Subtract the square of the divisor from the first segment and bring down the next segment to the right of the remainder to get the new dividend. (d) The new divisor is obtained by adding the digit of the previous quotient to the previous divisor and linking together a suitable digit chosen in such a way that the product of the new divisor and this digit is equal to or just less than the new dividend. This digit is also the next digit of the quotient. (e) Repeat steps (b), (c), and (d) till all the segments have been used for making new dividends. Now, the quotient so obtained is the required square root of the given number.
150 A to Z of Mathematics Example 2 The area of a square plot is 5329 m2. Find the side of the square plot. Solution Procedure Area of square plot = 5329 m2 ⇒ (side of the square plot)2 = 5329 m2 ∴ Side of the square plot =
5329 m = 73 m
Explanation Given
∵ Area of a square = (side)2 73 7 53 29 +7 –49 143 429 –429 0
Thus, the side of the square plot is 73 m. Example 3 Find the least number that must be added to 1385 to get a perfect square. Also, find the square root of the perfect square. Solution Procedure 37 3 1385 +3 –9 67 485 –469 16 The next perfect square number is 382 = 1444 ∴ The least number that must be added = 382 – 1385 = 1444 – 1385 = 59
Explanation Finding the square root of 1385 by using the long division method.
∵ From the above long division method, we get the remainder 16. It means to get the perfect square number (37)2, we have to subtract 16 from 1385. But, according to the question, we have to add a number in 1385 to get the perfect square number.
Hence, 59 should be added to 1385 to get 1444 whose square root is 38. Example 4 Evaluate:
9802
More about numbers 151 Solution Procedure 99 9802 9 +9 –81 189 1702 –1701 1 Hence,
Explanation Finding the square root of 9802 by using the long division method.
9802 ≈ 99
Finding square root of fractions24 Example 1 Find the value of Solution
15625 . 9
Procedure Given
15625 9
=
5×5×5×5×5×5 3× 3
Write the prime factors of 15625 and 9. Taking out one factor from each pair to get the square root.
5×5×5 3 125 = 3 =
Hence,
Explanation
15625 is 125 . 3 9
Example 2 Find the value of Solution
15552 . 108 Procedure
15552 108
2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3× 3 == 2 × 2 × 3× 3× 3 24
New learning outcome
Explanation Given Write the prime factors of 15552 and 108.
152 A to Z of Mathematics
= 2 × 2 × 2 × 2 × 3× 3
Reducing to simplifying the expression.
=2×2×3 = 12
Taking out one factor from each pair to get the square root.
Hence,
15552 = 12. 108
Finding square root of decimal numbers25 Example 1 Find the square root of 1341.7569. Solution Procedure 36.63 3 +3 66 +6 726 +6 7323
1341.7569 –9 441 –396 4575 –4356 21969 –21969 0
Explanation Here, the number of decimal places is already even. So, we place bars on the integral (starting from right before the decimal point) and decimal parts (starting from left after the decimal point) and then proceed.
Thus, 1341.7569 = 36.63 Example 2 Find the square root of 0.00286. Solution Procedure 0 0 0 5 +5
103
Hence, 25
0.053 0.002860 –0 000 –0 028 –25 360 –309 51
0.00286 ≈ 0.053
New learning outcome
Explanation Here the number of decimal places is odd. So, we add one more zero at the end of the number which makes the number of decimal places even. Then, find the square root.
More about numbers 153
23. Cube Root of a Number Finding cube root of a number using prime factorisation method26 Example 1 Find the value of 3 17576 . Solution Procedure 3
= =
Given
17576 3
2 × 2 × 2 ×13×13×13
3
23 × 133
∵ Prime Factors of 17576 = 2 × 2 × 2 ×13×13×13 Then, make the triplets of same factors. From each triplet, take out one factor to get cube root.
= 2 × 13 = 26 Example 2 Find the value of
Explanation
3
85750 .
Solution Procedure 3
Explanation Given
85750
=
3
7×7 ×7 ×5×5×5×2
=
3
73 ×53 ×2
∵ Prime factors of 85750 = 7 × 7 × 7 × 5 × 5 × 5 × 2 Then, make the triplets of same factors. From each triplet, take out one factor. The product of these and any incomplete triplet (which remains under cube root) is the cube root.
= 7× 5× 3 2 = 35 3 2
Estimating cube root of a number27 Example 1 Find the cube root of 164616 through estimation. Solution Procedure 164616 Second First group group
The unit digit of the cube root = 6 We know that, 53 = 125 and 63 = 216 ∴ 125 < 164 < 216 i.e., 53 < 164 < 63 26 27
New learning outcome New learning outcome
Explanation The given number is 164616. We will make groups of three digits starting from the right side of the number. ∵ The unit digit in the first group is 6. If a number ends in 6, it’s cube root will also end in 6. The second group is 164. Checking the perfect cube numbers between which the number (164) lies.
154 A to Z of Mathematics
The ten’s digit of the cube root = 5 The estimated cube root of 164616 is 56.
∵ We will consider the cube root of the smaller number, i.e., 53. ∵ The unit digit of the cube root = 6 The ten’s place digit of the cube root = 5
i.e., 3 164616 =≈ 56 56 Hence, the cube root of 164616 is approximately equal to 56. Example 2 Find the cube root of 51769 through estimation. Solution Procedure 51 769
Second First group group
The unit digit of the cube root = 9 We know that, 33 = 27 and 43 = 64 ∴ 27 < 51 < 64 i.e., 33 < 51 < 43
∴ The ten’s digit of the cube root = 3 ∴ The estimated cube root of 51769 is 39.
Explanation The given number is 51769. We will make groups of three digits starting from the right side of the number. The unit digit in the first group is 9. If a number ends in 9, it’s cube root will also end in 9. The second group is 51. Checking the perfect cube numbers between which the number (51) lies.
We will consider the cube root of the smaller number, i.e., 33. The unit digit of the cube root = 9 The ten’s digit of the cube root = 3
3 i.e., 51769 ≈ 39 Hence, the cube root of 51769 is approximately equal to 39.
Finding cube root of a negative and rational number28 Example 1 Find the cube root of (–729 × –4096). Solution Procedure 3
–729 × –4096
Explanation Given
= 3 [–(3× 3× 3× 3× 3× 3)]×[–(4 × 4 × 4 × 4 × 4 × 4)] Resolving –729 and –4096 into prime factors. = [–(3 × 3)] × [–(4 × 4)] = (–9 × –16) = 144 Therefore, the cube root of –729 × – 4096 is 144.
28
New learning outcome
After grouping the factors in triples of equal factors, taking one factor from each triple to get cube root. Product of –9 and –16 gives 144, as (–) × (–) = (+).
More about numbers 155 Example 2 Calculate the value of Solution
3
−448 . 2401
Procedure
Explanation Given
–448 2401 3
− ( 2×2×2×2×2×2×7 )
=
3
=
3
=
– (2 × 2) 7
=
−4 7
Resolving –448 and 2401 into prime factors.
7×7×7×7 Grouping the factors in triples of equal factors.
− (2 × 2 × 2) × (2 × 2 × 2) 7×7×7
Taking one factor from each triple to get the cube root.
Required value of
Therefore, the cube root of
3
−448 . 2401
−448 −4 is . 2401 7
Example 3 Find the cube root of (–262144). Solution Procedure 3
Explanation The cube root of a negative cube is the cube root of the absolute value multiplied by –1.
−262144 = −( 3 262144)
Making a group of three digits, from the right, starting from the unit’s digit of the number.
262144
Second First group group
Unit digit of cube root of first group 144 = 4 We know that 63 = 216 and 73 = 343 ∴ 63 < 262 < 73
The second group is 262. Checking the perfect cube numbers between which the number (262) lies.
∴ The tens digit of the cube root = 6. Hence, the estimated value
∵ The unit digit in the first group is 4. If a number ends in 4, it’s cube root will also end in 4.
3
262144 is 64.
∵ We will consider the cube root of the smaller number, i.e., 63. Estimated cube root of 262144.
Therefore, the estimated cube root of –262144 3 262144 × –1 –1 == –64 – 64 is –64.
156 A to Z of Mathematics
24. The Order of Operation Using BODMAS rule to simplify the expressions29 Example 1 Simplify: 25 – 56 ÷ 7 + 12 × 2 Solution Procedure 25 – 56 ÷ 7 + 12 × 2
= 25 – 8 + 12 × 2 = 25 – 8 + 24 = 49 – 8 = 41
Explanation Since there are no brackets and powers, we will use the BODMAS rule in which we first solve ‘Division’, followed by ‘Multiplication’, ‘Addition’, and then ‘Subtraction’. Simplifying ‘division’, 56 ÷ 7 = 8. Simplifying ‘multiplication’, 12 × 2 = 24. Simplifying ‘addition’, 25 + 24 = 49. Simplifying ‘subtraction’, 49 – 8 = 41.
Rules to simplify the expression using BODMAS rule: (a) First solve the vinculum or bars. (b) Then simplify the brackets. (c) Solve the exponents or roots. (d) Perform division or multiplication operation (from left to right). (e) Perform addition or subtraction operation (from left to right). Example 2 Simplify: 78 – [5 + 3 × (25 – 4 × 5)] Solution Procedure 78 – [5 + 3 × (25 – 4 × 5)]
Explanation Since there are brackets, the BODMAS rule will be used to solve. The BODMAS rule is followed inside the bracket. There are two brackets, the innermost are to be solved first and within it, the multiplication will be solved as the first step. = 78 – [5 + 3 × (25 – 20)] Simplifying ‘multiplication’, 4 × 5 = 20 in the inner bracket. = 78 – [5 + 3 × 5] Simplifying ‘subtraction’, 25 – 20 = 5. There is one bracket now and within it, the multiplication will be solved first. = 78 – [5 + 15] Simplifying ‘multiplication’, 3 × 5 = 15. = 78 – 20 Simplifying ‘addition’, 5 + 15 = 20 in the bracket. = 58
29
Refer P 303
Simplifying ‘subtraction’, 78 – 20 = 58.
More about numbers 157 Example 3 Simplify: 4 + 82 × 52 + (54 ÷ 9) of 3 Solution Procedure 4 + 82 × 5 + (54 ÷ 9) of 3 = 4 + 82 × 52 + 6 of 3 = 4 + 82 × 25 + 6 × 3 2
= 4 + 2050 + 18
Explanation Given First solve the brackets, 54 ÷ 9 = 6. Solve, 52 = 25 and Replace ‘of ’ with ‘×’ sign.
= 2054 + 18
Simplifying ‘multiplication’, 82 × 25 = 2050 and 6 × 3 = 18. Simplifying ‘addition’, 4 + 2050 = 2054.
= 2072
Simplifying ‘addition’, 2054 + 18 = 2072.
Example 4 (a) 45 – {9 – [14 – (18 – 10 + 2)]} 3 1 7 3 8 − + (c) 1 ÷ × (6 + 8 × 3 – 2) + ÷ 7 5 25 7 14 10 (e) 10 + 10 × 10 ÷ × 10 ÷ 20 10 Solution Procedure (a) 45 – {9 – [14 – (18 – 10 + 2)]} = 45 – {9 – [14 – (8 + 2)]} = 45 – {9 – [14 – 10]} = 45 – {9 – 4} = 45 – 5 = 40 (b) {89 – (–1) [7 – (4 – 3)]} ÷ [4 × (5 + 5)] = {89 – (–1) [7 – 1]} ÷ [4 × 10] = {89 + 1 × 6} ÷ 40 = {89 + 6} ÷ 40 = 95 ÷ 40 =
3 × (6 + 8 × 3 – 2) + 7
(d) 54 – {30 – [10 – (7 + 3 − 2 )]}
Explanation Given First solve the round brackets, 18 – 10 = 8. Simplifying ‘addition’, 8 + 2 = 10. Solve square brackets, 14 – 10 = 4. Solve curly brackets, 9 – 4 = 5. Simplifying ‘subtraction’, 45 – 5 = 40. Given First solve the inner most bracket – the round brackets. Then, solve the square brackets. Simplifying ‘multiplication’ in curly bracket, 1 × 6 = 6. Simplifying curly bracket, 89 + 6 = 95.
95 5 × 19 19 = = 40 5 × 8 8
19 8
(c) 1 ÷
(b) {89 – (–1) [7 – (4 – 3)]} ÷ [4 × (5 + 5)]
1 7 3 8 5 ÷ 25 − 7 + 14
Given
158 A to Z of Mathematics
=1÷
1 7 6 + 8 3 × (6 + 24 – 2) + ÷ − 7 5 25 14
First solve the round brackets.
=1÷
3 1 7 × (30 – 2) + ÷ – (1) 7 5 25
Again solve the round brackets.
=1÷
3 1 25 × 28 + × – 1 7 5 7
=1÷
3 5 × 28 + – 1 7 7
Simplifying ‘multiplication’ in square bracket.
=1÷
5 – 7 3 × 28 + 7 7
Take LCM in the square bracket.
=1÷
3 2 × 28 – 7 7
Solve the square brackets.
=1×
7 2 × 28 – 3 7
Writing the reciprocal of
= =
= =
196 2 – 3 7 196 × 7 3×7 1372 – 6
Simplifying ‘division’ in square bracket and ‘subtraction’ in round bracket. Simplifying ‘division’ by writing the reciprocal of 7 and changing division sign to multiplication 25 sign.
3 and changing division 7 sign to multiplication sign. Simplifying ‘multiplication’,
–
2×3 7×3
7 × 28 3
=
196 3
.
Take LCM of both sides.
21 1366 21
Simplifying ‘subtraction’,
(d) 54 – {30 – [10 – (7 + 3 – 2 )]} = 54 – {30 – [10 – (7 + 1)]} = 54 – {30 – [10 – 8]} = 54 – {30 – 2} = 54 – 28 = 26 (e) 10 + 10 × 10 ÷
10 × 10 ÷ 20 10
1372 – 6 21
=
1366 . 21
Given First solve ‘vinculum’, 3 – 2 = 1 Solve round brackets, 7 + 1 = 8 Solve square brackets, 10 – 8 = 2. Solve curly brackets, 30 – 2 = 28. Simplifying ‘subtraction’, 54 – 28 = 26. Given
More about numbers 159
Simplifying ‘division’,
= 10 + 100 ÷ 1 × 10 ÷ 20 = 10 + 100 × 10 ÷ 20 = 10 + 1000 ÷ 20 = 10 + 50 = 60
Simplifying ‘multiplication’, 10 × 10 = 100. Simplifying ‘division’, 100 ÷ 1 = 100. Simplifying ‘multiplication’, 100 × 10 = 1000. Simplifying ‘division’, 1000 ÷ 20 = 50. Simplifying ‘addition’, 10 + 50 = 60.
Example 5 Simplify: 3 +
1+
Solution 1+ 1 3 + 1+ 1 1 1+ 1 1 1-– 1 33 ⇒ 1+ 1 ⇒ 1+ = 3+ 11 1 3+ 1+ 1 111+++111 2 222 3 333 =3 + 1 3 1+ 2
⇒ 1+111 =⇒ + =3 1+ 3+ 555 222
=3+ 2 = 3+ 5 17 == 5 =3
10 = 1. 10
= 10 + 10 × 10 ÷ 1 × 10 ÷ 20
2 5
1
1 1 1– 3 Procedure
Explanation Given Start solving the expression from the bottom, i.e., denominator. 1 3 −1 2 = ∵1− = 3 3 3
1 3 = ∵ 2 2 3 ∵ 1+
3 2+3 5 = = 2 2 2
1 2 ∵ 5 =5 2 LCM of 1 and 5 is 5.
∵ 3+
2 15 + 2 17 2 = = =3 5 5 5 5
160 A to Z of Mathematics Example 6 Find the value of x. 1 +
1
1+ 1 1 1+ x
=3
Solution Procedure 1 =3 1+ 1 1+ 1 1+ x ⇒ 1+
Given
=3 1 1 + x+1 x 1 =3 ⇒ 1+ 1 + x x+ 1
Start solving the expression from the bottom, i.e., denominator. 1 x +1 1 + = x x
⇒1 +
1 =3 x+1+x x +1 1 ⇒1 + =3 2x + 1 x + 1 ⇒1 + x + 1 =3 2x + 1
LCM of 1 and (x + 1) is x + 1.
2x + 1 + x + 1 = 3 2x + 1 ⇒ 3x + 2 = 3 (2x + 1) ⇒ 3x + 2 = 6x + 3 ⇒ 2 – 3 = 6x – 3x ⇒ 3x = –1 −1 ⇒x = ⇒ 3 −1 Hence, the value ⇒ of x is . = 3
LCM of 1 and (2x + 1) is 2x + 1.
⇒
1
Explanation
1 x = x +1 x +1 x
x + 1 + x = 2x + 1
1 x +1 = 2x + 1 2x + 1 x +1
By cross multiplication.
N N for Negative Numbers
The numbers we created
25. Visualisation of Negative Numbers Visualising situations involving negative numbers1 Example 1 Which type of number (negative or positive) would represent the following situations. (a) Floors above the ground level (b) Money deposited in the bank (c) A loan taken from banks (d) Distance below sea level (e) Students dropping off from a class (f ) Deflation of balloons (g) Gains in stock market index (h) Underground car parking levels Solution Situation
Positive/Negative Numbers
(a) Floors above the ground level
Positive Numbers
(b) Money deposited in the bank
Positive Numbers
(c) A loan taken from banks
Negative Numbers
(d) Distance below sea level
Negative Numbers
(e) Students dropping off from a class
Negative Numbers
(f) Deflation of balloons
Negative Numbers
(g) Gains in stock market index
Positive Numbers
(h) Underground car parking levels
Negative Numbers
Visualising negative numbers on number lines2 Example 1 Draw a number line and answer the following. (a) The integer obtained on moving 6 units to the right of –5? (b) The integer obtained on moving 6 units to the left of –3? (c) If you are at –10 on the number line, in which direction should you move to reach –4? 1 2
Refer P 308 Refer P 309
161
162 A to Z of Mathematics
(d) If you are at –5 on the number line, in which direction should you move to reach the integers which are smaller than –5? Solution Procedure
Explanation
(a)
We will reach at 1 if we move 6 units to the right of –5. –6
–5
–4
–3
–2
–1
0
1
2
3
4
5
(b)
We will reach at –9 if we move 6 units to the left of –3. –9
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
(c) –10
–9
–8
–7
–6
–5
–4
–3
–2
–1
0
(d) Left –8
–7
–6
Right –5
Numbers smaller than (–5)
–4
–3
–2
–1
0
1
1
From the figure, we can see that –4 is to the right of –10. Therefore, we should move towards the right direction. We should move towards the left direction to reach integers which are smaller than –5.
2
Numbers greater than (–5)
Example 2 Observe the given vertical number line in which alphabet ‘O’ represents zero and find the numerical value of the following. (a) If point D is +10, then which point is –3? (b) Write the integers representing the points, A and F. (c) Which point marked on this number line has the least value? (d) Is point G a negative integer or a positive integer? Also, write the integer representing the point G.
D C B A O E F G H
The numbers we created 163 Solution Number line
Location of points
(a) Point E represents –3. (b) Point A represents 3, and point F represents –6. (c) Point H has the least value, i.e., –11. (d) Point G lies below 0. So, it is a negative integer. Point G represents –9.
D C B A O E F G H
Example 3 Using the number line, write the integer which is (a) 5 more than 2 (b) 6 less than 3
(c) 3 more than –4
10 9 8 7 6 5 4 3 2 1 0 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11
(d) 4 less than –3
Solution Procedure
Explanation To get an integer, start from 2 and move 5 units rightward on the number line.
(a) 5 more than 2
–2
–1
0
1
2 Start
3
4
5
6
From number line, we can see it is 7. (b) 6 less than 3
–4
–3
End
–2
–1
0
1
7 End
To get an integer, start from 3 and move 6 units leftward on the number line. 2
From the number line, we can see it is –3.
3 Start
4
164 A to Z of Mathematics
(c) 3 more than –4
–6
–5
–4
Start
To get an integer, start from –4 and move 3 units rightward on the number line. –3
–2
–1 End
0
1
From the number line, we can see it is –1. (d) 4 less than –3
–8
–7 End
–6
–5
–4
–3 –2 Start
To get an integer, start from –3 and move 4 units leftward on the number line. –1
0
1
From the number line, we can see it is –7.
26. Operations on Negative Numbers Addition of negative numbers3 Example 1 Find the value of 35 + (–18) + (–23) + 32. Solution Procedure
Explanation
35 + (–18) + (–23) + 32 = 35 + 32 + (–18) + (–23) = 67 + (–41)
Grouping positive integers and negative integers together. Adding all positive integers together and negative integers together.
= 26
Simplifying
Example 2 Find the sum of (–12), 42, 44 and (–18). Solution Procedure (–12) + 42 + 44 + (–18) = (–12) + (–18) + 42 + 44 = (–30) + 86 = 56
Explanation Grouping positive integers and negative integers together. Adding all positive integers together and negative integers together. Simplifying
Example 3 Amar had ₹61000. He gave ₹8750 to Alisha, ₹12638 to Arpit. How much money was left with him?
3
Refer P 310
The numbers we created 165 Solution Procedure Money with Amar = ₹61000 Money given to Alisha = ₹8750 Money given to Arpit = ₹12638
Explanation Given
Money left with Amar = ₹[61000 – (8750 + 12638)] = ₹[61000 – 21388] = ₹39612
Money left with Amar = Money with Amar – [(Money with Alisha) + (Money with Arpit)]
Therefore, money left with Amar was ₹39612.
Subtraction of negative numbers4 Example 1 Find the value of –6 – (–8) using a number line. Solution Procedure
Explanation Start from ‘–6’ on the number line.
+
– –7
–6
–5
–4
–3
–2
–1
0
1
2
3
Start
The minus sign ‘(–)’ after –6 indicates moving towards left from –6. However, 8 is also negative (–8). Thus, the direction changes, we need to move towards right. Therefore, start from (–6) and move towards the right on the number line by 8 units.
+
– –7
–6
Start
–5
–4
–3
–2
–1
0
1
2 End
3
∴ –6 – (–8) = –6 + 8 = 2 Example 2 Find the value of (a) –15 – (–15) – (5)
(b) 50 – (–45) – (–5) – 65
Solution Procedure (a) –15 – (–15) – (5) = –15 + 15 – 5 4
Refer P 314
Explanation Given Since, (–) × (–) = (+)
166 A to Z of Mathematics
= –20 + 15 = –5 (b) 50 – (–45) – (–5) – 65 = 50 + 45 + 5 – 65 = 100 – 65
On adding negative integers together. Simplifying Given ∵ (–) × (–) = (+) Adding positive integers together.
= 35
Simplifying
Multiplication of negative numbers5 Example 1 Find the product of (–10) × 31. Solution Procedure (–10) × 31 = –(10 × 31) = –310
Explanation Given ∵ (–) × (+) = (–)
Example 2 Find the product of (–12) × (–8). Solution Procedure (–12) × (–8) = 12 × 8 = 96
Explanation Given ∵ (–) × (–) = (+)
Example 3 Find the product of 6 × (–115). Solution Procedure 6 × (–115) = –(6 × 115) = –690
Explanation Given ∵ (+) × (–) = (–)
Example 4 Find the product of 15 × (–20) × (–8). Solution Procedure 15 × (–20) × (–8) = {(15 × (–20)} × (–8) 5
Refer P 318
Explanation Given Grouping first two terms.
The numbers we created 167
= –(15 × 20) × (–8) = (–300) × (–8) = 2400
∵ (+) × (–) = (–) ∵ (–) × (–) = (+)
Example 5 Find the product of (–6) × (–5) × (–7). Solution Procedure (–6) × (–5) × (–7) = {(–6) × (–5)} × (–7) = (6 × 5) × (–7) = 30 × (–7) = –(30 × 7) = –210
Explanation Given Grouping first two terms ∵ (–) × (–) = (+) ∵ (+) × (–) = (–)
Division of negative numbers6 Example 1 Divide (–2662) by 11. Solution Procedure Given
–2662 11
=
Explanation
–(11 × 11 × 11 × 2) 11
= –242
For simplification, writing prime factors of 2662. ∵ (–) ÷ (+) = (–)
Example 2 Divide 306 by (–18). Solution Procedure 306 –18 17 × 2 × 3 × 3 = –(2 × 3 × 3)
= –17 Example 3 Divide (–46332) by (–11).
6
Refer P 319
Explanation Given For simplification, writing prime factors of 306 and 18. ∵ (+) ÷ (–) = (–)
168 A to Z of Mathematics Solution Procedure
–46332 –11 –( 2 × 2 × 3 × 3 × 3 × 3 × 11 × 13 ) = –11 = 4212
Explanation Given For simplification, writing prime factors of 46332. ∵ (–) ÷ (–) = (+)
Example 4 An elevator descends into a mine shaft at the rate of 12 m/min. If the descent starts from 10 m above ground level, how long will it take to reach –350 m? Solution Distance descended is denoted by a negative integer. Procedure Initial height = + 10 m Depth/length = –350 m = 350 m
Explanation
Given ∵ Distance is always positive which has only magnitude. Total distance to be descended by the elevator = 350m + 10m = 360m Time taken by the elevator to descend 12 m = 1 min Since, rate of elevator descent (or speed) = 12 m/min 1 ⇒ Time taken by the elevator to descend 1 m = min 12
∴Time taken by elevator to descend 360 m ∵ Total distance to be covered by the elevator = 360 m 1 = 360 × min 12 = 30 minutes Hence, time taken by the elevator to reach –350 m from 10 m above ground level will be 30 minutes. Example 5 Write five pairs of numbers (a, b) such that a ÷ b = –3. One such pair is (6, –2). Solution The result of a ÷ b is –3, i.e., negative. And, we know that (–) ÷ (+) = (–) and (+) ÷ (–) = (–). Thus write pairs in such a way that either of the number in pair is negative and the quotient we get from division of the numbers –3. Procedure
Explanation
a ÷ b = –3
Given
6 ÷ (–2) = –3
Given
First pair = (3, –1)
3 ÷ (–1) = –3
Second pair = (–3, 1)
–3 ÷ 1 = –3
Third pair = (9, –3)
9 ÷ (–3) = –3
The numbers we created 169
Fourth pair = (–12, 4)
–12 ÷ 4 = –3
Fifth pair = (51, –17)
51 ÷ (–17) = –3
27. Commutative Property of Numbers Commutative law of addition7 Example 1 Which one of the following illustrates the commutative law of addition? (a) 8 + 9 = 9 + 8 (b) 8 + 9 = 14 + 3 (c) 8 + 9 = 10 + 7 Solution We know that in the commutative law of addition, the sum of two numbers remains the same, no matter the order in which they are added. Procedure (a) 8 + 9 = 9 + 8
Explanation Given
It illustrates the commutative law of addition. (b) 8 + 9 = 14 + 3 It does not illustrate the commutative law of addition. (c) 8 + 9 = 10 + 7 It does not illustrate the commutative law of addition.
The numbers on both sides of the equation are same, i.e., 8 and 9. Given The given numbers on both sides of the equation are not same. Given The given numbers on both sides of equation are not same.
When the sum of numbers on both sides of equation is same, however the given numbers on both sides of equation are not some, it will not illustrate commutative law of addition. Example 2 Which one of the following illustrates the commutative law of addition? (a) (–5) + (–7) = (–7) + (–5) (b) (–5) + (–7) = (–3) + (–9)
(c) (–5) + (–7) = (–10) + (–2)
Solution We know that in the commutative law of addition, the sum of two numbers remains the same, no matter the order in which they are added. Procedure (a) –5 + –7 = –7 + –5 It illustrates the commutative law of addition. (b) –5 + –7 = –3 + –9 It does not illustrate the commutative law of addition. (c) –5 + –7 = –10 + –2 7
Refer P 320
Explanation Given ∵ The integers on both sides are same, i.e., (–5) and (–7). Given
The given integers on both sides of the equation are not same. Given
170 A to Z of Mathematics
It does not illustrate the commutative law of addition.
The given integers on both sides are not same.
Example 3 Solve the following using commutative law of addition. (a) 1359 + 1500 + 141 (b) 738 + 965 + 262
(c) (–132) + 1524 + (–368)
Solution Procedure (a) 1359 + 1500 + 141
Explanation Given
= 1359 + 141 + 1500
Using commutative law of addition, 1500 + 141 = 141 + 1500
= 1500 + 1500 = 3000 (b) 738 + 965 + 262 = 738 + 262 + 965
1359 + 141 = 1500
= 1000 + 965 = 1965 (c) (–132) + 1524 + (–368) = (–132) + (–368) + 1524
738 + 262 = 1000
Given Using commutative law of addition, 965 + 262 = 262 + 965
= –500 + 1524 = 1024
Given Using commutative law of addition, 1524 + (–368) = (–368) + 1524
(–132) + (–368) = –500
Commutative law of multiplication8 Example 1 Which of the following illustrates the commutative law of multiplication? (a) 8 × 5 = 10 × 4 (b) 8 × 5 = 20 × 2 (c) 8 × 5 = 5 × 8 Solution In the commutative law of multiplication, the product of two numbers remains the same, no matter in which order they are multiplied. Procedure (a) 8 × 5 = 10 × 4 It does not illustrate the commutative law of multiplication. (b) 8 × 5 = 20 × 2 ⇒ It does not illustrate the commutative law of multiplication. (c) 8 × 5 = 5 × 8
8
Refer P 320
Explanation Given
The numbers involved in the multiplication on both sides of equation are not same. Given The numbers involved in the multiplication on both sides of equation are not same. Given
The numbers we created 171
It illustrates the commutative law of multiplication. The numbers involved in the multiplication on both sides of the equation are same. Example 2 Which of the following illustrates the commutative law of multiplication? (a) –3 × –4 = –6 × –2 (b) –3 × 4 = –12 × 1 (c) –3 × –4 = –4 × –3 Solution We know that, in the commutative law of multiplication, the product of two numbers remains the same, no matter in which order they are multiplied. Procedure (a) –3 × –4 = –6 × –2 It does not illustrate the commutative law of multiplication.
Explanation Given
The numbers involved in the multiplication on both sides of the equation are not same. (b) –3 × –4 = –12 × –1 Given It does not illustrate the commutative law of The numbers involved in the multiplication multiplication. on both sides of the equation are not same. (c) –3 × –4 = –4 × –3 Given It illustrates the commutative law of multiplication. The product of the two numbers is equal, and the numbers involved in the multiplication on both sides of the equation are same. Example 3 Solve the following using commutative law of multiplication. (a) 400 × 232 × 25 (b) 200 × 134 × 15
(c) (–400) × 125 × (–50)
Solution Procedure
Explanation
(a) 400 × 232 × 25 = 400 × 25 × 232
Given Using commutative law of multiplication, 232 × 25 = 25 × 232
= 10000 × 232 = 2320000 (b) 200 × 134 × 15
400 × 25 = 10000
= 200 × 15 × 134 = 3000 × 134 = 402000 (c) (–400) × 125 × (–50) = (–400) × (–50) × 125 = 20000 × 125 = 2500000
Given Using commutative law of multiplication, 134 × 15 = 15 × 134
200 × 15 = 3000 Given Using commutative law of multiplication, 125 × (–50) = (–50) × 125
(–400) × (–50) = 20000 and (–) × (–) = (+)
172 A to Z of Mathematics
28. Associative Property of Numbers Associative law of addition9 Example 1 Rewrite (11 + 9) + 8 using the associative law of addition. Show that the rewritten expression yields the same answer. Solution Associative law of addition states that when we add any three numbers, the grouping (or association) of the numbers does not affect the sum. Procedure (11 + 9) + 8 = 20 + 8 = 28 11 + (9 + 8) = 11 + 17 = 28 Therefore, (11 + 9) + 8 = 11 + (9 + 8) = 28
Explanation Given The original expression yields an answer, 28. Regrouping 9 and 8 instead of 11 and 9 results in the same answer, 28.
Example 2 Rewrite [(–5) + (–7)] + (–3) using the associative law of addition. Show that the rewritten expression yields the same answer. Solution Procedure [(–5) + (–7)] + (–3)
Explanation Given
= [(–5) + (–7)] + (–3) The original expression yields an answer, –15. = (–12) + (–3) = –15 = (–5) + [(–7) + (–3)] Regrouping –7 and –3 instead of –5 and –7 results in = (–5) + (–10) the same answer, –15. = –15 Therefore, [(–5) + (–7)] + (–3) = (–5) + [(–7) + (–3)] = –15 Example 3 Solve the following using associative law of addition. (a) (32 + 87) + 113 (b) [1450 + (–1675)] + (–325) Solution Procedure (a) (32 + 87) + 113
9
Refer P 323
Explanation Given
The numbers we created 173
= 32 + (87 + 113) = 32 + 200 = 232 (b) [1450 + (–1675)] + (–325) = 1450 + [–(1675 + 325)] = 1450 – 2000 = –550
Using Associative law of addition.
87 + 113 = 200 Given Using Associative law of addition.
[–(1675 + 325)] = –1675 – 325 = –2000
Associative law of multiplication10 Example 1 Rewrite (50 × 200) × 25 using the associative law of multiplication, and show that the rewritten expression yields the same product. Solution The associative law of multiplication states that when we multiply any three numbers, the grouping of the numbers does not affect the product. Procedure
Explanation
(50 × 200) × 25 Given = 10000 × 25 The original expression yields a value of 250000. = 250000 50 × (200 × 25) Regrouping 200 and 25 instead of 50 and 200 results = 50 × 5000 in the same answer, 250000. = 250000 Therefore, (50 × 200) × 25 = 50 × (200 × 25) = 250000 Example 2 Rewrite [(–10) × (–200)] × (–24) using the associative law of multiplication, and show that the rewritten expression yields the same product. Solution Procedure [(–10) × (–200)] × (–24) = 2000 × (–24) = –48000
Explanation Given
(–) × (–) = (+) and (+) × (–) = (–) The original expression yields a value of –48000. (–10) × [(–200) × (–24)] (–) × (–) = (+) and (–) × (+) = (–) = (–10) × 4800 = –48000 Regrouping –200 and –24 instead of –10 and –200 results in the same answer, –48000. Therefore, [(–10) × (–200)] × (–24) = (–10) × [(–200) × (–24)] = –48000 Example 3 Solve the following using associative law of multiplication. (a) (43 × 75) × 40 (b) [13895 × (–4)] × (–250) 10
Refer P 325
174 A to Z of Mathematics Solution Procedure (a) (43 × 75) × 40 = 43 × (75 × 40) = 43 × 3000 = 129000 (b) [13895 × (–4)] × (–250) = 13895 × [(–4) × (–250)] = 13895 × 1000 = 13895000
Explanation Given Using associative law of multiplication
75 × 40 = 3000 Given Using associative law of multiplication
[(–4) × (–250)] = 4 × 250 = 1000
29. Distributive Property of Numbers Distributive law of multiplication over addition11
Example 1 Solve 46 × 101 using the distributive law and show how it simplifies multiplication. Solution Distributive law of multiplication over addition states that multiplying the sum of two or more addends by a number will give the same product as multiplying each addend individually by the number and then adding the products together. Procedure 46 × 101 = 46 × (100 + 1) = (46 × 100) + (46 × 1) = 4600 + 46 = 4646
Explanation Given We can write 101 as (100 + 1). Using the distributive law.
Example 2 Solve 999 × 13 + 1 × 13 using the distributive law. Solution Procedure 999 × 13 + 1 × 13 = (999 × 13) + (1 × 13) = 13 × (999 + 1) = 13 × 1000 = 13000
Explanation Given Adding brackets to reorganise the expression. Taking 13 as common.
Example 3 Solve (–30) × (–1005) using the distributive law and show how it simplifies multiplication.
11
Refer P 326
The numbers we created 175 Solution Procedure –30 × –1005 = –30 × [(–1000) + (–5)] = [(–30) × (–1000)] + [(–30) × (–5)] = 30000 + 150 = 30150
Explanation Given We can write –1005 as [(–1000) + (–5)]. Using the distributive law
(–) × (–) = (+) On multiplying –30 × –1000, we get 30000 and on multiplying –30 × –5, we get 150.
Example 4 Solve (–94) × 122 + (–6) × 122 using the distributive law. Solution Procedure (–94) × 122 + (–6) × 122 = [(–94) × 122] + [(–6) × 122] = 122 × [(–94) + (–6)] = 122 × (–100) = –12200
Explanation Given Reorganise the expression using brackets. Taking 122 as common.
(+) × (–) = (–)
Distributive law of multiplication over subtraction12 Example 1 Solve 150 × 199 using the distributive law and show how it simplifies multiplication. Solution Distributive law of multiplication over subtraction states that subtract the numbers and then multiply or multiply and then subtract. Procedure 150 × 199 = 150 × (200 – 1) = (150 × 200) – (150 × 1) = 30000 – 150 = 29850
Explanation Given We can write 199 as (200 – 1). Using the distributive law.
Example 2 Solve 101 × 142 – 1 × 142 using the distributive law. Solution Procedure 101 × 142 – 1 × 142 = (101 × 142) – (1 × 142) 12
Refer P 326
Explanation Given Reorganise the expression using brackets.
176 A to Z of Mathematics
= 142 × (101 – 1) = 142 × 100 = 14200
Taking 142 as common.
Example 3 Solve –250 × –1998 using the distributive law and show how it simplifies multiplication. Solution Procedure –250 × –1998 = –250 × [–(2000 – 2)] = –250 × [–2000 + 2] = (–250 × –2000) + (–250 × 2) = 500000 – 500 = 499500
Explanation Given We can write –1998 as [–(2000 – 2)]. Using the distributive law.
(–) × (–) = (+) and (–) × (+) = (–).
Example 4 Solve (–120) × (–101) – (–1) × (–120) using the distributive law. Solution Procedure (–120) × (–101) – (–1) × (–120) = [–(120) × (–101)] – [(–1) × (–120)] = (–120) × [(–101) – (–1)] = –120 × (–101 + 1) = –120 × (–100) = 12000
Explanation Given Reorganise the expression using brackets. Taking –120 as common.
(–) × (–) = (+)
30. Additive Identity or Zero Property Additive identity law of positive and negative numbers13 Example 1 What is the additive identity of 121? Solution When zero is added to a number we get the number itself as sum. 0 is an additive identity. Procedure 121 + 0 = 121 Therefore, 0 is the additive identity of 121. Example 2 What is the additive identity of –5362? 13
Refer P 327
Explanation Adding 0 to 121. On adding 0 to 121, we get 121 again.
The numbers we created 177 Solution Procedure –5362 + 0 = –5362 Therefore, 0 is the additive identity of –5362.
Explanation Adding 0 to –5362. On adding 0 to –5362, we get –5362 again.
31. Multiplicative Identity Property Multiplicative identity law of positive and negative numbers14 Example 1 What is the multiplicative identity of 1521? Solution The multiplicative identity of a number is 1 as any number multiplied by 1, gives the same result as the number itself. Procedure
Explanation 1521 × 1 Multiplying 1521 with 1. = 1521 On multiplying 1521 and 1, we get 1521 again. Therefore, 1 is the multiplicative identity of 1521. Example 2 What is the multiplicative identity of –329? Solution Procedure
Explanation –329 × 1 Multiplying –329 with 1. = –329 On multiplying –329 and 1, we get –329 again. Therefore, 1 is the multiplicative identity of –329.
32. Inverse Property Additive inverse property of positive and negative numbers15 Example 1 What is the additive inverse of 3026? Solution Additive inverse means changing the sign of the number and adding it to the original number to get zero as a sum. Procedure
Explanation 3026 + (–3026) Adding –3026 to 3026. =0 On adding –3026 to 3026, we get 0. Therefore, –3026 is the additive inverse of 3026. 14 15
Refer P 327 Refer P 327
178 A to Z of Mathematics Example 2 What is the additive inverse of –7865? Solution Procedure
Explanation –7865 + (7865) Adding 7865 to –7865. =0 On adding 7865 to –7865, we get 0. Therefore, 7865 is the additive inverse of –7865.
Multiplicative inverse property of positive and negative numbers16 Example 1 What is the multiplicative inverse of
1 ? 365
Solution Multiplicative inverse is a reciprocal of a number, which when multiplied to the number gives 1 as the product. Procedure
Explanation 1 1 × 365 Multiplying 365 with . 365 365 1 =1 On multiplying 365 with , we get 1. 365 1 Therefore, 365 is the multiplicative inverse of . 365 Example 2 What is the multiplicative inverse of –8965? Solution Procedure –8965 ×
1 –8965
=1 Therefore,
16
Refer P 327
Explanation Multiplying –8965 with
1 . –8965
On multiplying –8965 with 1 is the multiplicative inverse of –8965. –8965
1 , we get 1. –8965
The numbers we created 179
33. Zero Property of Multiplication Zero property of multiplication of positive and negative numbers17 Example 1 Solve the following. (a) 87 × ? = 0
(b) –658 × 0 = ?
Solution Zero property of multiplication states that the product of any number and zero, is zero. Procedure (a) 87 × ? = 0 ⇒?=0 (b) –658 × 0 = ? ⇒ ? = –658 × 0 = 0
Explanation Given By zero property Given By zero property
34. Closure Property of Numbers
Closure property of addition of whole numbers and integers18 Example 1 With the given questions, show that whole numbers are closed under addition. (a) 7 + 9 =16 (b) 6 + 2 = 8 Solution Procedure
Explanation
(a) 7 + 9 = 16 Given LHS = 7 + 9 7 and 9 are whole numbers. RHS = 16 16 is also a whole number. Therefore, 7 + 9 = 16 shows that whole numbers are closed under addition because the addends and the sum are whole numbers. (b) 6 + 2 = 8 Given LHS = 6 + 2 6 and 2 are whole numbers. RHS = 8 8 is also a whole number. Therefore, 6 + 2 = 8 shows that whole numbers are closed under addition because the addends and the sum are whole numbers. Closure property of addition of whole numbers (or integers) states that the sum of two or more whole numbers (or integers) is always a whole number (or integer). Example 2 With the given questions, show that integers are closed under addition. (a) (–58) + (–12) = –70 (b) (–15) + 122 = 107 17 18
Refer P 328 Refer P 327
180 A to Z of Mathematics Solution Procedure Explanation (a) (–58) + (–12) = –70 Given LHS = (–58) + (–12) –58 and –12 are integers. RHS = –70 –70 is also an integer. Therefore, (–58) + (–12) = –70 shows that integers are closed under addition because the addends and the sum are integers. (b) (–15) + 122 = 107 Given LHS = (–15) + 122 –15 and 122 are integers. RHS = 107 107 is also an integer. Therefore, (–15) + 122 = 107 shows that integers are closed under addition because the addends and the sum are integers.
Closure property of subtraction of integers19 Example 1 With the given questions, show that integers are closed under subtraction. (a) (–95) – (–12) (b) (–100) – (–112) Solution According to closure property of subtraction, the difference between any two integers will always be an integer. Procedure (a) (–95) – (–12)
Explanation Given –95 and –12 are integers.
= –95 + 12 (–) × (–) = (+) = –83 –83 is also an integer. Thus, integers are closed under subtraction because the minuend, subtrahend, and difference are integers. (b) (–100) – (–112) Given –100 and –112 are integers. = –100 + 112 (–) × (–) = (+) = 12 12 is also an integer. Thus, integers are closed under subtraction because the minuend, subtrahend, and difference are integers. Whole numbers are not closed under subtraction. For example: 3 – 5 = –2, here 3 and 5 are whole numbers but –2 is an integer. Therefore, whole numbers are not closed under subtraction.
19
Refer P 327
The numbers we created 181
Closure property of multiplication of whole numbers and integers20 Example 1 Using the following expressions, show that whole numbers are closed under multiplication. (a) 8 × 15 (b) 135 × 12 Solution Closure property of multiplication of whole numbers (or integers) states that the product of whole numbers (or integers) will always be a whole number (or integer). Procedure
Explanation
(a) 8 × 15
Given 8 and 15 are whole numbers. = 120 120 is also a whole number. As the multiplier, multiplicand, and product are whole numbers, thus, whole numbers are closed under multiplication. (b) 135 × 12 Given 135 and 12 are whole numbers. =1620 1620 is also a whole number. As the multiplier, multiplicand, and product are whole numbers, thus, whole numbers are closed under multiplication. Example 2 With the given questions, show that integers are closed under multiplication. (a) (–105) × (–13) (b) (–60) × (30) Solution Procedure (a) (–105) × (–13)
Explanation
Given –105 and –13 are integers. = 1365 1365 is also an integer. As the multiplier, multiplicand, and product are integers, thus, integers are closed under multiplication. (b) (–60) × (30) Given –60 and 30 are integers. = –1800 –1800 is also an integer. As the multiplier, multiplicand, and product are integers, thus, integers are closed under multiplication.
20
Refer P 327
O O for Organising (data)
Numericalising geometry, exploring coordinate systems, graphs 35. Cartesian System: 1-Dimensional space Representing situations using integers1 Example 1 Represent the following situation with the help of integers. (a) A temperature which is 16 degrees below zero (b) An altitude, 550 feet above sea level Solution Procedure Explanation (a) A temperature which is 16 degrees below zero can Temperatures below zero are represented by negative integers. be written as −16 degrees. (b) 550 feet above sea level can be written as + 550 feet. An altitude, above sea level is represented by a positive integer.
Plotting points in 1-dimensional space2 Example 1 Which alphabet best represents −25? A
D
−52
C
B
0
Solution Explanation Procedure From the image, B best represents −25 as −25 is almost −25 is greater than −52. So, it lies to the right half of −52 and B is almost in the middle of 0 and −52. of −52 on the given number line. Example 2 Order the following integers from least to the greatest and plot them on a number line. 35, 48, −12, 0, 5, 10 Solution The smallest integer = −12; The largest integer = 48 Comparing the integers, and ordering them from the least to the greatest, we get −12 < 0 < 5 < 10 < 35 < 48 1 2
Refer P 308 Refer P 336
182
Numericalising Geometry 183 −12 −50 −40 −30 −20 −10
0 5 10 0
10
35 20
30
48 40
50
36. Cartesian System: 2-Dimensional space Abscissa, ordinates and quadrants3 Example 1 Identify the abscissa and ordinate in the following coordinates and find the quadrant to which the following points belong: K (1, 3); L (−4, −5); M (−3, 2); N (4, −2); O (0, 0) Y Solution A distance of a point from y-axis scaled with the x-axis is called abscissa or II (− , +) x-coordinate of the point. A distance of a point from x-axis scaled with the y-axis is called ordinate or y-coordinate of the point. Quadrant is an area contained by the x and y axes. There are four quadrants in III (− , −) a graph.
Point K (1, 3) L (−4, −5) M (−3, 2) N (4, −2) O (0, 0)
Abscissa 1 −4 −3 4 0
Ordinate 3 −5 2 −2 0
Plotting ordered pairs in quadrants4 Example 1 Plot the ordered pairs (3, 3), (−2, 4), and (0, −3) in a coordinate plane.
3 4
Refer P 340 Refer P 345
I (+ , +) X IV (+ , −)
Quadrant I III II IV Origin
184 A to Z of Mathematics Solution Visual Representation Y 5
(−2, 4)
4 (3, 3)
3 2 1
−5 −4 −3 −2 −1 −1
0
1
2
3
4
5 X
−2 −3
(0 −3)
−4 −5
Explanation Draw a coordinate plane with four quadrants. To plot the point (3, 3), begin at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y-direction. To plot the point (−2, 4), begin at the origin. The x-coordinate is –2, so move two units to the left of origin. The y-coordinate is 4, so move four units up in the positive y-direction. To plot the point (0, −3), begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y-direction.
Example 2 Plot the ordered pairs (0, 0), (–1, –4), (4, 0), and (3, –5) in a coordinate plane. Solution Visual representation Y 5 4 3 2 1
−5 −4 −3 −2 −1 −1
(4, 0)
(0, 0) 0
1
2
3
−2 −3 −4
(−1, −4)
−5
(3, −5)
4
5 X
Explanation Draw a coordinate plane with four quadrants. (0, 0) is the origin. Plot (0, 0) at the centre where two axis meet. To plot the point (–1, –4), begin at the origin. The x-coordinate is –1, so move 1 unit to the left of origin. The y-coordinate is –4, so move four units down in the negative y-direction. To plot the point (4, 0), begin at the origin. The x-coordinate is 4, so move 4 units to the right of origin. The y-coordinate is 0. This tells us not to move in either direction along the y-axis. To plot the point (3, –5), begin at the origin. The x-coordinate is 3, so move 3 units to the right of origin. The y-coordinate is –5, so move five units down in the negative y-direction.
Numericalising Geometry 185
37. Graph – A Cartesian Plane Dependent and independent variables in graph5 Example 1 Assume, you have to make a graph that represents money earned from a business of car wash. Identify the independent and dependent variables in this situation. Also, determine the axis on which these variables should be plotted. Solution Procedure Explanation The revenue is the dependent variable and the The revenue, or money earned depends upon number of cars is the independent variable. The the number of cars washed. revenue is plotted on the y-axis, and the number of cars washed is plotted on the x-axis. In graphs, there are two variables − Independent and dependent. Dependent variable is a variable whose quantity (value) varies with changes in the values of another (independent) variable. Typically the independent variable is represented on the x-axis and the dependent variable on the y-axis. Example 2 Identify the dependent and independent variables in the following situations. (a) Anisha wants to figure out which brand of microwave popcorn pops the most kernels so she can get the most value for her money. (b) Surface temperature is high during the afternoon. But in night, the temperature of the surface lowers down. (c) The number of toys Avy finds depends on how many hours she spends on looking for them. Solution Independent variable (a) Brand of popcorn bag is the independent variable because Anisha is actually deciding the popcorn brand. (b) Afternoon and night are the independent variables. (c) Number of hours is the independent variable as it depends on Avy that how much time she wants to spend looking for the toys.
Dependent variable Number of kernels popped is the dependent variable as it depends on the popcorn brand. The temperature is the dependent variable as it depends on what time of the day it is measured. Number of toys is the dependent variable because their number depends on the number of hours Avy spends looking for them.
Plotting points (or ordered pairs) to make graphs6 Example 1 The population of a small village is recorded every 10 years. Draw a line graph to show this data.
5 6
Refer P 353 Refer P 354
186 A to Z of Mathematics
Year Population (100000s)
1970 10.0
1980 10.69
1990 35.50
2000 42.10
2010 67.71
Solution Visual Representation
Explanation
Y 70
(2010, 67.71)
60
50 (2000, 42.10)
40 Population(100000s)
Passage of time is independent variable, hence ‘years’ would be on x axis. Population changes is dependent on passage of time, hence population (100000s) would be on y axis. Plot points correctly and join them with straight lines.
(1990, 35.50) 30 20
10 0
(1970, 10) 1970
(1980, 10.69) 1980
1990 Years
2000
2010
X
Example 2 Mobile phones sold by a shop in a certain week are as follows: Days Number of mobile phones sold
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
45
20
33
15
4
40
Numericalising Geometry 187
Solution Visual representation Y
50
(Mon, 45)
Number of mobile phones sold
45
(Sat, 40)
40 35
(Wed, 33)
30 25
(Tues, 20)
20
Explanation Passage of time (in days) is an independent variable, hence ‘Days’ would be on x-axis. ‘Number of phones sold’ is dependent on the ‘day’ of the week, hence it is a dependent variable, and plotted on y-axis. Plot points correctly and join them with straight lines.
(Thurs, 15)
15 10 5
(Fri, 4)
0
Mon
Tues
Days
Wed
Thurs
Fri
Sat X
Finding distance between two points7 Example 1 Find the distance between the points P (−6, 7) and Q (−1, −5). Solution Procedure x1 = −6, y1 = 7 and x2 = −1, y2 = −5 PQ =
(x 2 − x1 )2 + (y 2 − y 1 )2
Explanation The coordinates of the points P (x1, y1) = P (−6, 7) and Q (x2, y2) = Q (−1, −5) are given.
Distance between any two points =
( Difference of abscissa ) 2 2 ⇒ PQ = [–1 – (–6)] + (–5 – 7)
⇒ PQ = ⇒ PQ =
( 5 ) + ( −12 ) 2
2
25 + 144 = 169 = 13 units
Example 2 Find a point on x-axis, which is equidistant from A (2, −5) and B (−2, 9). Refer P 359
+ ( Difference of ordinates )
Putting the values of the respective coordinates.
Hence, the distance between points P and Q is 13 units.
7
2
2
188 A to Z of Mathematics Solution We know that a point on x-axis is of the form (x, 0), as it is the distance from the x axis, i.e. y − coordinate is always zero. So, let P (x, 0) be the point equidistant from A (2, −5) and B (−2, 9). Then, PA = PB. Procedure PA = PB ⇒
( x − 2)
2
+ ( 0 − ( − 5 )
2
Explanation Since the distance between any two points = (Difference of abscissa)2 + (Difference of ordinates)2
2 2 = [x – (–2)] + (0 – 9) ⇒ (x − 2)2 + 25 = (x + 2)2 + 81 Using (a − b)² = a² + b² − 2ab and (a + b)² = a² + b² + 2ab ⇒ x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81 Take all the terms consist of x2 and x along LHS and all the ⇒ x2 − x2 − 4x − 4x = 4 + 81 − 4 − 25 constants on RHS. ⇒ −8x = 56 ⇒ x = −7 Hence, the coordinates of point P is (−7, 0).
Example 3 If the distance between the points (7, –8) and (3, a) is 5, find the value of a. Solution Procedure x1 = 7, y1 = –8 and x2 = 3, y2 = a (x 2 – x1 )2 + (y 2 – y 1 )2 = 5 2 2 ⇒ (3 – 7) + [a – (–8)] = 5
Explanation The coordinate of two points are given. Distance between any two points = (difference of abscissa )2 + (difference of ordinates )2
Putting the values of the respective coordinates.
⇒ (–4)2 + (a + 8)2 = 5 ⇒ 16 + (a + 8)2 = (5)2 ⇒ (a + 8)2 = 25 – 16 ⇒ (a + 8)2 = 9 ⇒ (a + 8)2 = 32 ⇒a+8=3 ⇒ a = 3 – 8 = –5 Hence, the value of a is –5.
Defining and using section formula8 Example 1 Find the coordinates of the point, which divides the line segment joining the points A (6, 3) and B (−4, 5) in the ratio 3:2 internally.
8
Refer P 360
Numericalising Geometry 189
Solution Section formula is used to find the coordinates of a point that divides a line segment (externally or internally) into some ratio. Here, we use the section formula to find the coordinates of the point that divides the line segment in the ratio 3:2. Procedure
Explanation
Let P (x, y) be the required point that divides the line Assumption segment in the ratio 3:2 internally. The coordinates of the point P which divides the line segment joining the points A(x1, y1) and B(x2, y2) mx 2 + nx1 my 2 + ny 1 ,y = internally in the ratio m:n are given by x = m+n m+n m = 3 and n = 2 x1 = 6, y1 = 3; x2 = −4, y2 = 5
As it is given that the point P divides the line in the ratio 3:2. ∵ Two points A(6, 3) and B (−4, 5) are given.
(3)(− 4) +(2)(6) (3)(5) +(2)(3) and y = 3+2 3+2 −12 + 12 15 + 6 and y = ⇒x= 5 5 21 ⇒ x = 0 and y = 5
Putting the given values in the above formula.
21 So, the coordinates of P are 0, . 5
B (−4, 5)
x=
P 0,
21 5
A (6, 3)
When a point on a line segment divides it into the ratio m:n, it is said to be an internal point. When a line segment is extend to a point such that the new line segment and extended part are in ratio m:n, the point is said to be an external point. Example 2 A (4, 5) and B (6, –1) are two given points and the point Y divides the line-segment AB externally in the ratio 5:3. Find the coordinates of Y. Solution Procedure
Explanation
Given ∴ Y(x, y) is the required point that divides the line segment in the ratio 5:3 externally. As the coordinates of the point Y which divides the line segment joining the points A (x1, y1) and mx 2 − nx1 my 2 − ny 1 ,y = B(x2, y2) externally in the ratio m:n are given by x = m−n m−n m = 5 and n = 3
As it is given that the point Y divides the line segment in the ratio 5:3.
x1 = 4, y1 = 5; x2 = 6, y2 = −1.
Two points A(4, 5) and B(6, −1) are given.
190 A to Z of Mathematics
(5)(6) − (3)( 4) (5)(−1) − (3)(5) and y = 5−3 5−3 30 − 12 −5 − 15 ⇒x = and y = 2 2 ⇒ x = 9 and y = −10 The coordinates of Y are (9, −10).
x=
Putting the given values in the above formula.
A (4, 5)
B (6, –1) 5
3
Y (9, –10)
The point which divides the line segment externally in the ratio m:n lies outside the line segment and the line segment has to be extended to that point to get this ratio. Example 3 Determine the ratio in which the line 3x + y − 9 = 0 divides the segment joining the points (1, 3) and (2, 7). Solution Procedure Let the line 3x + y − 9 = 0 divides the line segment Assumption joining A (1, 3) and B (2, 7) in the ratio k:1 at point C.
Explanation
3x + y − 9 = 0 k A (1, 3)
mx 2 + nx1 my 2 + ny 1 ,y = x = m+n m+n
x= y=
mx 2 + nx1 m+n my 2 + ny 1 m+n
= =
2k + 1 k +1 7k + 3 k +1
1 C (x, y)
B (2, 7)
Using the section formula to find the coordinates of a point which divides a line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m:n. Here, m:n = k:1 And, x1 = 1, y1 = 3; x2 = 2, y2 = 7 Putting the above values in the section formula to find coordinates (x, y) of point C.
When two lines intersect, then the coordinates of their common point can be put in the equation of a given line to find their unknown value.
Numericalising Geometry 191
3x + y − 9 = 0
Equation of a line. (Given)
2k + 1 7k + 3 −9= 0 + k +1 k +1
Putting the values of coordinates of point C in the equation of line.
2k + 1 7k + 3 (9k + 9) − = 0 + k +1 k +1 k +1
LCM of (k + 1) and 1 is (k + 1).
⇒ 3
⇒ 3
⇒ 6k + 3 + 7k + 3 − 9k − 9 = 0 ⇒ 13k − 9k = 9 − 6 3 ⇒k= 4 Ratio assumed 3 k:1 = : 1 = 3:4 4 So, the required ratio in which the line 3x + y − 9 = 0 divides the given line segment internally is 3:4.
Mid-point formula9 Example 1 Find the mid-point of a line whose endpoints are (4, 5) and (6, 7). Solution We use the section formula to find the coordinates of the mid-point that divides the line segment in the ratio 1:1. Procedure Let (x, y) be the coordinates of a mid-point of a line whose endpoints are (x1, y1) = (4, 5) and (x2, y2) = (6, 7). x1 + x 2 y 1 + y 2 , 2 2
Mid-point (x, y) =
Explanation Assumption Mid-point formula
Putting the values of x1 = 4, x2 = 6, y1 = 5, and 4+6 5+7 y2 = 7 in the mid-point formula. ⇒ ( x, y ) = , 2 2 10 12 ⇒ ( x, y ) = , 2 2 ⇒ (x, y) = (5, 6) Hence, (5, 6) is the required coordinates of the mid-point of a line whose endpoints are (4, 5) and (6, 7).
Example 2 Prove that the diagonals of a rectangle are equal and bisect each other.
Solution Calculate the length of both the diagonals of a rectangle using distance formula and then compare. Then, find out the coordinates of mid-point of both the diagonals. If the coordinates of mid-point are same, then the diagonals bisect each other. 9
Refer P 364
192 A to Z of Mathematics
Procedure Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a units and OB = b units. Then, the coordinates of A and B are (a, 0) and (0, b) respectively.
Explanation Y
b (0,0) O
OA = BC = a units AC = OB = b units So, the coordinates of C are (a, b). x1 + x 2 y 1 + y 2 , 2 2
(x, y) =
C (a,b)
(0,b) B
a
A (a,0)
X
Since OACB is a rectangle. OA = a and AC = b
Mid-point formula
The coordinates of the mid-point of OC are Putting the values of the coordinates of O (x1, y1) = (0, 0) 0 + a 0 + b a b ....(i) and C (x2, y2) = (a, b) in the mid-point formula. , = ,
2
2
2
2
x1 + x 2 y 1 + y 2 , 2 2
(x, y) =
Mid-point formula
The coordinates of the mid-point of AB are Putting the values of the coordinates of A (x1, y1) = (a, 0) and B (x2, y2) = (0, b) in the mid-point formula. a + 0 b + 0 a b ....(ii) , = , 2 2 2 2 OC and AB bisect each other. OC =
AB =
a 2 + b2 units
(a − 0) + (0 − b) 2
2
= a 2 + b2 units
The coordinates of the mid-point of OC and AB are same. [From eq. (i) and eq. (ii)] Using distance formula, distance between any two points
= (Difference of abscissa)2 + (Difference of ordinates)2 Using distance formula, distance between any two points 2 2 = (Difference of abscissa) + (Difference of ordinates)
Diagonals of a rectangle are equal.
OC = AB =
a 2 + b2
Hence, the diagonals of a rectangle are equal and bisect each other.
Centroid of a triangle – application of section formula10 Example 1 Calculate the centroid of a triangle whose vertices are (15, 15), (22, 42), and (47, 12). 10
Refer P 364
Numericalising Geometry 193
Solution Procedure (x1, y1) = (15, 15) (x2, y2) = (22, 42) (x3, y3) = (47, 12) +2 x+2 x +3 x 3 y 1 y+1 y+2 y+2 y+3 y3 x x+1 x =, = 3 3 3 3
1 C=
15 + 22 + 47 15 + 42 + 12 , 3 3
⇒C=
Explanation Coordinates of vertices of a triangle (Given)
The formula for the coordinates of centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3). Putting the values of the coordinates in the formula.
84 69 ⇒C= , 3 3 ⇒ C = (28, 23) Hence, the coordinates of the centroid of a triangle is (28, 23). The point where the three medians of a triangle intersect is known as the centroid of a triangle. Example 2 Two vertices of a triangle are (4, –5) and (–7, 4). If its centroid is (2, –1), find the third vertex. Solution Procedure
Explanation
Let the coordinates of the third vertex of a Assumption triangle be (x1, y1). Coordinates of two vertices of a triangle are (x2, y2) = (4, –5) and (x3, y3) = (–7, 4). x1 + x 2 + x 3 y 1 + y 2 + y 3 , 3 3
⇒
x1 + 4 − 7 y −5+4 = 2 and 1 = −1 3 3
⇒ x1 – 3 = 6 and y1 – 1 = –3 ⇒ x1 = 9 and y1 = –2
Given The formula for the centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3). The coordinates of centroid = (2, –1) [given] Simplifying the equations.
Thus, the coordinates of the third vertex of a triangle are (9, –2).
Area of a triangle11 Example 1 Find the area of a triangle whose vertices are A (3, 2), B (11, 8), and C (8, 12).
11
Refer P 367
194 A to Z of Mathematics Solution When the coordinates of three vertices of a triangle are given, we can calculate the area of a triangle using determinants. x1 y 1 1 1 Area of traingle with vertices (x1, y1), (x2, y2), (x3, y3) = x 2 y 2 1 2 x3 y 3 1 =
1 x1 ( y 2 – y 3 ) + x 2 ( y 3 – y 1 ) + x 3 ( y 1 – y 2 ) 2
Procedure A (3, 2) = A (x1, y1) B (11, 8) = B (x2, y2) C (8, 12)= C (x3, y3) 11 Area of ΔABC = [x1 (y 2 − y 3 ) + x 2 (y 3 − y 1 ) + x 3 (y 1 − y 2 )] 22 1 ⇒ Area of ΔABC = [3(8 – 12) + 11(12 – 2) + 8(2 – 8)] 2 1 ⇒ Area of ΔABC = [(–12 + 110 – 48)] = 25 square units 2 Hence, the area of the triangle is 25 square units.
Explanation Coordinates of the vertices of a ΔABC. (Given) Formula of area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3). Substituting the values of the coordinates in the formula. On solving
Example 2 Find the area of the triangle formed by joining the mid-points of sides of the triangle whose vertices are (0, –1), (2, 1), and (0, 3). Find the ratio of the area of the triangle formed to the area of the given triangle. Solution Procedure Let A (0, –1), B (2, 1), and C (0, 3) be the vertices of ΔABC. Let D, E, F be the mid-points of sides BC, CA, and AB respectively.
Explanation D (1, 2)
(0, 3) C
F (1, 0)
(0, 1) E
A (0, –1)
x + x2 y1 + y 2 (x, y) = 1 , 2 2
Mid-point formula
B (2, 1)
Numericalising Geometry 195
The coordinates of D, E, and F can be calculated using the mid-point formula.
2 + 0 1+ 3 D= , = (1, 2) 2 2 0 + 0 –1 + 3 E= , = (0, 1) 2 2 0 + 2 –1+1 = (1, 0) F = , 2 2
∴ Área of ΔABC
Area of a triangle
Similarly, Area of ΔDEF
Area of a triangle
1 1 0 (1 – 3 ) + 2 ( 3 – ( –1) ) + 0 ( –1 – 1) = x1 ( y 2 – y 3 ) + x 2 ( y 3 – y 1 ) + x 3 ( y 1 – y 2 ) 2 2 1 = [[0 0+8+ + 0] 0] ==44sq. sq. units .....(i ) ....(i) Here, x1 = 0, x2 = 2, x3 = 0, y1 = –1, y2= 1, y3 = 3 2 =
1 = 1 (1 – 0 ) + 0 ( 0 – 2 ) + 1 ( 2 – 1) 2 1 = (1+1) = 1sq.units 1 sq.units ....(ii ) 2
=
....(ii)
1 x1 ( y 2 – y 3 ) + x 2 ( y 3 – y 1 ) + x 3 ( y 1 – y 2 ) 2
Here, x1 = 1, x2 = 0, x3 = 1, y1 = 2, y2= 1, y3 = 0
Area of ΔDEF : Area of ΔABC = 1:4 From eq. (i) and eq. (ii). Hence, the required ratio of the area of ΔDEF to area of ΔABC is 1:4.
Visualising collinearity of three points12 Example 1 Show that the points (0, –2), (2, 4), and (–1, –5) are collinear. Solution We know that three points are said to be collinear if they lie in a straight line. And, if the points lie in a straight line, then the area of the triangle formed by the three points is zero. Procedure
1 [x (y – y ) + x2(y3 – y1) + x3(y1 – y2)] = 0 2 1 2 3
The area of the triangle formed by joining the given points 1 = {0[4 – (–5)] + 2[–5 – (–2)] + (–1) (–2 – 4)} 2 1 = [(0 + 2(–5 + 2) – (1) (–6)] 2 1 = [ 0 – 6 + 6] 2 1 = (–6 + 6) 2 =0
Explanation Required condition of collinearity. Substituting the values of the coordinates x1 = 0, y1 = –2, x2 = 2, y2 = 4, x3 = –1, y3 = –5 in the given formula.
On solving, the area comes out to be zero.
Since the area of the triangle formed by joining the given three points is zero, therefore the given points are collinear. 12
Refer P 368
196 A to Z of Mathematics Example 2 Under what conditions the points (a, b), (b, a), and (a2, –b2) be in a straight line. Solution Since the three given points are in a straight line, hence, the area of the triangle formed by the three points must be zero. Procedure 1 [x (y – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 2 1 2
1 ⇒ {a [a – (–b2)] + b (–b2– b) + a2(b – a)} = 0 2 1 ⇒ [a (a + b2) – b3 – b2 + a2b – a3] = 0 2 1 ⇒ [a2 + ab2 – b3 – b2 + a2b – a3] = 0 2
⇒ a2 – b2 – (a3 + b3) + ab (a + b) = 0
⇒ (a + b) (a – b) – [(a + b) (a2 – ab + b2)] + ab (a + b) = 0
Explanation Required condition of collinearity. Substituting the values of the coordinates x1 = a, y1 = b, x2 = b, y2 = a, x3 = a2, y3 = –b2 in the given formula
Grouping (a2 – b2), (a3 + b3) and (a + b) together. Using (a2 – b2) = (a – b) (a + b), (a3 + b3) = (a + b) (a2 – ab + b2) Taking (a + b) common from all.
⇒ (a + b) [a – b – (a2 – ab + b2) + ab] = 0 ⇒ (a + b) [a – b – a2 + ab – b2 + ab] = 0 ⇒ (a + b)[(a – b) – (a2 + b2 – 2ab)] = 0 Using (a – b)2 = a2 + b2 – 2ab ⇒ (a + b) [(a – b) – (a – b)2] = 0 ⇒ (a + b) [(a − b) – (a – b) (a – b)] = 0 Taking (a – b) common from all. ⇒ (a + b) (a – b) [1 – (a – b)] = 0 Therefore, either a + b = 0 Using: If xyz = 0, then either x = 0, y = 0, or, a – b = 0 z = 0 because if any of the value is ‘0’ then or, 1 – (a – b) = 0 the product xyz = 0. Hence, either a + b = 0, a – b = 0 or 1 – (a – b) = 0 for the points (a, b), (b, a) and (a2, –b2) to be in a straight line.
38. Application of Coordinate System – Statistical Graphs Collecting and organising data13 Example 1 The age of 20 students of Class VII in a school is given below. Organise the data in a frequency distribution table. 15, 16, 16, 14, 17, 16, 15, 15, 16, 16, 17, 16, 16, 14, 16, 14, 15, 16, 15, 14 Solution Procedure 15, 16, 16, 14, 17, 16, 15, 15, 16, 16, 17, 16, 16, 14, 16, 14, 15, 16, 15, 14 13
Refer P 373
Explanation Given, age of 20 students
Numericalising Geometry 197
The required frequency distribution table is: Age 14 15 16 17
Tally marks
Total
Frequency 4 5 9 2 20
We will use tally marks to organise this data. Tally marks are useful for counting or tallying ongoing results.
Data is a collection of observations such as numbers, words, or measurement. The number of times an observation occurs in a given data is known as the frequency of the observation (data/datum). And, we use tally marks to organise the large frequency data. Example 2 Data gives the number of children in 40 families is as following. 1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2 Represent it in the form of a frequency table. Solution Procedure
Explanation
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2
Given, number of children in 40 families
The required frequency distribution table is:
Using tally marks to organise the data in the frequency distribution table.
No. of children Tally marks Frequency 0
5
1
7
2
11
3
5
4
6
5
3
6
3 Total
40
Example 3 The weight of 40 students in a class is given below. The measurement is to the nearest kg. 55 60 75 49 52
70 48 64 66 76
57 58 65 62 71
73 54 57 76 61
55 69 71 61 53
59 51 78 63 56
64 63 76 63 67
72 78 62 76 71
198 A to Z of Mathematics
Construct a frequency table for the data using an appropriate scale. Solution It is difficult to write each weight separately in a frequency table as there will be too many rows in the table. So, we group the data (weight) into class intervals (or groups) to organise the data. Procedure Range of data = (78 – 48) kg = 30 kg Number of classes =
30 =6 5
Frequency table using the selected scale and intervals: Weight (kg) 47 – 52 53 – 58 59 – 64 65 – 70 71 – 76 77 – 82
Tally Marks
Frequency
Explanation Greatest weight = 78 kg, smallest weight = 48 kg We decide the class size to be a convenient number. Let class size = 5 kg Beginning the class intervals with 47 and ending it with 82. And, we use tally marks to organise the data.
4 8 11 5 10 2 40
Total
Making pictograph of data14 Example 1 The colours of Refrigerators preferred by people living in a locality are given by the following pictograph. Colour
Number of people
Blue Green Red White = 10 People What is the ratio of the number of people preferring white colour to the number of people preferring blue colour? Solution Each picture in a pictograph represent one or more physical object.
14
Refer P 379
Numericalising Geometry 199
Procedure Explanation The ratio of the number of people preferring The number of people preferring white white colour to the number of people preferring colour = 2 × 10 = 20 blue colour The number of people preferring blue colour = 20:45 = (4 × 10) + 5 (as it is half) = 45 = 4:9 Simplest from of ratio. A pictograph is a graph that shows numerical information using picture symbols. Example 2 The pictograph given below expresses the number of persons who travelled from Varanasi to New Delhi by Shiva Ganga Express on each day of a week. Sunday Monday Tuesday Wednesday Thursday Friday Saturday = 100 travellers Find the difference between the number of travellers travelling on the days when there are maximum travellers and minimum travellers. Solution The key is the quantity that the picture symbol represents. Here, the key is 100 travellers. So, for each calculation, multiply the number of symbols with 100. Procedure Number of travellers on Thursday = 10 × 100 = 1000 travellers
Explanation The day on which maximum number of travellers
Number of travellers on Wednesday = 4 × 100 = 400 travellers
The day on which minimum number of travellers
travelled is Thursday, it has 10 travelled is Wednesday, it has only 4
200 A to Z of Mathematics
Required difference = 1000 – 400 = 600 travellers Required difference = –
Number of travellers travelled on Thrusday
Number of travellers travelled on Wednesday
Hence, the required difference is 600 travellers.
Making bar graph from data15 Example 1 The following table gives the number of students in Class VI in a school during the academic years 2011–12 to 2015–16. Academic year Number of students
2011–12 50
2012–13 75
2013–14 125
2014–15 150
2015–16 200
Make a bar graph to represent the data. Solution To represent the above data using a bar graph, we first draw the horizontal and vertical axes. Since five values of the numerical data are given, we mark five points on the horizontal x axis at equal distances and erect bars of the same breadth at these points. The heights of the bars are proportional to the numerical values of the frequency (number of students) as given in the question. Y 200 175
Number of students
150 125 100 75 50 25 0 2011–12
2012–13 2013–14 Academic years
2014–15
2015–16 X
In the above graph, y axis, the scale on the left shows frequency marked from 0-200. All kinds of graphs must have equal gaps between intervals (it is 25 in this example). A bar graph is a graph that represents the data with rectangular bars with heights proportional to the values that they represent. Example 2 The following table gives the number of vehicles passing through a busy crossing in Delhi at different time intervals on a particular day.
15
Refer P 380
Numericalising Geometry 201
Time interval
8–9 hrs
Number of vehicles
300
9–10 hrs 10–11 hrs 400
11–12 hrs
350
12–13 hrs 13–14 hrs 14–15 hrs
250
200
150
100
Represent the above data by using a horizontal bar graph. Solution Visual Representation Y 14–15 hrs 13–14 hrs
Time Interval
12–13 hrs 11–12 hrs
Explanation Here, 7 values of the data are given. So, we mark 7 points on the vertical axis at equal distances and erect bars of the same breadth, whose heights are as per the values of the frequency of the data given in the question. The vertical axis has labels such as ‘8−9 hrs’, ‘9−10 hrs’… as given in the data table. The horizontal axis has the scale as per the data that starts from 50 and goes up to 450 to cover the lowest and highest frequencies in the question.
10–11 hrs 9–10 hrs 8–9 hrs
0
50 100 150 200 250 300 350 400 450 X Number of Vehicles
Example 3 Answer the following questions after reading the bar graph. (a) What information is given by this bar graph? (b) Which state has maximum coal production in 2019–20? (c) Name the state having the same coal production in both the years? (d) Which state has minimum coal production?
70
30 20 10 0
Bihar
U.P.
States
Kerela
2019–20
2018–19 2019–20
40
2018–19
2018–19
50
2019–20
60
2019–20
80
2018–19
Coal production (million tonnes)
90
2018–19
Y 100
2019–20
202 A to Z of Mathematics
A.P.
M.P.
X
Solution (a) The given bar graph represents data about coal production in million tonnes over two consecutive years, namely 2018–19 and 2019–20 in various states. (b) M.P. has maximum coal production in 2019–20. (c) U.P. has the same coal production in both the years. (d) U.P. has minimum coal production.
Making histogram of data16 Example 1 Construct a histogram for the following frequency distribution table that shows the time spent by the children while playing outdoor games. Time spent (minutes) No. of students (frequency)
0–40 4
40–80 10
80–120 16
120–160 22
Solution We represent the class limits along x-axis on a suitable scale and the frequencies along y-axis on a suitable scale. Now, we construct rectangles to obtain the histogram of the given frequencies as shown in the figure.
16
Refer P 387
Numericalising Geometry 203
Y 24 20 16
No. of students
12 8 4 0
40
80 Time spent (minutes)
160 X
120
A Histogram is a bar graph – like representation that shows continuous data that are measured on an interval scale. Example 2 The histogram below shows the height (cm) distribution of 30 people. Y 10 9 8 7
Frequency
6 5 4 3 2 1 0
139.5
149.5
159.5
169.5
Height (cm)
179.5
189.5
199.5
X
204 A to Z of Mathematics
(a) How many people have height between 159.5 cm and 169.5 cm? (b) How many people have a height less than 159.5 cm? (c) How many people have a height more than 169.5 cm? (d) What percentage of people have height between 149.5 cm and 179.5 cm? Solution Procedure (a) 7 people have heights between 159.5 – 169.5 cm.
Explanation The rectangle corresponding to the interval 159.5 – 169.5 cm has a 7 as frequency. (b) Total number of people who have height less than Frequency of people with height between 159.5 cm 139.5 – 149.5 cm is 5 and frequency of people = 5 + 9 = 14 people with height between 149.5 – 159.5 cm is 9. (c) Total number of people with height more than 169.5 cm = 6 + 2 + 1 = 9 people
(d) The percentage of people with height between 149.5 – 179.5 cm Number of people with height between 149.5 – 179.5 cm = × 100 Total number of people =
22 × 100 = 73.33% ≈ 73% 30
Frequency of people with height between 169.5 – 179.5 cm is 6 and frequency of people whose heights between 179.5 – 189.5 cm is 2 and the frequency of people whose heights between 189.5 – 199.5 cm is 1. From the graph, frequency of people with height between 149.5 – 179.5 cm = Frequency of people with height between 149.5 – 159.5 cm + frequency of people with height between 159.5 – 169.5 cm + frequency of people with height between 169.5 – 179.5 cm = 9 + 7 + 6 = 22 people
Example 3 The histogram shows the cholesterol level (mg per dl) of 200 people. (a) How many people have cholesterol level between 205 and 210 mg per dl? (b) How many people have cholesterol level less than 205 mg per dl? (c) What percentage of people have cholesterol level more than 215 mg per dl? (d) What percentage of people have cholesterol level between 205 and 220 mg per dl?
Numericalising Geometry 205
Y 0.4 0.35 0.3 Relative frequency
0.25 0.2 0.15 0.1 0.05 0
195
200
205
210
215
220
225
X
Cholesterol level (mg per dl)
Solution Note that the relative frequency is shown on the vertical axis. Therefore, Relative frequency =
Number of people having particular cholesterol level Total number of people
(a) When the cholesterol level is between 205 – 210 mg per dl, relative frequency = 0.2 Total number of people = 200 Number of people having a cholesterol level between 205 – 210 mg per dl = 0.2 × 200 = 40 people (b) When the cholesterol level is between 195 – 200 mg per dl, relative frequency = 0.05 When the cholesterol level is between 200 – 205 mg per dl, relative frequency = 0.1 Number of people having cholesterol level less than 205 mg per dl = (0.05 + 0.1) × 200 = 30 people (c) When the cholesterol level is between 215 – 220 mg per dl, relative frequency = 0.25 When the cholesterol level is between 220 – 225 mg per dl, relative frequency = 0.05 Number of people having a cholesterol level more than 215 mg per dl = (0.25 + 0.05) × 200 = 0.30 × 200 = 60 people
From graph Given
∵ Number of people having a particular cholesterol level = Relative frequency × Total number of people From graph
∵ Number of people having a particular cholesterol level = Relative frequency × Total number of people From graph
∵ Number of people having a particular cholesterol level = Relative frequency × Total number of people
206 A to Z of Mathematics
Percentage of people having a cholesterol level more than 215 mg per dl 60 = ×100 = 30% 200
∵ Percentage = Particular cholesterol level people ×100 Total number of people
(d) When the cholesterol level is between From graph 205 – 210 mg per dl, relative frequency = 0.2 When the cholesterol level is between 210 – 215 mg per dl, relative frequency = 0.35 When the cholesterol level is between 215 – 220 mg per dl, relative frequency = 0.25 Number of people having a cholesterol level ∵ Number of people having a particular cholesterol between 205 – 220 mg per dl level = Relative frequency × Total number of = (0.2 + 0.35 + 0.25) × 200 people = 0.80 × 200 = 160 people Percentage of people having a cholesterol ∵ Percentage = level between 205 – 220 mg per dl Particular cholesterol level people ×100 160 = ×100 = 80% Total number of people 200 A relative frequency is the ratio of the frequency of a particular event to the total frequency. Example 4 Construct a histogram for the following frequency distribution: Heights (m) No. of Buildings
10–20 15
21–30 9
31–40 21
41–50 65
51–60 46
61–70 34
71–80 18
81–90 5
Solution Here, the class intervals are discontinuous. We cannot represent data with discontinuous class intervals in a histogram because histograms are made by placing bars next to each other without any gap, representing continuous data. Discontinuous class intervals would make bar graphs and not histograms. For representing discontinuous data in histograms, we convert the class intervals into the continuous form and then draw the histogram. To make the class intervals continuous, subtract 0.5 from the lower limit and add 0.5 to the upper limit. Now, we will construct the histogram with new class intervals, i.e., 9.5–20.5, 20.5–30.5, 30.5–40.5, and so on. Heights (m) No. of Buildings
9.5–20.5 20.5–30.5 30.5–40.5 40.5–50.5 50.5–60.5 60.5–70.5 70.5–80.5 80.5–90.5 15
9
21
65
46
34
18
5
Numericalising Geometry 207
Y 70
65
Number of Buildings
60 50
46
40
34
30
21
20
15
9
10 0
9.5
18
20.5
18 5
50.5 30.5 40.5 Heights (metre)
60.5
5 70.5
80.5
90.5 X
Making pie charts of data17 Example 1 Draw a pie diagram for the following data of the expenditure pattern in a family. Items Expenditure (₹)
Food 4000
Clothing 2000
Rent 1500
Education 1500
Miscellaneous 1000
Solution Pie charts represent data in a circle, with ‘slices’ corresponding to percentages or degrees. Step 1 Calculate the total expenditure of the family = ₹4000 + ₹2000 + ₹1500 + ₹1500 + ₹1000 = ₹10000
Step 2 Calculate the degrees of each sector of a pie chart using the formula, (Given data ÷ Total value of data) × 360°.
17
Items
Expenditure (₹)
Food
4000
4000 × 360° 10000
Clothing
2000
2000 × 360° = 72° 10000
Rent
1500
1500 × 360° = 54° 10000
Education
1500
1500 × 360° = 54° 10000
Refer P 390
Measure of central angles = 144°
208 A to Z of Mathematics
Miscellaneous
1000
Total
₹10000
Now, draw a circle of an appropriate radius. Construct sectors in the clockwise direction with descending order of magnitude of central angles.
1000 × 360° = 36° 10000
360°
Misc. 36° Education 54° 144° Food 54° Rent 72° Clothing
Each sector of a pie chart represents a proportion of the whole. The whole or 100% is represented by the 360° angle subtended by a circle. Thus, there are three main formulas in pie charts – (a) To calculate the percentage of the given data: (Frequency ÷ Total frequency) × 100 (b) To convert the data into degrees: (Given data ÷ Total value of data) × 360° (c) To convert the percentage into degrees: (Given percentage ÷ 100) × 360° Example 2 The favourite flavours of ice-cream among children in a locality is as follows. Flavours Vanilla Strawberry Chocolate Kesar pista Mango zap
% of children who prefer the flavours 25% 15% 10% 30% 20%
Draw a pie chart to represent the given information. Solution To draw the pie chart of favourite ice-cream flavours among children, we calculate the central angles of the various observations.
Flavours
% of children who prefer the flavours
Measure of central angles
Vanilla
25%
25 × 360o = 90o 100
Numericalising Geometry 209
Strawberry
15%
15 × 360o = 54 o 100
Chocolate
10%
10 × 360o = 36o 100
Kesar pista
30%
30 × 360o = 108o 100
Mango zap
20%
20 × 360o = 72o 100
Now, we shall represent these angles within a circle to obtain the required pie chart. Mango zap 20%
Vanilla 25% Strawberry 15%
Kesar pista 30%
Chocolate 10%
Favourite flavours of ice creams
39. Cumulative Frequency Distribution Tables and Curves Cumulative frequency table (ungrouped data)18 Example 1 The following table gives the frequency distribution of marks obtained by 30 students in a particular test. Marks No. of students
30 5
31 7
32 10
33 8
Construct a cumulative frequency table for the given data. Solution Cumulative frequency is the ‘running total’ of frequencies upto each class. Cumulative frequency distribution is a form of frequency distribution that represents the sum of frequency of a class and frequency of all classes above it. Marks 30 31 32 33 18
Refer P 394
No. of students 5 7 10 8
Cumulative Frequency 5 5 + 7 = 12 5 + 7 + 10 = 22 5 + 7 + 10 + 8 = 30
210 A to Z of Mathematics Example 2 The following table gives the frequency distribution of the age of children in a party. Age No. of children
5 5
6 4
7 9
8 2
9 3
10 1
Construct a cumulative frequency table for the given data. Solution Age 5 6 7 8 9 10
No. of children 5 4 9 2 3 1
Cumulative frequency 5 5+4=9 5 + 4 + 9 = 18 5 + 4 + 9 + 2 = 20 5 + 4 + 9 + 2 + 3 = 23 5 + 4 + 9 + 2 + 3 + 1 = 24
Example 3 Identify the situations which will generate ungrouped data. (a) Each of your friends makes a poster of their favorite comic character. The name of the comic character and the number of posters made for it are recorded in a table. (b) A list of shirt sizes worn by every student in a class. (c) A list of height of the students in a class. Solution (a) It is an example of ungrouped data because it is a qualitative data. (b) It is an example of ungrouped data because it is a qualitative data. (c) It is an example of grouped data because it is a quantitative data and it can be grouped into various range.
Cumulative frequency table (grouped data)19 Example 1 The distances (km) covered by 24 cars in 2 hours are given below: 125, 140, 128, 108, 96, 149, 136, 112, 84, 123, 130, 120, 103, 89, 65, 103, 145, 97, 102, 87, 67, 78, 98, 126 Represent them as a cumulative frequency table using 60 km as the lower limit of the first group and all the class intervals having the class size of 15 km. Solution We have, class size = 15 km Maximum distance covered = 149 km, Minimum distance covered = 65 km ∴ Range = (149 – 65) km = 84 km. 84 So, number of classes = = 5.6 ≈ 6 15 Thus, the class intervals are 60–75, 75–90, 90–105, 105–120, 120–135, 135–150. The cumulative frequency distribution is as given below. 19
Refer P 394
Numericalising Geometry 211
Class interval 60–75
Tally marks
Frequency 2
Cumulative frequency 2
75–90
4
2+4=6
90–105
6
2 + 4 + 6 = 12
105–120
2
2 + 4 + 6 + 2 =14
120–135
6
2 + 4 + 6 + 2 + 6 = 20
135–150
4
2 + 4 + 6 + 2 + 6 + 4 = 24
Example 2 The table given below shows the candies eaten by 80 children at the party. Construct less than and more than ogive. Candies No. of children
0–10 3
10–20 8
20–30 17
30–40 29
40–50 15
50–60 6
60–70 2
Solution Less than cumulative frequency for a particular value of the variable is obtained by adding its frequency to the frequencies of all the values smaller than that value. Here, less than cumulative frequency table is given below. No. of children 3 8 17 29 15 6 2
To draw less than ogive for the given data, plot the points with the upper limits of the class as abscissa and the corresponding less than cumulative frequencies as ordinates. The less than ogive graph can be drawn as given below:
Candies less than 10 20 30 40 50 60 70
Less than c.f. 3 3 + 8 = 11 3 + 8 + 17 = 28 3 + 8 + 17 + 29 = 57 3 + 8 + 17 + 29 + 15 = 72 3 + 8 + 17 + 29 + 15 + 6 = 78 3 + 8 + 17 + 29 + 15 + 6 + 2 = 80
Y 90 (60,78)
80 Less than cumulative frequencies
Candies 0–10 10–20 20–30 30–40 40–50 50–60 60–70
(70,80)
(50,72)
70 (40,57)
60 50 40 (30,28)
30 20 10 0
(20,11) (10,3) 10
20
30 Candies
40
50
60
70
X
212 A to Z of Mathematics
More than cumulative frequency for a particular value of the variable is obtained by adding its frequency to the frequencies of all the values greater than that value. More than cumulative frequency table: No. of children 3 8 17 29 15 6 2
To draw more than ogive for the given data, plot the points with the lower limits of the class as abscissa and the corresponding more than cumulative frequencies as ordinates. The more than ogive graph, can be drawn as given below.
Candies more than 0 10 20 30 40 50 60
More than c.f. 2 + 6 + 15 + 29 + 17 + 8 + 3 = 80 2 + 6 + 15 + 29 + 17 + 8 = 77 2 + 6 + 15 + 29 + 17 = 69 2 + 6 +15 + 29 = 52 2 + 6 + 15 = 23 2+6=8 2
Y 90 More than cumulative frequencies
Candies 0–10 10–20 20–30 30–40 40–50 50–60 60–70
80
(0,80)
(10,77) (20,69)
70 60
(30,52)
50 40 30
(40,23)
20
(50,8)
10 0
(60,2) 10
20
30 Candies
40
50
60
Example 3 The weight (kg) of 20 potato sacks are given below: 3.6, 6.2, 4, 0.7, 10.2, 7.5, 3.4, 12.6, 16.3, 14.3, 1.9, 2.7, 9.8, 7.3, 6, 0.5, 2.4, 11.3, 4.2, 6.3 Create a histrogram and a more than ogive. Solution Procedure Range = (16.3 – 0.5) kg = 15.8 kg Let class size = 3 kg
Explanation
Maximum weight of sack = 16.3 kg, Minimum weight of sack = 0.5 kg Assumption
Class size = 3 kg 15.8 Number of classes = = 5.266 ≈ 6 3 Thus, the class intervals are 0.5–3.5, 3.5–6.5, 6.5–9.5, 9.5–12.5, 12.5–15.5, and 15.5–18.5. Let us make the more than cumulative frequency table:
X
Numericalising Geometry 213
Weight of sacks (kg)
Number of sacks (Tally marks)
Number of sacks Sacks More than cumulative frequency (Frequency) more than
0.5 – 3.5 3.5 – 6.5 6.5 – 9.5 9.5 –12.5 12.5 – 15.5
6 6 2 3 2
0.5 3.5 6.5 9.5 12.5
1 + 2 + 3 + 2 + 6 + 6 = 20 1 + 2 + 3 + 2 + 6 = 14 1+2+3+2=8 1+2+3=6 1+2=3
15.5 – 18.5
1
15.5
1
Y Y 6
6
6
5
No. of sacks
4 3
3
2
2
2 1
1 0
0.5 3.5 6.5 9.5 12.5 15.5 18.5 X Weight of sacks (kg)
More than cumulative frequencies
7
24 (0.5,20)
20 16
(3.5,14)
12 (6.5,8) (9.5,6) (12.5,3) (15.5,1)
8 4 0
0.5 3.5 6.5 9.5 12.5 15.5 18.5 Weight of sacks (kg)
X
Introduction to Basics of Other Volumes 40. Comparing Quantities Idea of profit and loss Example 1 A shopkeeper bought a bicycle for ₹2800 and sold it for ₹3650. Find his profit or loss.
Solution Here the cost price (CP) of a bicycle is lesser than the selling price (SP), therefore the shopkeeper has gained a profit. For calculating profit, find the difference between CP and SP by subtracting CP from SP. Procedure Cost price (CP) of the bicycle = ₹2800
Explanation Given Given Formula for profit. Putting the value of CP and SP in the formula.
Selling price (SP) of the bicycle = ₹3650 ∴ Profit = SP – CP = ₹3650 – ₹2800 = ₹850 Hence, the shopkeeper earned a profit of ₹850.
The price at which an item is bought is known as its cost price. The price at which an item is sold is known as its selling price. Profit is a financial advantage or the advantage in terms of money. A person has more money than what he had initially. Whereas loss is a financial disadvantage or having lesser money than what he had initially.
Example 2 Arihant bought a used scooter for ₹15000 and his expenses to get it in the near-new condition is ₹5000. After few months, if he sells the scooter for ₹17000, determine his profit or loss. Solution Here, Arihant spent ₹15000 and ₹5000 on the scooter which means he spent ₹20000 (₹15000 + ₹5000) in total. Thus, the effective cost price is the total amount spent by Arihant. Now, if the effective CP is more than the SP, then Arihant incurred a loss in selling the scooter.
214
Basics of Other Volumes 215
Procedure Cost price (CP) of the scooter = ₹15000 Other expenses on the scooter = ₹5000
Effective cost price = ₹15000 + ₹5000 = ₹20000 Selling price (SP) of the scooter = ₹17000 ∴ Loss = Effective CP – SP = ₹20000 – ₹17000 = ₹3000 Hence, Arihant incurred a loss of ₹3000.
Explanation Given Given
∵ Effective cost price = CP + Other expenses Given ∵ Effective CP > SP ∴ Using loss formula.
Effective cost price is the total cost of purchasing the item and the other cost such as storing cost, insurance cost, delivery cost, repairing cost, shipment cost, etc. that are incurred before selling the item. The other costs incurred before selling the item are known as overhead expenses.
Example 3 A shopkeeper buys an umbrella for ₹450. He sells it for ₹549. Calculate the profit percentage.
Solution Calculate the profit earned by the shopkeeper by subtracting SP from CP and then calculate the profit percentage. For profit percentage, express the profit in terms of CP scaled down to 100. Procedure Cost price (CP) of the umbrella = ₹450
Selling price (SP) of the umbrella = ₹549 ∴ Profit = SP – CP = ₹549 – ₹450 = ₹99 ∴ The ratio of profit earned to CP = 99:450 ==
99 450
2 9 = 2 450 × 9 22 = 100 99 ×
Hence, the profit percentage is 22%.
Explanation Given Given Formula for profit. Putting the value of CP and SP in the formula. ∵ Profit earned on selling an umbrella = ₹99 CP of an umbrella = ₹450 Writing ratio in fractional form.
For percentage, scale down the precedent from 450 to 100. 100 50 × 2 2 = = ∵ 100 ÷ 450 = 450 50 × 9 9 2 is the multiplier for scaling the ratio. 9 Multiply the antecedent and precedent by 2 , so that the ratio remains equivalent. 9
216 A to Z of Mathematics
Similarly, we can calculate the loss percentage. The loss percentage is the loss that would be occurred when a CP is scaled up/down to 100.
Idea of discount Example 1 The marked price of a chair is ₹1200. Anita buys it at a discount of 10% on the marked price. Find the amount which Anita pays to the shopkeeper?
Solution First, calculate the discount given by the shopkeeper on a chair. Then, subtract it from marked price to calculate the selling price or the amount paid by Anita for a chair. Procedure Marked Price (MP) of the chair = ₹1200 Discount percentage given on a chair = 10% of marked price Discount given on a chair = 10% of ₹1200 10 ××1200 = ₹1200 100 = ₹120 ∴ Selling price of a chair = MP – Discount = ₹1200 – ₹120 = ₹1080 Hence, Anita paid ₹1080 to the shopkeeper.
Explanation ….(i) Given Given .…(ii) From (i) and (ii) ∵ In fractional form, 10% =
10 100
Formula for SP. Putting the value of MP and discount in the formula of SP after discount.
Discount means ‘do not count to the full’. When we buy something, on a discount means ‘not counting the selling price to the full’ or ‘cutting a part of the selling price of the things bought’. The price written/labelled on the item is called ‘marked price’. The amount by which the price of an item is reduced is called a discount. The new price at which an item is sold after a discount is called selling price.
Idea of simple interest Example 1 Karan takes a loan of ₹425700 for 3 years at a simple interest rate of 8%. How much interest will he pay?
Solution Interest is a fee paid for borrowing money and the reward earned for saving money in the bank. When we borrow money, we get the entire amount of money in one go. But when we return the money, we pay the fee and the amount in small installments over a duration of time (time can be in days, months, or years). The amount of fee for borrowing money depends upon the amount of money borrowed, the time duration in which the money would be returned, and the rate of interest.
Basics of Other Volumes 217
Procedure Principal (P) = ₹425700
Given
Rate (R) = 8% Time (T) = 3 years
Given Given
Let the interest paid by Karan be SI.
Assumption
∴ SI =
⇒ SI =
P×R×T 100
425700 × 8 × 3
100 ⇒ SI = ₹102168
Explanation
Formula for simple interest (SI) when interest is obtained on yearly basis. P = Principal or loan R = Interest rate (in percentage) T = Time duration (in years) Putting the values of principal, rate, and time in the formula to calculate the simple interest.
Hence, Karan will pay ₹102168 as an interest.
Principal is the money that is deposited or borrowed. The money borrowed is also known as a loan. Time is the duration for which the money has been borrowed or deposited. Simple Interest is the interest that is calculated on the principal amount (money borrowed or deposited) at particular rate throughout the loan period. Rate is rate of interest at which the principal amount is given to someone for a certain time. Rate of interest is the fee paid for deposit/ borrowing. It is expressed as percentage of fees for the amount borrowed or deposited. The total amount to be paid back equals the principal borrowed plus the interest charged.
Example 2 Roshan deposits ₹2000 in a bank for a period of 3 years. If the bank gives an interest of 10% per annum. Find the amount Roshan would get back at the end of 3 years.
Solution Interest on deposit is a fee a bank pays for keeping money with them. When a person deposits money in the bank, every month, the bank adds a small amount of the fee (reward) into his account. The amount of fee paid by the bank depends upon the amount of money deposited in the account, the time duration for which the money remains in the account and rate of interest. Procedure
Explanation
Principle = ₹2000 Time (T) = 3 years Rate of interest = 10% per annum
Given Given Given
Let the interest paid by the bank be SI. P×R×T ∴ SI = 100
Assumption Formula for simple interest (SI) when interest is obtained on yearly basis.
218 A to Z of Mathematics
Putting the values of principal, rate, and time in 2000 × 3 × 10 the formula to calculate the simple interest. 100 ⇒ SI = ₹600 Amount = SI + Principal Formula for calculating amount. Simplifying = ₹600 + ₹2000 = ₹2600 Therefore, Roshan will get back an amount of ₹2600 at the end of 3 years. ⇒ SI =
Idea of compound interest
Example 1 Calculate the compound interest on ₹12000 in 2 years at 10% per annum. Solution
Procedure Principal (P) = ₹12000 Rate (R) = 10% Number of years (n) = 2 ∴ CI = P 1 + R 100
n
Explanation Given Given Given Formula for Compound Interest (CI).
10 ⇒ CI = 12000 1+ 100
2
1 ⇒ CI = 12000 × 1+ 10
2
Putting the values of principal, rate, and number of years in the formula to calculate CI. Reducing
10 to its simplest form. 100
2
LCM of 1 and 10 is 10. 11 ⇒ CI = 12000 × 10 Simplifying 121 ⇒ CI = 12000 × 100 ⇒ CI = ₹14520 Hence, the required compound interest is ₹14520.
In simple interest, the principal amount always remains the same on which interest is calculated every year. Whereas in compound interest, the next year’s interest is calculated on the previous year’s total amount (amount borrowed + previous year’s interest).
Basics of Other Volumes 219
41. Triangle and its Properties Idea of different parts of a triangle Example 1 Write the name of the vertices, sides, and angle of the given triangle. Also, give the name of the triangle in three different ways.
A
B
Solution Procedure Triangle has three vertices. ⇒ Vertices of ∆ABC = A, B, and C Triangle has three sides. ⇒ Sides of ∆ABC = AB, BC, and CA Triangle has three angles. ⇒ Angles of ∆ABC = ∠A, ∠B, and ∠C = ∠BAC, ∠ABC, and ∠BCA = ∠CAB, ∠CBA, and ∠ACB ∆ABC can also be written as ∆BCA and ∆CAB.
C
Explanation ∵ Vertex is a point where two lines meet to form an angle. ∵ Side of any shape is a line segment that joins two vertices. ∵ Angle in any closed shape is formed by two line segments sharing a common vertex. Naming of angles in different ways. Naming of triangles in different ways.
Triangle is a closed plane (2-D) figure with three vertices and three sides. Example 2 Write the name of the interior and exterior angles of the given triangle.
f
B
e
A a
b
c
d C
Solution Interior angle is formed inside a triangle between any two sides of a triangle. While an exterior angle of a triangle is formed between one side of a triangle and a line extending from the adjacent side. Procedure Interior angles of ∆ABC = ∠a, ∠b, and ∠c
Explanation ∵ ∠a, ∠b, and ∠c lie inside the triangle.
220 A to Z of Mathematics
∵ ∠d, ∠e, and ∠f lie outside the triangle between one side of a triangle and a line extending from the adjacent side.
Exterior angles of ∆ABC = ∠d, ∠e, and ∠f
Classifying triangles on the basis of sides Example 1 Identify equilateral, isosceles, or scalene triangles. (a)
(b)
Y
(c) R
B
2c m A
5.6 cm
5.6 cm
V 7 cm
Z
5 cm
m
m
X
7c
Q
3 cm
U
8c
5 cm
5 cm
(d)
7 cm
C
10 cm
P W
Solution Procedure (a) ∆XYZ is an equilateral triangle.
Explanation ∵ In ∆XYZ, XY = YZ = ZX = 5 cm ∵ In ∆ABC, AB ≠ BC ≠ CA
(b) ∆ABC is a scalene triangle.
∵ In ∆PQR, PQ = PR = 5.6 cm ∵ In ∆UVW, UV ≠ VW ≠ WU
(c) ∆PQR is an isosceles triangle. (d) ∆UVW is a scalene triangle.
Classification of a triangle on the basis of the length of sides: A scalene triangle has all three sides of different lengths. An isosceles triangle has two sides of equal length. An equilateral triangle has all three sides of equal length.
Classifying triangles on the basis of angles Example 1 Identify obtuse-angled, acute-angled, or right-angled triangle. (a)
B
(b)
34°
A
49°
97°
C
(c)
Q
M
75°
P
52°
53°
(d) N
26° 64°
R
O
J
15°
135°
30°
L
Solution Procedure (a) ∆ABC is an obtuse-angled triangle.
Explanation ∵ In ∆ABC, ∠C = 97° which is more than 90°.
K
Basics of Other Volumes 221
∵ In ∆PQR, the measure of all the three angles is less than 90°. ∵ In ∆MNO, ∠N = 90°.
(b) ∆PQR is an acute-angled triangle. (c) ∆MNO is a right-angled triangle.
∵ In ∆JKL, ∠L = 135° which is more than 90°.
(d) ∆JKL is an obtuse-angled triangle
Classification of a triangle on the basis of measure of angles: An acute-angled triangle is a triangle in which the measure of all three angles is less than 90°. A right-angled triangle is a triangle in which the measure of one angle is exactly 90°. An obtuse-angled triangle is a triangle in which the measure of one angle is more than 90°.
Idea of a median and altitude of a triangle Example 1 In the given triangles, identify the dotted line as a median or an altitude. (a)
(b)
A
(c)
A
D
C D
B
(d)
C
C D
B
(e)
A
A
B
(f)
A
A D
B
D
C
D
.
B
B
E
C
Solution Procedure (a) AD is a median. (b) AD is an altitude. (c) AD is an altitude. (d) AD is a median. AD is an altitude. Thus, AD is a median as well as an altitude.
Explanation ∵ In ∆ABC, A is the vertex and point D is the midpoint of side BC. ∵ In ∆ABC, AD is perpendicular to side BC. ∵ In ∆ABC, CD is perpendicular to side AB. ∵ In ∆ABC, A is the vertex and point D is the midpoint of side BC. ∵ In ∆ABC, AD is perpendicular to side BC.
C
222 A to Z of Mathematics
(e) AD is an altitude.
(f) DE is not a median nor an altitude.
∵ ∆ABC is an obtuse-angled triangle. For such a tri angle, the base is extended, and then a perpendicular is drawn from the opposite vertex to the base. In ∆ABC, AD is perpendicular to the side CB which is extended to D. ∵ DE is perpendicular to side BC but D is not the vertex of the ∆ABC.
Median of a triangle – A line segment that joins a vertex of a triangle to the mid-point of the side that is opposite to that vertex. Altitude of a triangle – The perpendicular line segment drawn from the vertex of a triangle to the side opposite to it.
Angle sum property of a triangle Example 1 Find the measure of ∠C.
A 124°
?
C
34° B
Solution According to the angle sum property of a triangle, the sum of the interior angles of a triangle is 180°. Using this property to find out the third angle of the triangle. Procedure ∠A = 124° and ∠B = 34° ∠A + ∠B + ∠C = 180° ⇒ 124° + 34° + ∠C = 180°
⇒ 158° + ∠C = 180° ⇒ ∠C = 180° – 158° ⇒ ∠C = 22° Hence, the measure of ∠C is 22° in ∆ABC.
Explanation Given ∵ The sum of the interior angles of a triangle is 180°. Putting the values of ∠A and ∠B in the formula. Simplifying
Exterior angle property of a triangle Example 1 An exterior angle of a triangle is 105°, and one of the interior opposite angles is 60°. Find the other two angles of the triangle.
Basics of Other Volumes 223
Solution Let ABC be a triangle whose side BC is produced to form an exterior angle ∠ACD such that ext. ∠ACD = 105°. ∠BAC and ∠ABC are the opposite interior angles of ext. ∠ACD.
A
105° B
C
D
Procedure Explanation Let ∠ABC = 60° as one of the interior opposite angles is 60° and B is one of the interior opposite angles. Ext. ∠ACD = ∠ABC + ∠BAC Using exterior angle property, as ∠ACD is the exterior angle. ⇒ 105° = 60° + ∠BAC ∵ ∠ACD = 105° and ∠ABC = 60° ⇒ ∠BAC = 105° – 60° ⇒ ∠BAC = 45° ∠ABC + ∠BAC + ∠ACB = 180° Using angle sum property of a triangle. ⇒ 60° + 45° + ∠ACB = 180° ∵ ∠BAC = 45° (calculated) ⇒ ∠ACB = 180° – (60° + 45°) ∵ ∠ABC = 60° (Given) ⇒ ∠ACB = 180° – 105° Simplifying ⇒ ∠ACB = 75° Hence, the other two angles of the triangle are 45° and 75°.
According to the exterior angle property of a triangle, if a side of a triangle is produced then the exterior angle so formed is equal to the sum of the two interior opposite angles.
Triangle inequality property Example 1 Check whether is it possible to form a triangle with the following measures: 8 cm, 6 cm, and 11 cm. Solution Using triangle inequality property to check whether it is possible to form a triangle with the following measures or not. Procedure Explanation Let us assign the values as: a = 8 cm, b = 6 cm, and c = 11 cm. Now let us apply the triangle inequality theorem. Condition 1 a + b = 8 cm + 6 cm = 14 cm ∵ a = 8 cm and b = 6 cm (Given) c = 11 cm Given ∵ 14 > 11 Finding the sum of two sides and comparing it ∴a+b>c with the third side.
224 A to Z of Mathematics
Condition 2 a + c = 8 cm + 11 cm = 19 cm ∵ a = 8 cm and c = 11 cm (Given) b = 6 cm Given ∵ 19 > 6 Finding the sum of two sides and comparing it ∴a+c>b with the third side. Condition 3 b + c = 6 cm + 11 cm = 17 cm ∵ a = 8 cm and b = 6 cm (Given) a = 8 cm Given ∵ 17 > 8 Finding the sum of two sides and comparing it ∴b+c>a with the third side. Since all the three conditions are true, therefore it is possible to form a triangle with the given measurements: 8 cm, 6 cm, and 11 cm. According to the triangle inequality property, the sum of lengths of two sides of a triangle is always greater than the third side.
Pythagoras theorem Example 1 Find the length of the missing side of each right triangle. (a)
(b)
?
B
4m
R
C ?
6 cm
8 cm
Q
5m P A
Solution Using Pythagoras theorem to find out the length of the missing side as the given triangle is a right-angled triangle. Procedure (a) AC = AB + BC 2
2
2
⇒ (5)2 = (4)2+ BC2
Explanation ∵ ABC is a right angled triangle. ∴ Using Pythagoras theorem to find the length of the missing side, BC. Here, AB = Perpendicular, BC = Base, AC = Hypotenuse. AC = 5 m and AB = 4 m (Given)
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Simplifying ⇒ 25 = 16 + BC2 2 ⇒ BC = 25 – 16 ⇒ BC2 = 9 ⇒ BC2 = (3)2 ⇒ BC = 3 m Therefore, the length of the missing side (BC) is 3 m. (b) PR2 = PQ2 + QR2 ∵ PQR is a right angled triangle. ∴ Using Pythagoras theorem to find the length of the missing side, PR. Here, QR = Perpendicular, PQ = Base, PR = Hypotenuse. PQ = 8 cm and QR = 6 cm (Given) ⇒ PR2 = (8)2 + (6)2 Simplifying ⇒ PR2 = 64 + 36 2 ⇒ PR = 100 ⇒ PR2 = (10)2 ⇒ PR = 10 cm Therefore, the length of the missing side (PR) is 10 cm.
The hypotenuse is the largest side of the triangle. Also, it is opposite the right angle of the triangle. The base is a side on which the right-angle triangle stands. Perpendicular is a side which is perpendicular to the base.
42. Congruency of Triangles Idea of congruency Example 1 Which of the following statements is true? Give reasons. (a) A one-rupee coin is congruent to a five-rupee coin. (b) Two acute angles are congruent. (c) Two televisions of the same model of the same brand. (d) Two cards in a deck of playing cards
A Perpendicular
According to the Pythagoras theorem, the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. Hypotenuse2 = Base2 + Perpendicular2
B
Hypotenuse
Base
C
226 A to Z of Mathematics
Solution Procedure
Explanation
∵ One-rupee and five-rupee coins are similar in shape (circular), but are not of the same size. They cannot be placed exactly over each other. ∵ An acute angle is an angle whose measure is less than 90°. Thus, the two angles can be anything less than 90°. However, for two angles to be congruent their degree measure must be equal. ∵ Two televisions of the same model and same brand are identical in shape and size. ∵ All cards in a deck of playing cards are of the same size and shape.
(a) False
(b) False
(c) True (d) True
The word ‘congruent’ means ‘exactly equal’ in terms of shape and size. The figures are congruent when two figures can be placed exactly over each other. Example 2 If ∆ABC ≅ ∆DEF under the correspondence ABC ↔ DEF, write all the corresponding congruent parts of the triangles. A
F
B
C
E
D
Solution Congruent triangles and their corresponding vertices are given. First match their corresponding vertices and then write the corresponding congruent angles and sides of the triangles. Procedure ABC ↔ DEF ∴ A ↔ D; B ↔ E; C ↔ F A
B
F
C
D
∠A ≅∠D; ∠B ≅ ∠E; ∠C ≅ ∠F
AB ≅ DE; BC ≅ EF; AC ≅ DF
E
∵ ∆ABC ≅ ∆DEF
Explanation (Given)
∵ ABC ↔ DEF Matching the corresponding vertices of ∆ABC and ∆DEF. ∵ Corresponding angles of congruent triangles are congruent.
∵ Corresponding sides of congruent triangles are congruent.
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If two triangles are congruent, then their corresponding sides and corresponding angles are also congruent. Congruent angles imply the angles are equal and the line segments that form the angles are also equal.
Idea of SSS congruency Example 1 In the given figure, AD = CD and AB = CB. Is ∆ABD ≅ ∆CBD?
A D
B C
Solution Observe the ∆ABD and ∆CBD carefully and write their three corresponding equal pairs (sides or angles). Then, apply the appropriate congruency rule. Procedure
Explanation
In ∆ABD and ∆CBD, AB = CB Given AD = CD Given BD = BD Common in both. Since three corresponding sides of ∆ABD and ∆CBD are equal, By SSS congruency rule ⇒ ∆ABD ≅ ∆CBD
Two triangles are said to be congruent if all their corresponding sides are equal. This is called the SSS (side-side-side) congruency of triangles.
Idea of SAS congruency Example 1 In the given figure, PS bisects ∠QPR.
P
(a) Name the parts of the ∆PSQ and ∆PSR that are equal. (b) Is ∆PSQ ≅ SR? (c) Is ∠Q = ∠R? Q
Solution Procedure (a) In ∆PSQ and ∆PSR, the equal parts are PQ = PR ∠QPS = ∠RPS PS = PS
R
S
Explanation ….(i)
From the figure.
….(ii) ∵ PS bisects ∠QPR. ….(iii) Common side
(Given) (Given)
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(b) In ∆PSQ and ∆PSR, two corresponding sides and the angle included by them are equal. ⇒ ∆PSQ ≅ ∆PSR
From eq (i), (ii), (iii)
….(iv)
By SAS congruency rule.
From eq (iv)
(c) Since ∆PSQ ≅ ∆PSR ∴ ∠Q = ∠R
All corresponding angles of the congruent triangles are equal.
Two triangles are congruent if two corresponding sides and the angle included by the two sides are equal. This is called SAS (side – angle – side) congruency of triangles.
Idea of ASA congruency Example 1 Which of the following pairs of triangles are congruent? (a) In ∆ABC ; AB = 10 cm, ∠A = 40°, ∠B = 55° In ∆XYZ ; XY = 10 cm, ∠Y = 40°, ∠Z = 85° (b) In ∆PQR ; PR = 5 cm, ∠P = 37°, ∠R = 64° In ∆DEF ; DE = 5 cm, ∠D = 37°, ∠E = 64° Solution Procedure (a)
C
Explanation Drawing the two given triangles with their sides and angles.
Z 85°
A
40°
55° 10 cm
B
In ∆XYZ, ∠Y = 40° and ∠Z = 85° ∴ ∠X = 180° – (∠Y + ∠Z) ⇒ ∠x = 180° – (40° + 85°) ⇒ ∠x = 55° Thus, in ∆ABC and ∆XYZ, we have AB = XY = 10 cm ∠A = ∠Y = 40° ∠B = ∠X = 55° ⇒ ∆ABC ≅ ∆YXZ
Y
40° 10 cm
X
Given ∵ Sum of the interior angles of a triangle is 180°. ….(i) Given Given Using eq. (i) By ASA congruency rule.
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(b)
P
Q
37°
Drawing the two given triangles with their sides and angles
F
64° 5 cm
R
D
37°
64° 5 cm
E
In ∆PQR and ∆DEF, we have PR = DE = 5 cm ∠P = ∠D = 37° ∠R = ∠E = 64° ⇒ ∆PQR ≅ ∆DEF
Given Given Given By ASA congruency rule.
Two triangles are said to be congruent if two corresponding angles and the side included between the two angles are equal. This is called ASA (angle – side – angle) congruency of triangles.
Idea of AAS congruency Example 1 In the given figure, ∠Q = ∠R and PS is perpendicular to QR. Prove that ∆PQS ≅ ∆PRS.
Solution Procedure In ∆PSQ and ∆PSR, the corresp onding equal parts are ∠PSQ = ∠PSR = 90°
P
Q
S
R
Explanation
∵ PS⊥QR (Given) ∠PQS = ∠PRS Given PS = PS Common side In ∆PSQ and ∆PSR, two corresponding angles and a non-included side are equal. By AAS congruency rule. ⇒ ∆PSQ ≅ ∆PSR Two triangles are said to be congruent when two corresponding angles and a non-included sides are equal. This is called AAS (angle-angle-side) congruency.
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Idea of RHS congruencey Example 1 In the given figure, EF ⊥ AB and EG ⊥ BC such that EF = EG. Prove that ∆BEF ⊥ ∆BEG?
Solution
A F E
B
G
D
C
Procedure Explanation In ∆BEF and ∆BEG, the corresp onding equal parts are ∠BFE = ∠BGE = 90° Given BE = BE Common side EF = EG Given In ∆BEF and ∆BEG, the corresponding hypotenuse BE is common, and corresponding one side of the two right-angled triangles is equal. ⇒ ∆BEF ≅ ∆BEG
By RHS congruency rule.
Two right-angled triangles are said to be congruent when the hypotenuse and a side of the triangles are equal. This is called RHS triangle congruency.
43. Understanding Quadrilaterals Idea of quadrilaterals Example 1 Fill in the blanks: (a) A quadrilateral has two diagonals. (b) A quadrilateral has four sides. (c) A quadrilateral has four vertices, three of them are not collinear . (d) A quadrilateral has four angles. (e) A diagonal of a quadrilateral is a line segment that joins two opposite vertices of the quadrilateral.
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‘Quad’ means ‘four’ and ‘lateral’ means ‘side’, therefore, ‘quadrilateral’ means ‘four sides’. Thus, a quadrilateral is a closed 2-dimensional figure that has four sides, four vertices, four angles, and two diagonals.
B C
A
D
Example 2 Identify concave and convex quadrilaterals. (a)
D
C
(b)
(c)
D
(d)
B
A
C C
A A
B
D
B A
C
B
D
Solution Procedure (a) ABCD is a convex quadrilateral. (b) ABCD is a concave quadrilateral. (c) ABCD is a convex quadrilateral. (d) ABCD is a concave quadrilateral.
Explanation ∵ The measure of each interior angle is less than 180°.
∵ The measure of one interior angle is more than 180°. In this figure, ∠ABC is more than 180°. ∵ The measure of each angle is less than 180°.
∵ The measure of one interior angle is more than 180°. In this figure, ∠ADC is more than 180°.
Quadrilaterals can be concave and convex based on the measure of interior angles. Convex quadrilateral – The measure of all the four interior angles is less than 180° and the two diagonals lie inside the closed space of the quadrilateral.
U
A
Q Convex quadrilateral
D
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Concave quadrilateral – The measure of one interior angle is more B than 180° and one of its diagonals lies outside the quadrilateral. P
R
C Concave quadrilateral (∠BPR > 180O)
Idea of polygon’s interior angles Example 1 Find the measure of the interior angle of a regular polygon having 12 sides. Solution Procedure Number of sides (n) = 12 Interior angle =
180o (n - 2) n
180o (12 - 2) ⇒ Interior angle = 12 o 180 ×10 ⇒ Interior angle = 12 ⇒ Interior angle = 150°
Explanation Given The formula for the interior angle of a regular polygon. ∵ n = 12
Pictorially, we can show a regular polygon with 12 sides as follows: 150°
Hence, the measure of the each interior angle of a polygon having 12 sides is 150°. All interior angles of a regular polygon are equal and each interior angle can be obtained by using the following formula: 180o (n - 2) Interior angle = , where n is the number of sides of a polygon. n Example 2 What is the sum of all interior angles of a regular hexagon? Solution Find the measure of an interior angle of a regular hexagon and then multiply it with the number of interior angles in a polygon.
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Procedure Number of sides (n) = 6 Interior angle =
180o (n - 2) n
180o (6 - 2) 6 o 180 × 4 ⇒ Interior angle = 6 ⇒ Interior angle = 120° ⇒ Interior angle =
Explanation
∵ Hexagon has 6 sides.
Formula for the interior angle of a regular polygon, where ‘n’ is the number of sides. ∵n=6 Pictorially, we can show a regular hexagon as follows: 120° 120°
120°
120°
120° 120°
Sum of all interior angles of a hexagon Formula for finding the sum of all interior angles of = Number of interior angles × Measure of each a polygon. interior angle
= 6 × 120° ∵ Number of interior angles = Number of sides (n) = 6 = 720° The measure of each interior angle = 120° Hence, the sum of all interior angles of a regular hexagon is 720°. Number of sides of a polygon is equal to the number of interior angles in a polygon.
Angle sum property of quadrilaterals Example 1 Find the value of x in the given figure.
D 52o xo 126o A
C
106o B
Solution Using angle sum property of a quadrilateral to find out the fourth angle of the quadrilateral.
234 A to Z of Mathematics
Procedure ∠A + ∠B + ∠C + ∠D = 360° ⇒ 126° + 106° + x° + 52° = 360° ⇒ x° + 284° = 360° ⇒ x° = 360° – 284° ⇒ x° = 76° Hence, the value of xo is 76°.
Explanation
∵ The sum of interior angles of a quadrilateral is 360°. Putting the values of ∠A, ∠B, and ∠D in the formula. Simplifying
According to the angle sum property of a quadrilateral, the sum of the interior angles of a quadrilateral is 360°.
Exterior angles of quadrilaterals/polygons Example 1 Find the value of x in the given figure.
D
90o
C
50° x°
A 110°
B
Solution An exterior angle of a quadrilateral is an angle formed outside the quadrilateral between its side and the extension of its adjacent side. Using exterior angle sum property of a quadrilateral to find out the fourth exterior angle of the quadrilateral. Procedure Explanation ext. ∠A + ext. ∠B + ext. ∠C + ext. ∠D = 360° ∵ The sum of exterior angles of a quadrilateral is 360o. ⇒ 110° + x° + 90° + 50° = 360° ⇒ xo + 2 50° = 360° ⇒ xo = 360° – 2 50° ⇒ xo = 110° Hence, the value of x° is 110°.
Putting the values of all exterior angles in the formula. Simplifying
According to the exterior angle sum property of a quadrilateral, the sum of four exterior angles so formed is 360°. Example 2 Find the measure of each exterior angle of a regular octagon.
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Solution The regular octagon is an 8-sided polygon with equal sides, equal interior angles, and equal exterior angles. To find the measure of each exterior angle of a regular polygon, use the formula of an exterior angle. Procedure Number of sides (n) = 8
Explanation ∵ Octagon has 8 sides.
o
The formula for the exterior angle of a regular polygon.
360 Exterior Angle = n o 360 ⇒ Exterior Angle = 8 ⇒ Exterior Angle = 45°
Here, n = 8. Simplifying
Hence, the value of each exterior angle of an octagon is 45°. All exterior angles of a regular polygon are equal and each exterior angle can be obtained by using the following formula: o 360 Exterior angle = , where n is the number of sides of a polygon. n Also, the sum of the exterior angles of a polygon is always 360°.
Properties of parallelograms Example 1 In a parallelogram ABCD, ∠D = 68°. Determine the measure of ∠A and ∠B. Solution Procedure D
C
Explanation Drawing parallelogram ABCD with ∠D = 68°.
68O A
B
A parallelogram ABCD whose ∠D = 68°. ∠A + ∠D= 180° ⇒ ∠A + 68° = 180° ⇒ ∠A = 180° – 68° ⇒ ∠A = 112° ∠C = ∠A ⇒ ∠C = 112°
Given
∵ The sum of any two consecutive angles of a parallelogram is 180°. ∵ ∠D = 68°
∵ Opposite angles of a parallelogram are equal. ∵ ∠A = 112° (Calculated above)
236 A to Z of Mathematics
∵ Opposite angles of a parallelogram are equal. ∵ ∠D = 68°
∠B = ∠D ⇒ ∠B = 68°
Hence, the measure of ∠A = 112° and ∠B = 68°. The properties of a parallelogram are as follows: (a) Opposite sides are equal and parallel. (b) Opposite angles are equal; each pair of consecutive angles is supplementary. (c) Diagonals of a parallelogram bisect each other. (d) Each diagonal divides the parallelogram into two congruent triangles.
C
D
A
B
Properties of rectangles Example 1 The length and breadth of a rectangle ABCD are 32 cm and 24 cm respectively. Find the length of the diagonal of a rectangle. Solution We know that each interior angle of a rectangle is 90°. Draw a diagonal and use Pythagoras theorem to find the length of the diagonal (hypotenuse). Procedure A
B
Explanation Drawing a rectangle ABCD with breadth = 24 cm and length = 32 cm.
24 cm D
32 cm
∠C = 90° ⇒ ∆BCD is a right triangle. DB2 = DC2 + BC2
⇒ DB2 = (32)2 + (24)2
C
∵ Measure of each interior angle of a rectangle = 90° By applying Pythagoras theorem on ∆BCD. ∵ DC = 32 cm and BC = 24 cm
Simplifying ⇒ DB2 = 1024 + 576 2 ⇒ DB = 1600 ⇒ DB2 = (40)2 ⇒ DB = 40 cm Hence, the length of the diagonal of the rectangle is 40 cm.
(Given)
Basics of Other Volumes 237
Rectangle is a four-sided closed plane figure with straight sides. A square is a special type of rectangle. The properties of a rectangle are as follows: (a) Opposite sides are equal and parallel. (b) All angles are equal and right angles. (c) Diagonals of a rectangle are equal and bisect each other.
D
C
A
B
Properties of squares
Example 1 Find the length of a side of a square whose diagonal length is 32 m. Solution We know that measure of each angle of a square is 90°. Draw the diagonal and use Pythagoras theorem on the right triangle to find the length of the side. Procedure Q
R
Explanation Drawing a square PQRS with diagonal PR = 32 m.
32 m P
S
∠S = 90° ⇒ ∆PSR is a right triangle. PS2 + SR2 = PR2
⇒ PS2 + SR2 = (32)2 ⇒ y2 + y2 = (32)2 ⇒ 2y2 = 1024 1024 ⇒ y2 = 2 ⇒ y2 = 512 ⇒ y=
⇒ y=
∵ Measure of each interior angle of a square = 90° By applying Pythagoras theorem on ∆PSR. ∵ PR = 32 m
(Given)
∵ All sides of a square are equal. ∴ Let PS = PR = y (Assumption) Simplifying
512
2×2×2×2×2×2×2×2×2
⇒ y = 16 2 m
Prime factorising and making pairs to find the square root.
Hence, the length of the side of a square is 16 2 m.
238 A to Z of Mathematics
A square is a regular quadrilateral which means that it has four equal sides and four equal angles. The properties of a square are as follows: (a) All sides of a square are equal in length. D C (b) All angles are right angles. (c) Opposite sides of a square are parallel. (d) Diagonals of a square are equal in length and bisect each other A B at 90°.
Properties of rhombuses Example 1 In a rhombus, ∠ABD = 67°. Find the ∠ADC.
B 67° A
O
C
D Solution We know that the opposite angles of a rhombus are equal and the diagonals of a rhombus bisect the two pairs of opposite angles. Use these properties to find the ∠ADC.
Procedure
∠ABD = 67° ∠ABC = 2 ∠ABD ⇒ ∠ABC = 2 × 67° ⇒ ∠ABC = 134° ∠ADC = ∠ABC = 134° Hence, the measure of ∠ADC is 134°.
Explanation
Given ∵ The diagonals of a rhombus bisect the vertex angles. ∵ The opposite angles of a rhombus are equal.
A rhombus is a parallelogram with four equal sides. The properties of a rhombus are as follows: (a) All sides of a rhombus are equal. (b) Opposite angles are equal. (c) The diagonals of a rhombus bisect the vertex angles. (d) The diagonals of a rhombus bisect each other at right angles.
B
C E
A
D
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Properties of kites and trapeziums Example 1 In a given trapezium, SP||RQ. Find the measures of ∠P and ∠R.
R
Solution
S
Q 130°
P
Procedure Explanation Since PQRS is a trapezium in which SP || RQ. ∴ ∠P + ∠Q = 180° ∵ SP || RQ and PQ is a transversal, ∠P and ∠Q are adjacent angles. From figure, ∠Q = 130° ⇒ ∠P + 130° = 180° Simplifying ⇒ ∠P = 180° – 130° ⇒ ∠P = 50° ∴ ∠S + ∠R = 180° ∵ SP || RQ and SR is transversal, ∠S and ∠R are adjacent angles. ⇒ 90° + ∠R = 180° From figure, ∠S = 90° ⇒ ∠R = 180° – 90° ⇒ ∠R = 90° Hence, the measures of ∠P and ∠R are 50° and 90°, respectively. A trapezium is a quadrilateral in which at least one pair of opposite sides is parallel.
A
D
Example 2 In kite PQRS, ∠SRP = 24° and ∠TSP = 53°. Find ∠PSR.
E
B
C
F
Q
T
P
24°
R
53°
Solution
S
We know that the diagonals of a kite bisect each other at 90° and the sum of three interior angles of a triangle is 180°. Use these properties to find the ∠PSR.
240 A to Z of Mathematics
Procedure PQRS is a kite. ∠SRT = 24° and ∠TSP = 53° In right angled triangle ∆RTS, ∠RTS + ∠SRT + ∠TSR = 180° ⇒ 90° + 24° + ∠TSR = 180° ⇒ ∠TSR = 180° – 114° ⇒ ∠TSR = 66° In ∆PSR, ⇒ ∠PSR = ∠TSP + ∠TSR ⇒ ∠PSR = 53° + 66 ° ⇒ ∠PSR = 119° Hence, the measure of ∠PSR is 119°.
Explanation Given ....(i) Given ∵ The sum of the measures of the three angles of a triangles is 180°. From eq. (i) and eq. (ii). Simplifying ....(ii) From figure From eq. (i) and eq. (ii).
Kite has two pairs of equal adjacent sides and unequal opposite sides. (a) Two pairs of adjacent sides are equal. (b) One diagonal (the longer one) bisects the other and it is perpendicular to it. (c) The longer diagonal (called the main diagonal) also bisects the pair of opposite angles. (d) The angles at the end of the smaller diagonal (called cross D diagonal) are congruent.
A
B
C
44. Mensuration Finding perimeter and area of a triangle Example 1 Find the perimeter and area of the given triangle.
B
32 cm
32 cm 12 cm
A
D 23 cm
C
Solution A triangle has three sides. Therefore, the perimeter of the given triangle is the sum of the lengths of all three sides.
Basics of Other Volumes 241
And, the area of a triangle is half of the product of the base and height of the triangle. 1 Area of a triangle = × base × height 2 Procedure AB = 32 cm BC = 32 cm AC (Base) = 23 cm BD (height) = 12 cm Perimeter of ∆ABC = AB + BC + CA
Explanation Given Given Given Using the perimeter formula of a triangle.
⇒ Perimeter of ∆ABC = (32 + 32 + 23) cm Putting the values of AB, BC, and CA in the formula. ⇒ Perimeter of ∆ABC = 87 cm 1 Using the formula of area of a triangle. Area of ∆ABC = × base × height 2 1 Putting the values of AC and BD in the formula. ⇒ Area of ∆ABC = × AC × BD 2 æ1 ö ⇒ Area of ∆ABC = çç × 23×12÷÷÷ cm 2 çè 2 ø 2 ⇒ Area of ∆ABC = 138 cm Hence, the perimeter and area of the triangle are 87 cm and 138 cm2, respectively. Area – The total space that is occupied by a closed surface or two-dimensional geometric shape in a plane. Perimeter – The boundary of the closed shape (or figure) is termed as a perimeter. That is, the sum of the lengths of all sides is known as perimeter. Base – The base of a figure is usually a side or face on which the figure sits. However, this is not necessary that the bottom is always the base. Bases are often paired with a height that is perpendicular to the base. Height – The height of a figure is the perpendicular line dropped onto its base from the opposite vertex of the base.
Finding perimeter and area of a rectangle Example 1 The perimeter of a rectangle is 180 cm. If the length of the rectangle is 35 cm, find it’s breadth and area. Solution We know that the perimeter of a rectangle is the sum of the lengths of all sides of a rectangle. ∴ Perimeter of a rectangle = 2 (length + breadth) First, find the breadth using the perimeter formula, and then find the area using the below given formula: Area of a rectangle = length × breadth
breadth = ? Length = 35 cm
242 A to Z of Mathematics
Procedure Length of a rectangle = 35 cm Perimeter of a rectangle = 180 cm Perimeter of a rectangle = 2 (length + breadth)
Explanation Given Using the perimeter formula of a rectangle.
Putting the values of perimeter and length of ⇒ 180 = 2 (35 + breadth) the rectangle in the formula. ⇒ 180 = 70 + 2 (breadth) ⇒ 2 (breadth) = 180 – 70 110 ⇒ breadth = cm 2 ⇒ breadth = 55 cm Area of a rectangle = length × breadth Using the formula of area of a rectangle. 2 Putting the values of length and breadth in ⇒ Area of a rectangle = (35 × 55) cm 2 the formula. ⇒ Area of a rectangle = 1925 cm Hence, the breadth and area of the rectangle are 55 cm and 1925 cm2, respectively. We know that the diagonal of a rectangle, square, rhombus, or parallelogram divides it into two equal triangles. Thus, the area of these shapes can be obtained by finding the area of one triangle and multiplying it by 2. Area of a rectangle, square, rhombus, and parallelogram = 2 × Area of a triangle æ1 ö = 2 × ççç × base × height÷÷÷ è2 ø = base × height In case of rectangle, base = length and height = breadth.
Finding perimeter and area of a square Example 1 Find the perimeter of a square whose side is 8 cm. Solution A square has four sides of equal length. ∴ Perimeter of a square = 4 × (length of a side) And, the formula for the area of a square is as follows: Area of a square = (side)2 Procedure Side of a square = 8 cm Perimeter of a square = 4 × (Length of a side) ⇒ Perimeter of a square = 4 × 8 cm ⇒ Perimeter of a square = 32 cm Area of a square = (side)2
side = 8 cm
Explanation Given ∵ Using the perimeter formula of a square. Putting the value of side in the formula. Using the formula of area of a square.
Basics of Other Volumes 243
Putting the value of side in the formula. ⇒ Area of a square = (8)2 ⇒ Area of a square = 64 cm2 Hence, the perimeter and area of the square are 32 cm and 64 cm2, respectively.
Finding area of a parallelogram & rhombus
Example 1 Find the area of a parallelogram whose base is 7 cm and the altitude is 13 cm. Solution We know, Area of a parallelogram = base × height
Height = 13 cm Base = 7 cm
Procedure Base = 7 cm and altitude = 13 cm Area of a parallelogram = base × height
Explanation Given Using the formula of area of a parallelogram. Putting the value of base and altitude in the formula.
⇒ Area of a parallelogram = (7 × 13) cm2 ⇒ Area of a parallelogram = 91 cm2
Hence, the area of the parallelogram is 91 cm2. Example 2 Find the area of the given rhombus.
S
R PR = 24 cm SQ = 18 cm
O P
Q
Solution We know that the diagonal of a rhombus divides it into two equal triangles. Thus, the area of a rhombus can be obtained by finding the area of one triangle and multiplying it by 2. Procedure
Explanation
In ∆PRS. Base (PR) = 24 cm Height of ∆PRS = SO
1 ⇒ Height of ∆PRS = × SQ 2
1 × 18 cm 2 ⇒ Height of ∆PRS = 9 cm ⇒ Height of ∆PRS = Area of ∆PRS =
1 × base × height 2
From figure
(Given)
∵ Diagonals of a rhombus bisect each other at right angles.
∵ SQ = 18 cm
Using the formula of area of a triangle.
244 A to Z of Mathematics
⇒ Area of ∆PRS =
1 × 24 × 9 2
⇒ Area of ∆PRS = 108 cm2
∴ Area of the rhombus PQRS = 2 × Area of ∆PRS ⇒ Area of the rhombus PQRS = 2 × 108 cm2 = 216 cm2
Putting the value of base and height in the formula. Base = PR = 24 cm ….(i) Height = SO = 9 cm ∵ Sum of the area of ∆PRS and ∆PQR give the area of the rhombus. From eq. (i), Area of ∆PRS = 108 cm2.
Hence, the area of the given rhombus is 216 cm2. We can also find the area of a rhombus by direct formula: 1 Area of a rhombus = × length of diagonal1 × length of diagonal2 2 1 ⇒ Area of a rhombus = × d1 × d2 2
A
B d1 D
O
d2 C
Finding circumference and area of a circle
Example 1 Find the circumference and area of a circle whose diameter is 42 cm. Solution Procedure Let radius of the circle be ‘r’. Diameter = 42 cm 42 Radius = = 21 cm 2 Let C be the circumference of the circle. Then, C = 2πr
Explanation Assumption Given Diameter .... (i) Radius = 2 Using the formula of circumference of a circle. Putting the values of π and r. 22 Here, π = , r = 21 cm. 7 Using the formula of area of a circle. Putting the values of π and r in the formula.
22 ⇒ C = 2 × × 21 cm 7 ⇒ C = 132 cm Area of a circle = πr2 22 ⇒ A = × 21 × 21 cm2 7 2 = 1386 cm Hence, the circumference and area of the circle are 132 cm and 138 cm2 respectively.
Basics of Other Volumes 245
The circumference of a circle is a perimeter of a circle. The formula for the circumference of a circle is as follows: Circumference of a circle (C) = πD = 2πr Here, ‘r’ is a radius of the circle r ‘D’ is a diameter of the circle o 22 ‘π’ is a constant = 3.141 or 7 And, Area of a circle = πr2
Finding area between rectangles Example 1 A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path. Solution Let ABCD be a rectangular park and the shaded region represents the 2.5 m wide path to be constructed outside the park. P
Q A
45 m
30 m
2.5 cm C
D S
B
2.5 m
R
To find the area of the path, calculate the difference between the area of the rectangle PQRS and area of the rectangle ABCD. Procedure PQ = (45 + 2.5 + 2.5) m = 50 m
PS = (30 + 2.5 + 2.5) m = 35 m
Area of the path = Area of rectangle PQRS – Area of rectangle ABCD = (PQ × PS) – (AB × AD)
Explanation Sum of the length of one side of the park and twice the width of the path give the length of the outer rectangle PQRS. Sum of the breadth of one side of the park and twice the width of the path give the total breadth of the outer rectangle PQRS. Subtracting the area of rectangle ABCD from the area of rectangle PQRS to get the area of the path. ∵ Area of a rectangle = length × breadth
246 A to Z of Mathematics
= (50 × 35) m2 – (45 × 30) m2 = 1750 m2 – 1350 m2 = 400 m2 Hence, the area of the path is 400 m2.
∵ PQ = 50 m, PS = 35 m AB = 45 m, AD = 30 m
Finding area enclosed by two concentric circles Example 1 Find the area between two concentric circles whose radii are 21 cm and 14 cm. Solution Concentric circles can be defined as two or more circles of different radii but with same centre. Procedure r = 14 cm and R = 21 cm
r = 14 cm
O R = 21 cm
Area between concentric circles = Area of bigger circle – Area of smaller circle = πR2 – πr2 é 22 22 2 2ù = ê (21) - (14) ú cm2 êë 7 úû 7
22 (21)2 − (14)2 cm2 7 22 = [ 441 − 196] cm2 7 22 = × 245 cm2 7
=
= (22 × 35) cm2 = 770 cm2 Hence, the area between two concentric circles is 770 cm2.
Explanation Radii of two concentric circles are given. Drawing two circles of radii r = 14 cm and R = 21 cm with the same centre ‘O’. And, shading the region which represents the area between two concentric circles.
Subtracting the area of a smaller circle of radius (r) from the area of a bigger circle of radius (R) to get the shaded area between two concentric circles. ∵ Area of a circle = πr2 where ‘r’ is the radius of the circle. Putting the value of ‘π’, ‘r’ and ‘R’ in the formula. Here, 22 r = 14 cm, R = 21 cm, and π = 7 Simplifying
Basics of Other Volumes 247
Finding area of a trapezium Example 1 Find the area of a trapezium whose parallel sides are of length 10 cm and 12 cm and the distance between them is 4 cm. Solution Procedure Length of parallel sides = 10 cm and 12 cm Distance between parallel sides = 4 cm Area of a trapezium 1 = × (sum of its parallel sides) × (distance between its 2 parallel sides) 1 = × (10+12)× 4 cm2 2 1 2 = ×22× 4 cm 2 2 = 44 cm Hence, the area of a trapezium is 44 cm2.
Explanation Given Using the formula of area of a trapezium.
Putting the values of length of parallel sides and distance between the parallel sides in the formula. Simplifying
Finding area of a polygon Example 1 Find the area of a given regular octagon.
B A
C
5m
D
12 m
6m H
E G
F
Solution The area of a regular octagon is the sum of the areas of the two trapeziums and a rectangle. The area of two trapeziums is equal as the octagon is a regular polygon. Thus, Area of the octagonal surface = Area of ABCD trapezium + Area of ADEH rectangle + Area of EFGH trapezium = 2 (Area of trapezium ABCD) + Area of rectangle ADEH [Since the two trapeziums ABCD and EFGH are congruent.]
248 A to Z of Mathematics
Procedure
Explanation
DE = 6 m AD = 12 m BC = DE = 6 m
Given ∵ ABCDEFGH is a regular octagon. ∴ All sides are equal. Given
In trapezium ABCD, Length of parallel sides, BC = 6 m and AD = 12 m Distance between parallel sides = 5 m Area of a trapezium
Using the formula of area of a trapezium.
=
1 × (sum of its parallel sides) × (distance between its 2
=
1 × (BC + AD) × (distance between parallel sides) 2
parallel sides)
From figure Putting the values of length of parallel sides and distance between parallel sides in the formula. Simplifying
1 2 = × (6 + 12) × (5) m 2 1 = × 18 × 5 m 2 2 2 = 45 m Area of the rectangle = length × breadth
⇒ Area of the rectangle ADEH = AD × DE
= (12 × 6) m2 = 72 m2
….(i) Using the formula of the area of a rectangle. From figure Putting the values of AD and DE in the formula. ….(ii) Simplifying
Area of an octagon in terms of the small ∴ Area of the octagon ABCDEFGH = 2 × Area of trapezium ABCD + Area of rectangle ADEH polygons. = (2 × 45) m2 + 72 m2 From eq. (i) and (ii). = (90 + 72) m2 = 162 m2 Hence, the area of the regular octagon is 162 m2.
Simplifying
The area of a regular polygon can be calculated by breaking it into smaller polygons (triangles, squares, rectangles, or trapeziums).
Finding volume of a cuboid Example1 Find the volume of a cuboidal box whose length = 12 cm, breadth = 8 cm, and height = 3 cm.
Basics of Other Volumes 249
Solution Procedure length (l) = 12 cm breadth (b) = 8 cm height (h) = 3 cm Volume of a cuboidal box = length × breadth × height = (12 × 8 × 3) cm3 = 288 cm3 Hence, the volume of the cuboidal box is 288 cm3.
Explanation Given
Using the formula of volume of a cuboid. Putting the values of length, breadth, and height in the formula. Simplifying
Volume of a three-dimensional solid is the amount of space it occupies or the space enclosed by a boundary or the capacity to hold something. A cuboid is a three-dimensional box-like object having six rectangular faces. The volume of a cuboid is the amount of space contained by it. If the cuboid is a solid brick, then the volume is the space occupied by the brick. If the cuboid is a hollow shoebox, then the volume is the capacity of the shoe box which it can hold. If a cuboid has length ‘l’ units, breadth ‘b’ units and height ‘h’ units, then Volume of the cuboid = Area of base × height of cuboid = (length × breadth) × height = l × b × h cubic units
Height Length
Width
Finding volume of a cube Example1 Find the volume of a cube whose side is 5 cm. Solution Procedure Side = 5 cm Volume of a cube = (side)3 = (5)3 cm3 = 125 cm3
Explanation Given Using the formula of volume of a cube. Putting the value of side in the formula. Simplifying
Hence, the volume of the cube is 125 cm3. A cube is a three-dimensional box-like object with equal length, breadth (width), and height. Thus, we can also say that cube is a special type of cuboid whose length, breadth (width), and height are equal in measurement. Some examples of the cube are dice, sugar cube, Rubik cube, etc.
Finding volume of a cylinder Example 1 The area of the base of a cylinder is 102 cm2 and its height is 12 cm. Find the volume of the cylinder.
250 A to Z of Mathematics
Solution Procedure Area of the base of a cylinder (πr2) = 102 cm2 ⇒ πr2 = 102 cm2 ….(i) Height of a cylinder (h) = 12 cm Volume of a cylinder = πr2h
….(ii)
= (102 × 12) cm3
Explanation Given ∵ The base of a cylinder is circle. ∴ Area of a circle = πr2 Given Using the formula of volume of a cylinder. Putting the value of πr2 from eq.(i) and ‘h’ from eq.(ii) in the formula. Simplifying
= 1224 cm3 Hence, the volume of the cylinder is 1224 cm3.
A right circular cylinder is a three-dimensional shape that consists of two circular parallel bases linked by a curved surface. Some examples of the right circular cylinder are tube light, drum, cold drink can, pipe, etc. We can find the volume of any shape by multiplying the area of the base with the height (h) of the shape. The base of a right circular cylinder is a circle and the area of a circle of radius ‘r’ is πr2. Thus, volume of a right circular cylinder = Area of circular base × Height of a cylinder = πr2 h
Finding volume of a sphere
Example 1 Find the volume of a sphere whose radius is 21 cm. Solution Procedure Radius of a sphere (r) = 21 cm
Explanation Given
Using the formula of volume of a sphere. 4 3 πr 3 Putting the value of radius in the formula. æ 4 22 ö = çç × × 21× 21× 21÷÷ cm3 ÷ø çè 3 7
Volume of a sphere =
= 38808 cm3
Simplifying
Hence, the volume of the sphere is 38808 cm . 3
h r
Basics of Other Volumes 251
A sphere is a three-dimensional round solid figure in which every point on its surface is equidistant from its centre and this fixed distance is called the radius of the sphere.
r
4 Volume of a sphere = πr3 3
Some examples of the sphere are ball, globe, marbles, sun, etc.
Finding volume of a right circular cone Example 1 A right circular cone of volume 264 cm3 has a height 28 cm. Find the radius of the cone. Solution Procedure Volume of a cone (V) = 264 cm3 Height of cone (h) = 28 cm Volume of a cone = ⇒ 264 =
1 2 πr h 3
1 22 × × r 2 × 28 3 7
264 × 3 × 7 cm 22 × 28 ⇒ r2 = 9 cm ⇒ r2 = (3)2 cm ⇒ r = 3 cm Hence, the radius of the cone is 3 cm.
⇒ r2 =
Explanation Given Using the formula of volume of a cone. Putting the value of volume and radius in the formula. Simplifying
A right circular cone is a three-dimensional solid object which is having a circle at one end and a pointed end on the other and whose axis is perpendicular to the plane of the base. A right circular cone is obtained by a revolving right triangle about one of its legs. Some examples of cones are ice cream cone, funnel, birthday cap, etc.
1 × (Area of the base) × (height) 3 1 ⇒ Volume of a right circular cone = πr2h 3 Volume of a right circular cone =
axis
Base radius
Finding surface area of a cuboid
Example1 Find the total surface area of a chalk box whose length, breadth, and height are 6 cm, 8 cm, and 12 cm, respectively.
252 A to Z of Mathematics
Solution Procedure
Explanation
length (l) = 6 cm breadth (b) = 8 cm height (h) = 12 cm TSA of chalk box = 2 (l × b + b × h + h × l)
Given
⇒ TSA of chalk box = [2 (6×8 + 8×12 + 12×6)] cm2
⇒ TSA of chalk box = [2 (48 + 96 + 72)] cm2 ⇒ TSA of chalk box = (2 × 216) cm2 ⇒ TSA of chalk box = 432 cm2 Hence, the total surface area of the chalk box is 432 cm2.
Using the formula of total surface area of a cuboid. Putting the value of length, breadth, and height in the formula. Simplifying
Total surface area is the sum of the areas of all faces (or surfaces) of a three-dimensional shape. A cuboid has 6 rectangular faces. To find the total surface area of a cuboid, add the areas of all 6 faces. Total Surface Area of cuboid = 2 (length × breadth + breadth × height + height × length) ⇒ TSA of cuboid = 2 (l × b + b × h + h × l) l×b l h l
b
h
h×l
b×h
l
b
h
h×l l
l×b
b b×h h b
b
While lateral surface area is the sum of the areas of all faces of the object excluding the base and the top face. LSA of cuboid = 2 (l × h + b × h)
Finding surface area of a cube Example 1 Calculate the side and lateral surface area of a cube whose total surface area is 1944 m2. Solution Procedure Total surface area of cube = 1944 m2 TSA of cube = 6 × (side)
2
⇒ 1944 = 6 × (side)2
Explanation Given Using the formula of the total surface area of a cube. Putting the value of TSA in the formula.
Basics of Other Volumes 253
Simplifying
1944 ⇒ (side)2 = m 6 ⇒ (side)2 = 324 m ⇒ (side)2 = (18)2 m ⇒ side = 18 m LSA of a cube = 4 × (side)2
Using formula of a lateral surface area of a cube. Putting the value of side in the formula.
⇒ LSA of a cube = 4 × (18)2
Simplifying ⇒ LSA of a cube = 4 × 324 m2 2 ⇒ LSA of a cube = 1296 m Hence, the side and lateral surface area of the cube is 18 m and 1296 m2 respectively. A cube has 6 square faces. The total surface area of a cube is the sum of the areas of all 6 square faces. Total Surface Area of cube = 6 × (side)2 side2 side side
side2
side
side2
side2
side2
side2
And, lateral surface area of a cube = 4 × (side)2
Finding surface area of a cylinder Example 1 Find the total surface area of a right circular cylinder whose height is 21 cm and the radius of the base is 7 cm. Solution Procedure Radius of the cylinder (r) = 7 cm Height of the cylinder (h) = 21 cm TSA of cylinder = 2πr2 + 2πrh
Explanation Given Using the formula of the total surface area of a cylinder.
⇒ TSA of cylinder = 2πr (r + h) 22 Putting the value of π, radius, and height in ⇒ TSA of cylinder = 2 × × 7(7 + 21) cm2 the formula. 7 Simplifying ⇒ TSA of cylinder = 44(7 + 21) cm2 2 ⇒ TSA of cylinder = (44 × 28) cm ⇒ TSA of cylinder = 1232 cm2 Hence, the total surface area of the cylinder is 1232 cm2.
254 A to Z of Mathematics
A cylinder has two circular bases and a curved surface. The total surface area of the cylinder is the sum of the area of the two circular bases (of radius ‘r’) and the area of the curved surface. When we open the curved surface, then it forms a rectangle whose length is the circumference of the curved surface and width is the height of the cylinder. ∴ Total Surface Area of cylinder = (πr2 + πr2) + 2πrh = 2πr2 + 2πrh πr2
r
h
2πr
h πr2
And, lateral surface area of a cylinder = 2πrh
45. Visualising Solid Shapes Diagnosing solid shapes Example 1 Count the number of faces, edges, and vertices of the following solid shapes.
Solution
Sphere
Cone
Square pyramid
Triangular prism
Hexagonal prism
Cube
Cylinder
Cuboid
No. of Vertices
No. of Faces
No. of Edges
Sphere
0
1
0
Cone
1
2
1
Square based pyramid
5
5
8
Triangular prism
6
5
9
Name
Shapes
Shapes with Outline
Basics of Other Volumes 255
Hexagonal prism
12
8
18
Cube
8
6
12
Cylinder
0
3
2
Cubiod
8
6
12
The shape or the figure that has three dimensions (length, breadth, and height), are known as solids or 3-D shapes. Solid shapes can be drawn on a flat surface realistically. We call this a 2-D representation of a 3-D solid. The attributes of a 3-D figures are faces, edges and vertices. Faces are the flat surface of a solid shape. Vertices in 3-D shapes are the points where three or more line segments or edges meet (like a corner). Edges in 3-D figures are the line segments that join one vertex to another or where the shape’s faces meet. For example, Vertex Curved face
Curved edge
Flat face
vertex
Face
Edge
Drawing nets for building 3-D shapes Example 1 Choose the correct net for each solid shape. (a) (i)
(ii)
(iii)
(b)
(i)
(ii)
(iii)
(c)
(i)
(ii)
(iii)
256 A to Z of Mathematics
(d)
(i)
(ii)
(iii)
(e)
(i)
(ii)
(iii)
(f)
(i)
(ii)
(iii)
Solution A net is a 2-dimensional cutout that can be folded to form a 3-D shape or a solid. A solid may have different nets. To determine whether the net forms a solid or not, make sure that the solid and the net have the same number of faces and that the shapes of the faces of the solid match the shapes of the corresponding faces in the net. Then, visualise how the net is to be folded to form the solid. (a) option (i) (b) option (iii) (c) option (ii) (d) option (i) (e) option (ii) (f) option (iii) Nets are helpful when we need to find the surface area of the solids.
Euler’s relation for polyhedrons Example 1 If a polyhedron contains 5 vertices and 8 edges, then identify the name of the polyhedron. Solution To identify the name of any polyhedron, we must know its number of faces, vertices, and edges. For this, we can use Euler’s formula of polyhedrons. Procedure Number of vertices (V) = 5 Number of edges (E) = 8 F+V=E+2
Explanation Given Using Euler’s formula for polyhedrons. Putting the value of V and E in the formula. Simplifying
⇒F+5=8+2 ⇒F=8+2–5 ⇒F=5 ….(i) Since the number of faces = 5, we can say that the polyhedron could be a pentahedron – Square pyramid or Triangular prism.
Basics of Other Volumes 257
The square pyramid has 5 faces, 5 vertices, and 8 edges.
The triangular prism has 5 faces, 6 vertices, and 9 edges. Thus, the required pentahedron is a square pyramid as it has 5 faces, 5 vertices, and 8 edges.
Square Pyramid
Triangular Prism
∵F=5 [From eq.(i)] V= 5 and E = 8 (Given)
A figure or shape whose faces are polygons (triangles, quadrilaterals, etc.) is called a polyhedron. If F, E, and V denote respectively the number of faces, edges and vertices of a polyhedron, then F–E+V=2 This is called Euler’s Formula.
46. Symmetry Idea of symmetry Example 1 Cut and paste some pictures of symmetrical things in nature. Solution Symmetry means balance. Shapes, things, or figures, which can be equally divided into two along axes, are called symmetrical figures. Some pictures of symmetrical things in nature are as follows:
Symmetry means that one shape becomes exactly like another when it is moved by a turn, flip or slide.
Identifying lines of symmetry Example 1 Identify whether the dotted line on each shape is a line of symmetry or not.
258 A to Z of Mathematics
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Solution The axis or line of symmetry is an imaginary line that runs through the centre of a line or shape creating two perfectly identical halves. When the figure having a line of symmetry is folded along the line of symmetry, the two parts should coincide. Procedure (a) Dotted line is not a line of symmetry. (b) Dotted line is a line of symmetry. (c) Dotted line is a line of symmetry. (d) Dotted line is not a line of symmetry. (e) Dotted line is a line of symmetry. (f) Dotted line is not a line of symmetry. (g) Dotted line is not a line of symmetry. (h) Dotted line is a line of symmetry.
Explanation ∵ The dotted line does not divide the figure into two equal halves. ∵ The dotted line divides the figure into two equal halves. ∵ The dotted line divides the figure into two equal halves. ∵ The dotted line does not divide the figure into two equal halves. ∵ The dotted line divides the figure into two equal halves.
∵ The dotted line does not divide the figure into two equal halves.
∵ The dotted line divides the figure into two equal halves but the two parts does not coincide each other along the line. ∵ The dotted line divides the figure into two equal halves.
Idea of reflection symmetry Example 1 Identify which figures have reflectional symmetry. Also, draw the line of symmetry.
Basics of Other Volumes 259
(a)
(b)
(c)
(d)
Solution Reflection symmetry is also called line symmetry or mirror symmetry because there is a line in the figure where a mirror could be placed, and the combined figure would look the same as the original figure. Procedure Explanation (a) ∵ The dotted line is a line of symmetry as it divides the figure into two equal halves. ∴ The shape has a reflectional symmetry.
(b)
(c)
(d)
∵ Each dotted line is a line of symmetry as it divides the figure into two equal halves. ∴ The shape has a reflectional symmetry. ∵ The dotted line does not divide the figure into two equal halves. ∴ The shape do not have a reflectional symmetry.
∵ Each dotted line is a line of symmetry as it divides the figure into two equal halves. ∴ The shape has a reflectional symmetry
Idea of rotational symmetry Example 1 Write the order of rotational symmetry of the following shapes. Also, draw dotted lines to indicate the lines of symmetry.
260 A to Z of Mathematics
(a)
(b)
(c)
(d)
Solution If a shape is rotated around a centre point and it still looks the same after a rotation (by less than 360°), it is said to have rotational symmetry. And, the order of rotational symmetry is the number of times a shape fits onto itself when rotated by 360°. Procedure Explanation (a) The figure has 4 lines of symmetry (all dotted lines). When we rotate the figure about its centre The order of rotational symmetry is 4 (dotted lines point for 360°, the figure fits four times onto itself. with arrow heads).
(b) The figure has 5 lines of symmetry. The order of rotational symmetry is 5.
4
1
3
2
When we rotate the figure about its centre point for 360°, the figure fits five times onto itself. 5
1
2
4 3
(c) The figure has 2 lines of symmetry. The order of rotational symmetry is 2.
When we rotate the figure about its centre point for 360°, the figure fits two times onto itself.
2
1
Basics of Other Volumes 261
(d) The figure has 8 lines of symmetry. The order of rotational symmetry is 8.
When we rotate the figure about its centre point for 360°, the figure fits eight times onto itself. 8
1 2
7
3
6 5
4
The fixed point about which the object rotates is called the centre point or centre of rotation. And, the angle through which an object rotates about a fixed point is known as the angle of rotation.
47. Probability Idea of probability Example 1 Choose the appropriate option and give reasons. (a) How likely are you to grab a More likely Less likely
?
(b) How likely are you to grab a More likely Less likely
?
(c) How likely are you to grab a ? More likely Less likely (d) How likely are you to grab a More likely Less likely
?
Solution The probability of a certain event to occur depends on how many possible outcomes the event has. The higher the number of certain event from the possible outcomes, the more likely is it that the event will occur and thus, higher the probability.
262 A to Z of Mathematics
Procedure
Explanation ∵ Number of triangles < Number of squares
(a) Less likely
∵ Number of hearts < Number of stars ∵ Number of circles > Number of squares
(b) Less likely (c) More likely
∵ Number of triangles > Number of stars
(d) More likely
Probability means possibility. There are many real-life situations in which we may have to predict the outcome of an event. Example 2 Fill in the blanks: (a) Probability of a sure event is 1___. (b) Probability of an impossible event is 0___. (c) The probability of an event (other than sure and impossible event) lies between 0 and 1___. (d) Probability of an event A + Probability of event ‘not A’ = ___1___ The probability of happening of an event is always between 0 and 1. The probability of 1 means the event always happens. The probability of 0 means the event never happens and is impossible. Example 3 Write each letter in the correct place on the probability line. 0
1 4
1 2
3 4
1
(a) Next week, Wednesday will be the day after Tuesday. (b) There will be 33 days in February next year. (c) It will rain in Delhi in May. (d) It will snow in Chicago in January. (e) The next person to knock on the door will be a women. Solution The probabilities of the given situations are marked on the below given line. b
c
e
d
a
0
1 4
1 2
3 4
1
Procedure (a) Probability = 1 (b) Probability = 0
Explanation ∵ Wednesday always comes after Tuesday. ∵ February has either 28 days or 29 days.
Basics of Other Volumes 263
∵ There is a summer season in Delhi in May. Thus, the possibility of snow is less likely.
1 4 3 (d) Probability = 4 (c) Probability =
(e) Probability =
∵ There is a winter season in Chicago in January. Thus, the possibility of snow is more likely. ∵ There can be either man or a woman on the door. Thus, the possibility of a woman is equally likely.
1 2
The probability of 1 means the event has an equal chance of happening or not happening. 2 1 A probability value between 0 and means the event is less likely to happen. 2 1 A probability value of an event between and 1 means the event is more likely to happen. 2
Probability of throwing of dice Example 1 A die is rolled, what is the probability that (a) we get 5 on the die. (b) we get an even number on the die. Solution When a die is rolled, there are six possible outcomes: {1, 2, 3, 4, 5, 6}. So, the total number of possible outcomes is 6. Procedure (a) Total number of possible outcomes = 6 Number of favourable outcome = 1 P(getting 5) = ⇒ P(5) =
1 6
Explanation ….(i) ∵ Possible outcomes when a die is rolled = {1, 2, 3, 4, 5, 6} ….(ii) ∵ Favourable outcome is only 5.
Number of favourable outcomes Total number of possible outcomes
Finding the probability of getting 5 on the die. From eq. (i) and (ii)
Hence, the probability of getting 5 on the die is 1 . 6 (b) Total number of possible outcomes = 6 ....(i) ∵ Possible outcomes when a die is rolled = {1, 2, 3, 4, 5, 6}
264 A to Z of Mathematics
Number of favourable outcomes = 3 P(getting an even number) = Number of favourable outcomes Total number of possible outcomes
⇒ P(even number) = ⇒ P(even number) =
....(ii) ∵ Favourable outcome is getting an even number = {2, 4, 6} Finding the probability of getting an even number on the die.
3 6
From eq. (i) and (ii)
1 2
Simplifying
Hence, the probability that we get an even number on the die is
1 . 2
To find the probability of any event from the random experiment, divide the number of favourable outcomes of the event by the total number of possible outcomes. This will give us the probability of a single event occurring. P(A) =
Number of favourable outcomes of event A Total number of possible outcomes
Probability of tossing coins Example 1 We toss a fair coin twice. What is the probability that (a) the first toss turns up tails (b) we get two heads. Solution When a coin is tossed, there are only two possible outcomes: Head (H) or Tail (T). So, the total number of possible outcomes is 2 (Head or Tail). Procedure (a) Total number of possible outcomes = 2
….(i)
Number of favourable outcome = 1
….(ii)
P(getting a tail) = ⇒ P(T) =
1 2
Number of favourable outcomes Total number of possible outcomes
Explanation ∵ Possible outcomes which we will get on the first toss of the coin = {H, T} ∵ Favourable outcome is tail (T) Finding the probability of getting tail on the first toss of the coin. From eq. (i) and (ii)
Hence, the probability that the first toss turns up tails is
1 . 2
Basics of Other Volumes 265
(b) Total number of possible outcomes = 4 Number of favourable outcomes = 1 P(getting two heads) =
Number of favourable outcomes Total number of possible outcomes
⇒ P(HH) =
1 4
Hence, the probability that we get two heads is
....(i) ∵ Possible outcomes which we will get on tossing the coin twice = {HT, TH, TT, HH} ....(ii) ∵ Favourable outcome is getting two heads when the coin is tossed twice = (HH) Finding the probability of getting two heads when the coin is tossed twice. From eq. (i) and (ii)
1 . 4
A to Z Mathematics Series The three volumes of A to Z Mathematics reflect the distinction along the nature of content – Volume I: Elementary school (foundational, Grade KG-VIII) Volume II: Secondary school (Grade IX-X) Volume III: Senior Secondary school (Grade XI-XII) However, Volumes II and III cannot be transacted without extensively exploring the Volume I books. Math is rigidly hierarchal and it can only be transacted from the basics, the A to Z chapters. Volume II and Volume III are under preparation for print. Volume I has 15 chapters, from ‘A’ to ‘O’, and Volume II and III have 11 chapters, from ‘P’ to ‘Z’; the last chapter ‘Z’ ends with zero, i.e., limits, continuity, differentiation, integration. Each volume is further placed in three books to ensure ‘A to Z Mathematics’ is organised for most impact: • The Reading Book for comprehensive conceptual clarity • The Explained Examples Book for vivid visualisation and presentation of mathematical expressions and operations, • The Exercise(s) Book for just the right quantity and quality of practice for applying the concepts and arithmetical computations. This book is the third of eleven books: KG-VIII Explained Examples, Volume I (Part II) The other books in Volume I
KG-VIII Reading Book Volume I
KG-VIII Explained Examples Volume I – Part I
KG-VIII Exercise Book Volume I – Part I
KG-VIII Exercise Book Volume I – Part II
The books of ‘All that Matters in Math’ Volume II and Volume III 1. IX-X Reading Book, Volume II 2. IX-X Explained Examples, Volume II 3. IX-X Exercise Book, Volume II 4. XI-XII Reading Book, Volume III 5. XI-XII Explained Examples, Volume III 6. XI-XII Exercise Book, Volume III
Index of ‘All that Matters in Math’ Volume II and III Volume II
Volume III
Q for Quest – Mean, Median, Mode
P for Probability – Probability, Permutation, Combination
R for Roots – Polynomials, Quadratic Equations S for Shapes – Quadrilaterals, Circles T for Triangles – Triangle, Symmetry U for Uncut – Mensuration W for Watertight – Set Theory, Functions, Relations X for eXtra – Trigonometry, Linear Inequalities, LP, Standard Deviation, Variance
V for Vectors – Linear Equations and Solutions, Vectors, Matrices and Determinants X for eXtra – Conic Sections, 3D Geometry, Differential Equations Y for Yield – Commercial Math Z for Zero – Limits, Continuity, Differentiation, Integration
The Other Subject Books in A to Z Series
Story of Nations (Civics)
Science of History (3 Volumes)
All That Matters in Biology (2 Volumes)
Foundation of Physical Geography (3 Volumes)
All That Matters in Chemistry (2 Volumes)
Foundation of Human Geography (3 Volumes)
All That Matters in Physics (3 Volumes)
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