This book has been prepared by a group of faculties who are highly experienced in training GATE candidates and are also

*3,988*
*396*
*62MB*

*English*
*Pages [1163]*
*Year 2018*

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- Trishna Knowledge Systems

*Table of contents : Cover......Page 1Contents......Page 6Preface......Page 10Key Pedagogical Features......Page 11Syllabus: Computer Science and Information Technology......Page 13Chapter-wise Analysis of GATE Previous Years’ Papers......Page 14General Information about GATE......Page 15 Computer Science & Engineering......Page 17 General Aptitude......Page 33 Computer Science & Engineering......Page 35 General Aptitude......Page 50GATE 2018 Solved Paper......Page 52Part I: General Aptitude......Page 72 Part A: Verbal Ability......Page 74 Grammar—Overview......Page 76 Phrases and Clauses......Page 78 Conditionals......Page 80 Types of Sentences (Structural)......Page 81 Types of Sentences (Semantic)......Page 83 Sentences Correction......Page 84 Correct Use of Articles......Page 86 Correct use of Pronouns......Page 90 Correct Use of Adjectives......Page 93 Correct use of Adverbs......Page 96 Correct Use of the Verb......Page 98 Non-Finites......Page 103 Correct Use of Prepositions......Page 105 Correct Use of Conjunctions......Page 107 Active & Passive Voice......Page 109 Direct and Indirect Speech......Page 111 Diction (Words with Similar Pronunciation)......Page 113 Exercise......Page 114 Answer Keys......Page 119 Common Roots......Page 120 Exercises......Page 123 Previous Years’ Questions......Page 131 Answer Keys......Page 135 Test......Page 136 Answer Keys......Page 138 Part B: Numerical Ability......Page 140 Unit I: Quantitative Aptitude......Page 142 Two Equations in Two Unknowns......Page 144 Three Equations in Three Unknowns......Page 145 Additional Cases in Linear Equations......Page 146 Practice Exercise......Page 148 Answer Keys......Page 149 Ratio......Page 150 Proportion......Page 151 Variation......Page 152 Practice Exercise......Page 153 Answer Keys......Page 155 Classification of Real Numbers......Page 156 Rules for Divisibility......Page 157 Number of Factors of a Number......Page 159 LCM and HCF Models......Page 161 Successive Division......Page 162 Factorial......Page 163 Number Systems......Page 164 Conversions......Page 165 Binary Arithmetic......Page 166 Practice Exercise......Page 167 Answer Keys......Page 169 Percentage......Page 170 Profit and Loss......Page 172 Practice Exercise......Page 175 Answer Keys......Page 176 Interest......Page 177 Present Value......Page 179 Practice Exercise......Page 181 Answer Keys......Page 182 Points to Remember......Page 183 Mixtures......Page 184 Alligations......Page 186 Practice Exercise......Page 187 Answer Keys......Page 188 Work......Page 189 Practice Exercise......Page 192 Answer Keys......Page 194 Speed......Page 195 Races and Circular Tracks......Page 197 Practice Exercise......Page 198 Answer Keys......Page 200 Indices......Page 201 Logarithms......Page 204 Practice Exercise......Page 205 Answer Keys......Page 206 Quadratic Equations......Page 207 Sum and Product of Roots of a Quadratic Equation......Page 208 Practice Exercise......Page 210 Answer Keys......Page 212 Inequalities and Modulus......Page 213 Practice Exercise......Page 215 Answer Keys......Page 216 Progressions......Page 217 Geometric Progression (G.P.)......Page 219 Practice Exercise......Page 220 Answer Keys......Page 221 Combinations......Page 222 Circular Permutations......Page 224 Practice Exercise......Page 227 Answer Keys......Page 228 Data Table......Page 229 Two-Dimensional Graphs......Page 230 Practice Exercises......Page 232 Previous Years’ Questions......Page 239 Answer Keys......Page 244 Unit II: Reasoning......Page 248 Number Series......Page 250 Letter Series......Page 253 Practice Exercise......Page 254 Answer Keys......Page 255 Number Analogies......Page 256 Verbal Analogies......Page 257 Answer Keys......Page 258 Word Classification......Page 259 Practice Exercise......Page 260 Answer Keys......Page 261 Coding and Decoding......Page 262 Practice Exercise......Page 264 Answer Keys......Page 265 Blood Relations......Page 266 Practice Exercise......Page 269 Answer Keys......Page 270 Venn Diagram Type I......Page 271 Venn Diagram Type II......Page 272 Practice Exercise......Page 273 Answer Keys......Page 274 Linear Sequencing......Page 275 Circular Arrangement......Page 277 Practice Exercise......Page 280 Answer Keys......Page 282 Puzzles......Page 283 Order Sequence......Page 286 Selections......Page 288 Practice Exercise......Page 293 Answer Keys......Page 295 Clocks......Page 296 Calendar......Page 297 Practice Exercise......Page 299 Previous Years’ Questions......Page 300 Answer Keys......Page 303 Test......Page 304 Answer Keys......Page 305Part II: Engineering Mathematics......Page 306 Laws of First Order Logic......Page 308 First Order Logic (or) Predicate Calculus......Page 313 Exercises......Page 315 Previous Years’ Questions......Page 318 Answer Keys......Page 320 Probability......Page 321 Mathematical Expectation [E(X)]......Page 324 Exercises......Page 330 Previous Years’ Questions......Page 334 Answer Keys......Page 336 Some Definitions......Page 337 Basic Set Operations......Page 338 Sets in Number System......Page 339 Functions......Page 340 Algebra......Page 342 Partial Ordering and Lattices......Page 347 Lattice......Page 349 Exercises......Page 350 Previous Years’ Questions......Page 353 Answer Keys......Page 357 Combinations......Page 358 Generating Permutations......Page 363 Exercises......Page 366 Previous Years’ Questions......Page 368 Answer Keys......Page 369 Graph......Page 370 Planar Graphs......Page 376 Exercises......Page 379 Previous Years’ Questions......Page 384 Answer Keys......Page 387 Type of Matrices......Page 388 Exercises......Page 396 Previous Years’ Questions......Page 400 Answer Keys......Page 403 Limit of a Function......Page 404 Derivatives......Page 407 Improper Integrals......Page 411 Exercises......Page 412 Previous Years’ Questions......Page 415 Answer Keys......Page 417 Test......Page 418 Answer Keys......Page 419Part III: Computer Science and Information Technology......Page 420 Unit I: Digital Logic......Page 422 Digital Circuits......Page 424 Complements......Page 427 Exercises......Page 431 Previous Years’ Questions......Page 433 Answer Keys......Page 434 Logic Gates......Page 435 Boolean Algebra......Page 436 Boolean Functions, Min Terms and Max Terms......Page 438 Minimization of Boolean Functions......Page 440 Implementation of Function by Using NAND–NOR Gates......Page 443 EX-OR, EX-NOR Gates......Page 444 Exercises......Page 447 Previous Years’ Questions......Page 452 Answer Keys......Page 454 Arithmetic Circuits......Page 455 Code Converters......Page 460 Decoder......Page 461 Encoders......Page 463 Demultiplexer......Page 466 Exercises......Page 468 Previous Years’ Questions......Page 473 Answer Keys......Page 475 SequentiaL Circuits......Page 476 Latches......Page 477 Flip-flops......Page 480 Counters......Page 484 Registers......Page 488 Exercises......Page 492 Previous Years’ Questions......Page 497 Answer Keys......Page 499 Test......Page 500 Answer Keys......Page 502 Unit II: Computer Organizationand Architecture......Page 504 Computer......Page 506 Machine Instructions......Page 507 Addressing Modes......Page 510 Exercises......Page 513 Previous Years’ Questions......Page 516 Answer Keys......Page 518 ALU (Arithmetic and Logic Unit)......Page 519 CPU Control Design......Page 526 RISC and CISC......Page 529 Exercise......Page 530 Previous Years’ Questions......Page 533 Answer Keys......Page 534 Memory Interface......Page 535 Input–output Interfacing......Page 537 Exercises......Page 543 Previous Years’ Questions......Page 546 Answer Keys......Page 548 Pipelining......Page 549 Pipeline Hazards......Page 550 Exercises......Page 554 Previous Years’ Questions......Page 558 Answer Keys......Page 560 Characteristics of Memory System......Page 561 Cache Memory......Page 562 Secondary Storage......Page 567 Exercise......Page 569 Previous Years’ Questions......Page 572 Answer Keys......Page 574 Test......Page 575 Answers Keys......Page 576 Part A: Programming and Data Structures......Page 578 Data Types......Page 580 Operator......Page 582 Program Structure......Page 583 Control Statements......Page 584 Exercises......Page 587 Previous Years’ Questions......Page 589 Answer Keys......Page 590 Functions......Page 591 Recursion......Page 592 Parameter Passing......Page 593 Scope, Lifetime and Binding......Page 595 Exercises......Page 598 Previous Years’ Questions......Page 602 Answer Keys......Page 605 Arrays......Page 606 Two-dimensional Arrays......Page 607 Multidimensional Arrays......Page 608 Pointers......Page 609 Pointer to Function......Page 611 Structures......Page 612 Union......Page 614 Exercises......Page 615 Previous Years’ Questions......Page 619 Answer Keys......Page 621 SingLe-Linked List......Page 622 Uses of Stack......Page 627 Queue......Page 628 Exercises......Page 630 Previous Years’ Questions......Page 633 Answer Keys......Page 634 Binary Tree......Page 635 Binary Heap......Page 641 Exercises......Page 642 Previous Years’ Questions......Page 644 Answer Keys......Page 647 Test......Page 648 Answer Keys......Page 652 Part B: Algorithms......Page 654 Algorithm......Page 656 Tree Representation......Page 658 Data Structure......Page 659 Asymptotic Notations......Page 660 Notations and Functions......Page 663 Recurrences......Page 664 Exercises......Page 666 Previous Years’ Questions......Page 668 Answer Keys......Page 671 Merge Sort......Page 672 Insertion Sort......Page 673 Binary Search Trees......Page 674 Heap Sort......Page 675 Exercises......Page 677 Previous Years’ Questions......Page 679 Answer Keys......Page 680 Merge Sort......Page 681 Quick Sort......Page 682 Searching......Page 683 Exercises......Page 684 Answer Keys......Page 688 Knapsack Problem......Page 689 Prim’s Algorithm......Page 690 Tree and Graph Traversals......Page 691 Connected Components......Page 693 Huffman Codes......Page 694 Sorting and Order Statistics......Page 695 Graph Algorithms......Page 696 Exercises......Page 699 Previous Years’ Questions......Page 701 Answer Keys......Page 707 Multi-stage Graph......Page 708 All Pairs Shortest Path Problem (Floyd–Warshall Algorithm)......Page 710 Hashing Methods......Page 712 Matrix-chain Multiplication......Page 714 Longest Common Subsequence......Page 715 NP-hard and NP-complete......Page 716 Exercises......Page 717 Previous Years’ Questions......Page 720 Answer Keys......Page 722 Test......Page 723 Answers Keys......Page 726 Unit IV: Databases......Page 728 Schemas......Page 730 ER Model......Page 731 Relationship Sets......Page 732 Cardinality Ratio and Participation Constraints......Page 734 Relational Database......Page 735 Relational Model Constraints......Page 737 Triggers......Page 741 Exercises......Page 742 Previous Years’ Questions......Page 746 Answer Keys......Page 747 Relational Algebra......Page 748 Tuple Relational Calculus......Page 753 SQL (Structured Query Language)......Page 754 Exercises......Page 760 Previous Years’ Questions......Page 769 Answer Keys......Page 774 Normalization......Page 775 First Normal Form (1NF)......Page 776 Second Normal Form......Page 779 Third Normal Form......Page 780 Higher Normal Forms (Boyce–Codd Normal Form)......Page 781 Fourth Normal Form......Page 782 Fifth Normal Form......Page 783 Exercises......Page 784 Previous Years’ Questions......Page 788 Answer Keys......Page 789 Transaction......Page 790 Concurrency Control......Page 791 Transaction Processing Systems......Page 792 Concurrency Control with Locking Methods......Page 793 Concurrency Control with Time Stamping Methods......Page 795 Serializability......Page 796 Recoverability......Page 797 Equivalence of Schedules......Page 798 Exercises......Page 799 Previous Years’ Questions......Page 804 Answer Keys......Page 806 Memory Hierarchies......Page 807 File Records......Page 808 Hashing Techniques......Page 809 Indexing......Page 811 Exercises......Page 815 Previous Years’ Questions......Page 819 Answer Keys......Page 820 Test......Page 821 Answer Keys......Page 823 Unit V: Theory of Computation......Page 824 Fundamentals......Page 826 NFA with ∈-Moves......Page 829 Mealy and Moore Machines......Page 833 Regular Languages......Page 835 Constructing FA for Given RE......Page 836 Closure Properties of Regular Sets......Page 837 Types of Gramars......Page 838 Exercises......Page 840 Previous Years’ Questions......Page 842 Answer Keys......Page 846 Context Free Language (CFL)......Page 847 Minimization of Context Free Grammar......Page 848 Normal Forms......Page 849 Pumping Lemma for Context Free Languages......Page 851 Push Down Automata (PDA)......Page 852 Converting CFG to PDA......Page 853 Deterministic PDA (Deterministic CFL)......Page 854 Exercises......Page 855 Previous Years’ Questions......Page 857 Answer Keys......Page 858 Turing Machines......Page 859 Languages Accepted by a TM......Page 860 Types of Turing Machines......Page 861 Recursively Enumerable languages......Page 862 Undecidability......Page 863 Problems......Page 864 NP-Hard Problem......Page 865 Exercises......Page 866 Previous Years’ Questions......Page 869 Answer Keys......Page 870 Test......Page 871 Answers Keys......Page 873 Unit VI: Compiler Design......Page 874 Language Processing system......Page 876 Lexical Analysis......Page 878 Context Free Grammars and Ambiguity......Page 879 Topdown Parsing......Page 881 Bottom up Parsing......Page 883 Exercises......Page 892 Previous Years’ Questions......Page 895 Answer Keys......Page 898 Syntax Directed Definitions......Page 899 Synthesized Attribute......Page 900 Constructing Syntax Trees for Expressions......Page 901 Syntax Directed Transiation Schemes......Page 902 Bottom-up Evaluation of Inherited Attributes......Page 903 Exercises......Page 904 Previous Years’ Questions......Page 906 Answer Keys......Page 907 Introduciion......Page 908 Three-Address Code......Page 909 Assignment Statements......Page 911 Flow of Control Statements......Page 913 Procedure Calls......Page 915 Code Generation......Page 916 Runtime Storage Management......Page 918 DAG Representation of Basic Blocks......Page 921 Peephole Optimization......Page 922 Exercises......Page 923 Answer Keys......Page 926 Principle Sources of Optimization......Page 927 Loops in Flow Graphs......Page 932 Global Dataflow Analysis......Page 934 Exercises......Page 936 Previous Years’ Questions......Page 938 Answer Keys......Page 940 Test......Page 941 Answers Keys......Page 943 Unit VII: Operating System......Page 944 Basics of Operating System......Page 946 Processes......Page 947 Process Control Structures......Page 950 Process Control......Page 951 Threads......Page 952 Types of Threads......Page 953 Threading Issues......Page 954 Exercises......Page 955 Previous Years’ Questions......Page 958 Answer Keys......Page 959 PrIncIples of Concurrency......Page 960 Process Interaction (IPC)......Page 961 Mutual Exclusion......Page 962 Other Mechanisms for Mutual Exclusion......Page 964 Classical Problems of Synchronization......Page 966 Message Passing......Page 968 Exercises......Page 969 Previous Years’ Questions......Page 973 Answer Keys......Page 977 Deadlock......Page 978 Methods of Handling Deadlocks......Page 980 Dining Philosophers Problem......Page 983 Types of Processor Scheduling......Page 984 Scheduling Algorithms......Page 985 Exercises......Page 989 Previous Years’ Questions......Page 992 Answer Keys......Page 995 Memory Management Requirements......Page 996 Memory Mapping Techniques......Page 997 Virtual Memory......Page 1002 OS Software for Memory Management......Page 1006 Exercises......Page 1010 Previous Years’ Questions......Page 1013 Answer Keys......Page 1015 File Systems......Page 1016 File System Architecture......Page 1017 File Directories......Page 1019 Secondary Storage Management......Page 1020 Unix File Management......Page 1021 I/O Systems......Page 1023 Disk Scheduling......Page 1024 RAID (Redundant Array of Independent Disks)......Page 1027 Protection and Security......Page 1029 Exercises......Page 1032 Previous Years’ Questions......Page 1034 Answer Keys......Page 1036 Test......Page 1037 Answer Keys......Page 1039 Part A: Network......Page 1040 LAN......Page 1042 The OSI Reference Model......Page 1044 LAN technologies......Page 1047 Physical Layer......Page 1050 Types of Errors......Page 1053 Medium Access Control Sublayer......Page 1055 Exercises......Page 1057 Previous Years’ Questions......Page 1059 Answer Keys......Page 1061 Flooding......Page 1062 Link State Routing......Page 1063 OSPF......Page 1064 Congestion Control Techniques......Page 1065 Exercises......Page 1067 Previous Years’ Questions......Page 1070 Answer Keys......Page 1073 User Datagram Protocol (UDP)......Page 1074 TCP/IP......Page 1075 Application Layer......Page 1079 SMTP......Page 1081 HTTP......Page 1082 DNS......Page 1083 Networking Devices......Page 1084 Exercises......Page 1086 Answer Keys......Page 1088 IP Addressing......Page 1089 Subnet Mask......Page 1091 Classless Inter Domain Routing (CIDR)......Page 1092 IP-Protocol......Page 1094 Exercises......Page 1098 Previous Years’ Questions......Page 1101 Answer Keys......Page 1102 Network Security Basics......Page 1103 Cryptographic Techniques......Page 1104 Traditional Cipher Algorithms......Page 1105 Symmetric Key Encryption......Page 1106 Asymmetric Key Encryption......Page 1107 Exercises......Page 1109 Answer Keys......Page 1111 Test......Page 1112 Answer Keys......Page 1113 Part B: Information Systems......Page 1114 Software Components and Elements......Page 1116 Requirement Analysis......Page 1117 Feasibility Analysis......Page 1118 Input/Output DesIgn......Page 1119 Software Process Life Cycle......Page 1120 Software Process Models......Page 1121 Exercises......Page 1123 Answer Keys......Page 1125 Project Planning Tools......Page 1126 Software Design......Page 1127 Software Testing......Page 1131 Implementation and Maintenance......Page 1134 Exercises......Page 1137 Previous Years’ Questions......Page 1140 Answer Keys......Page 1141 Test......Page 1142 Answer Keys......Page 1144 Part C: Software Engineering and Web Technology......Page 1146 Tags......Page 1148 Cascading Style Sheets......Page 1154 Extensible Markup Language (XML)......Page 1155 Client–Server Computing......Page 1157 Exercises......Page 1158 Previous Years’ Questions......Page 1160 Answer Keys......Page 1161 Test......Page 1162 Answer Keys......Page 1163*

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GATE

(Graduate Aptitude Test in Engineering)

Computer Science and Information Technology Trishna Knowledge Systems

Copyright © 2017 Trishna Knowledge System Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128, formerly known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 978-93-528-6846-9 eISBN 978-93-530-6116-6 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India.

Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India. Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: [email protected]

Contents Preface ix Key Pedagogical Features x Syllabus: Computer Science and Information Technology xii Chapter-wise Analysis of GATE Previous Years’ Papers xiii General Information about Gate xiv Solved Papers 2017 xvi solved PaPers 2018 l i

PART I General Aptitude PART A Verbal Ability Chapter 1 Grammar

1.5

Chapter 2 Vocabulary

1.49

PART B Numerical Ability

UNIT I Quantitative Aptitude

1.71

Chapter 1 Simple Equations

1.73

Chapter 2 Ratio-Proportion-Variation

1.79

Chapter 3 Numbers

1.85

Chapter 4 Percentage, Profit and Loss

1.99

Chapter 5 Simple Interest and Compound Interest

1.106

Chapter 6 Average, Mixtures, Alligations

1.112

Chapter 7 Time and Work

1.118

Chapter 8 Time and Distance

1.124

Chapter 9 Indices, Surds, and Logarithms

1.130

Chapter 10 Quadratic Equations

1.136

Chapter 11 Inequalities

1.142

Chapter 12 Progressions

1.146

Chapter 13 Permutations and Combinations

1.151

Chapter 14 Data Interpretation

1.158

UNIT II Reasoning

1.177

Chapter 1 Number and Letter Series

1.179

Chapter 2 Analogies

1.185

Chapter 3 Odd Man Out (Classification)

1.188

Chapter 4 Coding and Decoding

1.191

Chapter 5 Blood Relations

1.195

Chapter 6 Venn Diagrams

1.200

vi | Contents Chapter 7 Seating Arrangements

1.204

Chapter 8 Puzzles

1.212

Chapter 9 Clocks and Calenders

1.225

PART I1 Engineering Mathematics Chapter 1 Mathematical Logic

2.237

Chapter 2 Probability

2.250

Chapter 3 Set Theory and Algebra

2.266

Chapter 4 Combinatorics

2.287

Chapter 5 Graph Theory

2.299

Chapter 6 Linear Algebra

2.317

Chapter 8 Calculus

2.333

PART III Computer Science and Information Technology UNIT 1 Digital Logic

3.351

Chapter 1 Number Systems

3.353

Chapter 2 Boolean Algebra and Minimization of Functions

3.364

Chapter 3 Combinational Circuits

3.384

Chapter 4 Sequential Circuits

3.405

UNIT II Computer Organization and Architecture

3.433

Chapter 1 Machine Instructions, Addressing Modes

3.435

Chapter 2 ALU and Data Path, CPU Control Design

3.448

Chapter 3 Memory Interface, I/O Interface

3.464

Chapter 4 Instruction Pipelining

3.478

Chapter 5 Cache and Main Memory, Secondary Storage

3.490

UNIT III Programming and Data Structures

3.507

PART A Programming and Data Structures Chapter 1 Programming in C

3.509

Chapter 2 Functions

3.520

Chapter 3 Arrays, Pointers and Structures

3.535

Chapter 4 Linked Lists, Stacks and Queues

3.551

Chapter 5 Trees

3.564

Contents | vii

PART B Algorithms Chapter 1 Asymptotic Analysis

3.585

Chapter 2 Sorting Algorithms

3.601

Chapter 3 Divide and Conquer

3.610

Chapter 4 Greedy Approach

3.618

Chapter 5 Dynamic Programming

3.637

UNIT 1V Databases

3.657

Chapter 1 ER Model and Relational Model

3.659

Chapter 2 Structured Query Language

3.677

Chapter 3 Normalization

3.704

Chapter 4 Transaction and Concurrency

3.719

Chapter 5 File Management

3.736

UNIT V Theory of Computation

3.753

Chapter 1 Finite Automata and Regular Languages

3.755

Chapter 2 Context Free Languages and Push Down Automata

3.776

Chapter 3 Recursively Enumerable Sets and Turing Machines, Decidability

3.788

UNIT V1 Compiler Design

3.803

Chapter 1 Lexical Analysis and Parsing

3.805

Chapter 2 Syntax Directed Translation

3.828

Chapter 3 Intermediate Code Generation

3.837

Chapter 4 Code Optimization

3.856

UNIT VII Operating System

3.873

Chapter 1 Processes and Threads

3.875

Chapter 2 Interprocess Communication, Concurrency and Synchronization

3.889

Chapter 3 Deadlock and CPU Scheduling

3.907

Chapter 4 Memory Management and Virtual Memory

3.925

Chapter 5 File Systems, I/O Systems, Protection and Security

3.945

viii | Contents

UNIT VIII Networks, Information Systems, Software Engineering and Web Technology

3.969

Part A Network Chapter 1 OSI Layers

3.971

Chapter 2 Routing Algorithms

3.991

Chapter 3 TCP/UDP

3.1003

Chapter 4 IP(V4)

3.1018

Chapter 5 Network Security

3.1032

Part B Information Systems Chapter 1 Process Life Cycle

3.1045

Chapter 2 Project Management and Maintenance

3.1055

Part C Software Engineering and Web Technology Chapter 1 Markup Languages

3.1077

Preface Graduate Aptitude Test in Engineering (GATE) is one of the preliminary tests for undergraduate subjects in Engineering/ Technology/Architecture and postgraduate subjects in Science stream only. The number of aspirants appearing for the GATE examination is increasing significantly every year, owing to multifaceted opportunities open to any good performer. Apart from giving the aspirant a chance to pursue an M.Tech. from institutions like the IITs /NITs, a good GATE score can be highly instrumental in landing the candidate a plush public sector job, as many PSUs are recruiting graduate engineers on the basis of their performance in GATE. The GATE examination pattern has undergone several changes over the years—sometimes apparent and sometimes subtle. It is bound to continue to do so with changing technological environment. GATE Computer Science and Information Technology, as a complete resource helps the aspirants be ready with conceptual understanding, and enables them to apply these concepts in various applications, rather than just proficiency with questions type. Topics are handled in a comprehensive manner, beginning with the basics and progressing in a step-by-step manner along with a bottom-up approach. This allows the student to better understand the concept and to practice applicative techniques in a focused manner. The content has been systematically organized to facilitate easy understanding of all topics. The given examples will not only help the students to understand the concepts involved in the problems but also help to get a good idea about the different models of problems on that particular topic. Due care has also been taken to cover a very wide range of problems including questions that have been appearing over the last few years in GATE examination. The practice exercises in every chapter, contain questions ranging simple to moderate to difficult level. These exercises are meant to hone the examination readiness over a period of time. At the end of each unit, practice tests have been placed. These tests will help the student assess their level of learning on a regular interval. This book has been prepared by a group of faculty who are highly experienced in training GATE preparations and are also subject matter experts. As a result, this book would serve as an effective tool for GATE aspirant to crack the examination. Salient Features 1. Elaborate question bank covering previous 12 years’ GATE question papers 2. 5 free online mock tests for practice 3. Detailed coverage of key topics 4. Complete set of solved 2017 GATE online papers with chapter-wise analysis 5. Exhaustive pedagogy: (a) More than 1300 Solved Examples (b) More than 6000 Practice Questions (c) Unit-wise time-bound tests (d) Modular approach for easy understanding We would like to thank the below mentioned reviewers for their valuable feedback and suggestions which has helped in shaping this book. R. Marudhachalam

Professor (Sr. Grade), Kumaraguru College of Technology Coimbatore, Tamil Nadu

Daya Gupta

Professor, Delhi Technological University, Main Bawana Road, Delhi

Manoj Kumar Gupta

Associate Professor, Delhi Technological University Main Bawana Road, Delhi

Gurpreet Kour

Lecturer, Lovely Professional University Phagwara, Punjab

Pinaki Chakraborty

Assistant Professor, Netaji Subhas Institute of Technology Dwarka, Delhi

Gunit Kaur

Lecturer, Lovely Professional University Phagwara, Punjab

Despite of our best efforts, some errors may have inadvertently crept into the book. Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully. Wishing you all the very best..!!! —Trishna Knowledge Systems

Chapter 1

Key Pedagogical Features Number Systems 3.536 | Digital Logic

BCD addition • BCD addition is performed by individually adding the corresponding digits of the decimal number expressed in 4-bit binary groups starting from the LSB. LEARNING OBJECTIVES • If there is no carry and the sum term is not an illegal code, no correction is needed. After reading this chapter, you will be able to understand: • If there is a carry out of one group to the next group or if • Digital circuits the sum term is an illegal code, the (6)10 is added to the •sum Number with different term system of that group, and thebase resulting carry is added •toConversion of number systems the next group.

Learning Objectives List of important topics which are covered in chapter. Solution:

• Complements +B 12 fExample {A ⊕43:B44⊕ ⊕ C } ⊕ {A ⊕ C •=Subtraction with complement 0100 0100 (44 in BCD) • Numeric codes

⊕ B ⊕ A}

Excess-3 (XS-3) code Excess-3 code is a non-weighted BCD code, where each digit binary code word is the corresponding 8421 code word plus 0011. Find the XS-3 code of Example 47: (3)10 → (0011)BCD = (0110)xS3

Chapter 2 (16) Boolean •Example Weighted non-weighted codes 48:and → (0001 Algebra 0110) 10 BCD (0100 1001) • Error detection and→correction code xS3

• Sequential, reﬂective and cyclic codes Gray code • Self complementing Y = (code A ⋅ B ⋅ AB) =

( A + Bnumber ) (A+ Each gray code number differs from the preceding by a single bit.

B) ( A ⋅ B = A + B)

= ( A + BGray ) + (Code A + B) = A ⋅ B + A ⋅ B

= {A ⊕ 0 ⊕ C} ⊕ {0(12 ⊕inCBCD) ⊕ B} 0001 0010 0101

and Minimization of Functions | 29

Decimal

0110 (56 in BCD)

0111 0110 . 1001 DIGITAL CIRCUITS 0101 0110 . 0110 Computers work with binary numbers, which (all use are onlyillegal the digits Solved Examples . 1111 are based ‘0’ and ‘1’. Since all the1100 digital1100 components on binary

0

0000

2

0011

= 1A ⋅ B + A0001 ⋅ B = A⊕ B

=A⊕C⊕C⊕B=A⊕0⊕B=A⊕B Example 44: 76.9+ 56.6

2 Example 5: (658)8 = 6 ×3 84: + 5Simplify × 81 0010 + 8 × 80 the Example 4 + 40 + 8 = 0110 = 384 (432)10

Boolean function A ⊕ A B ⊕ A

Solved Example Chapter 1 Number Systems

| 13

Solution: Asystem ⊕ A Bgiven ⊕ A topic-wise codes) number Solved problems 0110 0110 . 0110 operations, it is convenient to use binary numbers when analyzing Hexadecimal Binary to gray conversion Example 1: Simplify the Boolean function, x y + x′z + y z (A) 11001001 (B) 10011100 code. ForI: example, the system, base number 35 will rep0010 0010 . 0101 In hexadecimal number there areposition 16the numbers 0be toright, 9, and or designing digital circuits. Step Shift binary number one toconcepts the tothe learn to -6 apply Associativity digits from 10 to 15 number are represented by A to F, respectively. The +1 +1 (propagate carry)resented (C) 11010101 (D) +1 10101101 bytheits BCH code 011101. LSB of shifted is discarded. learned in a particular section as of hexadecimal number system is 16. with 0011 Different Solution:Number x y + x′Systems z+ y z 0001 0011 .Base 0101 Inbase numbering the BCH code Stepthis II: Exclusive or the system, bits = of1the binary number with 4. Let r denotes number system’s1radix. The3only value(s) ⊕ AB = A B 3 . 5 3 2 1 0 per exam pattern. Decimal number system Example 6: (1A5C) = 1 × 16 + Athe × 16following + 5 × 16 + C × 16 By using consensus property 001001101011 corresponds to number those of the binary number shifted. 16 3 (1331) = (11) is/are of r that satisfy thenumbers equation = 1 × 4096 + 10 × 256 + 5 × 16 + 12 × 1 r r we use in our day-toare usual which in base -6 system. Example 49: Convert (1001)2 to gray code BCD subtraction BCDnumbers subtraction is performed by sub= A + B ( De Morgan’s) xy + x′z +Decimal yz = xy + x′z = 4096 + 2560 + 80 + 12 = (6748)10 . daytracting life. Thethe base of the number system is 10. There are (A) 10 (B) 11 4-bit digits ofdecimal each group of the subtrahend → 1010 (A) Binary 4651 (B) 4562 Y(C)= xy + x′z ten numbers 0 to 9. Table 1 Different number systems 10 and 11 (D) any r >group 3 from the corresponding 4-bit of the minuend in Binary → 101 ⊕ (D) 1353 (C) Shifted 1153 The value of the nth digit of the number from the right side DecimalExample Hexadecimal AB binary starting from the LSB. Gray →Binary 1111 5: Octal = nth digit × (base)n–1 5

0111

5. Example X is 16-bit signed number. repre2: The outputThe of 2’s thecomplement given circuit is equal to signed 0 2’s complement 000 0 11. The representation (-589) 1010 00 010of 11 CD Gray to 1: 45: (99)1042 90100 × 10 9 × 10(042 in BCD 1 binary conversion 001 1 1 Example sentation ofExample X is (F76A) .→The 2’s1 +complement repre0010 ) 16 in Hexadecimal number system is 1 0 0 1of (a) Take the MSB of the binary number is same as MSB 00 = 90 + 9 = 99 A 8 × X is 2 010 2 2 sentation of −12 −0001 0010 (12 IN BCD) (A) (F24D) (B)3 (FDB3)16 3 gray code number. 2 1 0 16 Example 2: (332) → 3 × 10 + 3 × 10 + 2 × 10 3 011 (No borrow, so this is × × 10 0 1bit 01to (A) (1460) B16 30 (B) 0011(D643) 0000 X-OR the MSB the next16significant (F42D) (D) = 300 + 30 + 2 16the correct difference) (C) (b) 4 100of the binary 4 (F3BD) 4 16 (C) (4460)16 (D) (BB50) ×5 1 × × 11 5 the gray code. 101 5 Example 3: (1024)10 → 1 × 103 + 0 ×16102 + 2 × 101 + × 10012. The baseof of the system which the following Example 46: (c) X-OR thenumber 2nd110 bit of binary for 6 6to the 3rd bit 6of Gray code = 1000 + 0number + 20 + 4 system = 1024 is 6. The HEX number (CD.EF)16 in octal (Borrow 66 10 1 0 1 1 to7 is getto3rd binary and = so13 247.7 0010 0100 0111 . 0111 111 7on. 7 bebitcorrect (A) (315.736) (B) (513.637)8 are operation 8 (d) Continue this1000 till all the number system ABinary 5 gray 8 10 bits are exhausted. 8 − 156 . 9 0001 0101 0110 . 1001 present,(A) 6 The minimized expression for the given K–map is (C) (135.673) (D) there (531.367) (B)11 7 to binary 8 8 two digits ‘0’ and ‘1’. 9 1001 gray code 9 system, are only Example 50: Convert, 1010 BIn binary number 90.8 0000 0111 ⋅0000 . 1110 subtract(C) 8 (D)12 19 Since there are only two numbers, its base is 2. 1010 1 0AB 0 A 7. 8-bit 2’s complement representation a decimal number 0110) Gray 10 Solution: 11 1011 13 B = 1 × 23 + 1 ×−01001 22 + 0 × −210110 + 1 × 20 Example 4: (1101)in 2 decimal is 1010 is 10000000. The number 13 = 0 13. The solution to the 1100 quadratic x||2 - 11x ⇓ ⊕ || equation ⊕1400 ⊕ 01 || 11 +10 CD 12 C = 8 + 4 + 11001 = (13)10 000 ⋅ 1000 Corrected Solution: AB0 (A) +256 A B = AB +(B) (in number system with radix r) are x = 2 x = 4. difference 1 15 00 13 1100 1101 1 00 D1and 1 00 = (1100) (90.8) Then base (C) -128 Octal number system (D) -256 of the2 number system 14 1110 16 is (r) = E A Octal number system has eight numbers 0 to 7. The base of the × F× 1 01 1111 (A) 7 15 (B)17 0 6 1 numbers that can be rep8. The range of signed decimal B number system is 8. The number (8)10 is represented by (10)8.Exercises Exercises (Continued )× × × (C) 5 (D) 4 11 1 resented by 7-bit 1’s complement representation is 2. Ifcomplement (84)x (in base xofnumber Practice problems for (A) stu-64 to +Practice 63 (B) -63 Problems 1 to + 63 3 14.y The 16’s BADA is1 is0equal1to (64) 10 system) 1 y (in base y number system), then possible values of x and y (C) -127 to + 128 (D) -128 to +127 (A) 4525 (B) 4526 Directions for questions 1 to 15: Select the correct alternadents to master the concepts Chapter 01.indd 529 8/17/2015 8:25:39 PM are tive from the given choices. (C) ADAB (D)(B) 2141 studied in chapter. 9.Exercises Decimal A 54 in hexadecimal and BCD number system is (A) 12, 9A + B C 6, 8 = 2 1. Assuming all the numbers are in 2’s complement X OR gate rep(C)=9,(12CD) 12 (D) expression 12, 18 respectively 15. (11A1B) , in the above A and B consist of two levels of prob-B 8 16 resentation, which of the following is divisible by (A) 63, 10000111 (B) 36,01010100 represent digits octal system and C y , y , y will 6: In the shown, 3. Letpositive AExample = 1111 1011in and B = number 0000 figure 1011 be two 8-bit 11110110? lems “Practice Problem I” 2 1 0 (C) 66, 01010100 and D have their original meaning in Their Hexadecimal, signed 2’s complement numbers. product in the 2’s (A) 11101010(D) 36, 00110110 (B) 11100010 complement of x x x if z = ? and “Practice Problem II” A B = AB + (D) AB11100111 2 1 0 complement representation is (C) 11111010 values of A and B are? 10.diffi A culty new binary-coded hextary (BCH) number system based on increasing (A) 2, 5 (B) 2, 3 is proposed in which everycircuit digit ofisa ‘0’. base As -6 number So the output of above two inputs are same level. (C) 3, 2 (D)x03, 5 y0 system is represented by its corresponding 3-bit binary

at third gate. Output of XOR gate with two equal inputs is zero. Chapter 01.indd 536 x (A) 11111 (B) 110001 Practice Problems 2 \y=0 (C) 01111 (D) 10000 Directions for questions to 20: Select theincorrect alterna-is functionally Example 3: The 1circuit shown the figure 7. (0.25) in binary number system tive from the given choices. x2is? 10 equivalent to 1. The hexadecimal representation of (567)8 is (A) (B) D77 A 1AF (C) 177 (D) 133 B 2. (2326)8 is equivalent to (A) (B) (103112)4 A (14D6)16 (C) (1283)10 (D) (09AC)16 B 3. (0.46) equivalent in decimal is? 8

(A) 0.59375 (C) 0.57395 Solution:

(B) 0.3534 (D) 0.3435

4. The 15’s complement of (CAFA)16 is A (A) (2051)16 (B) (2050)16A · B A (A + B ) (C) (3506)16 (D) (3505)16 B

(A) (0.01) (C) 0.001

be 1s

8/12/2015 12:15:36 PM

(B) (0.11) (D) 0.101

y1 y2

8. The equivalent of (25)6 in number system base 7 z = with ? is? (A) 22 (B) 23 Solution: We(D) are26using X-OR gate (C) 24

∴ XOR out-put is complement of input only when other

9. The operation 35 + 26 = 63 is true in number system with radix input is high. ∴Z=1 (A) 7 (B) 8 (C) 9 (D) 11

Example 7: The output y of the circuit shown is the figure is

10. The hexadecimal equivalent of largest binary number A with 14-bits is? (A) 2FFF (B) 3FFFF B y=A⊕B

above code? (A) 5 (B) 6 (C) 8 (D) 7

What type of values is passed to the called procedure? (A) l-values (B) r-values (C) Text of actual parameters (D) None of these 19. Which of the following is FALSE regarding a Block? (A) The first statement is a leader. (B) Any statement that is a target of conditional / unconditional goto is a leader. (C) Immediately next statement of goto is a leader. (D) The last statement is a leader.

17. A basic block can be analyzed by (A) Flow graph (B) A graph with cycles (C) DAG (D) None of these

Previous Years’ Questions

previous yeArs’ QuesTions 1. The least number of temporary variables required to create a three-address code in static single assignment form for the expression q + r/3 + s – t * 5 + u * v/w is ________ [2015] 2. Consider the intermediate code given below. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11)

i=1 j=1 t1 = 5 * i t2 = t1 + j t3 = 4 * t2 t4 = t3 a[t4] = –1 j=j+1 if j < = 5 goto (3) i=i+1 if i < 5 goto (2)

The number of nodes and edges in the control-flowgraph constructed for the above code, respectively, are [2015]

(A) 5 and 7 (B) 6 and 7 (C) 5 and 5 (D) 7 and 8 3. Consider the following code segment. [2016] x = u – t; y = x * v; x = y + w; y = t – z; y = x * y; The minimum number of total variables required to convert the above code segment to static single assignment form is _____ . 4. What will be the output of the following pseudocode when parameters are passed by reference and dynamic scoping is assumed? [2016] a = 3; void n(x) { x = x* a; print (x);} void m(y) {a = 1; a = y – a; n(a) ; print (a)} void main( ) {m(a);} (A) 6,2 (B) 6,6 (C) 4,2 (D) 4,4

Answer keys

38 | Digital Logic exerCises

Hints/Solutions

Hints/solutions

Practice Problems 1 1. D 2. D 3. C 11. B 12. C 13. Practice Problems 1 A

4. C 14. C

5. B 15. B

6. A

7. B Then Z = A + B

8. A

9. A

10. A

Hence, the correct option is (B). 8. Error → transmits odd number of one’s, for both cases. 1. B 2. B 3. A 4. B 5. B 6. A 7. B 8. C 9. A 10. A option 11. B 12. D A 13. A A 14. C 15. C 16. A Hence, 17.theC correct18. B is (A). 19. D 9. ∑(0, 1, 2, 4, 6) P should contain minterms of each funcX-ORYears’ of two equal inputs will give you result as zero. Previous Questions tion of x as well as y correct option 1. 8Hence, the 2. B 3. 10 is (B).4. D Hence, the correct option is (B). 2. Positive level OR means negative level AND vice versa 10. AB + ACD + AC Hence, the correct option is (D). = AB(C + C )( D + D) + A( B + B )CD + A( B + B )C ( D + D) 3. AB⋅CD ⋅ EF ⋅ GH = AB (CD + CD + CD + C D) + ABCD + ABCD + (De Morgan’s law) = AC ( BD + BD + BD + BD) = ( A + B ) (C + D ) ( E + F ) (G + ( H )) ABCD + ABCD + ABCD + ABC D + ABCD + 1. Practice Problems0 2

A

A

y=1

Hence, the correct option is (B).

4.

A B

Contains previous 10 years GATE Questions at end of every chapter which help students to get an idea about the type of problems asked in GATE and prepare accordingly.

AB AB · B = AB + B (AB + B ) · B = A + B

Practice Tests

This section gives complete solutions of all the unsolved questions given in the chapter. The Hints/Solutions are included in the CD accompanying the book.

ABCD + ABCD + ABCD + ABCD Hence, the correct option is (A).

Test | 103

11. ABCD + ABCD + ABCD + ABCD

TesT DigiTal logic

= ABC + ABC y

= BC Time-bound Test provided at Hence, the correct option is (C). end of each unit for assessment 12. YZ C Directions for questions 1 to 01 30: Select of topics leaned in the(ABunit. 11 10the correct alterna00 WX + B ) · C = AC + BC

tive from the given choices. 1 00 1 1. What is the range of signed decimal numbers that can 1 1 1 01 1 be represented by 4-bit 1’s complement notation? = ( A + C + B ) = ABC 1 to +16 (A) –7 to + 7 11 1 (B) –16 (C) –7 to +8 10 1 (D) –15 1 to +16 Hence, the correct option is (C). 1 octet + 1 quad 2. Which following signed representation have a 5. The output should be high when at least two outputs are of the unique representation = z + wx of 0? high y = ABC + ABC + ABC + ABC (A) Sign-magnitude (B) 1’s complement Hence, the correct option is (D). The minimized output (C) 0’s complement (D) 2’s complement 13. A AB y = AB + AC + BC 3. Find the oddBone out among the following Hence, the correct option is (A). (A) EBCDIC (B) GRAY P = AB (C) Hamming (D) 6. f1(x, y, z) A ASCII x x + f3 4.z)Gray code for number 8 is Y f2(x, y, z) f (x, y, (A) 1100 (B) 1111 A + B (C) 1000 (D) B 1101 f3(x, y, z) 5. Find the equivalent expression for z = x + xy C = ABlogical ⋅(A + B ) (A) z = x y (B) Z = xy x consists of all min terms, so x = 0, and f = f3 = ( A + B ) ( A + B ) (C) Z = x + y (D) Z = x + y f3 (x1 y1 z) = (1, 4, 5) AB + AB 6. The number of=distinct Boolean expression of 3 variHence, the correct option is (A). ables is Hence, the correct option is (A). 7. J I (A) 256 (B) 16 14. AB (C) 1024 C (D) 65536 A Z Boolean expression 7. The 0 0 0 for0the 0truth table shown is

Time: 60 min. 9. The number of product terms in the minimized SOP from is 1

= ( A + B ) + AC + BC

1 0 X

Traal and error method I = 1, J = B

1 Y

0 Z

0

F

⇒ 0 B( A0+ C ) 0( A + C0). 0 0 correct 1 option 0 Hence, the is (A). 0

1

0

0

0

1

1

1

1

0

0

0

0

0 D

0

1 0

0

0 D

0 1

0

0

1

1

(A) 2 (B) 4 (C) 5 (D) 3 10. The minimum number of 2 input NAND gates needed to implement Z = XY + VW is (A) 2 (B) 3 (C) 4 (D) 5 11. The operation a ⊕ b represents (A) C ab + a b

(B) ab + ab

(C) ab + ab

(D) a − b

12. Find the dual of X + [Y + XZ] + U (A) X + [Y(X + Z)] + U (B) X(Y + XZ)U (C) X + [Y(X + Z)]U (D) X [Y(X + Z)]U 13. The simplified form of given function AB + BC + AC is equal to (A) AB + AC (B) AC + BC (C) AC + BC (D) AB + AC 14. Simplify the following YZ WX

1

1

0

1

1

1

0

0

1

1

0 0

0

0

1 0

Syllabus: Computer Science and Information Technology Computer Science and Information Technology Digital Logic: Boolean algebra. Combinational and sequential circuits. Minimization. Number representations and computer arithmetic (fixed and floating point). Computer Organization and Architecture: Machine instructions and addressing modes. ALU, data-path and control unit. Instruction pipelining. Memory hierarchy: cache, main memory and secondary storage; I/O interface (interrupt and DMA mode). Programming and Data Structures: Programming in C. Recursion. Arrays, stacks, queues, linked lists, trees, binary search trees, binary heaps, graphs. Algorithms: Searching, sorting, hashing. Asymptotic worst case time and space complexity. Algorithm design techniques: greedy, dynamic programming and divide-and-conquer. Graph search, minimum spanning trees, shortest paths. Theory of Computation: Regular expressions and finite automata. Context-free grammars and push-down automata. Regular and contex-free languages, pumping lemma. Turing machines and undecidability. Compiler Design: Lexical analysis, parsing, syntax-directed translation. Runtime environments. Intermediate code generation. Operating System: Processes, threads, inter1process communication, concurrency and synchronization. Deadlock. CPU scheduling. Memory management and virtual memory. File systems. Databases: ER1model. Relational model: relational algebra, tuple calculus, SQL. Integrity constraints, normal forms. File organization, indexing (e.g., B and B+ trees). Transactions and concurrency control. Computer Networks: Concept of layering. LAN technologies (Ethernet). Flow and error control techniques, switching. IPv4/IPv6, routers and routing algorithms (distance vector, link state). TCP/UDP and sockets, congestion control. Application layer protocols (DNS, SMTP, POP, FTP, HTTP). Basics of Wi-Fi. Network security: authentication, basics of public key and private key cryptography, digital signatures and certificates, firewalls.

Chapter-wise Analysis of GATE Previous Years’ Papers Subject General Aptitude 1 Marks Questions 2 Marks Questions Total Marks Engineering Mathematics 1 Marks 2 Marks Total Marks Theory of Computation 1 Marks 2 Marks Total Marks Digital Logic 1 Marks 2 Marks Total Marks Computer Organization and Architecture

2005 6 12 30 0 7 14 4 5 14

2006 3 11 25 2 6 14 1 5 11

2007 4 11 26 2 5 12 3 5 13

2008 5 11 27 3 6 15 4 1 6

2009 4 6 16 4 3 10 2 0 2

2010 5 5 15 6 5 16 1 3 7 3 2 7

2011 5 5 15 2 7 16 3 3 9 3 3 9

2012 5 5 15 3 3 9 4 1 6 2 2 6

2013 5 5 15 5 2 9 1 2 5 3 1 5

2014 5 5 15 5 5 15 5 6 17 3 5 13

2015 5 5 15 4 6 16 1 3 7 1 2 5

2016

2017

5 6 17

5 5 15

5 4 13

4 5 14

3 3 9

2 4 10

2 1 4

2 2 6

1 Marks Questions 2 Marks Questions Total Marks Programming and Data Structures

4 9 22

0 7 14

2 6 14

0 12 24

2 4 10

1 4 9

3 2 7

2 4 10

1 7 15

2 2 6

1 2 5

2 2 6

3 4 11

1 Marks Questions 2 Marks Questions Total Marks Algorithm 1 Marks Questions 2 Marks Questions Total Marks Compiler Design 1 Marks Questions 2 Marks Questions Total Marks Operating System 1 Marks Questions 2 Marks Questions Total Marks Database 1 Marks Questions 2 Marks Questions Total Mark Computer Networks 1 Marks Questions 2 Marks Questions Total Marks Software Engineering 1 Marks Questions 2 Marks Questions Total Marks Web Technology 1 Marks Questions 2 Marks Questions Total Marks

5 3 11 2 10 22 1 5 11 0 2 4 3 4 11 5 2 9

0 6 12 8 7 22 1 5 11 1 8 17 1 4 9 1 5 11

1 3 7 3 12 27 1 5 11 2 6 14 0 6 12 2 6 14

1 3 7 2 15 32 2 2 6 2 5 12 1 5 11 1 4 9

1 3 7 3 6 15 1 0 1 2 5 12 0 5 5 0 5 5 1 0 1 1 0 1

3 5 13 1 3 7 2 1 4 3 2 7 3 2 7 2 3 8 0 0 0 0 0 0

4 7 18 1 0 1 1 0 1 3 2 7 0 3 6 5 2 9 1 0 1 1 0 1

2 6 14 4 2 8 1 3 7 1 3 7 3 3 9 3 3 9 0 0 0 0 0 0

2 5 12 5 3 11 2 2 6 1 1 3 1 4 9 4 2 8 0 1 2 0 0 0

0 2 4 1 2 5 1 2 5 0 2 4 3 2 7 4 2 8 1 0 1 0 0 0

5 3 11 4 4 12 2 1 4 2 2 6 1 2 5 2 3 8 1 1 3 1 1 3

2 5 12

4 4 12

4 5 14

2 2 6

1 1 3

2 1 4

1 4 9

2 2 6

3 1 5

2 3 8

2 4 10

2 3 8

General Information about GATE Structure of GATE The GATE examination consists of a single online paper of 3-hour duration, in which there will be a total of 65 questions carrying 100 marks out of which 10 questions carrying a total of 15 marks are in General Aptitude (GA).

Section Weightage and Marks 70% of the total marks is given to the technical section while 15% weightage is given to General Aptitude and Engineering Mathematics each. Weightage

Questions (Total 65)

Respective 70 Marks EngineeringBranch EngineeringMaths 15 Marks General Aptitude

15 Marks

25—1markquestions30—2mark questions 5—1 mark questions 5—2 mark questions

Particulars For 1-mark multiple-choice questions, 1/3 marks will be deducted for a wrong answer. Likewise, for 2-mark multiple-choice questions, 2/3 marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions.

Question Types 1. Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four answers, out of which the candidate has to mark the correct answer. 2. Numerical Answer carrying 1 or 2 marks each in all papers and sections. For numerical answer questions, choices will not be given. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for this type of questions.

Design of Questions The fill in the blank questions usually consist of 35%– 40% of the total weightage. The questions in a paper may be designed to test the following abilities: 1. Recall: These are based on facts, principles, formulae or laws of the discipline of the paper. The candidate is expected to be able to obtain the answer either from his/her memory of the subject or at most from a one-line computation. 2. Comprehension: These questions will test the candidate’s understanding of the basics of his/her field, by requiring him/her to draw simple conclusions from fundamental ideas. 3. Application: In these questions, the candidate is expected to apply his/her knowledge either through computation or by logical reasoning. 4. Analysis and Synthesis: In these questions, the candidate is presented with data, diagrams, images etc. that require analysis before a question can be answered. A Synthesis question might require the candidate to compare two or more pieces of information. Questions in this category could, for example, involve candidates in recognising unstated assumptions, or separating useful information from irrelevant information.

About Online Pattern The examination for all the papers will be carried out in an ONLINE Computer Based Test (CBT) mode where the candidates will be shown the questions in a random sequence on a computer screen. The candidates are required to either select the answer (for MCQ type) or enter the answer for numerical answer-type question using a mouse on a virtual keyboard (keyboard of the computer will be disabled). The candidates will also be allowed to use a calculator with which the online portal is equipped with.

Important Tips for GATE The followings are some important tips which would be helpful for students to prepare for GATE exam 1. Go through the pattern (using previous year GATE paper) and syllabus of the exam and start preparing accordingly. 2. Preparation time for GATE depends on many factors, such as, individual’s aptitude, attitude, fundamentals, concentration level etc., Generally rigorous preparation for 4 to 6 months is considered good but it may vary from student to student. 3. Make a list of books which cover complete syllabus, contains solved previous year questions and mock tests for practice based on latest GATE pattern. Purchase these books and start your preparation. 4. Make a list of topics which needs to be studied and make priority list for study of every topic based upon the marks for which that particular topic is asked in GATE exam. Find out the topics which fetch more marks and give more importance to those topics. Make a timetable for study of topics and follow the timetable strictly. 5. An effective way to brush up your knowledge about technical topics is group study with your friends. During group study you can explore new techniques and procedures. 6. While preparing any subject highlight important points (key definitions, equations, derivations, theorems and laws) which can be revised during last minute preparation. 7. Pay equal attention to both theory and numerical problems. Solve questions (numerical) based on latest exam pattern as much as possible, keeping weightage of that topic in mind. Whatever topics you decide to study, make sure that you know everything about it. 8. Try to use short-cut methods to solve problems instead of traditional lengthy and time consuming methods. 9. Go through previous year papers (say last ten years), to check your knowledge and note the distribution of different topics. Also analyze the topics in which you are weak and concentrate more on those topics. Always try to solve papers in given time, to obtain an idea how many questions you are able to solve in the given time limit. 10. Finish the detail study of topics one and a half month before your exam. During last month revise all the topics once again and clear leftover doubts.

GATE 2017 Solved Paper CS: Computer Science and Information Technology Set – 1

Number of Questions: 65

Total Marks:100.0

Wrong answer for MCQ will result in negative marks, (-1/3) for 1 Mark Questions and (-2/3) for 2 Marks Question.

Computer Science & Engineering Number of Questions: 55

Section Marks: 85

Q.1 to Q.25 carry 1 mark each and Q.26 to Q.55 carry 2 marks each. Question Number: 1 Question Type: MCQ The statement (¬p) ⇒ (¬q) is logically equivalent to which of the statements below? I. p ⇒ q II. q ⇒ p III. (¬ q) ∨ p IV. (¬ p) ∨ q (A) I only (B) I and IV only (C) II only (D) II and III only Solution: We have (¬ p) ⇒ (¬ q) ≡q⇒p ( An implication and its contra positive are equivalent) ∴II is equivalent to (¬ p) ⇒ (¬ q) Also, (¬ p) ⇒ (¬ q) ≡ q ⇒ p ≡ (¬ q) ∨ p ( A ⇒ B ≡ ¬ A ∨ B) ∴ III is equivalent to (¬ p) ⇒ (¬ q)

∴ F: ∀ x (∃ y R (x, y)) implies ∃ y (∃ x R (x, y))

Hence, the correct option is (D). Question Number: 2 Question Type: MCQ Consider the first-order logic sentence F:∀x(∃yR(x,y)). Assuming non-empty logical domains, which of the sentences below are implied by F? I. ∃y(∃xR(x,y)) II. ∃y(∀xR(x,y)) III. ∀y(∃xR(x,y)) IV. ¬∃x(∀y ¬R(x,y)) (B) I and IV only (A) IV only (C) II only (D) II and III only Solution: Given F: ∀ x (∃ y R (x, y)) ≡ [¬(¬(∀ x (∃ y R (x, y)))] ( p ≡ ¬ p) ≡ [¬ (∃ x (¬(∃ y R (x, y))))] ( ¬ ((∀ t) P (t)) ≡ ∃ t (¬ P (t))) ≡ [¬ (∃ x (∀ y ¬ R (x, y)))] ( ¬ (∃ t P (t)) ≡ ∀ t ¬ P (t)) ∴ IV is implied by F. Also, F : ∀ x (∃ y R (x, y)) implies ∃ x (∃ y R (x, y)) (∀ x (∃ y P (x, y) implies ∃ x (∃ y P (x, y)) Also, ∃ x (∃ y R (x, y)) implies ∃ y (∃ x R (x, y))

Solution: As a1 , a2 ,...., an are column vectors, such that

∴ I is implied by F: ∀ x (∃ y R (x, y)) Hence, the correct option is (B). Question Number: 3

Question Type: MCQ n

Let c1,…,cn be scalers, not all zero, such that where ai are column vectors in Rn. Consider the set of linear equations Ax = b

∑c a

i i

=0

i =1

n

Where A = [a1,…,an] and b = ∑ ai . The set of equations i =1 has (A) a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1 (B) no solution (C) infinitely many solutions (D) finitely many solutions n

∑ c a = 0 for some set of scalars c , c ,...., c i i

1

2

n

not all zero,

i=1

a1 , a2 ,...., an are linearly dependent.

∴ The rank of the matrix A = [ a1 , a2 , a3 ...., an ] is less than n. n

Also, given that b = ∑ ai i=1

Consider the set of linear equations: AX = b x1 x2 i.e., [ a1 , a2 ,...., an ] . = b (1) . xn

By taking x1 = x2 ..... = xn = 1, equation (1) is satisfied As AX = b has a solution and rank of A is less than n (= The number of unknowns), AX = b has infinitely many solutions. Hence, the correct option is (C).

GATE 2017 Solved Paper CS: Set – 1 | xvii Question Number: 4 Question Type: MCQ Consider the following functions from positive integers to real numbers: 100 10, n , n, log2 n, . n The CORRECT arrangement of the above functions in increasing order of asymptotic complexity is:

log 2 n, (A)

100 ,10, n, n n

Note: The height of a tree with a single node is 0. (A) 4 and 15 respectively (B) 3 and 14 respectively (C) 4 and 14 respectively (D) 3 and 15 respectively Solution: Binary Search Tree (BST) with 15 nodes, the minimum height possible, when it is complete BST, which is of height 3.

100 (B) ,10, log 2 n, n , n n 10, (C)

100 , n , log 2 n, n n

100 (D) , log 2 n, 10, n , n n 100 n According to dominance ranking, for large values of ‘n’ increasing order of asymptotic complexity is Solution: 10, n , n, log n2 ,

100 , 10, log 2 n , n , n n

Maximum height is possible when BST is either left skewed or right skewed, which is of height 14. Hence, the correct option is (B).

(P) Kruskal

(i) Divide and Conquer

Question Number: 7 Question Type: MCQ The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n − f. The range of decimal values for X in this representation is (A) 2− f to 2i (B) 2− f to (2i − 2− f) i (C) 0 to 2 (D) 0 to (2i − 2− f)

(Q) Quicksort

(ii) Greedy

Solution: Given fixed-point representation has the form.

(R) Floyd-Warshall

(iii) Dynamic Programming

Hence, the correct option is (B). Question Number: 5 Consider the following table:

Question Type: MCQ

Algorithms

Design Paradigms

n-bits

Match the algorithms based on. (A) (P) ↔ (ii), (B) (P) ↔ (iii), (C) (P) ↔ (ii), (D) (P) ↔ (i),

to the design paradigms they are (Q) ↔ (iii), (Q) ↔ (i), (Q) ↔ (i), (Q) ↔ (ii),

(R) ↔ (i) (R) ↔ (ii) (R) ↔ (iii) (R) ↔ (iii)

Solution: The design strategy for Kruskals algorithm is greedy method. Quick sort is implemented using Divide and Conquer strategy. Floyd-warshall algorithm is implemented using dynamic programming. Hence, the correct option is (C). Question Number: 6

Question Type: MCQ Let T be a binary search tree with 15 nodes. The minimum and maximum possible heights of T are:

integral

fraction

i

f

These numbers are in unsigned form. The least number possible is 0.00 …. 0 (last bit of fraction is 1). = 0 The largest number possible is 1111 ….. 1.1111 …. 1 (i 1’s in integral and f 1’s in fractional part). The range of integral part is 2i −1 (for unsigned). The range of fraction part is

1 1 1 + 2 + ..... + f = 1 − 2−f 2 2 2

∴ The highest number possible is 2i −1 + 1− 2− f = 2i − 2-f ∴ Required range is 0 to 2i − 2-f Hence, the correct option is (D).

xviii | GATE 2017 Solved Paper CS: Set – 1 Question Number: 8 Question Type: MCQ Consider the C code fragment given below. typedef struct node { int data; node* next; } node; void join (node* m, node* n) { node* p = n; while (p − >next != NULL) { p = p − >next; } p − >next = m; } Assuming that m and n point to valid NULL-terminated linked lists, invocation of join will (A) append list m to the end of list n for all inputs. (B) either cause a null pointer dereference or append list m to the end of list n. (C) cause a null pointer dereference for all inputs. (D) append list n to the end of list m for all inputs. Solution: The above code appends list m to the end of list n (or) it may cause null pointer dereference error which may leads to segmentation fault, it happens when a page fault handler fails. Hence, the correct option is (B). Question Number: 9 Question Type: MCQ When two 8-bit numbers A7… A0 and B7… B0 in 2’s complement representation (with A0 and B0 as the least significant bits) are added using a ripple-carry adder, the sum bits obtained are S7…S0 and the carry bits are C7…C0. An overflow is said to have occurred if (A) the carry bit C7 is 1 (B) all the carry bits (C7,…,C0) are 1 ( A7 ⋅ B7 ⋅ S7 + A7 ⋅ B7 ⋅ S7 ) is 1 (C)

( A0 ⋅ B0 ⋅ S0 + A0 ⋅ B0 ⋅ S0 ) is 1 (D) Solution: Overflow occurs only when A7:0 A7:1 B7:1 (or) B7:0 S7:0

S7:1

∴ Overflow condition is given as

A7 B7 S 7 + A7 B 7 S7 .

Hence, the correct option is (C). Question Number: 10 Question Type: MCQ Consider the following context-free grammar over the alphabet Σ = {a, b, c} with S as the start symbol: S → abScT | abcT T → bT | b

Which one of the following represents the language generated by the above grammar? (A) {(ab)n(cb)n | n ≥ 1}

{( ab ) cb 1 cb 2 …cb n | n, m1 , m2 ,…, mn ≥ 1} (B) (C) {(ab)n(cbm)n | m, n ≥ 1} (D) {(ab)n(cbn)m | m, n ≥ 1} n

m

m

m

Solution: Given grammar S → abScT | abcT T → bT | b The strings of this grammar are, 1. S → abcT → abcb 2. S → abcT → abcbT → abcbb 3. S → ab ScT → ab abcTcT → ab ab cb cb 4. S → abScT → ab ab cTcT → ab abc bT cbT → ab ab cbbT cbb → ab ab cb bb cbb

{

n

n

}

Choice (A ): (ab) (cb) n ≥ 1

This generates equal number of ‘ab’ and ‘cb’’s. So, this is not equivalent to given grammar.

{

}

n

m m m Choice (B): (ab) cb 1 cb 2 .....cb n n, m1 , m2 ,..., mn ≥ 1

This generates strings like

abcb m1 (m1 ≥ 1), ababcb m1 cb m2 (m1 , m2 ≥ 1) . ∴ This is equivalent to given grammar.

{

}

Choice (C): (ab) (cb m ) m, n ≥ 1 . n

n

This generates strings like abcb m (m ≥ 1) , abab (cb m ) (m ≥ 1) . 2

(This cannot generate the strings like ab ab cbb cb). Choice (D): This generates strings of the form ab cb cb, ab ab cbb. Hence, the correct option is (B). Question Number: 11 Question Type: MCQ Consider the C struct defined below: struct data { int marks [100]; char grade;

GATE 2017 Solved Paper CS: Set – 1 | xix int cnumber; }; struct data student; The base address of student is available in register R1. The field student.grade can be accessed efficiently using (A) Post-increment addressing mode, (R1)+ (B) Pre-decrement addressing mode, − (R1) (C) Register direct addressing mode, R1 (D) Index addressing mode, X(R1), where X is an offset represented in 2’s complement 16-bit representation. Solution: Student grade can be accessed efficiently using index addressing mode. X (R1) R1 is base address of student. X is an offset. The effective address of student grade is X R1 + [(100) × size of (int )]

Hence, the correct option is (D). Question Number: 12 Question Type: MCQ Consider the following intermediate program in three address code p=a−b q=p*c p=u*v q=p+q Which one of the following corresponds to a static single assignment form of the above code? (A) p1 = a − b (B) p3 = a − b q1 = p1 * c q4 = p3 * c p1 = u * v p4 = u * v q1 = p1 + q1 q5 = p4 + q4 (C) p1 = a − b (D) p1 = a − b q1 = p2 * c q1 = p * c p2 = u * v p3 = u * v q2 = p4 + q3 q2 = p + q Solution: In static single assignment, each assignment to a temporary variable is given a unique name and all the uses reached by that assignment are renamed. Option (B) gives the static single assignment form for the above code. Hence, the correct option is (B). Question Number: 13 Consider the following C code:

Question Type: MCQ

# include int *assignval (int *x, int val) { *x = val; return x;

} void main ( ) { int *x = malloc (sizeof (int)); if (NULL == x) return; x = assignval (x, 0); if (x) { x = (int *) malloc (sizeof (int)); if (NULL == x) return; x = assignval (x, 10); } printf(“%d\n”, *x); free (x); } The code suffers from which one of the following problems: (A) compiler error as the return of malloc is not typecast appropriately (B) compiler error because the comparison should be made as x == NULL and not as shown (C) compiles successfully but execution may result in dangling pointer (D) compiles successfully but execution may result in memory leak Solution: The above program compiles successfully, but the execution may result in the memory leak. Hence, the correct option is (D). Question Number: 14 Question Type: MCQ Consider a TCP client and a TCP server running on two different machines. After completing data transfer, the TCP client calls close to terminate the connection and a FIN segment is sent to the TCP server. Server-side TCP responds by sending an ACK, which is received by the client-side TCP. As per the TCP connection state diagram (RFC 793), in which state does the client-side TCP connection wait for the FIN from the server-side TCP? (A) LAST-ACK (B) TIME-WAIT (C) FIN-WAIT-1 (D) FIN-WAIT-2

Solution: Fin-wait-1 state sends a segment, at which close-wait state sends an ACK to the Fin-wait-2 state. Hence, the correct option is (D). Question Number: 15 Question Type: MCQ A sender S sends a message m to receiver R, which is digitally signed by S with its private key. In this scenario, one or more of the following security violations can take place. (I) S can launch a birthday attack to replace m with a fraudulent message. (II) A third party attacker can launch a birthday attack to replace m with a fraudulent message. (III) R can launch a birthday attack to replace m with a fraudulent message.

xx | GATE 2017 Solved Paper CS: Set – 1 Which of the following are possible security violations? (A) (I) and (II) only (B) (I) only (C) (II) only (D) (II) and (III) only

Question Number: 17 Consider the following grammar: P → xQRS Q → yz | z R→w |ε S→y

Solution: Sender can launch a birthday attack, which replaces the message ‘m’ with fraud message. Hence, the correct option is (B). Question Number: 16 Question Type: MCQ The following functional dependencies hold true for the relational schema R {V, W, X, Y, Z}: V →W VW → X Y → VX Y→Z Which of the following is irreducible equivalent for this set of set of functional dependencies? (A) V → W (B) V →W V → X W→X Y → V Y →V Y → Z Y→Z (C) V → W (D) V →W V → X W→X Y → V Y →V Y → X Y→X Y → Z Y→Z Solution: Given FD’s : V→W VW → X Y → VX Y→Z Using pseudotransitive rule, {V → W, VW → X} | = V → X. Using projection rule, {Y → VX}| = Y → V, Y → X. ∴ Equivalent set of FD’s for given FD’s is V→W V→X Y→V Y→X Y→Z Irreducible equivalent set of FD’s is V→W V→X Y→V Y→Z ( X can be derived from either V or Y, so remove Y → X). Hence, the correct option is (A).

Question Type: MCQ

What is FOLLOW (Q)? (A) {R} (C) {w, y}

(B) {w} (D) {w, $}

Solution: Consider the grammar given P→xQRS Q→yz|z R→w|ε S→y Follow (Q) = First (RS) = First (R) – {ε} ∪ first (S) = {w} ∪ {y} = {w, y} Hence, the correct option is (C). Question Number: 18 Question Type: MCQ Threads of a process share (A) global variables but not heap. (B) heap but not global variables. (C) neither global variables nor heap. (D) both heap and global variables. Solution: Each program has 4 segments: Stack, Data, Code, Heap. All the threads of a process share all the segments except the stack. Global variables are stored in heap. So, threads of a process share both heap and global variable. Hence, the correct option is (D). Question Number: 19 Question Type: NAT Let X be a Gaussian random variable with mean 0 and variance σ2. Let Y = max (X, 0) where max (a, b) is the maximum of a and b. The median of Y is . Solution: Given that X is a Gaussian random variable Also given, mean of X = 0 Variance of X = s2 We know that in Gaussian distribution, Mean = Median 0; if X 0 Given Y = max. (X, 0) = X ; if X > 0 Hence, the median of Y = Median of X = 0 Hence, the correct answer is (0).

GATE 2017 Solved Paper CS: Set – 1 | xxi Question Number: 20 Question Type: NAT Let T be a tree with 10 vertices. The sum of the degrees of all the vertices in T is _______ Solution: Number of vertices (V) for given tree T is 10. Numbers of regions (R) for a tree is 1. we have R=E–V+2 For the above tree, numbers of vertices are 10, 1 = E – 10 + 2 E=9 Sum of all degrees of all vertices are twice the number of edges, i.e., 18 Hence, the correct answer is (18). Question Number: 21 Question Type: NAT Consider the Karnaugh map given below, where X represents “don’t care” and blank represents 0. ba dc

00

00

01

11

X

X

10

1

01

= ca Hence, we need one NOR gate. Hence, the correct answer is (1). Question Number: 22 Question Type: NAT Consider the language L given by the regular expression (a + b)*b (a + b) over the alphabet {a, b}. The smallest number of states needed in a deterministic finite-state automaton (DFA) accepting L is ________. Solution: Given regular expression:

R = (a + b) * b (a + b)

The NFA for given R is a, b

1 X

10

X

Assume for all inputs (a, b, c, d), the respective complements (a , b, c , d ) are also available. The above logic is implemented using 2-input NOR gates only. The minimum number of gates required is ________.

b

1

a, b

2

3

The DFA equivalent to this NFA is

X

1

11

c +a

q\δ

a

b

A{1}

{1}

{1, 2}

B{2}

{3}

{3}

C{3}

φ

φ

D{1, 2}

{1, 3}

{1, 2, 3}

E{1, 3}

{1}

{1, 2}

F{1, 2, 3}

{1, 3}

{1, 2, 3}

a

a b

a

A

D

E

Solution: ba

bc

b

b

00

01

11

10

X

X

1

a

00 01

1

X

11

1

1

F

b

(B, C states are not reachable from initial state). X

10

Simplified expression is ca a c

X

Hence, the correct answer is (4). Question Number: 23 Question Type: NAT Consider a database that has the relation schema EMP (EmpId, EmpName and DeptName). An instance of the schema EMP and a SQL query on it are given below.

xxii | GATE 2017 Solved Paper CS: Set – 1 EMP EmpId

EmpName

DeptName

1

XYA

AA

2

XYB

AA

3

XYC

AA

4

XYD

AA

5

XYE

AB

6

XYF

AB

7

XYG

AB

8

XYH

AC

9

XYI

AC

10

XYJ

AC

11

XYK

AD

12

XYL

AD

13

XYM

AE

SELECTIVE AVG ( EC.Num ) FROM EC WHERE ( DeptName, Num ) IN

(SELECT DeptName, COUNT ( EmpId ) AS

EC ( DeptName, Num )

FROM EMP GROUP BY DeptName)

The output of executing the SQL query is ______. Solution: Based on given query EC table has Dept Name, Count (EMPID) from EMP Table. EC

P1 0

P1 1

P1 2

P2 3

P1 :4

Dept Name

Num

AA

4

AB

3

AC

3

AD

2

AE

1

P2 : 3

4 + 3 + 3 + 2 +1 5 13 = 5 = 2.6

Average (EC. Num) =

Hence, the correct answer is (2.6). Question Number: 24 Question Type: NAT Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below : Process

Arrival time

Burst time

P1

0

7

P2

3

3

P3

5

5

P4

6

2

If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is milliseconds. Solution: For the given processes using preemptive SRTF, the Gantt chart will be as below:

P2 4

P2 5

P4 6

P4 7

P1 8

P2 :1 P3 : 5

P1 9

P3 :5 P4 : 2 P1 : 4

P1 10

P1 11

P3 12

13

P3 :5 P1 : 4

Waiting time of P1 = 5 Waiting time of P2 = 0 Waiting time of P3 = 12 − 5 = 7 Waiting time of P4 = 0 ∴ Average waiting time = 5 + 7 = 3 4 Hence, the correct answer is (3). Question Number: 25 Question Type: NAT Consider a two-level cache hierarchy with L1 and L2 caches. An application incurs 1.4 memory accesses per instruction on average. For this application, the miss rate of L1 cache is 0.1; the L2 cache experiences on average 7 misses per 1000 instructions. The miss rate of L2 expressed correct to two decimal places is . Solution: Miss rate =

Number of misses Total number of accesses

7 1000×1.4 = 0.05

=

Hence, the correct answer is (0.05). Question Number: 26 Question Type: MCQ Let G = (V,E) be any connected undirected edge-weighted graph. The weights of the edges in E are positive and distinct. Consider the following statements:

GATE 2017 Solved Paper CS: Set – 1 | xxiii (I) Minimum spanning Tree of G is always unique. (II) Shortest path between any two vertices of G is always unique. Which of the above statements is/are necessarily true? (A) (I) only (B) (II) only (C) both (I) and (II) (D) neither (I) nor (II)

minimum value of y together for which execution of P can result in a deadlock are: (A) x = 1, y = 2 (B) x = 2, y = 1 (C) x = 2, y = 2 (D) x = 1, y = 1 Solution: As the non reentrant locks not allow a thread to reacquire the lock, so only one thread and only one lock can lead to deadlock, if it tries to reacquire the lock.

Solution:

Question Number: 28 x 7 − 2 x5 +1 The value of lim 3 x →1 x − 3 x 2 + 2

• When the edge weights of a graph ‘G’ are unique, then the minimum spanning tree for graph ‘G’ is unique. • Shortest path between any 2 vertices of graph may not be always unique. Consider the following graph with above scenario. 1

A

B

Hence, the correct option is (D).

(A) is 0 (B) is −1 (C) is 1 (D) does not exist

Solution: We have lim x→1

2

3

= lim

The MST for the above graph ‘G’ is 1

A

Hence, the correct option is (C). Question Number: 29 Question Type: MCQ Let p, q, and r be propositions and the expression (p → q)→ r be a contradiction. Then, the expression (r → p)→ q is

B

4

2

D

C

The shortest path from vertex A to C is given by two paths. i.e., 1 A

A

2

B

3

7 −10 3− 6 =1

=

C

5

x7 − 2 x5 + 1 x3 − 3x 2 + 2

7 x 6 −10 x 4 (By L’ Hospital’s Rule) x→1 3 x 2 − 6 x

4

D

Question Type: MCQ

C

(A) (B) (C) (D)

a tautology a contradiction. always TRUE when p is FALSE. always TRUE when q is TRUE.

Solution: Given that (p → q) → r is a contradiction. This is possible only when p → q is TRUE and r is FALSE. p → q is TRUE only when p and q can have any combination of truth values except p is TRUE and q is FALSE. Consider the expression, (r → p) → q

C

Hence, the correct option is (A). Question Number: 27 Question Type: MCQ A multithreaded program P executes with x number of threads and uses y number of locks for ensuring mutual exclusion while operating on shared memory locations. All locks in the program are non-reentrant, i.e., if a thread holds a lock l, then it cannot re-acquire lock l without releasing it. If a thread is unable to acquire a lock, it blocks until the lock becomes available. The minimum value of x and the

As r is FALSE, r → p is always TRUE. ∴ (r → p) → q is TRUE when q is TRUE and (r → p) → q is FALSE when q is FALSE. So, (r → p) → q is always TRUE when q is TRUE. Hence, the correct option is (D). Question Number: 30 Question Type: MCQ 2 Let u and v be two vectors in R whose Euclidean norms satisfy ||u||= 2||v||. What is the value of α such that w = u +αv bisects the angle between u and v? (A) 2 (B) 1/2 (C) 1 (D) −1/2

xxiv | GATE 2017 Solved Paper CS: Set – 1 Solution: For two vectors u and v, || u || = 2 || v || ⇒ u and 2v will have the same length. We know that if two vectors v1 and v2 have the same length, then v1 + v2 bisects the angle between v1 and v2. As || u || = 2|| v ||, u + 2v bisects the angle between the vectors u and 2v. Also, we know that the angle between u and v is same as the angle between u and 2v. So, u + 2v bisects the angle between the vectors u and v.

x3 + x + 1 as the generator polynomial to generate the check bits. In this network, the message 01011011 is transmitted as (A) 01011011010 (B) 01011011011 (C) 01011011101 (D) 01011011100 Solution: Given polynomial generator as x3 + x + 1, then generator will be 1011. Given data = 01011011 Message transmitted will be:

∴ W = u + α v bisects the angle between u and v when α = 2.

1011

Hence, the correct option is (A). Question Number: 31 Question Type: MCQ Let A be n × n real valued square symmetric matrix of rank n

2 with

n

∑∑ A

2 ij

= 50 . Consider the following statements.

i =1 j =1

(I) One eigenvalue must be in [−5, 5] (II) The eigenvalue with the largest magnitude must be strictly greater than 5 Which of the above statements about eigenvalues of A is/are necessarily CORRECT? (A) Both (I) and (II) (B) (I) only (C) (II) only (D) Neither (I) nor (II)

The transmitted message is 01011011101. Hence, the correct option is (C). Question Number: 33 Question Type: MCQ Consider a combination of T and D flip-flops connected as shown below. The output of the D flip-flop is connected to the input of the T flip-flop and the output of the T flip-flop is connected to the input of the D flip-flop.

Solution: Case (i):Let n > 2.

∴ P(A) = 2 < order of A.

⇒ A is a singular matrix.

⇒ O is an eigen value of A and 06[-5, 5]

∴ I is correct. 5 0 0 Also for A = 0 0 0 , none of the eigen values have 0 0 −5 magnitude not greater than 5.

So, II is not correct.

T FilpFlop

Q0 D FilpFlop

Initially, both Q0 and Q1 are set to 1 (before the 1st clock cycle). The outputs

But none of its eigen values have magnitude greater than 5. Hence, the correct option is (B). Question Number: 32 Question Type: MCQ A computer network uses polynomials over GF(2) for error checking with 8 bits as information bits and uses

Q1

Clock

Case (II):- Let n = 2 −5 0 Consider the matrix B = 0 5 Clearly the eigen values of B lie in [-5, 5].

01011011000 01000011 0000 1011 1011 00000 0000 0001 0000 0011 0000 0110 0000 1100 1011 1110 1101 101

(A) Q1 Q0 after the 3rd cycle cycle are 00 respectively (B) Q1 Q0 after the 3rd cycle cycle are 01 respectively (C) Q1 Q0 after the 3rd cycle cycle are 11 respectively (D) Q1 Q0 after the 3rd cycle cycle are 01 respectively

are 11 and after the 4th are 11 and after the 4th are 00 and after the 4th are 01 and after the 4th

GATE 2017 Solved Paper CS: Set – 1 | xxv Solution:

Question Number: 35 Question Type: MCQ Consider the following two functions.

T FF

Q1

D FF

Q0

void fun1 (int n) { void fun2 (int n) { if (n == 0) return; if (n == 0) return; printf (“%d”, n); printf (“%d”, n); fun2 (n − 2) ; fun1(++n) printf (“%d”, n); printf (“%d”, n); } }

The output printed when fun1 (5) is called is

Clock

present state

next state

Q1

Q0

Q1

Q0

1

1

0

1

0

1

1

0

1

0

1

1

1

1

0

0

+

+

(A) 53423122233445 (C) 53423122132435

(B) 53423120112233 (D) 53423120213243

Solution: fun 1(5)

Hence, the correct option is (B). Question Number: 34 Question Type: MCQ If G is a grammar with productions S → SaS | aSb | bSa | SS |∈ Where S is the start variable, then which one of the following strings is not generated by G? (A) abab (B) aaab (C) abbaa (D) babba

fun 2(3) Print (5) fun 1(4) Print (3)

Print (4)

S → SaS | aSb | bSa | SS | E

abab:

Print (2)

Print (3) fun 2(1)

S → aSb → aSaSb → aSaSaSb → aaab

Print (3)

Print (3)

fun 1(2)

abbaa:

Print (4) fun 1(3)

S → aSb → abSab → abab

aaab:

Print (4) fun 2(2)

Solution: Given grammar

Print (5)

Print (1)

Print (2)

S → SS

→ aSbSS

→ abbSaSaS

→ abbaa

fun 2(0)

Print (2)

Print (2)

babba:

C → bSa

→ baSba

→ ba ba

babba not derivable from S. Hence, the correct option is (D).

It prints 53423122233445 Hence, the correct option is (A).

xxvi | GATE 2017 Solved Paper CS: Set – 1 Question Number: 36 Question Type: MCQ Consider the C functions foo and bar given below: int foo (int val) { int x = 0; while (val > 0) { x =x + foo (val−−); } return val; } int bar (int val) { int x = 0; while (val > 0) { x =x +bar (val − 1); } return val; } Invocations of foo (3) and bar (3) will result in : (A) Return of 6 and 6 respectively. (B) Infinite loop and abnormal termination respectively. (C) Abnormal termination and infinite loop respectively. (D) Both terminating abnormally. Solution: foo (3) terminates abnormally, while bar (3) gets into infinite loop. Hence, the correct option is (C). Question Number: 37 Question Type: MCQ Consider the context-free rammers over the alphabet {a, b, c} given below. S and T are non-terminals. G1 : S → aSb|T, T → cT|∈ G2 : S → bSa|T, T → cT|∈ The language L(G1) ∩ L(G2) is (A) Finite (B) Not finite but regular (C) Context-Free but not regular (D) Recursive but not context-free. Solution: The strings of L (G1) are {ε, c, cc … c, ab, acb, ac … cb, …} The strings of L (G2) are {ε, c, cc, … c, ba, bca, bc, …. ca, …} L (G1) ∩ L (G2) = {ε, c, cc, c, … c,} The resultant grammar can have zero or more c’s. This is regular and not finite. Hence, the correct option is (B). Question Number: 38 Question Type: MCQ Consider the following languages over the alphabet Σ = {a, b, c}. Let L1 = {anbncm| m, n ≥ 0} and L2 = {ambncn| m, n ≥ 0}.

Which of the following are context-free languages? I. L1 ∪ L2 II. L1 ∩ L2 (A) I only (B) II only (C) I and II (D) Neither I nor II Solution: Given :

L1 = {a nb n c m m, n ≥ 0}

L2 = {a mb n c n m, n ≥ 0}

Here both L1 and L2 are context-free languages. CFL’s are under union but not under intersection. L1 ∪ L2 needs to check the equality of number of a’s and b’s or equality of number of b’s and c’s. ∴ L1 ∪ L2 is CFL. L1 ∩ L2 requires equal number of a’s, b’s and c’s. A PDA cannot check this. ∴ L1 ∩ L2 is not CFL. Hence, the correct option is (A). Question Number: 39 Question Type: MCQ Let A and B be finite alphabets and let # be a symbol outside both A and B. Let f be a total function from A* to B*. We say f is computable if there exists a turning machine M which given an input x in A*, always halts with f(x) on its tape. Let Lf denote the language {x # f(x)| x ∈ A*}. Which of the following statements is true: (A) f is computable if and only if Lf is recursive. (B) f is computable if and only if Lf is recursively enumerable. (C) If f is computable then Lf is recursive, but not conversely. (D) If f is computable then Lf is recursively enumerable, but not conversely. Solution: A computable function is same as recursive function. Hence, choice (A) is correct. Hence, the correct option is (A). Question Number: 40 Question Type: MCQ Recall that Belady’s anomaly is that the page-fault rate may increase as the number of allocated frames increases. Now, consider the following statements: S1: Random page replacement algorithm (where a page chosen at random is replaced) suffers from Belady’s anomaly S2: LRU page replacement algorithm suffers from Belady’s anomaly Which of the following is CORRECT? (A) S1 is true, S2 is true (B) S1 is true, S2 is false (C) S1 is false, S2 is true (D) S1 is false, S2 is false Solution: Random page replacement algorithm simulates FIFO. So, there is a possibility of Belady’s anomaly.

GATE 2017 Solved Paper CS: Set – 1 | xxvii There is no possibility of Belady’s anomaly in LRU. ∴ S1 is true. S2 is false. Hence, the correct option is (B). Question Number: 41 Question Type: MCQ Consider a database that has the relation schemas EMP(EmpId, EmpName, DeptId), and DEPT(DeptName, DeptId), Note that the DeptId can be permitted to be NULL in the relation EMP. Consider the following queries on the database expressed in tuple relational calculus. (I) {t | ∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∀ v ∈ DEPT(t[DeptId] ≠ v[DeptId]))} (II) {t | ∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∃ v ∈ DEPT(t[DeptId] ≠ v[DeptId]))} (III) {t | ∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∃ v ∈ DEPT(t[DeptId] = v[DeptId]))} Which of the above queries are safe? (A) (I) and (II) only (B) (I) and (III) only (C) (II) and (III) only (D) (I), (II) and (III)

Question Number: 43 Question Type: NAT Consider the following grammar: stmt − > if expr then expr else expr; stmt | Ò expr − > term relop term | term term − > id | number id −>a|b|c number − > [0 − 9] where relop is a relational operator (e.g., ,…), Ò refers to the empty statement, and if, then, else are terminals. Consider a program P following the above grammar containing ten if terminals. The number of control flow paths in P is__________. For example, the program if e1 then e2 else e3 has 2 control flow paths, e1 → e2 and e1 → e3. Solution: The flow graph for the above grammar for ten if terminals is given below:

Solution: All the queries are safe expressions. ( All giving finite tuples). Hence, the correct option is (D). Question Number: 42 Question Type: MCQ In a database system, unique timestamps are assigned to each transaction using Lamport’s logical clock. Let TS(T1) and TS(T2) be the timestamps of transactions T1 and T2 respectively. Besides, T1 holds a lock on the resource R, and T2 has requested a conflicting lock on the same resource R. The following algorithm is used to prevent deadlocks in the database system assuming that a killed transaction is restarted with the same timestamp. if TS(T2) < TS(T1) then T1 is killed else T2 waits. Assume any transaction that is not killed terminates eventually. Which of the following is TRUE about the database system that uses the above algorithm to prevent deadlocks? (A) The database system is both deadlock-free and starvation-free. (B) The database system is deadlock-free, but not starvation-free. (C) The database system is starvation-free, but not deadlock-free. (D) The database system is neither deadlock-free nor starvation-free. Solution: Given Time Stamp algorithm is same as woundwait time stamp algorithm, which is free from deadlock and starvation-free. Hence, the correct option is (A).

The number of control flow paths will be 210. Hence, the correct answer is (1024).

xxviii | GATE 2017 Solved Paper CS: Set – 1 Question Number: 44 Question Type: NAT In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. If the public key of A is 35, then the private key of A is _________. Solution: Given two prime numbers, p = 13, q = 17 RSA algorithm: (1) p = 13, q = 17 (2) n=p×q = 13 × 17 ⇒221 z = (p − 1) × (q − 1) = 12 × 16 = 192 Given public key as 35 (i.e. e = 35) (3) (d * e) mod Z = 1 (d * 35) mod 192 = 1 d = 11 Hence, the correct answer is (11). Question Number: 45 Question Type: NAT The values of parameters for the Stop-and Wait ARQ protocol are as given below: Bit rate of the transmission channel = 1Mbps. Propagation delay from sender to receiver = 0.75 ms. Time to process a frame = 0.25 ms. Number of bytes in the information frame = 1980. Number of bytes in the acknowledge frame = 20. Number of overhead bytes in the information frame = 20. Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the stop-and-wait ARQ protocol for the above parameters is ________ (correct to 2 decimal places) length of data Solution: Transmission time = band width =

2000×8 bits

106 bps = 16 m sec Propagation time = 0.75 m sec 40×8 bits ACK transmission time = 106 bps = 0.32 m sec Time to process a frame = 0.25 sec Transmission efficiency (Stop an wait)

CR StudentName

200 × 100 229 = 87.3

Hence, the correct answer is (87.3).

CourseName

SA

CA

SA

CB

SA

CC

SB

CB

SB

CC

SC

CA

SC

CB

SC

CC

SD

CA

SD

CB

SD

CC

SD

CD

SE

CD

SE

CA

SE

CB

SF

CA

SF

CB

SF

CC

The following query is made on the database.

T 1 ← pCouraseName (sStudentName =' SA ' (CR))

T 2 ← CR ÷ T 1 The number of rows in T2 is

.

Solution: The resultant of studentName = ‘SA’ (CR) is : StudentName

CourseName

SA

CA

SA

CB

SA

CC

πCourseName (studentName = ‘SA’ (CR)) resultant is T1 and is equal to : Course Name CA CB

16 0.5 + 16 + 1.5 + 0.32 =

Question Number: 46 Question Type: NAT Consider a database that has the relation schema CR (studentName, CourseName). An instance of the schema CR is as given below.

CC

Now, T2 contains CR ÷ T1. T2 contains the StudentName’s who are enrolled for all the 3 courses CA, CB, CC. i.e.

GATE 2017 Solved Paper CS: Set – 1 | xxix T2 Student Name SA SC SD SF

∴ T2 has 4 rows. Hence, the correct answer is (4). Question Number: 47 Question Type: NAT The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is _______. Solution: Let, A, B and C denote the sets of divisors of 3, 5 and 7 among the integers 1 to 500 respectively. ∴ The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 = n (A ∪ B ∪ C)

Question Number: 48 Question Type: NAT Let A be an array of 31 numbers consisting of a sequence of 0’s followed by a sequence of 1’s. The problem is to find the smallest index i such that A[i] is 1 by probing the minimum number of locations in A. The worst case number of probes performed by an optimal algorithm is . Solution: Using binary search, number of comparisions required will be log n2 . Number of probes required is Èlog231˘ =5 Hence, the correct answer is (5). Question Number: 49 Question Type: NAT Consider a RISC machine where each instruction is exactly 4 bytes long. Conditional and unconditional branch instructions use PC-relative addressing mode Offset specified in bytes to the target location of the branch instruction. Further the Offset is always with respect to the address of the next instruction in the program sequence. Consider the following instruction sequence. Instr. No.

We know that

Instruction

i : i+1 : i + 2 :

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) − n (A ∩ B) − n (B ∩ C) − n (C ∩ A) + n (A ∩ B ∩ C) (1)

add R2, R3, R4 sub R5, R6, R7 cmp R1, R9, R10

i + 3 :

n (A) = The number of integers from 1 to 500 that are divisible by 3 = 166

beq R1, Offset

n (B) = The number of integers from 1 to 500 that are divisible by 5 = 100

If the target of the branch instruction is i, then the decimal value of the Offset is .

n (C) = The number of integers from 1 to 500 that are divisible by 7 = 71

Solution: Given instruction sequence

i : add R2, R3, R4

n (A ∩ B) = The number of integers from 1 to 500 that are divisible by 3 and 5 = The number of integers from 1 to 500 that are divisible by 15 = 33

i + 1: sub R5, R6, R7

i + 2 : cmp R1, R9, R10

i + 3 : beq R1, offset.

n (B ∩ C) = The number of integers from 1 to 500 that are divisible by 5 and 7 = The number of integers from 1 to 500 that are divisible by 35 = 14

In PC-relative addressing mode, effective address = PC + offset.

n (C ∩ A) = The number of integers from 1 to 500 that are divisible by 7 and 3 = The number of integers from 1 to 500 that are divisible by 21 = 23

PC always points to the address of next instruction.

n (A ∩ B ∩ C) = The number of integers from 1 to 500 that are divisible by 3, 5 and 7 = The number of integers from 1 to 500 that are divisible by 105 = 4

As each instruction requires 4 bytes, offset = − (4 × 4)

∴ Substituting these in (1), we get

∴ During execution of i + 3, the PC will point to address of i + 4.

= − 16 bytes.

If ‘i’ is at address 1000, then i + 4 is at 1016. EA = 1016 − 16 = 1000.

n (A ∪ B ∪ C) = 166 + 100 + 71 − 33 − 14 − 23 + 4 = 271

This gives address of ith instruction.

Hence, the correct answer is (271).

Hence, the correct answer is (−16).

xxx | GATE 2017 Solved Paper CS: Set – 1 Question Number: 50 Question Type: NAT Instruction execution in a processor is divided into 5 stages, Instruction Fetch (IF), Instruction Decode (ID), Operand Fetch (OF), Execute (EX), and Write Back (WB). These stages take 5, 4, 20, 10, and 3 nanoseconds (ns) respective. A pipelined implement action of the processor requires buffering between each pair of consecutive stages with a delay of 2 ns. Two pipelined implementations of the processor are contemplated: (i) A navie pipeline implementation (NP) with 5 stages and (ii) An efficient pipeline (EP) where the OF stage is divided into stages OF1 and OF2 with execution times of 12 ns and 8 ns respectively. The speedup (correct to two decimal places) achieved by EP over NP in executing 20 independent instructions with no hazards is . Solution: 5-stage pipeline:

In a 2-way set associative cache, a block is placed in the location, Block number % No. of sets in cache. set 0 0 % 128 = 0 → compulsory miss. set 0 128 % 128 = 0 → compulsory miss. set 0 256 % 128 = 0 → compulsory miss (1st time access) set 0 128 % 128 = 0 → hit. set 0 0 % 128 = 0 → replace block 256, conflict miss. set 0 128 % 128 = 0 → hit. set 0 256 % 128 = 0 → replace block 0, conflict miss. set 0 128 % 128 = 0 → hit. set 1 1 % 128 = 1 → compulsory miss. set 1 129 % 128 = 1 → compulsory miss. set 1 compulsory miss (1st time access) 257 % 128 = 1 →

Cycle time = 20 + 2 Execution time for 20 instructions = [5 + (20 − 1)] 22 6-stage pipeline:

set 1 129 % 128 = 1 → hit. set 1 1 % 128 = 1 → replace block 257, conflict miss. set 1 129 % 128 = 1 → hit.

Cycle time = 12 + 2 = 14 Execution time for 20 instructions = [6 + (20 − 1)] 14 5 + (20 −1) 22 Speed up of EP over NP = 6 + (20 −1) 14 = 1.51 Hence, the correct answer is (1.51). Question Number: 51 Question Type: NAT Consider a 2-way set associative cache with 256 blocks and uses LRU replacement. Initially the cache is empty. Conflict misses are those misses which occur due to contention of multiple blocks for the same cache set. Compulsory misses occur due to first time access to the block. The following sequence of accesses to memory blocks (0, 128, 256, 128, 0, 128, 256, 128, 1, 129, 257, 129, 1, 129, 257, 129) is repeated 10 times. The number of conflict misses experienced by the cache is . Solution: Cache has 256 blocks and is organized in a 2-way set associative manner. ∴ The cache has 256 = 128 sets. 2 set 0 set 0

set 127

set 1 257 % 128 = 1 → replace block 1, conflict miss.

129 % 128 = 1 hit. ∴ For 1st access, number of conflict misses = 4. The contents of cache before starting 2nd time access is 256 ...... set 0 128 257 ...... set 1 129

set 0 0 % 128 = 0 → conflict miss, replace block 256. set 0 128 % 128 = 0 → Hit. set 0 256 % 128 = 0 → conflict miss, replace block 0. set 0 128 % 128 = 0 → Hit. set 0 0 % 128 = 0 → conflict miss, replace block 256. set 0 128 % 128 = 0 → Hit. set 0 256 % 128 = 0 → conflict miss, replace block 0. set 0 128 % 128 = 0 → Hit.

Similarly for next 8 accesses, we will get 4 conflict misses. ∴ Number of conflict misses for second access = 8. Now, the cache state is same as before accessing the sequence for the second time.

GATE 2017 Solved Paper CS: Set – 1 | xxxi So, for remaining accesses also there will be 8 conflict misses.

Solution: a

∴ Total number of conflict misses to access the sequence for 10 times = 4 + 9 × 8 = 76. Hence, the correct answer is (76). Question Number: 52 Question Type: NAT ∗ Consider the expression (a—1) (((b + c)/3) + d)). Let X be the minimum number of registers required by an optimal code generation (without any register spill) algorithm for a load/store architecture, in which (i) only load and store instruction can have memory operands and (ii) arithmetic instructions can have only register or immediate operands. The value of X is . Solution: Consider the expression (a – 1) * ((b + c) / 3) + d)

b

d

s

e

f

g

h

t

strlen (t) = 5

strlen (s) = 3

strlen( ) function return an unsigned value, computation among unsigned variables will result in unsigned value i.e., positive value. strlen(s) − strlen(t) ⇓ (3 − 5) ⇓ | −2 | > 0 ⇒ 2 > 0

Consider 2 registers R1 and R2. R1 = b R2 = c R1 = R1 + R2 R1 = R1/3 R2 = d R1 = R1 + R2 R2 = a R2 = R2 – 1 R1 + R1 × R2

as condition is true, it stores strlen(s) into variable len. It prints 3. Hence, the correct answer is (3). Question Number: 54 Question Type: NAT A cache memory unit with capacity of N words and block size of B words is to be designed. If it is designed as a direct mapped cache, the length of the TAG field is 10 bits. If the cache unit is now designed as a 16-way set-associative cache, the length of the TAG filed is _________ bits.

Only 2 registers are required Hence, the correct answer is (2). Question Number: 53 Consider the following C program.

c

Question Type: NAT

# include > = 1; } return count; } void main ( ) { static int x = 0; int i = 5; for (; i > 0,i−−) { x = x + total (i); } printf (“%d\n”, x); } Solution: For the above code, the total( ) function calls will be total(5), total(4), total(3), total(2) and total(1). As the variables ‘count’ and ‘x’ are static, it restores the previous value. The values of count and x at each function call is shown below.

Total (5) V = 5 count 0 1 2 while (5) Count = count + V & 1 ; ⇒ 0 + (101) and (001) ⇒1 V >> = 1 ⇒ (101) >> 1 ⇒ 2 while (2) Count = count + V & 1 ; ⇒ 1 + (010) and (001) ⇒ 1 V >> = 1 ⇒ (010) >> 1 ⇒ 1 while (1) Count = count + V & 1 ; ⇒ 1 + (001) and (001) ⇒ 2 V >> = 1 ⇒ (001) >> 1 ⇒ 0 total (5) Similarly, total (4) total (3) total (2) total (1)

returns

value 2

and

x=2

returns returns returns returns

value 3 value 5 value 6 value 7

and and and and

x=5 x = 10 x = 16 x = 23

The value printed is 23. Hence, the correct answer is (23).

General Aptitude Number of Questions: 10 Q.56 to Q.60 carry 1 mark each and Q.61 to Q.65 carry 2 marks each. Question Number: 56 Question Type: MCQ After Rajender Chola returned from his voyage to Indonesia, he ________ to visit the temple in Thanjavur. (A) was wishing (B) is wishing (C) wished (D) had wished Ans: (C) Question Number: 57 Question Type: MCQ Research in the workplace reveals that people work for many reasons . (A) money beside (B) beside money (C) money besides (D) besides money Ans: (D) Question Number: 58 Question Type: MCQ Rahul, Murali, srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is setting to the right of Arul. Which of the following paris are seated opposite each other? (A) Rahul and Murali (B) Srinivas and Arul (C) Srinivas and Murali (D) Srinivas and Rahul Ans: (C)

Section Marks: 15 Question Number: 59 Question Type: MCQ Find the smallest number y such y × 162 is a perfect cube. (A) 24 (B) 27 (C) 32 (D) 36 Ans: (D) Question Number: 60 Question Type: MCQ The probability that a k-digit number does NOT contain the digits 0, 5, 0 or 9 is (A) 0.3k (B) 0.6k k (C) 0.7 (D) 0.9k Ans: (C) Question Number: 61 Question Type: MCQ “The hold of the nationalist imagination on our colonial past is such that anything inadequately or improperly nationalist is just not history.” Which of the following statements best reflects the author’s opinion? (A) Nationalists are highly imaginative. (B) History is viewed through the filter of nationalism. (C) Our colonial past never happened. (D) Nationalism has to be both adequately and properly imagined. Ans: (B)

GATE 2017 Solved Paper CS: Set – 1 | xxxiii Question Number: 62 Question Type: MCQ Six people are seated around a circular table. There are at least two men and two women. There are at least three righthanded persons. Every woman has a left-handed person to her immediate right. None of the women are right- handed. The number of women at the tables is (A) 2 (B) 3 (C) 4 (D) Cannot be determined Ans: (A) Question Number: 63 The expression

Question Type: MCQ

( x − y) − x − y

is equal to 2 (A) the maximum of x and y (B) the minimum of x and y (C) 1 (D) none of the above

Ans: (B) Question Number: 64 Question Type: MCQ Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts coloured red, pink, blue and white respectively. Arun dislikes the colour red and Shweta dislikes the colour white. Gulab and Neel like all the colours.

In how many different ways can they choose the shirts so that no one has a shirt with a colour he or she dislikes? (A) 21 (B) 18 (C) 16 (D) 14 Ans: (D) Question Number: 65 Question Type: MCQ A contour lines joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25m intervals in this plot. If in a flood, the water level rises to 525m, which of the villages P, Q, R, S, T get submerged ? 425

Q R

450

550

P 550 T

500 450

500

(A) P, Q (C) R, S, T Ans: (C)

S

(B) P, Q, T (D) Q, R, S

GATE 2017 Solved Paper CS: Computer Science and Information Technology Set – 2

Number of Questions: 65

Total Marks:100.0

Wrong answer for MCQ will result in negative marks, (-1/3) for 1 Mark Questions and (-2/3) for 2 Marks Question.

Computer Science & Engineering Number of Questions: 55

Section Marks: 85

Q.1 to Q.25 carry 1 mark each and Q.26 to Q.55 carry 2 marks each. Question Number: 1 Question Type: MCQ The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is (A) 571244 (B) 736251 (C) 571247 (D) 136251 Solution: (BCA9)16 octal ( x )8 (BCA9)16 → (x)2 (BCA9)16 → (1011 1100 1010 1001) 2

⇓ (x)8 001 011 110 0 10 10 1 001 → (136251)8 1 3 6 2 5 1

Hence, the correct option is (D). Question Number: 2 Match the following:

Question Type: MCQ

(P) static char var;

(i) Sequence of memory locations to store addresses

(Q) m = malloc (10); m = NULL;

(ii) A variable located in data section of memory

(R) char *ptr [10];

(iii) Request to allocate a CPU register to store data

(S) register int var1;

(iv) A lost memory which cannot be freed

(A) P → (ii), Q → (iv), R → (i), S → (iii) (B) P → (ii), Q → (i), R → (iv), S → (iii) (C) P → (ii), Q → (iv), R → (iii), S → (i) (D) P → (iii), Q → (iv), R → (i), S → (ii)

Solution: • Static variables are stored in the program data segment. • The statement m = malloc (10) is assigned with 10 memory units and when m is assigned with NULL value, then memory allocated is lost and it can’t be freed. • char *ptr [10]; declaration specifies a 10 pointers which are used to store memory addresses. • register int var1; declaration request the CPU for allocation of CPU registers for storing data. Hence, the correct option is (A).

Question Number: 3

Question Type: MCQ Match the algorithms with their time complexities: Algorithm (P) Towers of Hanoi with n disks

Time complexity (i) Θ (n2)

(Q) B inary search given n sorted (ii) Θ (n log n) numbers (R) Heap sort given n numbers at the (iii) Θ (2n) worst case (S) Addition of two n × n matrices

(iv) Θ (log n)

(A) P → (iii), Q → (iv), R → (i), S → (ii) (B) P → (iv), Q → (iii), R → (i), S → (ii) (C) P → (iii), Q → (iv), R → (ii), S → (i) (D) P → (iv), Q → (iii), R → (ii), S → (i) Solution: To solve Towers of Hanoi puzzle of ‘n’ disks, it takes θ(2n) time. Binary search over ‘n’ sorted numbers takes θ(log n) time. ‘n’ numbers can be sorted in θ(n log n) time using Heap sort even in worst case. Addition of two matrices of order n × n takes θ(n2) time, since it requires n2 computations. Hence, the correct option is (C). Question Number: 4 Question Type: MCQ Let L1 L2 be any two context-free languages and R be any regular language. Then which of the following is/are CORRECT? I. L1 ∪ L2 is context-free. II. L1 is context-free. III. L1 − R is context-free. IV. L1 ∩ L2 is context-free. (A) I, II and IV only (B) I and III only (C) II and IV only (D) I only Solution: Given L1, L2 are CFL. R is regular. CFL’s are closed under union. CFL’s are not closed under complement. L1 − R = L1 ∩ R R is closed under complement. Intersection of regular and CFL is CFL.

GATE 2017 Solved Paper CS: Set – 2 | xxxv CFL’s are not closed under intersection. ∴ I and III are correct. Hence, the correct option is (B). Question Number: 5 Question Type: MCQ Match the following according to input (from the left column) to the compiler phase (in the right column) that processes it:

(P) Syntax tree

(i) Code generator

(Q) Character stream

(ii) Syntax analyzer

(R) Intermediate representation

(iii) Semantic analyzer

(S) Token stream

(iv) Lexical analyzer

(A) (B) (C) (B)

P → (ii), Q → (iii), R → (iv), S → (i) P → (ii), Q → (i), R → (iii), S → (iv) P → (iii), Q → (iv), R → (i), S → (ii) P → (i), Q → (iv), R → (ii), S → (iii)

Solution: Phases of compiles with inputs Source code ↓ Character stream Lexical analysis (lexical analyzer) ↓ Token stream Syntax analysis (syntax analyzer) ↓ Syntax tree Semantic analysis (semantic analyzer) ↓ Intermediate representation Code generator ↓ Output Hence, the correct option is (C). Question Number: 6 Question Type: MCQ Which of the following statements about parser is/are CORRECT? I. Canonical LR is more powerful than SLR. II. SLR is more powerful than LALR. III. SLR is more powerful than Canonical LR. (A) I only (B) II only (C) III only (D) I and III only Solution: Relation among parsers (with respect to accepting languages) is shown with below Venn diagram: CLR (1) LALR (1) SLR (1) LR (0) LL (1)

Hence, the correct option is (A).

Question Number: 7 Question Type: MCQ Which of the following is/are shared by all the threads in a process? I. Program counter II. Stack III. Address space IV. Registers (A) I and II only (B) III only (C) IV only (D) III and IV only Solution: Each thread has its own program counter, stack, Registers and state. Address space is shared by all threads of a process Hence, the correct option is (B). Question Number: 8 Question Type: MCQ In a file allocation system, which of the following allocation scheme(s) can be used if no external fragmentation is allowed? I. Contiguous II. Linked III. Indexed (A) I and III only (B) II only (C) III only (D) II and III only Solution: In contiguous allocation, there is a possibility of external fragmentation. Hence, the correct option is (D). Question Number: 9 Question Type: MCQ Consider the following statements about the routing protocols. Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) in an IPv4 network. I: RIP uses distance vector routing II: RIP packets are sent using UDP III: OSPF packets are sent using TCP IV: OSPF operation is based on link-state routing Which of the statements above are CORRECT? (A) I and IV only (B) I, II and III only (C) I, II and IV only (D) II, III and IV only Solution: •

RIP uses distance vector routing.

• RIP packets are sent using UDP as RIP uses UDP as transport protocol. • OSPF packets are not sent using TCP and its operation is based on link state routing. Hence, the correct option is (C). Question Number: 10 Question Type: MCQ 1 1 p x 2R , If f(x) = R sin + S , f ' = 2 and ∫ f ( x) dx = 2 2 0 p then the constants R and S are, respectively.

xxxvi | GATE 2017 Solved Paper CS: Set – 2 (C) (¬ p ∧ r) ∨ ((p ∧ q) → ¬ r) (D) (¬ p ∧ r) ∨ ( r → (p ∧ q))

(A)

2 16 and p p

(B)

2 and 0 p

(C)

4 and 0 p

(D)

4 16 and p p

Solution: Given statements are q: It is cold

πx Solution: Given f ( x ) = R sin + S 2 1 f ′ = 2 and 2

f 1 ( x) =

1

∫

f ( x ) dx =

0

r: It is pleasant

2R π

π π R cos x 2 2

1 π 1 π f = 2 ⇒ R cos × = 2 2 2 2 2 ⇒

⇒

π π R cos = 2 4 2

∫

f ( x ) dx =

0

2R π

4 2 π π ⇒ ∫ R sin x + S dx = 2 π 0 1

1

⇒

4

π

8

∫ π sin 2 x + S dx = π

2

0

1

“It is not raining and it is pleasant, and it is not pleasant, only if it is raining and it is cold”. “It is not raining and it is pleasant, and if it is not pleasant then it is raining and it is cold”. Which can be written in symbolic form as (¬p ∧ r) ∧ (¬r → (p ∧ q)) Hence, the correct option is (A).

1 π R. = 2 2 2 4 ⇒ R= π 1

Also,

Given compound statement is

is same as,

1

p: It is raining

π − cos x 2 4 8 ⇒ + Sx = 2 π π π 2 0

−8 −8 π 8 ⇒ 2 cos + S − 2 cos 0 + S × 0 = 2 2 π π π 8 8 ⇒ S+ 2 = 2 π π ⇒S=0 4 ∴ R = and S = 0 π Hence, the correct option is (C).

Question Number: 11 Question Type: MCQ Let p. q. r denote the statements “It is raining”, “It is cold”, and “It is pleasant”, respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by (A) (¬ p ∧ r) ∧ (¬ r → (p ∧ q)) (B) (¬ p ∧ r) ∧ ((p ∧ q) → ¬ r)

Question Number: 12 Question Type: MCQ Given the following binary number in 32-bit (single precision) IEEE-754 format: 00111110011011010000000000000000 The decimal value closest to this floating-point number is (A) 1.45 × 101 (B) 1.45 × 10−1 −1 (C) 2.27 × 10 (D) 2.27 × 101 Solution: Given 32-bit IEEE 754 number is 0 0111110 0

11011010000000000000000

Sign-bit = 0 ⇒ number is positive Biased exponent = 22 + 23 + 24 +25 + 26 =124 ∴ Exponent = 124−127 = −3 Mantissa = 1.1101101 ∴ Number in binary = 1.1101101 × 2−3 = 0.0011101101 = 0.2314453125 ≃ 2.3 × 10−1 This is approximately equal to choice (C). Hence, the correct option is (C). Question Number: 13 Question Type: MCQ A circular queue has been implemented using a singly linked list where each node consists of a value and a single pointer pointing to the next node. We maintain exactly two external pointers FRONT and REAR pointing to the front node and the rear node of the queue, respectively. Which of the following statements is/are CORRECT for such a circular queue, so that insertion and deletion operations can be performed in O (1) time?

GATE 2017 Solved Paper CS: Set – 2 | xxxvii I. II.

Next pointer of front node points to the rear node. Next pointer of rear node points to the front node. (A) I only (B) II only (C) Both I and II (D) Neither I nor II

Solution: Circular Queue with Linked List: a

b

(A) (B) (C) (D)

MNOPQR NQMPOR QMNROP POQNMR

Solution: Given graph M

N

O

R

Q

P

c

FRONT

REAR

In above, implementation of circular queue using linked list, Rear pointer always points Front as it is done always for every new insertion. FRONT node next pointer points to REAR only when they are only 2 elements in a queue. Insertion and Deletion can be done in O(1) time. Hence, the correct option is (B). Question Number: 14 Question Type: MCQ Consider the following function implemented in C: void printxy (int x, int y) { int ptr; x = 0; ptr = &x; y = ptr; ptr = 1; printf (“%d, %d” x, y); }

In BFS traversal, it first traverse all the neighbours of start node and traverse corresponding neighbours iteratively till all the nodes are visited. Option (D) is the correct BFS traversal of above graph ‘G’. Hence, the correct option is (D). Question Number: 16 Question Type: MCQ Identify the language generated by the following grammar, where S is the start variable. S → XY X → aX|a Y → aYb|∈

The output of invoking printxy (1, 1) is (A) 0, 0 (B) 0, 1 (C) 1, 0 (D) 1, 1

(A) {ambn| m ≥ n, n > 0} (C) {ambn| m >n, n ≥ 0}

(B) {ambn| m ≥ n, n ≥ 0} (D) {ambn| m > n, n > 0}

Solution: Given grammar S → XY X →aX | a

Solution: x 1 01 y 1 0

Y → aYb | ε

ptr y = *ptr ⇒ y = 0 when * ptr = 1 ⇒ only value of x will change, i.e., 1 when x and y are printed, it prints 1, 0. Hence, the correct option is (C). Question Number: 15 Question Type: MCQ The Breadth First Search (BFS) algorithm has been implemented using the queue data structure. Which one of the following is a possible order of visiting the nodes in the graph below? M

N

O

R

Q

P

For given grammar, there need to atleast one a. There can be zero b’s. n≥0 Y generals equal number of a’s, and b’s. X generates one or more a’s so S generates {am bn | m > n, n ≥ 0} Hence, the correct option is (C). Question Number: 17 Question Type: MCQ An ER model of a database consists of entity types A and B. These are connected by a relationship R which does not have its own attribute, Under which one of the following conditions, can the relational table for R be merged with that of A? (A) Relationship R is one-to-many and the participation of A in R is total. (B) Relationship R is one-to-many and the participation of A in R is partial.

xxxviii | GATE 2017 Solved Paper CS: Set – 2

(C) Relationship R is many-to-one and the participation of A in R is total. (D) Relationship R is many-to-one and the participation of A in R is partial.

Solution: m

A

I

R

B

The relationship R table can be merged with entity A’s table. If R is many-to-one and A is total Participated in relation R. Hence, the correct option is (C). Question Number: 18 Question Type: MCQ Consider socket API on a Linux machine that supports connected UDP sockets. A connected UDP socket is a UDP socket on which connect function has already been called. Winch of the following statements is/are CORRECT? I. A connected UDP socket can be used to communicate with multiple peers simultaneously. II. A process can successfully call connect function again for an already connected UDP socket. (A) I only (B) II only (C) Both I and II (D) Neither I nor II

〈3, 8〉 from table T1, the number of additional records that need to be deleted from table T1 is __________. Solution: Based on given description, S is depend on P; Q is depend on R. After deleting row (3, 8) from T1, set S = NULL at R = 8 in T2. ∴ zero rows will get deleted in T1, other than (3, 8).

Hence, the correct answer is (0). Question Number: 20 Question Type: NAT The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is__________. Solution: In IPV4, maximum number of router addresses specified will be 9. Hence, the correct answer is (9). Question Number: 21 Question Type: NAT Consider the set X = {a,b,c,d,e} under the partial ordering R = {(a, a).(a, b). (a, c), (a, d). (a, e),(b, b), (b, c), (b, e), (c, c). (c, e), (d, d), (d, e), (e, e)}. The Hasse diagram of the partial order (X, R) is shown below. e

Solution: A process can use Connection function on a UDP socket, if the communication is point to point. Connected UDP simulates like TCP connection (like peer to peer connection) Hence, the correct option is (B). Question Number: 19 Question Type: NAT Consider the following tables T1 and T2. T1

c

d

b

T2

P

Q

R

S

2

2

2

2

3

8

8

3

7

3

3

2

The minimum number of ordered pairs that need to be added to R to make (X, R) a lattice is __________.

5

8

9

7

Solution: Given poset (X, R) is itself a lattice.

6

9

5

7

So, the minimum number of ordered pairs to be added = 0.

8

5

7

2

9

8

In table T1, P is the primary key and Q is the foreign key referencing R in table T2 with on-delete cascade and onupdate cascade. In table T2, R is the primary key and S is the foreign key referencing P in table Tl with on-delete set NULL and on-update cascade. In order to delete record

a

Hence, the correct answer is (0). Question Number: 22 Question Type: NAT 1 1 −1 −1 −2 −1 Let P = 2 −3 4 and Q = 6 12 6 be two 3 −2 3 5 10 5 matrices. Then the rank of P + Q is __________.

GATE 2017 Solved Paper CS: Set – 2 | xxxix 1 1 −1 −1 −2 −1 Solution: Let P = 2 −3 4 and Q = 6 12 6 3 −2 3 5 10 5

0 −1 −2 ∴ P + Q = 8 9 10 8 8 8

0 −1 −2 Det ( P + Q ) = 8 9 10 = 0 8 8 8 0 −1 of P + Q is and determinant of a 2 × 2 sub-matrix 8 9 8 ≠ 0.

Question Number: 25 Question Type: NAT The minimum possible number of states of a deterministic finite automaton that accepts the regular language L = {w1aw2| w1, w2 ∈ {a, b}*, |w1| = 2, |w2| ≥ 3} is __________. Sulution: Given language L = {w1aw2 | w1, w2 ∈ {a, b}*, |w1| = 2 |w2| ≥ 3} The minimum DFA for L can be a,b a,b

a,b

a

b

∴ Rank of P + Q = 2 Hence, the correct answer is (2).

Dead siale

a,b

Question Number: 23 Question Type: NAT G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is __________.

a,b

Solution: Number of edges of G = n (E) = 25 Given number of vertices of G = n (V) = n As the degree of each vertex is atleast 3, we have sum of the degrees of all the vertices will be of the form 3n + k, where k is a positive integer. We know that, sum of the degrees of all the vertices = 2 × Number of edges ⇒ 3 n + k = 2 × 25 ⇒ 3 n + k = 50 For, n to be maximum, k should be minimum. ∴ 3 n + k = 50 and k is minimum only when n = 16. Hence, the correct answer is (16). Question Number: 24

Question Type: NAT

Consider a quadratic equation x − 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b = __________. 2

Solution: x2 − 13x +36 = 0 5.6 = (36)b 30 = 3b + 6 3b = 24 b=8 ∴ base = 8 Hence, the correct answer is (8.0 to 8.0).

a,b

∴ Minimal DFA which accepts L has 8 states. (Including dead state) Hence, the correct answer is (8). Question Number: 26 Question Type: MCQ P and Q are considering to apply for a job. The probability 1 that P applies for the job is , the probability that P applies 4 1 for the job given that Q applies for the job is and the 2 probability that Q applies for the job given that P applies for 1 the job is . Then the probability that P does not apply for 3 the job given that Q does not apply for the job is 5 4 (A) (B) 6 5 7 11 (C) (D) 8 12 Solution: Let A and B denote the events of the persons P and Q applying for a job respectively.

B 1 1 A 1 ∴ P ( A) = , P = and P = A 3 4 B 2

xl | GATE 2017 Solved Paper CS: Set – 2 B P ( A ∩ B) We have P = A P ( A)

= 1−

1 6

1 1− 6

B ⇒ P(A ∩ B) = P ⋅ P (A) A

1 1 = × 3 4 1 ⇒ P ( A ∩ B ) = 12 From the diagram, we have

1 = 1− 6 5 6

= 1−

1 5

4 5 Hence, the correct option is (A). =

B

A A∩B A∩B

A∩B

A = ( A ∩ B) ∪ ( A ∩ B)

Question Number: 27 Question Type: MCQ If w, x, y, z are Boolean variables, then which one of the following is INCORRECT? (A) wx + w(x + y) + x(x + y) = x + wy

⇒ P ( A) = P ( A ∩ B ) + P ( A ∩ B )

(B) w x ( y + z ) + w x = w + x + yz

( A ∩ B and A ∩ B are mutually exclusive).

(C)

⇒ P ( A ∩ B) = P ( A) − P ( A ∩ B )

(D) (w + y) (wxy + wyz) = wxy + wyz

1 1 = − 4 12 1 ∴ P ( A ∩ B ) = 6 Also, we know that

Solution: By observation, we can say that option (D) is correct.

(C): (wx ( y + xz ) + w x ) y

( wxy + 0 + w x ) y wxy + w xy = x y ≠ x y Hence option (C) is wrong.

A P ( A ∩ B) P = B P ( B)

(B): ( wx y + z ) + wx

P ( A ∩ B) A P B

⇒ P ( B ) =

w + x + yz + w x = w + x + yz Thus; option (B) is also correct. (A): wx + w ( x + y ) + x( x + y ) = x + wy

1 12 = 1 2 1 ∴ P ( B ) = 6

A = 1− P B = 1−

P ( A ∩ B) P ( B)

wx + wx + wy + x + xy

wx + wy + x = x + wy Hence, the correct option is (C).

∴ Probability that P does not apply for the job given that Q does not apply for the job.

A = P B

(w x ( y + x z ) + w x ) y = x y

Question Number: 28 Question Type: MCQ Given f(w, x, y, z) = ∑m(0,l,2,3,7,8,10) + ∑d(5,6,ll,15), where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-ofsums (POS) form of f(w, x, y, z)? (A) f = ( w + z )( x + z )

= 1−

P ( A ∩ B) 1− P ( B )

(B) f = ( w + z )( x + z )

f = ( w + z )( x + z ) (C)

(D) f = ( w + z )( x + z )

GATE 2017 Solved Paper CS: Set – 2 | xli Solution: Given f (w, x, y, z) = ∑m (0, 1, 2, 3, 7, 8, 10) + ∑d(5, 6, 11, 15). yz

00

01

11

10

00

1

1

1

1

01

0

X

1

X

11

0

0

X

0

10

1

0

X

1

wx

2T T (n) = 2,

n>2 02 0E−T|T T−>T+F|F F − > (E) | id Which of the following grammars is not left recursive, but is equivalent to G? (A) E − >E − T | T (B) E − > TE T − > T + F | F E′ − > −TE | ∈ F − > (E) | id T−>T+F|F F − > (E) | id (C) E − > TX (D) E − > TX | (TX) X − > −TX | ∈ X − > −TX | +TX | ∈ T − > FY T − > id Y − > + FY | ∈ F − > (E) | id Solution: The given grammar is : E → E –T |T T →T + F | F F → (E) | id

Hence, the correct option is (B). Question Number: 34 Question Type: MCQ Consider a binary code that consists of only four valid codewords as given below: 00000,01011,10101,11110 Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are (A) p=3 and q=l (B) p=3 and q=2 (C) p=4 and q=1 (D) p=4 and q=2 00000, 01011, 10101, 11110

E → TE1 E1 → -TE1| T → FT1 T1 → +FT1/E F → (E)|id

The hamming distance is 00000 01011 ± 01011 Hamming distance = 3

It is similar to option (C) in which E is replaced with X and T1 is replaced with Y. 1

Hence, the correct option is (C). Question Number: 33 Question Type: MCQ A system shares 9 tape drives. The current allocation and maximum requirement of tape drives for three processes are shown below: Current Allocation

Hence, the system is in safe state and there is no deadlock.

Solution: Given code words

Removal of left recursion of above grammar is:

Process

After allocating 3 + 1 + 3 = 7 tape drives, 9 − 7 = 2 tape drives. With these 2 tape drives, the safe sequences possible are < P3, P2, P1> or

Similarly, the hamming distance with 00000 and other code words 10101, 11110 is 3, 4. The hamming distance between 01011 and 10101 is 01011 10101 ±11110 Hamming distance = 4

Maximum Requirement

Similarly, hamming distance 01011 and 11110 is 3. Hamming distance between 10101 and 11110 is 3.

P1

3

7

P2

1

6

P3

3

5

∴ p = 3.

GATE 2017 Solved Paper CS: Set – 2 | xliii For correcting code, the condition is

12

2d+1=3 d=1 Therefore, the value of q = 1.

8, 6, 2, 7, 9, 10

16, 15, 19, 17, 20

Hence, the correct option is (A).

fl

Question Number: 35 Question Type: MCQ Consider two hosts X and Y connected by a single direct link of rate 106 bits/sec. The distance between the two hosts is 10.000 km and the propagation speed along the link is 2 × 108 m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the values of p and q are (A) p=50 and q=100 (B) p=50 and q=400 (C) p=100 and q=50 (D) p=400 and q=50

12

Solution:

2

15

9

6

7

10

19

17

20

The post order transversal of above tree is 2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12 Hence, the correct option is (B).

10,000kms X

16

8

10 bps 6

2 × 108 m/s

Transmission time =

Y

length of packet band width

50000×8 106 = 400 ms distance propagation time = velocity =

10000×103 2×108 = 50 msec

=

Hence, the correct option is (D). Question Number: 36 Question Type: MCQ The pre-order traversal of a binary search tree is given by 12,8,6,2,7,9,10,16,15,19,17,20. Then the post-order traversal of this tree is: (A) 2, 6, 7, 8, 9, 10, 12, 15, 16, 17, 19, 20 (B) 2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12 (C) 7, 2, 6, 8, 9,10, 20, 17, 19, 15, 16, 12 (D) 7, 6, 2, 10, 9, 8, 15, 16, 17, 20, 19, 12 Solution: Given pre order transversal of binary search tree is 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20 As the transversal is on binary search tree, it follows ordering property. From the given pre order transversal, 12 is the root.

Question Number: 37 Question Type: MCQ Consider the C program fragment below which is meant to divide x by y using repeated subtractions. The variables x, y, q and r are all unsigned int. while (r >= y) { r = r − y; q = q + 1; } Which of the following conditions on the variables x, y, q and r before the execution of the fragment will ensure that the loop terminates in a state satisfying the condition x == (y*q + r)? (A) (q == r) && (r == 0) (B) (x > 0) && (r == x) &&(y > 0) (C) (q == 0) && (r == x) && (y > 0) (D) (q == 0) && (y > 0) Solution: Given code : while (r >= y) { r = r – y; q = q + 1; } the pre-condition for x = = (y * q + r). Should be r = x, as x ÷ y is computed as r ÷ y. Similarly, q = 0 as the ‘x’ should be equal to ‘r’, i.e., (q = = 0), ‘y’ should be greater than 0 as it is unsigned variable (y > 0 ⇒ r > 0 ⇒ x > 0). Hence, the correct option is (C).

xliv | GATE 2017 Solved Paper CS: Set – 2 Question Number: 38 Question Type: MCQ Consider the following C function.

δˆ (q2 , aba ) gives all states reachable from q2, while reading aba. So, it can reach q0, q1, q2.

int fun (int n) { int i, j; for(i = 1; i array[i−l]) { swap(&array[i], & array[i−1]); done = 0; } } } printf{“%d”, array[3]);

The output of the program is ________. Solution: The above code sorts the elements in the decreasing order. The output array after the code execution is 0 1 2 3 4 5 6 5 4 3 2 1 So, array [3] contains an element ‘3’, so, it outputs 3. Hence, the correct answer is (3). Question Number: 44 Question Type: NAT Two transactions T1 and T2 are given as T1 : r1(X)w1 (X)r1 (Y)w1 (Y) T2 : r2 (Y)w2 (Y)r2 (Z)w2 (Z) where ri (V) denotes a read operation by transaction Ti on a variable V and wi (V) denotes a write operation by transaction Ti on a variable V. The total number of conflict serializable schedules that can be formed by T1 and T2 is _______. Solution: Number of conflict serializable schedules possible such that T1 is depend on T2 is 1. Number of conflict serislizable schedules possible such that T2 is depend on T1 is 53. ∴ Total possible conflict serializable schedules is 54

Hence, the correct answer is (54). Question Number: 45 Question Type: NAT The read access times and the hit ratios for different caches in a memory hierarchy are as given below. Cache

Read access time (in nanoseconds)

Hit ratio

I-cache

2

0.8

D-cache

2

0.9

L2-cache

8

0.9

The read access time of main memory is 90 nanoseconds. Assume that the caches use the referred- word-first read policy and the write back policy. Assume that all the caches are direct mapped caches. Assume that the dirty bit is always 0 for all the blocks in the caches. In execution of a program.

xlvi | GATE 2017 Solved Paper CS: Set – 2 60% of memory reads are for instruction fetch and 40% are for memory operand fetch. The average read access time in nanoseconds (up to 2 decimal places) is __________. Solution: Given there are 60% instruction fetch, 40% operand fetch.

AND ta.goals >ANY (SELECT tc.goals FROM top_scorer AS tc WHERE tc. country = ‘Germany’) The number of tuples returned by the above SQL query is _________.

For instruction fetch, average memory access time

Solution: Select tb. goals FROM top-scorer AS tb where tb. ‘country’ = Spain

= 0.6 [HI * RI + (1 − HI) HL (RI + RL ) + (1 − HI) (1 − HL ) 2 2 2 (RI + RL + Rm)]

This query returns zero tuples as no country has name ‘Spain’.

= 0.6[0.8 * 2 + (1−0.8) 0.9 (2 + 8) + (1 − 0.8) (1 − 0.9) (2 + 8 + 90)]

TC table has below rows:

2

tc

= 0.6[1.6 + 1.8 + 2]

tc. goals

= 3.24 ns

16

For operand fetch,

14

Average memory access time

11

= 0.4[HD * RD + (1 − HD) HL (RD + RL ) + (1 − HD) (1 − HL ) 2 2 2 (RD + RL + Rm)]

10

2

ta consists the player name whose goals is > ANY (tc.goals)

= 0.4[0.9 * 2 + 0.1 * 0.9 * (2 + 8) + 0.1 * 0.1 * (2 + 8 + 90)]

ta

= 0.4 [1.8 + 0.9 + 1]

ta.player Klose

(∵ 16 >15)

Ronaldo

(∵ 15 >14)

G muller

(∵ 14 >11)

Question Number: 46 Question Type: NAT Consider the following database table named top_scorer.

Foundtaine

(∵ 13 >10)

Pele

(∵ 12 >10)

top_scorer

Klinsmann

(∵ 11 >10)

Kocsis

(∵ 11 >10)

= 3.7 * 0.4 = 1.48 ns ∴ Average memory access time = 3.24 + 1.48 = 4.72 Hence, the correct answer is (4.72).

player

country

goals

Klose

Germany

16

Ronaldo

Brazil

15

G Miiller

Germany

14

Fontaine

France

13

Pelé

Brazil

12

Klinsmann

Germany

11

Kocsis

Hungary

11

Batistuta

Argentina

10

Cubillas

Peru

10

Lato

Poland

10

Lineker

England

10

T Muller

Germany

10

Rahn

Germany

Consider the following SQL query: SELECT ta.player FROM top_scorer AS ta WHERE ta.goals >ALL (SELECT tb.goals FROM top_scorer AS tb WHERE tb.country = ‘Spain’)

10

∴ 7 rows returned by given query.

Hence, the correct answer is (7). Question Number: 47 Question Type: NAT If the ordinary generating function of a sequence 1+ z ∞ then a3 −a0 is equal to __________. {an }n=0 is 3 (1− z ) Solution: Given, the generating function of a sequence ∞ {an }n=0 is =

1+ z 3

(1− z )

= (1 + z )

1 3

(1− z )

∞ = (1 + z )∑ C (3 −1 + r , r ) z r r =0 ∞ 1 C (n −1 + r , r ) X r (1− X ) n = ∑ r =0

GATE 2017 Solved Paper CS: Set – 2 | xlvii ∞

= (1 + z ) ∑ C (2 + r , r ) z r r =0

= (1 + z ) C (2, 0) z 0 + C (3,1) z + C (4, 2) z 2 + C (5,3) z 3 + ... = (1 + z ) 1 + 3 z + 6 z 2 + 10 z 3 + ......

Question Number: 50 Question Type: NAT A message is made up entirely of characters from the set X = {P, Q, R, S,T}. The table of probabilities for each of the characters is shown below: Character

Probability

= (1 + 3 z + 6 z + 10 z + ...) + ( z + 3 z + 6 z + 10 z + ...)

P

0.22

= 1 + 4 z + 9 z 2 + 16 z 3 + .......

Q

0.34

∴ a0 = constant term of the generating function

R

0.17

⇒ a0 = 1

S

0.19

T

0.08

Total

1.00

2

3

2

3

4

a3 = The coefficient of z in the generating function. 3

⇒ a3 = 16 ∴ a3 −a0 = 16 −1 = 15 Hence, the correct answer is (15). Question Number: 48 Question Type: NAT If a random variable X has a Poisson distribution with mean 5. then the expectation E[(X + 2)2] equals _________. Solution: Given X is a Poisson random variable and mean of X = E(X) = 5

If a message of 100 characters over X is encoded using Huffman coding, then the expected length of the encoded message in bits is _________. Solution: Message of 100 characters contains letters P, Q, R, S, T with their frequencies 22, 34, 17, 19, 8. The Huffman tree for the above is: I4

∴ Variance of X = Var (X) = 5 2 Consider E ( X + 2) = E X 2 + 4 X + 4

= E ( X 2 ) + 4E ( X ) + 4

2 = var ( X ) + ( E ( X )) + 4 E ( X ) + 4

= 5 + 52 + 4×5 + 4 = 54

1

0

Hence, the correct answer is (54).

I3 (59)

Question Number: 49 Question Type: NAT In a B+ tree, if the search-key value is 8 bytes long, the block size is 512 bytes and the block pointer size is 2 bytes, then the maximum order of the B+ tree is __________.

1

0

Solution: Given,

I2 (41)

Q

Search key value V = 8 B

I1 (25)

(34)

Block Size B = 512 B 0

Block pointer size P = 2B

1

0

1

If the maximum order is m, then (m × P) + ((m – 1) × V) ≤ B ⇒ m * 2 + ((m – 1) × 8) ≤ 512 ⇒ 2 m + 8m – 8 ≤ 512 ⇒ 10m ≤ 520 ⇒ m ≤ 52

∴ maximum order is 52.

Hence, the correct answer is (52).

P

S

R

(22)

(19)

(17)

Number of bits required for each character is P – 00 – 2 bits × 22 ⇒ 44 Q – 10 – 2 bits × 34 ⇒ 68 R – 110 – 3 bits × 17 ⇒ 51

T (8)

xlviii | GATE 2017 Solved Paper CS: Set – 2 S – 01 – 2 bits × 19 ⇒ 38 T – 111 – 3 bits x 8 ⇒ 24

one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is__________.

Total bits required = 225

Solution: Given the characteristic polynomial of a 3 × 3 matrix M is λ3 − 4λ2 +aλ + 30, a ∈

Hence, the correct answer is (225). Question Number: 51 Question Type: NAT Consider the set of processes with arrival time (in milliseconds). CPU burst time (in milliseconds). and priority (0 is the highest priority) shown below. None of the processes have I/O burst time. Process

Arrival Time

Burst Time

Priority

P1

0

11

2

P2

5

28

0

P3

12

2

3

P4

2

10

1

P5

9

16

4

Also given one eigen value of M is 2. Let λ1 and λ2 be the other two eigen values of M. The characteristic equation of M is

λ3 − 4λ2 + aλ + 30 = 0 We know that, Sum of the eigen values of M = − (The coefficient of λ2 in its characteristic equation) ⇒ 2 × λ1 +λ2 = − (−4) ⇒ λ1 + λ2 = 2

(1)

and product of the eigen values of M = −(constant term of the characteristic equation of M). ⇒ 2 + λ1 × λ2 = −30 ⇒ λ1λ2 = −15(2)

The average waiting time (in milliseconds) of all the processes using preemptive priority scheduling algorithm is________.

We know that the quadratic equation with λ1 and λ2 as roots is

Solution: Using preemptive priority scheduling algrithm, the Gantt chart is as below:

⇒ λ2 − (2) λ − 15 = 0

λ2 − (λ1 + λ2) λ + λ1 λ2 = 0 (From (1) and (2))

⇒ λ − 2λ − 15 = 0 2

P1

P1 0

1

P4 2

P4 3

P4 4

P2 5

P4 33

P1 40

P4 (Priority 1)

P3 49

P5 51 67

P2 (Priority 0)

At time 33 P2 completed, P4 Left with 7 units of burst time. At time 40 P4 completed, P1 Left with 9 units of burst time After completion of process execution: waiting time of P1 = 40 – 2 = 38 waiting time of P2 = 0

⇒ (λ − 5) (λ + 3) = 0 ⇒ λ = 5; λ = −3 i.e., λ1 = 5 and λ2 = −3 ∴ The eigen values of M are −3, 2 and 5. ∴ The largest among the absolute values of the eigen values of M is 5. Hence, the correct answer is (5). Question Number: 53 Question Type: NAT Consider a machine with a byte addressable main memory of 232 bytes divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is_______.

waiting time of P3 = 49 – 12 = 37

Solution: Main memory capacity = 232 B

Waiting time of P4 = 33 – 5 = 28

Block size = 32B = 25B

waiting time of P5 = 51 – 9 = 42 38 + 0 + 37 + 28 + 42 ∴ Average waiting time = 5 = 29 ms

Number of lines in cache = 512 = 29

Hence, the correct answer is (29).

Question Number: 52 Question Type: NAT If the characteristic polynomial of a 3 × 3 matrix M over (the set of real numbers) is λ3 − 4 λ2 + aλ + 30, a ∈ , and

9

5

Tag Line offset 32−bits

Tag = 32 − (9 + 5) = 18 Hence, the correct answer is (18).

GATE 2017 Solved Paper CS: Set – 2 | xlix Question Number: 54 Consider the following C Program. #include int main () { int m = 10; int n, nl ; n = ++m; nl = m++; n−−; −−nl; n −= nl; printf (“%d”, return 0; }

Question Type: NAT

n 11 0

The output will result in 0. Hence, the correct answer is (0). Question Number: 55 Consider the following C Program. n) ;

The output of the program is _____________. Solution:

n = n − n1; ⇓ n = 10 − 10; n = 0;

m 10 n = ++m; n 11 m 11 n 1= m ++; n1 11 m 11 12 n - -; n 11 10 - - n1;

Question Type: NAT

#include #include int main () { char* c = “GATECSIT2017”; char* p = c; printf{“%d”, (int) strlen(c+2[p]-6[p]-1)) ; return 0; }

The output of the program is __________. Solution: The expression (c + 2[p] − 6 [p] − 1) will map to the sub string 17, i.e.

G A T E C S

I T 2 0 1 7 expression maps to above character

n1 11 10

n− =n1; ⇓

The string length for the expression will result in 2. Hence, the correct answer is (2).

General Aptitude Number of Questions: 10 Q.56 to Q.60 carry 1 mark each and Q.61 to Q.65 carry 2 marks each. Question Number: 56 Question Type: MCQ Choose the option with words that are not synonyms. (A) aversion, dislike (B) luminous, radiant (C) Plunder, loot (D) yielding, resistant Ans: (D) Question Number: 57 Question Type: MCQ Saturn is ________ to be seen on a clear night with the naked eye. (A) enough bright (B) bright enough (C) as enough bright (D) bright as enough Ans: (B) Question Number: 58 Question Type: MCQ There are five buildings called V. W. X. Y and Z in a row (not necessarily in that order). V is to the West of W. Z is to the East of X and the West of V. W is to the West of Y. Which is the building in the middle?

Section Marks: 15

(A) V (C) X

(B) W (D) Y

Ans: (A) Question Number: 59 Question Type: MCQ A test has twenty questions worth 100 marks in total. There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have? (A) 12 (B) 15 (C) 18 (D) 19 Ans: (B) Question Number: 60 Question Type: MCQ There are 3 red socks. 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is

l | GATE 2017 Solved Paper CS: Set – 2

Ans: (D) Question Number : 61 Question Type: MCQ “We lived in a culture that denied any merit to literary works, considering them important only when they were handmaidens to something seemingly more urgent - namely ideology. This was a country where all gestures, even the most private, were interpreted in political terms.” The author’s belief that ideology is not as important as literature is revealed by the word: (A) ‘culture’ (B) ‘seemingly’ (C) ‘urgent’ (D) ‘political’

Ans: (A)

R 0.65

0.9

0.1 S

5

Ans: (B) Question Number: 63 Question Type: MCQ X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X3 will have (A) 90 digits (B) 91 digits (C) 92 digits (D) 93 digits

Ans: (C) Question Number: 65 Question Type: MCQ An air pressure contour line joins locations in a region having the same atmospheric pressure. The following is an air pressure contour plot of a geographical region. Contour lines are shown at 0.05 bar intervals in this plot.

0.9

Ans: (B) Question Number: 62 Question Type: MCQ There are three boxes. One contains apples, another contains oranges and the last one contains both apples and oranges. All three are known to be incorrectly labelled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes? (A) The box labelled ‘Apples’ (B) The box labelled ‘Apples and Oranges’ (C) The box labelled ‘Oranges’ (D) Cannot be determined

Question Number: 64 Question Type: MCQ x 2 The number of roots of e + 0.5x − 2 = 0 in the range [−5, 5] is (A) 0 (B) 1 (C) 2 (D) 3

0.95

(B) 7/30 (D) 4/15

P

0.9

0.8 Q

8

(A) 1/5 (C) 1/4

0.

0.75

If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region. which of the following regions is most likely to have a thunderstorm? (A) P (B) Q (C) R (D) S Ans: (C)

GATE 2018 Solved Paper CS: Computer Science and Information Technology Set – 1 Number of Questions: 65

Total Marks: 100.0

Wrong answer for MCQ will result in negative marks, (-1/3) for 1 Mark Questions and (-2/3) for 2 Marks Question.

General Aptitude Number of Questions: 10 Q.1 to Q.5 carry one mark each. Question Number: 1 Question Type: MCQ “From where are they bringing their books? ______ bringing ______ books from ______.” The words that best fill the blanks in the above sentence are: (A) Their, they’re, there (B) They’re, their, there (C) There, their, they’re (D) They’re, there, there Solution: The words that are apt for the three blanks are “they’re” (which is a contraction of “they are”), “their” (which means “belonging to the people previously mentioned) and “there” (which means in that place). Hence, the correct option is (B). Question Number: 2 Question Type: MCQ “A ___ investigation can sometimes yield new facts, but typically organized ones are more successful.” The word best fills the blank in the above sentence is: (A) meandering (B) timely (C) consistent (D) systematic Solution: A process or activity that does not seem to have a clear purpose or direction is said to meander. Thus, “meandering” is the apt word; the sentence suggests that a meandering investigation might yield new facts, but organized ones are more successful. Hence, the correct option is (A). Question Number: 3 Question Type: MCQ The area of a square is d. What is the area of the circle which has the diagonal of the square as its diameter? (A) πd (B) πd 2 1 2 1 (C) pd (D) pd 4 2 Solution: The area of the square is d. Its side is d . Its diagonal is 2d . This is the diameter of the circle. Therefore, π(2d ) πd the area of the circle = = ⋅ 4 2 Hence, the correct option is (D). Question Number: 4 Question Type: MCQ What would be the smallest natural number which when divided either by 20 or by 42 or by 76 leaves a remainder of 7 in each case?

GATE_CSIT-2018.indd 51

Section Marks: 15 (A) 3047 (C) 7987

(B) 6047 (D) 63847

Solution: The LCM of 20, 42, 76 = 20 (3) (7) (19) = 7980. The required number is 7987. Hence, the correct option is (C). Question Number: 5 Question Type: MCQ What is the number missing in the following sequence? 2, 12, 60, 240, 720, 1440, ______, 0 (A) 2880 (B) 1440 (C) 720 (D) 0 Solution: 2, 12, 60, 240, 720, 1440, ______, 0 2

12

60

2×6 12×5 60×4

240

720 1440

1440

240×3 720×2 1440×

The blank value will be 1440. Hence, the correct option is (B). Q.6 to Q.10 carry two marks each. Question Number: 6 Question Type: MCQ In appreciation of the social improvements completed in a town, a wealthy philanthropist decided to gift `750 to each male senior citizen in the town and `1000 to each female senior citizen. Altogether, there were 300 senior citizens eligible for this gift. However, only 8/9th of the eligible men and 2/3rd of the eligible women claimed the gift. How much money (in Rupees) did the philanthropist give away in total? (A) 1,50,000 (B) 2,00,000 (C) 1,75,000 (D) 1,51,000 Solution: Let the number of men be 9x and the number of women be 3y. (8/9 of the number of men and 2/3 of the number of women are integers) ∴ 9x + 3y = 300 ⇒ 3x + y = 100 (1) 8x men and 2y women claimed the gift. Amount given = 750(8x) + 1000(2y) = 6000x + 2000y (2) = 6000x + 2000(100 – 3x) = 200,000x

5/2/2018 3:33:20 PM

lii | GATE 2018 Solved Paper CS: Set – 1 (A) ∠BCD – ∠BAD (B) ∠BAD + ∠BCF (C) ∠BAD + ∠BCD (D) ∠CBA + ∠ADC

If x is an integer, among the options only B satisfies the condition. Hence, the correct option is (B). Question Number: 7

Question Type: MCQ 1 1 = , r − z = , what is the r p

1 −y ,q q value of the product xyz? If pqr ≠ 0 and p − x =

Solution: E

1 (B) pqr (D) pqr

(A) –1 (C) 1

1 1 1 p–x = , q − y = , r − z = r p q x y z ∴ p = q, q = r, r = p p = rz = (qy)z = pxyz If p ≠ –1, 0 or 1, then xyz = 1 The condition that p = –1 or 1 is not given, only with this condition we can conclude that xyz = 1. But as cannot be determined is not an option, we select 1. Hence, the correct option is (C). Question Number: 8 Question Type: MCQ In a party, 60% of the invited guests are male and 40% are female. If 80% of the invited guests attended the party and if all the invited female guests attended, what would be the ratio of males to females among the attendees in the party? (A) 2:3 (B) 1:1 (C) 3:2 (D) 2:1

Question Number: 9 Question Type: MCQ In the figure below, ∠DEC + ∠BFC is equal to ______.

A

∠E + ∠F + ∠A = ∠ECF = ∠BCD \ ∠E + ∠F = ∠BCD − ∠A = ∠BCD − ∠BAD Hence, the correct option is (A). Question Number: 10 Question Type: MCQ A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment? (A) Three green faces and four red faces. (B) Four green faces and three red faces. (C) Five green faces and two red faces. (D) Six green faces and one red face. Solution: The most likely outcome in the one where the ratio of the number of green and red faces in closest to 2:1. This is 5g, 2r. We may want to consider 4g, 3r. 2 5 1 2 P(5g, 2r) = 7C5 = 32(21)/37 3 3 = 16(42)/37

E D A

C

F

F B

Solution:

Solution: Let the number of invited men and women be 6 and 4. All the women attended. Overall 80% attended. Therefore, 4 men and 4 women attended. The required ratio is 1:1. Hence, the correct option is (B).

C

D

2 4 1 3 P(4g, 3r) = 7C4 = 16(35)/37 3 3

We see that 5g, 2r is more probable. Hence, the correct option is (C).

B

Computer Science and Information Technology Number of Questions: 55 Q.1 to Q.25 carry one mark each. Question Number: 1 Question Type: MCQ Which one of the following is a closed form expression for the generating function of the sequence {an}, where an = 2n + 3 for all n = 0, 1, 2, ...?

GATE_CSIT-2018.indd 52

Section Marks: 85 3 2 (1 − x )

(B)

2−x 2 (1 − x )

(D)

(A) (C)

3x 2 (1 − x ) 3− x 2 (1 − x )

5/2/2018 3:33:21 PM

GATE 2018 Solved Paper CS: Set – 1 | liii Solution: The generating function of the given sequence {an}, where an = 2n + 3 for all n = 0, 1, 2, ... is ∞

F(x) = ∑ (2n + 3) x n n=0 = 3 + 5x + 7x2 + 9x3 + 11x4 + ... = (3 + 6x + 9x2 + 12x3 + 15x4 + ...) – (x + 2x2 + 3x3 + 4x4 + ...) = 3(1 + 2x + 3x2 + 4x3+ ...) – x(1 + 2x + 3x2 + 4x3 + ...) 1 1 3− x = 3⋅ −x = 2 2 − − ( 1 x ) ( 1 x ) ( 1 − x )2 Hence, the correct option is (D). Question Number: 2 Question Type: MCQ Consider the following C program. #include

struct Ournode {

} ;

char x, y, z;

Int main () {

struct Ournode p = {‘1’, ‘0’, ‘a’+2};

printf (“%c, %c”, *( (char*)q+1),

struct Ournode *q = &p;

* ( (char*)q+2) );

return 0; }

The output of this program is: (A) 0, c (B) 0, a+2 (C) ‘0’, ‘a+2’ (D) ‘0’, ‘c’ Solution: P→

1

0

‘a’ + 2

q

q+1

q+2

*((char*) q + 1)⇒0 *((char*) q + 2)⇒‘a’+2 ⇒c

The output of program is 0, c Hence, the correct option is (A). Question Number: 3 Question Type: MCQ A queue is implemented using a non-circular singly linked list. The queue has a head pointer and a tail pointer, as shown in the figure. Let n denote the number of nodes in the queue. Let enqueue be implemented by inserting a new node at the head, and dequeue be implemented by deletion of a node from the tail.

GATE_CSIT-2018.indd 53

head

tail

Which one of the following is the time complexity of the most time-efficient implementation of enqueue and dequeue, respectively, for this data structure? (A) q(1), θ(1) (B) θ(1), θ(n) (C) θ(n), θ(1) (D) θ(n), θ(n) Solution: As the enqueue operation is done at header of linked list, if will take θ(1) time. Dequeue() takes θ(n) time—as tail points last node of linked list, after deletion it takes ‘n’ computations to reach last node. Hence, the correct option is (B). Question Number: 4 Question Type: MCQ Let ⊕ and denote the Exclusive OR and Exclusive NOR operations, respectively. Which one of the following is NOT CORRECT? (A) P ⊕ Q = P Q (B) P ⊕ Q = P Q (C) P ⊕ Q = P ⊕ Q (D) ( P ⊕ P ) ⊕ Q = ( P P ) Q Solution: ( P ⊕ P ) ⊕ Q = ( P P ) Q 1 ⊕ Q ≠ 0 Q Q ≠ Q Hence, the correct option is (D). Question Number: 5 Question Type: MCQ Consider the following processor design characteristics. I. Register-to-register arithmetic operations only II. Fixed-length instruction format III. Hardwired control unit Which of the characteristics above are used in the design of a RISC processor? (A) I and II only (B) II and III only (C) I and III only (D) I, II and III Solution: Characteristics of RISC: OO The operations are done in the register of the CPU. OO It has fixed length, easily decoded instruction format OO It uses Hard wired control unit rather than micro programmed control. Hence, the correct option is (D). Question Number: 6 Question Type: MCQ Let N be an NFA with n states. Let k be the number of states of a minimal DFA which is equivalent to N. Which one of the following is necessarily true? (A) k ≥ 2n (B) k≥n (C) k ≤ n2 (D) k ≤ 2n

5/2/2018 3:33:22 PM

liv | GATE 2018 Solved Paper CS: Set – 1 Solution: The maximum number of states minimal DFA can have is 2n, when we convert NFA → DFA. ∴ k ≤ 2n Hence, the correct option is (D). Question Number: 7 Question Type: MCQ The set of all recursively enumerable languages is: (A) Closed under complementation. (B) Closed under intersection. (C) A subset of the set of all recursive languages. (D) An uncountable set. Solution: The set of all recursively enumerable language is closed under intersection. Hence, the correct option is (B). Question Number: 8 Question Type: MCQ Which one of the following statements is FALSE? (A) Context-free grammar can be used to specify both lexical and syntax rules. (B) Type checking is done before parsing. (C) High-level language programs can be translated to different Intermediate Representations. (D) Arguments to a function can be passed using the program stack. Solution: The parsing is implemented at syntax analysis phase of complier, whereas type checking is done by semantic analysis phase.

Syntax analyser

Semantic analyser

Type checking is done after parsing, as syntax analysis is done before semantic analysis. Hence, the correct option is (B). Question Number: 9 Question Type: MCQ The following are some events that occur after a device controller issues an interrupt while process L is under execution. (P) The processor pushes the process status of L onto the control stack. (Q) The processor finishes the execution of the current instruction. (R) The processor executes the interrupt service routine. (S) The processor pops the process status of L from the control stack.

GATE_CSIT-2018.indd 54

(T) The processor loads the new PC value based on the interrupt. Which one of the following is the correct order in which the events above occur? (A) QPTRS (B) PTRSQ (C) TRPQS (D) QTPRS Solution: Option (A) gives the correct sequence of the events that occur after a device controller issues an interrupt while process ‘L’ is under execution. Hence, the correct option is (A). Question Number: 10 Question Type: MCQ Consider a process executing on an operating system that uses demand paging. The average time for a memory access in the system is M units if the corresponding memory page is available in memory and D units if the memory access causes a page fault. It has been experimentally measured that the average time taken for a memory access in the process is X units. Which one of the following is the correct expression for the page fault rate experienced by the process? (A) (D – M)/(X – M) (B) (X – M/(D – M) (C) (D – X/(D – M) (D) (X – M/(D – X) Solution: Effective memory access time = (1 – α) ∗ memory access time + α (page fault overhead) α = page fault rate From given data, X = (1 – α) × M + α × D X = M – αM + αD X −M α = D−M Hence, the correct option is (B). Question Number: 11 Question Type: MCQ In an Entity-Relationship (ER) model, suppose R is a manyto-one relationship from entity set E1 to entity set E2. Assume that E1 and E2 participate totally in R and that the cardinality of E1 is greater than the cardinality of E2. Which one of the following is true about R? (A) Every entity in E1 is associated with exactly one entity in E2. (B) Some entity in E1 is associated with more than one entity in E2. (C) Every entity in E2 is associated with exactly one entity in E1. (D) Every entity in E2 is associated with at most one entity in E1.

5/2/2018 3:33:22 PM

GATE 2018 Solved Paper CS: Set – 1 | lv Solution

Field

E1

M

R

I

E2

Length in bits

R.

IPv6 Next Header

III.

32

S.

TCP Header’s Sequence Number

IV.

16

(A) P-III, Q-IV, R-II, S-I (B) P-II, Q-I, R-IV, S-III (C) P-IV, Q-I, R-II, S-III (D) P-IV, Q-I, R-III, S-II E1

>

E2

Every entity in E1 is associated with exactly one entity in E2. Hence, the correct option is (A). Question Number: 12 Question Type: MCQ Consider the following two tables and four queries in SQL. Book (isbn, bname), Stock (isbn, copies) Query 1: SELECT B.isbn, S.copies FROM Book B INNER JOIN Stock S ON B.isbn = S.isbn; Query 2: SELECT B.isbn, S.copies FROM Book B LEFT OUTER JOIN Stock S ON B.isbn = S.isbn; Query 3: SELECT B.isbn, S.copies FROM Book B RIGHT OUTER JOIN Stock S ON B.isbn = S.isbn; Query 4: SELECT B.isbn, S.copies FROM Book B FULL OUTER JOIN Stock S ON B.isbn = S.isbn; Which one of the queries above is certain to have an output that is a superset of the outputs of the other three queries? (A) Query 1 (B) Query 2 (C) Query 3 (D) Query 4 Solution: 1. The natural join will return only those values which are common in both the tables. 2. The left outer join will gives all rows from left table with null values, if right side table is not having match. 3. The right outer join will gives all rows from table on right hand side with null values, if left hand side table is not having the match. 4. Full outer join will gives all rows in output which are common in both tables, present in left table only or right table only. Hence, the correct option is (D). Question Number: 13 Match the following:

Question Type: MCQ

Field

Length in bits

P.

UDP Header’s Port Number

I.

48

Q.

Ethernet MAC Address

II.

8

GATE_CSIT-2018.indd 55

Solution: UDP Header port Number is 16 bits OO The Ethernet MAC Address = 48 bits OO The length of IPV next header is 8 bits 6 OO The sequence number field in TCP header is 4 bytes, which is 32 bits. OO

Hence, the correct option is (C). Question Number: 14 Question Type: MCQ Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS denotes the Maximum Segment Size. (i) The cwnd increases by 2 MSS on every successful acknowledgment. (ii) The cwnd approximately doubles on every successful acknowledgement. (iii) The cwnd increases by 1 MSS every round trip time. (iv) The cwnd approximately doubles every round trip time. Which one of the following is correct? (A) Only (ii) and (iii) are true (B) Only (i) and (iii) are true (C) Only (iv) is true (D) Only (i) and (iv) are true Solution: Slow start phase of TCP congestion control algorithm: Each time an ACK is received the congestion window is increased by one segment. So, cwnd approximately doubles every round trip time. Hence, the correct option is (C). Question Number: 15 Question Type: NAT Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ______.

5/2/2018 3:33:23 PM

lvi | GATE 2018 Solved Paper CS: Set – 1 Solution: Let A = The event of both P and Q getting same number on the dice in one trial and B = The event of P and Q getting different numbers on the dice in one trial. 6 1 30 5 \ P(A) = = and P ( B = ) = 36 6 36 6 ∴ The probability that one of them wins on the third trial = P(A ∩ A ∩ B) = P(A) P(A) P(B) (∴ A and B are independent) 1 1 5 = × × 6 6 6 5 = = 0.023 216 Hence, the correct answer is 0.021 to 0.024. Question Number: 16 π/4

The value of

∫

Question Type: NAT

x cos ( x 2 ) dx correct to three decimal

Question Number: 17

Question Type: NAT 1 1 Consider a matrix A = uvT where u = , v = ⋅ Note 2 1

that vT denotes the transpose of v. The largest eigen value of A is ______. 1 1 u = and v = 2 1 1 A = uvT = (1 1) = 2

Solution: Given

∴

1 1 2 2

As det (A) = 0, one of the eigen values of A is 0. Let λ be the other eigen value of A ∴ 0 + λ = Trace (A) ⇒ λ = 1 + 2 = 3 So, the largest eigen value of A is 3. Hence, the correct answer is (3). Question Number: 18 Question Type: NAT The chromatic number of the following graph is ______.

0

places (assuming that π = 3.14) is ______. a

π/4

Solution: Let Put ⇒ ⇒ At x = 0;

t = 02 = 0

(1) c

I =

∫

1 a

0

= ∫ (cos ( x )) ( x dx ) 0 π2 /16

= ∫ t =0

1 (cos (t )) dt 2

π 1 = sin t 0 2

2

/16

1 π2 = sin − sin 0 2 16 = 0.289 Hence, the correct answer is 0.27 to 0.30.

2

c d 2

2

e 3

x cos ( x 2 ) dx

π/4

GATE_CSIT-2018.indd 56

f

Solution: The chromatic number of the given graph is 3.

π ; 4

π/4

d

b

2 π2 π t = = 16 4 ∴ Equation (1) becomes,

and at x =

I = ∫ x cos ( x 2 ) dx 0 2 x = t 2x dx = dt 1 xdx = dt 2

e

b

f

3

Hence, the correct answer is (3). Question Number: 19 Question Type: NAT Let G be a finite group on 84 elements. The size of a largest possible proper subgroup of G is ______. Solution Given order of G = O(G) = 84 Any proper subgroup of G will have order less than 84. Also, we know that the order of a subgroup of a finite group divides the order of the group. ∴ The size of a largest possible proper subgroup of G. = The largest divisor of 84 that is less than 84 = 42 Hence, the correct answer is (42).

5/2/2018 3:33:24 PM

GATE 2018 Solved Paper CS: Set – 1 | lvii Question Number: 20 Question Type: NAT The postorder traversal of a binary tree is 8, 9, 6, 7, 4, 5, 2, 3, 1. The inorder traversal of the same tree is 8, 6, 9, 4, 7, 2, 5, 1, 3. The height of a tree is the length of the longest path from the root to any leaf. The height of the binary tree above is ______. Solution: Postorder: 8, 9, 6, 7, 4, 5, 2, 3, 1 In order: 8, 6, 9, 4, 7, 2, 5, 1, 3

} } int main () { calc (4, 81); printf (“%d”, counter); }

The output of this program is ______. Solution: calc (4, 81)

8694725 1 3 Counter = 1 1

calc (4, 27)

Counter = 2

calc (4, 9)

3

86947 2 5

calc (4, 3)

Counter = 3 1

Counter = 4

2

3

The output of this program is 4. Hence, the correct answer is (4).

5

869 4 7

1 2 4

3

Question Number: 22 Question Type: NAT Consider the sequential circuit shown in the figure, where both flip-flops used are positive edge-triggered D flip-flops.

5

out

in

7

8 6 9

D Q

D Q

Clock 1 2 4 6 8

- - - - - - - - - - - - root level 0 3 - - - - - - - - - - - level 1

5 - - - - -- - - - - - - - - - - level 2 7 - - - - - - - - - -- - - - - - - - - - - level 3

Solution:

9 - - - - - - - - - - - - - -- - - - - - - - - - - level 4

Hence, the correct answer is (4). Question Number: 21 Consider the following C program:

In

D

A

Q

D

B

Q

Out

Question Type: NAT

#include int counter = 0; int calc (int a, int b) { int c; counter++; if (b==3) return (a*a*a); else { c = calc (a, b/3); return (c*c*c);

GATE_CSIT-2018.indd 57

The number of states in the state transition diagram of this circuit that have a transition back to the same state on some value of “in” is ______.

Clock p.s

x

A

B

0

0

0

DA 0

DB 0

n.s

Out

A

B

B

0

0

0

0

0

1

1

0

1

0

0

0

1

0

0

0

0

0

0

0

1

1

1

0

1

0

0

1

0

0

0

1

0

1

1

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lviii | GATE 2018 Solved Paper CS: Set – 1 p.s A

B

1

0

x

DA

DB

1

1

1

n.s

Out

A

B

B

1

1

1

1

1

0

0

1

0

1

1

1

1

1

1

1

1

1

1

States transition diagram of the above sequential circuit is as shown below: 0/0 00

P1→1 P2→1 + 1 P3→1

0/0

1/1 0/1

1/0

01

0/1

11 1/1

10

1/0

\ the number of states in the state transition diagram of this circuit that have a transition back to the same states on some value of “in” is 2. Hence, the correct answer is (2). Question Number: 23 Question Type: NAT A 32-bit wide main memory unit with a capacity of 1 GB is built using 256 M × 4-bit DRAM chips. The number of rows of memory cells in the DRAM chip is 214. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closest integer) of the time available for performing the memory read write operations in the main memory unit is ______. Solution: Given main memory capacity = 1 GB Having 256 × 4-bit DRAM chips Number of memory cells rows in DRAM chip is 214 = 16 K Total time for refreshing, if one refreshing takes 50 nanoseconds = 16 K × 50 ns = 800 µ sec = 0.8 m sec. Given refresh period = 2 m sec Out of 2 m sec, 0.8 m sec is used for refreshing and remaining 1.2 msec is used for RD/WR operation. Percentage of time available for Read/Write operation 1.2 m sec = × 100 2 m sec = 60%. Hence, the correct answer is 59 to 60.

GATE_CSIT-2018.indd 58

Question Number: 24 Question Type: NAT Consider a system with 3 processes that share 4 instances of the same resource type. Each process can request a maximum of K instances. Resource instances can be requested and released only one at a time. The largest value of K that will always avoid deadlock is ______. Solution: There are 3 processes and 4 resources, one process can be able to obtain two resources. Maximum number of instances a process can request will be 2.

Hence, the correct answer is (2). Question Number: 25 Question Type: NAT Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _______. Solution: TCP sequence number field has 4 bytes i.e., 32 bits. In one second, 109 bits are transferred. Amount of time required to lapse sequence numbers are 232 × 8 = 34.35 sec. 109 Hence, the correct answer is (34). Q.26 to Q.55 carry two marks each. Question Number: 26 Question Type: MCQ Consider a matrix P whose only eigenvectors are the mul1 tiples of ⋅ 4 Consider the following statements. (I) P does not have an inverse (II) P has a repeated eigen value (III) P cannot be diagonalized Which one of the following options is correct? (A) Only I and III are necessarily true (B) Only II is necessarily true (C) Only I and II are necessarily true (D) Only II and III are necessarily true Solution: Given that any eigen vector of a matrix P is a 1 multiple of ⋅ 4 1 As every eigen vector of P is a multiple of , 4 P has only one linearly independent eigen vector.

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GATE 2018 Solved Paper CS: Set – 1 | lix As P is a 2 × 2 matrix, if the two eigen values of P are distinct, then P should have two linearly independent eigen vectors. So, P has a repeated eigen value Hence, statement (II) is correct. A 2 × 2 matrix is diagonalizable if and only if it has two linearly independent eigen vectors. But P has only one linearly independent eigen vector. Hence P is not diagonalizable. So, statement (III) is correct. P need not be a singular matrix. So, statement (I) is not correct. ∴ Only statements (II) and (III) are necessarily correct. Hence, the correct option is (D). Question Number: 27 Question Type: MCQ Let N be the set of natural numbers. Consider the following sets. P: Set of Rational numbers (positive and negative) Q: Set of functions from {0, 1} to N R: Set of functions from N to {0, 1} S: Set of finite subsets of N. Which of the sets above are countable? (A) Q and S only (B) P and S only (C) P and R only (D) P, Q and S only Solution: We know that the set of rational numbers is countable. So, P is countable. Q: set of functions from {0, 1} to N. As 0 can be mapped to a number in N ways and 1 can be mapped to a number in N ways, The number of elements in the set of functions from {0, 1} to N = The number of elements in the Cartesian product N×N We know that the Cartesian product of two countable sets is countable. As N is countable, N × N is countable So, Q is countable. R: Set of functions from N to { 0, 1} In a function from N to {0, 1}, Every element of N is mapped to 0 or 1 So, the number of ways of mapping any element of N is 2. ∴ The number or elements in the set of functions from N to {0, 1} = The number of elements in the power set of N But the power set of N is uncountable because the power set of a infinite countable set is uncountable. So R is uncountable. S: Set of finite subsets of N. As we are considering only the finite subsets of N, S is a countably infinite set. So, only P, Q and S are countable. Hence, the correct option is (D).

GATE_CSIT-2018.indd 59

Question Number: 28 Question Type: MCQ Consider the first-order logic sentence j = ∃ s ∃ t ∃ u∀v∀w∀x∀yψ( s, t , u, v, w, x, y ) where ψ( s, t , u, v, w, x, y ) is a quantifier-free first-order logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose ϕ has a model with a universe containing 7 elements. Which one of the following statements is necessarily true? (A) There exists at least one model of ϕ with universe of size less than or equal to 3. (B) There exists no model of ϕ with universe of size less than or equal to 3. (C) There exists no model of ϕ with universe of size greater than 7. (D) Every model of ϕ has a universe of size equal to 7. Solution: There exists atleast one model of ϕ with universe of size less than or equal to 3. Hence, the correct option is (A). Question Number: 29 Question Type: MCQ Consider the following C program: #include void fun1 (char *s1, char * s2) { char *tmp; tmp = s1; s1 = s2 s2 = tmp; } void fun2 (char **s1, char **s2) { char *tmp; tmp = *s1; *s1 = *s2; *s2 = tmp; } int main () { char *str1 = “Hi”, *str2 = “Bye”; fun1 (str1, str2); printf (“%s %s “, str1, str2); fun2 (&str1, &str2); printf (“%s %s”, str1, str2); return 0; }

The output of the program above is: (A) Hi Bye Bye Hi (B) Hi Bye Hi Bye (C) Bye Hi Hi Bye (D) Bye Hi Bye Hi Solution: 100 Str 1

H

i

10

y

e

100

200

B

Str 2

200

10

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lx | GATE 2018 Solved Paper CS: Set – 1

main ( ) fun 1 ( )

(C) F3F4 only (D) F1F2 and F4F5 only Str 1

Str 2

100

200

S1

S2

fun 2 ( ) Str 1 Str 2 Str 2 Str 1 S1

S2

The output of the program is Hi Bye Bye Hi Hence, the correct option is (A). Question Number: 30 Question Type: MCQ Let G be a simple undirected graph. Let TD be a depth first search tree of G. Let TB be a breadth first search tree of G. Consider the following statements. (I) No edge of G is a cross edge with respect to TD. (A cross edge in G is between two nodes neither of which is an ancestor of the other in TD.) (II) For every edge (u, v) of G, if u is at depth i and v is at depth j in TB, then |i – j| = 1. Which of the statements above must necessarily be true? (A) I only (B) II only (C) Both I and II (D) Neither I nor II Solution: In DFS of an undirected graph, we will not get cross edge since all edges that are incident on a vertex are explored. In directed graph, if we apply DFS, then we may get cross edge. Hence, the correct option is (A). Question Number: 31 Question Type: MCQ Assume that multiplying a matrix G1 of dimension p × q with another matrix G2 of dimension q × r requires pqr scalar multiplications. Computing the product of n matrices G1G2G3, ..., Gn can be done by parenthesizing in different ways. Define Gi Gi+1 as an explicitly computed pair for a given paranthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization (G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs. Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4, and F5 are of dimensions 2 × 25, 25 × 3, 3 × 16, 16 × 1 and 1 × 1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/are: (A) F1F2 and F3F4 only (B) F2F3 only

GATE_CSIT-2018.indd 60

Solution: Use Parenthesization ((F1(F2(F3F4)))F5) F3F4 = 48 [16 × 3 × 1] (F2(F3F4)) = 75 [25 × 3 × 1] (F1(F2(F3F4))) = 50 [2 × 25 × 1] ((F1(F2(F3F4)))F5) = 2000 [2 × 1 × 1000] ∴ Total no. of multiplication required = 2000 + 50 + 75 + 48 = 2173 (minimum number of multiplications). Hence, the correct option is (C). Question Number: 32 Question Type: MCQ Consider the following C code. Assume that unsigned long int type length is 64 bits. unsigned long int fun (unsigned long int n) { unsigned long int i, j = 0, sum = 0; for (i = n; i > 1. i = i/2) j++; for (; j > 1; j = j/2) sum++; return (sum); }

The value returned when we call fun with the input 240 is: (A) 4 (B) 5 (C) 6 (D) 40 Solution: For input 240, j get incremented to ‘40’ j = 40 sum=1

j = 20 sum=2

j = 10 sum=3

j=5 sum=4

j=2 sum=5

Hence, the correct option is (B). Question Number: 33 Question Type: MCQ Consider the unsigned 8-bit fixed point binary number representation below: b7 b6 b5 b4 b3 ⋅ b2 b1 b0 where the position of the binary point is between b3 and b2. Assume b7 is the most significant bit. Some of the decimal numbers listed below cannot be represented exactly in the above representation: (i) 31.500 (ii) 0.875 (iii) 12.100 (iv) 3.001 Which one of the following statements is true? (A) None of (i), (ii), (iii), (iv) can be exactly represented (B) Only (ii) cannot be exactly represented (C) Only (iii) and (iv) cannot be exactly represented (D) Only (i) and (ii) cannot be exactly represented Solution: Binary representation of given numbers are: (i) (31.500)10 = (11111.100)2 (ii) (0.875)10 = (00000.111)2 (iii) (12.100)10 = (01100.000110011...)2 (iv) (3.001)10 = (00011.0000000001...)2

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GATE 2018 Solved Paper CS: Set – 1 | lxi Fraction part in the given representation is of only 3-bits, (iii) and (iv) can’t be represented accurately due to lack of storage. Hence, the correct option is (C). Question Number: 34 Question Type: MCQ The size of the physical address space of a processor is 2P bytes. The word length is 2W bytes. The capacity of cache memory is 2N bytes. The size of each cache block is 2M words. For a K-way set-associative cache memory, the length (in number of bits) of the tag field is: (A) P – N – log2 K (B) P – N + log2 K (C) P – N – M – W – log2 K (D) P – N – M – W + log2 K Solution: Given main memory space = 2M bytes Physical address size = ‘P’ bits Cache memory size = 2N bytes Then size of the block = (2M+W) B (2M words × 2W bytes/word) Number of lines = 2N–M–W Number of Cache memory Number of sets = P -way 2 N − M −W = log K 2 ∴ The Tage size will be = P – (M + W – log2 K + M + W) = P – N + log2 K. Hence, the correct option is (B). Question Number: 35 Question Type: MCQ Consider the following languages: I. {ambncpdq | m + p = n + q, where m, n, p, q ≥ 0} II. {ambncpdq | m = n and p = q, where m, n, p, q ≥ 0} III. {ambncpdq | m = n = p and p ≠ q, where m, n, p, q ≥ 0} IV. {ambncpdq | mn = p + q, where m, n, p, q ≥ 0} Which of the languages above are context-free? (A) I and IV only (B) I and II only (C) II and III only (D) II and IV only Solution: I. {am bn cp dq | m + p = n+ q} The given language is context free language, as it can be implemented by using stack II. {am bn cp dq | m = n and p = q} The given language is CFL, as it can be implemented by using stack. III. Not CFL, more than one comparison at a time. IV. Not CFL, multiplication is not possible with stack. Hence, the correct option is (B).

GATE_CSIT-2018.indd 61

Question Number: 36 Question Type: MCQ Consider the following problems. L(G) denotes the language generated by a grammar G. L(M) denotes the language accepted by a machine M. (I) For an unrestricted grammar G and a string w, whether w ∈ L(G) (II) Given a Turing machine M, whether L(M) is regular (III) Given two grammars G1 and G2, whether L(G1) = L(G2) (IV) Given and NFA N, whether there is a deterministic PDA P such that N and P accept the same language. Which one of the following statements is correct? (A) Only I and II are undecidable (B) Only III is undecidable (C) Only II and IV are undecidable (D) Only I, II and III are undecidable Solution I. Membership problem of unrestricted grammar (i.e. REL) is undecidable as there is a possibility of etering into infinite loop. II. Every regular language is REL, but REL may or may not be regular. (undecidable) III. As the type of grammar is not given, the equivalence problem is undecidable in REL. IV. Every regular language is Deterministic CFL. (decidable) Hence, the correct option is (D). Question Number: 37 Question Type: MCQ A lexical analyzer uses the following patterns to recognize three tokens T1, T2, and T3 over the alphabet {a, b, c}. T1: a?(b|c)*a T2: b?(a|c)*b T3: c?(b|a)*c Note that ‘x?’ means 0 or 1 occurrence of the symbol x. Note also that the analyzer outputs the token that matches the longest possible prefix. If the string bbaacabc is processed by the analyzer, which one of the following is the sequence of tokens it outputs? (A) T1T2T3 (B) T1T1T3 (C) T2T1T3 (D) T3T3 Solution Given string bbaacabc The longest possible prefix will be bbaac that can be obtained from T3. a b c is the another possible prefix obtained from T3. bbaac abc ↓ ↓ T3 T3 Hence, the correct option is (D).

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lxii | GATE 2018 Solved Paper CS: Set – 1 Question Number: 38 Question Type: MCQ Consider the following parse tree for the expression a#b$c$d#e#f, involving two binary operators $ and #.

(C) $ has higher precedence and is left associative; # is left associative (D) # has higher precedence and is right associative; $ is left associative Solution:

# a

#

# $ d

$

a

# e

f

#

$

c

b

#

d e

$

Which one of the following is correct for the given parse tree? (A) $ has higher precedence and is left associative; # is right associative (B) # has higher precedence and is left associative; $ is right associative

f

c

b

The operator which appears at lower level of tree has higher precedence. Therefore ‘$’ has higher precedence. # operator is right associative, as right subtree is extended. Hence, the correct option is (A).

Question Number: 39 Question Type: MCQ In a system, there are three types of resources: E, F and G. Four processes P0, P1, P2 and P3 execute concurrently. At the outset, the processes have declared their maximum resource requirements using a matrix named Max as given below. For example, Max[P2, F] is the maximum number of instances of F that P2 would require. The number of instances of the resources allocated to the various processes at any given state is given by a matrix named Allocation. Consider a state of the system with the Allocation matrix as shown below, and in which 3 instances of E and 3 instances of F are the only resources available. Allocation

Max

E

F

G

E

P0

1

0

1

F

G

P0

4

3

P1

1

1

1

2

P1

2

1

4

P2

1

P3

2

0

3

P2

1

3

3

0

0

P3

5

4

1

From the perspective of deadlock avoidance, which one of the following is true? (A) The system is in safe state. (B) The system is not in safe state, but would be safe if one more instance of E were available. (C) The system is not in safe state, but would be safe if one more instance of F were available. (D) The system is not in safe state, but would be safe if one more instance of G were available. Solution: Max

Allocation

Need

Available

E

F

G

E

F

G

E

F

G

E(3)

F(3)

G(0)

P0

4

3

1

1

0

1

3

3

0

3 ↓

3 ↓

0 ↓

(On ‘P1’ execution)

P1

2

1

4

1

1

2

1

0

2

4 ↓

3 ↓

1 ↓

(On ‘P2’ execution)

P2

1

3

3

1

0

3

0

3

0

5 ↓

3 ↓

4 ↓

(On ‘P1’ execution)

P3

5

4

1

2

0

0

3

4

1

6 ↓

4 ↓

6 ↓

(On ‘P3’ execution)

8

4

6

The system is in safe state, with safe sequence P0, P2, P1, P3. Hence, the correct option is (A).

GATE_CSIT-2018.indd 62

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GATE 2018 Solved Paper CS: Set – 1 | lxiii Question Number: 40 Question Type: MCQ Consider the following solution to the producer-consumer synchronization problem. The shared buffer size is N. Three semaphores empty, full and mutex are defined with respective initial values of 0, N and 1. Semaphore empty denotes the number of available slots in the buffer, for the consumer to read from. Semaphore full denotes the number of available slots in the buffer, for the producer to write to. The placeholder variables, denoted by P, Q, R, and S, in the code below can be assigned either empty or full. The valid semaphore operations are: wait () and signal (). Producer

do { wait(P); wait (mutex); //Add item to buffer signal (mutex); signal (Q); } while (1);

Solution: Consider a data set r

A

B

s

B

C

12

2

2

A

13

6

3

B

15

4

4

C

12

6

5

A

13

4

6

C

16

7

Consumer

do { wait(R); wait (mutex); //Consume item from buffer signal (mutex); signal (S); } while (1);

Which one of the following assignments to P, Q, R and S will yield the correct solution? (A) P: full, Q: full, R: empty, S: empty (B) P: empty, Q: empty, R: full, S: full (C) P: full, Q: empty, R: empty, S: full (D) P: empty, Q: full, R: full, S: empty Solution: In the producer-consumer problem, producer keeps the item in buffer when the slots on the buffer are empty. While, it has to wait when the buffer is full. In the given problem, full denotes number of available slots in buffer, i.e. full will be 0, when there are no empty slots in buffer. So, P: full. When producer places an item in buffer, it implies there is increase in the number of available slots in the buffer, for the consumer to read from. This can be done by signal (empty). So, Q: empty. Similarly, consumer has to wait when there are no items to consume in buffer, which can be shown by wait (empty). So, R: empty. Once the consumer consumes item, it implies, there is increase in-the available slots in the buffer, for the producer to write to. So, S: full. Hence, the correct option is (C). Question Number: 41 Question Type: MCQ Consider the relations r(A, B) and s(B, C), where s ⋅ B is a primary key and r ⋅ B is a foreign key referencing s ⋅ B. Consider the query Q: r (σB 1? (A) R1 > R2 (B) 0.5R2 < R1 < R2 (C) R1 = R2 (D) R2 3 < R1 < R2 19. A sum takes two years to become 40% more under simple interest at a certain rate of interest. If it was lent at the same interest rate for the same time under compound interest, interest being compounded annually, it would amount to x% more than itself. Find x. (A) 36 (B) 48 (C) 40 (D) 44 20. A sum was divided into two equal parts. One part was lent at 20% p.a. simple interest. The other part was lent at 20% p.a. compound interest, interest being compounded annually. The difference in the interests fetched by the parts in the second year is `400. Find the difference in the interests fetched by the parts in the fourth year (in `). (A) 1414 (B) 1442 (C) 1456 (D) 1484

Answer Keys

Exercise Practice Problems 1. C 11. A

2. B 12. B

3. D 13. D

4. C 14. D

5. D 15. A

6. A 16. C

7. D 17. C

8. B 18. A

9. A 19. D

10. D 20. C

Chapter 6 Averages, Mixtures, and Alligations LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Averages • Points to remember • An easy method to calculate averages

AVerAges ‘Average is a very simple but effective way of representing an entire group by a single value.’ ‘Average’ of a group is defined as: Average =

Sum of all items in the group Number of items in the group

‘Sum of all the items in the group’ means ‘sum of the values of all the items in the group’. A batsman’s performance can be expressed as the average number of runs scored per innings rather than giving the scores in individual innings. For example, let us say a cricketer scored the following runs in 9 different innings in a year: 35, 56, 124, 29, 0, 87, 98, 45 and 75. Then his average score (per innings) for the year is 35 + 56 + 124 + 29 + 0 + 87 + 98 + 45 + 75 = 61 9 Similarly, if there are 60 students in a class, instead of talking of the height of each individual student, we can talk of ‘average’ height of the class. The average height of the class of students is equal to the sum of the heights of all the students of the class divided by the number of students in the class. Average is also called the ‘mean’ or mean value of all the values.

Points to reMeMBer 1. If the value of each item is increased by the same value p, then the average of the group or items will also increase by p.

• Weighted average • Mixtures • Alligations

2. If the value of each item is decreased by the same value p, then the average of the group or items will also decrease by p. 3. If the value of each item is multiplied by the same value p, then the average of the group or items will also get multiplied by p. 4. If the value of each item is divided by the same value p (p ≠ 0), then the average of the group or items will also get divided by p. 5. The average of a group of items will always lie between the smallest value in the group and largest value in the group i.e., the average will be greater than the smallest value and less than the largest value in the group.

An Easy Method to Calculate Averages As already discussed, the average of a group of items whose values are given can be found out by the rule given at the beginning of this section. However, in most of the cases, we do not need to perform such elaborate additions and divisions. The calculation of averages can be simplified greatly by taking some arbitrary number (P) as a starting point, take the deviations (differences) of the given items (Qi) from this arbitrary number, find the average of all these deviations (Qi – P) and algebracially add it to the arbitrary number (P) to give the correct average of the given items. If there are n items and they are denoted by Q1, Q2, Q3, …, Qn, then the average of these n items is given by Average = P +

1 n

n

∑ (Q − P ) i =1

i

For example, the cricketer that we considered above scored the following runs in seven innings: 35, 56, 45, 43, 67, 70 and 48.

Chapter 6 Averages, Mixtures, and Alligations | 1.113 Now, to find his average, we take an arbitrary figure, say 50 and first find the deviations of each of the scores from this figure. The deviations of the scores from 50 are –15, +6, –5, –7, +17, +20 and –2. The sum of these deviations is +14. Hence, the average of the cricketer’s scores is 14 50 + = 52 7 Please note that the number P (= 50 above) can be any value. Let us work out the same example taking a different value for P. Let us take P equal to 45. The deviations of the scores from P are –10, +11, 0, –2, +22, +25 and +3. The sum of these deviations is 49. Hence, the average is 45 + 49/7 = 45 + 7 = 52.

Weighted Average When two groups of items are combined together, then we can talk of the average of the entire group. However, if we know only the average of the two groups individually, we cannot find out the average of the combined group of items. For example, there are two sections A and B of a class where the average height of section A is 150 cm and that of section B is 160 cm. On the basis of this information alone, we cannot find the average of the entire class (of the two sections). As discussed earlier, the average height of the entire class is Total height of the entire class Total number of students inn the entire class Since we do not have any information regarding the number of students in the two sections, we cannot find the average of the entire class. Now, suppose that we are given that there are 60 students in the section A and 40 students in section B, then we can calculate the average height of the entire class which, in this case will be equal to 60 × 150 + 40 × 160 = 154 cm. 60 + 40 This average height 154 cm of the entire class is called ‘weighted average’ of the class. The above step in calculating the weighted average of the class can be rewritten:

by the same method. For example, if three sections in a class have their average marks as 75, 76 and 79 respectively and their respective strengths are 30, 35 and 35, then the average mark of the entire class is given by 30 × 75 + 35 × 76 + 35 × 79 = 76.75 30 + 35 + 35 The method of deviations we used for calculating averages can be applied to calculate weighted average also. Here, that method will involve finding out deviations from the arbitrarily chosen number and calculating the weighted average of these deviations. In the above example, if we take 70 as the arbitrary figure, then the deviations of the three observed values given from 70 are +5, +6 and +9. The weighted average of these deviations is 30 × 5 + 35 × 6 + 35 × 9 675 = = 6.75. 30 + 35 + 35 100 Hence, the weighted average will be 70 + 6.75 = 76.75 The arbitrary figure chosen can be any figure and if it is selected, as in the previous case, between the smallest and largest observed figures, some of the deviations will be positive and some negative making the final division relatively simpler. For example, in the above case, if we take with 76 as the arbitrary figure, the deviations are –1, 0 and +3. Then, the weighted average will be 30 − ( −1) + 35 − 0 + 35 − ( +3) 75 = = 0.75 30 + 35 + 35 100 Hence, the weighted average will be 76 + 0.75 = 76.75 ‘Weighted Average’ can be defined or calculated for any MIXTURE.

Mixtures

Mixing of two or more qualities of things produces a mixture. When two items of different qualities are thus mixed, the quality of the resultant mixture lies in between the qualities of the original constituent items, i.e., it will be higher than the lowest quality and lower than the highest 60 × 150 + 40 × 160 60 40 quality of the items being mixed. = 150 + 160 In the above example that we took, the ‘quality’ that we 60 + 40 100 100 looked at was the height of the students. We could also have 3 2 taken their weights or the marks scored by them or any other = 150 + 160 5 5 ‘quality’ or ‘parameter’ and calculated the ‘weighted aver It is clear from the above step that we would have been able age’ value of that particular ‘quality’ for the entire group. Similarly, if two types of a product of different prices to calculate the average height of the entire class even if we per unit are mixed, the unit price of the resultant mixture had not been given the number of students in the individual will lie between the prices of the two types that form the sections but only the ratio of the number of students in the mixture. two sections (which in this case is 3 : 2). Here, the average quality is essentially the weighted Even if there are more than two groups of items to be average of the two constituent items. combined, then also the weighted average can be calculated

1.114 | Quantitative Aptitude If q1 is the quantity (or number of items) of one particular item of quality p1, and q2 be the quantity (or number of items) of the second item of quality p2 are mixed together to give a new mixture, then the weighted average value (p) of the quality of the mixture is given by p=

p1 q1 + p2 q2 q1 + q2

Even if there are more than two groups of items mixed, the weighted average rule can be applied. We will only have to take figures (as shown in the formula for the two groups) for all the groups in the numerator as well as the denominator and calculate the weighted average. For example, if there are four groups of quantities q1, q2, q3 and q4 whose respective qualities are p1, p2, p3 and p4, then the weighted average quality of the group can be written as p=

p1 q1 + p2 q2 + p3 q3 + p4 q4 q1 + q2 + q3 + q4

A mixture can also be a solution—that is, a liquid mixed with another liquid which is normally water. The concentration of the solution is expressed as the proportion (or percentage) of the liquid in the total solution. For example, if 10 litres of pure alcohol is mixed with 40 litres of water, then in a total solution of 50 litres, there is 10 litres of alcohol. Hence, the concentration of this solution is 0.2 (= 10/50) or 20%. Similarly, if 30 litres of pure milk is mixed with 10 litres of water, the concentration of this solution can be expressed as 75% (= 30/40) milk or 25% water. We can also have two solutions mixed together to give a new solution. Such problems can also be handled in the same manner as other mixtures. In the weighted average rule, the quality of the constituents (p1, p2, etc.) will then be the concentrations of various solutions mixed together.

Solved Examples Example 1: Rajiv purchased three dozen mangoes at `10 per dozen, two dozen mangoes at `15 per dozen and five dozen mangoes at `16 per dozen. Find the average cost per dozen of the mangoes that he purchased. Solution: The cost of first three dozen mangoes = (3) (10) = `30 The cost of next two dozen mangoes = (2) (15) = `30 The cost of next five dozen mangoes = (5) (16) = `80 Total cost of the mangoes purchased = `140. Average cost per dozen

=

Total cost of mangoes 140 = = `14. Number of dozens 10

Example 2: The average age of 5 men is 20 years. Their average age increased by 1 year when a new man joined them. Find the age of the new man. Solution: Total age of 5 men = (5) (20) = 100 years Total age of 6 men = (6) (21) = 126 years The age of the new man = 126 – 100 i.e. 26 years Example 3: Six kilograms of wheat costing of `18 per kg is mixed with nine kilograms of wheat costing of `12 per kg. Find the price per kg of the mixture. Solution: Total cost of 6 kg = (6) (18) = `108.

Total cost of 9 kg = (9) (12) = `108.

Average cost of the mixture =

Total cost Total quantity

108 ( 2) = = ` 14.40 per kg. 15

Example 4: The average marks of three sections in the tenth class were 90, 120 and 150. If the number of students in these sections are 30, 40 and 50 respectively, find the average mark of the tenth class. Solution: Total mark of the first section = (90) (30) = 2700 Total mark of the second section

= (120) (40) = 4800

Total mark of the third section = (150) (50) = 7500

Total mark Number of students 2700 + 4800 + 7500 15000 = = = 125. 30 + 40 + 50 120

Average mark of class X =

Example 5: Tarun earned an average of `1500 per month from January to April in a year. He earned an average of `1600 per month from May to October in that year. His earning in the month of December of that year was `300 more than his earning in the month of November of that year. His average monthly earnings for that year was `1675. Find his earnings in the month of November. Solution: Total earnings of Tarun from January to April = (1500) (4) = `6000 Total earnings of Tarun from May to October = (1600) (6) = `9600. Let his earnings in November be `x His earnings in December = ` (x + 300) 6000 + 9600 + x + x + 300 = 1675 12 x = 2100. Example 6: The average age of a group of friends is 37 years. If 6 new friends whose average age is 35 years join

Chapter 6 Averages, Mixtures, and Alligations | 1.115 them, the average age of the entire group becomes 36 years. How many people were there in the group initially? Solution: Let the initial number of people in the group be n. The total age of the initial group of friends = 37n years The total age of the six friends who joined the group = 35 × 6 i.e., 210 years. Given that, 37n + 210 = 36 (n + 6) ⇒ 37n – 36n = 216 – 210 \

n = 6.

left hand quantity (Dearer Price) and Average Price has to be written at the bottom right hand corner. Similarly, the difference between the top right hand corner (Cheaper Price) and the Average Price has to be written at the bottom left hand corner. Now the ratio of the two quantities in the bottommost line will give us the ratio of the quantities of Dearer and Cheaper varieties. Please note that since we took Dearer Price on the top left corner, the ratio of the bottom left figure to that of the bottom right figure will give the ratio of Dearer Quantity to Cheaper Quantity} Example 7: A bag contains a total of 120 coins in the denominations of 50 p and `1. Find the number of 50 p coins in the bag if the total value of the coins is `100.

Alligations We will take the Weighted Average rule discussed in the previous section and rewrite the formula such that the quantity terms come on one side and the price terms come on the q p − p2 other side. If we do this we get the rule 1 = q2 p1 − p

Solution: Let the number of 50 p coins be x. ⎤ ⎡ 50 x + 1 (120 − x )⎥ Total value of the coins = ` ⎢ 100 ⎣ ⎦

50 x + 120 − x = 100 100

This is called the RULE OF ALLIGATION. This rule connects quantities and prices in mixtures. This can also be q p −p written as 1 = 2 q2 p − p1

⇒

In a descriptive manner, the Rule of Alligation can be Quantity of Cheaper written as Quantity of Dearer

Average value per coin =

=

Rate of Dearer − Average Rate Average Rate − Rate of Cheaper

This rule is a very powerful rule and is useful in problems on weighted averages and mixtures. This rule is also useful in a number of problems which can be treated as mixtures and applied to parameters other than price also. We will take examples where alligation rule can be applied. In actual practice, to apply alligation rule, we do not need to remember the above formula at all. It can be made very simple by representing the rule pictorially. The above formula can be represented as follows: Dearer Price

Cheaper Price Average Price

(Average Price – Cheaper Price)

(Dearer Price – Average Price)

{We write the dearer and cheaper prices in one line with some gap in between them. Then, we write the average price in between these two but in the line below the line in which dearer and cheaper prices are written. Then, take the differences of quantities as shown along the arrows and write along the same direction of the arrows continued, i.e., in a diagonally opposite corner. The difference between the top

x = 40

Alternative method: (100) (100) 120

500 = paise. 6 x Using allegation rule, 120 − x 500 100 − 6 = 1 = 500 2 − 50 6 ⇒ 2x = 120 – x

⇒

x = 40.

Example 8: A vessel has 60 litres of solution of milk and water having 80% milk. How much water should be added to it to make it a solution in which milk forms 60%? Solution: Let the quantity of water to be added be x litres. Quantity of milk in the vessel

⎛ 80 ⎞ = ⎜⎝ ⎟ (60) = 48 litres 100 ⎠

⇒ 48 = 0.6 (60 + x) \

x = 20

Example 9: In what ratio must Anand mix two varieties of barley costing `24 per kg and `38 per kg so that by selling 1 the mixture at `40 per kg he would make 11 % profit? 9

1.116 | Quantitative Aptitude Solution: Cost price of the mixture =

40 (100 ) 1 100 + 11 9

This gives the concentration (k) of the liquid as a PROPORTION of the total volume of the solution. If the concentration has to be expressed as a percentage, then it will be equal to 100 k. If the volume of the liquid is to be found out at the end of n operations, it is given by kP, i.e., the concentration k multiplied by the total volume P of the solution.

= `36 per kg

Cost of cheaper variety

Cost of dearer variety

24

38 36

2

Cost of the mixture 12

\ The ratio of the costs of the cheaper to the dearer variety = 2 : 12 = 1 : 6. If there is P volume of pure liquid initially and in each operation, Q volume is taken out and replaced by Q volume of water, then at the end of n such operations, the concentration (k) of the liquid in the solution is given n ⎧P −Q⎫ by ⎨ ⎬ =k ⎩ P ⎭

Example 10: A vessel has 400 litres of pure milk. 40 litres of milk is removed from the vessel and replaced by water. 40 litres of the mixture thus formed is replaced by water. This procedure is repeated once again. Find the percentage of milk in the resultant solution. Solution: Let v litre be volume of milk with a concentration of c1%. If x litres of the solution is removed and replaced with ⎛v−x ⎞ × c1 ⎟ % water, the new concentration is ⎜ ⎝ v ⎠ Given that initial concentration is 100%. (pure milk), v = 400, x = 40 and the replacement is done thrice. \ Concentration of milk in the resultant solution 3

⎛ 400 − 40 ⎞ × 100 = 72.9% = ⎜ ⎝ 400 ⎟⎠

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. Find the average of all the two digit numbers divisible by 10. (A) 40 (B) 50 (C) 45 (D) 60 2. Find the average of all odd numbers less than 50. (A) 26.5 (B) 25.5 (C) 26 (D) 25 3. Find the average of all the multiples of 12 less than 100. (A) 48 (B) 54 (C) 60 (D) 66 4. The average salary per month of a man for the first four months, next four months and the last four months of a year are `6000, `8000 and `13000 respectively. Find his average salary per month in that year (in `) (A) 7500 (B) 9000 (C) 10500 (D) 6600 5. In an office there are 20 employees. The average heights of the male employees is 180 cm. The average height of the female employees is 170 cm. Find the average height of all the employees (in cm). (A) 172 (B) 174

(C) 176 (D) Cannot be determined

6. The average age of the boys in a class is ten years. The average age of the girls in the class is eight years. There are 50% more boys than girls in the class. Find the average age of the class (in years). (A) 8.4 (B) 8.8 (C) 9.2 (D) 9.6 7. A vessel has 20 litres of a mixture of milk and water having 60% milk. Five litres of pure milk is added to the vessel. Find the percentage of milk in the new solution. (A) 34% (B) 51% (C) 68% (D) None of these 8. In what ratio must two kinds of coffee which cost `80 per kg and `108 per kg be mixed such that the resultant mixture costs `96 per kg? (A) 1 : 2 (B) 2 : 3 (C) 3 : 4 (D) 2 : 1 9. Vessel A has 20 litres of a mixture of milk and water having 75% milk. Vessel B has x litres of a mixture of milk and water having 60% milk. The contents of the vessels are mixed to form a mixture having 66% milk. Find x. (A) 25 (B) 30 (C) 20 (D) 40

Chapter 6 Averages, Mixtures, and Alligations | 1.117 10. A milkman has 15 litres of pure milk. How many litres of water have to be added to it so that he gets a 60% profit by selling at cost price? (A) 9 (B) 10 (C) 8 (D) 12 11. From 90 litres of pure milk, 9 litres is withdrawn and replaced by water. 9 litres of the mixture is then withdrawn and replaced by water. Find the ratio of milk and water in the present mixture. (A) 19 : 81 (B) 19 : 100 (C) 81 : 19 (D) 81 : 100 12. Just before the last match in a season, the total number of runs scored by Sachin Tendulkar added up to 2100. In his last match he scored 101 runs. As a result his average score for the season went up by one run. Find the total number of matches he played in that season if he got out in every match. (A) 31 (B) 5 (C) 71 (D) Either 31 or 71 13. The average weight of all the students of classes I and II equals the average of the average weight of the students of the two classes. There are twice as many students in class II as in class I. The sum of twice the average weight of the students of class I and the average weight of the students of class II is 60 kg. Find the average weight of class I (in kg). (A) 10 (B) 15 (C) 20 (D) 25 14. Two varieties of wheat are mixed in the proportion of 3 : 4 and the mixture is sold at `28 per kg at a profit of 40%. If the second variety of wheat costs `3 more than the first variety of wheat, find the cost price of the first variety of wheat. (A) `128/7 per kg (B) `120/7 per kg (C) `141/7 per kg (D) `149/7 per kg 15. A man buys milk at `4 per litre, mixes it with water and sells the mixture at the same price. If his profit is 25%, find the amount of water mixed with each litre of milk. (A) 0.25 litres (B) 0.5 litres (C) 0.75 litres (D) 0.6 litres 16. In what proportion can three varieties of sugar priced at `10 per kg, `12 per kg and `18 per kg, be mixed so that the price of the mixture is `14 per kg? (A) 2 : 2 : 5 (B) 2 : 3 : 4 (C) 1 : 3 : 4 (D) 3 : 4 : 5 17. The ratio of alcohol and water in three mixtures of alcohol and water is 3 : 2, 4 : 1 and 7 : 3. If equal quantities of the mixtures are drawn and mixed, the

concentration of alcohol in the resulting mixture will be_____. (A) 65% (B) 70% (C) 75% (D) 80% 18. In what proportion should milk and water be mixed to reduce the cost of litre of milk from `18 per litre to `16? (A) 8 : 1 (B) 6 : 1 (C) 10 : 1 (D) 7 : 1 19. A’s weight equals the average weight of B, C and D. B’s weight equals the average weight of A, C and D. The average weight of C and D is 30 kg. Find the average weight of A and B. (A) 15 kg (B) 30 kg (C) 60 kg (D) 45 kg 20. Of five numbers, the first number is thrice the third, the fourth number is two less than the first, the fifth number is one-seventh of the second and the second number is three less than thrice the first. Find the fifth number, if the average of the numbers is 16.2. (A) 3 (B) 4 (C) 5 (D) 6 21. There are nine two-digit numbers with distinct tens digits. The units digit of each number is one less than its tens digit. Find the average of the units digits. (A) 3 (B) 4 (C) 5 (D) 6 22. A sum of `7.75 is made up of 100 coins, which are in the denominations of 5 paise and 10 paise. Find the number of 5 paise coins. (A) 50 (B) 55 (C) 75 (D) 45 23. A businessman lends `1800 in two parts, one at 10% and the other at 12% interest. At the end of the year, the average interest he obtained worked out to be 10.5%. Find the interest earned by the businessman from the part which was lent at 10%. (A) `135 (B) `150 (C) `200 (D) `250 24. A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold? (A) 27 litres (B) 18 litres (C) 36 litres (D) 40 litres 25. Three varieties of rice, A, B and C costing `6/kg, `9/kg and `12/kg are mixed together in a certain ratio. 2 The mixture is sold at 66 % profit for `15 / kg. Of the 3 total of 100 kg of the mixture, 50 kg is variety B. Find the quantity of variety A (in kgs) (A) 15 (B) 25 (C) 20 (D) 10

Answer Keys

Exercise Practice Problems 1. B 11. C 21. B

2. D 12. D 22. D

3. B 13. C 23. A

4. B 14. A 24. A

5. D 15. A 25. B

6. C 16. D

7. C 17. B

8. C 18. A

9. B 19. B

10. A 20. D

Chapter 7 Time and Work LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • •

Work and time Unitary method

wOrK Work to be done is usually considered as one unit. It may be constructing a wall or a road, filling up or emptying a tank or cistern or eating certain amount of food. There are some basic assumptions that are made in the problems on Time and Work. These are taken for granted and are not specified in every problem. 1. If a person (or one member of the workforce) does some work in a certain number of days, then we assume (unless otherwise explicitly stated in the problem) that he does the work uniformly, i.e., he does the SAME amount of work everyday. For example, if a person can do some work in 15 days, he does 1/15th of the work in one day. If a person completes the work in 4 days, he does 1/4th of the work on each day and conversely, if a person can complete 1/4th of the work in one day, he can complete the work in 4 days. If a tap can fill a tank in 20 minutes, then in one minute, it can fill 1/20th part of the tank. 2. If there is more than one person (or members of ‘workforce’) carrying out the work, it is assumed that each person (or members of the workforce), unless otherwise specified, does the same amount of work each day. This means they share the work equally. If two people together can do the work in 8 days it means that one man can do it in 16 days. This, in turn means, each person can do 1/16th of the work per day. If a man works three times as fast as a boy does, the man takes one-third of the time the boy takes to complete the work. If the boy takes 12 days to complete the work, then the man takes 4 days to complete the work.

•

Pipes and cisterns

This method is known as ‘UNITARY METHOD’, i.e., the time taken per ‘Unit Work’ or number of persons required to complete ‘Unit Work’ or work completed by ‘Unit Person’ in ‘Unit Time’, etc., is what is first calculated. We should recollect the fundamentals on variation (direct and inverse) here. 1. Time remaining constant, Work and Men are directly proportional to each other, i.e., if the work increases the number of men required to complete the work in the same number of days increases proportionately and vice-versa. 2. Work remaining constant, Men and Days are inversely proportional, i.e., if the number of men increases, the number of days required to complete the same work decreases in inverse proportion and vice-versa. 3. The number of workingmen remaining constant, Work and Days are directly proportional i.e., if the work increases, the number of days required to complete the work with the same number of working men also proportionately increases and vice-versa. The concept of MANDAYS is very important and useful here. The number of men multiplied by the number of days that they take to complete the work will give the number of mandays required to do the work. The total number of mandays required to complete a specific task will remain a constant. So, if we change one of the variables – men or days – the other will change accordingly so that their product will remain constant (remember from our knowledge of VARIATION, two variables whose product is a constant are said to be inversely proportional to each other). The two variables – men and days – are inversely proportional to each other, when work is constant.

Chapter 7 Time and Work | 1.119 Solved Examples

From this we can find out the number of days that they take to complete the work.

Example 1: If 15 men take 60 days to complete a job, find the time taken by 45 men to complete it.

If A can do a piece of work in p days and B can do it in q days then A and B together can complete the same in pq days. p+q

Solution: Number of mandays required to complete the job = 900 mandays. Time taken by 45 men to complete it 900 = i.e., 20 days. 45 Example 2: 18 men take 20 days to complete a job working 12 hours a day. Find the number of days that 15 men will take to complete it if they work 9 hours a day. Solution: Total time for which 18 men work = 240 hours. Number of man hours required to complete the job = (18) (240) man hours. Number of days taken by 15 men working 9 hours a day (18)(240) = 32 . to complete it = (15)(9) Hence, in general we can say that If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then M1 D1 H1 M 2 D2 H 2 = W1 W2 Example 3: 20 men take 10 days to complete a job working 12 hours a day. Find the number of men required to complete a job, twice as large, in 30 days working 8 hours a day. Solution: Number of man hours required to complete the job = (20) (10) (12) = 2400 Number of men required to complete a job twice as large 2400 × 2 by 240 hours = i.e., 20 days. 240 Alternative method:

M1 = 20, D1 = 10, H1 = 12

D2 = 30, H2 = 8

D2 = 2W1 M2 =

M1 D1 H1 W2 ( 20 )(10 )(12) ( 2W1 ) = 20. = W1 D2 H 2 W1 (30 )(8)

If two persons A and B can individually do some work in p and q days respectively, we can find out how much work can be done by them together in one day. Since A can do 1/ pth part of the work in one day and B can do 1/qth part of the work in one day, the two of them together do (1/p + 1/q)th part of the work in one day.

Example 4: A and B can complete a job in 10 days and 12 days respectively. Find the time taken to complete it, if both A and B work together. Solution: Time taken by them to complete it

=

(10)(12) = 60 days. 10 + 12

11

Example 5: A and B together can complete a job in 12 days. A alone can complete it in 24 days. Find the time taken by B to complete it. Solution: Part of the job that A and B can complete in a `1320 day = 6 1 Part of the job that A can complete in a day = 24 Part of the job that B can complete in a day 1 1 1 = − = 12 24 24 \ B can complete it in 24 days.

Example 6: Ajay and Bala working together can complete a job in 16 days. Ajay alone can complete it in 18 days. Both work together for 4 days and then Bala leaves. Find the time taken by Ajay to complete the remaining work. Solution: Part of the job that can be done by both in a day 1 = 16 1 Part of the job that can be done by them in 4 days = 4 × 16 1 = 4 3 Remaining part of the job = 4 Time taken by Ajay to complete it 3 = (18) = 13.5 days. 4 Example 7: A can complete a job in 16 days. He started the work and after 4 days, B joined him. They completed the job in 4 more days. Find the number of days in which B alone can complete it. 1 Solution: Part of the job done by A in a day = 16 A worked for a total of 8 days

1.120 | Quantitative Aptitude 1 1 \ A completed (8) ⎛⎜ ⎞⎟ = of the job. ⎝ 16 ⎠ 2 1 of the job in Hence, B can complete the remaining 2 4 days. \ B alone can complete the entire job in 8 days.

Solution: Part of the job completed in the first 2 days 1 1 1 = + = 18 36 12 \ To complete the job, 12 cycles of 2 days i.e., a total of 24 days will be required.

Example 8: P and Q together can complete a job in 2 4 14 days . Q and R together can complete it in 20 days. 5 7 P and R together can complete it in 16 days. Find the time taken by each of them to complete the job. 5 Solution: Part of the job that P and Q can do in a day = 72 7 Part of the job that Q and R can do in a day = 144 1 Part of the job P and R can do in a day = 16

Example 11: P and Q together can complete a job in 8 days and 16 days respectively. They work on alternate days with Q starting the job. In how many days will the job be completed?

Let the time taken by P, Q and R to complete the job be p days, q days and r days respectively.

1 1 5 + = (5) p q 72 1 1 7 + = (6) q r 144 1 1 1 + = (7) p r 16

Adding (5) and (6) and subtracting (7), ⇒

2 5 7 1 8 = + − = q 72 144 16 144 q = 36

substitute q = 36, in (5), we get p = 24 substi q = 36 in (6), we get r = 48. Example 9: A contractor decided to complete a job in 30 days for which he employed 20 men in the beginning. After 10 days he released that the job could not be completed on time. Hence, he employed 15 more men and thus completed the job on time. Find the number of extra days it would have taken to complete the job if the additional men were not employed. Solution: Number of mandays required to complete the job = (20) (10) + (20 + 15) (20) = 900 mandays. If the additional men were not employed, number of 900 extra days = − 30 = 15 days. 20 Example 10: A and B together can complete a job in 18 days and 36 days respectively. They work on alternate days with A starting the job. In how many days will the job be completed?

Solution: Part of the job completed by P and Q in the first 1 1 3 2 days = + = 8 16 16 15 th After 5 cycles of 2 days, i.e., after 10 days, of the 16 job will be completed. 1 th Remaining part = . Q will work on the 11th day and 16 he takes exactly one day to complete the remaining part. \ The job will be completed in 11 days. In general, money earned should be shared by people doing the work together in the ratio of the SHARE OF WORK done by each of them. For example, if A does two-fifth of the work, then he should get two-fifth of the total earnings for the work. If the remaining three-fifth of the work is done by B and C in the ratio of 1 : 2, then the remaining three-fifth of the earnings (after paying A) should be shared by B and C in the ratio of 1 : 2. Suppose `500 is paid to A, B and C together for doing the work, then A will get `200 (which is 2/5 of `500), B will get `100 and C, `200 (because the remaining `300 after paying A is to be divided in the ratio 1 : 2 between B and C). When people work for the same number of days each, then the ratio of the total work done will be the same as the work done by each of them PER DAY. Hence, if all the people involved work for the same number of days, then the earnings can directly be divided in the ratio of work done per day by each of them. Example 12: P, Q and R can together earn `3100 in 10 days. Q and R together can earn `1320 in 6 days. P and R together can earn `1050 in 5 days. Find R’s daily earning. Solution: Total daily wage of P, Q and R = Total daily wage of Q and R =

`3100 = `310 10

`1320 = `220 6

`1050 = `210 5 Total daily wage of P, Q and 2R = `430 \ R’s daily wage = `120.

Total daily wage of P and R =

Example 13: Two men under take a job for `960. They can complete it in 16 days and 24 days. They work along with a third man and take 8 days to complete it. Find the share that the third man should get.

Chapter 7 Time and Work | 1.121 Solution: The amount payable should be proportional to the fraction of work done. Part of the job done by the third man 8⎞ 1 ⎛ 8 = 1− ⎜ + ⎟ = . ⎝ 16 24 ⎠ 6 \ The third man should get

i.e., pipe X is in operation for n minutes and pipe Y for all the 30 minutes. n 30 So, + = 1 30 60 ⇒

n = 15.

Example 16: Pipes P, Q and R together can empty a full tank in 6 hours. All the three pipes are opened simultaneously and after 2 hours, P is closed. The tank is emptied in another 6 hours. Find the time in which P can empty the tank.

`960 = `160. 6

Pipes and Cisterns There can be pipes (or taps) filling (or emptying) tanks with water. The time taken by different taps (to fill or empty the tank) may be different. Problems related to these can also be dealt with in the same manner as the foregoing problems on Work have been dealt with. There is only one difference between the problems on regular Work (of the type seen earlier on in the chapter) and those in Pipes and Cisterns. In Pipes and Cisterns, a filling pipe or tap does positive work and an emptying pipe or a leak does negative work.

Solution: Part of the tank that can be emptied by P, Q and 1 R per hour = . 6 Part of the tank that was emptied by P, Q and R in 2 1 hours = 3 Part of the tank which was emptied by Q and R per hour 1 1− 3=1 = 6 9 Time in which P can empty the tank 1 = i.e. 18 hours. 1 1 − 6 9

Example 14: Pipes P and Q take 24 minutes and 36 minutes respectively to fill an empty tank. If both take 18 minutes to fill a tank along with an outlet pipe R, find the time R would take to empty the full tank. Solution: Let the time taken by R to empty the tank be r minutes. 1 1 1 1 + − = ; r = 72. 24 36 r 18 Example 15: Pipes X and Y can fill a tank in 30 minutes and 60 minutes respectively. Both pipes are opened simultaneously. After how much time should X be closed so that the tank is filled in 30 minutes? Solution: Let us say pipe X should be closed after n minutes.

Example 17: A tank has a leak at its bottom which empties it at 6 litres/minutes. It also has a filling tap which can fill the tank in 6 hours. The tank takes 18 hours to become full. Find the capacity of the tank. Solution: Let the time that would be taken by the leak to empty the full tank be x hours. \

1 1 1 − = ; x = 9 6 x 18

\ Capacity of the tank = (6) (9) (60) = 3240 litres.

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. X men can complete a work in 120 days. If there were 10 men more, the work would be completed in 20 days less. Find the value of X. (A) 75 (B) 50 (C) 90 (D) 60 2. Nine men can complete a job in 15 days. If a man works thrice as fast as a woman, find the number of days taken by 15 women to complete the job. (A) 20 (B) 24 (C) 27 (D) 36

3. The ratio of the time taken by A, B and C to complete a job is 3 : 4 : 6. Find the ratio of the work they can complete in an hour. (A) 6 : 4 : 3 (B) 4 : 3 : 2 (C) 2 : 3 : 4 (D) 3 : 4 : 6 4. Amar, Bharat and Charu can complete a job in 12, 24 and 24 days respectively. If they all work together, how long will they take to complete the same work? (A) 18 days (B) 6 days (C) 20 days (D) 16 days

1.122 | Quantitative Aptitude 5. Adam can complete a job in 25 days. Adam and Chris 3 together can complete it in 9 days. In how many days 8 can Chris alone complete the job? (A) 125/8 (B) 10 (C) 25 (D) 15 6. P and Q can complete a job in 10 days. Q and R can complete it in 12 days. P and R can complete it in 20 days. Who is the slowest of the three workers? (A) P (B) Q (C) R (D) Cannot be determined 7. Ten men can do a piece of work in 15 days. How many men are needed to complete a work which is five times as large as the first one, in 10 days? (A) 60 (B) 75 (C) 70 (D) 85 8. Tap X can fill a tank in 10 hours. Tap Y can fill it in 15 hours. If the two taps fill the tank together, what fraction of the tank is filled by X? (A) 1/10 (B) 1/6 (C) 2/3 (D) 3/5 9. Pipe A can fill an empty tank in 9 hours. Pipe B can empty a full thank in 18 hours. If both pipes are opened simultaneously when the tank is empty, find the time taken to fill the tank. (in hours) (A) 24 (B) 27 (C) 18 (D) 36 10. Raj can build a wall in 18 days and Kiran can do the same in 30 days. After Raj had built half the wall, Kiran joins him. What is the total number of days taken to build the wall? 5 1 1 (A) 24 (B) 14 (C) 15 (D) 16 8 2 2 11. Kaushik is one and a half times more efficient than Ravi. Kaushik can do a piece of work in 20 days. What portion of the total work can both of them together complete in 10 days? (A) 3/10 (B) 4/5 (C) 9/10 (D) 7/10 12. Had then been one men less, then the number of days required to do a piece of work would have been one more. If the number of mandays required to complete the work is 56, how many workers were there? (A) 6 (B) 8 (C) 9 (D) 14 13. In 8 days, Peter can do as much work as Pan can do in 12 days. To do a certain job both together take 36 days. In how many days can Pan, working alone, complete the job? (A) 60 days (B) 80 days (C) 108 days (D) 90 days 14. X can complete a job in 36 days and Y can complete it in 45 days. Z can complete the job in z days. Z started the job. After 28 days, X and Y joined. The job was completed in 4 more days. Find z. (A) 40 (B) 35 (C) 30 (D) 50

15. Working in pairs, PQ, QR and RP can complete a job in 24 days, 20 days and 30 days respectively. Find the respective times taken by P, Q and R individually to complete the same job (in days). 240 240 (A) 48, 80, (B) 80, 48, 7 7 240 240 (C) 80, , 48 (D) 48, , 80 7 7 16. A frog was at the bottom of a 80 m deep well. It attempted to come out of it by jumping. In each jump it covered 1.15 m but slipped down by 0.75 m. Find the number of jumps after which it would out of the well. (A) 198 (B) 201 (C) 200 (D) 199 17. A man, a woman and a boy can do a piece of work in 2, 4 and 8 days respectively. How many boys must work together with one man and one woman to complete the work in one day? (A) 5 (B) 4 (C) 2 (D) 1 18. A machine of type A which has to produce a set of 1500 bolts, can do so in 30 days. The machine breaks down after 10 days. A machine of type B completes the remaining work in 10 days. In 30 days how many bolts can both of them together produce? (A) 3,000 (B) 4,500 (C) 6,000 (D) 2,500 19. In a farm, each cow eats twice as much grass as each sheep. The cost of grass for 10 cows and 40 sheep for 20 days is ` 900. Find the cost of grass for 20 cows and 10 sheep for 18 days (in `). (A) 600 (B) 675 (C) 750 (D) 800 20. The cost of grass for 20 cows and 30 sheep for 30 days is `720. If the 30 sheep eat double the grass eaten by the 20 cows, then what is the cost of grass eaten by 20 sheep in 15 days? (A) `200 (B) `160 (C) `240 (D) `100 21. George and Gagan together repair a bridge in 45 days and receive `13,500. If Gagan is three times as efficient as George, what is the amount of money he earns in 10 days? (A) `2,000 (B) `2,250 (C) `2,500 (D) `2,750 22. Two pipes A and B which can fill a tank in 20 and 30 hours respectively were opened simultaneously. But there was a leak and it took 3 hours more to fill the tank. In how many hours can the leak empty the tank? (A) 60 (B) 50 (C) 30 (D) 40 23. Gokul, Govardhan and Ganesh can do a piece of work in 10, 20 and 30 days respectively. They begin a new job of similar nature and each of them works on it for one third of the total period of work. If they get `6,600 for the new job, how much should Govardhan get, given that the amounts distributed are in proportion to the work done by them? (A) `1,800 (B) `2,200 (C) `3,300 (D) `2,400

Chapter 7 Time and Work | 1.123 24. Rakesh and Ramesh take 30 days and 60 days respectively to complete a job. They work on alternate days to complete it with Rakesh starting the job. Find the time in which the job is completed (in days). (A) 60 (B) 80 (C) 40 (D) 90 25. If Rakesh and Ramesh had instead taken 10 days and 12 days respectively to complete the job, find the

time in which the job would have been completed (in days). 1 5 (A) 10 (B) 10 3 6

(C) 11

(D) 10

1 2

Answer Keys

Exercise Practice Problems 1. B 11. D 21. B

2. C 12. B 22. A

3. B 13. D 23. A

4. B 14. A 24. C

5. D 15. C 25. B

6. C 16. D

7. B 17. C

8. D 18. B

9. C 19. B

10. B 20. B

Chapter 8 Time and Distance LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • General problems on time, speed and distance • Speed • Relative speed

introduction In this chapter, we will look at problems in the following different areas: 1. 2. 3. 4.

General problems on Time, Speed and Distance Relative Speed Boats and Streams Races and Circular Tracks

Before we look at problems in various areas, let us first look at some basic concepts pertaining to speed, time and distance.

sPeed Distance covered per unit time is called speed. i.e.,

Speed = Distance/time

The above relationship between the three variables distance, speed and time can also be expressed as follows: Distance = Speed × Time or Time = Distance/Speed 1. If two bodies travel with the same speed, distance covered ∝ time (Direct Variation). 2. If two bodies travel for the same period of time, distance covered ∝ speed (Direct Variation). 3. If two bodies travel the same distance, Time ∝

1 (Inverse Variation) Speed

• Average speed • Boats and streams • Races and circular tracks

Distance is normally measured in kilometres, metres or miles; time in hours or seconds and speed in km/hr (also denoted by kmph), miles/hr (also denoted by mph) or metres/second (denoted by m/s). To convert speed in kmph to m/sec, multiply it with 5/18. To convert speed in m/sec to kmph, multiply it with 18/5. In the case of moving trains, three different situations need to be considered. When a train passes a stationary point, the distance covered (in the passing) is the length of the train. If the train is crossing a platform (or a bridge), the distance covered by the train (in the crossing) is equal to the length of the train plus the length of the platform (or bridge). If two trains pass each other (travelling in the same direction or in opposite directions), the total distance covered (in the crossing or the overtaking, as the case may be) is equal to the sum of the lengths of the two trains.

Average Speed Average speed of a body travelling at different speeds is defined as follows: Average speed =

Total distance travelled Total time taken

Please note that the Average speed of a moving body is Not equal to the Average of the speeds. If a body travels from point A to point B with a speed of p and back to point A (from point B) with a speed of q, then the average

Chapter 8 Time and Distance | 1.125 speed of the body can be calculated as 2pq/(p + q). Please note that this does not depend on the distance between A and B. If a body covers part of the journey at speed p and the remaining part of the journey at speed q and the distances of the two parts of the journey are in the ratio m : n, then the average speed for the entire journey is (m + n)pq/(mq + np). Solved Examples

Solution: Let the distance be x km. Time taken by Ashok if x he walked at 4 kmph = hours. 4 x Time taken by Ashok if he walked at 3 kmph = hours. 3 In this case he would take one hour more to reach his office compared to the time taken if he had walked at 4 kmph. x x \ − =1 3 4 ⇒

Example 1: Ashok covered a certain distance at a certain speed. If his speed was 20% more, he would take 10 minutes less to cover the same distance. Find the time he takes to cover the distance. Solution: If his speed was 20% more, it would be 120% 6 i.e., times his actual speed. 5 5 \ Time taken would be times his actual time. 6 1 Reduction in time = (actual time) = 10 minutes 6 \ Actual time = 60 minutes. Example 2: A car covered a certain distance at 90 kmph and returned back at 60 kmph. Find its average speed for the entire journey. Solution: Let x km be the distance to be covered, each way. Total time of travel (in hours) x x 5x x = = + = 90 60 180 36 Average speed (in km/hr)

2x Total disance travelled = 72 = = x Total time taken 36

Example 3: Find the time taken by a train, 100 m long, travelling at a speed of 63 kmph to cross a platform of length 250 m. ⎛ 5 ⎞ 35 Solution: Speed of the train = (63) ⎜ ⎟ = m/sec ⎝ 18 ⎠ 2 Distance to be travelled by the train to cross the platform = length of the train + length of the platform. Time taken to cross the platform =

100 + 250 = 20 seconds 35 2

Example 4: Ashok would reach his office 15 minutes early if he walked at 4 kmph from his house. He would reach it 45 minutes late if he walked at 3 kmph from his house. Find the distance between his house and office.

x = 12.

In general, if a person travelling between two points reaches p hours late travelling at a speed of u kmph and reaches q hours early travelling at v kmph, the distance vu between the two points is given by ( p + q). v−u

Relative Speed The speed of one (moving) body in relation to another moving body is called the relative speed of these two bodies, i.e., it is the speed of one moving body as observed, from the second moving body. If two bodies are moving in the same direction, the relative speed is equal to the difference of the speeds of the two bodies. If two bodies are moving in opposite directions, the relative speed is equal to the sum of the speeds of the two bodies. Example 5: Find the time taken by a train 175 m long running at a speed of 54 kmph to overtake another train 75 m long running at a speed of 36 kmph. Solution: Relative speed = 18 kmph = 5 m/sec Time taken for the faster train to overtake the slower train =

( Length of the faster train ) + ( Length of the slower train ) Their relative speed

175 + 75 = 50 seconds 5 Example 6: A train overtakes two persons, cyling at 9 kmph and 18 kmph in 40 seconds and 48 seconds respectively. Find its length and speed.

=

Solution: Let the length and the speed of the train be ℓ m and s kmph respectively. ⇒ ⇒ \

= 40 ( s − 9) s−9 48 = 40 s − 18 s = 63

5 5 = 48 ( s − 18) 18 18

= 40 (63 – 9) ×

5 = 600 m 18

1.126 | Quantitative Aptitude

Boats and Streams Problems related to boats and streams are different in the computation of relative speed from those of trains/cars. When a boat is moving in the same direction as the stream or water current, the boat is said to be moving WITH THE STREAM OR CURRENT. When a boat is moving in a direction opposite to that of the stream or water current, it is said to be moving AGAINST THE STREAM OR CURRENT. If the boat is moving with a certain speed in water that is not moving, the speed of the boat is then called the SPEED OF THE BOAT IN STILL WATER. When the boat is moving upstream, the speed of the water opposes (and hence reduces) the speed of the boat. When the boat is moving downstream, the speed of the water aids (and thus adds to) the speed of the boat. Thus, we have Speed of the boat against stream = Speed of the boat in still water – Speed of the stream Speed of the boat with the stream = Speed of the boat in still water Speed of the stream These two speeds, the speed of the boat against the stream and the speed of the boat with the stream, are RELATIVE SPEEDS. If u is the speed of the boat down the stream and v is the speed of the boat up the stream, then we have the following two relationships.

Speed of the boat in still water = (u + v)/2

Speed of the water current = (u – v)/2

In problems, instead of a boat, it may be a swimmer but the approach is exactly the same. Instead of boats/swimmers in water, it could also be a cyclist cycling against or along the wind. The approach to solving the problems still remains the same. Example 7: A boat travels 30 km upstream in 5 hours and 100 km downstream in 10 hours. Find the speed of the boat in still water and the speed of the stream. 30 Solution: Upstream speed = = 6 kmph 5 100 Downstream speed = = 10 kmph 10 6 + 10 Speed in still water = = 8 kmph 2 10 − 6 Speed of the stream = = 2 kmph 2 Example 8: Anand can row 20 km in 2 hours in still water. If the speed of the stream is 6 kmph, he would take 3.75 hours to cover a round trip journey. Find the distance that he would then cover each way.

20 Solution: Speed of the boat in still water = = 10 kmph 2 Let the total distance covered be 2x km. Given that,

x x + = 3 ⋅ 75 10 + 6 10 − 6 x = 12

Races and Circular Tracks When two persons P and Q are running a race, they can start the race at the same time or one of them may start a little later than the other. In the second case, suppose P starts the race and after 5 seconds, Q starts. Then we say P has a ‘start’ of 5 seconds. Alternatively, in a race between P and Q, P starts first and then when P has covered a distance of 10 metres, Q starts. Then we say that P has a ‘start’ of 10 metres. In a race between P and Q where Q is the winner, by the time Q reaches the winning post, if P still has another 15 metres to reach the winning post, then we say that Q has won the race by 15 metres. Similarly, if P reaches the winning post 10 seconds after Q reaches it, then we say that Q has won the race by 10 seconds. In problems on RACES, we normally consider a 100 m race or a 1 km race. The length of the track. NEED NOT necessarily be one of the two figures mentioned above but can be as given in the problem. When two or more persons running around a circular track (starting at the same point and at the same time), then we will be interested in two main issues: 1. When they will meet for the first time and 2. When they will meet for the first time at the starting point To solve the problems on circular tracks, you should keep the following points in mind. When two persons are running around a circular track in OPPOSITE directions

1. The relative speed is equal to the sum of the speeds of the two individuals and 2. From one meeting point to the next meeting point, the two of them TOGETHER cover a distance equal to the length of the track.

When two persons are running around a circular track in the SAME direction 1. The relative speed is equal to the difference of the speeds of the two individuals and 2. From one meeting point to the next meeting point, the faster person covers one COMPLETE ROUND more than the slower person. We can now tabulate the time taken by the persons to meet for the first time ever or for the first time at the starting point in various cases.

Chapter 8 Time and Distance | 1.127 When TWO people are running around a circular track Let the two people A and B with respective speeds of a and b(a > b) be running around a circular track (of length L) starting at the same point and at the same time. Then, When the two persons are running in the SAME direction Time taken to meet for the FIRST TIME EVER Time taken to meet for the first time at the STARTING POINT

When the two persons are running in OPPOSITE directions

L (a − b ) ⎧L L ⎫ LCM of ⎨ , ⎬ ⎩a b ⎭

L (a + b ) ⎧L L ⎫ LCM of ⎨ , ⎬ ⎩a b ⎭

Time taken by B to complete the race

200 m = 40 seconds = 5 m/s

\ Time taken by A to complete the race = 38 seconds A’s speed =

200 100 = m/sec 38 19

Example 10: Ramu is 50% faster than Somu. In a race, Ramu gave Somu a head start of 200 m. Both finished the race simultaneously. Find the length of the race. Solution: Let the length of the race be x m. ⇒

x 150 = x − 200 100 x = 600.

Please note that when we have to find out the time taken by the two persons to meet for the first time at the starting point, what we have to do is to find out the time taken by each of them to complete one full round and then take the LCM of these two timings (L/a and L/b are the timings taken by the two of them respectively to complete on full round).

Example 11: In a 1200 m race, Ram beats Shyam by 300 m. In the same race, Shyam beats Tarun by 400 m. Find the distance by which Ram beats Tarun.

Example 9: In a 200 m race, A beats B by 10 m or 2 seconds. Find B’s speed and A’s speed.

r 4 1200 = = s 1200 − 300 3

Solution: A beat B by 10 m or 2 seconds. ⇒ When A reached the finishing line B was 10 m behind the finishing line and took 2 seconds to cover it.

s 3 1200 = = t 1200 − 400 2

\

B’s speed =

10 = 5 m/sec 2

Solution: Let the speeds of Ram, Shyam and Tarun be r m/sec, s m/sec, and t m/sec respectively

r ⎛ r ⎞ ⎛ s⎞ =⎜ ⎟⎜ ⎟ =2 t ⎝ s⎠ ⎝ t ⎠ ⇒ By the time Ram covers 1200 m, Tarun covers 600 m. \ Ram beats Tarun by (1200 – 600) i.e. by 600 m.

Practice Exercise Practice Problems Directions for questions 1 to 30: Select the correct alternative from the given choices. 1. Convert the following speeds into meters per second (a) 36 km/hr (A) 10 (B) 12 (C) 15 (D) 20 (b) 12.6 km/hr (A) 3.5 (B) 4 (C) 0.35 (D) 6 (c) 252/35 km/hr (A) 2.2 (B) 2.4 (C) 2 (D) 2.6 2. If a man runs at 6 metres per second, what distance (in km) will he cover in 3 hours and 45 minutes? (A) 81 (B) 96 (C) 91 (D) 27

3. Travelling at 5/6th of his usual speed a man is 10 minutes late. What is the usual time he takes to cover the same distance? (A) 50 minutes (B) 70 minutes (C) 1 hour (D) 75 minutes 4. X and Y are 270 km apart. At 9:00 a.m, buses A and B left X and Y for Y and X respectively. If the speeds of A and B are 50 kmph and 40 kmph respectively, find their meeting time. (A) 11:00 a.m (B) 12:00 p.m (C) 1: 00 p.m (D) 2:00 p.m 5. Car A left X for Y at 9:00 a.m. Car B left Y for X at 10:00 a.m. XY = 180 km. Speeds of A and B are 30 kmph and 20 kmph respectively. Find their meeting time.

1.128 | Quantitative Aptitude (A) 12:36 p.m. (B) 1:36 p.m. (C) 1:00 p.m (D) 2:00 p.m 6. Ashok left X and reached Y in 4 hours. His average speed for the journey was 90 kmph. Find the distance between X and Y (in km). (A) 180 (B) 360 (C) 720 (D) 900 7. Alok travelled from Hyderabad to Tirupati at 60 kmph and returned at 90 kmph. Find his average speed for the journey (in kmph) (A) 72 (B) 75 (C) 66 (D) 78 8. What is the time taken by a train 650 m long travelling at 72 km/hr to cross a 750 m long platform? (A) 60 sec (B) 65 sec (C) 70 sec (D) 75 sec 9. What is the time taken by a 750 m long train travelling at 99 km/hr to cross a boy running at 9 km/hr towards the train? (A) 30 sec (B) 33 sec (C) 36 sec (D) 25 sec 10. In a 200 m race, Eswar gives Girish a start of 10 m and beats him by 10 m. Find the ratio of their speeds (A) 1 : 1 (B) 9 : 10 (C) 10 : 9 (D) 19 : 20 11. In a 100 m race, Ganesh beats Harish by 10 m or 2 seconds. Find Harish’s speed (in m/sec). 5 (A) 5 (B) 5 9 1 (C) 4 (D) 6 2 12. In a 100 m race, Akbar gives Birbal a start of 2 seconds. Birlbal covers 10 m by the time Akbar starts. If both of them finish together, find Akbar’s speed. (in m/sec)

18.

19.

20.

2 1.

5

(A) 5 (B) 5 9 1 (C) 4 (D) 4 2 13. In a race, P beats Q by 20 seconds. Q beats R by 30 seconds. By how many seconds did P beat R? (A) 44 (B) 25 (C) 50 (D) 36 14. In a 100 m race, A beats B by 10 m and B beats C by 20 m. Find the distance by which A beats C. (in m) (A) 30 (B) 28 (C) 32 (D) 36 15. Anand can row a boat in still water at a speed of 5 kmph. The speed of the stream is 3 kmph. Find the time taken by him to row 40 km downstream (in hours). (A) 5 (B) 20 (C) 8 (D) 10 16. Ram, Shyam and Tarun started cycling from a point on a circular track 600 m long with speeds of 10 m/sec, 15 m/sec and 20 m/sec respectively. Find the time taken by them to meet at the starting point for the first time (in seconds). (A) 120 (B) 60 (C) 240 (D) 600 17. Ashwin and Bhaskar started running simultaneously from a point on a 300 m long circular track. They ran in opposite directions with speeds of 6 m/sec and

22.

23.

2 4.

4 m/sec respectively. After meeting for the first time, they exchange their speeds. Who will reach the starting point first? (A) Ashwin (B) Bhaskar (C) Both reach simultaneously (D) Cannot be determined A man reaches his destination which is 16 km away, 9 min late, if he travels at 8 kmph. What should his speed be if he wishes to reach 15 minutes ahead of the right time? (A) 10 kmph (B) 3 m/sec (C) 20/9 m/sec (D) 12 kmph The distance between two points P and Q is 84 km. Two persons start at the same time but one travelling from P towards Q and the other travelling from Q towards P. If their respective speeds are 36 kmph and 27 kmph, where do they meet each other? (A) 48 km from Q (B) 24 km from P (C) 36 km from P (D) 48 km from P Towns P and Q are 80 km apart. Cars A and B are stationed at towns P and Q respectively. If they start simultaneously towards each other, they would meet in an hour. If both start simultaneously in the same direction, the faster car would overtake the slower car in 4 hours. Find the speed of the faster car (in kmph). (A) 50 (B) 55 (C) 60 (D) 65 A cat on seeing a dog 100 m away turns around and starts running away at 24 kmph. The dog spots him one minute later and starts chasing the cat at a speed of 33 kmph. After how much time, from the start of the cat’s run, will the chase end? (A) 160 s (B) 220 s (C) 260 s (D) 280 s Train A starts at 6 a.m. from city P towards city Q at a speed of 54 kmph. Another train ‘B’ starts at 9 a.m. from P towards Q at 72 kmph. If the distance between P and Q is 1440 km, find at what distance from Q would the two trains meet each other? (A) 648 km (B) 792 km (C) 486 km (D) 954 km Mahesh travelled from Hyderabad to Tirupati at a certain speed and returned at a certain speed. His average speed for the entire trip was the average of his onward and return speeds. He travelled a total distance of 1200 km in 12 hours. Find his onward speed (in kmph). (A) 100 (B) 80 (C) 60 (D) 40 Two cars left simultaneously from two places P and Q, and headed for Q and P respectively. They crossed each other after x hours. After that, one of the cars took y hours to reach its destination while the other took z hours to reach its destination. Which of the following always holds true?

Chapter 8 Time and Distance | 1.129 2 yz y+z (B) x= y +z 2 2 y + z2 (C) x = yz (D) x= y+ z

(A) x=

25. A boat travels 30 km upstream in 5 hours and 24 km downstream in 3 hours. Find the speed of the boat in still water and the speed of the water current (A) 7 kmph, 2 kmph (B) 14 kmph, 1 kmph (C) 7 kmph, 1 kmph (D) 8 kmph, 2 kmph 26. Amar, Akbar and Anthony start running in the same direction and from the same point, around a circular track with speeds 7 m/sec, 11 m/sec and 22 m/sec respectively. If Akbar can complete 5 revolutions around the track in 40 sec, when will they meet for the first time after they start? (A) 56 s (B) 88 s (C) 118 s (D) 79 s 27. If Ashok travelled at 4/5th of his usual speed, he would reach his destination 15 minutes late. By how many minutes would he be early if he travelled at 6/5th of his usual speed? (A) 12 (B) 10 (C) 15 (D) 20

28. In a 500 ft race, Habib beats Akram by 60 ft. If Habib takes 5 paces for every 4 paces taken by Akram, what is the ratio of the length of Habib’s pace to that of Akram? (A) 10 : 11 (B) 11 : 10 (C) 25 : 22 (D) 22 : 25 29. Girish takes 1 minute to complete a round around a circular track. Harish is twice as fast as Girish, Suresh is thrice as fast as Harish. All three start at the same point. Find the time taken by them to meet at the starting point for the first time (in minutes). (A) 1 (B) 2 (C) 6 (D) 12 30. Two cars C and D start from a junction along two perpendicular roads at 8:00 a.m. and 9:00 a.m. respectively. If at 12 noon, the cars, which travel at the same speed, are 150 km apart, then, find the speed of each car. (A) 15 kmph (B) 45 kmph (C) 60 kmph (D) 30 kmph

Answer Keys

Exercise Practice Problems 1. (a) A (b) A (c) C 9. D 10. C 11. A 19. D 20. A 21. C 29. A 30. D

2. A 12. B 22. B

3. A 13. C 23. A

4. B 14. B 24. C

5. C 15. A 25. C

6. B 16. A 26. B

7. A 17. C 27. B

8. C 18. A 28. A

Chapter 9 Indices, Surds, and Logarithms LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Indices • Surds • Rationalisation of a surd

indiceS If a number ‘a’ is added three times to itself, then we write it as 3a. Instead of adding, if we multiply ‘a’ three times with itself, we write it as a3. We say that ‘a’ is expressed as an exponent. Here, ‘a’ is called the ‘base’ and 3 is called the ‘power’ or ‘index’ or ‘exponent’. Similarly, ‘a’ can be expressed to any exponent ‘n’ and accordingly written as an. This is read as ‘a to the power n’ or ‘a raised to the power n.’ an = a × a × a × a × … n times For example, 23 = 2 × 2 × 2 = 8 and 34 = 3 × 3 × 3 × 3 = 81 While the example taken is for a positive integer value of n, the powers can also be negative integers or positive or negative fractions. In the sections that follow, we will also see how to deal with numbers where the powers are fractions or negative integers. If a number raised to a certain power is inside brackets and quantity is then raised to a power again, {i.e., a number of the type (am)n—read as ‘a raised to the power m whole raised to the power n’ or ‘a raised to power m whole to the power n’}, then the number inside the brackets is evaluated first and then this number is raised to the power which is outside the brackets.

• Square root of a surd • Comparison of surds • Logarithms

For example, to evaluate (23)2, we first find out the value of the number inside the bracket (23) as 8 and now raise this to the power 2. This gives 82 which is equal to 64. Thus (23)2 is equal to 64. If we have powers in the manner of ‘steps’, then such a number is evaluated by starting at the topmost of the ‘steps’ and coming down one ‘step’ in each operation. 3 For example, 24 is evaluated by starting at the topmost level ‘3’. Thus we first calculate 43 as equal to 64. Since 2 is raised to the power 43, we now have 264. 2 Similarly, 23 is equal to ‘2 raised to the power 32’ or ‘2 raised to the power 9’ or 29 which is equal to 512. There are certain basic rules/formulae for dealing with numbers having powers. These are called Laws of Indices. The important ones are listed below but you are not required to learn the proof for any of these formulae/rules. The students have to know these rules and be able to apply any of them in solving problems. Most of the problems in indices will require one or more of these formulae. These formulae should be internalised by the students to the extent that after some practice, application of these rules should come naturally and the student should not feel that he is applying some specific formula.

Chapter 9 Indices, Surds, and Logarithms | 1.131 Table 1 Table of Rules/Laws of Indices

Solved Examples

Rule/Law

Example

(1)

am × an = am+n

52 × 57 = 59

(2)

am = am–n an

75 = 72 = 49 73

(3)

(a ) = a

(4 ) = 4

(4)

a–m =

(5)

m n

m

mn

1 am

a = a1/m

2 3

2–3 = 3

6

1 1 = = 0.125 23 8

64 = 641/3 = 4

(6)

(ab)m = a m ⋅ b m

(7)

⎛a⎞ ⎜ ⎟ ⎝b ⎠

(8)

a0 = 1 (where a ≠ 0)

30 = 1

(9)

a1 = a

41 = 4

n

an = n b

⎛ 729 ⎞ Solution: ⎜ ⎝ 1728 ⎟⎠

2

9 32 = 2 = 16 4

−2 3

1

⎛ 93 ⎞ ⎜⎝ 123 ⎟⎠

−2 3

×

32 324 144 32 324 = 256 × = × × 3 24 81 3 24

Rule 1: When the bases of two EQUAL numbers are equal, then their powers also will be equal. (If the bases are neither zero nor ±1.) For example: If 2n = 23, then it means n = 3 Rule 2: When the powers of two equal numbers are equal (and not equal to zero), two cases arise:

⎛ 5⎞ Solution: Given, ⎜ ⎟ ⎝ 7⎠

⎛ 5⎞ ⎜⎝ ⎟⎠ 7

x +1

x +1

=

125 343 3

9 ⎡⎛ 5 ⎞ 3 ⎤ ⎛ 5⎞ = ⎢⎜ ⎟ ⎥ = ⎜ ⎟ . ⎝ 7⎠ ⎢⎣⎝ 7 ⎠ ⎥⎦

By equating their indices. x + 1 = 9

x = 8.

⎛ 49 ⎞ Example 3: If ⎜ ⎝ 2401⎟⎠

4− x

= 492 x − 6 , find x.

1. If the power is an odd number, then the bases are equal. For example, if a3 = 43 then a = 4. 2. If the powers are even numbers, then the bases are numerically equal but can have different signs. For example, if a4 = 34 then a = +3 or –3.

⎛ 49 ⎞ Solution: ⎜ ⎝ 2401⎟⎠

The problems associated with indices are normally of THREE types:

x – 4 = 2x – 6

Simplification: Here, the problem involves terms with different bases and powers which have to be simplified using the rules/formulae discussed in the table above. Solving for the value of an unknown: Here, the problem will have an equation where an unknown (like x or y) will appear in the base or in the power and using Rule 1 and Rule 2 discussed above, values of unknown are to be determined. Comparison of numbers: Here two or more quantities will be given—each being a number raised to a certain power. These numbers have to be compared in magnitude—either to find the largest or smallest of the quantities or to arrange the given quantities in ascending or descending order. The following examples will make clear the different types of problems that you may be asked.

Example 2: In the equation given below, solve for x 3

These rules/laws will help you in solving a number of problems. In addition to the above, the student should also remember the following rules:

1

⎛ 92 ⎞ ⎛ 322 ⎞ 2 ⎛ 324 ⎞ = × ⎜⎝ 122 ⎟⎠ ⎜⎝ 32 ⎟⎠ × ⎜⎝ 24 ⎟⎠

1

⎛ 1024 ⎞ 2 ⎛ 24 ⎞ ×⎜ ÷⎜ ⎝ 9 ⎟⎠ ⎝ 324 ⎟⎠

⎛ 1024 ⎞ 2 ⎛ 24 ⎞ ×⎜ ÷⎜ ⎝ 9 ⎟⎠ ⎝ 324 ⎟⎠ −1

(2 × 3)4 = 24 ⋅ 34 ⎛ 3⎞ ⎜⎝ ⎟⎠ 4

−2 3

⎛ 729 ⎞ Example 1: Simplify: ⎜ ⎝ 1728 ⎟⎠

4− x

(

= 49 −1

)

4− x

= 49x – 4

Given, 49x – 4 = 492x – 6 x = 2 Example 4: Arrange the following in ascending order 6256, 1257 and 2510 Solution: 6256 = (54)6 = 524 1257 = (53)7 = 521 2510 = (52)10 = 520 2510 < 1257 < 6256

Surds Any number of the form p/q, where p and q are integers and q ≠ 0 is called a rational number. Any real number which is not a rational number is an irrational number. Amongst irrational numbers, of particular interest to us are SURDS.

1.132 | Quantitative Aptitude Amongst surds, we will specifically be looking at ‘quadratic surds’—surds of the type a + b and a + b + c , where the terms involve only square roots and not any higher roots. We do not need to go very deep into the area of surds—what is required is a basic understanding of some of the operations on surds. If there is a surd of the form (a + b ), then a surd of the form ±(a – b ) is called the conjugate of the surd (a + b ). The product of a surd and its conjugate will always be a rational number.

Rationalisation of a Surd When there is a surd of the form

1

, it is difficult to a + b perform arithmetic operations on it. Hence, the denominator is converted into a rational number thereby facilitating ease of handling the surd. This process of converting the denominator into a rational number without changing the value of the surd is called rationalisation. To convert the denominator of a surd into a rational number, multiply the denominator and the numerator simultaneously with the conjugate of the surd in the denominator so that the denominator gets converted to a rational number without changing the value of the fraction. That is, if there is a surd of the type a + b in the denominator, then both the numerator and the denominator have to be multiplied with a surd of the form a – b or a surd of the type (–a + b ) to convert the denominator into a rational number. If there is a surd of the form (a + b + c ) in the denominator, then the process of multiplying the denominator with its conjugate surd has to be carried out TWICE to rationalise the denominator.

Square Root of a Surd If there exists a square root of a surd of the type a + b, then it will be of the form x + y . We can equate the square of x + y to a + b and thus solve for x and y. Here, one point should be noted—when there is an equation with rational and irrational terms, the rational part on the left hand side is equal to the rational part on the right hand side and, the irrational part on the left hand side is equal to the irrational part on the right hand side of the equation. However, for the problems which are expected in the entrance exams, there is no need of solving for the square root in such an elaborate manner. We will look at finding the square root of the surd in a much simpler manner. Here, first the given surd is written in the form of ( x + y)² or ( x – y)². Then the square root of the surd will be ( x + y) or ( x – y) respectively.

surds in ascending/descending order. The surds given in such cases will be such that they will be close to each other and hence we will not be able to identify the largest one by taking the approximate square root of each of the terms. In such a case, the surds can both be squared and the common rational part be subtracted. At this stage, normally one will be able to make out the order of the surds. If even at this stage, it is not possible to identify the larger of the two, then the numbers should be squared once more. 1 Example 5: Rationalize the denominator: 1+ 6 − 7 Solution: The rationalizing factor of 1+ 6 – 7 is 1 + 1

1+ 6 − 7

=

(1 + 6 − 7 ) (1 + 6 + 7 )

1+ 6 + 7 1+ 6 + 7 = = 2 2 (1 + 6 ) − ( 7 ) 2 6 6 is

6

6 + 6 + 42 = 12

Example 6: Find the value of Solution: Let

7

(1 + 6 + 7 )

The rationalizing factor of

6 +

62 + 480 .

62 + 480 = a + b

Squaring both sides, 62 + 480 = a + b + 2 ab 62 + 480 = a + b + 4 ab Equating the corresponding rational and irrational parts on both sides, a + b = 62 4ab = 480 ⇒ ab = 120 As a + b = 60 + 2 and ab = (60) (B) it follows that a = 60 and b = 2 or vice versa. \

a + b = 60 + 2

Example 7: Which of the surds given below is greater? 3 + 23 and 6 + 19 Solution: ( 3 + 23 ) 2 = 26 + 2 69 69 lies between

64 and

81

\ 26 + 2 69 lies between 26 + 2(8) and 26 + 2(9) i.e., 42 and 44. Similarly ( 6 + 19 ) 2 lies between 45 and 47.

Comparison of Surds

\

Sometimes we will need to compare two or more surds either to identify the largest one or to arrange the given

\

( 3 + 23 ) 2 < ( 6 + 19 ) 2 6 + 19 > 3 + 23

Chapter 9 Indices, Surds, and Logarithms | 1.133

Logarithms In the equation ax = N, we are expressing N in terms of a and x. The same equation can be re-written as, a = N1/x. Here we are expressing a in terms of N and x. But, among a, x and N, by normal algebraic methods known to us, we cannot express x in terms of the other two parameters a and N. This is where logarithms come into the picture. When ax = N, then we say x = logarithm of N to the base a, and write it as x = logaN. The definition of logarithm is given as: ‘the logarithm of any number to a given base is the index or the power to which the base must be raised in order to equal the given number.’ Thus, if ax = N then x = loga N This is read as ‘log N to the base a’. In the above equation, N is a POSITIVE NUMBER and a is a POSITIVE NUMBER OTHER THAN 1. This basic definition of logarithm is very useful in solving a number of problems on logarithms. Example of a logarithm: 216 = 63 can be expressed as log6 216 = 3. Since logarithm of a number is a value, it will have an ‘integral’ part and a ‘decimal’ part. The integral part of the logarithm of a number is called the CHARACTERISTIC and the decimal part of the logarithm is called the MANTISSA. Logarithms can be expressed to any base (positive number other than 1.) Logarithms from one base can be converted to logarithms to any other base. (One of the formulae given below will help do this conversion). However, there are two types of logarithms that are commonly used. (i) Natural Logarithms or Napierian Logarithms: These are logarithms expressed to the base of a number called ‘e.’ (ii) Common Logarithms: These are logarithms expressed to the base 10. For most of the problems under LOGARITHMS, it is common logarithms that we deal with. In examinations also, if logarithms are given without mentioning any base, it can normally be taken to be logarithms to the base 10. Given below are some important rules/ formulae in logarithms: (i) loga a = 1 (logarithm of any number to the same base is 1) (ii) loga 1 = 0 (log of 1 to any base other than 1 is 0) (iii) loga (mn) = loga m + loga n (iv) loga (m/n) = loga m – loga n (v) loga mp = p × loga m 1 (vi) log a b = log b a

(vii) log a m =

log b m log b a

(viii) log q m p = a

p log a m q

(ix) a loga N = N (x) a log b = b log a You should memorize these rules/formulae because they are very helpful in solving problems. Like in the chapter on INDICES, in LOGARITHMS also there will be problems on (i) Simplification using the formulae/rules listed above and (ii) Solving for the value of an unknown given in an equation In solving problems of the second type above, in most of the cases we take recourse to the basic definition of logarithms (which is very important and should be memorized). The following examples will give problems of both the above types and also some problems on common logarithms. The following rules also should be remembered while solving problems on logarithms: Given an equation logaM = logbN, (i) if M = N, then a will be equal to b; if M ≠ 1 and N ≠ 1. (ii) if a = b, then M will be equal to N. The examples that follow will explain all the above types of problems. Please note that unless otherwise specified, all the logarithms are taken to the base 10). Example 8: Solve for x : log10 20x = 4 Solution: Given that log10 20x = 4 ⇒ 20x = 104 = 10000 \

x = 500

Example 9: Solve for x : log (x + 3) + log (x – 3) = log 72 Solution: log (x + 3) + log (x – 3) = log 72 log (x + 3) (x – 3) = log 72 (x + 3) (x – 3) = 72 x2 = 81 x = 9 (If x = – 9, log (x – 3) would be undefined) Example 10: If log 2 = 0.301, find the value of log 1250, log 0.001250 and log 125000. 10000 8 = 4 log 10 – 3 log 2 = 4 – 3 (0.3010) = 3.097

Solution: log 1250 = log

1.134 | Quantitative Aptitude 1250 106 = 3.097 – 6 = –2.903 log 125000 = log (1250) (100) = log 1250 + 2 = 5.097

Example 12: Find the value of log 3 2 32 3 16 .

log 0.001250 = log

( )

Solution: log 3 2 32 3 16 = log 21/3 25 24

1 3

19

Example 11: Find the number of digits in 29420 given that log 6 = 0.778 and log 7 = 0.845 Solution: log 29420 = 20 log (72 6) = 20 (2 log 7 + log 6) = 20 (2 (0.845) + 0.778) = 20 (1.69 + 0.778) 49.36 Characteristic = 49. \ 29420 has 50 digits.

⎛ 1⎞ = log 21/3 ⎜ 2 3 ⎟ = 19 ⎝ ⎠ Example 13: Find the number of zeros after the decimal 500 ⎛ 3⎞ point in ⎜ ⎟ , given that log 3 = 0.4771 and log 2 = 0.3010. ⎝ 4⎠ 500 3⎞ ⎛ 3⎞ ⎛ Solution: log ⎜ ⎟ = 500 ⎜ log ⎟ ⎝ 4⎠ ⎝ 4⎠ = 500 (log 3 – 2 log 2) = 500 (0.4771 – 2 (0.3010) = – 62.4500. \ Number of zeros after the decimal point is 62.

Practice Exercise 6. Solve for x: 92 x + 1 = 275 x − 3 (A) 1 (B) 2

Practice Problems Directions for questions 1 to 30: Select the correct alternative from the given choices. 1. Simplify the following: ⎛ 243 ⎞ ⎜⎝ ⎟ 1024 ⎠

−2 5

⎛ 144 ⎞ ×⎜ ⎝ 49 ⎟⎠

−1 2

(A) 25 × 3–1 × 7 (C) 24 × 3–3 × 7–1 2. Simplify the following: −2

3

⎛ 8 ⎞ ÷⎜ ⎝ 343 ⎟⎠

−2 3

(B) 25 × 3–3 × 7–2 (D) 24 × 3–1 × 7–2 −1

⎛ x ⋅y ⎞ ⎛ x ⋅ y⎞ ⎛ x ⋅ y ⎞ ⎜⎝ z 4 ⎟⎠ × ⎜⎝ z −2 ⎟⎠ ÷ ⎜⎝ z −8 ⎟⎠ (A) x10. y16 . z–22 (B) x7. y–16 . z–22 –7 16 –22 (C) x . y . z (D) x–10. y16 . z22 −3

2

2

−12

7

−1

3. Simplify the following: (A)

y

(1 − y ) 2

1 − [1 − {1 − (1 + y ) }] (1 − y )

y (B) (1 − y )2

1+ y 1+ y 2 (D) (C) 1− y2 (1 − y )2 4. ( x

a − b ( a2 + ab + b 2 )

)

×(x

b − c ( b 2 + bc c 2 )

)

×(x

c − a ( c 2 + ac + a 2 )

)

(A) 0 (B) 1 3 3 3 (C) xa + b + c

(

3 a2 + b2 + c 2 + ac + bc + ca

(D) x

(D) –2

p r 1 1 = and pa = qb = r c = s d, then − = q s a b 1 1 1 1 (A) − (B) + d c c d 7. If

1 1 ⎛ 1 1⎞ (C) − (D) −⎜ + ⎟ ⎝ c d⎠ c d 8. Which of the following is the largest in value? (A) 61/2 (B) 71/3 (C) 81/4 (D) 91/5 9. 2

5 2 −5 + 10 + 1000 = 2 5

(A) 9 10 (B) 8 10 (C) 8 10 (D) 11 10 ⎛ 1 0. ⎜ ⎝

−4

p − 4 pq ⎞ ⎟ = 4 pq − q ⎠ p p (A) (B) q q q q (C) (D) p p 11. If y = 12 + 2 35 , then

y−

1 y

=

7+ 5 3 5− 7 (A) (B) 2 2

)

5. 3430.12 × 24010.08 × 490.01 × 70.1 = (A) 7 (B) 74/5 (C) 78

(C) –1

(D) 73/5

2 5+ 7 3 5+ 7 (C) (D) 2 2

Chapter 9 Indices, Surds, and Logarithms | 1.135

(a + b + c ) + 2

ac + bc .

12. Arrange the following in ascending order.

20. Simplify:

a=

2+

11 , b =

(A) a + b + c (B) a+b + c

c=

3+

10 and d =

6+

7,

5+

8

(A) abcd (B) abdc (C) acdb (D) acbd 13. Arrange the following in descending order. a = 13 + 11 , b = d=

15 + 9 , c =

a2

18 + 6 ,

7 + 17 .

(A) abdc (B) dcab (C) adcb (D) acdb 14. Solve for x and y: 3.5x + 2y + 2 = 107, 5x + 1 + 8.2y = 189 (A) 3, 2 (B) 5, 7 (C) 7, 5

( )

15. Solve for x, if 5 7

(A) 2

5x−4

(D) 2, 3

= (35) ( 25) . 3/ 2

3

(B) 5/4

(C) 7/2

(D) 3

16. If 5x + 3 – 5x – 3 = 78120, find x. (A) 4 (B) 3 (C) 5

(D) 6

17. If aa ⋅ bb ⋅ cc = ab ⋅ bc ⋅ ca = ac ⋅ ba ⋅ cb and a, b, c are positive integers greater than 1, then which of the following can NOT be true for any of the possible values of a, b, c? (A) abc = 8 (B) a + b + c = 8 (C) abc = 27 (D) a + b + c = 27 7

3

2

18. The ascending order of 1612 , 818 , 625 3 is _____. 3

7

2

2

7

3

(A) 1612 , 818 , 625 3 (B) 1612 , 625 3 , 818 , 2

7

3

3

7

(C) ab + bc (D) abc 21. Find the value of x² – y², if logy(x – 1) + logy (x + 1) = 2. (A) 2 (B) 2y (C) 1 (D) 2xy 22. If a > 1, loga a + log 1 a + log 1 a + … + log 1 a =

2

a3

a 20

(A) 420 (B) 210 (C) 380 (D) 190 2 3. If log7(x – 7) + log7(x² + 7x + 49) = 4, then x = (A) 196 (B) 7 (C) 49 (D) 14 log a log b log c = = 24. If , then b² = 5 6 7 (A) ac (B) a² (C) bc (D) ab 25. What is the value of log(1/5) 0.0000128? (A) –7 (B) –5 (C) 5 (D) 7 26. If (log tan5°) (log tan10°) (log tan 15°) … (log tan 60°) = x, what is the value of x? (A) log(sin5°)12 (B) 1 (C) 0 (D) log(cos60°) 27. Solve for x, if log x [log5 ( x + 5 + x )] = 0. (A) 1 (B) 9 (C) 12 (D) 4 28. If a, b, c are distinct values, what is the value of abc if (logb a) (logc a) + (loga b) (logc b) + (loga c) (logb c) – 3 = 0? (A) 2 (B) 1 (C) 1 – log a – log b – loc c (D) 0

11 + 4 6 19. If 2 2 + 3 = x, what is the value of in 2 2− 3 terms of x? x2 (A) (B) x3 2

29. If log6 161 = a, log6 23 = b, what is the value of log7 6 in terms of a and b? (A) a/b (B) a+b (C) 1/(a – b) (D) b/a 30. x = y2 = z3 = w4 = u5, then find the value of logxxyzwu. 111 47 (A) 1 (B) 60 120

x3 x3 (C) (D) 8 5

17 13 (C) 2 (D) 2 60 60

(C) 625 3 , 1612 , 818 (D) 818 , 1612 , 625 3

Answer Keys

Exercise Practice Problems 1. C 11. D 21. C

2. D 12. C 22. B

3. A 13. A 23. D

4. B 14. D 24. A

5. B 15. A 25. D

6. A 16. A 26. C

7. C 17. B 27. D

8. A 18. A 28. B

9. D 19. D 29. C

10. D 20. B 30. C

Chapter 10 Quadratic Equations LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • •

Quadratic equations Finding the roots by factorisation Finding the roots by using the formula Sum and product of roots of a quadratic equation

• • • •

Quadratic EQuationS ‘If a variable occurs in an equation with all positive integer powers and the highest power is two, then it is called a Quadratic Equation (in that variable).’ In other words, a second degree polynomial in x equated to zero will be a quadratic equation. For such an equation to be a quadratic equation, the co-efficient of x² should not be zero. The most general form of a quadratic equation is ax2 + bx + c = 0, where a ≠ 0 (and a, b, c are real) Some examples of quadratic equations are x2 – 5x + 6 = 0

(1)

x2 – x – 6 = 0

(2)

2x2 + 3x – 2 = 0

(3)

2x2 + x – 3 = 0

(4)

Like a first degree equation in x has one value of x satisfying the equation, a quadratic equation in x will have TWO values of x that satisfy the equation. The values of x that satisfy the equation are called the ROOTS of the equation. These roots may be real or imaginary. For the four quadratic equations given above, the roots are as given below: Equation (1) Equation (2) Equation (3) Equation (4)

: : : :

x = 2 and x = 3 x = –2 and x = 3 x = 1/2 and x = –2 x = 1 and x = –3/2

Nature of the roots Signs of the roots Constructing a quadratic equation Maximum or minimum value of a quadratic expression

In general, the roots of a quadratic equation can be found out in two ways. 1. By factorising the expression on the left-hand side of the quadratic equation. 2. By using the standard formula. All the expressions may not be easy to factorise whereas applying the formula is simple and straightforward.

Finding the Roots by Factorisation If the quadratic equation ax2 + bx + c = 0 can be written in the form (x – a)(x – b) = 0, then the roots of the equation are a and b. To find the roots of a quadratic equation, we should first write it in the form of (x – a)(x – b) = 0, i.e., the left hand side ax2 + bx + c of the quadratic equation ax2 + bx + c = 0 should be factorised into two factors. For this purpose, we should go through the following steps. We will understand these steps with the help of the equation x2 – 5x + 6 = 0 which is the first of the four quadratic equations we looked at as examples above. 1. First write down b (the co-efficient of x) as the sum of two quantities whose product is equal to ac. In this case –5 has to be written as the sum of two quantities whose product is 6. We can write –5 as (–3) + (–2) so that the product of (–3) and (–2) is equal to 6.

Chapter 10 Quadratic Equations | 1.137 2. Now rewrite the equation with the ‘bx’ term split in the above manner. In this case, the given equation can be written as x2 – 3x – 2x + 6 = 0 3. Take the first two terms and rewrite them together after taking out the common factor between the two of them. Similarly, the third and fourth terms should be rewritten after taking out the common factor between the two of them. In other words, you should ensure that what is left from the first and the second terms (after removing the common factor) is the same as that left from the third and the fourth term (after removing their common factor). In this case, the equation can be rewritten as x(x – 3) – 2(x – 3) = 0; Between the first and second terms as well as the third and fourth terms, we are left with (x – 3) is a common factor. 4. Rewrite the entire left-hand side to get the form (x – a) (x – b ). In this case, if we take out (x – 3) as the common factor, we can rewrite the given equation as (x – 3) (x – 2) = 0 5. Now, a and b are the roots of the given quadratic equation. \ For x2 – 5x + 6 = 0, the roots of the equation are 3 and 2. For the other three quadratic equations given above as examples, let us see how to factorise the expression and get the roots. For equation (2), i.e., x2 – x – 6 = 0, the co-efficient of x which is –1 can be rewritten as (–3) + (+2) so that their product is –6 which is equal to ac (1 multiplied by –6). Then we can rewrite the equation as (x – 3) (x + 2) = 0 giving us the roots as 3 and –2. For equation (3), i.e., 2x2 + 3x – 2 = 0, the co-efficient of x which is 3 can be rewritten as (+4) + (–1) so that their product is –4 which is the value of ac (–2 multiplied by 2). Then we can rewrite the equation as (2x – 1)(x + 2) = 0 giving the roots as 1/2 and –2. For equation (4), i.e., 2x2 + x – 3 = 0, the co-efficient of x which is 1 can be rewritten as (+3) + (–2) so that their product is –6 which is equal to ac (2 multiplied by –3). Then we can rewrite the given equation as (x – 1) (2x + 3) = 0 giving us the roots as 1 and –3/2.

Finding the Roots by Using the Formula If the quadratic equation is ax2 + bx + c = 0, then we can use the standard formula given below to find out the roots of the equation. x=

−b ± b 2 − 4 ac 2a

The roots of the four quadratic equations we took as examples above can be taken and their roots found out by using the above formula. The student is advised to check it out for himself/herself that the roots can be obtained by using this formula also.

Sum and Product of Roots of a Quadratic Equation For the quadratic equation ax2 + bx + c = 0, the sum of the roots and the product of the roots can be given by the following: Sum of the roots = –b/a Product of the roots = c/a These two rules will be very helpful in solving problems on quadratic equation.

Nature of the Roots We mentioned already that the roots of a quadratic equation with real co-efficients can be real or complex. When the roots are real, they can be equal or unequal. All this will depend on the expression b2 – 4ac. Since b2 – 4ac determines the nature of the roots of the quadratic e quation, it is called the ‘DISCRIMINANT’ of the quadratic equation. If b2 – 4ac > 0, then the roots of the quadratic equation will be real and distinct. If b2 – 4ac = 0, the roots are real and equal. If b2 – 4ac < 0, then the roots of the quadratic equation will be complex conjugates. Thus we can write down the following about the nature of the roots of a quadratic equation when a, b and c are all rational. when b2 – 4ac < 0

the roots are complex and unequal

when b – 4ac = 0

the roots are rational and equal

when b – 4ac > 0 and a perfect square

the roots are rational and unequal

when b2 – 4ac > 0 but not a perfect square

the roots are irrational and unequal

2 2

Whenever the roots of the quadratic equation are irrational, (a, b, c being rational) they will be of the form a + b and a – b , i.e. whenever a + b is one root of a quadratic equation, then a – b will be the second root of the quadratic equation and vice versa.

Signs of the Roots We can comment on the signs of the roots, i.e., whether the roots are positive or negative, based on the sign of the sum of the roots and the product of the roots of the quadratic equation. The following table will make clear the relationship between the sum and the product of the roots and the signs of the roots themselves.

1.138 | Quantitative Aptitude Sign of product of the roots

Sign of sum of the roots

Therefore, 11x2 – 37x + 30 = 0 Sign of the roots

⇒ 11x2 – 22x – 15x + 30 = 0 ⇒ 11x (x – 2) – 15 (x – 2) = 0

+ ve

+ ve

Both the roots are positive.

+ ve

- ve

Both the roots are negative.

⇒ (11x – 15) (x – 2) = 0 x =

- ve

+ ve

The numerically larger root is positive and the other root is negative.

Example 2: Discuss the nature of the roots of the equation 8x2 – 2x – 4 = 0.

- ve

- ve

The numerically larger root is negative and the other root is positive.

Solution: For the quadratic equation ax2 + bx + c = 0 the nature of the roots is given by the discriminant b2 – 4ac. Discriminant of 8x2 – 2x – 4 = 0 is

Constructing a Quadratic Equation We can build a quadratic equation in the following three cases: 1. When the roots of the quadratic equation are given. 2. When the sum of the roots and the product of the roots of the quadratic equation are given. 3. When the relation between the roots of the equation to be framed and the roots of another equation is given. If the roots of the quadratic equation are given as a and b, the equation can be written as (x – a) (x – b) = 0 i.e., x2 – x(a + b) + ab = 0 If p is the sum of the roots of the quadratic equation and q is the product of the roots of the quadratic equation, then the equation can be written as x2 – px + q = 0.

Maximum or Minimum Value of a Quadratic Expression An equation of the type ax2 + bx + c = 0 is called a quadratic equation. An expression of the type ax2 + bx + c is called a ‘quadratic expression’. The quadratic expression ax2 + bx + c takes different values as x takes different values. As x varies from –∞ to +∞, (i.e., when x is real) the quadratic expression ax2 + bx + c 1. Has a minimum value whenever a > 0 (i.e., a is positive). The minimum value of the quadratic expression is (4ac – b2)/4a and it occurs at x = –b/2a. 2. Has a maximum value whenever a < 0 (i.e. a is negative). The maximum value of the quadratic expression is (4ac – b2)/4a and it occurs at x = –b/2a. Solved Examples Example 1: Find the roots of the equation 11x2 – 37x + 30 = 0. Solution: We have to write –37 as the sum of two parts whose product should be equal to (11) × (30) (–22) + (–15) = –37 and (–22) (–15) = 11 × 30

15 or 2. 11

(–2)2 – 4(8) (–4) = 132. Since the discriminant is positive but not a perfect square, the roots of the equation are irrational and unequal. Example 3: If the sum of the roots of the equation Rx2 + 5x – 24 = 0 is 5/11, then find the product of the roots of that equation. Solution: For a quadratic equation ax2 + bx + c = 0, the sum of the roots is (–b/a) and the product of the roots is (c/a). The sum of the roots of the equation 5 ⎛ − 5⎞ Rx2 + 5x – 24 = 0 is ⎜ ⎟ which is given as ⎝ R⎠ 11 \ R = –11 In the given equation, product of the roots = − 24 −24 24 = =+ . R −11 11 Example 4: Find the value of k, so that the roots of 6x2 – 12x – k = 0 are reciprocals of each other. Solution: If the roots of the equation are reciprocals of each other, then the product of the roots should be equal to 1. −k ⇒ = 1. 6 Therefore k = –6. Example 5: If 4 + 7 is one root of a quadratic equation with rational co-efficients, then find the other root of the equation. Solution: When the coefficients of a quadratic equation are rational and the roots are irrational, they occur only in pairs like p ± q i.e., if p + q is one root, then the other root of the equation will be p – q. So, in this case, the other root of the equation will be 4 – 7. Example 6: Form a quadratic equation with rational coefficients, one of whose roots is 5 + 6. Solution: If 5 + 6 is one root, then the other root is 5 – 6 (because the coefficients are rational). The sum of the roots = 5 + 6 + 5 – 6 = 10. The product of the roots = (5 + 6) (5 – 6) = 25 – 6 = 19. Thus the required equation is x2 – 10x + 19 = 0.

Chapter 10 Quadratic Equations | 1.139 Example 7: If the price of each book goes up by `5, then Atul can buy 20 books less for `1200. Find the original price and the number of books Atul could buy at the original price. Solution: Let the original price of each book be x. Then the new price of each book will be x + 5. The number of books that can be bought at the original 1200 price = x The number of books that can be bought at the new price 1200 = x +5 Given that Atul gets 20 books less at new price i.e., 1200 1200 − = 20 x x+5 60 60 − =1 ⇒ x x+5 ⇒

=±

= ± 9 + 720

= ± 729 = ± 27; as a < b, a – b = –27

Example 9: If x + 4 + x + 8 = 7, then find the value of x. Solution: Given x + 4 + x + 8 = 7 Squaring on both sides, we get

(

x+4+x+8+2

⇒ 2x + 12 + 2

)

x + 4 x + 8 = 49

x 2 + 12 x + 32 = 49

⇒ 2x – 37 = –2 x 2 + 12 x + 32 Squaring again on both sides, we have (2x – 37)2 = 4 (x2 + 12x + 32)

x2 + 5x – 300 = 0

⇒

⇒

⇒ 1241 = 196x

x = –20 or 15

As the price cannot be negative, the original price is `15. Example 8: If a and b are the roots of the equation x2 – 3x – 180 = 0 such that a < b, then find the values of 1 1 (i) a 2 + b 2 (ii) + (iii) a–b α β

1241 196

Solution: Given 42x + 1 + 4x + 1 = 80 ⇒ 42x × 4 + 4x × 4 = 80 42x + 4x = 20 Substituting 4x = a, we get ⇒

(i) a + b = (a + b) – 2ab = (3)2 – 2 (–180) = 369

⇒

2

x=

Example 10: If 42x + 1 + 4x + 1 = 80, then find the value of x.

Solution: From the given equation, we get a + b = 3 and ab = –180 2

(+3)2 − 4 ( −180)

⇒ 4x2 – 148x + 1369 = 4x2 + 48x + 128

⇒ (x + 20) (x – 15) = 0

2

(α + β )2 − 4αβ

60( x + 5 − x ) =1 x 2 + 5x

⇒ 300 = x2 + 5x ⇒

⇒ a – b = ±

a2 + a = 20 a2 + a – 20 = 0

⇒ (a + 5) (a – 4) = 0 a = –5 or 4

If 4x = –5, there is no possible value for x as no power of 4 gives negative value.

1 1 α +β +3 −1 = = (ii) + = α β αβ −180 60 (iii) (a – b)2 = (a + b)2 – 4 ab

If 4x = 4, then x = 1.

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. The roots of the quadratic equation 2x2 – 7x + 2 = 0 are (A) Rational and unequal (B) Real and equal (C) Imaginary (D) Irrational 2. Find the nature of the roots of the quadratic equation 2x2 + 6x – 5 = 0.

(A) (B) (C) (D)

Complex conjugates Real and equal Conjugate surds Unequal and rational

3. Construct a quadratic equation whose roots are one third of the roots of x2 + 6x + 10 = 0. (A) x2 + 18x + 90 = 0 (C) 9x2+ 18x + 10 = 0

(B) x2 + 16x + 80 = 0 (D) x2 + 17x + 90 = 0

1.140 | Quantitative Aptitude 4. A quadratic equation in x has its roots as reciprocals of each other. The coefficient of x is twice the coefficient of x2. Find the sum of the squares of its roots. (A) 5 (B) 4 (C) 3 (D) 2 2 5. If one root of the quadratic equation 4x – 8x + k = 0, is three times the other root, find the value of k. (A) 3 (B) 9 (C) –3 (D) –6 6. The roots of the quadratic equation (m – k + l) x2 – 2mx + (m – l + k) = 0 are l + m − k 2m (B) 1, 1, (A) k + m −l l+m−k k +m−l 2k (D) 1, 1, (C) l+m−k k −m+l 4 ac − b 2 7. The expression represents the maximum/ 4a minimum value of the quadratic expression ax2 + bx + c. Which of the following is true? (A) It represents the maximum value when a > 0. (B) It represents the minimum value when a < 0. (C) Both (A) and (B) (D) Neither (A) nor (B) 8. Find the signs of the roots of the equation x2 + x – 420 = 0. (A) Both are positive. (B) Both are negative. (C) The roots are of opposite signs with the numerically larger root being positive. (D) The roots are of opposite signs with the numerically larger root being negative. 9. If k is a natural number and (k2 – 3k + 2) (k2 – 7k + 12) = 120, find k. (A) 7 (B) 6 (C) 5 (D) 9 10. Both A and B were trying to solve a quadratic equation. A copied the coefficient of x wrongly and got the roots of the equation as 12 and 6. B copied the constant term wrongly and got the roots as 1 and 26. Find the roots of the correct equation. (A) 6, 16 (B) -6, -16 (C) 24, 3 (D) -3, -24 11. If the roots of the equation (x – k1) (x – k2) + 1 = 0, k1 and k2 are integers, then which of the following must be true? (A) k1, k2 are two consecutive integers. (B) k2 – k1 = 2 (C) k1 – k2 = 2 (D) Either (B) or (C) 12. The roots of the equation ax2 + bx + c = 0 are k less than those of the equation px2 + qx + r = 0. Find the equation whose roots are k more than those of px2 + qx + r = 0. (A) ax2 + bx + c = 0 (B) a(x – 2k)2 + b(x – 2k) + c = 0 (C) a(x + 2k)2 + b(x + 2k) + c = 0 (D) a(x – k)2 + b(x – k) + c = 0

13. If one root of the equation x2 – 10x + 16 = 0 is half of one of the roots of x2 – 4Rx + 8 = 0. Find R such that both the equations have integral roots. (A) 1 (B) 2/3 (C) 3/2 (D) 4 14. If x + y = 4, find the maximum/minimum possible value of x2 + y2. (A) Minimum, 8 (B) Maximum, 8 (C) Maximum, 16 (D) Minimum, 16 15. Find positive integral value(s) of p such that the equation 2x2 + 8x + p = 0 has rational roots. (A) 8 (B) 4 (C) 6 (D) (A) or (C) 16. Two equations have a common root which is positive. The other roots of the equations satisfy x2 – 9x + 18 = 0. The product of the sums of the roots of the two equations is 40. Find the common root. (A) 1 (B) 2 (C) 3 (D) 4 3 2 17. If one root of the equation x –11x + 37x – 35 = 0 is 3 − 2 , then find the other two roots. 5, 3 − 2 (B) −5, 3 + 2 (A) (C) 5, 3 + 2 (D) −5, 3 − 2 18. The roots [the values of x (and not |x|)] of the equation |x|2 + 6|x| – 55 = 0 are a and b. One of the roots of py2 + qy + r = 0 is ab times the other root. Which of the following can be concluded? (A) 25q2 = –576pr (B) 25pr = –576q2 2 (C) 25q = 576pr (D) 25pr = 576q2 19. The sides of a right-angled triangle are such that the sum of the lengths of the longest and that of the shortest side is twice the length of the remaining side. Find the longest side of the triangle if the longer of the sides containing the right angle is 9 cm more than half the hypotenuse. (A) 30 cm (B) 25 cm (C) 20 cm (D) 15 cm 20. Solve for x: 2{32(1 + x)} – 4(32+x) + 10 = 0

(A) –1, log3 ⎛ 5 ⎞ ⎜⎝ ⎟⎠ 3 5 (C) –1, 3

(B) –1, log32 (D) –1, log3 ⎛ 3 ⎞ ⎜⎝ ⎟⎠ 5

then 21. If x 2 − 2 x − 3 + x 2 + 5 x − 24 = x 2 + 7 x − 30 , find x. (A) 2 (B) 3 (C) 4 (D) 6 22. Two software professionals Ranjan and Raman had 108 floppies between them. They sell them at different prices, but each receives the same sum. If Raman had sold his at Ranjan’s price, he would have received `722 and if Ranjan had sold his at Raman’s price, he would have received `578. How many floppies did Ranjan have? (A) 51 (B) 57 (C) 68 (D) 40

Chapter 10 Quadratic Equations | 1.141 23. The sum and product of the roots of a quadratic equation E are a and b respectively. Find the equation whose roots are the product of first root of E and the square of the second root of E, and the product of the second root of E and the square of the first root of E. (A) x2 – abx + b3 = 0 (B) x2 + abx + b3 = 0 (C) x2 + abx – b3 = 0 (D) x2 – abx – b3 = 0

24. Which of the following options represent(s) a condition for the equations x² + ax + b = 0 and x² + bx + a = 0 to have exactly one common root, given that the roots of both the equations are real? (A) a – b = 1 (B) b – a = 1 (C) 1 + a + b = 0 (D) Either (A) or (B) 2 25. If the roots of 2x + (4m + 1)x + 2(2m – 1) = 0 are reciprocals of each other, find m. (A) –1 (B) 0 (C) 1 (D) 3/4

Answer Keys

Exercise Practice Problems 1. D 11. D 21. B

2. C 12. B 22. A

3. C 13. C 23. A

4. D 14. A 24. C

5. A 15. D 25. C

6. C 16. B

7. D 17. C

8. D 18. A

9. B 19. A

10. C 20. A

Chapter 11 Inequalities LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Inequalities and modulus • Symbols and notations

ineQUaLities anD MODULUs If ‘a’ is any real number, then ‘a’ is either positive or negative or zero. When ‘a’ is positive, we write a > 0 which is read ‘a is greater than zero’. When ‘a’ is negative, we write a < 0 which is read ‘a is less than zero’. If ‘a’ is zero, we write a = 0 and in this case, ‘a’ is neither positive nor negative.

Symbols and Notations ‘>’ means ‘greater than’ ‘ b when a – b > 0 and 2. a < b when a – b < 0. For example, 3 is greater than 2 because 3 – 2 = 1 and 1 is greater than zero. –3 is less than –2 because –3 – (–2) = –1 and –1 is less than zero. Certain properties and useful results pertaining to inequalities are given below. A thorough understanding of these properties/ results is very essential for being able to solve the problems pertaining to inequalities. [In the following list of properties and results, numbers like a, b, c, d, etc. are real numbers] 1. For any two real numbers a and b, either a > b or a < b or a = b.

• Absolute value • Properties of modulus

If a > b, then b < a. If a b, then a ≤ b. If a > b and b > c, then a > c. If a < b and b < c, then a < c. If a > b, then a ± c > b ± c. If a > b and c > 0, then ac > bc. If a < b and c > 0, then ac < bc. If a > b and c < 0, then ac < bc. If a < b and c < 0, then ac > bc. If a > b and c > d, then a + c > b + d. If a < b and c < d, then a + c < b + d. Let A, G and H be the Arithmetic mean, Geometric mean and Harmonic mean of n positive real numbers. Then A ≥ G ≥ H, the equality occurring only when the numbers are all equal. 14. If the sum of two positive quantities is given, their product is the greatest when they are equal; and if the product of two positive quantities is given, their sum is the least when they are equal. 15. If a > b and c > d, then we cannot say anything conclusively about the relationship between (a – b) and (c – d); depending on the values of a, b, c and d, it is possible to have (a – b) > (c – d), (a – b) = (c – d) or (a – b) < (c – d) 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Absolute Value (written as |x| and read as ‘modulus of x’) For any real number x, the absolute value is defined as follows: ⎧ x, if x ≥ 0 and x =⎨ ⎩ − x, if x < 0

Chapter 11 Inequalities | 1.143

Properties of Modulus

Example 4: Solve for x if 4x2 – 21x + 20 > 0

For any real number x and y,

Solution: 4x2 – 21x + 20 > 0 ⇒ (4x – 5) (x – 4) > 0 Both factors are positive (i.e., the smaller is positive) or both 5 are negative (i.e. the greater is negative) i.e., x > 4 or x < 4 or it can be expressed in the interval notation as (4, ∞) ∪

1. x = 0 ⇔ |x| = 0 2. |x| ≥ 0 and –|x| ≤ 0 3. |x + y| ≤ |x| + |y| 4. ||x| – |y || ≤ |x – y| 5. –|x| ≤ x ≤ |x| 6. |x · y| = |x| · |y|

5⎞ ⎛ ⎜⎝ −∞, ⎟⎠ 4

x 2 + 5 x − 24 0 always lies between 2 and 2.8. x b \ lies between a 2 2.8 = 0.056 = 0.04 and 50 50 \

b < 1 a

\

a > b.

Solution: The modulus of any number is always non negative. \ |2x + 4| ≥ 0 \ The given inequality will never satisfy. The solution is null set. Example 8: Solve for x: |2x – 3| = 5 Solution: 2x – 3 = 5 or 2x – 3 = –5 (If | y| = a, y = ± a) ⇒ x = 4 or x = –1. Example 9: Find the maximum value of g(x) = 16 – |–x – 6|; x ∈ R. Solution: g(x) is maximum when |–x – 6| is minimum. The minimum value of the modulus of all numbers is 0. \ The maximum value of g(x) = 16 – 0 = 16.

1.144 | Quantitative Aptitude

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. If a < b and c < 0 then which of the following is true? a b (A) ac < bc (B) < c c (C) ac > bc (D) None of these 2. If p and q are two real numbers, then which of the following statements is always true? p (A) < 1 ⇒ p < q q p (B) p > 0, q > 0 and > 1 ⇒ p > q q p (C) > 1 ⇒ p > q q (D) All the above 3. If 5x – 8 < 2x + 9 and 4x + 7 > 7x – 8, then the range of the values of x that satisfies the inequalities is (A) (5, ∞) (B) (–∞, 5) ⎛ 17 ⎞ 17 ⎞ ⎛ (C) ⎜⎝ 5, ⎟⎠ (D) ⎜⎝ −∞, ⎟⎠ 3 3 4. Solve for real values of x; 5x2 – 3x – 2 ≥ 0. ⎡ −2 ⎤ ⎛ −2 ⎞ (A) R – ⎜ ,1⎟ ⎢ 5 ,1⎥ (B) ⎝ 5 ⎠ ⎣ ⎦ (C) [1, ∞) (D) R – (0, 1) 5. If x2 – 9x – 36 is negative, then find the range of x. (A) (-3, 12) (B) [-3, 12] (C) (-12, 3) (D) [-12, 3] 6. Which of the following is true? (A) x+ y ≤ x + y

(D) All the above

7. If 6x + 8 > 7x – 9 and 4x – 7 < 6x – 3, then the values of x is (A) (– 17, 2) (B) (2, 17) (C) (– 2, 17) (D) (– ∞, 17) 8. The solution set of the inequality x − 5 < 9 is (A) (0, 14) (B) (–4, 14) (C) (–4, 0) (D) (9, 14) 9. The number of integral values of x, that do not satisfy x+5 ≥ 0 is the inequation x−2 (A) 7 (C) 6

12. Which of the following is true? (A) 3031 < 3130 (B) 7169 > 7070 29 30 (C) (155) < (150) (D) Both (B) and (C) 21 maximum? 2 21 (B) 2 (D) 3

13. At what value of x is –|x – 3|+

(A) –3

(C) 0

14. Find the range of all real values of x if |3x + 5| < 5x – 11. (A) (8, ∞) (B) (– ∞, –5/3) ∪ (8, ∞) (C) (–5/3, 8) (D) (–5/3, ∞) 15. If ac = bd = 2, then the minimum value of a2 + b2 + c2 + d2 is (A) 4 (B) 6 (C) 8 (D) 16 16. If x, y > 0 and x + y = 3 then (A) xy ≤ 0.72 (B) xy ≤ 1.8 (C) xy ≤ 2.25 (D) xy ≤ 1.25 17. Find the complete range of values of x that satisfies |x – 16| > x2 – 7x + 24. ⎛ 3 5⎞ (A) (0, 2) (B) ⎜ , ⎟ ⎝ 2 2⎠ (C) (1, 3)

(D) (2, 4)

18. For which of the following range of values of x is x2 + x less than x3 + 1? (A) (–∞, –1) (B) (1, ∞) (C) (–1, 1) ∪ (1, ∞) (D) [–1, 1]

(C) x− y ≥ x − y

11. Find the range of the real values of x satisfying 8 – 3x ≤ 5 and 4x + 5 ≤ –7. (A) [–3, 1] (B) (– ∞, -3] ∪ [1, ∞) (C) (–3, 1) (D) f

x x (B) = , y ≠ 0 y y

10. If (x + 5) (x + 9) (x + 3)2 < 0, then the solution set for the inequality is (A) (–9, – 3) (B) (–9, –5) (C) (– 3, ∞) (D) (–9, ∞)

(B) 5 (D) 4

19. If x, y, z are positive, then the value of ( 4 x 2 + x + 4) (5 y 2 + y + 5) (7 z 2 + z + 7) can be xyz (A) 400 (B) 500 (C) 1000 (D) 1500 20. The range of x for which 2x2 – 5x – 8 ≤ |2x2 + x| is ⎡ 4 ⎞ ⎛ 4 ⎞ (A) ⎜⎝ − , −1⎟⎠ ⎢ − 3 , ∞⎟⎠ (B) 3 ⎣ A=

(C) [–1, ∞) (D) [–1, 2] 21. For how many integral values of x, is the inequation x −5 > 4 satisfied? x+7 (A) 5 (B) 4 (C) 2 (D) 3

Chapter 11 Inequalities | 1.145 22. If 1 ≤ x ≤ 3 and 2 ≤ y ≤ 5, then the minimum value of x+ y is y 3 1 6 5 (A) (B) (C) (D) 5 5 5 6

24. Find the number of solutions of the equation x − x − 2 = 6.

23. If b ≥ 5 and x = a b, which of the following is true?

x 2 y + y 2 z + z 2 x + xy 2 + yz 2 + zx 2 is xyz (A) 6 (B) 9 (C) 12

(A) a – xb > 0 (C) a + xb > 0

(B) a + xb < 0 (D) a – xb ≤ 0

(A) 2

(B) 1

(C) 3

(D) 4

25. If x, y and z are positive real numbers, then the minimum value of

(D) 14

Answer Keys

Exercise Practice Problems 1. C 11. D 21. D

2. B 12. C 22. C

3. B 13. D 23. D

4. B 14. A 24. B

5. A 15. C 25. A

6. D 16. C

7. C 17. D

8. B 18. C

9. A 19. D

10. B 20. A

Chapter 12 Progressions LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Progressions • Arithmetic progression (A.P.) • Geometric progression (G.P.)

• Inﬁnite geometric progression • Some important results

ProGressions In this chapter, we will look at the problems on sequences or progressions of numbers, where the terms of the sequence follow a particular pattern either addition of a constant (Arithmetic Sequence or Arithmetic Progression) or multiplication by a constant (Geometric Sequence or Geometric Progression). A third type of progression–Harmonic Progression–has also been defined later.

Arithmetic progression (A.p.) An arithmetic progression is a sequence of numbers in which any number (other than the first) is more (or less) than the immediately preceding number by a constant value. This constant value is called the common difference. In other words, any term of an arithmetic progression can be obtained by adding the common difference to the preceding term. Let a be the first term of an arithmetic progression, d the common difference and n the number of terms in the progression. The nth term is normally represented by Tn and the sum to n terms of an arithmetic progression is denoted by Sn Tn = nth term = a + (n – 1) d n × [2a + (n – 1) d], then the progression 2 can be represented as a, a + d, a + 2d, …, [a + (n – 1)d ]. Here, quantity d is to be added to any chosen term to get the next term of the progression. The sum to n terms of an arithmetic progression can also be written in a different manner. Sn = Sum of n terms =

Sum of first n terms =

n × [2a + (n – 1)d] 2

n = × [a + {a + (n – 1)d}] 2

But, when there are n terms in an arithmetic progression, a is the n first term and {a + (n – 1) d} is the last term. Hence, Sn = × [First 2 Term + Last Term] The average of all the terms in an arithmetic progression is called their Arithmetic Mean (A.M.). Since average is equal to {sum of all the quantities/number of quantities}, arithmetic progression must be equal to the Sum of the terms of the arithmetic progression divided by the number of terms in the arithmetic progression. Arithmetic Mean of n terms in arithmetic progression

Sn 1 = = {2a + (n – 1)d} n 2

1 = × (First Term + Last Term) 2

(First Term + Last Term) = 2

i.e., A.M. is the average of the first and the last terms of the A.P. Arithmetic Mean can also be obtained by taking the average of any two terms which are EQUIDISTANT from the two ends of the A.P. i.e., 1. The average of the second term from the beginning and the second term from the end will be equal to the A.M.

Chapter 12 Progressions | 1.147 2. The average of the third term from the beginning and the third term from the end will also be equal to the A.M. and so on. In general, the average of the kth term from the beginning and the kth term from the end will be equal to the A.M. Conversely, if the A.M. of an A.P. is known, the sum to n terms of the series (Sn) can be expressed as

Subtracting (3) from (4), 2d = 6 d = 3 Substituting d = 3 in (3) or (4), a = –8 given, tn = –8 + (n – 1)3 = 37

Sn = n × A.M.

n = 16

If three numbers are in arithmetical progression the middle number is called the Arithmetic Mean, i.e., if a, b, c are in a+c A.P., then b is the A.M. of the three terms and b = . 2

Example 3: Three terms in arithmetic progression have a sum of 45 and a product of 3240. Find them.

If a and b are in Arithmetic Progression (A.P.), then their ( a + b) A.M. = . 2 If three numbers are in A.P., we can represent the three numbers as (a – d), a and (a + d). If four numbers are in A.P., we can represent the four numbers as (a – 3d), (a – d), (a + d) and (a + 3d); (in this case, 2d is the common difference). If five numbers are in A.P., we can represent the five numbers as (a – 2d), (a – d), a, (a + d) and (a + 2d). Solved Examples Example 1: The sixth and the tenth terms of an arithmetic progression are 22 and 38 respectively. Find the first term and the common difference. Solution: Let the first term and the common difference be a and d respectively.

a + 5d = 22

(1)

a + 9d = 38

(2)

a – d + a + a + d = 45 a = 15 (a – d) a (a + d) = 3240 152 – d 2 = 216 d = ±3 If d = 3 the terms are 12, 15 and 18. If d = -3, the terms are same but in the descending order. Example 4: The first term and the last term of an arithmetic progression are 9 and 69 respectively. If the sum of all the terms is 468, find the number of terms and the common difference. Solution: Let the number of terms and the common difference be n and d respectively, n Sn = [9 + 69] = 468 2 ⇒ 39n = 468 n = 12 tn = 9 + 11d ⇒ 11d = 60 ⇒

Subtracting (1) from (2), 4d = 16, d = 4

d=

60 11

Example 5: The sum of three numbers which are in arithmetic progression is 24. The sum of their square is 200. Find the numbers.

Substituting d in (1) or (2), we get

Solution: Let the terms be a – d, a and a + d.

a = 2

Example 2: The 12th term, the 14th term and the last term of an arithmetic progression are 25, 31 and 37 respectively. Find the first term, common difference and the number of terms. Solution: Let the first term, the common difference and the number of terms be a, d and n respectively. Given that

a + 11d = 25

(3)

a + 13d = 31

(4)

Solution: Let the numbers be a – d, a and a + d. Given, a – d + a + a + d = 24 \ a = 8 (a – d)2 + a2 + (a + d)2 = 200 3a2 + 2d2 = 200 ⇒ \

d 2 = 4 d = ±2

If d = 2, the numbers are 6, 8 and 10. If d = –2, the numbers are same, but in the descending order.

1.148 | Quantitative Aptitude

Geometric Progression (G.p.) Numbers taken in a certain order, are said to be in Geometrical Progression, if the ratio of any (other than the first number) to the preceding one is the same. This ratio is called the Common Ratio. In other words, any term of a geometric progression can be obtained by multiplying the preceding number by the common ratio. The common ratio is normally represented by r. The first term of a geometric progression is denoted by a. A geometric progression can be represented as a, ar, 1ar 2, … where a is the first term and r is the common ratio of the geometric progression. nth term of the geometric progression is arn – 1. Sum to n terms:

(

a 1 − rn

)

or

(

)

a rn −1

1− r r −1 xar n−1 − a r × Last term − First term = = r −1 r −1 Thus the sum to n terms of a geometric progression can also be written as

Sn =

r × Last term − First term r −1

Harmonic progression: If the reciprocals of the terms of a sequence are in arithmetic progression, the sequence is said 1 1 1 to be a harmonic progression, For example, 1, , , , … 2 3 4 is a harmonic progression. In general, the sequence 1 1 1 , , , … is a harmonic progression. a a + d a + 2d If a, b, c are in harmonic progression, b is said to be the harmonic mean of a and c. In general, if x1, x2, …, xn are in harmonic progression x2, x3, …, .xn – 1 are the n – 2 harmonic means between x1 and xn.

Some Important Results The results of the sums to n terms of the following series are quite useful and hence should be remembered by students. n( n + 1) 2 Sum of squares of the first n natural numbers Sum of the first n natural numbers = ∑ n =

∑n

2

=

n( n + 1)( 2n + 1) 6

Sum of cubes of first n natural numbers

If n terms a1, a2, a3, …, an are in G.P., then the Geometric Mean (G.M.) of these n terms is given by = n a1 ⋅ a2 ⋅ a3 ⋅… ⋅ an If three terms are in geometric progression then the middle term is a Geometric Mean of the other two terms, i.e., if a, b and c are in G.P., then b is the geometric mean of the three terms and b² = ac. If there are two terms a and b, their geometric mean (G.M.) is given by ab. For any two unequal positive numbers a and b, their Arithmetic Mean is always greater than their Geometric Mean, i.e.

Example 6: Find the 7th term of the geometric progression whose first term is 6 and common ratio is 2.

For any two unequal positive numbers a and b. a+b > ab ; (a + b) > 2 ab b) > 2

1 64 and its common ratio as . Find the sum of its first five 2 terms.

2

2 n2 ( n + 1) 2 ⎡ n( n + 1) ⎤ ∑ n3 = ⎢⎣ 2 ⎥⎦ = 4 = ⎡⎣∑ n⎤⎦

Solution: nth term of a G.P. = ar n – 1 7th term = 6 (26) = 384 Example 7: A geometric progression has its first term as

a(1 − r n ) 1− r

When there are three terms in geometric progression, we can represent the three terms to be a/r, a and ar When there are four terms in geometric progression, we a a can represent the four terms as 3 , , ar and ar3. r r (In this case r2 is the common ratio)

Solution: Sum of the first n terms of a G.P. =

Infinite Geometric Progression

Example 8: Find the common ratio of the geometric pro1 gression whose first and last terms are 5 and respec25 624 tively and the sum of its terms is . 100

If –1 < r < +1 or | r | < 1, then the sum of a geometric progression does not increase infinitely; it ‘converges’ to a particular value. Such a G.P. is referred to as an infinite geometric progression. The sum of an infinite geometric progression a is represented by S∞ and is given by the formula S∞ = . 1− r

⎛ ⎛ 1⎞ 5 ⎞ 64 ⎜1 − ⎜ ⎟ ⎟ ⎝ ⎝ 2⎠ ⎠ = 124 Sum of its first five terms = 1 1− 2

Solution: Sum of the terms of a geometric progression whose common ratio is r, is given by

r (last term) − (first term) r −1

Chapter 12 Progressions | 1.149

a , a and ar. r

r = 4 or

729 (1 − r )[2(1 − r ) − (1 + r )] = 0 2 (1 – r) (1 – 3r) = 0

⇒

r ≠ 1 (∵ | r | < 1)

r=

1 3

Example 11: If | x | < 1, find the value of 3 + 6x + 9x2 + 12x3 + … Solution: Let S = 3 + 6x + 9x2 + 12x3 + …

(7)

xS = 3x + 6x + 9x +(8) 2

3

Subtracting (8) from (7) S(1 – x) = 3 (1 + x + x2 + …)

⎛ 1 ⎞ 3⎜ ⎝ 1 − x ⎟⎠ 3 = As |x| < 1, S = 1− x (1 − x ) 2

1 4

1 If r = 4, the numbers are 2, 8 and 32. If r = , the numbers 4 are same, but in the descending order. Example 10: The sum of the terms of an infinite geometric progression is 27. The sum of their squares is 364.5. Find the common ratio. Solution: Let the first term and the common ratio be a and r respectively. 2 a ⎛ a ⎞ = 27 ⇒ ⎜ Given that = 729 ⎝ 1 − r ⎟⎠ 1− r And

729 (1 − r )(1 + r ) = 0 2

729 (1 – r)2 –

\

a + a + ar = 42 r ⎛ a⎞ ⎜⎝ ⎟⎠ ( a)( ar ) = 512 r a = 8 8 + 8 + 8r = 42 r 8r2 – 34r + 8 = 0 8r2 – 32r – 2r + 8 = 0 (r – 4) (4r – 1) = 0

= 364.5 (1 – r2)

Example 9: Three numbers in geometric progression have a sum of 42 and a product of 512. Find the numbers. Solution: Let the numbers be

a2 = 729 (1 – r)2

⇒

⎛ 1⎞ r⎜ ⎟ −5 ⎝ 25 ⎠ 624 = . r − 1 100 ⇒ 4r – 500 = 624r – 624 1 ⇒ r= . 5

a2 = 364.5 1 − r2

Example 12: Evaluate Solution:

1 1 1 = – 1( 2) 1 2 1 1 1 = – 2(3) 2 3

Finally

1 1 1 1 + + +…+ . 1( 2) 2(3) 3( 4) 99(100)

1 1 1 = – 99(100) 99 100

The given expression is 1 –

1 99 = 100 100

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. The sixth term and the eleventh term of an arithmetic progression are 30 and 55 respectively. Find the twentyfirst term of the series. 1 1 (A) 88 (B) 105 (C) 110 (D) 92 2 3 2. What is the 15th term of an arithmetic progression whose first term is equal to its common difference and whose 3rd term is 9? (A) 15 (B) 30 (C) 45 (D) 60

3. If x + 4, 6x – 2 and 9x – 4 are three consecutive terms of an arithmetic progression, then find x. (A) 2 (B) 4 (C) 6 (D) 8 4. Find the number of terms and the sum of the terms of the arithmetic progression 32, 28, … 4. (A) 8; 144 (B) 7; 126 (C) 14; 252 (D) 15; 270 5. Find the sum of the first 31 terms of the arithmetic progression whose first term is 6 and whose common 8 difference is . 3 (A) 1410 (B) 1418 (C) 1426 (D) 1434

1.150 | Quantitative Aptitude 6. The sum of five terms of an arithmetic progression is 70. The product of the extreme terms is 132. Find the five terms. (A) 4, 8, 12, 16, 20 (B) 10, 12, 14, 16, 18 (C) 6, 10, 14, 18, 22 (D) 8, 12, 16, 20, 24 7. The sum to n terms of an arithmetic progression is 5n2 + 2n. Find the nth term of the series. (A) 10n + 5 (B) 10n – 3 (C) 5n – 1 (D) 5n – 2 8. Which term of the geometric progression 4, 4 2 , 8 … is 64 2 ? (A) 8 (B) 9 (C) 10 (D) 12 9. Find the sixth term of the geometric progression whose first term is 2 and common ratio is 3. (A) 96 (B) 486 (C) 1458 (D) 162 10. Find the sum of the first 4 terms of a geometric progression whose first term is 6 and whose common ratio is 2. (A) 90 (B) 84 (C) 96 (D) 102 11. What is the sum of the first 7 terms of a geometric progression whose first term is 1 and 4th term is 8? (A) 129 (B) 128 (C) 127 (D) None of these 12. If the sum to 37 terms of an arithmetic progression is 703, then find the middle term of the arithmetic progression. (A) 34 (B) 17 (C) 38 (D) 19 13. Find the sum of the 20 terms of the series 1, (1 + 2), (1 + 2 + 3), (1 + 2 + 3 + 4), (1 + 2 + 3 + 4 + 5), … (A) 1540 (B) 1435 (C) 1450 (D) 1345 2 14. If the real numbers a, c and b as well as a + b2, a2 + c2 and b2 + c2 are in geometric progression, then which of the following is necessary true? (A) a = b (B) b=c (C) a = c (D) a=b=c 15. How many numbers between 450 and 950 are divisible by both 3 and 7? (A) 20 (B) 24 (C) 30 (D) 35 2 3 16. S = 2 + 4x + 6x + 8x … where | x | < 1. Which of the following is the value of S ? 4 3 (A) (B) (1 − x ) 2 (1 − x ) 2 2 1 (C) (D) (1 − x ) 2 (1 − x ) 2

17. The sum of the first eight terms of a geometric progression. is 510 and the sum of the first four terms of the geometric progression. is 30. Find the first term of the geometric progression., given that it is positive. (A) 2 (B) 4 (C) 6 (D) 8 18. Find the integer value of y, if –x, 2y and 2(y + 3) are in arithmetic progression and (x + 2), 2(y + 1) and (5y – 1) are in geometric progression. (A) 2 (B) 3 (C) 4 (D) 5 19. Find the number of terms common to the progressions 2, 8, 14, 20, …, 98 and 6, 10, 14, 18, …, 102. (A) 7 (B) 6 (C) 8 (D) 9 20. Find the sum of the series 2 + 3x + 4x2 + 5x3 + … to infinity, if | x | < 1. 2− x 2+ x (B) (A) 2 (1 − x ) (1 + x )2 2− x 2+ x (C) (D) 2 (1 + x ) (1 − x )2 21. The mean of the sequence 3, 8, 17, 30, …, 1227 is ______. (A) 531 (B) 431 (C) 314 (D) 315 22. Find the value of –12 + 22 – 32 + 42 – 52 + 62 + … –192 + 202 (A) 210 (B) 420 (C) 630 (D) 720 23. Find the sum of the given terms in the following series: 1 1 1 1 + + + ... + 3 +1 3+ 5 5+ 7 119 + 121 (A) 2 3 +1

(B) 5

(D) 10

(C) 11 – 2 3

24. If log3 x + log 3 3 x + log 5 3 x + … + log 23 3 x = 432, then find x. (A) 9 (B) 27 (C) 3 3 (D) 81 25. The sum of the first n terms of two arithmetic progressions S1 and S2 are in the ratio 11n – 17 : 5n – 21. Find the ratio of the 16th terms of S1 and S2. (A) 3 : 2 (B) 162 : 67 (C) 9 : 4 (D) 27 : 8

Answer Keys

Exercise Practice Problems 1. B 11. C 21. B

2. C 12. D 22. A

3. A 13. A 23. B

4. A 14. A 24. B

5. C 15. B 25. B

6. C 16. C

7. B 17. A

8. C 18. A

9. B 19. C

10. A 20. A

Chapter 13 Permutations and Combinations LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Permutations • Combinations • Number of linear permutations of ‘n’ dissimilar items taken ‘r’ at a time without repetition • Number of arrangements of n items of which p are of one type, q are of a second type and the rest are distinct

introDuCtion Permutations and Combinations is one of the important areas in many exams because of two reasons. The first is that solving questions in this area is a measure of students’ reasoning ability. Secondly, solving problems in areas like Probability requires thorough knowledge of Permutations and Combinations. Before discussing Permutations and Combinations, let us look at what is called as the ‘fundamental rule’. ‘If one operation can be performed in ‘m’ ways and (when, it has been performed in any one of these ways), a second operation then can be performed in ‘n’ ways, the number of ways of performing the two operations will be m × n’. This can be extended to any number of operations. If there are three cities A, B and C such that there are 3 roads connecting A and B and 4 roads connecting B and C, then the number of ways one can travel from A to C is 3 × 4, i.e., 12. This is a very important principle and we will be using it extensively in Permutations and Combinations. Because we use it very extensively, we do not explicitly state every time that the result is obtained by the fundamental rule but directly write down the result.

Permutations Each of the arrangements which can be made by taking some or all of a number of items is called a Permutation. Permutation implies ‘arrangement’ or that ‘order of the items’ is important.

• Number of arrangements of n distinct items where each item can be used any number of times • Total number of combinations • Dividing given items into groups • Circular permutations • Rank of a word • Arrangements

The permutations of three items a, b and c taken two at a time are ab, ba, ac, ca, cb and bc. Since the order in which the items are taken is important, ab and ba are counted as two different permutations. The words ‘permutation’ and ‘arrangement’ are synonymous and can be used interchangeably. The number of permutations of n things taking r at time is denoted by nPr (and read as ‘nPr’).

ComBinations Each of the groups or selections which can be made by taking some or all of a number of items is called a Combination. In combinations, the order in which the items are taken is not considered as long as the specific things are included. The combination of three items a, b and c taken two at a time are ab, bc and ca. Here, ab and ba are not considered separately because the order in which a and b are taken is not important but it is only required that a combination including a and b is what is to be counted. The words ‘combination’ and ‘selection’ are synonymous. The number of combinations of n things taking r at time is denoted by nCr (and read as ‘nCr’). Number of linear permutations of ‘n’ dissimilar items taken ‘r’ at a time without repetition (nPr) Consider r boxes each of which can hold one item. When all the r boxes are filled, what we have is an arrangement of r items

1.152 | Quantitative Aptitude taken from the given n items. So, each time we fill up the r boxes with items taken from the given n items, we have an arrangement of r items taken from the given n items without repetition. Hence the number of ways in which we can fill up the r boxes by taking things from the given n things is equal to the number of permutations of n things taking r at a time. Boxes

.......................... 1 2 3 4 r

The first box can be filled in n ways (because any one of the n items can be used to fill this box). Having filled the first box, to fill the second box we now have only (n – 1) items; any one of these items can be used to fill the second box and hence the second box can be filled in (n – 1) ways; similarly, the third box in (n – 2) ways and so on the rth box can be filled in {n – (r – 1)} ways, i.e. [n – r + 1] ways. Hence, from the Fundamental Rule, all the r boxes together be filled up in n × (n – 1) × (n – 2) … (n – r + 1) ways So,

Pr = n × (n – 1) × (n – 2) … (n – r + 1)

n

This can be simplified by multiplying and dividing the right hand side by (n – r) (n – r – 1) … 3.2.1 giving us nPr = n(n – 1) (n – 2) … [n – (r – 1)] =

( n − 1)( n − 2)…[n − ( r − 1).( n − r )… 3.2.1] n! = ( n − r )……… 3.2.1 ( n − r )!

The number of permutations of n distinct items taking r items at a time is n! n pr = ( n − r )! If we take n items at a time, then we get nPn. From a discussion similar to that we had for filling the r boxes above, we can find that nPn is equal to n! The first box can be filled in n ways, the second one in (n – 1) ways, the third one in (n – 2) ways and so on, then the nth box in 1 way; hence, all the n boxes can be filled in n(n – 1) (n – 2) … 3.2.1 ways, i.e., n! ways. Hence, n

Pn = n!

But if we substitute r = n in the formula for nPr, then we n! get nPn = ; since we already found that nPn = n!, we can 0! conclude that 0! = 1 Number of combinations of n dissimilar things taken r at a time. Let the number of combinations nCr be x. Consider one of these x combinations. Since this is a combination, the order of the r items is not important. If we now impose the condition that order is required for these r items, we can get r! arrangements from this one combination. So each combination can give rise to r! permutations. x combinations will thus give rise to x . r! permutations. But since these are

all permutations of n things taken r at a time, this must be equal to nPr. So, n! x.r! = nPr = ( n − r )! n! n ⇒ Cr = r !.( n − r )! The number of combinations of n dissimilar things taken all at a time is 1. Out of n things lying on a table, if we select r things and remove them from the table, we are left with (n – r) things on the table – that is, whenever r things are selected out of n things, we automatically have another selection of the (n – r) things. Hence, the number of ways of making combinations taking r out of n things is the same as selecting (n – r) things out of n given things, i.e., Cr = nCn - r

n

When we looked at nPr, we imposed two constraints which we will now release one by one and see how to find out the number of permutations. Number of arrangements of n items of which p are of one type, q are of a second type and the rest are distinct When the items are all not distinct, then we cannot talk of a general formula for nPr for any r but we can talk of only n Pn (which is given below). If we want to find out nPr for a specific value of r in a given problem, we have to work on a case to case basis (this has been explained in one of the solved examples). The number of ways in which n things may be arranged taking them all at a time, when p of the things are exactly alike of one kind, q of them exactly alike of another kind, r of them exactly alike of a third kind, and the rest all disn! . tinct is p! q! r ! Number of arrangements of n distinct items where each item can be used any number of times (i.e., repetition allowed) You are advised to apply the basic reasoning given while deriving the formula for nPr to arrive at this result also. The first box can be filled up in n ways; the second box can be filled in again inn ways (even though the first box is filled with one item, the same item can be used for filling the second box also because repetition is allowed); the third box can also be filled in n ways and so on … the rth box can be filled in n ways. Now all the r boxes together can be filled in {n.n.n.n … r times} ways, i.e., nr ways. The number of permutations of n things, taken r at a time when each item may be repeated once, twice, … up to r times in any arrangement is nr. What is important is not this formula by itself but the reasoning involved. So, even while solving problems of this type, you will be better off if you go from the basic reasoning and not just apply this formula.

Chapter 13 Permutations and Combinations | 1.153 Total number of combinations: Out of n given things, the number of ways of selecting one or more things is where we can select 1 or 2 or 3 … and so on n things at a time; hence the number of ways is nC1 + nC2 + nC3 + … + nCn This is called ‘the total number of combinations’ and is equal to 2n – 1 where n is the number of things. The same can be reasoned out in the following manner also. There are n items to select from. Let each of these be represented by a box. 1 2 3 4 n No. of ways of dealing ............... with the boxes 2 2 2 2 2 The first box can be dealt with in two ways. In any combination that we consider, this box is either included or not included. These are the two ways of dealing with the first box. Similarly, the second box can be dealt with in two ways, the third one in two ways and so on, the nth box in two ways. By the Fundamental Rule, the number of ways of dealing with all the boxes together in 2 × 2 × 2 × … n times ways, i.e., in 2n ways. But out of these, there is one combination where we ‘do not include the first box, do not include the second box, do not include the third box and so on, do not include the n th box.’ That means, no box is included. But this is not allowed because we have to select one or more of the items (i.e., at least one item). Hence this combination of no box being included is to be subtracted from the 2n ways to give the result of Number of ways of selecting one or more items from n given items is 2n – 1 Dividing given items into groups: Dividing (p + q) items into two groups of p and q items respectively. Out of (p + q) items, if we select p items (which can be done in p+qCp ways), then we will be left with q items, i.e., we have two groups of p and q items respectively. So, the number of ways of dividing (p + q) items into two groups of p and q items respectively is equal to p+qCp which is equal ( p + q)! to p! q! The number of ways of dividing (p + q) items into two ( p + q)! groups of p and q items respectively is p !⋅ q ! If p = q, i.e., if we have to divide the given items into two EQUAL groups, then two cases arise

1. When the two groups have distinct identity and 2. When the two groups do not have distinct identity.

In the first case, we just have to substitute p = q in the above formula which then becomes

The number of ways of dividing 2p items into two equal ( 2 p )! groups of p each is where the two groups have ( p !) 2 distinct identity. In the second case, where the two groups do not have distinct identity, we have to divide the above result by 2!, i.e., it then becomes The number of ways of dividing 2p items into two equal (2 p)! where the two groups do not groups of p each is 2 !( p !) 2 have distinct identity. Dividing (p + q + r) items into three groups consisting of p, q and r items respectively The number of ways in which (p + q + r) things can be divided into three groups containing p, q and r things ( p + q + r )! respectively is p!q!r ! If p = q = r, i.e., if we have to divide the given items into three EQUAL groups, then we have two cases where the three groups are distinct and where the groups are not distinct. When the three groups are distinct, the number of ways (3 p)! is ( p !)3 When the three groups are not distinct, then the number (3 p)! of ways is 3!( p !)3

Circular Permutations When n distinct things are arranged in a straight line taking all the n items, we get n! permutations. However, if these n items are arranged in a circular manner, then the number of arrangements will not be n! but it will be less than that. This is because in a straight line manner, if we have an arrangement ABCDE and if we move every item one place to the right (in cyclic order), the new arrangement that we get EABCD is not the same as ABCDE and this also is counted in the n! permutations that we talked of. However, if we have an arrangement ABCDE in a circular fashion, by shifting every item by one place in the clockwise direction, we still get the same arrangement ABCDE. So, if we now take n! as the number of permutations, we will be counting the same arrangement more than once. The number of arrangements in circular fashion can be found out by first fixing the position of one item. Then the remaining (n – 1) items can be arranged in (n – 1)! ways. Now even if we move these (n – 1) items by one place in the clockwise direction, then the arrangement that we get will not be the same as the initial arrangement because one item is fixed and it does not move.

1.154 | Quantitative Aptitude Hence, the number of ways in which n distinct things can be arranged in a circular arrangement is (n – 1)! The number of circular arrangements of n distinct items is (n – 1)! if there is DIFFERENCE between clockwise and anticlockwise arrangements and (n – 1)!/2 if there is NO DIFFERENCE between clockwise and anticlockwise arrangements. The number of diagonals in an n-sided regular polygon An n-sided regular polygon has n vertices. Joining any two vertices we get a line of the polygon which are nC2 in number. Of these nC2 lines, n of them are sides. Hence n( n − 3) diagonals are nC2 – n = 2 Number of integral solution of the equation x1 + x2 + … + xn = s Consider the equation x1 + x2 + x3 = 10 If we consider all possible integral solutions of this equation, there are infinitely many. But the number of positive (or non-negative) integral solutions is finite. We would like the number of positive integral solutions of this equation, i.e., values of (x1, x2, x3) such that each xi > 0. We imagine 10 identical objects arranged on a line. There are 9 gaps between these 10 objects. If we choose any two of these gaps, we are effectively splitting the 10 identical objects into 3 parts of distinct identity. Conversely, every split of these 10 objects corresponds to a selection of 2 gaps out of the 9 gaps. Therefore, the number of positive integral solutions is 9 C2. In general, if x1 + x2 + … + xn = s where s ≥ n, the number of positive integral solutions is s–1Cn–1. If we need the number of non negative integral solutions, we proceed as follows. Let a1, a2, … be a non-negative integral solution. Than a1 + 1, a2 + 1, …, an + 1 is a positive integral solution of the equation x1 + x2 + … + xn = s + n. Therefore, the number of non-negative integral solutions of the given equation is equal to the number of positive integral solutions of x1 + x2 + … + xn = s + n, which is s+n–1Cn-1. For x1 + x2 + x3 + … + xn = s where s ≥ 0, the number of positive integral solutions (when s ≥ n) is s–1Cn-1 and the number of non-negative integral solutions is n+s–1Cn–1

D(n) = n! ⎡1 − 1 + 1 − 1 + … + ( −1) n 1 ⎤ ⎢ 1! 2 ! 3! n ! ⎥⎦ ⎣ For example, when n = 3, the number of derangements is

⎡ 1 1 1⎤ D(3) = 3! ⎢1 − + − ⎥ = 2 and when n = 4, ⎣ 1! 2 ! 3! ⎦

⎡ 1 1 1 1⎤ D(4) = 4! ⎢1 − + − + ⎥ = 9 ⎣ 1! 2 ! 3! 4 ! ⎦ 2. The total number of ways in which a selection can be made by taking some or all out of p + q + r + … things where p are alike of one kind, q alike of a second kind, r alike of a third kind and so on is [{(p + 1)(q + 1)(r + 1) …} – 1].

n+1 3. Cr = nCr + nCr–1 and nPr = r.n – 1Pr – 1 + n – 1Pr

Solved Examples Example 1: Consider the word PRECIPITATION. Find the number of ways in which (i) a selection (ii) an arrangement of 4 letters can be made from the letters of this word. Solution: The word PRECIPITATION has 13 letters I, I, I, P, P, T, T, E, R, C, A, O, N of 9 different sorts. In taking 4 letters, the following are the possibilities to be considered.

(a) all 4 distinct. (b) 3 alike, 1 distinct. (c) 2 alike of one kind, 2 alike of other kind. (d) 2 alike, 2 other distinct.

Selections (a) 4 distinct letters can be selected from 9 distinct letters (I, P, T, E, R, C, A, O, N) in 9C4 = 126 ways. (b) As 3 letters have to be alike, the only possibility is selecting all the I’s. Now the 4th letter can be selected from any of the remaining 8 distinct letters in 8C1 = 8 ways. (c) Two pairs of two alike letters can be selected from I’s, Q’s and T ’s in 3C2 = 3 ways. (d) The two alike letters can be selected in 3C1 = 3 ways and the two distinct letters can now be selected from the 8 distinct letters in 8C2 = 28 ways. Hence required number of ways are 3 × 28 = 84.

Some additional points

Hence, the total selections are 126 + 8 + 3 + 84 = 221.

1. Suppose there are n letters and n corresponding addressed envelopes. The numbers of ways of placing these letters into the envelopes such that no letter is placed in its corresponding envelope is often referred as derangements. The number of derangements of n objects is given by

Arrangements: For arrangements, we find the arrangements for each of the above selections and add them up. (a) As the 4 letters are distinct, there are 4! arrangements for each selection. Hence required arrangements are 126 × 4! = 3024.

Chapter 13 Permutations and Combinations | 1.155 4! 3! arrangements for each of the selection. Hence 4! required arrangements are 8 × = 32. 3! 4! (c) The required arrangements here are 3 × = 18 2! 2! 4! (d) The required arrangements are 84 × = 1008. 2! Total number of arrangements are 3024 + 32 + 18 + 1008 = 4082. (b) Since 3 of the 4 letters are alike, there are

Example 2: How many four letter words can be formed using the letters of the word ‘ROAMING’? Solution: None of the letters in the word are repeated. \ The number of four letter words that can be formed = 7P4 7! = (7) (6) (5) (4) = 840. = 3! Example 3: In a party, each person shook hands with every other person present. The total number of hand shakes was 28. Find the number of people present in the party. Solution: Let the number of people present in the party be n. Method 1: The first people shakes hands with a total of (n – 1) persons, the second with (n – 2) other people and so on. The total number of hand shakes is (n – 1) + (n – 2) + … + 2 + 1 ⇒

n( n −1) = 28 (given) 2 n = 8

Method 2: Number of hand shakes = Number of ways of selecting 2 people out of n = nC2. n

⇒

C2 = 28

n( n −1) = 28 2! n = 8

Directions for Examples 4 to 7: The following examples are based on the data below. The letters of NESTLE are permuted in all possible ways. Example 4: How many of these words begin with T? Solution: NESTLE has 6 letters of which the letter E occurs two times. There fore the required number of words = Number of ways of filling N, E, S, E and L in the second to 5! sixth positions = = 60. 2! Example 5: How many of these words begin and end with E? Solution: The required number of words = The number of ways of filling N, S, T and L in the second to fifth positions = 4! = 24.

Example 6: How many of these words begin with S and end with L? Solution: The required number of words = The number of ways of filling N, E, T and E in the second to fifth positions 4! = = 12. 2! Example 7: How many of these words neither begin with S nor end with L? Solution: The required number of words = The total number of words which can be formed using the letters N, E, S, T and E – (Number of words which begin with S or 6! end with L) = – (Number of words beginning with S + 2! Number of words ending with L – Number of words beginning with S and ending with L)

6 ! ⎛ 5 ! 5 ! 4 !⎞ −⎜ + − ⎟ = 2 ! ⎝ 2 ! 2 ! 2 !⎠ = 360 – (60 + 60 – 12) = 252.

Example 8: How many of these words begin with T and do not end with N? Solution: The required number of words = The number of words beginning with T – The number of words beginning 5! 4 ! with T and ending with N = − = 48. 2! 2! Directions for Examples 8 to 11: The following examples are based on the data below. The letters of FAMINE are permuted in all possible ways. Example 9: How many of these words have all the vowels occupying odd places? Solution: FAMINE has 3 vowels and 3 consonants. The vowels can be arranged in the odd places in 3! or 6 ways. The consonants would have to be arranged in even places. This is possible in 3! or 6 ways as well. \ The required number of words = 62 = 36. Example 10: How many of these words have all the vowels together? Solution: If all the vowels are together, the vowels can be arranged in 3! ways among themselves. Considering the vowels as separate a unit and each of the other letters as a unit, we have a total of 4 units which can be arranged in 4! ways. \ The required number of words = 4! 3! = 144 Example 11: How many of these words have at least two of the vowels separated? Solution: The required number of words = The total number of words which can be formed using the letters F, A,

1.156 | Quantitative Aptitude M, I, N and E – The number of words with all the vowels together = 6! – 4! 3! = 576. Example 12: How many of these words have no two v owels next to each other? Solution: To ensure that no two vowels are together, we first arrange the 3 consonants say –c1 – c2 – c3 – and place the vowels in the gaps between the consonants or the initial or final position. For each arrangement of the consonants, there are 4 places where the vowels can go. The vowels can be dealt with in 4 (3) (2) ways. \ The total number of words is 3! 4! = 144. Directions for Examples 13 and 14: The following examples are based on the data below. A committee of 5 is to be formed from 4 women and 6 men. Example 13: In how many ways can it be formed if it consists of exactly 2 women? Solution: The committee must have 2 women and 3 men. \ The required number of ways = 4C2 6C3 = 120. Example 14: In how many ways can it be formed if it consists of more women than men?

Solution: The committee must have either 4 women and 1 man or 3 women and 2 men. \ The required number of ways = 4C4 6C1 + 4C3 6C2 = 6 + 60 = 66. Example 15: Find the number of four-digit numbers which can be formed using four of the digits 0, 1, 2, 3 and 4 without repetition. Solution: The first digit has 4 possibilities (1, 2, 3 and 4). The second digit has 4 possibilities (0 and any of the three digits not used as the first digit). The third digit has 3 possibilities. The last digit has 2 possibilities. \ The required number of numbers = (4) (4) 3 (2) = 96. Example 16: The number of diagonals of a regular polygon is four times the number of its sides. How many sides does it have? Solution: Let the number of sides in the polygon be n. \ \

n( n − 3) = 4n 2 n(n – 11) = 0; n > 0 n – 11 = 0; n = 11

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. A man has 12 blazers, 10 shirts and 5 ties. Find the number of different possible combinations in which he can wear the blazers, shirts and ties. (A) 27 (B) 300 (C) 240 (D) 600 2. How many different words can be formed by using all the letters of the word INSTITUTE? 9! 9! 9! (B) 9 ! (C) (D) (A) 2! 3! 3! 2 ! 3. In how many ways can a cricket team of 11 members be selected from 15 players, so that a particular player is included and another particular player is left out? (A) 216 (B) 826 (C) 286 (D) 386 4. A group contains n persons. If the number of ways of selecting 6 persons is equal to the number of ways of selecting 9 persons, then the number of ways of selecting four persons from the group is (A) 1365 (B) 273 (C) 455 (D) 285 5. The number of ways of arranging 10 books on a shelf such that two particular books are always together is (A) 9! 2! (B) 9! (C) 10! (D) 8

6. Find the number of ways of inviting at least one among 6 people to a party. (A) 26 (B) 26 – 1 (C) 62 (D) 62 – 1 7. An eight-letter word is formed by using all the letters of the word ‘EQUATION’. How many of these words begin with a consonant and end with a vowel? (A) 3600 (B) 10800 (C) 2160 (D) 720 8. A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can the committee be formed such that it contains more men than women? (A) 180 (B) 186 (C) 126 (D) 66 9. In how many ways can 10 boys and 10 girls be arranged in a row so that all the girls sit together? (A) 10! (B) 11! (C) 20! (D) 10! 11! 10. In how many ways can 6 boys and 5 girls be arranged in a row so that boys and girls sit alternately? (A) (6!)2 (B) (5!)2 (C) 6! 5! (D) 2.5! 6!

Chapter 13 Permutations and Combinations | 1.157 11. There are seven letters and corresponding seven addressed envelopes. All the letters are placed randomly into the envelopes – one in each envelope. In how many ways can exactly two letters be placed into their corresponding envelopes? (A) 44 (B) 924 (C) 308 (D) 189 12. We are given 3 different green dyes, 4 different red dyes and 2 different yellow dyes. The number of ways in which the dyes can be chosen so that at least one green dye and one yellow dye is selected is (A) 336 (B) 335 (C) 60 (D) 59 13. Prahaas attempts a question paper that has 3 sections with 6 questions in each section. If Prahaas has to attempt any 8 questions, choosing at least two questions from each section, then in how many ways can he attempt the paper? (A) 18000 (B) 10125 (C) 28125 (D) 9375 14. Find the number of selections that can be made by taking 4 letters from the word INKLING. (A) 48 (B) 38 (C) 28 (D) 18 15. A man has (2n + 1) friends. The number of ways in which he can invite at least n + 1 friends for a dinner is 4096. Find the number of friends of the man. (A) 11 (B) 15 (C) 17 (D) 13 16. How many four-digit numbers are there between 3200 and 7300, in which 6, 8 and 9 together or separately do not appear? (A) 1421 (B) 1420 (C) 1422 (D) 1077 17. Raju has forgotten his six-digit ID number. He remembers the following: the first two digits are either 1, 5 or 2, 6, the number is even and 6 appears twice. If Raju uses a trial and error process to find his ID number at the most, how many trials does he need to succeed? (A) 972 (B) 2052 (C) 729 (D) 2051

18. A matrix with four rows and three columns is to be formed with entries 0, 1 or 2. How many such distinct matrices are possible? (D) 212 (A) 12 (B) 36 (C) 312 19. In how many ways can 4 postcards be dropped into 8 letter boxes? 8 (A) P4 (B) 48 (C) 84 (D) 24 20. In how many ways can 12 distinct pens be divided equally among 3 children? 12 ! 12 ! (A) 4 (B) (3!) ( 4 !)3 3! 12 ! 12 ! (C) (D) 3! 4 ! ( 4 !)3 21. If all possible five-digit numbers that can be formed using the digits 4, 3, 8, 6 and 9 without repetition are arranged in the ascending order, then the position of the number 89634 is (A) 91 (B) 93 (C) 95 (D) 98 22. Manavseva, a voluntary organisation has 50 members who plan to visit 3 slums in an area. They decide to divide themselves into 3 groups of 25, 15 and 10. In how many ways can the group division be made? 50 ! (A) 25! 15! 10! (B) 25! 15! 10 ! (C) 50! (D) 25! + 15! + 10! 23. In how many ways is it possible to choose two white squares so that they lie in the same row or same column on an 8 × 8 chessboard? (A) 12 (B) 48 (C) 96 (D) 60 24. The number of four digit telephone numbers that have at least one of their digits repeated is (A) 9000 (B) 4464 (C) 4000 (D) 3986 25. There are 4 identical oranges, 3 identical mangoes and 2 identical apples in the basket. The number of ways in which we can select one or more fruits from the basket is (A) 60 (B) 59 (C) 57 (D) 55

Answer Keys

Exercise Practice Problems 1. D 11. B 21. C

2. D 12. A 22. B

3. C 13. C 23. C

4. A 14. D 24. B

5. A 15. D 25. B

6. B 16. D

7. B 17. B

8. B 18. C

9. D 19. C

10. C 20. D

Chapter 14 Data Interpretation LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Methods of presenting data • Data table

introDuCtion Not a day passes without our coming across figures and statistics. Study and manipulation of such data leads us to an important area – Data Interpretation. Data can be organised in a number of ways so that larger volume of data can be presented in a more compact and precise form. Data thus presented has to be deciphered correctly by the user of the data. This process of deciphering the data from its compactly presented form is called Data Interpretation.

Methods of Presenting Data Numerical data can be presented in one or more of the following ways 1. 2. 3. 4. 5. 6. 7. 8. 9.

Data Tables Pie Charts 2-Dimensional Graphs Bar Charts 3-Dimensional Graphs Venn Diagrams Geometrical Diagrams Pert Charts Others

The ‘Others’ category covers miscellaneous forms like descriptive case format, etc. customised for the situation. Data can also be presented by using a combination of two or more of the above forms. While some data can be presented in many different forms, some other data may be amenable to be presented only in a few ways. In real life situations, the style of data presentation is based

• Pie-charts • Two-dimensional graphs

on the end-objective. In certain situations data has to be presented as a combination of two or more forms of data presentation. Let us understand each of the above forms of data presentation with an example.

Data taBle Here data is presented in the form of simple table. While any type of data can be presented in tabular form, that too in a very accurate manner, interpreting the data in tabular form is more difficult and time consuming than the other modes, all of which are basically pictorial or graphical in presentation. Data tables can be of a number of types. They can be of a singletable variety or combination of tables. Some examples of tables are given below. Table 1 Movement of goods by different modes of transport (in 000’s of metric-ton-kms) Year

Road

Rail

Air

Water

Total

1985

1000

1500

120

20

2640

1986

1600

2000

129

24

3753

1987

2907

3090

139

28

6164

1988

4625

5200

152

27

10004

1989

6346

7540

174

33

14093

1990

7920

10250

212

40

18422

1991

9540

13780

266

50

23636

Chapter 14 Data Interpretation | 1.159 From the table we can deduce the following: 1. Rate of growth by each mode of transport in successive years as well as cumulative annual growth. 2. Rate of growth of total haulage by all modes of transport together in any year. 3. Contribution by each mode of transport to the total haulage in any given year. 4. Trends of growth over time for various modes of transport. 5. Given the cost of transportation for each mode, we can calculate the total annual cost of transportation over the years for various modes of transport as well make a cost comparison. 6. Finding out the mode of transportation in any given year that forms the largest percentage of total haulage. 7. For a given mode of transport, finding out the year in which the percentage increase in haulage over the previous year was the highest.

East 18%

West 22%

Central 12% South 25%

North 23%

Figure 1 Distribution of population in geographical zones

From the above pie chart, we can calculate the following: 1. Population in any zone given the total population. 2. Population of any zone as a percentage of that of another zone. 3. Percentage increase in the total population given the percentage increase in the population of one or more zones. Pie Charts are also very frequently used in combination with other forms of data or along with other Pie-Charts.

Pie-Charts This is probably the simplest of all pictorial forms of data presentation. Here, total quantity to be shown is distributed over one complete circle or 360 degrees. In pie-charts, data is essentially presented with respect to only one parameter (unlike in two and 3-dimensional graphs described later). This form essentially presents shares of various elements as proportion or percentage of the total quantity. Each element or group in the pie-chart is represented in terms of quantity (or value, as the case may be) or as the angle made by the sector representing the elements or as a proportion of the total or as a percentage of the total. Figure 1 gives distribution of the population in different geographical zones.

Two-Dimensional Graphs This is essentially used for continuous data but can also be used for depicting discrete data provided we understand the limitation. Also known as Cartesian Graphs, they represent variation of one parameter with respect to another parameter each shown on a different axis. These types of graphs are useful in studying the rate of change or understanding the trends through extrapolations. These graphs can be of various types and a few of them are shown below (Figures 2 and 3):

FOREIGN EXCHANGE (CRORES OF RS.) 6000 5000 4000 3000 2000 1000 0 Figure 2 Foreign exchange reserves of India

The graph in Figure 2 shows the changes in the foreign exchange reserves of our country during a period of time. One can find out trends and the growth rates of foreign exchange reserves.

1.160 | Quantitative Aptitude (000's of UNITS) 140 120 100 80 60 40 20 0 1988

1989

300 ltr Model

Year

1990

165 ltr Model

1991

Double Door Model

Figure 3 Refrigerator sales of company ABC

Figure 3 shows model wise sales of refrigerators during four years. From this graph we can obtain the following: 1. Percentage contribution of each model to the company’s total sales for four years. 2. Relative increase or decrease in the share of each model. 3. Sales trend of various models. Using this bar chart one can carry out a detailed performance evaluation of the company with respect to the sales Class

Number of students

Cricket

Volleyball

Basketball

Football

6

120

60%

70%

50%

60%

7

140

50%

60%

60%

50%

8

160

40%

65%

55%

45%

9

180

65%

75%

65%

55%

10

240

70%

80%

75%

45%

Solved Examples Example 1: How many students in the school like cricket? (A) 436 (B) 432 (C) 491 (D) 511 Solution: (C) Number of students who like Cricket

of the four year period 1988 to 1991 for any given model. These bar charts can also be depicted horizontally. Another variation could be showing each product at one place (rather than each year at one place). Example: These questions are based on the following table, which gives the details of the sports which students in all the classes of a school like. The table gives the number of students in each class and the percentage of students in it who like Cricket, Volleyball, Basketball and Football.

60 50 40 (160) + (120) + (140) + 100 100 100 65 70 (180) + ( 240) = 491 100 100 =

Example 2: By what percentage is the number of students who like Volleyball in class 6 more/less than those who like Basketball in class 10? (A) 40% (B) 50% (C) 53.33% (D) 56.67%

Solution: (C) Number of students who like Volleyball in class 6 70 = (120) = 84 100 Number of students who like Basketball in class 10 75 = (240) = 180 100 180 − 84 84 is less than 180 by (100) 180 = 53.33% . Example 3: The number of students who like Cricket in class 7 is what percentage of the number of students who like Football in class 8? (A) 88% (B) 93.5% (C) 95.6% (D) 97.2%

Chapter 14 Data Interpretation | 1.161 Solution: (D) Number of students who like Cricket in class 7 = 50 (140) = 70 100 Number of students who like Football in class 8 45 = (160) = 72 100 70 Required percentage = × 100 72 = 97.2% Example 4: In how many of the given classes can more than 90 students like all the three games? (A) 2 (B) 3 (C) 1 (D) 0 Solution: (A) In any class, the maximum value of the number of students who like all the three games would be the number of students who like the game liked by the least number of students. In class 6, the percentage of students who like a game is the least for Basketball. Number of those who like basket50 ball = (120) = 60 < 90 100 In class 7, the percentage of students who like a game is the least for cricket and football. Number of students who like cricket 50 = (140) = 70 < 90 100

In class 8, the number of students who like a game is the least for cricket. Number of students who like cricket = 40 (160) = 64 < 90 100 In class 9, the percentage of students who like a game is the least for Football. Number of those who like Football =

55 (180) = 99 > 90 100

In class 10, the percentage of students who like a game is the least for Football. Number of students who like Football =

45 (240) = 108 > 90 100

\ In two classes, more than 90 students can like all the games. Example 5: What can be the maximum percentage of students in class 6 who do not like any of the given games? (A) 40% (B) 10% (C) 50% (D) 30% Solution: (D) In class 6, the maximum percentage of students who like a game = Percentage of students who like Volleyball i.e., 70%. Percentage of students who like at least one game would be minimum when all students who like other games are the ones who like Volleyball. \ Maximum percentage required = 100 – 70 = 30%.

Practice Exercises Practice Problems 1 Directions for question 1: Select the correct alternative from the given choices. 1. The table shows the total marks of four students P, Q, R and S in all their subjects for the two years 2012 and 2013. Students

2012

2013

P

997

1295

Q

664

876

R

585

732

S

480

689

How many students had a percentage Increase in their total marks of more than 35% from 2012 to 2013? (A) 1 (B) 2 (C) 3 (D) 4 Directions for questions 2 to 4: These questions are based on the following data which gives some details of new states joining the United States of America across time. Union rank

Population

Number of representatives in the house of representatives

State

Capital

Joined the union

Washington

Olympia

Nov, 11, 1889

42

62,87,759

9

Texas

Austin

Dec, 29, 1845

28

2,28,59,968

32

Delaware

Dover

Dec, 7, 1781

1

8,43,524

1

Virginia

Raleigh

Nov, 21, 1789

12

86,83,242

13

Minnesota

St. Paul

May, 11, 1832

32

51,32,799

8

Kansas

Topeka

Jan, 29, 1861

34

27,44,687

4

Illinois

Springfield

Dec, 3, 1818

21

1,27,63,371

19

(Continued)

1.162 | Quantitative Aptitude

Union rank

Number of representatives in the house of representatives

State

Capital

Joined the union

Population

New Hampshire

Concord

June 21, 1788

9

13,09,940

2

Arizona

Phoenix

Feb, 14, 1912

48

59,39,292

8

Hawai

Honolulu

Aug 21, 1959

50

12,75,194

2

Indiana

Indianapolis

Dec 11, 1816

19

62,71,973

9

Vermont

Montpelier

March 14, 1791

14

6,23,050

1

Nebraska

Lincoln

March 1, 1867

37

17,58,787

3

Georgia

Atlanta

Jan 2, 1788

4

9,07,256

13

Union rank is the chronological order in which the states joined the Union. 2. How many states joined the Union from March 1, 1867 to Feb 14, 1912? (A) 11 (B) 12 (C) 13 (D) 14 3. If it is known that the House of Representatives of USA has a strength of 535 members, then the number of representatives in the House of Representatives of the

given states will form what approximate percentage of the total strength of the House of Representatives? (A) 16 (B) 19 (C) 21 (D) 23 4. In how many of the given states is the population less than 15 million but the number of representatives is not less than six? (A) 4 (B) 5 (C) 6 (D) 7

Directions questions 5 to 7: These questions are based on the following graphs. Number of projects undertaken in Mexico

450 400 350 300 250 200 150 100 50 0 1995

1996

1997

1998

1999

2000

2001

Num ber of projects undertaken in Texas 700 600 500 400 300 200 100 0 1995

1996

1997

5. In how many years was the number of projects undertaken in Mexico greater than that in Texas? (A) 3 (B) 6 (C) 5 (D) 4

1998

1999

2000

2001

6. How many projects were undertaken in the year 1998 in both places together? (A) 1000 (B) 900 (C) 750 (D) 500

Chapter 14 Data Interpretation | 1.163 7. In which of the following years was the average (arithmetic mean) number of projects undertaken in both places the highest? (A) 2000 (B) 1996 (C) 1998 (D) 1999 Directions for questions 8 to 10: The table below shows the percentages of colleges offering the courses mentioned in medicine in four states New York, New Jersey, Illinois and Michigan in a certain year. SI. No.

Course

New York

New Jersey

Illinois

Michigan

1.

Biochemistry

86

80

74

68

2.

Biophysics

74

92

88

64

3.

Biomechanics

59

82

84

68

4.

Biostatistics

56

84

86

70

The total number of colleges offering courses in medicine in the four states is 2000. The percentage wise distribution of the number of colleges in the four states is as shown below. New Jersey 15%

(C) Exactly two states. (D) Exactly one state. 9. What percentage of the colleges in the four states do not offer Biophysics as well as Biochemistry? (A) 41% (B) 36% (C) 34% (D) Cannot the determined 10. What is the total number of colleges offering Biostatistics in all the four states? (A) 1392 (B) 1408 (C) 1432 (D) 1476 Directions for questions 11 and 12: Study the given table and answer the questions that follow. Percentage break-up of the workers working in six different factories J, K, L, M, N and P. Factory

Men

Women

Boys

J

4800

50

37.5

12.5

New York 35%

Illinois 20%

Michigan 30%

8. The number of colleges offering the Biochemistry course is more than 230 in (A) All the four states. (B) Exactly three states.

Percentage

Total no. of workers

K

8750

40

36

24

L

5250

24

56

20

M

12000

35

25

40

N

8500

38

30

32

P

2700

45

40

15

11. By what percent is the number of women working in factory P is more than the number of boys working in factory J? 4 (A) 20% (B) 44 % 9 (C) 80% (D) 180% 12. What is the ratio of the number of men working in factory M to the number of women working in factory L? (A) 7 : 10 (B) 10 : 7 (C) 7 : 5 (D) 5 : 7

Directions for questions 13 to 15: These questions are based on the following line graph. Numbers of employees of two companies each years over the period 2000 to 2005 (in thousands) 60 50 40 30 20 10 0 2000

2001

2002

2003

2004

Years Moon Macro Systems

Equlibrium Inc.

2005

1.164 | Quantitative Aptitude 13. What is the ratio of the average number of employees of company Moon Macro Systems for the period 2001, 2002 and 2003 and the average number of employees of company Equilibrium Inc. for the same period? (A) 2 : 5 (B) 3 : 5 (C) 3 : 4 (D) 4 : 3 14. During which of the following years was the percentage increase in the number of employees of company Equilibrium Inc. over that in the previous year the highest?

15.

(A) 2000 (B) 2001 (C) 2002 (D) 2004 What is the approximate percentage increase in the total number of employees of the two companies from 2004 to 2005? (A) 8.5% (B) 9% (C) 9.5% (D) 10.5%

Directions for questions 16 and 17: These questions are based on the following bar graph which gives the production and sales of a company across five years from 2004 to 2008. 7000 5750

6000 5000

5100

4750

4500

4000

3750

5000 4250

4500

4400

3500

3000 2000 1000 0 2004

2005 Production

16. In the given period what percentage was the average production more than the average sales?

(A) 20%

(B) 23%

(C) 25%

(D) 28%

Practice Problems 2 Directions for questions 1 to 3: These questions are based on the following table which shows the Btu content of common types of energy. Type of energy

Btu content

1 barrel of Crude oil (42 gallons)

5,838,000

1 gallon of gasoline

124,000

1 gallon of heating oil

139,000

1 barrel of residual fuel oil

6,287,000

1 cubic foot of natural gas

1,026

1 gallon of propane

91,000

1 short ton of coal

20,681,000

1 kilowatt hour of electricity

3,412

Note 1: British Thermal Unit (Btu) represents the amount of energy required to raise the temperature of 1 pound of water by 1°F. Also 1 calorie represents the amount of energy

2006

2007

2008

Sales

17. The percentage increase/decrease in the total sales of the company in a given year with respect to that in the previous year was highest in which of the following years? (A) 2005 (B) 2006 (C) 2007 (D) 2008

required to raise the temperature of 2.2 pounds (1 kg) of water by 1.8°F (1°C). Note 2: 1 Joule = 0.2390 calories and 1 Btu = 1.06 kilojoules 1. The Btu content in 42 gallons of heating oil is equal to the Btu content in which of the following? (A) 60 gallons of propane (B) 42 gallons of crude oil (C) 48 gallons of gasoline (D) 57 cubic feet of natural gas 2. The Btu content in 60 gallons of propane is less than that in one short ton of coal by (A) 15,221,000 units (B) 16,290,000 units (C) 17,820,000 units (D) 18,630,000 units 3. 1 Btu is equal to _____ calories (A) 228.72 (B) 244.68 (C) 253.34 (D) 262.44

Chapter 14 Data Interpretation | 1.165 Directions for questions 4 to 6: These questions are based on the following data which shows the number of representatives of different political parties in the Parliament of UK which has three wings – The Northern Ireland Assembly, The Scottish Parliament and the National Assembly of Wales. Northern ireland assembly

Political parties

Males

Females

DUP

36

12

UUP

10

SDLP

24

Others

18

Political parties

Males

Females

Labour

48

8

Labour

12

4

3

Democrats

16

4

SDLP

15

2

2

Conservative

24

6

Conservative

6

5

3

Republicans

14

9

Others

9

7

6. The number of parliamentarians who are graduates in the National Assembly of Wales is 45. How many parliamentarians in the National Assembly of Wales are either females or graduates? (A) 27 (B) 30 (C) 32 (D) Cannot be determined Directions for questions 7 to 9: These questions are based on the table and the pie chart given below. Details of Airports in USA

Atlanta

Political parties

Females

5. If there are six Catholic parliamentarians for each party in the Scottish Parliament, then how many Scottish parliamentarians are not members of the Conservative party nor are Catholics? (A) 27 (B) 42 (C) 60 (D) 81

Name of the Airport

National assembly of wales

Males

4. Only among parties with at least 25 members in any wing, the number of young parliamentarians is more than eight. What is the maximum number of parliamentarians in the two wings – Northern Ireland Assembly and Scottish Parliament who are not young? (A) 201 (B) 144 (C) 165 (D) 176

Flights per day

Scottish parliament

Arrivals

Departures

Number of passengers per day (in hundreds)

1420

1380

560

JFK

1380

1420

700

Florida

1300

1500

360

Chicago

1440

1560

720

National

1200

1300

360

Los Angeles

920

880

468

Dallas

800

700

405

Detroit

690

710

392

Seattle

1100

1000

441

Classification of passengers in each of of the flights First class 10%

Business class 15% Economy class 75%

7. How many airports handle more than 1500 flights per day? (A) 5 (B) 6 (C) 7 (D) 8 8. Which of the following airports carries the maximum number of passengers per flight per day? (A) Chicago (B) JFK (C) Detroit (D) Atlanta 9. How many passengers (in hundreds) travel in the Economy class at JFK Airport? (A) 310 (B) 490 (C) 525 (D) 595

1.166 | Quantitative Aptitude Directions for questions 10 to 12: These questions are based on the graph and the table given below. Distribution of students who passed in different subjects in the year 2000 in a school Biological Science Mathematics History Literature Physics Chemistry 0

5

10

15 (in hundreds)

20

25

30

Percentage change in the number of students who passed over the previous year in the school 2000

2001

Biological Science

10%

20%

Mathematics

15%

20%

History

–25%

–10%

Literature

30%

–20%

Physics

20%

20%

Chemistry

15%

25%

Note: The negative sign indicates a decrease. 10. What is the ratio of the number of students who passed in History in 1999 to that in 2001? (A) 9 : 4 (B) 8 : 5 (C) 40 : 27 (D) 6 : 5 11. By what percent did the number of students who passed in Chemistry increase from 1999 to 2001? (A) 35% (B) 37.5% (C) 40% (D) 43.75%

12. In which of the following pairs of subjects did an equal number of students pass in the year 2001? (A) Mathematics and Literature (B) Literature and Chemistry (C) Mathematics and Physics (D) History and Physics

Directions for questions 13 to 15: These questions are based on the graph, pie chart and the table given below.

Percentage Rate

Interest rates in U.S.A. across 1995 to 1999. 7 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 1995

1996

1997

1998

1999

Chapter 14 Data Interpretation | 1.167 Distribution of loans availed of by various sectors of the industry ofor each of the years from 1995 to 1999 Others 30°

Services 60°

Power 90°

Agriculture 45°

Manufacturing 135°

Total loan amount availed of by the industry (In Billions of US$) 1995

36

1996

48

1997

72

1998

108

1999

144

13. In the years 1998 and 1999 combined the total interest paid by exactly one sector as a percentage of the total interest paid by all the sectors of the industry cannot be which of the following? 1 2 (A) 8 % (B) 16 % 3 3 (C) 25% (D) 40% 14. ENRON Inc. has availed 25% of the loan pertaining to one of the sectors in 1997. What is the minimum

possible amount of interest that they could have paid in 1997 in millions of dollars? [Consider others to be a single sector] (A) 60.75 (B) 66.75 (C) 72.75 (D) 78.75 15. The total interest paid in 1995 and in 1996 was more than $390 million for (A) at least three sectors (B) exactly one sector (C) exactly two sectors (D) exactly three sectors

Directions for questions 16 to 18: Answer the questions on the basis of the information given below. In the bubble graph below, the horizontal axis shows the weight of the product in pounds and the vertical axis shows the price in dollars. The relative size of the circles indicates how many of the products were sold. The largest circle indicates 4000 products and there is a decrease of 10% in the number of products for an incremental decrease in the size of the circles. 30000

P

Price in dollars

Q

R

S

25000

T

U

20000 V

15000

10000

2.1

2.2

2.3

2.4 2.5 2.6 Weight in pound

2.7

2.8

1.168 | Quantitative Aptitude 16. A shop sells one unit each of all products having a weight of 2.2 to 2.5 pound. The sales generated in dollars is ______ (A) 42500 (B) 70000 (C) 47500 (D) 72500 17. The number of the heaviest product sold was ______ that of any product shown and was the _______ (A) greater than; most expensive (B) greater than; least expensive (C) lesser than; most expensive (D) lesser than; most expensive 18. The total numbers of products of type U and T sold were ______. (A) 5400 (B) 6000 (C) 6840 (D) 8000 Directions for questions 19 to 21: These questions are based on the information given below. Six activities – A, B, C, D, E and F – need to be performed. Time required for each activity is given in the table below, along with the list of activities that need to be completed before that activity is started.

Activity

Time to complete (in min.)

Activities to be performed before

A

50

–

B

60

A, C

C

45

–

D

40

B, C

E

30

D

F

20

B

Assume that two or more activities can be performed simultaneously. 19. Find the minimum time in which all the activities can be completed. (A) 180 minutes (B) 175 minutes (C) 195 minutes (D) 225 minutes 20. If the work is to be completed in the least possible time, then what is the maximum possible time gap between starting the activities D and F? (A) 20 minutes (B) 40 minutes (C) 50 minutes (D) 80 minutes 21. Now, two new activities – Y, taking 10 minutes, and Z, taking 30 minutes – are included. If Y and Z are to be completed before A and D respectively, then what is the least possible time in which the entire work is completed? (A) 205 minutes (B) 210 minutes (C) 180 minutes (D) 190 minutes

Previous Years’ Questions 1. If 137 + 276 = 435, how much is 731 + 672? [2010] (A) 534 (B) 1403 (C) 1623 (D) 1513 2. 5 skilled workers can build a wall in 20 days, 8 semi-skilled workers can build the wall in 25 days, 10 unskilled workers can build the wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? [2010] (A) 20 (B) 18 (C) 16 (D) 15 3. From the digits 2, 2, 3, 3, 3, 4, 4, 4, 4, how many distinct 4-digit numbers greater than 3000 can be formed? [2010]

(A) 50 (C) 52

(B) 51 (D) 54

4. If log(P) = (1/2) log (Q) – (1/3) log (R) then which of the following options is TRUE? [2011] 2 3 2 2 (A) P = Q R (B) Q = PR (C) Q2 = R3P (D) R = P3Q2 5. P, Q, R and S are four types of dangerous microbes recently found in a human habitat. The area of each circle with its diameter printed in brackets represents the growth of a single microbe surviving human immunity system within 24 hours of entering the body. The danger to human beings varies proportionately with the toxicity, potency and growth attributed to a microbe shown in the figure below:

Toxicity (miligrams of microbe required to destroy half of the body mass in kilograms)

Chapter 14 Data Interpretation | 1.169 1000 P (50 mm) 800 600 Q (40 mm)

400

R (30 mm)

S (20 mm)

200 0 0.2

0.4

0.6

0.8

1

Potency (Probability that microbe will overcome human immunity system) A pharmaceutical company is contemplating the development of a vaccine against the most dangerous microbe. Which microbe should the company target in its first attempt? [2011] (A) P (B) Q (C) R (D) S 6. The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quantity produced. The fixed cost (F) of production of same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V + F)? [2011] (A) 5 (B) 4 (C) 7 (D) 6 7. A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day?[2011] (A) 4 (B) 5 (C) 6 (D) 7 8. A container originally contains 10 litres of pure spirit. From this container 1 litre of spirit is replaced with 1 litre of water. Subsequently 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container?[2011] (A) 7.58 litres (B) 7.84 litres (C) 7 litres (D) 7.29 litres 9. The cost function for a product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of `50 per unit. The number of units to be produced by the firm such that the profit is maximized is[2012] (A) 5 (B) 10 (C) 15 (D) 25 10. Which of the following assertions are CORRECT? P: Adding 7 to each entry in a list adds 7 to the mean of the list

Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list R: Doubling each entry in a list doubles the mean of the list S: Doubling each entry in a list leaves the standard deviation of the list unchanged [2012] (A) P, Q (B) Q, R (C) P, R (D) R, S 11. A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x – 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is[2012] (A) 8 meters (B) 10 meters (C) 12 meters (D) 14 meters 12. An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X ’s shock absorbers, 96% are reliable. Of Y ’s shock absorbers, 72% are reliable. The probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is [2012] (A) 0.288 (B) 0.334 (C) 0.667 (D) 0.720 13. What will be the maximum sum of 44, 42, 40, …? [2013] (A) 502 (B) 504 (C) 506 (D) 500 14. A tourist covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and the rest by cycle at 10 km/h. The average speed of the tourist in km/h during his entire journey is [2013] (A) 36 (B) 30 (C) 24 (D) 18

1.170 | Quantitative Aptitude

25. If x is real and | x 2 − 2 x + 3 = 11, then possible values of | − x 3 + x 2 − x | include[2014]

[2014]

19. The roots of ax2 + bx + c = 0 are real and positive. a, b and c are real. Then ax2 + b|x| + c = 0 has [2014] (A) no roots (B) 2 real roots (C) 3 real roots (D) 4 real roots 20. Round–trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In addition, groups of 4 or more get a discount of 5% on the total fare. If the one way signle person fare is `100, a group of 5 tourists purchasing round - trip tickets will be charged `_______. [2014] 21. In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were further asked to mention whether they own a car or scooter or both. Their responses are tabulated below. What percent of respondents do not own a scooter? [2014]

Own vehicle

Men

Women

Car

40

34

Scooter

30

20

Both

60

46

20

50

Do not own vehicle

22. When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines? _______. [2014] 23. What is the average of all multiples of 10 from 2 to 198? [2014] (A) 90 (B) 100 (C) 110 (D) 120 2 4. The value of (A) 3.464 (C) 4.000

12 + 12 + 12 +… is[2014] (B) 3.932 (D) 4.444

(B) 2, 14 (D) 14, 52

26. The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students doubled in 2009, by what percent did the number of male students increase in 2009? [2014]

2

1⎞ 1⎞ ⎛ ⎛ 18. If ⎜ z + ⎟ = 98, compute ⎜ z 2 + 2 ⎟ . ⎝ ⎠ ⎝ z z ⎠

(A) 2, 4 (C) 4, 52

Ratio of male to female students

15. The current erection cost of a structure is `13,200. If the labour wages per day increase by 1/5 of the current wages and the working hours decrease by 1/24 of the current period, then the new cost of erection in ` is [2013] (A) 16,500 (B) 15,180 (C) 11,000 (D) 10,120 16. Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7? [2013] (A) 13/90 (B) 12/90 (C) 78/90 (D) 77/90 17. Find the sum of the expression 1 1 1 1 + + +…+ 1+ 2 2+ 3 3+ 4 80 + 81 [2013] (A) 7 (B) 8 (C) 9 (D) 10

3.5 3 2.5 2 1.5 1 0.5 0 2008

2009

2010

(A) 1 : 1 (C) 1.5 : 1

2011

2012

(B) 2 : 1 (D) 2.5 : 1

27. The table below has question – wise data on the performance of students in an examination. The marks for each questions are also listed. There is no negative or partial marking in the examination. Q.No.

Marks

Answered correctly

Answered wrongly

Not attempted

1

2

21

17

6

2

3

15

27

2

3

2

23

18

3

What is the average of the marks obtained by the class in the examination? [2014] (A) 1.34 (B) 1.74 (C) 3.02 (D) 3.91 28. The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012–2013. For international comparison, the GDP is compared in US Dollars (USD) after conversion based on the market exchange rate. During the period 2012–2013 the exchange rate for the USD increased from `50/- USD to `60/USD. India’s GDP in USD during the period 2012–2013 [2014] (A) increased by 5% (B) decreased by 13% (C) decreased by 20% (D) decreased by 11% 29. Consider the equation : (7256)8 – (Y)8 = (4364)8, where (X)N stands for X to the base N. Find Y. [2014] (A) 1634 (B) 1737 (C) 3142 (D) 3162

Chapter 14 Data Interpretation | 1.171 30. Based on the given statements, select the most appropriate option to solve the given question. If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building? Statements:[2015] 3 (I) Each step is foot high. 4 (II) Each step is 1 foot wide. (A) Statement I alone is sufficient but statement II alone is not sufficient. (B) Statement II alone is sufficient, but statement I alone is not sufficient. (C) Both statements together are sufficient, but neither statement alone is sufficient. (D) Statement I and II together are not sufficient. 31. Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals 16?[2015] (A) 0.20 (B) 0.25 (C) 0.30 (D) 0.33 32. The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5:4. There are 40 males in Electrical Engineering. What is the difference between the numbers of female students in the Civil department and the female students in the Mechanical department?[2015]

Electrical 20% Computer Science 40%

Mechanical 10%

Civil 30%

(A) Only relation I is true. (B) Only relation II is true. (C) Relations II and III are true. (D) Relations I and III are true. 3 4. The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking. Q.No.

Marks

Answered Correctly

Answered Wrongly

Not Attempted

1

2

21

17

6

2

3

15

27

2

3

1

11

29

4

4 5

2

23

18

3

5

31

12

1

What is the average of the marks obtained by the class in the examination?[2015] (A) 2.290 (B) 2.970 (C) 6.795 (D) 8.795 35. Based on the given statements, select the most appropriate option to solve the given question. What will be the total weight of 10 poles each of same weight?[2015] Statements: (I) One fourth of the weight of a pole is 5 kg. (II) The total weight of these poles is 160 kg more than the total weight of two poles. (A) Statement I alone is not sufficient. (B) Statement II alone is not sufficient. (C) Either I or II alone is sufficient. (D) Both statements I and II together are not sufficient. 36. Consider a function f (x) = 1 – | x | on –1 ≤ x ≤ 1. The value of x at which the function attains a maximum and the maximum value of the function are:[2015] (A) 0, –1 (B) –1, 0 (C) 0, 1 (D) –1, 2 37. In a triangle PQR, PS is the angle bisector of ∠QPR and ∠QPS = 60o. What is the length of PS ?[2015] P

33. The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c:[2015] (I) p + m + c = 27/20 (II) p + m + c = 13/20 (III) (p) × (m) × (c) = 1/10

r

Q

q s p

R

(q + r ) qr (B) qr (q + r )

(A)

(q + r ) 2 qr

(C) (q 2 + r 2 ) (D)

1.172 | Quantitative Aptitude 38. If p, q, r, s are distinct integers such that:[2015] f (p, q, r, s) = max(p, q, r, s) g(p, q, r, s) = min(p, q, r, s) h(p, q, r, s) = remainder of (p × q)/(r × s) if (p × q) > (r × s) or remainder of (r × s)/(p × q) if (r × s) > (p × q) Also a function fgh(p, q, r, s) = f (p, q, r, s) × g(p, q, r, s) × h(p, q, r, s) Also the same operations are valid with two variable functions of the form f (p, q) What is the value of fg(h(2, 5, 7, 3), 4, 6, 8)? 39. If the list of letters, P, R, S, T, U is an arithmetic sequence, which of the following are also in arithmetic sequence?[2015] I. 2P, 2R, 2S, 2T, 2U II. P–3, R–3, S–3, T–3, U–3 III. P2, R2, S 2, T 2, U 2 (A) I only (B) I and II (C) II and III (D) I and III 40. A function f(x) is linear and has a value of 29 at x = -2 and 39 at x = 3. Find its value at x = 5.[2015] (A) 59 (B) 45 (C) 43 (D) 35 41. The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given in the following bar chart. In which year is the combined percentage increase in imports and exports the highest?[2015] 120 110 100 90 80 70 60 50 40 30 20 10 0

Exports

Imports

Quarter/ Product Q1 Q2 Q3 Q4

Elegance

Smooth

Soft

Executive

27300 25222 28976 21012

20009 19392 22429 18229

17602 18445 19544 16595

9999 8942 10234 10109

Which product contributes the greatest frication to the revenue of the company in that year? (A) Elegance (B) Executive (C) Smooth (D) Soft 7 44. If f (x) = 2x +3x– 5, which of the following is a factor of f (x)? [2016] (A) (x3+8) (B) (x – 1) (C) (2×-5) (D) (x+1) 45. In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ___. [2016] (A) 40.00 (B) 46.02 (C) 60.01 (D) 92.02 46. In a quadratic function, the value of the product of the α n + βn [2016] roots (α β) is 4. Find the value of − n α + β− n (A) n4 (B) 4n 2n-1 (C) 2 (D) 4n-1 47. In a 2 × 4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?[2016] (A) 21 (B) 27 (C) 30 (D) 36 48.

2000 2001 2002 2003 2004 2005 2006 2007

42. A cube is built using 64 cubic blocks of side one unit. After it is built, one cubic block is removed from every corner of the cube. The resulting surface area of the body (in square units) after the removal is _____. [2016] (A) 56 (B) 64 (C) 72 (D) 96 43. A shaving set company sells 4 different types of razors, Elegance, smooth, soft and Executive. Elegance sells at Rs. 48, Smooth at Rs. 63, soft at Rs.78 and Executive at Rs. 173 per price. The table below shows the numbers of each razor sold in each quarter of a year.

Choose the correct expression for f(x) given in the graph.[2016] (A) f(x) = 1 – |x – 1| (B) f(x) = 1 + |x – 1| (C) f(x) = 2 – |x – 1| (D) f(x) = 2 + |x – 1|

Chapter 14 Data Interpretation | 1.173

Answer Keys

Exercises Practice Problems 1 1. A 11. C

2. B 12. B

3. D 13. C

4. D 14. C

5. A 15. D

6. C 16. B

7. A 17. B

8. A

9. D

10. B

4. A 14. D

5. D 15. A

6. D 16. B

7. C 17. B

8. C 18. C

9. C 19. A

10. C 20. C

4. B 14. C 24. C 34. C 44. B

5. D 15. B 25. D 35. C 45. B

6. A 16. D 26. C 36. C 46. B

7. C 17. B 27. C 37. B 47. C

8. D 18. 96 28. D 38. 8 48. C

9. A 19. D 29. C 39. B

10. C 20. 850 30. A 40. C

Practice Problems 2 1. B 11. D 21. D

2. A 12. C

3. C 13. D

Previous Years’ Questions 1. C 11. B 21. 48% 31. A 41. 2006

2. D 12. B 22. 6 32. 32 42. D

3. B 13. C 23. B 33. D 43. B

Test Quantitative Aptitude Directions for questions 1 to 30: Select the correct alternative from the given choices. 1. P, Q, R and S have a total amount of `220 with them. P has `30 more than Q. S has half the amount with Q. R has `10 more than S. Find the amount with S (in `). (A) 20 (B) 40 (C) 30 (D) 50 2. In a room, there are some girls and some benches. If 5 girls sit on each bench, three girls will have no bench to sit on. If there is one bench less, 6 girls can sit on each bench. Find the number of benches. (A) 6 (B) 8 (C) 7 (D) 9 3. A test has 60 questions. Each correct answer fetches 1 mark. For each wrong answer and each unanswered question 1 mark is deducted. A candidate who wrote this test scored 20 marks. Find the number of questions he correctly answered. (A) 50 (B) 45 (C) 35 (D) 40 a+b−c a−b+c c+b−a = = then x (b – a) + y (a – c) + x y z z (c – b) = (A) 0 (B) 2 (C) 3 (D) 1 5. The value of a diamond varies directly with the cube of its weight. It broke into two pieces whose weights are in the ratio 3 : 4. The loss due to breakage is `504000. Find its initial value (in `). (A) 1029000 (B) 686000 (C) 1372000 (D) 1715000 6. The average of 25 observations is 120. By mistake one of the observations, 144, is taken as 169. Find the average of the 25 observations, after the mistake is corrected. (A) 120 (B) 119 (C) 125 (D) 132

4. If

7. The average of 13 observations is 50. The average of first seven observations is 45 and the average of last seven observations is 52. Find the value of seventh observation. (A) 41 (B) 30 (C) 29 (D) 62 8. A container contains 100 litres of milk. 10 litres of milk is replaced by 10 litres of water. From the solution formed, 10 litres of solution is replaced by 10 litres of water and this process is repeated one more time. Find the percentage of water in the resulting solution. (A) 33.3% (B) 67% (C) 36.5% (D) 27.1% 9. In a 729 ml of solution, the ratio of acid to water is 7 : 2. How much more water should be mixed so that the resulting mixture contains acid and water in the ratio 7 : 3 (in ml)? (A) 100 (B) 40 (C) 37 (D) 81 10. A merchant buys sulphuric acid at a certain rate per gallon and after mixing it with water, sells it at the same rate. If the merchant makes a profit of 20%, how many gallons of water is there per gallon of acid? (A) 0.2 (B) 0.5 (C) 0.7 (D) 0.25

Time: 30 min. 11. If A travelled a certain distance at 6 kmph, he would have reached his destination 10 minutes early. If he travelled it at 4 kmph, he would have reached his destination 10 minutes late. Find the speed at which he must travel to reach his destination on time (in kmph). (A) 5 (B) 5.4 (C) 4.8 (D) 4.5 12. A car travelled the first hour of its journey at 30 kmph, the next 5 hours of its journey at 50 kmph and the remaining 4 hours of its journey at 75 kmph. Find its average speed for its journey (in kmph). (A) 56 (B) 60 (C) 58 (D) 62 13. Without stoppages, a train can cover 54 km in an hour. With stoppages it can cover 36 km in an hour. Find its stoppage time per hour in a journey it covers with stoppages (in minutes). (A) 15 (B) 18 (C) 20 (D) 12 14. A and B can complete a job in 40 days. B and C can complete it in 30 days. A and C can complete it in 20 days. Find the time taken by A to complete it (in days). 180 240 (A) (B) 30 (C) 48 (D) 7 7 15. 3 men and 4 women can complete a job in 10 days. 24 men and 2 women can complete it in 2 days. Find the time taken by five men and 10 women to complete it (in days). (A) 4 (B) 5 (C) 3 (D) 6 16. Abhilash spends 25% of his income towards rent, 20% of the remaining income towards food, 8% of the remaining towards medical expenses, and 25% of the remaining towards miscellaneous expenses. If he saves `82,800, what is his income? (A) `2,00,000 (B) `2,25,000 (C) `2,40,000 (D) `2,50,000 17. By selling 30 articles, a shopkeeper gained the selling price of 10 articles. Find the profit percent. (A) 20% (B) 30% (C) 50% (D) 40% 18. When 1036 is divided by N, the remainder is 12 and when 1545 is divided by N, the remainder is 9. Find the greatest possible value of N. (A) 128 (B) 512 (C) 250 (D) 64 19. Five bells toll at regular intervals of 10, 15, 20, 25 and 30 seconds respectively. If they toll together at 8:00 a.m., then at what time will they toll together for the first time after 8:00 a.m.? (A) 8:04 a.m. (B) 8:06 a.m. (C) 8:05 a.m. (D) 8:07 a.m. 20. Find the value of 3 26 + 29

3 2+ 5

+

3 5+ 8

+

.

(A) 29 − 2 (B) 26 + 5 (C) 29 + 26 (D) 26 − 8

3 8 + 11

+…+

Test | 175 21. If a + b + c = 0, find the value of

(3 ) . (3 ) . (3 ) 1

a8

1

a6 bc

b8

ab6 c

c8

1 abc6

.

(A) 2 (B) 6 (C) 27 2 2. If 3x+3 – 3x+2 = 486, then find x. (A) 3 (B) 5 (C) 6

(D) 81 (D) 2

5 5 , log r q = and log r p = 3 x, then find x. 4 6 1 2 1 2 (A) (B) (C) (D) 9 3 18 9 23. log p q =

24.

5 5 5 + + = 1 + log p qr 1 + log q pr 1 + log r pq

(A) 0 (B) 1 (C) 5 (D) 10 2 5. If log102 = 0.3010 then find the number of digits in 255. (A) 17 (B) 11 (C) 18 (D) 16 26. The maximum sum of the arithmetic progression 45, 41, 37 … is (A) 256 (B) 274 (C) 276 (D) 264 2 27. The greatest value of n such that 1 + 3 + 3 + 33 + … + 3n, which is less than 3000 will be: (A) 6 (B) 8 (C) 7 (D) 9

28. 3 + 33 + 333 + … + upto n terms = (10 n − 1) 10(10 n − 1) – n (B) – n (A) 3 27 (10 n − 1) n 10(10 n − 1) n (C) – (D) − 3 9 27 3 29. What is the minimum value of the function f (x) = x2 – 15x + 9? 289 −200 295 −189 (A) (B) (C) (D) 4 9 2 4 30. Find the area of the shaded region, where PQR is a triangle and QRS is a quadrant. PQ = 6 cm and QR = 8 cm. S P 6

Q

8

(A) 4p – 8 sq. cm. (C) 8p – 16 sq. cm.

R

(B) 2p – 3 sq. cm. (D) 16p – 24 sq. cm.

Answer Keys 1. C 11. C 21. C

2. D 12. C 22. A

3. D 13. C 23. D

4. A 14. C 24. C

5. B 15. B 25. A

6. B 16. A 26. C

7. C 17. C 27. C

8. D 18. B 28. D

9. D 19. C 29. D

10. A 20. A 30. D

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Reasoning Chapter 1: Number and Letter Series

1.179

Chapter 2: Analogies1.185 Chapter 3: Odd Man Out (Classification)

1.188

Chapter 4: Coding and Decoding

1.191

Chapter 5: Blood Relations1.195 Chapter 6: Venn Diagrams

1.200

Chapter 7: Seating Arrangements

1.204

Chapter 8: Puzzles1.212 Chapter 9: Clocks and Calenders

1.225

U N I t II

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Chapter 1 Number and Letter Series LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • •

Number series Difference series Product series Squares/cubes series

• Miscellaneous series • Combination series • Letter series

intrODUctiOn

nUMBer SerieS

Number and Letter Series form an important part of the Reasoning section in various competitive examinations. There are two or three broad categories of questions that appear in various exams from this particular chapter. In the first category of questions, a series of numbers/letters is given with one number/letter (or two numbers/letters) missing, represented by a blank or a question mark. The given series of numbers/letters will be such that each one follows its predecessor in a certain way, i.e., according to a definite pattern. Students are required to find out the way in which the series is formed and hence work out the missing number/numbers or letter/letters to complete the series. For the purpose of our discussion, we will refer to this category of questions as Number Series Type I or Letter Series Type I questions. Under Type I questions, there are a large variety of patterns that are possible and the student requires a proper understanding of various patterns to be able to do well in these types of questions. In the second category of questions, a series of numbers/letters is given and the student is required to count how many numbers/ letters in that series satisfy a given condition and mark that as the answer. For the purpose of our understanding, we will refer to this category of questions as Number Series Type II or Letter Series Type II questions. These questions will mainly involve counting of numbers/letters satisfying a given condition.

For better understanding, we will classify this into the following broad categories. 1. 2. 3. 4. 5.

Difference series Product series Squares/Cubes series Miscellaneous series Combination series

Difference Series The difference series can be further classified as follows. 1. Number series with a constant difference. 2. Number series with an increasing or decreasing difference. In the number series with a constant difference, there is always a constant difference between two consecutive numbers. For example, the numbers of the series 1, 4, 7, 10, 13, … are such that any number is obtained by adding a constant figure of 3 to the preceding term of the series. If we have to find the next number in the above series, we need to add a 3 to the last term 13. Thus, 16 is the next term of the series. Under the series with constant difference, we can have series of odd numbers or series of even numbers also.

1.180 | Reasoning In the series with increasing/decreasing difference, the difference between consecutive terms keeps increasing (or decreasing, as the case may be). For example, let us try to find out the next number in the series 2, 3, 5, 8, 12, 17, 23, … Here, the difference between the first two terms of the series is 1; the difference between the second and third terms is 2; the difference between the third and the fourth terms is 3 and so on. That is, the difference between any pair of consecutive terms is one more than the difference between the first number of this pair and the number immediately preceding this number. Here, since the difference between 17 and 23 is 6, the next difference should be 7. So, the number that comes after 23 should be (23 + 7) = 30. We can also have a number series where the difference is in decreasing order (unlike in the previous example where the difference is increasing). For example, let us find out the next term of the series 10, 15, 19, 22, 24, … 10,

15, +5

19,

22,

+4

24 +3

+2

Here the differences between 1 and 2 , 2nd and 3rd, 3rd and 4th numbers, etc., are 5, 4, 3, 2, and so on. Since the difference between 22 and 24 is 2, the next difference should be 1. So, the number that comes after 24 should be 25. st

nd

Product Series A product series is usually a number series where the terms are obtained by a process of multiplication. Here also, there can be different types of series. We will look at these through examples. Consider the series 2, 4, 8, 16, 32, 64, … 2,

4, ×2

8, ×2

16, ×2

32, ×2

64 ×2

Here, each number in the series is multiplied by 2 to get the next term. So, the term that comes after 64 is 128. So, each term is multiplied by a fixed number to get the next term. Similarly we can have a series where we have numbers obtained by dividing the previous term with a constant number. For example, in the series 64, 32, 16, 8, …, each number is obtained by dividing the previous number by 2 1 (or in other words, by multiplying the previous term by ). 2 So, here, the next term will be 4 (obtained by dividing 8 with 2). Consider the series 4, 20, 80, 240, … 4

20, ×5

80, ×4

240 ×3

Here, the first term is multiplied by 5 to get the second term; the second term is multiplied by 4 to get the third term; the third term is multiplied by 3 to get the fourth term. Hence, to get the fifth term, we have to multiply the fourth term by 2, i.e., the fifth term is 480. So, each term is multiplied by a decreasing factor (or it could also be an increasing

factor) to get the next term. That is, with whatever number a particular term is multiplied to get the next term, this latest term is multiplied by a number different from the previous multiplying factor to get the next term of the series. All the multiplying factors follow a certain pattern (normally of increasing or decreasing order). Consider the series 2, 6, 12, 20, 30, … 2,

6, +4

12, +6

20, +8

30 +10

This can be looked at a series of increasing differences. The differences of consecutive pairs of terms are 4 (between 2 and 6), 6 (between 6 and 12), 8 (between 12 and 20), 10 (between 20 and 30) and so on. Hence, the difference between 30 and the next term should be 12 and so the next term will be 42. But this series can also be looked at as a product series. 2, 6, 12, 20, 30 ↓ ↓ ↓ ↓ ↓ 1 × 2 2 × 3 3 × 4 4 × 5 5 × 6 The first term is the product of 1 and 2; the second term is the product of 2 and 3; the third term is the product of 3 and 4; the fourth term is the product of 4 and 5; the fifth term is the product of 5 and 6. Hence, the next term will be the product of 6 and 7, that is 42.

Squares/Cubes Series There can be series where all the terms are related to the squares of numbers or cubes of numbers. With squares/ cubes of numbers as the basis, there can be many variations in the pattern of the series. Let us look at various possibilities of series based on squares/cubes. Each term of the series may be the square of a natural number, such as 1, 4, 9, 16, … 1, 4, 9, 16 ↓ ↓ ↓ ↓ 12 22 32 42 The numbers are squares of 1, 2, 3, 4 … respectively. The number which follows 16 (which is the square of 4) will be 25 (which is the square of 5). The terms of the series may be the squares of odd numbers (for example, 1, 9, 25, 49, …) or even numbers (for example, 4, 16, 36, 64, …). The terms of the series could be such that a number and its square are both given one after the other and such pairs are given in some specific pattern. For example, take the series 2, 4, 3, 9, 4, 16, … 2,

4, +1

3,

9,

4,

16

+1

Here, 2 is followed by its square 4; then comes the number 3 (which is one more than 2) followed by its square 9 and so on. Hence, the next number in the series is 5 and the one after that is its square i.e., 25.

Chapter 1 Number and Letter Series | 1.181 Similarly, each term could be the square root of its predecessor. For example, in the series 81, 9, 64, 8, 49, 7, 36, …, 81 is the square of 9, 64 the square of 8, and so on. Therefore the next number which follows in the series should be the square root of 36, i.e., 6. The terms of the series could be the squares of natural numbers increased or reduced by certain number. For example, in the series 3, 8, 15, 24, … 3, ↓ 22 – 1

8, ↓ 32 – 1

15, ↓ 42 – 1

24 ↓ 52 – 1

We have {Squares of natural numbers – 1} as the terms. The first term is 22 – 1; the second term is 32 – 1; the third term is 42 – 1 and so on. Hence, the next term will be 62 – 1, i.e., 35 [Please note that the above series can also be looked at as a series with increasing differences. The differences between the 1st and 2nd terms, the 2nd and 3rd terms, and so on are 5, 7, 9, and so on. Hence, the next difference should be 11 giving us the next term as 35]. There could also be a series with {squares of natural numbers + some constant}. Like we have seen series with squares of numbers, we can have similar series with cubes of numbers. For example, take the series 1, 8, 27, 64, … 1, 8, 27, 64 ↓ ↓ ↓ ↓ 13 23 33 43 Here, all the terms are cubes of natural numbers. So, the next term will be 53, i.e., 125. Consider the series 2, 9, 28, 65, … 2, ↓ 13 + 1

9, ↓ 23 + 1

28, ↓ 33 + 1

65 ↓ 43 + 1

Here, the terms are {Cubes of natural numbers + 1}. The first term is 13 + 1; the second term is 23 + 1; the third term is 33 + 1 and so on. Hence the next term will be 53 + 1, i.e., 126.

Miscellaneous Series There are series that do not come under the other patterns and are of general nature but are important and are fairly common. Even here, some times, there can be a specific pattern in some cases. Take the series 3, 5, 7, 11, 13, … . This is a series of consecutive PRIME NUMBERS. It is an important series and the student should look out for this as one of the patterns. The next term in this series is 17. There can also be variations using prime numbers. Take the series 9, 25, 49, 121, … . In this series, the terms are squares of prime numbers. Hence, the next term is 132, i.e., 169. Take the series 15, 35, 77, … . The first term is 3 × 5; the second term is 5 × 7; the third term is 7 × 11; here the terms

are product of two consecutive prime numbers. So, the next term will be the product of 11 and 13, i.e., 143. Take the series 8, 24, 48, 120, 168, … Here, the 2nd term is 3 times the first term and the 3rd term is 2 times the 2nd term, but after that it does not follow this pattern any more. If you look at the terms carefully, you will find that the terms are {one less than squares of prime numbers}. Hence, the next term will be 172 – 1, i.e., 288. Consider the series 1, 4, 9, 1, 6, 2, 5, 3, … At first sight there is nothing we can say about the series. This is actually a series formed by squares of natural numbers. However, if any of the squares is in two or more digits, each of the digits is written as a separate term of the series. Thus, the first terms are 1, 4 and 9, the squares of 1, 2 and 3 respectively. After this, we should get 16 (which is the square of 4). Since this has two digits 1 and 6, these two digits are written as two different terms 1 and 6 in the series. Similarly, the next square 25 is written as two different terms 2 and 5 in the series. So, the next square 36 should be written as two terms 3 and 6. Of these, 3 is already given. So, the next term of the series is 6. Consider the series 1, 1, 2, 3, 5, 8, … 1, 1, 2, 3, 5, 8 ↓ ↓ ↓ ↓ 1 + 1 1 + 2 2 + 3 3 + 5 Here, each term, starting with the third number, is the sum of the two preceding terms. After taking the first two terms as given (1 and 1), then onwards, to get any term, we need to add the two terms that come immediately before that position. Hence, to get the next term of the series, we should take the two preceding terms 5 and 8 and add them up to get 13. So, the next term of the series is 13. The term after this will be 21 (= 8 + 13).

Combination Series A number series which has more than one type of (arithmetic) operation performed or more than one series combined together is a combination series. The series that are combined can be two series of the same type or could be different types of series as described above. Let us look at some examples. First let us look at those series which are formed by more than one arithmetic operation performed on the terms to get the subsequent terms. Consider the series: 2, 6, 10, 3, 9, 13, 4, 12, … Here, the first term 2 is multiplied by 3 to get the second term, and 4 is added to get the third term. The next term is 3 (one more than the first term 2) and it is multiplied by 3 to get 9 (which is the next term) and then 4 is added to get the next term 13. The next term 4 (which is one more than 3) which is multiplied with 3 to get 12. Then 4 is added to this to get the next number 16.

1.182 | Reasoning Consider the series: 1, 2, 6, 21, 88, … Here, we can observe that 88 is close to 4 times 21. It is in fact 21 × 4 + 4. So, if we now look at the previous term 21, it is related to the previous term 6 as 6 × 3 + 3. Now we get the general pattern: to get any term, multiply the previous term with k and then add k where k is a natural number with values in increasing order from 1. So, to get the second term, the first term has to be multiplied with 1 and then 1 is added. To get the third term, the second term is multiplied with 2 and then 2 is added and so on. Hence, after 88, the next term is 88 × 5 + 5, i.e., 445. Now, let us look at a series that is formed by combining two (or more) different series. The two (or more) series can be of the same type or of different types described above. Consider the series: 8, 12, 9, 13, 10, 14, … Here the 1st, 3rd, 5th, … terms which are 8, 9, 10, … form one series whereas the 2nd, 4th, 6th, etc. terms which are 12, 13, 14 form another series. Here, both series that are being combined are two simple constant difference series. Therefore the missing number will be the next term of the first series 8, 9, 10, … which is equal to 11. Consider the series: 0, 7, 2, 17, 6, 31, 12, 49, 20, … Here, the series consisting of 1st, 3rd, 5th, … terms (i.e., the series consisting of the odd terms) which is 0, 2, 6, 12, 20, … is combined with another series consisting of 2nd, 4th, 6th, … terms (i.e., the series consisting of the even terms) which is 7, 17, 31, 49, … . The first series has the differences in increasing order 2, 4, 6, 8, 10 and so on. The second series also has the difference in increasing order 10, 14, 18, … Since, the last term 20 belongs to the first series, a number from the second series should follow next. The next term of the second series will be obtained by adding 22 to 49, that is 71. Consider the series: 1, 1, 2, 4, 3, 9, 4, 16, … Here, one series consisting of odd terms, which is 1, 2, 3, 4, …, is combined with the series of even terms which is 1, 4, 9, 16, … The first series is a series of natural numbers. The second series is the squares of natural numbers. Hence, the next term is 5. Consider the series: 1, 1, 4, 8, 9, 27, … Here, the series of squares of natural numbers is combined with the series of cubes of natural numbers. The next term in the series will be 4. Consider the series: 2, 4, 5, 9, 9, 16, 14, ? , 20, … Here, we have to find out the term that should come in place of the question mark. The odd terms form one series 2, 5, 9, 14, 20, … where the difference is increasing. The differences are 3, 4, 5, 6, … This series is combined with the series of even terms 4, 9, 16, … where the terms are squares of numbers 2, 3, 4, … . Hence, the term that should come in place of the question mark is the next term of the second series which is 52, i.e., 25. A general approach to Number Series: The best way of approaching the number series questions is to first observe the difference between terms. If the difference is constant,

it is a constant difference series. If the difference is increasing or decreasing by a constant number, then it is a series with a constant increasing or decreasing difference. If there is no constant increasing or decreasing difference, then try out the product series approach. For this, first divide the second term with the first term, third with the second, and so on. If the numbers obtained are the same, then it is a product series. Alternatively, try writing each term of the series as a product of two factors and see if there is any pattern that can be observed. If still there is no inference, but the difference is increasing or decreasing in a rapid manner, then check out the square series. If the increase is very high, and it is not a square series, then try out the cube series. If the difference is alternately decreasing and increasing (or increasing for some time and alternately decreasing), then it should most probably be a mixed series. Therefore test out the series with alternate numbers. If still the series is not solved, try out the general series.

Letter Series The questions here are similar to the questions in Number Series Type I. Instead of numbers we have letters of the alphabet given here. We have to first identify the pattern that the series of letters follow. Then, we have to find the missing letter based on the pattern already identified. In Number Series, we saw different patterns that the numbers in the series can follow – like squares, cubes. In letter series, obviously, patterns like squares, cubes will not be possible. In Letter Series, in general, we have a series with constant or increasing or decreasing differences. The position of the letters in the English alphabet is considered to be the value of the alphabet in questions on Letter Series. Also, when we are counting, after we count from A to Z, we again start with A, i.e., we treat the letters as being cyclic in nature. Like in Number Series, in this type of Letter Series also, we can have a ‘combination’ of series, i.e., two series are combined and given. We need to identify the pattern in the two series to find out the missing letter. Sometimes, there will be some special types of series also. Let us look at a few examples to understand questions on Letter Series. Solved Examples Example 1: Find the next letter in the series D, G, J, M, P, ______. (A) Q (B) R (C) S (D) T Solution: (C) Three letters are added to each letter to get the next letter in the series. i.e., D+3, G+3, J+3, M+3, P+3, S P + 3 and P = 16 and 16 + 3 = 19 and the 19th letter in the alphabet is S.

Chapter 1 Number and Letter Series | 1.183 Example 2: Find the next letter in the series A, B, D, H, ______. (A) L (B) N (C) R

(D) P

Solution: (D) Each letter in the given series is multiplied with 2 to get the next letter in the series. A × 2 ⇒ 1 × 2 = 2 and the 2nd letter is B, B × 2 ⇒ 2 × 2 = 4 and the 4th letter is D. Similarly, H × 2 ⇒ 8 × 2 = 16 and the 16th letter is P. Example 3: What is the next letter in the series? B, D, G, K, P, ______ (A) S (B) V (C) W (D) X Solution: (B) B+2, D+3, G+4, K+5, P+6, ______ P + 6 = 16 + 6 = 22 and the 22nd letter is V. Example 4: I, X, J, W, K, V, L, ______. (A) M (B) U (C) S

(D) T

Solution: (B) The given series is an alternate series.

I+1, J+1, K+1, L is one series and X–1, W–1, V–1, ______ is the other series. X – 1 = W, W – 1 = V and V – 1 = 22 – 1 = 21 and the 21st letter is U. Example 5: 97, 83, 73, 67, 59, _____ (A) 53 (B) 49 (C) 47 (D) 51 Solution: (C) The given numbers are alternate prime numbers in decreasing order, starting with 97. Hence, the next number in the series is 47. Example 6: 75, 291, 416, 480, 507, _____ (A) 515 (B) 532 (C) 511 (D) 521 Solution: (A) 75+216, 291+125, 416+64, 480+27, 507, _____ The differences are cubes of consecutive natural numbers in decreasing order. Hence, the next number in the series in 507 + (B)3 = 515.

Practice Exercise Practice Problems Directions for questions 1 to 25: Complete the following series.

1 120 , 20, 24, ______ 10. 13 , 15, 3 7

(A) 30

(B) 36

(C) 40

(D) 371/3

1. 17, 19, 23, 29, 31, 37, ______ (A) 41 (B) 43 (C) 40

(D) 42

2. 225, 196, 169, ______, 121, 100, 81 (A) 156 (B) 144 (C) 136

11. 6, 15, 35, 77, 143, 221, ______ (A) 357 (B) 437 (C) 323

(D) 125

3. 64, 125, 216, 343, ______ (A) 64 (B) 424

12. 29, 29, 27, 23, 25, 19, 23, 17, ______, ______ (A) 19, 13 (B) 19, 15 (C) 21, 13 (D) 19, 13

(C) 317

(D) 512

4. 54, 66, 82, 102, 126, ______ (A) 146 (B) 130 (C) 154

(D) 144

14. 3731, 2923, 1917, 1311, ______

5. 7, 11, 20, 36, 61, ______, 146 (A) 25 (B) 91 (C) 97

(D) 92

15. 11, 28, 327, 464, ______

13. 24, 625, 26, 729, 28, 841, ______

(A) 30 (A) 117

(B) 119 (B) 5625

(C) 900

(D) 961

(C) 917

(D) 75

(C) 5125

(D) 5250

(C) 368

(D) 322

7. 125, 375, 377, 1131, 1133, ______ (A) 3399 (B) 1136 (C) 1135 (D) 1234

17. 132, 182, 306, 380, 552, 870, ______

9. 2, 4, 7, 35, 42, 462, ______ (A) 5016 (B) 470 (C) 4712 (D) 475

(A) 525

(B) 29

6. 8, 16, 48, 96, 288, 576, ______ (A) 1152 (B) 1728 (C) 1052 (D) 1428

8. 12, 35, 106, 317, 952, ______ (A) 2851 (B) 2855 (C) 1851 (D) 1849

(D) 383

16. 6, 24, 60, 120, 210, ______

(A) 336 (A) 930

(B) 343 (B) 1010

(C) 992

(D) 1142

18. KPD, LOE, MNF, NMG, ______

(A) ONF

(B) OLH

(C) MLH

(D) MNH

19. BEP, CIQ, DOR, FUS, GAT, ______

(A) HEV

(B) HIT

(C) IET

20. GKF, IPC, LTY, PWT, UYN, ______ (A) ABZ (B) XBX (C) XAH

(D) IEU (D) AZG

1.184 | Reasoning 21. 22. 23.

QLR, JPD, RNU, GNC, SPX, DLB, ______ (A) TRA (B) AJA (C) BTU (D) KJE GTB, CYV, YDP, ______, QND (A) DIV (B) UIJ (C) DDV (D) UVV ABDH, BDHP, CFLX, DHPF, ______ (A) EKNT (B) TNEK (C) EJTN (D) JNTE

24. 25.

TCFK, RADI, OXAF, JSVA, ______ (A) DMPU (B) DMOT (C) CMOT (D) CLOT KJAM, GGWJ, ______, YAOD, UXKA (A) CDUI (B) DFTC (C) DCTF (D) CDSG

Answer Keys

Exercise Practice Problems 1. A 11. C 21. A

2. B 12. C 22. B

3. D 13. A 23. C

4. C 14. D 24. D

5. C 15. C 25. D

6. B 16. A

7. A 17. C

8. B 18. B

9. D 19. A

10. A 20. D

Chapter 2 Analogies LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Analogy • Number analogies

AnALogy Analogy means ‘similarity’ or ‘similar relation ship’. In questions on number or letter analogies, a pair, that has a certain relationship between them, is given. This number/letter pair is followed by a third number/letter. The student is expected to identify the relationship between the pair given and find out a FOURTH number such that the relationship between the third and the fourth is similar to the relationship that exists between the first and the second. (In some cases, it may not be the fourth one that has to be found out. The fourth one will be given and the student has to find out one of the other three, whichever is not given).

nUMber AnALogies

• Letter analogies • Verbal analogies

There can be many more combinations that one can think of but the student has to note an important point in solving questions on Number Analogies. In Number Series related questions, since a series of numbers (more than two numbers) will be given, the relationship or pattern can be identified uniquely. In Number Analogies, since only two numbers are given, it may be possible to think of more than one relationship existing between the two numbers in the given pair. But, it should be kept in mind that generally, simple addition of one number or subtraction of one number is not what is given in Number Analogies. The questions try to test the insight that the student has got into the relationship between the numbers. Let us take a few examples and understand the questions on Number Analogies.

Typical relationships between the numbers in a given pair can be any of the following: • • • •

One number is a multiple of the other. One number is the square or square root of the other. One number is the cube or cube root of the other. The two numbers are squares of two other numbers which themselves are related. For example, the two numbers are squares of two consecutive integers or squares of two consecutive even integers or squares of two consecutive odd integers. • The two numbers are such that they are obtained by subtracting a certain number from the squares or cubes of the two related numbers. • The two numbers are such that they are obtained by adding a certain number to the squares or cubes of the two related numbers. • The two numbers can be consecutive, even, odd or prime numbers.

Solved Examples Example 1: Find the missing number 25 : 36 : : 49 : _____. (A) 61 (B) 63 (C) 65 (D) 60 Solution: (D) When the numbers in the question are considered the students tend to consider 25 and 36 as squares of two consecutive natural numbers. But the answer choices does not consist of an answer suitable to the above logic. Hence, it is important that, the student keeps the answer choices in view in arriving at the logic. 25 + 11 = 36 Similarly, 49 + 11 = 60 Example 2: Find the missing number 27 : 51 : : 83 : _____. (A) 102 (B) 117 (C) 123 (D) 138

1.186 | Reasoning Solution: (C) The given analogy can be written as 52 + 2 : 72 + 2 : : 92 + 2 : _____. 5 and 7 are successive odd numbers. Similarly, next odd number to 9 is 11 and 112 + 2 = 121 + 2 = 123.

Example 6: SUWY : LPTX : : PRTV : _____. (A) INRU (B) INQU (C) IMRU (D) IMQU

Example 3: Find the missing number. 11 : 25 : : 17 : _____. (A) 33 (B) 28 (C) 41 (D) 37

Solution: (D)

11 × 2 + 3 = 22 + 3 = 25 Similarly, 17 × 2 + 3 = 34 + 3 = 37.

Letter Analogies

Solution: (D) S U W Y; Similarly, P R T V –7 –5 –3 –1 –7 –5 –3 –1 L P T X I M Q U

Example 7: BCDE : DFHH : : FGHI : _____. (A) LJPL (B) LKPL (C) JKPJ (D) IKPL Solution: (A)

The questions in this area are similar to Verbal Analogies. Here, the questions are based on the relationship between two groups of letters (instead of two words as in Verbal Analogies). Typically, three sets of letters are given followed by a question mark (where a fourth set of letters is supposed to inserted). The student has to find the relation or order in which the letters have been grouped together in the first two sets of letters on the left hand side of the symbol : : and then find a set of letters to fit in place of the question mark so that the third and the fourth set of letters will also have the same relationship as the first and the second. The sequence or order in which the letters are grouped can be illustrated by the following examples.

Example 4: BDEG : DFGI : : HKMO : _____. (A) ILNP (B) JMOP (C) JMOQ (D) JNOQ

Example 8: Gum : Stick : : Needle : _____ (A) Cloth (B) Prick (C) Taylor (D) Stitch

Solution: (C) Two letters are added to each letter to get the next letters in the analogy. B D E G; Similarly, H K M O +2 +2 +2 +2 +2 +2 +2 +2 D F G I J M O Q

Example 9: Socks : Feet : _____ : Hands (A) Arms (B) Shirt (C) Gloves (D) Fingers

Example 5: ACDF : CGJN : : BEHI : _____. (A) DJNQ (B) DINQ (C) DINR (D) DHNQ

B C D E; Similarly, F G H I ×2 +3 ×2 +3 ×2 +3 ×2 +3 D G H H L J P L

Verbal Analogies Here, the questions are based on relationship between two words. In these kind of questions three words are followed by a blank space, which the student has to fill up in such a way that the third and the fourth words have the same relationship between them as the first and the second words have. The following examples help in understanding the concepts.

Solution: (D) Gum is used to Stick and Needle is used to Stitch.

Solution: (C) Socks are worn on Feet, similarly Gloves are worn on Hands.

Solution: (B)

Example 10: Soft : Hard : : Cold : _____ (A) Hot (B) Shirt (C) Gloves (D) Fingers

Solution: (A) Soft and Hard are antonyms similarly the antonym of Cold is Hot.

A C D F; Similarly, B E H I +2 +4 +6 +8 +2 +4 +6 +8 C G J N D I N Q

Chapter 2 Analogies | 1.187

Practice Exercise Practice Problems Directions for questions 1 to 25: Find the missing term. 1. 97 : 89 : : 43 : ______ (A) 37 (B) 31 (C) 39 (D) 41 2. 196 : 256 : : 324 : ______ (A) 361 (B) 400 (C) 411 (D) 484 3. 121 : 169 : : 361 : ______ (A) 529 (B) 400 (C) 484 (D) 576 4. 125 : 343 : : 343 : ______ (A) 512 (B) 1331 (C) 1728 (D) 81 5. 4 : 256 : : 5 : ______ (A) 625 (B) 1025 (C) 525 (D) 875 6. 12 : 144 : : 18 : ______ (A) 160 (B) 180 (C) 190 (D) 150 7. 25 : 21 : : 59 : ______ (A) 42 (B) 46 (C) 76 (D) 56 8. 8 : 72 : : 10 : ______ (A) 95 (B) 106 (C) 99 (D) 90 9. 8 : 0.125 : : 4 : ______ (A) 0.5 (B) 0.4 (C) 0.35 (D) 0.25 10. 11 : 143 : : 19 : ______ (A) 443 (B) 450 (C) 420 (D) 437 11. 568 : 352 : : 732 : ______ (A) 516 (B) 496 (C) 526 (D) 536 12. 6 : 222 : : 9 : ______ (A) 738 (B) 720 (C) 729 (D) 744 13. 5 : 120 : : 8 : ______ (A) 520 (B) 504 (C) 448 (D) 512 14. 16 : 68 : : 36 : ______ (A) 216 (B) 210 (C) 222 (D) 226

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

10 : 95 : : 16 : ______ (A) 218 (B) 318 (C) 248 (D) 102 3829 : 3851 : : 2987 : ______ (A) 301 (B) 3007 (C) 3017 (D) 3023 47 : 121 : : 89 : ______ (A) 183 (B) 187 (C) 193 (D) 195 NATURE : PEVASI : : ISOMERS : ______ (A) OTUNJTV (B) OTUNIST (C) PUVNJST (D) OVTNJST BAD : BBL : : JDFE : ______ (A) JHRI (B) JHPX (C) JFTV (D) JHRT FIELD : LRJXH : : CRICKET : ______ (A) FHRDXLJ (B) FJPDTLN (C) FJRDXAL (D) FJRFVJN TAP : SUZBOQ : : RED : ______ (A) QTDGDE (B) PSDEDF (C) QSDFCE (D) QRDGBE Train : Track : : Bus : ______ (A) Driver (B) Road (C) Petrol (D) Passengers Earth : Planet : : Carrot : ______ (A) Vegetable (B) Plant (C) Cooking (D) Nut Wood : Carpenter : : Iron : ______ (A) Goldsmith (B) Instrument (C) Melting (D) Blacksmith Pen : Write : : Knife : ______ (A) Vegetable (B) Cut (C) Sharp (D) Shoot

Answer Keys

Exercise Practice Problems 1. D 11. A 21. C

2. B 12. A 22. B

3. A 13. B 23. A

4. B 14. C 24. D

5. A 15. C 25. B

6. D 16. A

7. D 17. B

8. D 18. B

9. D 19. D

10. D 20. D

Chapter 3 Odd Man Out (Classification) LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Odd man out • Alphabet classiﬁcation

iNtroDuCtioN

• Word classiﬁcation • Number classiﬁcation

Solution: (D)

Finding the odd man out from the given alternatives is a very common type of questions that one comes across in different competitive examinations. In the questions on odd man out, all the items – except one – follow a certain pattern (in their formation) or belong to a group. The item that does not follow the pattern or does not belong to the group has to be marked as the answer choice. The problems of this variety often fall under the category of CLASSIFICATION. When a given set of elements is classiﬁed under a single head, one of the items will not fall into that group to which the rest belong, i.e. it will not have the common property, which the others will have. Hence it becomes the odd man out. Questions on classiﬁcation can be asked in any form. Some of the commonly asked ones are given below.

alPHabet ClassifiCatioN In this type, a group of jumbled letters typically consisting of three letters, (but can be four or two or just a single letter) are put together. The pattern or order in which they are grouped is to be studied and we need to ﬁnd out which groups have the same pattern or relationship between the letters. There will be one choice, which will have a pattern different from the rest and that is our answer. Solved Examples Example 1: Find the odd one among the following. (A) ZW (B) TQ (C) SP (D) NL

Z–3W, T–3Q, S–3P, N–2L Hence, NL is the odd one. Example 2: Find the odd one among the following. (A) CFD (B) GJH (C) KNM (D) JMK Solution: (C) C+3F−2D, G+3J−2H, K+3N−1M, J+3M−2K Hence, KNM is the odd one.

worD ClassifiCatioN Here, different items are classiﬁed based on common properties like names, places, professions, parts of speech, etc. A few examples are illustrated below. Example 3: Find the odd one among the following. (A) Mercury (B) Moon (C) Jupiter (D) Saturn Solution: (B) All others except Moon are planets whereas Moon is a satellite. Example 4: Find the odd one among the following. (A) SORE (B) SOTLU (C) NORGAE (D) MEJNIAS

Chapter 3 Odd Man Out (Classification) | 1.189 Solution: (C) The words are jumbled. The actual words are ROSE, LOTUS, ORANGE and JASMINE. All, except ORANGE, are flowers whereas ORANGE is a fruit. Example 5: (A) Cow (B) Goat (C) Horse (D) Dog

Example 7: Find the odd one among the following. (A) 17 (B) 27 (C) 37 (D) 47 Solution: (B) All the given numbers except 27 are prime n umbers whereas 27 is a composite number. Example 8: (A) 441 (C) 361

Solution: (D) All except Dog are herbivorous animals.

(B) 289 (D) 343

Solution: (D) The given numbers can be written as (21)2, (17)2, (19)2, (7)3. All except 343 are the squares whereas 343 is a cube.

Example 6: (A) Shoe (B) Spectacle (C) Scissor (D) Shirt

Example 9: (A) 10 (C) 120

(B) 50 (D) 290

Solution: (D) All expect 120 can be expressed as n2 + 1

Solution: (D) All except shirt are in prises.

10 = 32 + 1, 50 = 72 + 1

Number Classification

290 = 172 + 1 but 120 = 112 – 1.

In this case, we need to choose the odd number from the given alternatives. The numbers may belong to a particular set, i.e., they may be odd, even, prime, rational, squares, cubes, and they may also be coded into binary digits (involving 0’s and 1’s) etc. and only one of the choices will not follow the rule which others do and that is our answer. A few illustrations are given below.

Example 10: (A) 235 (C) 523

(B) 352 (D) 253

Solution: (B) All expect 352 are odd numbers but whereas 352 is an even number.

Practice Exercise Practice Problems Directions for questions 1 to 25: Find the odd man out. 1. (A) 16 (B) 28 (C) 36 (D) 64 2. (A) 27 (B) 37 (C) 47 (D) 67 3. (A) 8 (B) 27 (C) 64 (D) 125 4. (A) 42624 (B) 37573 (C) 84284 (D) 93339 5. (A) 30 (B) 630 (C) 10 (D) 520 6. (A) 8 : 9 (B) 25 : 25 (C) 64 : 81 (D) 16 : 16 3 7 7. (A) (B) 4 + 25 36 + 64 11 5 (C) (D) 9 + 49 49 + 169

73 79

45 49 89 (D) 83 97

9. (A) 4422 (C) 4242 10. (A) 350 (C) 30 11. (A) N (C) B V 12. (A) E R R (C) I V

(B) 2442 (D) 2244 (B) 70 (D) 520 (B) O (D) K L (B) O B B (D) U L

13. (A) ABD (C) CEJ

(B) BDH (D) DFL

14. (A) BCDE (C) RSTU

(B) FGHI (D) WXYZ

8. (A) 13

(C) 71

17 23

(B) 41

1.190 | Reasoning 15. 16. 17. 18. 19. 20.

(A) DFRTH (C) NBEJM (A) Cat (C) Tiger (A) Chameleon (C) Turtle (A) Trivandrum (C) Calicut (A) Part (C) Cart (A) Rocket (C) Planet

21. 22. 23. 24. 25.

(B) ABEJM (D) DHKVY (B) Dog (D) Elephant (B) Crocodile (D) Allegator (B) Hyderabad (D) Bangalore (B) Trap (D) Dart (B) Star (D) Comet

(A) Skin (C) Leg (A) Baseball (C) Chess (A) Walk (C) Drink (A) Ganga (C) Yamuna (A) HEWAT (C) ROWAJ

(B) Tongue (D) Nose (B) Boxing (D) Wrestling (B) Talk (D) Plank (B) Nagarjuna sagar (D) Sutlez (B) CERI (D) EECRALS

Answer Keys

Exercise Practice Problems 1. B 11. B 21. C

2. A 12. D 22. A

3. C 13. A 23. D

4. C 14. D 24. B

5. B 15. A 25. D

6. B 16. D

7. B 17. A

8. B 18. C

9. C 19. B

10. B 20. A

Chapter 4 Coding and Decoding LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Coding

• Decoding

cODinG anD DecODinG Before looking at the different types of questions and some of the codes that can be used with the help of examples, let us first understand what we mean by coding and decoding. When we say coding, a particular code or pattern is used to express a word in English language as a different word or in a different form. The coded word itself does not make any sense unless we know the pattern or code that has been followed. Decoding refers to the process of arriving at the equivalent English word from the code word given. In the questions, a particular code is given and on the basis of this given code, we have to find out how another word (in English language) can be coded. The correct code for the given word has to be selected from the answer choices on the basis of the code given in the question. Solved Examples Example 1: In a certain code language, if the word ‘PARTNER’ is coded as OZQSMDQ, then what is the code for the word ‘SEGMENT’ in that language? (A) TFHNFOU (B) RDFLDMS (C) RDELDMS (D) RDFEDNS Solution: (B) Word : P A R T N E R Logic : –1 –1 –1 –1 –1 –1 –1 Code : O Z Q S M D Q Similarly the code for SEGMENT is Word : S E G M E N T Logic : –1 –1 –1 –1 –1 –1 –1 Code : R D F L D M S

Example 2: In a certain code language, if the word RECTANGLE’ is coded as TGEVCPING, then how is the word ‘RHOMBUS’ coded in that language? (A) TJOQDWV (B) TJQNDWU (C) TJQODWU (D) TJQOEWU Solution: (C) Word : R E C T A N G L E Logic : +2 +2 +2 +2 +2 +2 +2 +2 +2 Code : T G E V C P I N G Similarly, the code for RHOMBUS is Word : R H O M B U S Logic : +2 +2 +2 +2 +2 +2 +2 Code: T J Q O D W U Example 3: In a certain code language, if the word ‘SPHERE’ is coded as EREHPS, then how is the word ‘EXHIBITION’ coded in that language? (A) NOTITBIHXE (B) NOITIDIHXE (C) NOITIBIHWE (D) NOITIBIHXE Solution: (D) Word: S P H E R E Logic: The letters in the given word are reversed. Code: E R E H P S Similarly, the code for EXHIBITION, is Word: E X H I B I T I O N Logic: The letters in the given word are reversed. Code: N O I T I B I H X E

1.192 | Reasoning Example 4: In a certain code language, if the word ‘REJECTION’ is coded as SGMIHZPWW, then how is the word ‘MECHANIC’ coded in that language? (A) NGFLFTPK (B) NGPLFTPK (C) NGFKFTPK (D) NGPTPKIL Solution: (A) Word: R E J E C T Logic : +1 +2 +3 +4 +5 +6 Code : S G M I H Z Similarly, the code for MECHANIC is Word : M E C H A N Logic : +1 +2 +3 +4 +5 +6 Code : N G F L F T

I O N +7 +8 +9 P W W I C +7 +8 P K

Example 5: In a certain code language, if the word ‘PLAYER’ is coded as AELPRY, then how is the word ‘MANAGER’ coded in that language? (A) AEAGMNR (B) AAGEMNR (C) AAEGMNR (D) AAEGNMR Solution: (C) Word: P L A Y E R Logic: The letters in the word are arranged in the increasing order of their value as in the alphabet. Code: A E L P R Y Similarly, the code for MANAGER is AAEGMNR. Example 6: In a certain code language, if the number 1 is assigned to all the letters in odd numbered places in the alphabet and the remaining letters are assigned the number 2, then what is the code for the word ‘INDIAN’? (A) 121212 (B) 111222 (C) 112212 (D) 122112 Solution: (D) The code for the word INDIAN is 122112. Example 7: In a certain code language, if CRICKET is coded as 3923564, ROCKET is coded as 913564 and KETTLE is coded as 564406, then how is LITTLE coded in that language? (A) 244060 (B) 024406 (C) 020446 (D) 200446 Solution: (B) As we observe that the letters and their corresponding codes are given in order i.e., the code for C is 3, R is 9, I is 2 and so on. Hence, the code for LITTLE is 024406.

Directions for questions 8 to 10: In a certain code language, the codes for some words are as follow. NATION - agvnab REMOTE - rzgrbe STAIR - efgnv FORMAL - bensyz COMMON - zabzpb FOR - ebs Based on the above coding pattern answer the following questions. 8. What is the code for ‘SCREEN’? (A) fepcra (B) fpersa (C) fpreba (D) fperra 9. What is the code for ‘RATION’? (A) ensvba (B) engvba (C) engrba (D) engvca 10. What is the code for ‘CREATOR’? (A) prengbc (B) persbgc (C) perngbe (D) pebryc Solutions for questions 8 to 10: The given words and their codes are as follow (A) NATION - agvnab (B) REMOTE - rzgrbi (C) STAIR - efgnv (D) FORMAL - bensyz (E) COMMON - zabzpb (6) FOR - ebs In the 1st word the letter N is repeated and so is the code a. Hence, for N, the code is a. Similarly, from the 2nd word, the code for E is ‘r’. In 1st and 6th words the letter o is common and so is the code b. Hence, the code for o is b. In the 5th word the letter m is repeated and so is the code z. Hence, the code for m is z. Similarly the codes for the remaining letters can be determined. The letters and their respective codes are as follows Letter A

C

E

F

I

L M N O

R

S

T

Code letter

p

R

s

v

y

e

f

G

n

z

a

8. The code for ‘SCREEN’ is fperra. Hence, the correct option is (D). 9. The code for ‘RATION’ is engvba. Hence, the correct option is (B). 10. The code for ‘CREATOR’ is perngbe. Hence, the correct option is (C).

b

Chapter 4 Coding and Decoding | 1.193

Practice Exercise Practice Problems Directions for questions 1 to 12: Select the correct alternative from the given choices. 1. In a certain code language, if the word CIRCUMSTANCE is coded as CRUSACICMTNE, then how is the word HAPPINESS coded in that language? (A) HPEISAPNS (B) HPISEAPNS (C) HPIESPANS (D) HPIESAPNS 2. In a certain code language, if the word REGISTRATION is coded as TSIGERNOITAR, then how is the word ACCURATE coded in that language? (A) UCCAETAR (B) UACCETAR (C) UCACETAR (D) UCCATEAR 3. In a certain code language, if the word LIBERAL is coded as MJCFSBM, then how is the word REDUCTION coded in that language? (A) EDCTBSHNM (B) SFEVDUJPO (C) SFEVCTJPO (D) SFDUCTJPO 4. In a certain code language, if the word STRUCTURE is coded as TVUYHZBZN, then how is the word REMEDY coded in that language? (A) SGPIJE (B) SGPEJD (C) SGPIHE (D) SGPIIE 5. In a certain code language, if the word SEARCH is coded as IDSBFT, then how is the word FURNISH coded in that language? (A) ITKNSVG (B) ITJORWG (C) ITJOSVG (D) ITHNRVG 6. In a certain code language ‘two’ is called ‘three’, ‘three’ is called ‘four’, ‘four’ is called ‘one’, ‘one’ is called ‘five’, ‘five’, is called ‘six’ and ‘six’ is called ‘nine’, then what in the code language is the sum of one and three? (A) six (B) two (C) nine (D) one 7. In a certain code language if ‘pink’ means ‘black’, ‘black’ means ‘white’, ‘white’ means ‘yellow’, ‘yellow’ means ‘orange’, ‘orange’ means ‘red’ and ‘red’ means ‘green’, then which colour stands for peace in that code? (A) Red (B) Black (C) Orange (D) Green 8. In a certain code language, if MENTION = 49 and NEUROTIC = 64, then MARVELLOUS = ? (A) 81 (B) 88 (C) 64 (D) 100 9. In a certain code language, if CABINET = 70 and BEAUTY = 60, then PRODUCTION = ? (A) 90 (B) 100 (C) 110 (D) 120 10. In a certain code language, if IMPEND = 61 and DISH = 40, then FRUIT = ? (A) 86 (B) 68 (C) 74 (D) 76 11. In a certain code language, if BUG = 90 and ALMS = 180, then CADET = ? (A) 153 (B) 165 (C) 175 (D) 148

12. In a certain code language, if INFER = 25 and JERSEY = 28, then CHOICE =? (A) 34 (B) 39 (C) 41 (D) 47 Directions for questions 13 to 15: These questions are based on the following data. In a certain code language, if the word ROUTINE is coded as JMPRRLJ and the word FIDELITY is coded as LGHCXGNW, then how will you code the following words in that language? 13. PREVAIL (A) FPLRDGX (B) FPJTBGX (C) FTJBNKX (D) FPJVBIX 14. LANGUAGE (A) XYBDPXNC (B) XYBDPXMC (C) XYCEPXNC (D) XYBEPYNC 15. TOBACCO (A) NMDXEAF (B) NMDYEBF (C) NMCYFBD (D) NMDYFAD Directions for questions 16 to 20: For the following groups of letters given in column I, the codes are given in column II. Answer the following questions by finding the codes for the groups from the given columns. Column I

Column II

(A)

lit kit bit dit

brpd

(B)

fit git mit kit

tdsv

(C)

rit bit git tit

xpvw

(D)

nit dit fit rit

rsxj

16. What is the code for lit? (A) v (B) r 17. What is the code for tit? (A) w (B) x 18. What is the code for rit? (A) j (B) s 19. What is the code for nit? (A) x (B) s 20. What is the code for kit? (A) r (B) p

(C) p

(D) b

(C) p

(D) v

(C) r

(D) x

(C) j

(D) r

(C) x

(D) d

Directions for questions 21 to 25: For the words given in column I, the codes are given in column II. Answer the following questions by finding the codes for the letters from the words and their codes given in the columns. Column I

Column II

(A)

PRETEND

4396408

(B)

COMMON

615715

(C)

HOUSE

4*2&1

(D)

SUPPORT

3*21839

(E)

DRUM

5*08

1.194 | Reasoning 21. What is the code for the word PROTECT? (A) 3895479 (B) 3846978 (C) 3819479 (D) 3814978 22. What is the code for the word HORMONE? (A) &385364 (B) &176561 (C) &175184 (D) &185164 23. What is the code for the word EMPEROR? (A) 5495717 (B) 4534818 (C) 3453919 (D) 4537178

24. What is the code for the word DETHRONE? (A) 049&7264 (B) 049&8164 (C) 059&7164 (D) 059&8164 25. What is the code for the word COMPOUND? (A) 71531*60 (B) 72532*80 (C) 91531*70 (D) 72542*60

Answer Keys

Exercise Practice Problems 1. D 11. B 21. C

2. A 12. A 22. D

3. B 13. B 23. B

4. D 14. D 24. B

5. C 15. D 25. A

6. D 16. D

7. B 17. A

8. D 18. D

9. B 19. C

10. C 20. D

Chapter 5 Blood Relations LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Blood relations • Grandparents • Parents and in-laws

• Siblings, spouse and in-laws • Children and in-laws • Grand children

BLOOD reLatiOns There are two types of questions based on blood relations that are given in different competitive examinations. For the sake of convenience we will refer to the two types of questions as Type I and Type II. (Please note that the questions on blood relationships are not categorised as above in the actual exam papers. It is being done purely from the point of better understanding).

In the exams, the success of a candidate in the questions on blood relations depends upon his knowledge about various blood relations. Some of the relationships given below help in solving the problems. The easiest and non-confusing way to solve these types of problems would be to draw a family tree diagram and increase the levels in the hierarchy as shown below:

(Grandfather, Grandmother, Granduncle, Grandaunt)

1ststage :

GRANDPARENTS

nd 2 stage :

PARENTS and IN-LAWS

(Father, Mother, Uncle, Aunt, Father-in-law, Mother-in-law)

3rdstage :

SIBLINGS, SPOUSE and IN-LAWS

(Brother, Sister, Cousin, Wife, Husband, Brother-in-law, Sister-in-law)

4 stage :

CHILDREN and IN-LAWS

(Son, Daughter, Niece, Nephew, Son-in-law, Daughter-in-law)

th 5 stage :

GRAND CHILDREN

th

Mother’s or Father’s son Mother’s or Father’s daughter Mother’s or Father’s brother Mother’s or Father’s sister Mother’s or Father’s mother

: : : : :

Brother Sister Uncle Aunt Grandmother

(Grandson, Granddaughter)

Mother’s or Father’s father Grandmother’s brother Grandmother’s sister Grandfather’s brother Grandfather’s sister

: : : : :

Grandfather Granduncle Grandaunt Granduncle Grandaunt

1.196 | Reasoning Sister’s or Brother’s son Sister’s or Brother’s daughter Uncle or Aunt’s son or daughter Son’s wife Daughter’s husband Husband’s or Wife’s sister Husband’s or Wife’s brother Sister’s husband Brother’s wife Children of same parents

Children Children’s Children

: :

Nephew Niece

: Cousin : Daughter-in-law : Son-in-law : Sister-in-law : Brother-in-law : Brother-in-law : Sister-in-law : Siblings (could be all brothers, all sisters or some brothers and some sisters) : Son, Daughter : Grandchildren (Grandson, Granddaughter)

In addition, remember the word spouse which means either husband or wife. Grandfather and grandmother will come in the first stage; mother, father, uncle and aunt will come in the second stage; sister, brother and cousin will come at the third stage; son, daughter, niece and nephew will come in the fourth stage and finally, granddaughters and grandsons will come. The above stages are made from the point of view of an individual. In Type-I questions, the relationship between two people is given through a roundabout way of relating them through other people. We have to go through the series of relationships and finally determine the relationship between the two people given in the question. The relationship can be given as a simple statement or as a statement made by a person. In the first example given below, a person is involved in making a statement whereas in the second question, there is no person involved in making a statement.

Solution: (C) A’s father is B and mother is D. Therefore D is B’s wife and C is the father of B. Hence C is D’s father-in-law. C father B

wife

D

father A

Example 3: A + B means A is the son of B. A – B means A is the daughter of B. A × B means A is the father of B. A ÷ B means A is the mother of B. If M × N + O – P ÷ Q, then how is M related to Q? (A) Husband (B) Cousin (C) Brother-in-law (D) Uncle Solution: (C) M × N + O – P ÷ Q means M is the father of N, N is the son of O, O is the daughter of P, P is the mother of Q. M is the father of N and N is the son of O means M is the husband of O. O is the daughter of P and P is the mother of Q means O is the sister of Q. M is the husband of O and O is the sister of Q means M is the brother-in-law of Q. P mother M

wife

O

sister

Q

father N

Solved Examples Example 1: A’s father’s mother-in-law’s only daughter’s son is B. How is A related to B? (A) Brother (B) Sister (C) Nephew (D) Cannot be determined Solution: (D) A’s father’s mother-in-law’s only daughter is A’s mother. A’s mother’s son is A’s brother. But A can be either brother or sister to B. Example 2: If A’s father is B, C is the father of B and D is A’s mother, then How is C related to D? (A) Father (B) Grandfather (C) Father-in-law (D) Uncle

Example 4: A + B means A is the son of B. A – B means A is the daughter of B. A × B means A is the father of B. A ÷ B means A is the mother of B. Which of the following means S is the son-in-law of P? (A) P + Q ÷ R × S – T (B) P × Q ÷ R – S + T (C) P + Q × R – S ÷ T (D) P × Q – R ÷ S × T Solution: (B) P + Q ÷ R × S – T means P is the son of Q. Q is the mother of R, R is the father of S and S is the daughter of T. Hence S is the nephew of P. P × Q ÷ R – S + T means P is the father of Q, Q is the mother of R, R is the daughter of S and S is the son of T. Hence S is the son-in-law of P.

Chapter 5 Blood Relations | 1.197 P + Q × R – S ÷ T means P is the son of Q, Q is the father of R, R is the daughter of S and S is the mother of T. Hence S is the mother of P. P × Q – R ÷ S × T means P is the father of Q, Q is the daughter of R, R is the mother of S and S is the father of T. Hence S is the son of P. Example 5: Pointing to a person, Raju said, ‘He is the only brother of my father’s mother’s daughter.’ How is the person related to Raju? (A) Brother (B) Father (C) Uncle (D) Nephew Solution: (B) Raju’s father’s mother’s daughter is Raju’s father’s sister. Raju’s father’s sister’s only brother is Raju’s father. Hence the person is Raju’s father. Example 6: A’s mother’s father is the husband of B’s mother. How is A related to B, if A and B are both male. (A) Uncle (B) Father (C) Nephew (D) Son Solution: (C) A’s mother’s father is the husband of B’s mother. That means A’s mother is the sister of B. Hence all the nephew of B. Example 7: Pointing to a photograph Ramesh said, ‘she is the sister of my father’s mother’s only child’s son.’ How is the person in the photograph related to Ramesh? (A) Sister (B) Aunt (C) Mother (D) Cousin Solution: (A) My father’s mother’s only child is my father. My father’s son’s sister is in the photograph. Hence she is Ramesh’s sister. (

)

mother ( ) father (Ramesh)

sister

(

)

Example 8: B’s father’s father is the husband of C’s mother’s mother. How is B related to C? (A) Brother (B) Sister (C) Cousin (D) Cannot be determined

Solution: (D) B’s father’s father is B’s grandfather. B’s grandfather is the husband of C’s mother’s mother i.e., grandmother. It is possible that B and C are sibling’s and the persons mentioned are their paternal/maternal grand parents. It is also possible that B and C are cousins. Hence the relationship cannot be determined. (

)

mother

husband

(

) father

(

)

(

)

mother/father

father/mother (C)

(B)

Directions for questions 9 and 10: A × B means A is the daughter of B A * B means A is the son of B A + B means A is the mother of B A – B means A is the brother of B A ÷ B means A is the sister of B A = B means A is the father of B Example 9: Which of the following means S is the nephew of P? (A) P – Q ÷ R = S (B) P – Q × R * S (C) S * R – Q ÷ P (D) P + Q – R × S Solution: (C) P – Q ÷ R = S means P is the brother of Q, Q is the sister of R and R is the father of S. As we do not know whether S is the son or daughter of R, we cannot determine that S is the nephew. P – Q × R * S means P is the brother of Q, Q is the daughter of R and R is the son of S. Here S is the grandfather of P. S * R – Q ÷ P means S is the son of R. R is the brother of Q and Q is the sister of P. Hence, S is the nephew of P. Example 10: Which of the following means T is the husband of V? (A) T = S – R * V (B) T + S – R × V (C) R × V – T * S (D) R + S – T × V Solution: (A) T = S – R * V means T is the father of S, S is the brother of R and R is the son of V. Therefore, T is the husband of V.

1.198 | Reasoning

Practice Exercise Practice Problems Directions for questions 1 to 15: Select the correct alternative from the given choices. 1. A person who is the husband of my son’s sister is my (A) Nephew (B) Son-in-law (C) Son (D) Brother 2. Y is the daughter of X’s brother’s wife’s father-in-law. Y is X’s _______. (A) Niece (B) Daughter (C) Sister (D) Sister-in-law 3. Showing a photograph P said,’ She is my mother’s mother’s son’s daughter’. How is the person in the photograph related to P? (A) Sister (B) Cousin (C) Niece (D) Mother 4. How is my father’s mother’s only daughter-in-law’s sister related to me? (A) Aunt (B) Sister (C) Cousin (D) Niece 5. How is my grandmother’s only child’s husband’s mother related to me? (A) Mother (B) Grandmother (C) Aunt (D) Sister 6. How is Ramu’s mother-in-law’s only daughter’s son related to Ramu? (A) Nephew (B) Brother (C) Son (D) Uncle 7. How is my son’s mother’s daughter related to me? (A) Niece (B) Granddaughter (C) Daughter (D) Aunt 8. How is my father’s brother’s only sibling’s mother related to me? (A) Mother (B) Cousin (C) Daughter (D) Grandmother 9. A is the father of B. C is the son of D. E is the brother of C while D is the sister of B. How is B related to E? (A) Uncle (B) Aunt (C) Mother (D) Either (A) or (B) 10. My mother’s sister’s son’s father’s mother-in-law is related to me as (A) Mother (B) Grandmother (C) Mother-in-law (D) Aunt 11. How is David’s father’s only daughter-in-law’s son’s wife related to David? (A) Daughter (B) Daughter-in-law (C) Niece (D) Granddaughter 12. How is Ravi’s mother’s father’s son related to Ravi’s father? (A) Cousin (B) Uncle (C) Brother-in-law (D) Son-in-law

13. Divya’s father, pointing towards a person, said, ‘He is the brother of my father’s only sibling’. How is the person related to Divya? (A) Father (B) Uncle (C) Brother (D) Grandfather 14. Tinku, introducing a person to Rinku, said ‘He is the father of your sister’s son and he is also my mother’s husband’. How is Tinku’s father related to Rinku’s mother? (A) Nephew (B) Uncle (C) Son-in-law (D) Father 15. A is B’s father, B is C’s daughter, E is D’s only sibling. C is D’s only daughter. How is B related to E’s niece? (A) Niece (B) Granddaughter (C) Daughter (D) Mother Directions for questions 16 to 20: Use the relations defined below and answer the following questions. A + B means A is the mother of B. A – B means A is the sister of B. A × B means A is the father of B. A ÷ B means A is the son of B. A = B means A is the brother of B. A ≠ B means A is the daughter of B. 16. Which of the following means P is the aunt of Q? (A) P – R ÷ Q (B) P + R × Q (C) P ≠ R × Q (D) P – R + Q 17. Which of the following means, S is the son of T’s daughter? (A) T × M + S + N (B) T × M + S = N (C) T + M × S – N (D) S ÷ M ÷ T – N 18. Which of the following means W is the uncle of Z? (A) W × A - B + Z (B) W = A + B - Z (C) W = A + B + Z (D) W × A × B = Z 19. Which of the following means C is the grandfather of both D and E? (A) C × A ÷ D - E (B) C + A + D ≠ E (C) C ÷ A ≠ D = E (D) C × A × D - E 20. Which of the following means I is the mother of L? (A) I + B - C ≠ D × L (B) I ≠ B + C × L (C) I + B × C ≠ D - L (D) I + B - C × L Directions for questions 21 to 25: These questions are based on the information given below. A, B, C, D, E and F are six members of a family. A is the mother of B, who is the husband of D. F is the brother of one of the parents of C. D is the daughter-in-law of E and has no siblings. C is the son of D. 21. How is C related to A? (A) Nephew (B) Son-in-law (C) Grandson (D) Father

Chapter 5 Blood Relations | 1.199 22. 23.

How is F related to D? (A) Cousin (C) Brother How is E related to F? (A) Mother (B) Son (C) Father (D) Father-in-law

(B) Brother-in-law (D) Father

24. 25.

If F is married to G, then how is G related to B? (A) Sister (B) Sister-in-law (C) Cousin (D) Mother How many male members are there in the family? (A) Two (B) Three (C) Four (D) Cannot be determined

Answer Keys

Exercise Practice Problems 1. B 11. B 21. C

2. C 12. C 22. B

3. B 13. D 23. C

4. A 14. C 24. B

5. B 15. C 25. C

6. C 16. D

7. B 17. B

8. D 18. B

9. D 19. D

10. B 20. A

Chapter 6 Venn Diagrams LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Venn Diagrams • Venn Diagram Type I

• Venn Diagram Type II

Venn Diagrams Venn Diagrams are diagrammatic representation of sets, using geometrical ﬁgures like circles, triangles, rectangles, etc. Each geometrical ﬁgure represents a group as shown in the examples. The area common to two or more ﬁgures represent those elements which are common to two or more groups. There are various models in Venn Diagrams which we will discuss with examples.

Venn Diagram Type i In this type, two, three or four different groups could be given with some elements common to two or more groups. Let us observe the diagram given below. m A

B

d

a f

r

b e

c n

C

Here, A, B and C are three different groups, and the various regions can be explained as given below. Only A = a Only B = b

Only C = c A and B only = d B and C only = e C and A only = f All the three (A, B and C) = r Both A and B = d + r Both B and C = e + r Both C and D = f + r Neither A, nor B, nor C = n A, B or C and none = m Also, m = (A ∪ B ∪ C) + n Here, the rectangle represents the sample space, which consists of three groups A, B and C, and also n, which is the number of people belonging to neither A, nor B, nor C. Some more formulae are as given under: A′ = (b + e + c) + n; where A′ = A complement (not in A) B′ = (a + f + c) + n; where B′ = B complement (not in B) C′ = (a + d + b) + n; where C′ = C complement (not in C) A − B = A – (A ∩ B) A ∆ B = (A – B) ∪ (B − A) Number of people (or things) belonging to at least one out of the three groups = A ∪ B ∪ C = (a + b + c) + (d + e + f ) + r ↑ ↑ ↑ exactly one exactly two exactly three 7. A + B + C = (A ∪ B ∪ C) + (d + e + f ) + 2r = (a + b + c) + 2 (d + e + f ) + 3r 1. 2. 3. 4. 5. 6.

Chapter 6 Venn Diagrams | 1.201 µ = 150

Solved Examples Directions for questions 1 to 3: These questions are based on the data given below. In a class of 165 students, 45 students are passed in Maths a well as in English, whereas 60 students are failed in Maths and 65 students are failed in English. Example 1: How many students are passed in exactly one subject? (A) 160 (B) 100 (C) 115 (D) 165 Example 2: How many students are failed in both the subjects? (A) 25 (B) 20 (C) 45 (D) 5 Example 3: How many students are failed only in Maths? (A) 55 (B) 60 (C) 65 (D) 70 Solutions for questions 1 to 3: Since, 60 students are failed in Maths. \ 165 – 60 = 105 students passed in Maths similarly, 65 students failed in English. \ 165 – 65 = 100 students passed in English. The respective Venn – diagrams is as follows which shows the number of students who passed the subject. m = 165 English = 100

Maths = 105

a

45

b n

45 students passed in Maths as well as English. \ a = 105 – 45 = 60 students passed only in Maths and b = 100 – 45 = 55 students passed only English. Number of students passed in atleast one subject = 60 + 55 + 45 = 160 Hence, n = 165 – 160 = 5 students failed in both. Example 1: a + b = 60 + 55 = 115 students passed exactly in one subject. Hence, the correct option is (C). Example 2: 5 students failed in both the subjects. Hence, the correct option is (D). Example 3: As 55 students passed only in English which implies that 55 students failed only in Maths. Hence, the correct option is (A). Directions for questions 4 to 7: These questions are based on the given diagram.

P

Q

9

12 15

8 16

14 18 R

Example 4: How many elements are there in Q′ (complement of Q)? (A) 100 (B) 49 (C) 101 (D) 50 Example 5: How many elements are there in P′ ∩ Q′ ∩ R′? (A) 35 (B) 8 (C) 58 (D) 48 Example 6: How many elements are there in R? (A) 16 (B) 57 (C) 41 (D) 8 Example 7: How many elements are there in P ∩ (Q ∪ R)? (A) 32 (B) 48 (C) 54 (D) 44 Solutions for questions 4 to 7: Example 4: Number of elements in Q′ = (m) – (number of elements in Q) = 150 – (14 + 18 + 8 + 9) ⇒ 150 – 49 = 101. Hence, the correct option is (C). Example 5: Number of elements in P′ ∩ Q′ ∩ R′ = m – (P ∪ Q ∪ R) = 150 – (12 + 15 + 9 + 8 + 18 + 16 + 14) = 150 – 92 = 58. Hence, the correct option is (C). Example 6: Number of elements in R = 16 + 15 + 8 + 18 = 57 Hence, the correct option is (B). Example 7: Number of elements in Q ∪ R = (14 + 9 + 8 + 15 + 18 + 16) Number of elements in P = (12 + 9 + 15 + 8) P ∩ (Q ∪ R) is the region common to P and Q ∪ R Number of elements in P ∩ (Q ∪ R) = 9 + 8 +15 = 32. Hence, the correct option is (A).

Venn Diagram Type II In this type, Venn diagrams are used to establish relationship between the given groups. In other words, two or more groups are given and the Venn diagram, which most correctly establishes a relation between them, has to be chosen out of the various Venn diagrams given in the choices. Let us look at some of the examples given below.

1.202 | Reasoning Example 8: Animals, Cat, Dog

We know that day is a part of the week and week is a part of the month. So the above diagram is the most appropriate representation of the given groups.

Animals

Example 10: Mars, Earth, Jupiter Cat

Dog

Here, in animals we have many species of which cat and dog are two different kinds of species, having nothing in common. So the above diagram is the most appropriate representation of the given groups. Example 9: Month, week, day Month

We know that Mars, Earth and Jupiter are three independent entities having nothing in common. So the above diagram is the most appropriate representation of the given groups.

Week Day

Practice Exercise Practice Problems Directions for questions 1 to 5: There are 1500 students in a college. Each student can be a member of three student communities namely P, Q, and R. Now using the data mentioned and the diagram given below answer the questions that follow. P 240

Q 20 10

30

R

•• Total members in community P is 300. •• Total members in community Q is 420. •• Total members in community R is 490. 1. How many students are part of only community R? (A) 360 (B) 420 (C) 210 (D) 350 2. How many students is not part of any community? (A) 390 (B) 420 (C) 410 (D) 490 3. How many students are part of at least two communities? (A) 10 (B) 30 (C) 80 (D) 90 4. How many students are part of at least one community? (A) 1000 (B) 1090 (C) 1110 (D) 1100 5. How many students are part of exactly two communities? (A) 90 (B) 80 (C) 100 (D) 120

Directions for questions 6 to 10: These questions are based on the data given below. In a class of 95 students, 40 play cricket, 50 play football and 10 play both cricket and football. 6. How many students play only football? (A) 45 (B) 30 (C) 40 (D) 28 7. How many students play at least one game? (A) 80 (B) 70 (C) 60 (D) 50 8. How many students play only cricket? (A) 30 (B) 35 (C) 40 (D) 25 9. How many students play exactly one game? (A) 85 (B) 80 (C) 70 (D) 75 10. How may students play neither cricket nor football? (A) 12 (B) 15 (C) 18 (D) 20 Directions for questions 11 to 15: Study the following data and answer the questions given below. In a certain college, 37% of the students write EAMCET exam, 47% of the students write IIT-JEE exam and 50% of the students write AIEEE exam. Also known that, 11% of the students write both EAMCET and IIT-JEE, 11% of the students write both EAMCET and AIEEE, 15% of the students write both IIT-JEE and AIEEE, while 15 students write all the three exams. Each student in the college writes at least one of the three exams. 11. How many students appear for the exams from the college? (A) 400 (B) 200 (C) 500 (D) 600 12. How many students write exactly two exams? (A) 120 (B) 110 (C) 140 (D) 150

Chapter 6 Venn Diagrams | 1.203 13. The number of students who write only EAMCET as a percentage of the number of students who write only AIEEE is 1 2 (A) 33 % (B) 66 % 3 3 2 1 (C) 33 % (D) 66 % 3 3 14. How many students write exactly one exam? (A) 345 (B) 395 (C) 198 (D) 398 15. What is the ratio of the number of students who write only AIEEE to that of those who write only IIT JEE? (A) 3 : 2 (B) 2 : 3 (C) 8 : 9 (D) 9 : 8 Directions for questions 16 to 20: These questions are based on the data given below. In a library maintained by a student, there are books on different subjects. It was found that 35 books are on sports, 45 books are on business and 15 books are on current affairs. 14 books are on at least two subjects among sports, business and current affairs. 3 books have sports, business as well as current affairs in them. Every book in the library is assumed to contain at least one of sports, business or current affairs in them. 16. How many books are there which contain information regarding only one subject? (A) 58 (B) 64 (C) 60 (D) 62 17. What are the total number of books in his library? (A) 78 (B) 72 (C) 68 (D) 80 18. How many books contained information regarding exactly two subjects? (A) 11 (B) 10 (C) 9 (D) 14

19. How many books are there which contain information regarding at most two subjects? (A) 11 (B) 64 (C) 72 (D) 75 20. If the number of books on only sports is equal to 26, then how many books are there in the library, which are on both business and current affairs but not sports? (A) 5 (B) 3 (C) 2 (D) 8 Directions for questions 21 to 25: These questions are based on the data given below. In a colony, it is known that three brands of mobile phones are used, namely Nokia, Sony Ericsson and Motorola. 70 families use only one brand, 47 families use exactly two brands and 8 use all the three brands. It is assumed that each family uses at least one of these three brands. 21. How many families are there in the colony? (A) 75 (B) 100 (C) 105 (D) 125 22. How many families use at least two brands? (A) 117 (B) 55 (C) 47 (D) 125 23. If 10 families stop using Nokia and start using Motorola, then what is the maximum number of families who use exactly two brands? (A) 57 (B) 37 (C) 47 (D) 67 24. What is the ratio of the number of families which use exactly one brand to that which use at least one brand? (A) 14 : 25 (B) 14 : 11 (C) 11 : 25 (D) 11 : 14 25. How many families do not use all the three brands? (A) 125 (B) 117 (C) 0 (D) 8

Answer Keys

Exercise Practice Problems 1. B 11. C 21. D

2. A 12. C 22. B

3. D 13. B 23. A

4. C 14. A 24. A

5. B 15. D 25. B

6. C 16. B

7. A 17. A

8. A 18. A

9. C 19. D

10. B 20. A

Chapter 7 Seating Arrangements LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Seating arrangement • Linear sequencing

linear seQuenCing Linear sequencing is essentially arranging the items given in a sequence (in a single line). The questions of this type are also referred to as ‘Seating Arrangement’. The word ‘seating arrangement’ should not be misconstrued – it should not be treated as consisting of questions involving only persons sitting as per specified conditions. Essentially, these questions involve arranging subjects (people or things) satisfying the given conditions. The arrangement is done only on one ‘axis’ and, hence, the position of the subjects assumes importance here in terms of order like first position, second position, etc. Let us look at the examples: Directions for questions 1 to 5: Read the data given below carefully and answer the questions that follow. Seven persons Paul, Queen, Rax, Sam, Tom, Unif and Vali are sitting in a row facing us. Rax and Sam sit next to each other. There must be exactly four persons between Queen and Vali. Sam sits to the immediate right of Queen. Solved Examples Example 1: If Paul and Tom are separated exactly by two persons, then who sits to the immediate left of Vali? (A) Paul (B) Tom (C) Unif (D) Rax Example 2: If Queen is not sitting at either extreme of the row, then who among the following has as many persons on his left as on his right?

• Circular arrangement

(A) Sam (C) Rax

(B) Unif (D) Vali

Example 3: If Queen sits at one extreme, then who is at the other extreme? (A) Paul (B) Tom (C) Vali (D) Cannot be determined Example 4: Tom sits to the right of Queen, and Paul is separated from Tom by exactly three persons. Then, who is sitting to the immediate left of Vali? (A) Unif (B) Paul (C) Tom (D) Rax Example 5: In how many different ways can the seven persons sit in a row? (A) 3 (B) 2 (C) 10 (D) 12 Solutions for questions 1 to 5: Let us write down the conditions given in short form and then represent them pictorially. Also, let us treat the left of the persons sitting as ‘left’ and their right as ‘right’ for interpreting the conditions. Rax and Sam sit next to each other → RS or SR. There are exactly 4 persons between Queen and Vali → Q — — — — V or V — — — — Q. Sam sits to the immediate right of Queen → SQ.

Chapter 7 Seating Arrangements | 1.205 Now let us analyse the data/conditions that we are given and then put the three conditions together. Let us number the seats from our left to right as Seat 1 to Seat 7. Since S is to the right of Q and since R and S have to be next to each other, R can come only to the immediate right of S. Thus, R, S and Q, will be in the order RSQ. Since there are four persons between Q and V, Q can be placed in seats 1, 2, 6 or 7. But if Q is in Seat 1or 2, then there are no seats for R and S. Hence, there are only two seats available for Q. Let us fix the positions of R, S and V in each of these two positions of Q and write them down. The directions Left and Right are as shown below. R

L

Arrangement I:

1 2 3 4 5 6 7 V R S Q

Arrangement II:

1 2 3 4 5 6 7 V R S Q

V, U, T, R, S, Q, P The person sitting to the immediate left of Vali is Unif. Hence, the correct option is (A). Example 5: We have two possible arrangements— Arrangement I and Arrangement II that we looked at already. In each arrangement, the remaining three people can sit in the remaining three seats in 6 ways. Thus, a total of 12 ways of seating the seven persons is possible. Hence, the correct option is (D). Directions for questions 6 to10: Read the data given below carefully and answer the questions that follow. Seven boys – Rajan, Shyam, Vardhan, Mithra, Vimal, Raj and Kishan – are sitting in a row. Shyam sits to the immediate left of Vardhan and third to the right of Rajan, whereas Mithra, who sits at the left extreme, is next to Kishan. Example 6: Who is sitting to the immediate right of Shyam? (A) Mithra (B) Kishan (C) Vimal (D) Vardhan

These are the only two arrangements possible for the four persons V, R, S and Q. The other three persons Paul, Tom and Unif can sit in the three vacant seats in any order, as no information is given about them. Now let us look at each of the questions.

Example 7: If Vardhan and Kishan exchange places with each other without changing the rest of the arrangement that is already done, who will be sitting to the immediate left of Rajan? (A) Kishan (B) Raj (C) Vimal (D) Vardhan

Example 1: Paul and Tom are separated by exactly two persons. Arrangement I is the only one possible as in Arrangement II, Paul and Tom cannot have exactly two persons between them. So, we have the arrangement as follows:

Example 8: If only Shyam sits between Raj and Vardhan, who is exactly in the middle of the row? (A) Raj (B) Vardhan (C) Vimal (D) Rajan

T/P, V, U, P/T, R, S, Q So, Unif must be sitting to the immediate left of Vali. Hence, the correct option is (C). Example 2: If Queen is not at the extreme right, then only Arrangement II above is possible. The person who has as many persons on his left as on his right can only be the person who is sitting in the middle seat, i.e., seat 4. In this arrangement, Rax is sitting in seat 4. Hence, the correct option is (C). Example 3: ‘Queen sits at one extreme’ means that we should look at arrangement I. In this arrangement, any one out of the three persons Paul, Tom and Unif can be in seat 1, i.e., extreme right. Hence, the correct option is (D). Example 4: If Tom and Paul are separated by exactly three persons, then only Arrangement II is possible. So, Tom and Paul have to be in seats 3 and 7, Since, we are also given that Tom is to the right of Queen, Tom has to be in seat 3 and Paul, in seat 7. So, the arrangement must be as follows:

Example 9: Which of the following cannot confirm the seating arrangement of all the boys? (A) Raj is to the immediate right of Rajan, whereas Vimal is to the left of Shyam. (B) Mithra and Raj have two persons between them. (C) Raj and Kishan have two persons between them. (D) Rajan and Shyam have two persons in between them. Example 10: After arranging all the boys as per the conditions given in the data, if Rajan now exchanges his place with Mithra, and Vardhan exchanges his place with Vimal, then how many persons will be there between Vimal and Rajan? (A) Three (B) Two (C) Five (D) Cannot be determined Solutions for question 6 to10: Let us denote Left and Right as shown below: L

R

Now, let us represent the data given in pictorial form (We use R for Raj and Rn for Rajan; Va for Vardhan; Vi for Vimal; S for Shyam; M for Mithra and K for Kishan).

1.206 | Reasoning Mithra sits at the left extreme – next to Kishan → M K — — — — —. Shyam sits to the immediate left of Vardhan and third to the right of Rajan → Rn — — S Va. Putting both the above together, Va can go only to extreme right position. Thus, we have the arrangement as M K Rn — — S Va. Raj and Vimal occupy the two vacant seats between Rajan and Shyam. Example 6: From the seating arrangement figure above, Vardhan is to the immediate right of Shyam. Hence, the correct option is (D). Example 7: If Kishan and Vardhan exchange places, as can be seen from the arrangement, the person to the immediate left of Rajan will be Vardhan. Hence, the correct option is (D). Example 8: If Shyam sits between Raj and Vardhan, then the seating arrangement is as follows: Mithra, Kishan, Rajan, Vimal, Raj, Shyam, Vardhan. Then, Vimal will be exactly in the middle of the row. Hence, the correct option is (D). Example 9: Statement (A) makes the arrangement as: Mithra, Kishan, Rajan, Raj, Vimal, Shyam, Vardhan Statement (B) gives the seating arrangement as: Mithra, Kishan, Rajan, Raj, Vimal, Shyam, Vardhan. Statement (C) makes the seating arrangement as: Mithra, Kishan, Rajan, Vimal, Raj, Shyam, Vardhan. So, only statement (D) cannot make the seating arrangement unique while others can. Hence, the correct option is (D). Important point to note is that on the basis of the given data, we know that the places of only Raj and Vimal have not been fixed. Hence, if there is an additional statement that we are considering to determine the arrangement uniquely, it SHOULD have at least one of the two people Raj and Vimal. In this case, choice (D) does not have either one of the two names and hence, this statement cannot help us determine the arrangement uniquely. So, this becomes the answer choice. Hence, the correct option is (D). Example 10: The arrangement is M K Rn R/Vi Vi/R S Va Rajan exchanges his place with Mithra, and Vimal with Vardhan, then we have the following arrangement: Rn K M R/Va Va/R S Vi. While we still do not know the exact position of Vardhan (or which place Vimal sits), we can see that there are five persons between Rajan and Vimal. Hence, the correct option is (C).

In addition to the questions that we saw above, where a set of questions are based on the data given, there are also ‘stand-alone’ questions. In these questions, on the basis of the data given, only one question is asked. Given below is an example of this type. Directions for question 11: Select the correct alternative from the given choices. Example 11: Four persons A, B, C and D arrive to attend a meeting. D arrives 10 minutes after B and twenty minutes before A, who arrives 10 minutes before C. Who is the first person to arrive at the meeting? (A) A (B) B (C) C (D) D Solution for question 11: Example 11: C arrived after A. A arrived after D. D arrived after B. This implies that B arrived first. Hence, the correct option is (B).

Circular Arrangement Questions on circular arrangement involve seating of people around a table or arrangement of things in a circular manner (for example, different coloured beads strung to form a necklace). In case of people sitting around a table, the table could be of any shape – rectangular, square, circular or any other. The data given in such sets of questions specify the positions of some or all of the individuals (or things) in the arrangement. The positions are specified through conditions involving specified persons sitting (or not sitting) opposite each other or a particular person sitting to the right or left of another person, etc. Once you read the data, first draw the shape specified in the data and then draw the slots in the seating arrangement. Six people around a circular table

Eight people around a circular table

Statements like ‘A and B are sitting farthest from each other’ or ‘A and B sit across the table’ imply that A and B sit opposite each other. On the other hand, you should remember that, unlike in straight-line arrangement, the words ‘immediately’ and ‘directly’ do not play any role in circular arrangement. In general, there is no left side or right side (unless we are talking of ‘immediate right’ or ‘immediate left’). So if it is given that C sits to right of B, then it is clear that C must be to the immediate right of B. Go ‘anti-clockwise’ if anybody’s right has to be located, and go ‘clockwise’ if somebody’s left has to be located.

Chapter 7 Seating Arrangements | 1.207 Let us take some examples. Directions for questions 12 to 16: Read the following information and answer the questions that follow. P, Q, R, S and T sit around a table. P sits two seats to the left of R and Q sits two seats to the right of R. Example 12: If S sits in between Q and R, who sits to the immediate right of P? (A) T (B) S (C) Q (D) R

As can be seen from the diagram, T is to the immediate right of P. Hence, the correct option is (A). Example 13: We will take each choice and see whether it fits in the arrangement that we represented through a diagram in the analysis of the data (the same diagram is reproduced below). S or T

Q

R

Example 13: Which of the following cannot be the correct seating arrangement of the five persons in either the clockwise direction or the anti-clockwise direction? (A) P, Q, R, S, T (B) P, S, R, T, Q (C) P, Q, S, R, T (D) P, T, R, S, Q Example 14: If S is not sitting next to Q, who is sitting between Q and S? (A) R (B) P (C) T (D) Both (R) and (P) Example 15: If a new person U joins the group such that the initial conditions for the seating arrangement should be observed and also a new condition that U does not sit next to R be satisfied, then which of the following statements is true? (A) U sits to the immediate right of S. (B) U sits to the immediate left of T. (C) U sits to the immediate left of P. (D) Either (A) or (B) above Example 16: If a new person U joins the group such that the initial conditions for the seating arrangement should be observed and also a new condition that U does not sit next to P, S or T be satisfied, then who will be the neighbours of P (one on either side)? (A) S and T (B) S and Q (C) T and R (D) R and Q

T or S P

We can see that the arrangement given in choice (A) is not possible and hence the answer choice is (A). Hence, the correct option is (A). Example 14: If S is not next to Q, then the seating arrangement is fixed as follows. T R Q S P

Now P is between Q and S. Hence, the correct option is (B). Example 15: On the basis of the diagram that we drew, we find that to accommodate U we have to create a new slot between P and Q. S or T Q

R

Solutions for questions 12 to 16: U

S or T Q

T or S

R

P

Hence, choice (C) is the correct answer. T or S P

P sits two seats to the left of R, and Q sits two seats to the right of R. We can represent this information in the diagram below. Example 12: If S sits between Q and R, then the arrangement is as follows.

Example 16: We create a new slot for the sixth person. But since U will not sit next to P, S or T, he will have to sit between R and Q. The arrangement will then look as follows: U Q

R

S

Q

R

T or S

T or S P

T P

As we can see from the diagram, the neighbours of P will be T and S. Hence, the correct option is (A).

1.208 | Reasoning Directions for questions 17 to 21: Read the following information and answer the questions that follow. There are 10 persons at a round table conference, consisting of a Professor, a Lawyer, a Doctor, a Scientist, an Accountant, a Grocer, two Computer Specialists and two Marketing Executives. The Professor sits opposite to the Lawyer. The Scientist and the Doctor sit opposite each other. The two Marketing Executives sit opposite each other with one of them sitting to the immediate left of the scientist. The Professor sits to the immediate right of the Scientist. Example 17: If the two Computer Specialists sit opposite each other but neither of them is immediately next to any Marketing Executive, who sits to the immediate right of the professor? (A) Computer Specialist (B) Marketing Executive (C) Grocer (D) Accountant Example 18: If the Grocer and Accountant do not sit opposite each other, then which of the following must be TRUE? (A) The Computer Specialist cannot sit beside the Lawyer. (B) One of the Computer Specialists is next to a Marketing Executive. (C) The Professor cannot have the Scientist and a Computer Specialist on his either side. (D) The Computer Specialists must sit next to one another. Example 19: If the Grocer and the Accountant each have a Marketing Executive as his immediate neighbour, then which of the following is definitely FALSE? (A) The two Computer Specialists are opposite each other. (B) A Computer Specialist is an immediate neighbour of the Scientist. (C) The Grocer is next to a Computer Specialist. (D) A Computer Specialist is an immediate neighbour of the Lawyer. Example 20: If a Computer Specialist is the immediate neighbour of a Marketing Executive and the Grocer is the immediate neighbour of the Lawyer, how many different kinds of seating arrangements are possible? (Assume that the two Computer Specialists are indistinguishable from each other and the two Marketing Executives are indistinguishable from each other.) (A) 3 (B) 6 (C) 16 (D) 8 Example 21: The maximum number of persons you can count if you start counting with the Scientist and end with a Marketing Executive (excluding both) is (A) 0 (B) 8 (C) 5 (D) 6 Solutions for questions 17 to 21: The Professor sits to the immediate right of the Scientist and opposite to the Lawyer. The Scientist sits opposite to

the Doctor and one Marketing Executive is to the immediate left of the Scientist. Choosing to place the Scientist in one of the 10 seats, we have the arrangement as follows. Doctor ME

Lawyer

Professor

ME

Scientist

The vacant seats are one each for the two Computer Specialists, one for the Grocer and one for the Accountant. Example 17: The two Computer Specialists sit opposite each other. Neither of them is next to any Marketing Executive. So, the arrangement must be as follows Doctor ME

Lawyer

Grocer or Accountant

CS

CS Grocer or Accountant

Professor

ME Scientist

So, a Computer Specialist sits to the immediate right of the professor. Hence, the correct option is (A). Example 18: The Grocer and the Accountant do not sit opposite each other. Then the arrangements can be as follows: Doctor ME

Lawyer

3 2 1 ME

4 Professor Scientist

The Grocer and the Accountant can occupy the following pairs of seats: 3 and 4, 1 and 4, 1 and 2 or 2 and 3. Then, the two computer specialists may occupy one of the pairs of seats 1 and 2, 2 and 3, 3 and 4 or 4 and 1. We check for the choices given in the question, one by one, and find that whichever combination is taken, there is a Computer Specialist in seat 1 or seat 3, both of, which are next to the Marketing Executives seats. So, choice (B), which states that one of the Computer Specialists is next to a Marketing Executive, is true. Hence, the correct option is (B).

Chapter 7 Seating Arrangements | 1.209 Example 19: The Grocer and the Accountant have one each of the Marketing Executives on their immediate side. So, the arrangement must be as follows. Doctor ME

Lawyer 1 Accountant or Grocer ME

2

has to be next to a Marketing Executive, he should be in slot 1 or 3. By fixing the Accountant in any one of the three slots 1, 2 or 3, we can ensure that there is a Computer Specialist next to a Marketing Executive. Hence, there are three possible seating arrangements. Doctor

Grocer or Accountant

ME

Lawyer

3

Grocer

Professor

2

Scientist

1

Now the Computer Specialists must sit in chairs 1 and 2 only. But no Computer Specialist can be the immediate neighbour of the Scientist. Choice (B) is definitely FALSE. (Note that choices (A), (C) and (D) are true). Hence, choice (B) is the correct answer. Hence, the correct option is (B). Example 20: Given that the Grocer is the immediate neighbour of the Lawyer, we have the three slots numbered 1, 2 and 3 (in the following diagram) free for the two Computer Specialists and the Accountant. Since a Computer Specialist

ME

Professor Scientist Scientist

Hence, the correct option is (A). Example 21: Based on the seating arrangement that we discussed, the number of persons between the Scientist and a Marketing Executive can be 3 or 8 (counted clockwise) or 0 or 5 (counted anti-clockwise). Maximum number that can be counted is 8. Hence, the correct option is (B).

Practice Exercise Practice Problems Directions for questions 1 to 3: These questions are based on the following information. Five boys – Anil, Charan, David, John and Kamal sit in a row facing north, not necessarily in the same order. I. John sits exactly in between Anil and David. II. John sits exactly in between Charan and Kamal. 1. Who sits exactly at the middle of the row? (A) John (B) Kamal (C) David (D) Cannot be determined 2. In how many different ways these five boys can sit? (A) 2 (B) 4 (C) 8 (D) 16 3. If Anil sits to the immediate left of John and if a boy sits to the immediate right of Kamal then who is that boy? (A) David (B) Anil (C) Charan (D) None of these Directions for questions 4 to 6: These questions are based on the following information. Seven girls – A, B, C, D, E, F and G sit in a row facing north, not necessarily in the same order. It is also known that, I. Two girls sit in between B and F. II. Three girls sit in between C and G. III. Four girls sit in between A and D.

4. Who sits exactly at the middle of the row? (A) B (B) D (C) E (D) Cannot be determined 5. If B sits to the immediate right of D then who sits in between A and E? (A) F (B) C (C) G (D) Cannot be determined 6. If F and G sit on either sides of E then who sits at the right end of the row? (A) A (B) C (C) D (D) Cannot be determined Directions for questions 7 to 9: These questions are based on the following information. Five persons P, Q, R, S and T sit in a row facing North not necessarily in the same order. The following information is known about them: I. Either P or S sits at the one end of the row. II. Either Q and T or S and T sit on either sides of P. III. R sits to the left of S and to the immediate left of Q. 7. In how many different ways can these five people sit? (A) 2 (B) 3 (C) 1 (D) 4

1.210 | Reasoning 8. If Q sits to the immediate left T then who sits exactly at the middle of the row? (A) P (B) R (C) T (D) Cannot be determined 9. If P is not sitting adjacent to S, then who sits to the immediate right of Q? (A) Q (B) P (C) R (D) Cannot be determined Directions for questions 10 to 12: These questions are based on the following information. Each of six persons – Pavan, Raman, Kiran, Charan, Shravan and Rajan stay in a different floor of a six storied (1st, 2nd, 3rd, 4th, 5th, and 6th from bottom to top respectively) building. I. Raman stays above Kiran but below Charan. II. Pavan stays below Rajan but above Shravan. III. Kiran stays above Pavan but below Raman who stays above Rajan. 10. Who stays in the 2nd floor? (A) Pavan (B) Shravan (C) Rajan (D) Cannot be determined 11. Who stays in the 4th floor? (A) Raman (B) Rajan (C) Kiran (D) Cannot be determined 12. If one person stays in between Pavan and Kiran then who stays in the 3rd floor? (A) Shravan (B) Pavan (C) Rajan (D) Charan Directions for questions 13 to 15: These questions are based on the following information. There are five buildings of different heights in a row. These houses are painted with a different colour among – Red, Blue, White, Green and Yellow such that each house is painted with exactly one colour. The following information is know about them: I. Yellow and green coloured buildings are on either sides of the white coloured building. II. The shortest building is painted in red colour but it is neither at any end of the row nor adjacent to the tallest building. III. The white coloured building is exactly in between the tallest and the second tallest buildings. 13. Which among the following buildings is definitely at one end of the row? (A) Yellow coloured building (B) Green coloured building (C) The tallest building (D) The third tallest building

14. Which among the following is definitely false? (A) The White coloured building is the third tallest (B) The third tallest and the shortest buildings are together (C) Blue and Yellow and coloured buildings are at either ends of the row. (D) Yellow and Green coloured buildings are at either ends of the row. 15. If the yellow coloured building is to the immediate left of the third tallest building then what could be the order of these buildings in the descending order of their heights? (A) Blue, Yellow, Red, White, Green (B) Blue, Green, White, Yellow, Red (C) Green, White, Blue, Yellow, Red (D) Green, Yellow, White, Blue, Red Directions for questions 16 to 18: These questions are based on the following information given below. Each of the six persons – John, Ted, Humpty, Dumpty, Jack and Jill is from one different country among India, Japan, China, Australia, America and England and are sitting around a circular table, may not be in the same order. John, who is from China is sitting adjacent to American, who is not Humpty. Ted is not an Indian and Chinese is not sitting adjacent to Indian. The person from England is sitting one place away to the left of the Australian. Humpty is sitting opposite the Indian, who is adjacent to the Japanese. Australian and Dumpty are sitting opposite each other. Jack is not from India and Ted is not from Japan but both are not adjacent to each other. 16. Who among them is from India? (A) Jill (B) Dumpty (C) Humpty (D) None of these 17. If Jack is the Japanese, then who is sitting opposite the American? (A) Jill (B) Ted (C) Jack (D) Dumpty 18. Which country does Humpty belong to? (A) Japan (B) Australia (C) America (D) England Directions for questions 19 to 22: These questions are based on the following information given below. Eight persons - Ram, Ramesh, Mohan, Sohan, Seema, Saroj, Sakshi and Saloni are sitting around a circular table. Each of them is one among Doctor, Engineer, Dancer, Singer, Teacher, Lawyer, Accountant and Pilot, not necessarily in the given order. Further it is known that I. Pilot is sitting opposite Ramesh, who is adjacent to the Accountant. II. Dancer is sitting opposite the Lawyer and is not adjacent to Sakshi who is not sitting adjacent to the Lawyer. III. Saloni is sitting opposite the Engineer, Ramesh is not a Lawyer or Doctor or Engineer.

Chapter 7 Seating Arrangements | 1.211 IV. Sakshi, the Singer, is sitting one place away to the right of Saroj. V. Seema is sitting opposite the Lawyer and Ram is sitting opposite the Dancer. VI. Ramesh is sitting three places to the right of Singer. Mohan is neither the Accountant nor adjacent to the Dancer. 19. Who among the following is the Doctor? (A) Ramesh (B) Saloni (C) Saroj (D) Cannot be determined 20. What is the profession of Mohan? (A) Accountant (B) Pilot (C) Engineer (D) Cannot be determined 21. Who is sitting opposite Ramesh? (A) Seema (B) Sakshi (C) Saroj (D) None of these 22. Who is sitting opposite the Accountant? (A) Sakshi (B) Mohan (C) Seema (D) Saroj

Directions for questions 23 to 25: These questions are based on the following information given below. Eight persons – Arun, Pankaj, Rohan, Veda, Suman, Shanu, Dimple and Pinky are sitting around a circular table for a group discussion. Suman is not sitting opposite Pinky and Shanu is sitting three places away to the right of Pankaj. Dimple is sitting in between Pankaj and Suman. Rohan is sitting adjacent to Pankaj who is sitting opposite Arun. 23. Who is sitting opposite Dimple? (A) Piniky (B) Shanu (C) Rohan (D) Cannot be determined 24. Who is sitting opposite Veda? (A) Suman (B) Pinky (C) Shanu (D) Cannot be determined 25. If Rohan is sitting to the left of Veda, then who is sitting opposite Shanu? (A) Rohan (B) Dimple (C) Suman (D) Cannot be determined

Answer Keys

Exercise Practice Problems 1. A 11. D 21. C

2. C 12. C 22. A

3. B 13. C 23. D

4. C 14. D 24. A

5. D 15. D 25. B

6. D 16. B

7. A 17. A

8. C 18. B

9. B 19. B

10. A 20. C

Chapter 8 Puzzles LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • Distribution • Order sequence

• Selections • Questions on routes/networks

PuZZleS In this type of problems, you have to match two or more ‘variables’ (Variable means a ‘subject’ as used in the discussion of Linear Arrangement). In double line-up, the data given may talk of four people living in four houses each of a different colour. What we need to find out is the colour of the house of each of the four persons. There is no first position or second position of the houses. Sometimes, Double line-up is also called as ‘Distribution.’ An example of data given for this variety of questions is: ‘Each of the four persons A, B, C and D wears a different coloured shirt – Red, Pink, Blue and White. A has a Red shirt and D does not have a Pink shirt.’ From the above statement, it becomes clear that no person among A, B, C and D can have shirts of two different colours among Red, Pink, Blue and White. As discussed in the questions on Single Line-up, questions can be solved easily by representing the given data pictorially. In case of Double Line-up, it will help us if we represent the data in the form of a matrix or a table. Let us see how to draw a matrix for the data given above. Colours Names

Red

A

✓

Pink

B C D

✕

Blue

White

As it is given that A has red colour shirt, it is clear that he does not have any other colour shirt. Similarly B, C, D do not Red colour shirt. So, in all the other cells in the row belonging to A, we put a cross (‘✕’). Then, the table will look as follows: Colours Names

Red

Pink

Blue

White

✕

✕

✕

A

✓

B

✕

C

✕

D

✕

✕

In this manner, we can fill up the cells on the basis of the data given to us. Once, we use up all the data, we will draw any conclusions that can be drawn and then answer the questions given in the set. Let us Take a Few Examples Directions for questions 1 to 5: These questions are based on the following information. P, Q, R, S, T, U, V and W are eight employees of a concern. Each is allotted a different locker, out of eight lockers numbered 1 to 8 in a cupboard. The lockers are arranged in four rows with two lockers in each row. Lockers 1 and 2 are in the top row from left to right respectively while lockers 7 and 8 are in the bottom row – arranged from left to right respectively. Lockers 3 and 4 are in the second row from the top – arranged from right to left respectively. So are lockers 5 and

Chapter 8 Puzzles | 1.213 6 – arranged from right to left respectively – in the second row from the bottom. P has been allotted locker 1 while V has been allotted locker 8. T’s locker is just above that of Q which is just above that of R, whereas W’s locker is in the bottom row. Solved Examples Example 1: Which of the following cannot be the correct locker number–occupant pair? (A) 3-Q (B) 7-W (C) 4-U (D) 6-R Example 2: If U’s locker is not beside Q’s locker, whose locker is just above that of W? (A) U (B) S (C) R (D) Q Example 3: Which of these pairs cannot have lockers that are diagonally placed? (A) P-Q (B) S-R (C) U-R (D) Either (B) or (C) Example 4: Which of the following groups consists only occupants of odd numbered lockers? (A) Q, R, W (B) R, V, W (C) T, R, Q (D) P, T, Q Example 5: If U’s locker is in the same row as that of R, and S exchanges his locker with V, then who is the new neighbour of V in the same row? (Assume that nothing else is disturbed from the original arrangement) (A) P (B) Q (C) R (D) U Solutions for questions 1 to 5: Let us first try to locate the lockers in the cupboard as per the conditions given. Then, we will do the allotment to the persons. Lockers 1 and 2 are in the top row and lockers 7 and 8 are in the bottommost row. In these two rows, the lockers are numbered from left to right. In the other two rows, the lockers are numbered from right to left. L

R

1

2

4

3

6

5

7

8

4 6 7-W

3 5 8-V

T’s locker is just above that of Q, which is just above that of R → The lockers of T, Q and R must be 2, 3 and 5 respectively (there are no other group of lockers which satisfy this condition). 1-P 4 6 7-W

2-T 3-Q 5-R 8-V

S and U have lockers 4 and 6 left for them. Thus, on the basis of the data given to us, we can show the final arrangement of lockers as below: 1-P 4-S/U 6-U/S 7-W

2-T 3-Q 5-R 8-V

Now we can answer the questions easily on the basis of the above. Example 1: By looking at the final arrangement of lockers above, we find that choice (D) does not represent the correct combination of locker number-occupant pair. Hence, the correct option is (D). Example 2: If U’s locker is not beside Q’s locker, then U’s locker must be locker 6. So, it is U’s locker that will be immediately above W’s. Hence, the correct option is (A).

Top Row

Example 3: R’s locker is in the same row as that of exactly one of S or U and diagonally placed to the other one. Hence, ‘either S-R or U-R’ is the answer. Hence, the correct option is (D).

Bottom Row

Example 4: The odd-numbered lockers 1, 3, 5 and 7 which belong to P, Q, R and W respectively. Of the choices, we find that Q, R, W appear in choice (A). Hence, this is the correct choice. Hence, the correct option is (A).

Now let us look at the conditions given for the allotment of the lockers. P has locker 1. V has locker 8. 1-P

2

4

3

6

5

7

8-V

Locker of W is in the bottom row → W’s locker must be 7. 1-P 2

Example 5: U’s locker is in the same row as that of R which means that locker 6 belongs to U. So locker 4 belongs to S. Now V and S exchange lockers. Then the new neighbour of V is Q. Hence, the correct option is (B). Directions for questions 6 to 10: These questions are based on the following information. There are four trees – Lemon, Coconut, Mango and Neem – each at a different corner of a rectangular plot. A well is located at one corner and a cabin at another corner.

1.214 | Reasoning Lemon and Coconut trees are on either side of the Gate which is located at the centre of the side opposite to the side at whose extremes, the well and the cabin are located. The mango tree is not at the corner where the cabin is located. Example 6: Which of the following pairs can be diagonally opposite to each other in the plot? (A) Neem tree and Lemon tree (B) Cabin and Neem tree (C) Mango tree and Well (D) Coconut tree and Lemon tree Example 7: If the Lemon Tree is diagonally opposite to the well, then the Coconut tree is diagonally opposite to the (A) Mango tree (B) Well (C) Cabin (D) Gate Example 8: If the Coconut tree and the Neem tree cannot be at adjacent corners of the plot, then which of the following will necessarily have to be at diagonally opposite corners of the plot? (A) Coconut tree and Well (B) Lemon tree and Cabin (C) Lemon tree and Coconut tree (D) Lemon tree and Well Example 9: Which of the following must be TRUE? (A) Cabin and Well are not at adjacent corners. (B) Cabin and Coconut tree cannot be at the adjacent corners. (C) Neem tree and Well are at adjacent corners. (D) Neem tree and Well are not at adjacent corners. Example 10: Which of the following is definitely FALSE? (A) Mango Tree is adjacent to the well at one corner. (B) Neem Tree is adjacent to the Cabin at one corner. (C) Coconut Tree is at the corner adjacent to the Well. (D) Lemon Tree is not on the same side of the plot as the gate. Solutions for questions 6 to 10: Lemon and Coconut are on either sides of the gate. Lemon Or Coconut

Gate

Coconut Or Lemon

(Well, Mango) Or (Cabin, Neem) Lemon or Coconut

(Cabin, Neem) or (Well, Mango) Coconut or Lemon

Gate

Example 6: Let us take each choice and check with the above diagram to see if it is possible or not. Neem and Lemon can be diagonally opposite each other. Hence, this is the correct answer choice. (In an exam, you do not need to check the other choices since the first choice is correct. But, for the sake of clarity and proper understanding, we will check all the choices). From the diagram given above, we can see that Cabin and Neem cannot be located diagonally opposite each other. Mango and Well cannot be located diagonally opposite to each other. Coconut and Lemon cannot be located diagonally opposite each other. Hence, the correct option is (A). Example 7: If Lemon tree is diagonally opposite to the Well, then we can have the following two possible a rrangements. Neem Cabin Lemon

Well Mango Gate

Coconut

OR Well Mango Coconut

Gate

Cabin Neem Lemon

The Coconut tree is diagonally opposite the Cabin and Neem. Hence, the correct option is (C). Example 8: Since Coconut and Neem trees cannot be at adjacent corners, the following arrangements are possible. (Well, Mango)

(Cabin, Neem)

Coconut

Lemon OR

The Well and the Cabin are at either end of the Wall opposite to the Gate. Mango tree and Cabin are not at the same corner. So, Neem tree and Well are not at the same corner. This means that Mango tree and the Well are at the same corner and Neem tree and the Cabin are at the same corner.

(Cabin, Neem)

(Well, Mango)

Lemon

Coconut

From the above diagrams, we find that choice (D) is the correct answer. Hence, the correct option is (D).

Chapter 8 Puzzles | 1.215 Example 9: We can check each statement with the diagram that we drew initially. We find that the statement in Choice (C) which says that Neem tree and the Well are at the adjacent corners is true. Hence, choice (C) is the correct answer. Hence, the correct option is (C). Example 10: We check each statement with the diagram that we drew initially to find out which of the statements has to be false. We find that choice (D) has to be false. Hence, the correct option is (D). Directions for question 11: Select the correct answer from the given choices. Example 11: A, B, C and D play four different games among Baseball, Cricket, Kabaddi and Volley ball. A does not play Baseball or Cricket. B does not play Kabaddi or Volleyball. C plays Volleyball and D plays either Baseball or Volleyball. Who plays Cricket? (A) A (B) B (C) C (D) D Solution for question 11: Example 11: C plays Volleyball. A does not play Cricket and D does not play Cricket as he plays either Baseball or Volleyball. \ B should play Cricket. Hence, the correct option is (B).

Order Sequence The term ‘Order Sequence’ is self-explanatory. In questions for this category, you will be asked to deal with relative positions of subjects. The absolute values of the subjects is not what you should be interested in. It is comparison between different subjects that you have to deal with. The data also specifies the relationships like ‘A is greater than B’ or ‘C is not less than D’ and so on. You have to decide the positions of the subjects in ascending or descending order on the parameters given. The subjects of comparison can be people or things. In short, data will be given to compare the quality or quantity. The parameters on which the subjects are compared can be heights or weights of people, the money with them, complexion, sizes of things, etc. In such questions, you will come across typical statements like ‘A is taller than B,’ ‘B is not shorter than C’, and so on. You may use the following symbols to symbolically represent the conditions given and then later, represent all the subjects pictorially. Greater than Less than Greater than or equal Less than or equal

> < ≥ ≤.

‘Not greater than’ is the same as ‘less than or equal to.’ Similarly, ‘not less than’ is the same as ‘greater than or equal to’.

Words like ‘Who, And, Which, But’ used in the data play a significant role in analysing the data. ‘AND’ and ‘BUT’ play the same role whereas ‘Who’ and ‘Which’ play the same role. Let us take one statement. ‘A is taller than B, who is shorter than C and taller than D but shorter than E, who is taller than F and G but shorter than H’. By using appropriate symbols, the above statement can be represented as follows. A > B; B < C; B > D; B < E; E > F; E > G; E < H Questions on the above data can be as follows.

1. Who is the tallest? 2. Who is the shortest? 3. Who is the second tallest in the group?, etc.

Let us take some examples. Directions for questions 12 to 16: Read the information given below and answer the questions that follow. A, B, C, D and E are five cars while P, Q and R are three motorcycles. A is the fastest of the cars and R is the slowest of the motorcycles. C is costlier than D and Q but cheaper than B. Among cars, A is not the costliest. D is cheaper than E and there is no car whose cost lies between the cost of these two. E is faster than three of the cars and all the motorcycles. Q is costlier than R but cheaper than P, who is faster than Q. Example 12: Which of the following cars cannot stand exactly in the middle position among cars as far as their cost is concerned? (A) A (B) C (C) E (D) D Example 13: Which of the following statements is true about the motorcycles? (A) P is the costliest as well as the fastest motorcycle. (B) The fastest motorcycle is not the costliest motorcycle. (C) The slowest motorcycle is also the cheapest motorcycle. (D) Both (A) and (C) Example 14: If P is costlier than E, how many cars are cheaper than P? (A) 1 (B) 2 (C) 3 (D) Cannot be determined Example 15: If P is cheaper than A which is not costlier than E, which of these is the cheapest of all the cars and motorcycles put together? (A) R (B) Q (C) E (D) Cannot be determined

1.216 | Reasoning Example 16: Which of these is the slowest of the cars, if B and C are faster than D? (A) B (B) D (C) E (D) A Solutions for questions 12 to 16: Let us first write down all the comparisons given for costs and speeds. Then we will tabulate them. Speed A → fastest car E → Faster than three of the cars E is the second fastest car R → slowest motorcycle P>Q Cost C>D C>Q B>C A → Not the costliest among cars E > D → No other car lies between these two Q>R P>Q Now let us tabulate this data. Speed Cars Fastest

A

E

Slowest

Motorcycles Fastest

P

Q

R

Slowest

Cost Cars Costliest

BCED

Cheapest

Here, we know that A is not the costliest car but we do not know where it will fit in. It can come anywhere after B except between E and D. Motorcycles Costliest

P

Q

R

Cheapest

In addition to the above, we have to also keep in mind that C > Q in cost. (From this we can conclude that B > Q, B > R, C > R in cost). Example 12: In terms of cost of the cars, A can come between B and C or between C and E or to the right of E. In each of the above cases, the middle car will be C, A and E respectively. Hence, among the cars given, D cannot be in the middle. Hence, the correct option is (D). Example 13: By looking at the tables above, we can make out that choices (A) and (C) are both correct and hence, the correct answer is choice (D). Hence, the correct option is (D).

Example 14: If P is costlier than E, we can also conclude that it is costlier than D but we cannot conclude anything about the relationship between the cost of P and that of B, C and A. Hence, the correct option is (D). Example 15: Since A is not costlier than E, it means that A is at the same level of E or cheaper than E. We cannot conclude which of these two positions A is in. Hence, we cannot conclude which the cheapest of all the vehicles is. {Please note that if A is the cheapest car, then R will be the cheapest of all the vehicles. However, if A is at the same level as E in cost, then there is a possibility of R or D being the cheapest of all the vehicles.} Hence, the correct option is (D). Example 16: If B and C are faster than D, then the order will be as follows: 1

2

3

4

5

A

E

B/C

C/B

D

Hence, D is the slowest of all the cars. Hence, the correct option is (B). Directions for questions 17 to 21: Read the information given below and answer the questions that follow. J, K, L, M and N are five boys in a class. They are ranked in the order of heights – from the tallest to the shortest – and in order of cleverness – from the cleverest to the dullest. K is taller than N, but not as clever as J and L, whereas M is the cleverest of all but shorter than J. While L is shorter than M but taller than K, L is not as clever as J. No two persons got the same ranks in any of these parameters. Example 17: Who is the third in the order of heights? (A) J (B) N (C) K (D) L Example 18: If N is not the last in at least one of the two comparisons, which of the following is the dullest of all the five? (A) K (B) L (C) M (D) J Example 19: If L is the third in order of cleverness, who is the dullest of all? (A) M (B) N (C) L (D) Cannot be determined Example 20: Who among the following is cleverer as well as taller than K? (A) L and J only (B) N (C) L and N (D) J, L and M Example 21: How many people are definitely shorter than K? (A) 1 (B) 2 (C) 3 (D) None of these

Chapter 8 Puzzles | 1.217 Solutions for questions 17 to 21: Let us first write down all the conditions given and then tabulate the data. Clever ness J>K L>K M is the cleverest. J>L Height K>N J>M M>L L>K

Solution for question 22: Example 22: R and P have at least two girls before them R and P have to be in two out of 3rd, 4th and 5th positions. T and P have not more than one girl behind each of them T and P have to be in the 4th or 5th positions. The above two statements together mean that R will have to be in the third position. Hence, the correct option is (D).

Selections

Now let us put together all the information we have. Cleverness Cleverest

MJLK

Dullest

We do not know where N will come in the order of cleverness but he will definitely be after M. Height Tallest

JMLKN

Shortest

Example 17: From the table above, we can clearly see that L is ranked third in order of heights. Hence, the correct option is (D). Example 18: N is the last in terms of height. Since we are given that he is not the last in at least one of the lists, he CANNOT be the last in cleverness. So, K is the dullest of all. Hence, the correct option is (A). Example 19: If L is the third in the order of cleverness, as can be seen from the table above, either N or K can be the dullest. Hence, the correct option is (D). Example 20: By looking at the tables we made above and from the answer choices, we find that L, J and M are taller as well as cleverer than K. Hence, the correct option is (D). Example 21: Only N is shorter than K. Hence, the correct option is (A). Directions for question 22: Select the correct alternative from the given choices. Example 22: P, Q, R, S, and T are five girls competing in a running race. R and P have at least two girls ahead of each of them. T and P do not have more than one girl behind each of them. Who arrives at the finishing line after two girls as well as before two other girls, if no two girls finish the race at the same time? (A) Q (B) S (C) T (D) R

In this category of questions, a small group of items or persons has to be selected from a larger group satisfying the given conditions. The conditions will specify as to when a particular item or person can be included or cannot be included in the subgroup. For example, the condition may specify that two particular persons should always be together or that two particular persons should not be together. Sometimes, the conditions given for selection or nonselection of items or persons may be based on logical connectives like if-then, either-or, unless, etc. You should be careful in interpreting the logical connectives used in the conditions. Directions for questions 23 to 27: These questions are based on the following information. Amit, Bittu, Chintu, Dumpy, Falgun, Hitesh, Ronit, Purav and Saurav are nine players from among whom three teams consisting respectively of 4 members, 3 members and 2 members must be formed subject to the following conditions. Chintu must have three more players with him while Dumpy must have only two more with him. Chintu and Saurav cannot be in the same team. Purav and Bittu cannot be in the same team. Ronit and Hitesh must be in the same team. Example 23: If Dumpy, Falgun, Purav form the team of 3 members, then which of the following must be TRUE? (A) Hitesh must be in a team with Bittu. (B) Saurav must form a two-member team with Amit or Chintu. (C) Saurav must form a two-member team with Bittu or Amit. (D) Chintu should form a team of 4 members with Hitesh, Ronit and Amit. Example 24: If Dumpy takes Amit as a part of his threemember team, which of the following must go into Chintu’s team? (A) Bittu and Hitesh (B) Hitesh and Ronit (C) Purav and Ronit (D) Purav and Falgun Example 25: If Chintu and Falgun are together and Saurav is in the team of two members, then how many sets of different teams are possible? (A) 4 (B) 3 (C) 2 (D) 1

1.218 | Reasoning Example 26: If Chintu does not have Purav in his team and the two member team consists of Saurav and Amit, then Chintu should take (A) Hitesh, Bittu and Ronit. (B) Bittu but not Ronit. (C) Bittu and Falgun. (D) Hitesh and Ronit. Example 27: If Purav is in the same team as Chintu and Falgun, then Saurav must be in the same team as (A) Bittu (B) Bittu and Amit. (C) Amit (D) Bittu and Dumpy. Solutions for questions 23 to 27: It is given that: Chintu must form a team of 4 members only Dumpy must form a team of 3 members only. Since Chintu and Dumpy are in two different teams, let us, for convenience, denote the two teams as the respective teams of these two persons. Let us call the team with four members as the first team and the team with three members as the second team. The third team should have two persons. Number of members 4 3 2 Chintu Dumpy Saurav Saurav Now let us take the other conditions and fill them up in the table above. Chintu and Saurav cannot be in the same team. Saurav will be in the second or the third team. Purav and Bittu cannot be in the same team. Hitesh and Ronit must be in the same team. We cannot represent these two conditions right now in the table above but we will use them as we go along. Example 23: If Dumpy, Falgun, Purav form the team of 3 members, then Saurav should be in the third team. Since Hitesh and Ronit must be in the same team, they have to be in the first team. That leaves only Amit or Bittu to be with Saurav in the third team. Hence, the correct option is (C). (Also, note that we can eliminate choice (B) easily.) Example 24: Dumpy takes Amit as a member of his team. If we take Hitesh and Ronit as the two members of the third team, then Saurav has to be in the second team, in which case we will have both Purav and Bittu coming into the same team – the first team – which is not possible. Since Saurav cannot be in Chintu’s team and Purav and Bittu cannot be in the same team, the three people required for Chintu’s team will have to be Hitesh and Ronit checkfont Falgun or Purav or Bittu. Hence, the correct option is (B).

Example 25: Let us analyse the conditions. It is given that Chintu and Falgun are together, whereas Saurav is in the team of two members. Let us fill up these details in the box that we made above and then see in how many ways we can fill up the remaining cells in the box. Chintu

Dumpy

Saurav

Falgun

First let us look at Hitesh and Ronit who must be in the same team. They can go into the first team or the second team. Let us consider these two cases. Case 1: Hitesh and Ronit go into the first team. Then, one out of Bittu and Purav will to go into the third team and the other into the second team. This gives rise to two ways of forming the teams – one with Bittu in the second team and the other with Bittu in the third team. Case 2: Hitesh and Ronit go into the second team. In this case too, one out of Bittu and Purav will go into the third team and the other into the second team. Hence, this will also give rise to two ways of forming the teams. Hence, there are total four ways of forming the teams. Hence, the correct option is (A). Example 26: Let us use the table that we built in the initial analysis and fill up the details that we have in this problem. Since the two member team is already formed and Chintu does not take Purav, hence Purav will have to go into the second team. Chintu

Dumpy

Saurav

Purav

Amit

Since Ronit and Hitesh have to be in the same team, they should go into the first team. Since Bittu cannot go with Purav, he should also be in the first team. This leaves Falgun for the second team. Thus, we can fill up the table as follows: Chintu

Dumpy

Saurav

Ronit

Purav

Amit

Hitesh

Falgun

Bittu Hence, the correct option is (A). Example 27: If Purav is with Chintu and Falgun, then Bittu cannot be with them. Since Ronit and Hitesh should be together, the only other person left is Amit. These four members form the first team.

Chapter 8 Puzzles | 1.219 If Hitesh and Ronit together form the two member team, then Bittu and Saurav will be part of the three member team. Instead, if Hitesh and Ronit are in the three-member team, then Saurav and Bittu will form the two-member team. In either case, Saurav and Bittu are together in one team. Hence, the correct option is (A). Directions for questions 28 to 31: These questions are based on the following information. A, B, C, D, E, F and G are seven players. They form two teams of two players each and one team of three players. A and B cannot be in the same team. B and C cannot be in the same team whereas E and F must be in the same team. G and D cannot be in the same team. Example 28: If C, D and A form a team of three players, which of the following can be the members of one of the other teams? (A) A and E (B) G and B (C) E and F (D) Both (B) and (C) Example 29: If E, F and G form a team of three members, then in how many ways can the remaining two teams of two players each be formed? (A) 2 (B) 4 (C) 3 (D) 1 Example 30: If D and A are not in the same team, then altogether in how many ways can the teams of two members be formed? (A) 4 (B) 7 (C) 8 (D) 5 Example 31: If B, E and F form a team of three members, which of the following cannot be the two teams of two members each? (A) AC, GD (B) AD, CG (C) AG, CD (D) Both (A) and (B) Solutions for questions 28 to 31: Let Team I be of 3 players, Team II be of 2 players and Team III be of 2 players. It is given that A and B cannot be together. We will represent it as A × B. Similarly, we have B × C and G × D. E and F must be in the same team. So E and F can form a team of 2 members on their own or can form a team of 3 members with another person. Let us now take up the questions and work them out. Example 28: Given that C, D, A form a team of 3 members, one of the other teams has to have E and F together. Hence, B and G should form one team. Hence, the correct option is (D). Example 29: Given that E, F, G form a team of 3 players. Since A and B or B and C cannot be in the same team, we must necessarily have A and C together in one team and B and D in the other team. So the teams can be formed only in one way. Hence, the correct option is (D).

Example 30: Given that A and D are not in the same team. Hence A × B, B × C, G × D and A × D. We already know that E and F must be in the same team. They may form a team of 3 members or they themselves be a team of 2 members. Let us consider the above two possibilities and then fill up the other teams. They can be formed as follows: Team I

Team II

Team III

1.

AEF

BD

CG

2.

AEF

CD

BG

3.

BEF

AG

CD

4.

CEF

AG

BD

5.

DEF

AC

BG

6.

GEF

AC

BD

7.

ACG

BD

EF

Thus the teams can be formed in 7 ways. Hence, the correct option is (B). Example 31: If B, E, F form a team of 3 members, then the two members teams must be formed from A, C, D, G. The teams can be AD and CG or AG and CD. As D and G cannot form a team, AC and GD cannot be formed. Choice (A). {Please note that we can answer this question from the answer choices – from choice (A), we find that G and D are together in one team which is not possible. Thus, choice (A) is the answer. Hence, the correct option is (A). Directions for question 32: Select the correct alternative from the given choices. Example 32: At least two boys out of A, B, C and D and at least two girls out of P, Q, R and S have to be chosen to form a group of 5 members. Neither A nor C can go with Q. Neither P nor S can go with B. Q and R cannot be together. Which of the following is an acceptable team? (A) ARCQP (B) ASQPD (C) ASQRP (D) PSRAD Solution for question 32: Example 32: The required group of 5 members must be formed with at least two boys from A, B, C, D and at least 2 girls from P, Q, R, S. Answers 1, 2 and 3 can be ruled out as A and Q cannot be together. In choice (D), P, S, R, A, D can be together without violating any of the given conditions. Hence, the correct option is (D).

1.220 | Reasoning Questions on Routes/Networks involve different points or locations between which there is some movement or communication. The way the movement or communication is effected is described in the data/conditions. Sometimes, these are also referred to as ‘Maps’ because the routes given resemble a map. The data given in these types of questions may not always have the word ‘route’ or ‘network’ in them but a network is indicated by some sort of connectivity between two ‘points.’ The way the statements are worded is important. The wording includes statements like 1. Some poles are connected through wires. 2. Some towers send signals to one another. 3. Some cabins, market, cities, etc. are connected via. passages or roads and so on. The connectivity between the two ‘points’ can be only oneway or two-way. In one-way connectivity, the flow will be in only one direction whereas in two-way connectivity, the flow will be in both directions between the points. Read the data carefully and then draw the diagram or network. The words ‘from’ and ‘to’ play an important role in these questions and hence, care should be taken while interpreting the data. While drawing a diagram, arrow marks can be used very effectively to indicate the direction of connectivity as explained below.

1. If the statement mentions that there is a one-way route from city A to B then it can be represented as follows. A B 2. If the statement mentions that cities X and Y have roads on which you can travel in either direction, it means that it is a two-way connectivity. Then it can be represented as follows. X Y 3. If the statement mentions that all the projected roads are one-ways and there is a route from point K to P and then from P to K, then it should be represented as follows. P K i.e., from P to K is one route and from K to P is another route. Now, consider the following network.

K

L

JPO, JPKO, JMLPO, JMLPKO, JLPO, JLPKO This gives us a total of six distinct ways of reaching O from J. Directions for questions 33 to 37: Read the following information and answer the questions given below. P, Q, R, S, T, U, V are seven places on a map. The following places are connected by two-way roads: P and Q; P and U; R and U; R and S; U and V; S and T; Q and R; T and V. No other road exists. Example 33: The shortest route (the route with the least number of intermediate places) from P to V is (A) P-R-V (B) P-T-V (C) P-Q-R-U-V (D) P-U-V Example 34: How many distinct routes exist from S to U (without touching any place more than once)? (A) 3 (B) 2 (C) 1 (D) 4 Example 35: The route covering the maximum number of places and going from P to R does not pass through (A) U (B) T (C) S (D) Q Example 36: If U to V and S to R are only one-way routes, then which of the following places lose contact with P? (A) V (B) T (C) S (D) No place loses contact with P

P

O

In this network, let us say that a person starts from J and he wants to reach K. We want to find out the number of distinct routes he can take without touching any point twice. Starting from J, the possible directions of movement are from J to N, J to M and J to P, but if he goes from J to N then coming back or travelling in some other direction is not possible. Hence, the person has only two options for movement from J (J to M and J to P). Now, if he goes to M, he has to go to L from M. At L, it appears that he has two options – he can go to P or to J. But, since he started from J and as he cannot touch any point twice, he cannot go to J. So, there is only one option at L – that is going to P. So, to reach P from J, there are two options – one directly to P from J and the other via. M and L. Once he reaches P, he has only one way of reaching K – along the diagonal PK. If he goes to O from P, then he cannot travel to K from O (the route is one way in the O to K). Thus the total number of ways from J to K is two (JPK and JMLPK). Thus one has to look at all the possible routes carefully in the manner discussed above. In the above example, if the route between J and L is two-way and then we have to find out the number of ways to reach O stating from J, the routes we have will be as follows:

J N

M

Example 37: If the number of places to which a place is connected directly considered as the measure of importance, then which of the following places is of the highest importance? (A) Q (B) P (C) R (D) S

Chapter 8 Puzzles | 1.221 Solutions for questions 33 to 37: The route map of the places is as follows: Q

R S

P T

U

V

Example 33: As is seen, P-U-V is the shortest way (with only one intermediate point) from P to V. Hence, the correct option is (D). Example 34: To travel from S to U, the routes available are: S-R-U; S-T-V-U and S-R-Q-P-U – a total of 3 routes. Hence, the correct option is (A).

Example 40: If a person visits each of the places starting from P and gets back to P, which of the following places must he visit twice? (A) Q (B) R (C) S (D) T Example 41: Between which two cities among the pairs of cities given below are there maximum travel options available? (If there is more than one route possible between two cities, consider the route with least number of cities enroute.) (A) Q and S (B) P and R (C) P and T (D) Q and R Solutions for questions 38 to 41: Example 38:

air boat, rail

Example 35: First let us write down the route from P to R with the maximum number of intermediate points. By observation, we find that it is P-U-V-T-S-R. It does not touch Q. Hence, the correct option is (D). Example 36: If U to V and S to R are only one-way routes, from the figure, we find that all places can still be reached from P. Hence, none of the places loses contact with P. Hence, the correct option is (D). Example 37: For each of the places given in the choices, Q, P, R and S, let us see how many places are directly connected. Q is directly connected to 2 places. P is directly connected to 2 places. R is directly connected to 3 places. S is directly connected to 2 places. Hence, the correct option is (C). Directions for questions 38 to 41: Read the following information and answer the questions given below. Five cities P, Q, R, S and T are connected by different modes of transport as follows: P and Q are connected by boat as well as by rail. S and R are connected by bus and by boat. Q and T are connected only by air. P and R are connected only by boat. T and R are connected by rail and by bus. Example 38: Which of the following pair of cities are connected by any of the routes directly [without going through any other city]? (A) P and T (B) T and S (C) Q and R (D) None of these Example 39: Which mode of transport would help one to reach R starting from Q but without changing the mode of transport? (A) Boat (B) Rail (C) Bus (D) Air

T

Q

P

rail, bus boat

S

bus, boat

R

P and T are connected through Q. T and S are connected through R Q and R are connected through T or P. Q and S are connected through R and P or T. So none of the pairs in the choices are directly connected. Hence, the correct option is (D). Example 39: From Q to P, he can reach by boat. From P to R also he can travel by boat. So, a person should travel by boat to reach R from Q without changing the mode of transport. Hence, the correct option is (A). Example 40: If a person wants to visit all the places and again return to P, then he can go in the order of P→ Q → T → R → S and then S → R → P (OR) P → R → S and then S→R→T→Q→P He must visit R twice. Hence, the correct option is (B). Example 41: One has to travel between any of the two cities with a restriction that if there is more than one possible route, he has to go by the least number of cities enroute. It is better to take the pair of cities given in each of the choices. Choice (A) Q and S: A person can go from Q to T to R to S (or) Q to P to R to S or vice-versa. In both the routes, there are two cities enroute. We can calculate the number of options in the entire route by multiplying the options available in each segment of the route.

1.222 | Reasoning For the route Q-T-R-S, the number of options = 1 × 2 × 2 = 4 For the route Q-P-R-S, the number of options = 2 × 1 × 2 = 4 Choice (B) P and R: The route between P and R has only one mode of travel, that is boat. Choice (C) P and T: A person can go by PQT in either way or by PRT in either way. If it is by PQT, then the options are boat-air or rail-air i.e. two ways. If it is by PRT, then the options are boat-rail or boat-bus in either way. Hence, they are only two options. Choice (D) Q and R: If a person travels between Q and R, then he can go by QTR or by QPR. If he goes by QTR, then the options are air-rail or air-bus, that gives two options or if he goes by QPR, then the options are rail-boat or boatboat, that will again give us two options. So, it is very clear that Q and S have maximum number of travel options available between them. Hence, the correct option is (A). Directions for question 42: Select the correct alternative from the given choices.

Example 42: Four computers P, Q, A and B are interconnected for the transmission of data. A and B each can send data to both P and Q but B cannot receive data from A. P and Q can have data flow in both directions between them but they cannot transmit the data so received to B but can otherwise send the data directly to B. Which of the following routes can be followed if B has to receive data from A? I. AQPB II. APB III. AQB IV. APQB (A) I and III (B) II and III (C) III and IV (D) All four Solution for question 42: Example 42: AQPB and APQB can be eliminated because P and Q cannot send the data to B. The paths AQB and APB do not violate any conditions and hence can be possible routes to send data from A to B. Hence, the correct option is (B).

Practice Exercise Practice Problems Directions for questions 1 to 3: These questions are based on the following information. Each of the seven delegates A through G came to India to attend a conference from seven different countries – China, Japan, Malaysia, England, Australia, Germany and Poland. (i) China, Japan, and Malaysia are the only Asian countries. (ii) A and B are from Asian countries whereas D is neither from England nor from Australia. (iii) E and F are from non Asian countries but neither of them came from either Australia or England. (iv) C is not from England and the person from Poland is not F. (v) A is from China. 1. Who is from Germany? (A) E (B) C (C) F (D) G 2. Who is from Malaysia? (A) B (B) D (C) A (D) Either (A) or (B) 3. Which country did G come? (A) England (B) Australia (C) Poland (D) Cannot be determined Directions for questions 4 to 6: These questions are based on the following information. Bingo, Pingo, Tingo, Hingo and Mingo are five friends, each of whom is working in a different company among – C1, C2, C3, C4 and C5 and they belongs to the same city but a different locality – l1, l2, l3 and l4 and l5.

(i) The persons who are working with C1 and C2 are from l3 and l4 (ii) Bingo is from l5 but does not work for C5. (iii) Tingo is not from l4 but works for C2. (iv) Pingo works neither for C5 nor in C3 and is not form l2. (v) The person working for C3 is from l1. (vi) Mingo does not work for C3. 4. For which company does Hingo work? (A) C3 (B) C4 (C) C5 5. Who is from l4? (A) Mingo (B) Hingo (C) Tingo (D) Pingo 6. Who works for C4? (A) Bingo (B) Mingo (C) Pingo (D) Hingo

(D) C2

Directions for questions 7 to 9: These questions are based on the following information. A team of three is to be selected from six persons – Amar, Bhavan, Chetan, Dawan, Ekta and Farheen under the following constraints. (i) If Amar or Bhavan is selected, then Chetan must not be selected. (ii) If Chetan or Dawan is selected, then at least one of Ekta and Farheen must be selected. 7. If Dawan is selected, then who must not be selected? (A) Amar (B) Bhavan (C) Chetan (D) None of these

Chapter 8 Puzzles | 1.223 8. If Amar is selected, then in how many ways the team can be selected? (A) 5 (B) 6 (C) 4 (D) 7 9. If Bhavan is selected, then who must be selected? (A) Dawan (B) Ekta (C) Farheen (D) Either (B) or (C) Directions for questions 10 to 12: These questions are based on the following information. Three girls – Anjali, Bharathi and Chandrika and four boys – Kiran, Lala, Manoj and Naveen are to be divided into two teams under the following constraints. (i) Each team must have at least one girl and at least one boy and at least three persons in total. (ii) If Anjali and Bharathi are selected in a team, then the team must have only one boy. (iii) Kiran and Lala cannot be in the same team. (iv) Chandrika and Naveen can be in the same team, only if Bharathi is selected in that team. 10. If Kiran and Chandrika are in the same team, then in how many ways can the other team be selected? (A) 6 (B) 3 (C) 4 (D) 5 11. If Manoj is not in the same team as Bharathi, then in how many ways can the teams be selected? (A) 3 (B) 4 (C) 5 (D) 6 12. If three boys are selected into one team, then in how many ways can the teams be selected? (A) 4 (B) 5 (C) 3 (D) 6 Directions for questions 13 to 15: These questions are based on the following information. Seven persons – P, Q, R, S, T, U and V, who are of different ages, are comparing their ages. We know the following information. (i) P is younger than R, who is not older than S. (ii) S is younger than only two persons. (iii) Q is not the oldest but older than fourth youngest person. (iv) T is older than only U. 13. Who is oldest? (A) S (B) T (C) U 14. Who is the third youngest? (A) V (B) P (C) R 15. Who is the fourth eldest? (A) R (B) P (C) S

(D) V (D) S (D) V

Directions for questions 16 to 19: These questions are based on the diagram given below: Five cities P, Q, R, S and T are connected by one way rail routes as shown below. One takes one hour duration to travel between any two directly connected cities.

T

P

Q

S

R

At station S, for every 2 hours one train departs and the departure time of the first train is 6:00 a.m. Similarly at station R, for every 3 hours one train departs in each route and the departure time of the first train is 4 a.m. At station Q, one train departs for every 2 hour in each route and the departure time of the first train is 7 a.m. At station P, train departs for every 1 hour in each route and the departure time of the first train is 8 a.m. At station T, for every 3 hours one train departs and the departure time of the first train is 5:30 a.m. 16. What is the least time will it take to reach P from R, if one takes the longest route without visiting any station more than once? (A) 6 hours (B) 3 hours (C) 5 hours (D) 4 hours 17. If a person reaches Q at 1:00 p.m. from R, which of the following can be the time at what time he must have started from R if that person takes the shortest route? (A) 10:00 a.m. (B) 9:00 a.m. (C) 11:00 a.m. (D) None of these 18. A person wants to travel from R to T and he takes the longest route without visiting any station more than once. If he starts at 4:00 a.m. then for how much time he has to wait for the trains altogether in all stations before reaching T. (A) 2 hours (B) 3 hours (C) 2 hours (D) 1 hour 19. If a person starts from P at 10:00 a.m. to reach S and he takes the longest route without visiting any station more than once then at what time will he be reachings? (A) 1:00 p.m. (B) 4:00 p.m. (C) 5:00 p.m. (D) 2:00 p.m. Directions for question 20: This question is based on the information given below. Five cities - Ahmedabad, Bangalore, Calicut, Delhi and Indore are connected by one way routes from Ahmedabad to Bangalore, Delhi to Ahmedabad, Indore to Delhi, Delhi to Calicut, Ahmedabad to Calicut, Bangalore to Calicut, Calicut to Indore, Indore to Bangalore and Ahmedabad to Indore. 20. In how many ways a person can travel from Delhi to Indore without visiting any city more than once? (A) 5 (B) 3 (C) 6 (D) 4

1.224 | Reasoning Directions for questions 21 and 22: Select the correct alternative from the given choices. 21. There are 15 identical coins out of which fourteen are of equal weights and one coin lighter than each of the other coins. What is the minimum number of weighings required using a common balance to definitely identify the counterfeit coin? (A) 3 (B) 4 (C) 5 (D) None of these 22. Beside a lake, there are three temples and a flower garden. Whenever some flowers are dipped into the lake, the flowers gets triplet. A person brought some flowers from the garden and dipped then into the lake. He placed x flowers in front of the first temple and dipped the remaining flowers into the lake. He placed x flowers in front of the second temple and dipped the remaining flowers into the lake. Now, he placed x flowers in front of the third temple and has no flowers. Which of the following numbers can be the value of x? (A) 9 (B) 18 (C) 27 (D) 36 Directions for questions 23: These questions are based on the following letter – multiplication in which each letter is represents a unique non-zero digit. A B C × C B A C D E F C Also, it is known that D = 3C and F = 4B 23. What is the value of D? (A) 3 (B) 6 (C) 9 (D) Cannot be determined

Directions for questions 24 and 25: These questions are based on the following data. Each individual of a city called ‘Josh’ belongs exactly to one of the two types, viz., Yes-type or No-type. Yes-type people always give the true reply while the No-type always lies. Answer the following questions based on the above information. 24. You met three residents A, B and C, of the city and asked them, ‘who among you are married?’ and got the following replies. A: I am married to B. B: I am married to C. C: I am not married to A. If it is further known that A is married to one of B and C and there is exactly one married couple among the three, then which of the following is definitely true? (A) C is married to A. (B) B is married to A. (C) A is of Yes-type. (D) B is of No-type. (E) A is of No-type. 25. You approached three inhabitants A, B and C of the city and asked them, ‘Who is of No-type among you?’, and got the following replies. A: B is of No-type. B: C is of No-type. C: A is of No-type. It can be concluded that: (A) A is a No-type. (B) B is a No-type. (C) C is a No-type. (D) Either A or B is of No-type. (E) Data inconsistent.

Answer Keys

Exercise Practice Problems 1. C 11. D 21. A

2. D 12. A 22. C

3. A 13. D 23. C

4. A 14. B 24. D

5. D 15. A 25. C

6. A 16. B

7. D 17. A

8. A 18. D

9. D 19. C

10. D 20. D

Chapter 9 Clocks and Calendars LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • •

Clocks Calendar Leap and non-leap years Counting the number of odd days

CloCKs The hour hand and the minute hand of a clock move in relation to each other continuously and at any given point of time, they make an angle between 0° and 180° with each other. If the time shown by the clock is known, the angle between the hands can be calculated. Similarly, if the angle between two hands is known, the time shown by the clock can be found out. When we say angle between the hands, we normally refer to the acute/obtuse angles (upto 180°) between the two hands and not the reﬂex angle (> 180°). For solving the problems on clocks, the following points will be helpful. • Minute hand covers 360° in 1 hour, i.e., in 60 minutes. Hence, MINUTE HAND COVERS 6° PER MINUTE. • Hour hand covers 360° in 12 hours. Hence, hour hand covers 30° per hour. Hence, HOUR HAND COVERS 1/2° PER MINUTE. The following additional points also should be remembered. In a period of 12 hours, the hands make an angle of • 0° with each other (i.e., they coincide with each other), 11 times. • 180° with each other (i.e., they point exactly in opposite directions), 11 times. • 90° or any other angle with each other, 22 times. Note: We can also solve the problems on clocks using the method of ‘Relative Velocity’

• Counting of number of odd days, when only one date is given • Counting number of odd days, when two dates are given

In 1 minute, Minute Hand covers 6° and Hour hand covers 1/2°. 1 Therefore, Relative Velocity = 6 – 1/2 = 5 ° per minute. 2 Alternately, in 1 hour, the minute hand covers 60 minute divisions whereas the hour hand covers 5 minute divisions. \

Relative Speed = 60 – 5 = 55 minutes per hour.

However, adopting the approach of actual angles covered is by far the simplest and does not create any confusion.

Points to Note • Any angle other than (0° and 180°) is made 22 times in a period of 12 hours. • In a period of 12 hours, there are 11 coincidences of the two hands, when the two hands are in a straight line facing opposite directions. • The time gap between any two coincidences is 12/11 hours or 5 65 minutes. 11 • If the hands of a clock (which do not show the correct time) coincide every p minutes, then 5 If p > 65 , then the watch is going slow or losing time. If 11 5 p < 65 , then the watch is going fast or gaining time. 11 To calculate the angle ‘q ’ between the hands of a clock, we use the following formula (where m = minutes and h = hours)

1.226 | Reasoning 1. q=

11 11 ⎞ ⎛ m > 30 h ⎟ m − 30 h ⎜ when 2 2 ⎝ ⎠

2. q = 30 h −

11 ⎛ 11 ⎞ m m ⎜ when 30 h > 2 2 ⎟⎠ ⎝ Solved Examples

Example 1: What is the angle between the minute hand and the hour hand of a clock at 3 hours 40 minutes? (A) 20° (B) 70° (C) 90° (D) 130° Solution: (D) The angle between the hands can be calculated by q = 11 m − 30 h , where m is minutes and h is hours. Here, 2 m = 40 and h = 3 \

q=

11 × 40 − 30 × 3 = |220 – 90| = 130° 2

The angle between the two hands is 130°. Example 2: Find the time between 2 and 3 O’clock at which the minute hand and the hour hand make an angle of 60° with each other. 11 Solution: In the formula q = m − 30 h , 2 q = 60° and h = 2 11 \ 60 = m – 30 × 2 2 11 m = 120 2 240 9 m= = 21 min past 2 11 11 11 (or) 60 = 30 × 2 – m 2 11 \ m = 0 2 m = 0 Therefore, the angle between the hour hand and the minute 9 hand is 60° at 2 O’clock and at 21 minutes past 2 O’clock. 11 Example 3: Find the time between 2 and 3 O’clock at which the minute hand and the hour hand overlap. Solution: When the two hands overlap, the angle between them is 0°. 11 q= m − 30 h 2

\

q = 0° and h = 2

11 m = 30 × 2 2 120 10 m= = 10 min past 2. 11 11

Example 4: Find the time between 2 and 3 O’clock at which the minute hand and the hour hand are perpendicular to each other. Solution: When two hands are perpendicular, q = 90° and h=2 11 ⎞ ⎛ 11 ⎞ ⎛ \ q = ⎜ m − 30h ⎟ or ⎜ 30 h − m ⎟ 2 ⎠ ⎝2 ⎠ ⎝

11 11 m – 30 × 2; m = 150 2 2 300 3 m= = 27 minutes past 2 11 11 11 (or) 90 = 30 × 2 – m 2 11 ⇒ m = –30 2 As m cannot be negative, this case is not possible. So, the hands are perpendicular to each other only once i.e., 3 at 27 minutes past 2 O’clock. 11 90 =

Example 5: Find the time between 2 and 3 O’clock at which the minute hand and the hour hand are on the same straight line but are pointing in opposite directions. Solution: When two hands are pointing opposite directions and are on a straight line the angle between them would be 180°. i.e. q = 180° and h = 2. 11 11 m – 30 h; m = 180 + 60 = 240 2 2 480 7 m= = 43 11 11 7 So, at 43 minutes past 2 O’clock the hands will be at 180°. 11 180°=

Calendar Suppose you are asked to find the day of the week on 30th June, 1974, it would be a tough job to find it if you do not know the method. The method of finding the day of the week lies in the number of ‘odd days’. Note: Every 7th day will be the same day count wise, i.e. if today is Monday, then the 7th day counting from Tuesday onwards will once again be Monday. Odd days is the days remaining after completion of an exact number of weeks. Odd days is the reminder obtained on dividing the total number of days with seven. Example: 52 days ÷ 7 = 3 odd days. Leap and Non-leap Years: A Non-leap year has 365 days whereas a leap year has one extra day because of 29 days in the month of February. Every year which is divisible by 4 is called a leap year. Leap year consists of 366 days, (52 complete weeks + 2 days), the extra two days are the odd days. So, a leap year has two odd days.

Chapter 9 Clocks and Calendars | 1.227 An non-leap year consists of 365 days (52 complete weeks + 1 day). The extra one day is the odd day.

Hence, 14th April 1993 is one day after Sunday, i.e., Monday.

Note: Every century, year which is a multiple of 400, is a leap year. A century year which is not divisible by 400 is a non-leap year.

Example 7: If 1st Jan, 1992 is a Tuesday then on which day of the week will 1st Jan, 1993 fall? (A) Wednesday (B) Thursday (C) Friday (D) Saturday

Example: 400, 800, 1200, 1600 … are leap years. 500, 700, 900, 1900 … are non-leap years. Counting the number of Odd Days: 100 years consist of 24 leap years + 76 ordinary years. (100 years when divided by 4, we get 25. But at the 100th year is not a leap year, hence only 24 leap years).

= 2 × 24 odd days + 1 × 76 odd days = 124 days

= 17 weeks + 5 days

The extra 5 days are the odd days. So, 100 years contain 5 odd days. Similarly, for 200 years we have 10 extra days (1 week + 3 days). \ 200 years contains 3 odd days. Similarly, 300 years contain 1 odd day and 400 years contain 0 odd days. Counting of number of odd days, when only one date is given: Here we take January 1st 1 AD as the earlier date and we assume that this day is a Monday. We take its previous day, i.e., Sunday as the reference day. After this the above mentioned method is applied to count the number of odd days and find the day of the week for the given date. Counting number of odd days, when two dates are given: Any month which has 31 days has 3 odd days. ( ∵ 31 ÷ 7 leaves 3 as remainder) and any month which has 30 days has 2 odd days (30 ÷ 7 leaves 2 as remainder). Then, the total number of odd days are calculated by adding the odd days for each month. The value so obtained is again divided by 7 to get the final number of odd days. The day of the week of the second date is obtained by adding the odd days to the day of the week of the earlier date. Example 6: If you were born on 14th April, 1992, which was a Sunday, then on which day of the week does your birthday fall in 1993? (A) Monday (B) Tuesday (C) Wednesday (D) Cannot be determined Solution: (A) 14th April 1992 to 14th April 1993 is a complete year, which has 365 days. Hence, the number of odd days from 14th April 1992 to 14th April 1993 is 1.

Solution: (B) Since 1992 is a leap year there are 2 odd days. Hence, 1st January 1992 is two days after Tuesday, i.e., Thursday. Example 8: If 1st April, 2003 was Monday, then which day of the week will 25th December of the same year be? (A) Tuesday (B) Wednesday (C) Thursday (D) Friday Solution: (B) The number of days from 1st April to 25th December (29 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 25) days = 268 days 268 = = 38 + 2 odd days. 7 Hence, 25th December is two days after Monday, i.e., Wednesday. Example 9: Which year will have the same calendar as that of 2005? (A) 2006 (B) 2007 (C) 2008 (D) 2011 Solution: (D) Year: 2005 + 2006 + 2007 + 2008 + 2009 + 2010 Odd days : 1+ 1+ 1+ 2+ 1 + 1 Total number of odd days from 2005 to 2010 are 7 ≅ 0 odd days. Hence, 2011 will have the same calendar as that of 2005. Example 10: What day of the week was 18th April 1901? (A) Monday (B) Tuesday (C) Wednesday (D) Thursday Solution: (D) 18th April 1901 ⇒ (1600 + 300) years + 1st January to 18th April 1901. 1600 years have – 0 odd days 300 years have – 1 odd day The number of days from 1st January, 1901 to 18th April 1901 is (31 + 28 + 31 + 18) days 108 days @ 3 odd days \ Total number of odd days = 3 + 1 = 4 Hence, 18th April 1901 is Thursday.

1.228 | Reasoning

Practice Exercise Practice Problems Directions for questions 1 to 25: Select the correct alternative from the given choices. 1. What is the angle covered by the minute hand in 22 minutes? (A) 66° (B) 110° (C) 132° (D) 220° 2. By how many degrees will the minute hand move, in the same time, in which the hour hand moves 6°? (A) 54° (B) 84° (C) 72° (D) 60° 3. What is the angle between the hands of the clock, when it shows 40 minutes past 6? (A) 40° (B) 70° (C) 80° (D) 90° 4. What is the angle between the two hands of a clock when the time is 25 minutes past 7 O’clock? 1 1 (A) 62 ° (B) 66 ° 2 2 1 1 (C) 72 ° (D) 69 ° 2 2 5. At what time between 9 and 10 O’clock, will the two hands of the clock coincide? 3 (A) 43 minutes past 9 O’clock 11 6 (B) 45 minutes past 9 O’clock 11 1 (C) 49 minutes past 9 O’clock 11 6 (D) 49 minutes past 9 O’clock 11 6. At what time between 4 and 5 O’clock are the two hands of a clock in the opposite directions? 3 (A) 52 minutes past 4 O’clock 11 6 (B) 54 minutes past 4 O’clock 11 7 (C) 51 minutes past 4 O’clock 11 9 (D) 53 minutes past 4 O’clock 11 7. The angle between the two hands of a clock is 20° and the hour hand is in between 2 and 3. What is the time shown by the clock? 3 (A) 7 minutes past 2 11 6 (B) 14 minutes past 2 11 5 (C) 15 minutes past 2 11 (D) Both (A) and (B)

8. Which of the following can be the time shown by the clock, when the hour hand is in between 4 and 5 and the angle between the two hands of the clock is 60°? 4 9 16 min past 4 (B) 18 min past 4 (A) 11 11 8 5 (C) 32 min past 4 (D) 36 min past 4 11 11 9. How many times, the two hands of a clock will be at 30° with each other in a day? (A) 36 (B) 40 (C) 44 (D) 48 10. If the time in a clock is 10 hours 40 minutes, then what time does its mirror image show? (A) 2 hours 20 minutes (B) 1 hour 15 minutes (C) 1 hour 10 minutes (D) 1 hour 20 minutes 11. There are two clocks on a wall, both set to show the correct time at 5:00 p.m. The clocks lose 2 minutes and 3 minutes respectively in an hour. When the clock which loses 2 minutes in one hour shows 9:50 p.m. on the same day, then what time does the other clock show? (A) 9:30 p.m. (B) 9:40 p.m. (C) 9:45 p.m. (D) 10:15 p.m. 12. A watch which gains uniformly was observed to be 1 minute slow at 8:00 a.m. on a day. At 6:00 p.m. on the same day it was 1 minute fast. At what time did the watch show the correct time? (A) 12:00 noon (B) 1:00 p.m. (C) 2:00 p.m. (D) 3:00 p.m. 13. A watch, which gains uniformly was observed to be 6 minutes slow at 9:00 a.m. on a Tuesday and 3 minutes fast at 12:00 noon on the subsequent Wednesday. When did the watch show the correct time? (A) 9:00 p.m. on Tuesday (B) 12:00 a.m. on Wednesday (C) 3:00 a.m. on Wednesday (D) 6:00 a.m. on Wednesday 14. The number of odd days in a non-leap year is (A) 0 (B) 1 (C) 2 (D) 3 15. What will be next leap year after 2096? (A) 2100 (B) 2101 (C) 2104 (D) 2108 16. If 21st March 2000 was a Monday, what day of the week will 21st March 2003 be? (A) Tuesday (B) Friday (C) Thursday (D) Wednesday 17. If 5th January 2001 was a Friday then what day of the weak will 25th December 2001 be? (A) Monday (B) Tuesday (C) Wednesday (D) Thursday 18. If 14th February 2001 was a Wednesday, then what day of the week will 14th February 2101 be (i.e., after a century)?

Chapter 9 Clocks and Calendars | 1.229 (A) Friday (B) Saturday (C) Sunday (D) Monday 19. If 8th February 1995 was a Wednesday, then what day of the week will 8th February 1994 be? (A) Wednesday (B) Thursday (C) Tuesday (D) Monday 20. If holidays are declared only on Sundays and in a particular year 12th March is a Sunday, is 23rd September in that year a holiday? (A) Yes (B) No (C) Yes, if it is a leap year. (D) No, if it is a leap year. 21. Which day of the week was 1601, Jan 15? (A) Monday (B) Tuesday (C) Wednesday (D) Thursday 22. In a year, if 23rd November is a Friday then what day of the week will 14th March in that year be? (A) Monday (B) Wednesday (C) Sunday (D) Cannot be determined

23. The calendar of which of the following years is the same as that of the year 2001? (A) 2005 (B) 2006 (C) 2007 (D) 2008 24. Pankaj met his friend three days ago. He told his friend that he has his last exam five days later. He met his friend again, three days after the last exam. Six days after he met his friend after the last exam, they left for a vacation. The day on which they left for a vacation is a Saturday. What is today? (A) Saturday (B) Tuesday (C) Sunday (D) Cannot be determined 25. Five days ago Shweta lost her phone. Two days after loosing the phone she lodged a complaint with the police. Six days after lodging the complaint she bought a new phone. Four days after buying a new phone i.e., on a Thursday she found her old phone. On which day did she loose her phone? (A) Friday (B) Saturday (C) Thursday (D) None of these

Previous Years’ Questions 1. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. The number of persons playing neither hockey nor football is _____. [2010] (A) 2 (B) 17 (C) 13 (D) 3 2. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed: Worker [2010] (A) fallow : land (B) unaware : sleeper (C) wit : jester (D) renovated : house 3. Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters). All were born on 1st January, in different years. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts: (i) Hari’s age + Gita’s age > Irfan’s age + Saira’s age (ii) The age difference between Gita and Saira is 1 year. However Gita is not the oldest and Saira is not the youngest. (iii) There are no twins. Which of the following in a possible order in which they were born? [2010] (A) HSIG (B) SGHI (C) IGSH (D) IHSG

4. Given the sequence of terms AD, CG, FK, JP, ? the next term is [2012] (A) OV (B) OW (C) PV (D) PW 5. A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x – 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is [2012] (A) 8 meters (B) 10 meters (C) 12 meters (D) 14 meters 6. Few school curricula include a unit on how to deal with bereavement and grief, and yet all students at some point in their lives suffer from losses through death and parting. Based on the above passage which topic would not be included in a unit on bereavement? [2012] (A) How to write a letter of condolence (B) What emotional stages are passed through in the healing process (C) What the leading causes of death are (D) How to give support to a grieving friend. 7. After several defeats in wars, Robert Bruce went in exile and wanted to commit suicide. Just before committing suicide, he came across a spider attempting tirelessly to have its net. Time and again, the spider failed but that did not deter it to refrain from

1.230 | Reasoning making attempts. Such attempts by the spider made Bruce curious. Thus, Bruce started observing the near-impossible goal of the spider to have the net. Ultimately, the spider succeeded in having its net despite several failures. Such act of the spider encouraged Bruce not to commit suicide. And then, Bruce went back again and won many a battle, and the rest is history. Which one of the following assertions is best supported by the above information? [2013] (A) Failure is the pillar of success (B) Honesty is the best policy (C) Life begins and ends with adventures (D) No adversity justifies giving up hope 8. The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the Western Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons for the formation of this gap are not clear. It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighboring regions of Kerala having higher summer temperatures. [2014] What can be inferred from this passage?

(A) The Palghat gap is caused by high rainfall and high temperatures in Southern Tamil Nadu and Kerala. (B) The regions in Tamil Nadu and Kerala that are near the Palghat Gap are near the low – lying. (C) The low terrain of the Palghat Gap has a significant impact on weather patterns in neighbouring parts of Tamil Nadu and Kerala. (D) Higher summer temperatures result in higher rainfall near the Palghat Gap area.

9. Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and Schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely? [2014] (A) Strategies are now available for eliminating psychiatric illnesses. (B) Certain psychiatric illnesses have a genetic basis. (C) All human diseases can be traced back to genes and how they are expressed. (D) In the future, genetics will become the only relevant field for identifying psychiatric illness. 10. The old city of Koenigsberg, which had a German majority population before World War 2, is now called Kaliningrad. After the events of the war, Kaliningrad is now a Russian territory and has a predominantly

Russian population. It is bordered by the Baltic Sea on the north and the countries of Poland to the South and West Lithuania to the east respectively. Which of the statements below can be inferred from this passage? [2014] (A) Kaliningrad was historically Russian in its ethnic make up (B) Kaliningrad is part of Russia despite it not being contiguous with the rest of Russia. (C) Koenigsberg was renamed Kaliningrad, as that was its original Russian name (D) Poland and Lithuania are on the route from Kalinigrad to the rest of Russia. 11. The number of people diagnosed with dengue fever (contracted from the bite of a mosquito) in north India is twice the number diagnosed last year. Municipal authorities have concluded that measures to control the mosquito population have failed in this region. Which one of the following statements, if true, does not contradict this conclusion? [2014] (A) A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent. (B) More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency. (C) Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test (D) The number of people with malarial fever (also contracted from mosquito bites) has increased this year. 12. At what time between 6 a.m. and 7 a.m, will the minute hand and hour hand of a clock make an angle closest to 60°?[2014] (A) 6:22 a.m (B) 6:27 a.m (C) 6:38 a.m (D) 6:45 a.m 13. Which number does not belong in the series below? 2, 5, 10, 17, 26, 37, 50, 64 [2014] (A) 17 (B) 37 (C) 64 (D) 26 14. A dance programme is scheduled for 10.00 a.m. Some students are participating in the programme and they need to come an hour earlier than the start of the event. These students should be accompanied by a parent. Other students and parents should come in time for the programme. The instruction you think that is appropriate for this is[2014] (A) Students should come at 9.00 a.m. and parents should come at 10.00 a.m. (B) Participating students should come at 9.00 a.m. accompanied by a parent, and other parents and students should come by 10.00 a.m.

Chapter 9 Clocks and Calendars | 1.231

(C) Students who are not participating should come by 10.00 a.m. and they should not bring their parents. Participating students should come at 9.00 a.m. (D) Participating students should come before 9.00 a.m. Parents who accompany them should come at 9.00 a.m. All others should come at 10.00 a.m.

15. By the beginning of the 20th century, several hypotheses were being proposed, suggesting a paradigm shift in our understanding, of the universe. However, the clinching evidence was provided by experimental measurements of the position of a star which was directly behind our sun. Which of the following inference(s) may be drawn from the above passage? (i) Our understanding of the universe changes based on the positions of stars. (ii) Paradigm shifts usually occur at the beginning of centuries. (iii) Stars are important objects in the universe. (iv) Experimental evidence was important in confirming this paradigm shift. [2014] (A) (i), (ii) and (iv) (B) (iii) only (C) (i) and (iv) (D) (iv) only 16. The given statement is followed by some courses of action. Assuming the statement to be true, decide the correct option. Statement: There has been significant drop in the water level in the lakes supplying water to the city. Course of action:[2015] (I) The water supply authority should impose a partial cut in supply to tackle the situation. (II) The government should appeal to all the residents through mass media for minimal use of water. (III) The government should ban the water supply in lower areas. (A) Statements I and II follow. (B) Statements I and III follow. (C) Statements II and III follow. (D) All statements follow. 17. Four branches of a company are located at M, N, O and P. M is north of N at a distance of 4 km; P is south of O at a distance of 2 km; N is southeast of O by 1 km. What is the distance between M and P in km? [2015]

(A) 5.34 (C) 28.5

(B) 6.74 (D) 45.49

18. If ROAD is written as URDG, then SWAN should be written as:[2015] (A) VXDQ (B) VZDQ (C) VZDP (D) UXDQ 19. The head of a newly formed government desires to appoint five of the six selected members P, Q, R, S, T and U to portfolios of Home, Power, Defense, Telecom and Finance. U does not want any portfolio if S gets one of the five. R wants either Home or Finance or no portfolio. Q says that if S gets either Power or Telecom, then she must get the other one. T insists on a portfolio if P gets one. Which is the valid distribution of portfolios?[2015] (A) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance. (B) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance. (C) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance. (D) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance. 20. If ‘relftaga’ means carefree, ‘otaga’ means careful and ‘fertaga’ means careless, which of the following could mean ‘aftercare’? [2016] (A) zentaga (B) tagafer (C) tagazen (D) relffer 21. Pick the odd one from the following options. [2016] (A) CADBE (B) JHKIL (C) XVYWZ (D) ONPMQ 22. Among 150 faculty members in an institute, 55 are connected with each other through Facebook® and 85 are connected through WhatsApp®. 30 faculty members do not have Facebook® or WhatsApp® accounts. The number of faculty members connected only through Facebook® accounts is ______. [2016] (A) 35 (B) 45 (C) 65 (D) 90 2 3. All hill – stations have a lake. Ooty has two lakes. Which of the statement(s) below is/are logically valid and can be inferred from the above sentences? [2016] (i) Ooty is not a hill- station. (ii) No hill – station can have more than one lake. (A) (i) only (B) (ii) only (C) both (i) and (ii) (D) neither (i) nor (ii)

1.232 | Reasoning

Answer Keys

Exercise Practice Problems 1. C 11. C 21. A

2. C 12. B 22. B

3. A 13. C 23. C

4. C 14. B 24. B

5. C 15. C 25. B

6. B 16. C

7. D 17. B

8. C 18. D

9. C 19. C

10. D 20. B

4. A 14. B

5. B 15. D

6. C 16. A

7. D 17. A

8. C 18. B

9. B 19. B

10. B 20. C

Previous Years’ Questions 1. D 11. D 21. D

2. A 12. A 22. A

3. B 13. C 23. D

Test Reasoning Directions for questions 1 to 7: Complete the following series. 1. 11, 26, 51, 76, ____ (A) 101 (B) 115 (C) 125 (D) 133 2. 23, 57, 1113, 1719, ____ (A) 2329 (B) 2931 (C) 3137 (D) 3743 3. VIQ, TAC, WJR, VCE, XKS, XEG, _____ (A) YGL (B) ZFH (C) YLT (D) YNR 4. 25 : 343 : : 49 : _____ (A) 121 (B) 343 (C) 512 (D) 1331 5. BIDM : DLPR : : HSBC : _____ (A) PVEH (B) PXDH (C) PVHH (D) RVHD 6. 2Y5 : 4W9 : : 3J6 : _____ (A) 4W9 (B) 6L4 (C) 8C1 (D) 6N4 7. Aeroplane : Pilot : : Ship : _____ (A) Driver (B) Chef (C) Captain (D) Marshal Directions for questions 8 to 11: Find the odd man out. 8. (A) 38 – 121 (B) 48 – 144 (C) 68 – 196 (D) 98 – 361 9. (A) BDGC (B) DHKR (C) FLOH (D) EJMZ 10. (A) 6V12 (B) 2H4 (C) 9F18 (D) 3R6 11. (A) Mercury (B) Mars (C) Moon (D) Venus Directions for questions 12 to 14: Choose the correct alternative from the given choices. 12. In a certain code language the word PRIVATE is coded as AEIPRTV then how is the word PRESENT coded in that language? (A) EEPNRST (B) EENPRST (C) EPSNERT (D) EENRPST 13. In a certain code, if the word CHLORATE is written as DFOKWUAW then how is the word PHOSPHATE written in that code? (A) QFRUOBHLN (B) QFROBUHLN (C) QFHROUBLN (D) QFROUBHLN 14. In a certain code, if the word PRESSURE is written as KIVHHFIV then how is the word SOLUTION written in that code? (A) HLOUTRLM (B) HLPGFRLM (C) HLOFGRLM (D) HLOGTROM Directions for questions 15 and 16: These questions are based on the following information. Six persons – P through U – are standing in a queue in the increasing order of their heights so that the shortest is at the front of the queue and the tallest is at the back. Further,

Time: 25 min. (i) U is the shortest. (ii) Exactly two persons are taller than T. (iii) P is taller than S and exactly two persons stand between P and S. (iv) Q is taller than P. 15. 16.

Who is the second tallest person? (A) T (B) R (C) S (D) P Who is/are the persons in between P and R? (A) Only T (B) Q and S (C) T and U (D) Only S

Directions for questions 17 to 19: These questions are based on the following data. Eight persons – A, B, C, D, E, F, G and H attended a conference and are sitting around a circular table. Among them, there are C.E.Os of 4 companies who came along with one assistant each. Each C.E.O. has his assistant sitting to his right. (i) Assistants of C and A are sitting opposite each other. (ii) E, who is the assistant of B, is sitting opposite F. (iii) E was not sitting adjacent to A. (iv) G is neither adjacent nor opposite to D. 17. 18. 19.

Who is to the left of A? (A) D (B) G (C) F (D) H Who is the assistant of C? (A) D (B) G (C) E (D) H If H is opposite to G, then D is to the right of ____. (A) A (B) B (C) F (D) H

Directions for questions 20 to 24: These questions are based on the following data. In a class, 50 students failed in Maths. 40 students failed in Physics. 30 students failed in Chemistry. 10 students failed in Physics and Chemistry. 10 students failed in Maths and Physics. No student failed in both Maths and Chemistry. None of the students failed in all the three subjects. 20. How many students failed atleast in one subject? (A) 50 (B) 100 (C) 75 (D) 125 21. What is the ratio of the number of students who failed in Maths and Physics to that who failed in Physics and Chemistry? (A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 4 : 3 22. How many students failed in exactly two subjects? (A) 10 (B) 20 (C) 30 (D) 40 23. The number of students who failed in only Maths, in only Physics and in only Chemistry respectively is (A) 40, 20, 20 (B) 20, 40, 20 (C) 20, 20, 40 (D) 50, 40, 30

1.234 | Reasoning 24. Which of the following statements is true? (A) The number of students who failed in only Maths equals to that of the students who failed in only Physics. (B) The number of students who failed in only Maths equals to that of the students who failed in only Physics or in only Chemistry. (C) The number of students who failed in all the three subjects is 10. (D) None of these. Directions for question 25 to 30: select the correct alternative from the given choices. 25. In the following addition each latter represents a different digit from of O to 9. which of the following is a possible number represented by FAN ? H A N F A F N 5 5 8 8 (A) 434 (B) 534 (C) 345 (D) 135 26. At what time between 10 O’ clock and 11 O’ clock are the hands of the clock together?

6 8 past 10 (B) 27 past 10 11 11 (C) Both A and B (D) None of these 27. How many times the hands of a clock are at right angles in a day? (A) 24 (B) 22 (C) 44 (D) 48 28. On a particular day if it is found that a clock is showing 10 minutes less at 1:00 pm and 5 minutes more at 6:00 pm on the same day. At what time did the clock show the correct time? (A) 3 hr 20 min (B) 4 hr 20 min (C) 5 hr 40 min (D) 6 hr 40 min st 29. If 21 March 2000 was a Monday, which day of the week will be 21st March 2003? (A) Tuesday (B) Friday (C) Thursday (D) Wednesday 30. The movie of my favorite hero is going to be released on Wednesday. To watch the movie on the first day of release, I booked my ticket the day before yesterday. If I watch the movie on the fourth day from today, on which day of the week did I book my ticket? (A) Wednesday (B) Thursday (C) Friday (D) Saturday 54 (A)

Answer Keys 1. B 11. C 21. C

2. A 12. B 22. B

3. C 13. D 23. A

4. D 14. C 24. B

5. C 15. D 25. B

6. B 16. A 26. A

7. C 17. C 27. C

8. D 18. D 28. B

9. A 19. D 29. C

10. A 20. B 30. B

Engineering Mathematics Chapter 1: Mathematical Logic

2.237

Chapter 2: Probability2.250 Chapter 3: Set Theory and Algebra

2.266

Chapter 4: Combinatorics2.287 Chapter 5: Graph Theory2.299 Chapter 6: Linear Algebra

2.317

Chapter 7: Calculus2.333

P A R T II

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Chapter 1 Mathematical Logic LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • • • • • • •

Logic Laws of ﬁrst order logic Propositional variable Truth table One variable case Two variable case Three variable case Negation (NOT) Connectives or compound statements Conjunction (AND) Disjunction (OR)

intrOductiOn Logic is basically the science of reasoning, proof, thinking or inference. Logic involves a set of principles within which one makes statements which sound acceptable or objectionable within a context. Coming to the Mathematical Logic, the type of sentences which are made in Mathematics often be evaluated for truth or falsity. If one looks at a sentence ‘the diagonals of a square are equal’, it is accepted as true. If one makes a statement ‘All odd numbers are prime numbers’, we would not accept this or we would say that it is false, as 9 is odd but not a prime. So we look into those sentences for which we can assign either ‘True’ or ‘False’ as values. This type of logic is termed as First Order Logic. We note that the statements should not be looked for meaning in English language, but the truth values that will be given to them are important. Proposition: A declarative sentence for which one can meaningfully assign true or false in a given context is called a proposition. Note: The words proposition and statement mean the same in the rest of this chapter. (Statements are represented by symbols p, q, r, s, t). Some propositions are: 1. 2. 3. 4.

A pigeon is a bird. Every set is a subset of itself. 0 is a natural number. The sun rises in the east.

• • • • • • • • • •

Conditional statements Converse Inverse Contrapositive Biconditional Other connectives Logical inferences Contingency Normal forms First order logic (or) predicate calculus

Statements 1, 2 and 4 are true, whereas statement 3 is false. Some sentences which are not propositions are 1. 2. 3. 4. 5.

Please switch off the fan. Which way shall we go? God bless you! Who did this? Sit down and rest a while.

The imperatives, interrogatives or exclamations are not classified as propositions.

Laws OF First Order LOgic 1. Law of Excluded Middle: A statement in Logic is either True or False, but not both. 2. Law of Contradiction: A statement cannot be True as well as False.

Propositional Variable Propositional variable is a variable which takes True or False as values.

Truth Table As the values taken by these variables are finite, one can present them in a tabular form.

2.238 | Engineering Mathematics

One Variable Case

We now look into basic connectives. p

Conjunction (AND)

T

If p and q are two statements, the compound statement ‘p and q’ denoted by (p ∧ q) is called Conjunction of p and q.

F

Two Variable Case

Table 2 Truth table p

q

p

q

p∧q

T

T

T

T

T

T

F

T

F

F

F

T

F

T

F

F

F

F

F

F

We see that unless both p and q are TRUE, p ∧ q is not TRUE.

Three Variable Case p

q

r

T

T

T

T

T

F

T

F

T

T

F

F

F

T

T

F

T

F

F

F

T

F

F

F

Negation (NOT) If p is a proposition, then the statement ‘not p’ denoted by ~p is called the negation of p. Thus ~p is the statement, ‘It is not the case that p’. Table 1 Truth table p

~p

T

F

F

T

Example: (i) p:a+b>1 ~ p : a + b ≤ 1 (ii) p : The coffee is hot. ~ p : The coffee is not hot. (iii) p : All birds can fly. ~ p : It is not the case that all birds can fly. (OR) Not all birds can fly. (OR) There is at least one bird which cannot fly.

Connectives or Compound Statements Just as in any language, it is possible to combine two or more sentences. We form new propositions starting from one or more propositions. Such propositions are termed as connectives.

Example: (a) p : 5 > 4 q : 7 is a positive integer p ∧ q would be true as both p and q are true. (b) p : 5 < 4 q : 7 is a positive integer p ∧ q is false as p is false. (c) p : 4 < 5 q:7>9 As q is false, it is clear that p ∧ q is false. (d) p : 5 < 4 q:7>9 p ∧ q is false as both p and q are false.

Disjunction (OR) There are two types of disjunctions, namely

1. Inclusive OR 2. Exclusive OR

(a) Inclusive OR: If p and q are 2 propositions then ‘p or q’ denoted by p ∨ q is called a disjunction of p, q. Example: p : 5 is a factor of 10. q : 5 is a factor of 15 p ∨ q : 5 is a factor of 10 or 15. In this case 5 happens to be a factor of both 10 and as well as 15. That is why we call this as an inclusive disjunction. Table 3 Truth table q

p∨q

T

T

T

T

F

T

p

F

T

T

F

F

F

It is clear that at least one of the statements p, q has to take TRUE as truth value, for the disjunction to be TRUE.

Chapter 1 Mathematical Logic | 2.239 (b) Exclusive (OR): If p and q are two statements, the compound proposition ‘either p or q but not both’ denoted by p • q is called Exclusive OR. Table 4 Truth table p

3. If p is false, then p ⇒ q is true for any q. This is often quoted as ‘A false antecedent implies any conclusion’. Example: The implication ‘If 2 + 5 = 100, then there are 8 days in a week’ will have truth value as TRUE. 4. p ⇒ q is logically equivalent to ~p ∨ q. This would facilitate in writing the truth table.

q

p•q

T

T

F

T

F

T

Converse

F

T

T

F

F

F

If p ⇒ q is a conditional, then the converse of p ⇒ q is the implication q ⇒ p. We say ‘p only if q’.

p and q must take different truth values for p • q to be TRUE. For instance 1. p : 20 is a multiple of 5 (T). q : 3 is a factor of 9 (T). \ p ∨ q is True whereas p • q is False. 2. p : Hyderabad is in Tamilnadu (F). q : Mumbai is the capital of India. (F). \ p ∨ q is False and p • q is also False.

Conditional Statements If p and q are two statements, the compound statement ‘if p then q’ denoted by p ⇒ q, is called a conditional statement. The statement p is called antecedent while the statement q is called consequent. p → q is also a symbol found in literature.

Table 6 Truth table p

q

p⇒q

q⇒p

T

T

T

T

T

F

F

T

F

T

T

F

F

F

T

T

Example: p ⇒ q : If Rahul is smart, then he is intelligent. The converse of p ⇒ q is q ⇒ p: If Rahul is intelligent, then he is smart. Note: ‘q only if p’ is same as p ⇒ q. Example: ‘I like a place, only if I like the surroundings’ means ‘If I like the surroundings, then I like the place’.

Inverse

Table 5 Truth table p

q

p⇒q

T

T

T

T

F

F

F

T

T

F

F

T

Examples: (a) p : I am hungry. q : I will eat something. p ⇒ q : If I am hungry, then I will eat something. The above conditional sounds quite reasonable in general sense. However, consider the statements given below. (b) p : It is hot. q:2+5=7 p ⇒ q : If it is hot, then 2 + 5 = 7 is meaningful in logic even though, it is not convincing in general terms. Thus we see that implication is used in a much weaker sense. Note: 1. One should not look at p as the cause and q as the effect. 2. p ⇒ q is FALSE only when p is TRUE and q is FALSE.

If p ⇒ q is a conditional, then the compound proposition ~p ⇒ ~q is called its inverse. Table 7 Truth table p

q

~p ⇒ ~q

T

T

T

T

F

T

F

T

F

F

F

T

Example: (a) p ⇒ q: If Rahul is smart, then he is intelligent. The inverse of p ⇒ q is ~p ⇒ ~q: If Rahul is not smart, then he is not intelligent. (b) p ⇒ q: If wishes were horses, then beggers would ride them. The inverse of p ⇒ q is ~p ⇒ ~q : If wishes were not horses, then beggers would not ride them.

Contrapositive If p ⇒ q is an implication, then ~q ⇒ ~p is called contrapositive of the implication.

2.240 | Engineering Mathematics Table 8 Truth table p

q

~q⇒~p

T

T

T

Example: Consider a compound statement defined as s : (p ∨ q) ∧ ~r. As there are three variables, ‘s’ has to be evaluated at 23 = 8 rows.

T

F

F

p

q

r

s : (p ∨ q) ∧ ~r

F

T

T

T

T

T

F

F

F

T

T

T

F

T

T

F

T

F

T

F

F

T

Example: p ⇒ q : If you want to top the college, then you study hard. ~q ⇒ ~p : If you do not go study hard then you will not top the college.

F

T

T

F

F

T

F

T

F

F

T

F

F

F

F

F

Explanation: Identify p and q in the above p : You will top the college. q : You study hard. Write down ~q and ~p ~q : You do not study hard. ~p : You will not top the college. Put If (~q) then (~p).

Similarly it is possible to evaluate the values of new propositions that are obtained from one or more variables.

This is how one can write a contrapositive of an implication.

There are two other connectives NAND and NOR which have useful applications in the designing of computers.

Note: The contrapositive is logically equivalent to implication, but the construct is different.

Biconditional If p ⇒ q is an implication, then the statement ‘p if and only if q’ denoted by p ⇔ q is called biconditional. Table 9 Truth table

Other Connectives

NAND: The word NAND is a combination of NOT and AND where ‘NOT’ stands for negation and ‘AND’ stands for conjunction. It is denoted by ↑. If p and q are two statements p ↑q = ~(p ∧ q). The truth table is as follows: p

q

p∧q

p↑q

T

T

T

F

F

F

T

p

q

p⇔q

T

T

T

T

F

T

F

T

T

F

F

F

F

F

T

F

T

F

F

F

T

Other representations are ‘p iff q’ and ‘p if and only if q’. Note: 1. (p ⇒ q) ∧ (q ⇒ p), (p ∧ q) ∨ (~p ∧ ~q) are the equivalent forms. 2. For p ⇔ q to be true both p and q take same truth values. Example: 1. x² + 1/x² ≥ 2 ⇔ (x – 1/x)² ≥ 0, where x ∈ R. 2. Three angles of a triangle are equal ⇔ Three sides of it are equal. In general, a compound statement may have many component parts, each of which, is in itself a statement, denoted by some variable say p1, p2, p3, … pn. If there are n-variables involved in a statement ‘s’ i.e. s = s (p1, p2, …, pn), then s has to be evaluated at 2n rows.

NAND is Commutative but not Associative. NOR: The word NOR is combination of NOT and OR where NOT stands for negation and OR stands for disjunction. It is denoted by ↓. If p and q are two statements p ↓ q = ≅ (p ∨ q). The truth table is as follows: p

q

p∨q

p↓q

T

T

T

F

T

F

T

F

F

T

T

F

F

F

F

T

NOR is commutative but not Associative. Properties: 1. p↑p=≅p 2. p↓p=≅p

Chapter 1 Mathematical Logic | 2.241

Logical Inferences

2. c : p ∧ (~ p ∧ q)

Logic is the Study of inferences. Every argument has two things: 1. The premises (or evidences). 2. The inference (or the conclusion). We start from premises and proceed to the conclusions. The process is called inference.

q

c : p ∧ (~ p ∧ q)

T

T

F

T

F

F

F

T

F

F

F

F

p

Fundamental Rule 1

Contingency

Rule of detachment or Modus Ponens. If p and p → q is true then q is true. This is symbolically denoted as follows p

A statement that can be either TRUE or FALSE, depending on the truth values of its propositional variable, is called a contingency or a statement which is neither a tautology nor a contradiction is called contingency.

p⇒q ∴ q,

Example: 1. s : (p ⇒ q) ∧ (p ∧ q)

Fundamental Rule 2 Rule of transitivity If p ⇒ q and q ⇒ r are true then p ⇒ r is also true. This is denoted by p⇒q q⇒r ______ p ⇒ r. Tautology: A compound statement that takes ‘True’ as truth value for all possible values of propositional variables is called a tautology. Example: 1. t:p∨~p

p

(p ⇒ q) ∧ (p ∧ q)

q

T

T

T

T

F

F

F

T

F

F

F

F

2. s : (p ⇒ q) ∧ (q ⇒ p) p

q

(p ⇒ q) ∧ (q ⇒ p)

T

T

T

T

F

F

F

T

F

F

F

T

p

~p

p ∨ ~p

T

F

T

Logical Equivalences

F

T

T

If two statements s and t are such that s ⇔ t is a tautology, then we say s and t are logically equivalent. Generally s and t will be compound propositions. Such situations are written in the form of properties. We denote equivalence by ‘≡’

2. t : (p ∧ q) ⇒ p p

q

p∧q

(p ∧ q) ⇒ p

T

T

T

T

T

F

F

T

F

T

F

T

F

F

F

T

At the end of this chapter, some tautologies are listed.

Contradiction A compound statement that always takes False as its value is called a contradiction or absurdity. Example: 1. c : p ∧ ~p p

~p

p ∧ ~p

T

F

F

F

T

F

Some Logical Equivalences 1. Commutative Properties: p∨q≡q∨p;p∧q≡q∧p 2. Associative Properties: p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r ; p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r 3. Distributive Properties: p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) ; p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) 4. Idempotent Properties: p∨p≡p;p∧p≡p 5. Properties of Negation: (a) Involution ~(~p) ≡ p (b) ~ (p ∨ q) ≡ ~p ∧ ~q (De Morgan’s law) (c) ~ (p ∧ q) ≡ ~p ∨ ~q (De Morgan’s law)

2.242 | Engineering Mathematics 6. Absorption: p ∨ (p ∧ q) ≡ p p ∧ (p ∨ q) ≡ q Note: The structure [propositions, ∧, ∨, ~] has properties same as [sets, ∪, ∩, ].

List of Equivalences Based on Implications 1. p ⇒ q ≡ (~p) ∨ q 2. p ⇒ q ≡ ~q ⇒ ~p (contrapositive) 3. p ⇔ q ≡ ((p ⇒ q) ∧ (q ⇒ p)) 4. ~(p ⇒ q) ≡ (p ∧ ~q) 5. ~(p ⇔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

List of Tautologies 1. p ∧ q ⇒ p or p ∧ q ⇒ q 2. p ⇒ p ∨ q or q ⇒ p ∨ q 3. ~p ⇒ (p ⇒ q) or ~(p ⇒ q) ⇒ p 4. (p ∧(p ⇒ q)) ⇒ q 5. (~p ∧ (p ∨ q)) ⇒ q 6. (~q ∧ (p ⇒ q)) ⇒ ~p 7. ((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r) hypothetical syllogism or transitive rule

Solution: We have [P → (Q ∧ R)] ∨ ˥ (P ∨ R) ⇔ [˥P ∨ (Q ∧ R)] ∨ (˥P ∧ ˥R) ⇔ (˥P ∧ T) ∨ [(Q ∧ R) ∧ T] ∨ [(˥P ∧ ˥R ) ∧ T] Where T is tautology ⇔ (˥P ∧ (Q ∨ ˥Q)) ∨ [(Q ∧ R) ∧ (P ∨ ˥P)] ∨ [(˥P ∧ ˥R) ∧ (Q ∨ ˥Q)] ⇔ [˥P ∧ Q) ∨ (˥P ∧ ˥Q )] ∨ [(Q ∧ R ∧ P) ∨ (Q ∧ R ∧ ˥P)] ∨ [(˥P ∧ ˥R ∧ Q) ∨ (˥P ∧ ˥R ∧ ˥Q) ⇔ [˥P ∧ Q) ∧ T] ∨ [(˥P ∧ ˥Q ) ∧ T] ∨ [(P ∧ Q ∧ R) ∨ (˥P ∧ Q ∧ R) ∨ (˥P ∧ Q ∧ ˥R ) ∨ (˥P ∧ ˥Q ∧ ˥R) ⇔ [˥P ∧ Q) ∧ (R ∨ ˥R)] ∨ [(˥P ∧ ˥Q ) ∧ (R ∨ ˥R)] ∨ (P ∧ Q ∧ R) ∨ (˥P ∧ Q ∧ R ) ∨ (˥P ∧ Q ∧ ˥R) ∨ (˥P ∧ ˥Q ∧ ˥R) \ The PDNF of the given statements formula is (P ∧ Q ∧ R) ∨ (˥P ∧ Q ∧ R ) ∨ [ (˥P ∧ Q ∧ ˥R) ∨ [(˥P ∧ ˥Q ∧ ˥R)

First Order Logic (or) Predicate Calculus This section introduces the notion of predicates and quantifiers

Predicates

1. p ∨ ~p ≡ T 2. p ∧ ~p ≡ F 3. T∨p≡T 4. F∧p≡F 5. F∨p≡p 6. T∧p≡p 7. ~F ≡ T 8. ~T ≡ F

The statement x < 5, x = y + 7 and x = y + z involve variables x, y, z. These statements are neither true nor false when the values of the variables are not specified. Consider the statement x > 5. There are two parts (a) the variable x which is the subject and (b) ‘is less then 5’ which is the predicate. We use the notation P(x); x < 5, where P(x) in termed as prepositional function (Likewise, we could talk of P(x, y) and P(x, y, z) P(1): 1 < 5 (True) P(2): 4 < 5 (True) P(7): 7 < 5 (False)

Open Sentence

Quantifier

A sentence which involves one or more variables is termed as an Open Sentence, if it becomes TRUE or FALSE when the variables are replaced by some specific value from a given set.

These allow statements about entire collections of objects rather than having to enumerate the objects by name. These are two types

Operations with F and T (T – Tautology and F – Contradiction)

1. x + 5 = 10 is an Open Sentence For x = 5, it becomes TRUE, whereas for other real values it is FALSE (Existential) 2. |x| ≥ 0 is TRUE for all real values of x. (Universal) Solved Examples Example 1: Find the principal Disjunctive Normal Form (PDNF) of the statement formula [P → (Q ∧ R)] ∨ ˥ (P ∨ R)

1. Universal quantifiers and 2. Existential quantifiers

Universal Quantifiers (∀): The universal quantification of P(x) is the proposition ‘P(x) is true for all values of x in the universe of discourse’ NOTATION: ∀ x P(x); read as ‘For all x P(x)’ or ∀ for every x P(x)’ Existential Quantifier ($): The existential quantification of P(x) is the proposition ‘there exists an element x in the universe of discourse such that P(x) is true’

Chapter 1 Mathematical Logic | 2.243 NOTATION: $ x P(x) read as ‘there exist x such that P(x)’ or there is an x such that P(x)’ or for some x P(x)’ Statement

Truth

Falsity

∀ x P(x)

P(x) is true for every x

There is an x for which P(x) is false

$ x P(x)

There is an x for which P(x) is true

P(x) is false for every x

Relation between quantifiers: Universal and existential quantifiers are related through negation. This is tabulated as follows; Negation

Equivalent Statement

Truth of Negation

Falsity of Negation

~ ($x) P(x)

(∀ x) (~ P(x))

P(x) is false for every x

There is an x for which P(x) is true

~ (∀x) P(x)

($ x) (~ P(x)

There is an x for which P(x) is false

P(x) is true for every x

Rules for Quantified Propositions 1. Universal specification: If a statement of the form ∀ x, P(x) is assumed to be true, there the universal quantifier could be removed to obtain P(x) is true for an arbitrary object k in the universe of discourse. NOTATION

∀ x ; P( x) ∴ P (k ) for all k

2. Universal Generalisation: if a statement P(k) is true for each object k in the universe, then the universal quantifier may be inducted, to obtain ∀ x, P(x) P ( k ) for all k NOTATION ∴ ∀ x , P( x) 3. Existential Specification: If there exists an x such that P(x) is true, then there is an object k in the universe such that P(k) is true (k is not arbitrary) ∃ x , P( x) NOTATION ∴ P ( k ) for some k 4. Existential Generalization: If P(k) is true for some k in the universe, then there exist x such that P(x) is true. P ( k ) for some k NOTATION ∴ ∃ x , P( x)

Standard Results 1. ~ (∀ x P(x) ) ≡ $ x ~ P(x) 2. ~ ($ x ~ P(x) ) ≡ ∀ x P(x) 3. $ x (P(x) ⇒ Q(x)) ≡ $ x P(x) ⇒ $ x Q(x) 4. ∀ x P(x) ⇒ ∀ x Q(x) ≡ ∀ x (P(x) ⇒ Q(x)) 5. $ x (P(x) ∨ Q(x)) ≡ $ x P(x) ∨ $ x Q(x)) 6. ∀ x (P(x) ∧ Q(x)) ≡ ∀ x P(x) ∧ ∀ x Q(x)

Proof: If any statement is made in Mathematics we would like to convince ourselves about its truth or validity. The sequence of steps involved in the process is referred as proof. Axiom: The statements which do not require proofs or their truth is obvious are termed as axioms. Example: When a assertion or a statement has to be disproved, we look for opposing example or counter example. ‘Every prime number is an odd number’. The above statement is false. We can convince ourselves by looking at number 2 which is not an odd though it is prime. Thus 2 is an prime number which is not odd. So 2 is a counter example. Example 3: The statements p and q are defined as follows. p : Moon is white q : Mars is red Write the statements for the following symbolic forms : ~ p, ~q, p ∧ ~ q and ~ p ∨ q Solution: ~ p : Moon is not white. ~ q : Mars is not red. p ∧ ~ q : Moon is white and Mars is not red. ~ p ∨ q : Either moon is not white or Mars is red. Example 4: Write the negation of the following statements. (i) No man in city A is rich. (ii) All squares are rectangles. (iii) The glass was hot. (iv) 5 is an odd number. Solution: (i) Some men in city A are rich. (ii) Some squares are not rectangles. (iii) The glass was not hot. (iv) 5 is not an odd number. Example 5: Using De Morgan’s law, negate the following statements. (i) Either Ram went to a hotel or his friend’s house. (ii) Shyam wore a chain and a ring. Solution: (i) Let p: Ram went to a hotel. q: Ram went to his friend’s house. The symbolic form of the given statement is p ∨ q. The negation of (i) in symbolic form is ~ (p ∨ q). According to De Morgan’s law, ~(p ∨ q) ≡ ~p ∧ ~q \ Negation of the statement is ‘Neither Ram went to a hotel nor he went to his friend’s house’ (ii) Let statement p and q be defined as shown below. p : Shyam wore a chain. q : Shyam wore a ring. The symbolic form of the given statement is p ∧ q. The negation of the statement in Symbolic form is ~ (p ∧ q).

2.244 | Engineering Mathematics According to De Morgan’s law, ~ (p ∧ q) ≡ ~p ∨ ~q. \ Negation of the statement is either shyam did not wear a chain or he did not wear a ring. Example 6: Evaluate the expressions below when p = F and q = T (i) p ∨ (p ∧ q) (ii) p ∧ (p ∨ q)

Negation: Raju is clever and he is not the topper in the college. Example 10: Write the inverse and negation of (q ∨ ~ r) ⇒ p. Solution: (q ∨ ~r) ⇒ p We know that inverse of p ⇒ q = ~p ⇒ ~q. The inverse of (q ∨ ~r) ⇒ p is ~(q ∨ ~r) ⇒ ~p

Solution: (i) p ∧ q = F ∧ T = F \ p ∨ (p ∧ q) = F ∨ F = F (ii) p ∧ (p ∨ q) (p ∨ q) = F ∨ T = T p ∧ (p ∨ q) = F ∧ T = F

We know that the negation of

Example 7: Show that the expression ~p ∧ q ⇒ p is not a tautology.

⇒

(~q ∧ r) ⇒ ~p. p ⇒ q is p ∧ ~q \ The negation of (q ∨ ~r)

Solution: When p is F and q is T ~p ∧ q is T (~p ∧ q) ⇒ p is False \ (~p ∧ q) ⇒ p is not a tautology.

p is (q ∨ ~r) ∧ ~p

Example 11: Evaluate the expression p ⇔ (~p ∧ q) for all possible values of p and q. Solution:

Example 8: Show that the statement p ∧ q ⇒ p is a tautology.

p

~p

q

~p ∧ q

p ⇔ (~p ∧ q)

Solution:

T

F

T

F

F

p

q

p∧q

(p ∧ q) ⇒ p

T

T

T

T

T

F

F

T

F

T

F

T

F

F

F

T

T

F

F

F

F

F

T

T

T

F

F

T

F

F

T

\ p ⇔ (~ p ∧ q) is a contingency. Example 12: Show that implication is logically equivalent to contrapositive.

From the above table, (p ∧ q) ⇒ p is always true \ (p ∧ q) ⇒ p is a tautology

Solution:

Example 9: Write the converse, inverse, contrapositive and negation of the implication. If Raju is clever, then he is the topper in the college. Solution: Converse: If Raju is the topper in the college, then he is clever. Inverse: If Raju is not clever, then he is not the topper in the college. Contrapositive: If Raju is not the topper in the college, then he is not clever.

p

q

~p

~q

p⇒q

~q ⇒ ~p

T

T

F

F

T

T

T

F

F

T

F

F

F

T

T

F

T

T

F

F

T

T

T

T

\ observing the last two columns, implication and contrapositive, they have same truth values. Hence, implication is logically equivalent to contropositive.

Exercises Practice Problems 1 Directions for questions 1 to 40: Select the correct alternative from given choices. 1. Let p : Prasad does not play carroms. q : Raju watches a movie. The symbolic form of ‘Prasad plays carroms or Raju does not watch a movie’ is (A) (~p) ⇒ (~q) (B) ~ (p ∧ q) (C) ~ (p ∨ q) (D) (~p) ∧ (q)

2. The truth value of ~ q ⇒ (~ p ∨ ~ F) is (A) F (B) T (C) Either (A) or (B) (D) None of these 3. The negation of a statement, ‘If you are lazy, then you won’t get the marks’, is (A) ‘You are not lazy or you wont get the marks.’ (B) ‘You are lazy or you will get the marks.’ (C) ‘You are lazy and you will get the marks.’ (D) ‘You are not lazy and you will get the marks.’

Chapter 1 Mathematical Logic | 2.245 4. The negation of the statement ‘All A’s are B’s’ is (A) All A’s are not B’s. (B) Only one A which is not B. (C) There is atleast one A which is B. (D) Some A’s are not B’s. 5. The negation of the statement ‘Pasha is handsome or he is intelligent’ is (A) Pasha is either handsome or intelligent. (B) Pasha is not handsome and he is not intelligent. (C) Pasha is not handsome and he is intelligent. (D) Pasha is not handsome or he is intelligent. 6. The negation of the statement ‘Book is good if and only if print is good’ is (A) ‘Book is not good if and only if print is good’. (B) ‘Book is good if and only if print is not good’. (C) Either (A) or (B) (D) None of these 7. Which of the following is tautology? (A) ~p ⇒ (p ∧ q) (B) ~ [p ∧ (~ p)] (C) p ∧ q (D) ~ p ∧ ~ q 8. Which of the following is a contingency? (A) p ∨ (~p) (B) [(p ⇒ q) ∧ (q ⇒ r)] ⇒ (p ⇒ r) (C) p ⇒ (p ∨ q) (D) p∧q 9. Which of the following is a contradiction? (A) p ⇒ q (B) (p) ∧ (~ p ∧ q) (C) (p ∧ q) ⇒ q (D) None of these 10. If p and q are two statements, then which of the following is the converse of ~ q ⇒ ~p (A) p ⇒ ~q (B) p⇒q (C) ~p ⇒ q (D) ~p ⇒ ~q 11. The inverse of (p ∧ q) ⇒ (~ q ∧ ~p) is (A) (~p ∨ ~q) ⇒ (q ∨ p) (B) (~p ∨ ~q) ⇒ (~q ∧ ~p) (C) p ∧ q ⇒ (~q ∧ ~p) (D) ~ (p ∧ q) ⇒ (~q ∧ ~p) 12. Write the contrapositive of p ⇒ (p ∧ q). (A) (~ (p) ∧~q) ⇒ ~p (B) ((~p) ∨ q) ⇒ p (C) ((~p) ∧ q) ⇒ p (D) (~p ∨ ~q) ⇒ ~p 13. If the inverse of a conditional statement is ~p ⇒ q, then the contrapositive of that conditional statement is (A) ~ p ⇒ ~q (B) q⇒~p (C) ~q ⇒ ~p (D) p⇒q Common data for questions 14 and 15: Let p : All sides of the triangle are equal. q : The triangle is equilateral. 14. The inverse of p ⇒ q is (A) If all sides of the triangle are equal, then it is not equilateral triangle. (B) If all sides of the triangle are not equal, then it is not equilateral triangle.

(C) If all sides of the triangle are not equal, then it is equilateral triangle. (D) None of these.

15. The contrapositive of p ⇒ q is (A) If the triangle is equilateral, then all sides are not equal (B) If the triangle is not equilateral, then all sides of it are equal. (C) If the triangle is equilateral, then all sides are equal. (D) If the triangle is not an equilateral triangle, then all sides are not equal. 16. If p and q are two statements, then p ∧ (p ∨ q) is equivalent to (A) p ∨ q (B) p∧q (C) p (D) q 17. The equivalent of p ⇒ q is (A) (~p) ∧ (~q) (B) p ∧ (~q) (C) p ∨ (~q) (D) (~p) ∨ q 18. The converse of an implication is equivalent to (A) contropositive. (B) inverse. (C) conditional statement (D) None of these 19. If p and q are two statements, then the equivalent of ~ (p ⇔ q) is (A) p ∧ ~q (B) q ∧ ~p (C) (p ∧ ~q) ∨ (q∧ ~p) (D) (p ∧ ~q) ∧ (q∧ ~p) 20. p ↓ q is true when (A) p is true q is true (B) p is false q is true (C) p is false q is false (D) p is true q is false 21. p ↑ q is false when (A) p false, q false (B) p false, q true (C) p true, q true (D) p true q, false 22. If p is true and q is false, then which of the following is true? (A) p ↑ ~ q ≡ F (B) p↑q≡T (C) ~ p ↓ q ≡ T (D) All the above 23. Which of the following is equivalent to (~ p ∨ q)? (A) ( p ↓ q) ↑ ( p ↓ q) (B) (~ p ↓ q) ↓ (~ p ↓ q) (C) (~ p ↑ q) ↓ (~ p ↓ q) (D) None of the above 24. Which of the following is equivalent to (p ∧ ~ q)? (A) (p ↑ q) ↑ (p ↑ q) (B) (p ↓ q) ↑ (p ↓ q) (C) (p ↑ ~ q) ↑ (p ↑ ~ q) (D) (p ↑ q) ↑ (p ↑ q) 25. If P(x) denotes the statement ‘x > 12’, then which of the following is true? (x ∈ R) (A) P (10) (B) P (7) (C) P (13) (D) P (9) 26. If P(x) denotes the statement ‘| x | > 3’, then which of the following is false? (x ∈ R) (A) P (6) (B) P (5) (C) P (4) (D) P (3) 27. Let Q(x, y) be the statement ‘x is greater than y’, then which of the following is false?

2.246 | Engineering Mathematics

(A) Q(1, 2) (C) Q(3, 2)

(B) Q(– 1, – 2) (D) Q(5, 4)

28. Let Q(x, y) be the statement ‘x is the capital of y’, then which of the following is true? (A) Q(New Delhi, Iran) (B) Q(New Delhi, Afghanistan) (C) Q(New Delhi, India) (D) Q(New Delhi, Russia) 29. If Q(x, y) denotes ‘x = y + 3’, then which of the following in false (x, y ∈ R)? (A) Q(5, 2) (B) Q(6, 3) (C) Q(4, 1) (D) Q(2, 1) 30. The negation of the statement ‘there is a day in a week which is not a working day’ (A) All days in a week are working days (B) Some days in a week are working days (C) Some days in a week are not working days (D) None of the days in a week are working days 31. Which of the following is appropriate representation for the statement. ‘No positive integer is negative’ Px: x is a positive integer Nx : x is a negative integer (A) x (Px ⇒ Nx ) (B) x (~ Px ⇒ Nx) (C) x (~ Nx ⇒ ~ Px) (D) x (Px ⇒ ~ Nx) Common data for questions 32 to 35: Let L (x, y) be the statement ‘x likes y’ ‘where the universal of discourse is set of all people in the world. Choose the appropriate quantifier for the statements given in these questions 32. Every body likes Ranbir. (A) $ x L (x, Ranbir) (C) ∀ x L ( x, Ranbir)

(B) ∀ x L ( Ranbir, x) (D) $ x L ( Ranbir, x)

33. There is somebody whom every body likes. (A) $ y $ x L (x , y) (B) $ y ∀ x L (x, y) (C) ∀ x ∀ y L (x , y) (D) None of these

34. Everybody likes himself or herself. (A) ∀ x ∀ y L (x, y) (B) $ x $ y L (x, y) (C) ∀ x $ y L (x, y) (D) None of these 35. There is somebody whom nobody likes. (A) $ y [(∀ x) ~L (x, y)] (B) ∀ x [~$ y L (x , y) (C) $ x [∀ ~L (x, y) (D) None of these 36. P(x, y): x is taller than y’ Q(x, y): x is heavier than y’ R(x): x is bold Write the symbolic form of the statement ‘If x is taller than y and not heavier than y, then he is bold’. (A) ∀ x (∀ y P(x, y) ∧ ~ Q(x, y) ⇒ R(x)) (B) $ x (∀ y P(x, y) ∧ Q(x, y) ⇒ R(x)) (C) ∀ x ( $y P (x, y) ∧ ~ Q(x, y) ⇒ R(x)) (D) None of these 37. A universal quantification is false if (A) All its substitution instances are true (B) Some of its substitution instances are true (C) None of its substitution instances are false (D) At least one substitution instance is false 38. The notation $! x P(x) denotes the proposition ‘there exists a unique x such that P(x) is true.’ If the universe of discourse is the set of integers, identify true statement. (A) $ ! x ( x > 1) (B) $ ! x (x2 = 1) (C) $ ! x (2x + 5 = 3x) (D) $ ! x (x2 = x) 39. Which of the following is correct (x, y ∈R)? (A) ∀ x $ y (xy = 100) (B) $ x ∀ y (xy = 100) (C) $ x $ y (xy = 100) (D) None of these 40. Which of the following states the principle of mathematical induction on the universe of positive integers? (A) (P(1) ∧ ∀ k [P(k) ⇒ P(k + 1)]) ⇒ ∀ x P(x) (B) P(1) ∧ $ k [P(k) ⇒ P(k + 1)]) ⇒ ∀ n P(x) (C) P(1) ∧ ∀ k [P(k +1) ⇒ P(k)]) ⇒ ∀ n P(x) (D) P(1) ∧ $ k [P(k+1) ⇒ P(k +1)]) ⇒ ∀ x P(x)

Directions for questions 41 to 44: Match the following. Universe of discourse as set of quadrilaterals 41.

The sum of interior angles in a convex quadrilaterals and in a concave quadrilaterals (A) Universal equals 360°. Therefore sum of interior angles in any quadrilateral is 360° Generalisation

42.

There exist a quadrilateral whose diagonals bisect each other. In a square diagonals bisect each other.

(B) Existential generalization

43.

The angle between the diagonals in a rhombus is a right angle. So there exists a quadrilateral whose diagonals are at right angles.

(C) Universal specification

44.

All quadrilaterals have four sides. Therefore a square has four sides.

(D) Existential specification

Chapter 1 Mathematical Logic | 2.247

Previous Years’ Questions 1. Identify the correct translation into logical notation of the following assertion. Some boys in the class are taller than all the girls Note: taller (x, y) is true if x is taller than y. [2004] (A) ($x)(boy (x) → (∀y)(girl(y) ∧ taller(x,y)))

(B) ($x)(boy (x) ∧ (∀y)(girl(y) ∧ taller(x,y)))

(C) ($x)(boy (x) → (∀y)(girl(y) → taller(x,y)))

(D) ($x)(boy (x) ∧ (∀y)(girl(y) → taller(x,y))) 2. Let P, Q and R be three atomic prepositional assertions. Let X denote (P ∨ Q) → R and Y denote (P → R) ∨ (Q → R). Which one of the following is a tautology?[2005] (A) X ↔ Y (B) X→Y (C) Y → X (D) ¬Y → X 3. What is the first order predicate calculus statement equivalent to the following? Every teacher is liked by some student [2005] (A) ∀(x) [teacher (x) → $(y) [student (y) → likes(y,x)]] (B) ∀(x) [teacher (x) → $(y) [student (y) ∧ likes(y,x)]] (C) $(y)∀(x) [teacher (x) → [student (y) ∧ likes(y,x)]] (D) ∀(x) [teacher (x) ∧ $(y) [student (y) → likes(y,x)]] 4. Which one of the first order predicate calculus statements given below correctly expresses the following English statement? Tigers and lions attack if they are hungry or t hreatened [2006] (A) ∀x[(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}] (B) ∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) ∧ attacks(x)}] (C) ∀x[(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry(x)) ∨ threatened(x)}] (D) ∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}] 5. Consider the following propositional statements: P1: ((A ∧ B) → C) ≡ ((A → C) ∧ (B → C)) P2: ((A ∨ B) → C) ≡ ((A → C) ∨ (B → C)) Which one of the following is true? [2006] (A) P1 is a True, but not P2 (B) P2 is a True, but not P1 (C) P1 and P2 are both True (D) Both P1 and P2 are not True 6. A logical binary relation Q is defined as follows: A

B

AQB

True

True

True

True

False

True

False

True

False

False

False

True

Let ~ be the unary negation (NOT) operator, with higher precedence than Q. Which one of the following is equivalent to A ∧ B? [2006] (A) (~A Q B) (B) ~(A Q ~B) (C) ~(~A Q ~B) (D) ~(~A Q B) 7. Let Graph(x) be a predicate which denotes that x is a graph. Let Connected(x) be a predicate which denotes that x is connected. Which of the following first order logic sentences DOES NOT represent the statement: ‘Not every graph is connected’? [2007] (A) ¬ x(Graph(x) Connected(x)) (B) x(Graph(x) ∧ ¬ Connected(x)) (C) ¬ ∀ x(¬ Graph(x) ∨ Connected(x)) (D) ∀x(Graph(x) ⇒ ¬ Connected(x)) 8. Let fsa and pda be two predicates such that fsa(x) means x is a finite state automaton, and pda(y) means that y is a pushdown automaton. Let equivalent be another predicate such that equivalent (a, b) means a and b are equivalent. Which of the following first order logic statements represents the following: Each finite state automaton has an equivalent pushdown automaton [2008] (A) (∀ x fsa(x)) ⇒ ($y pda(y) ∧ equivalent (x, y)) (B) ~ ∀ y ($x fsa(x) ⇒ pda(y) ∧ equivalent (x, y)) (C) ∀ x $ y (fsa(x) ∧ pda(y) ∧ equivalent (x, y)) (D) ∀ x $ y (fsa(y) ∧ pda (x) ∧ equivalent (x, y)) 9. P and Q are two propositions. Which of the following logical expressions are equivalent? I. P ∨ ~ Q II. ~ (~ P ∧ Q) III. (P ∧ Q) ∨ (P ∧ ~ Q) ∨ (~ P ∧ ~ Q) IV. (P ∧ Q) ∨ (P ∧ ~ Q) ∨ (~ P ∧ Q) [2008] (A) Only I and II (B) Only I, II and III (C) Only I, II and IV (D) All of I, II III and IV 10. Which one of the following is the most appropriate logical formula to represent the statement? ‘Gold and silver ornaments are precious’. The following notations are used: G(x): x is a gold ornament S(x): x is a silver ornament P(x): x is precious [2009] (A) ∀ x (P(x) → (G(x) ∧ S(x))) (B) ∀ x ((G(x) ∧ S(x)) → P(x)) (C) $ x ((G(x) ∧ S(x)) → P(x)) (D) ∀ x ((G(x) ∨ S(x)) → P(x))

2.248 | Engineering Mathematics 11. The binary operation ⊂ is defined as follows P

Q

P⊂Q

T

T

T

T

F

T

F

T

F

F

F

T

Which one of the following is equivalent to P ∨ Q? [2009] (A) ¬Q ⊂ ¬P (B) P ⊂¬Q (C) ¬P ⊂ Q (D) ¬P ⊂ ¬Q 12. Consider the following well-formed formulae: I. ¬∀ x (P(x)) II. ¬$ x (P(x)) III. ¬$ x (¬P(x)) IV. $ x (¬P(x)) Which of the above are equivalent? [2009] (A) I and III (B) I and IV (C) II and III (D) II and IV 13. The following propositional statement is (P → (Q ∨ R)) → ((P ∧ Q) → R)[2009] (A) satisfiable but not valid (B) valid (C) a contradiction (D) None of the above 14. Which one of the following options is CORRECT given three positive integers x, y and z and a p redicate: [2010] P(x) = ¬(x = 1) ∧ ∀y ($z (x = y * z) ⇒ (y = x) ∨ (y = 1)) (A) P(x) being true means that x is a prime number (B) P(x) being true means that x is a number other than 1 (C) P(x) is always true irrespective of the value of x (D) P(x) being true means that x has exactly two factors other than 1 and x 15. Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x$y$t(¬F(x, y, t))? [2010] (A) Everyone can fool some person at some time (B) No one can fool everyone all the time (C) Everyone cannot fool some person all the time (D) No one can fool some person at some time 16. What is the correct translation of the following statement into mathematical logic? [2012] ‘Some real numbers are rational’ (A) $ x (real(x) ∨ rational(x)) (B) ∀ x (real(x) → rational(x)) (C) $ x (real(x) ∧ rational(x)) (D) $ x (rational(x) → real(x)) 17. Consider the following logical inferences. I1: If it rains then the cricket match will not be played. [2012]

The cricket match played. Inference: There was no rain. I2: If it rains then the cricket match will not be played. It did not rain. Inference: The cricket match was played. Which of the following is TRUE? (A) Both I1 and I2 are correct inferences (B) I1 is correct but I2 is not a correct inference (C) I1 is not correct but I2 is a correct inference (D) Both I1 and I2 are not correct inferences 18. What is the logical translation of the following statement? [2013] ‘None of my friends are perfect.’ (A) $ x (F(x)∧ ¬P(x)) (B) $ x (¬F(x) ∧ P(x)) (C) $ x (¬F(x) ∧ ¬P(x)) (D) ¬ $ x (F(x) ∧ P(x)) 19. Which one of the following is NOT logically equivalent to ¬$x(∀y(a)∧ ∀ z(β))? [2013] (A) ∀ x ($z(¬β ) → ∀ y(a)) (B) ∀ x (∀z(β ) → $y(¬a)) (C) ∀ x (∀y(a) → $z(¬β)) (D) ∀ x ($y(¬a) → $z(¬β)) 20. Consider the statement ‘Not all that glitters is gold’ Predicate glitters(x) is true if x glitters and predicate gold (x), is true if x is gold. Which one of the following logical formulae represents the above statements? [2014] (A) ∀x : glitters (x) ⇒ ¬ gold (x) (B) ∀x : gold (x) ⇒ glitters (x) (C) $x : gold (x) ∧ ¬ glitters (x) (D) $x : glitters (x) ∧ ¬ gold (x) 21. Which one of the following propositional logic formulas is TRUE when exactly two of p, q and r are TRUE? [2014] (A) ((p ↔ q) ∧ r) ∨ (p ∧ q ∧ ~ r) (B) (~ (p ↔ q) ∧ r) ∨ (p ∧ q ~ r) (C) ((p → q) ∧ r) ∨ (p ∧ q ∧ ~ r) (D) (~ (p ↔ q) ∧ r) ∧ (p ∧ q ∧ ~ r) 22. Which one of the following Boolean expressions is NOT a tautology?[2014] (A) ((a → b) ∧ (b → c)) → (a → c) (B) (a ↔ c) → (~b → (a ∧ c)) (C) (a ∧ b ∧ c) → (c ∨ a) (D) a → (b → a) 23. Consider the following statements: P: Good mobile phones are not cheap Q: Cheap mobile phones are not good L: P implies Q M: Q implies P N: P is equivalent to Q Which one of the following about, L, M and N is CORRECT? [2014]

Chapter 1 Mathematical Logic | 2.249 2 4.

(A) Only L is TRUE (B) Only M is TRUE (C) Only N is TRUE (D) L, M and N are TRUE The CORRECT formula for the sentence, ‘not all rainy days are cold’ is [2014] (A) ∀ d (Rainy(d) ∧ ~ Cold(d)) (B) ∀ d (~Rainy(d) → Cold(d)) (C) $ d (~Rainy(d) → Cold(d)) (D) $ d (Rainy(d) ∧ ~ Cold(d)) 25. Which one of the following is NOT equivalent to p ↔ q?[2015] (A) (ℸ p ∨ q) ∧ (p ∨ ℸq) (B) (ℸ p ∨ q) ∧ (q → p) (C) (ℸp ∧ q) ∨ (p ∧ ℸq) (D) (ℸ p ∧ q) ∨ (p ∧ q) 26. Consider the following two statements. S1: If a candidate is known to be corrupt, then he will not be elected. S2: If a candidate is kind, he will be elected. Which one of the following statements follows from S1 and S2 as per sound inference rules of logic? [2015] (A) If a person is known to be corrupt, he is kind (B) If a person is not known to be corrupt, he is not kind (C) If a person is kind, he is not known to be corrupt. (D) If a person is not kind, he is not known to be corrupt. 27. Which one of the following well formed formulae is a tautology?[2015] (A) ∀x ∃ y R(x, y) ↔ ∃y ∀x R(x, y) (B) (∀x [∀y R(x, y) → S(x, y)] ∀x ∃y S(x, y) (C) [∀x ∃y (P(x, y) → R(x, y)] ↔ [∀x ∃y (P(x, y) V R(x, y)] (D) ∀x ∀y P(x, y) → ∀x ∀y P( y, x)

28. In a room there are only two types of people, namely Type 1 and Type 2. Type 1 people always tell the truth and Type 2 people always lie. You give a fair coin to a person in that room, without knowing which type he is from and tell him to toss it and hide the result from you till you ask for it. Upon asking, the person replies the following “The result of the toss is head if and only if I am telling the truth.” Which of the following options is correct?[2015] (A) The result is head (B) The result is tail (C) If the person is of Type 2, then the result is tail (D) If the person is of Type 1, then result is tail 29. Let p, q, r, s represent the following propositions. p : x ∈{8,9,10,11,12} q : x is a composite number r : x is a perfect square s : x is a prime number The integer x ≥ 2 which satisfies ¬ ((p ⇒ q) ∧ (¬ r ∨ ¬s)) is ____.[2016] 30. Consider the following expressions (i) false (ii) Q (iii) true (iv) PVQ (v) QVP

The number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is ______.[2016] 31. Which one of the following well - formed formulae in predicate calculus is NOT valid?[2016]

(A) (∀x p(x) ⇒ ∀xq(x)) ⇒(∃x¬p(x) ∨ ∀xq(x)) (B) (∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))) (C) ∃x(p(x) ∧ q(x)) ⇒ (∃xp(x) ∧ ∃xq(x))

(D) ∀x(p(x) ∨ q(x)) ⇒ (∀xp(x) ∨ ∀xq(x))

Answer Keys

Exercises Practice Problems I 1. B 11. A 21. C 31. D 41. A

2. B 12. D 22. D 32. C 41. D

3. C 13. B 23. B 33. B 42. B

4. D 14. B 24. C 34. D 44. C

5. B 15. D 25. C 35. A

6. C 16. C 26. D 36. A

7. B 17. D 27. A 37. D

8. D 18. B 28. C 38. C

9. B 19. C 29. D 39. C

10. D 20. C 30. A 40. A

4. D 14. A 23. D

5. D 15. B 24. D

6. D 16. C 25. C

7. D 17. B 26. C

8. A 18. D 27. C

9. B 10. D 19. A and D 28. A 29. 11

Previous Years’ Questions 1. D 11. B 20. D 30. 4

2. B 12. B 21. B 31. D

3. B 13. A 22. B

Chapter 2 Probability LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • • • • •

Probability Random experiment Mutually exclusive events Equally likely events Independent events Compound events Deﬁnition of probability Addition theorem of probability Conditional probability

ProbAbility The word PROBABILITY is used, in a general sense, to indicate a vague possibility that something might happen. It is also used synonymously with chance.

Random Experiment If the result of an experiment conducted any number of times under essentially identical conditions, is not certain but is any one of the several possible outcomes, the experiment is called a trial or a random experiment. Each of the out comes is known as an event. Example: 1. Drawing 3 cards from a well shuffled pack is a random experiment while getting an Ace or a King are events. 2. Throwing a fair die is a random experiment while getting the score as ‘2’ or an odd number’ are events.

Mutually Exclusive Events If the happening of any one of the events in a trial excludes or prevents the happening of all others, then such events are said to be mutually exclusive. Example: The events of getting a head and that of getting a tail when a fair coin is tossed are mutually exclusive.

• • • • • • • •

Multiplication theorem Advanced probability Mathematical expectation [E(X)] Binomial distribution Normal distribution Poisson distribution Exponential distribution Statistics

Equally Likely Events Two events are said to be equally likely when chance of occurrence of one event is equal to that of the other. Example: When a die is thrown, any number from 1 to 6 may be got. In this trial, getting any one of these events are equally likely.

Independent Events Two events E1 and E2 are said to be independent, if the occurrence of the event E2 is not affected by the occurrence or non-occurrence of the event E1. Example: Two drawings of one ball each time are made from a bag containing balls. Here, we have two events drawing a ball first time (E1) and drawing a ball second time (E2). If the ball of the first draw is replaced in the bag before the second draw is made, then the outcome of E2 does not depend on the outcome of E1. In this case E1 and E2 are Independent events. If the ball of the first draw is not replaced in the bag before the second draw is made, then the outcome of E2 depends on the outcome of E1. In this case, events E1 and E2 are Dependent events.

Chapter 2 Probability | 2.251

Compound Events When two or more events are in relation with each other, they are known as Compound Events. Example: When a die is thrown two times, the event of getting 3 in the first throw and 5 in the second throw is a Compound Event.

Definition of Probability If an event E can happen in m ways and fail in k ways out of a total of n ways and each of them is equally likely, then the probability of happening E is m/(m + k) = m/n where n = (m + k). In other words, if a random experiment is conducted n times and m of them are favourable to event E, then the probability of happening of E is P(E) = m/n. Since the event does not occur (n - m) times, the probability of non-occurrence of E is P ( E ).

P( E ) =

n−m m == 1 - P(E) n n

Therefore,

P(E) + P ( E ) = 1.

Note: 1. Probability [P(E)] of the happening of an event E is known as the probability of success and the probability [P ( E )] of the non-happening of the event is the probability of failure. 2. If P(E) = 1, the event is called a certain event and if P(E) = 0 the event is called an impossible event. 3. Instead of saying that the chance of happening of an event is m/n, we can also say that the odds in favour of the event are m to (n – m) or the odds against the event are (n – m) to m.

Addition Theorem of Probability If A and B are two events, then P(A ∪ B) = P(A) + P(B) - P(A ∩ B) This result follows from the corresponding result in set theory. If n (X) represents the number of elements in set X, n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y) Example: If a die is rolled, what is the probability that the number that comes up is either even or prime? A = The event of getting an even number = {2, 4, 6} B = The event of getting a prime = {2, 3, 5} A ∪ B = {2, 3, 4, 5, 6} A ∩ B = {2} 3 3 5 1 P(A) = , P(B) = , P(A ∪ B) = and P(A ∩ B) = . 6 6 6 6 We can verify that P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Solved Examples Example 1: When a cubical dice is rolled, find the probability of getting an even integer. Solution: When a dice is rolled, the number of possible out comes is 6. The number of favourable outcomes of getting an even integer is 3. 3 1 The required probability = = 6 2 Example 2: If a card is drawn from a pack of cards, find the probability of getting a queen. Solution: When a card is drawn, the number of possible outcomes is 52. The number of favourable outcomes of getting a queen card is 4. 4 1 The required probability = = 52 13 Example 3: A bag contains 5 green balls and 4 red balls. If 3 balls are picked from it at random, then find the odds against the three balls being red. Solution: The total number of balls in the bag = 9. Three balls can be selected from 9 balls in 9C3 ways. Three red balls can be selected from 4 red balls in 4C3 ways. Probability of picking three red balls 4

=

P( E ) =

Odd

against

9

C3 4 1 = = ; C3 84 21

20 21

the three balls 20 1 P( E ) : P( E ) = : = 20 : 1 21 21

being

red

are

=

Example 4: When two dice are rolled together, find the probability of getting at least one 4. Solution: Let E be the event that at least one dice shows 4. E be the event that no dice shows 4. The number of favour25 able outcomes to E is 5 × 5 = 25. P ( E ) = 36 25 11 \ P(E) = 1 – P ( E ) = 1 – = 36 36 Example 5: When two dice are rolled together find the probability that total score on the two dice will be 8 or 9? Solution: When two dice are rolled, the total number of outcomes = 6 × 6 = 36. Favourable outcomes for getting the sum 8 or 9 are {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (3, 6), (6, 3), (4, 5), (5, 4)} i.e., the total number of favourable outcomes = 9. 9 1 The required probability = = 36 4

2.252 | Engineering Mathematics Example 6: If two cards are drawn simultaneously from a pack of cards, what is the probability that both will be jacks or both are queens? Solution: Here two events are mutually exclusive, P(J ∪ Q) = P(J) + P(Q). Probability of drawing two jacks is 4 C P(J) = 52 2 C2 4 C Probability of drawing two queens is P(Q) = 52 2 C2 P(J ∪ Q) = P(J) + P(Q) 4

4 C2 C2 + = 2. 52 52 C2 C2

=

4

C2 2 = 52 C 2 221

26

2

The probability of drawing two black kings is \ The required probability 4

=

C2 + C2

52

26 52

C2 2 C2 55 − = C2 52 C2 221

52

C2 C2

C2 C2

52

Conditional Probability Let S be a finite sample space of a random experiment and A, B are events, such that P(A) > 0, P(B) > 0. If it is known that the event B has occurred, in light of this we wish to complete the probability of A, we mean conditional probability of A given B. The occurrence of event B would reduce the sample space to B, and the favourable cases would now be A ∩ B. A ∩ B (new favourable set) A

Sample space

So,

i.e.,

⎛ A ⎞ n ( A ∩ B) P⎜ ⎟ = ⎝ B⎠ n ( B)

n ( A ∩ B) n (S) 1 = P ( A ∩ B) × =× P ( B) n (S) n ( B) ⎛ A⎞ P ⎜ ⎟ . P ( B) = P ( A ∩ B) ⎝ B⎠

This is called the multiplication theorem on probability. Example 8: A letter is selected at random from the set of English alphabet and it is found to be a vowel. What is the probability that it is e? Solution: Let A be the event that the letter selected is e and B be the event that the letter is a vowel. Then, A ∩ B = {e} and B = {a, e, i, o, u} ⎛ 1 ⎞ ⎜ ⎟ P A B ∩ ) = ⎝ 26 ⎠ = 1 ( ⎛ A⎞ P⎜ ⎟ = So, ⎝ B⎠ P ( B) ⎛ 5 ⎞ 5 ⎜ 26 ⎟ ⎝ ⎠

Independent Events In a random experiment if A, B are event such that P(A) > 0, P(B) > 0 and if P(A/B) = P(A) or P(B/A) = P(B) (conditional probability equals to unconditional probability) then we say A, B are independent events. If A, B are independent P(A ∩ B) = P(A) P(B) Example 9: Two coins are tossed one after the other and let A be the event of getting tail on second coin and B be the event of getting head on first coin then find P(A/B) Solution: Sample space = {HH, HT, TH, TT}, A = {HT, TT} and B = {HH, HT}, (A ∩ B) = {HT}

B B (new sample space)

Notation: The conditional probability of A given B is denoted by P(A/B) \ P(A/B) =

Let A and B be two events of certain random experiment such that A occurs only when B has already occurred. Then, A for the conditional event , the total possible outcomes are B the outcomes favourable to the event B and its favourable outcomes are the outcomes favourable to both A and B.

Example 7: When two cards are drawn from a pack of cards, find the probability that the two cards will be kings or blacks. 4 C Solution: The probability of drawing two kings = 52 2 C2 The probability of drawing two black cards is =

Multiplication Theorem

n( A ∩ B) n( A ∩ B) /n(S) P( A ∩ B) = = n( B) n( B) /n(S) P( B)

Note: 1. This definition is also valid for infinite sample spaces. 2. The conditional probability of B given A is denoted P( A ∩ B) by P(B/A) and P(B/A) = P( A )

\ P(A) =

2 1 = and 4 2

P(A/B) =

P( A ∩ B) 1 / 4 1 = = P( B) 1/ 2 2

Thus P(A/B) = P(A) \ Logically too we understand that occurrence or nonoccurrence of tail in 2nd coin. Baye’s rule: Suppose A1, A2, …, An are n mutually exclusive and exhaustive events such that P(Ai) ≠ 0. Then for i = 1, 2, 3, …, n,

Chapter 2 Probability | 2.253

⎛A ⎞ P⎜ i ⎟ = ⎝ A⎠

⎛ A⎞ P (A i ) . P ⎜ ⎟ ⎝ Ai ⎠ ⎛

n

⎞

A ∑ P (A ) P ⎜⎝ A ⎟⎠ k =1

K

K

Where A is an arbitrary event of S. Example 10: Akshay speaks the truth in 45% of the cases, In a rainy season, on each day there is a 75% chance of raining. On a certain day in the rainy season, Akshay tells his mother that it is raining outside. What is the probability that it is actually raining?

1. p(x) ≥ 0; 2. Σp(x) = 1.

Probability Density Function (P.D.F) If X is a continuous random variable then a function f(x), x ∈ I (interval) is called a probability density function. The probability statements are made as P(x ∈ I) = ∫ f ( x )dx I We also have,

1. f(x) ≥ 0

2.

∞

∫ f ( x ) dx = 1

−∞

Solution: Let E denote the event that it is raining and A denote the event that Akshay tells his mother that it is raining outside. 3 1 Then, P(E) = , P E = 4 4

The probability P(X ≤ x) is called the cumulative distribution function (c.d.f) of X and is denoted by F(x). It is a point function. It is defined for discrete and continuous random variables. The following are the properties of probability distribution function F(x),

9 ⎛ A ⎞ 11 ⎛ A ⎞ 45 = P⎜ ⎟ = and P ⎜ ⎟ = 100 20 E ⎝ ⎠ ⎝ E ⎠ 20 By Baye’s Rule, we have ⎛ A⎞ P (E) P ⎜ ⎟ ⎝E⎠ ⎛E⎞ P⎜ ⎟ = ⎝ A⎠ P E P⎛ A⎞ + P E P⎛ A⎞ ( ) ⎜ ⎟ ⎜ ⎟ ⎝E⎠ ⎝E⎠

1. F(x) ≥ 0. 2. F(x) is non-decreasing i.e., for x > y, F(x) ≥ F(y). 3. F(x) is right continuous. 4. F(–∞) = 0 and F(+∞) = 1. Also, 5. P(a < x ≤ b) = F(b) – F(a). For a continuous random variable 6. Pr{x < X ≤ x + dx} = F(x + dx) – F(x) = f(x) dx; where dx is very small d 7. f(x) = [ F ( x)] where; dx (a) f(x) ≥ 0 ∀ x ∈ R.

( )

( )

3 9 × 27 4 20 = = 3 9 1 11 38 × + × 4 20 4 20

Advanced Probability

(b) ∫ f ( x )dx = 1 . R

Random Variable

Mathematical Expectation [E(X)]

A random variable is a real valued function defined over the sample space. (discrete or continuous). A discrete random variable takes the values that are finite or countable. For example when we consider the experiment of tossing of 3 coins, the number of heads can be appreciated as a discrete random variable (X). X would take 0, 1, 2 and 3 as possible values. A continuous random variable takes values in the form of intervals. Also, in the case of a continuous random variable P(X = c) = 0, where c is a specified point. Heights and weights of people, area of land held by individuals, etc., are examples of continuous random variables.

Mathematical Expectation is the weighted mean of values of a variable. If X is a random variable which can assume any one of the values x1, x2, …, xn with the respective probabilities p1, p2, …, pn, then the mathematical expectation of X is given by E(X) = p1x1 + p2 x2 + … + pn xn For a continuous random variable,

Probability Mass Function (p.m.f ) If X is a discrete random variable, which can take the values x1, x2, … and f(x) denote the probability that X takes the value xi, then p(x) is called the probability mass function (p.m.f) of X. p(xi) = P(x = xi). The values that X can take and the corresponding probabilities determine the probability distribution of X. We also have

E(X ) =

+∞

∫ x f ( x ) dx

−∞

where f(x) is the p.d.f. of X.

Some Special Discrete Distributions Discrete Uniform Distribution A discrete random variable defined for values of x from 1 to n is said to have a uniform distribution if its probability mass function is given by ⎧1 ⎪ ; for x = 1, 2, 3..., n f (x) = ⎨ n ⎪⎩0, otherwise

2.254 | Engineering Mathematics The cumulative distribution function F(x) of the discrete uni⎧0, for x < 1 ⎪x ⎪ form random variable x is given by F(x) = ⎨ ; for 1 ≤ x ≤ n ⎪n ⎪⎩1; for x > 1 n +1 Mean of x = m = 2 n2 − 1 Variance of x = s2 = 12

Binomial Distribution An experiment which is made of n independent trials, each of which resulting in either ‘success’ with probability ‘p’ or ‘failure’ with probability ‘q’ (q = 1 – p), then the probability distribution for the random variable X when represents the number of success is called a binomial distribution. The probability mass function. p(x) = b(x; n, p) = nCx px qn-x; x = 0, 1, 2, … n Example: Hitting a target in 5 trials Here the random variable (X) represents the number of trials made for hitting the target, i.e., x = 0 or 1 or 2 or 3 or 4 or 5. We have a set of 5 trials n = 5. Each trial may hit the target termed to be success (p) or not termed to be failure (q) which are independent. \ This is an example for Binomial distribution.

Properties of Binomial Distribution 1. E(X) = np (mean) 2. V(X) = E(X2) – (E(X))2 = npq; (variance) (mean > variance.) 3. S.D (X) = npq 4. Mode of a binomial distribution lies between (n + 1) p – 1 ≤ x ≤ (n + 1)p 5. If X1 ~ b(n1, p) and X2 ~ b (n2, p) and if X1 and X2 are independent, then X1 + X2 ~ b (n1 + n2, p) where (n, p) is the p.m.f of binomial distribution.

Poisson Distribution A random variable X is said to follow a Poisson distribution with parameter l, l > 0 if it assumes only non–negative values and its probability mass function is given by ⎧e λ ⎪ p( x ) = p( x; λ ) = ⎨ x ! ⎪ 0 ⎩ −λ

x

: x = 0, 1, 2, ..... λ>0 otherwise

In a binomial distribution if n is large compared to p, then np approaches a fixed constant say l. Such a distribution is called poisson distribution (limiting case of binomial distribution)

Properties of Poisson Distribution

e−λ λ x =l x! x 2. V(X) = E(X2) – (E(X))2 = l 1. E(X) =

∑x .

S.D(X) = λ \ Mean = l = Variance 3. Mode of a Poisson Distribution lies between l – 1 and l 4. If X1 ~ P (l1) and X2 ~ P (l2), and X1, X2 independent then X1 + X2 ~ P (l1 + l2).

Some Special Continuous Distributions Continuous Uniform Distribution (or) Rectangular Distribution A continuous random variable x defined on [a, b] is said to have a uniform distribution, if its probability density function is given by ⎧ 1 ; for x ∈[a, b] ⎪ f(x) = ⎨ b − a ⎪⎩0; otherwise The cumulative distribution function of the continuous uniform random variable x is given by ⎧0; if x ≤ a ⎪x −a ⎪ ; if a < x < b f(x) = ⎨ ⎪b − a ⎪⎩1; if x ≥ b a+b 2 ( b − a) 2 Variance of x = s 2 = 12

Mean of x = m =

Normal Distribution A continuous random variable X is said to have a normal distribution with parameters m and s 2 if its density function is given by the probability density function ⎧ ( x − μ )2 ⎪ 1 e − 2σ 2 ⎪ f ( x ) = ⎨ σ 2π ⎪ ⎪⎩0

−∞ < x < ∞ ⎫ −∞ < μ < ∞⎪⎪ ⎬ σ >0 ⎪ otherwise ⎪⎭

It is denoted as X ~ N (m, s 2). The graphical representation of normal distribution is as given below.

m−s m m+s

Chapter 2 Probability | 2.255

Properties of Normal Distribution 1. The function is symmetrical about the value m. 2. It has a maximum at x = m. 3. The area under the curve within the interval (m ± s) is 68%. i.e. P(m – s ≤ x ≤ m + s) = 0.68. 4. A fairly large number of samples taken from a ‘Normal’ population will have average, median and mode nearly the same, and within the limits of average ±2 × S.D., there will be 95% of the values. 5. E(X ) =

X

1

2

3

4

5

6

P(X = x)

1 6

1 6

1 6

1 6

1 6

1 6

E(X) =

∑ x P( X = x)

x

1 1 1 1 1 1 = 1× + 2 × + 3 × + 4 × + 5 × + 6 × 6 6 6 6 6 6 1 6×7 7 = = 3.5 = (1 + 2 + 3 + 4 + 5+ 6) = 6 6×2 2

+∞

−∞

Example 12: In a city 5 accidents take place in a span of 25 days. Assuming that the number of accidents follows the Poisson distribution, what is the probability that there will be 3 or more accidents in a day? (Given e–0.2 = 0.8187)

∫ x . f ( x ) dx = μ .

6. V (X) = s 2; S.D (X) = s 7. For a normal distribution, Mean = Median = Mode. 8. All odd order moments about Mean vanish for a Normal Distribution. i.e. μ 2 n+1 = 0 ∀ = n = 0, 1, 2, … 9. If X1 ~ N (m1, s12) and X2 ~ N (m2, s22), X1, X2 independent, then, X1 + X2 ~ N (m1 + m2, s12 + s22) Also, X1 – X2 ~ N (m1 – m2, s12 + s22) 10. If m = 0 and s ² = 1, we call it as standard normal distribution. The standardisation can be obtained by the transformation X −μ x−μ z= . Also, ~ N (0, 1) σ σ

5 Solution: Average number of accidents per day = = 25 0.2; \l = 0.2 Probability (3 or more accidents per day) = 1 – P(2 or less accidents) = 1 – [P(X = 0) + P(X = 1) + P (X = 2)] = 1 – [e–0.2 + 0.2e–0.2 + 0.02e–0.2] = 1 – e–0.2[1.22] = 1 – 0.99814 = 0.001186 Example 13: What is the area under the normal curve to the left of Z = –1.54 (given area between 0 and –1.54 = 0.4382)? Solution: Required area = 0.5 – 0.4382 = 0.0618

Exponential Distribution

f(z)

A continuous random variable x is said to have an exponential distribution if its probability density function f(x) is given by ⎧λ e − λ x ; for x > 0 f(x) = ⎨ ⎩0; otherwise Here l is the parameter of the exponential distribution and l > 0. The cumulative distribution function F(x) of an exponential distribution with l as parameter is

–1.54

0

+1.54

z

Example 14: A family consists of five children. If the random variable (X) represents the number of boys in that family then

⎧1 − e − λ x ; if x > 0 f(x) = ⎨ ⎩0, otherwise

1 λ2

1 ⎛ 1⎞ X ~ b ⎜ 5, ⎟ ; E ( X ) = np = 5 × = 2.5 2 ⎝ 2⎠

Example 11: An unbiased die is thrown at random. What is the expectation of the number on it?

1 1 V(X) = npq = 5 × × = 1.25 2 2

Solution: Let X denote the number on the die, which can take the values 1, 2, 3, 4, 5 or 6; 1 Probability of each will be equal to 6

Example 15: Ram and Shyam play a game in which their chances of winning are in the ratio 2 : 3. Find Shyam’s chance of winning at least 3 games out of five games played.

Mean = m =

1 λ

Variance = s 2 =

1. Find the expected value E(X) of X 2. Find the variance of X.

Solution: This situation can be modelled as binomial distribution.

2.256 | Engineering Mathematics Solution: P(Shyam wins) =

3 ; 5

Solution: P(X ≤ 10) = ∫ f ( x ) dx = ∫ 10

2 5 Let x denote the number of games won by Shyam. P(Shyam wins at least 3 games) = P(X ≥ 3) x

5− x

=

33 ⎡5C3 22 + 5C4 × 3 × 2 + 1 × 32 × 1⎤⎦ 55 ⎣

=

27 × 79 = 0.68 3125

5

0

(

)

(

)

⎧⎪ P x = xi , y = y j , for xi , y j ∈ Rx × Ry Pxy (xi, yj) = ⎨ ⎪⎩0, otherwise This joint probability mass function can be represented in the form of a table as follows

Example 16: The p.d.f. of a random variable X is ⎧(1 / 10) e ; x>0 f(x) = ⎨ otherwise ⎩0 ( − x /10 )

0

1 −10x e 10

Joint distribution of random variables joint probability mass function: Let x and y be two discrete random variables on the same sample space S with the range space of x as Rx = {x1, x2, …, xm} and the range space of y as Ry = {y1, y2, …, yn} and Px(x) and Py(y) as the probability mass functions of x and y. Then the joint probability mass function Pxy(x, y) of the two dimensional random variable (x, y) on the range space Rx × Ry is defined as

3 x 25 − x = ∑ 5C x 55 x =3

⎛3⎞ ⎛ 2⎞ 5C x ⎜ ⎟ ⎜ ⎟ = ∑ ⎝5⎠ ⎝ 5⎠ x =3

10

⎛ −x ⎞ 1 ⎜ e 10 ⎟ dx = ⎜ ⎟ = 1 – e–1 = 0.6321 10 ⎜ 1 ⎟ ⎜− ⎟ ⎝ 10 ⎠0

P(Shyam loses) =

5

10

What is P(X ≤ 10)? (given e–1 = 0.3679) y

n

x

y1

y2

y3 …

yn

∑ P (x , y ) j =1

xy

i

X1

Pxy(x1, y1)

Pxy(x1, y2)

Pxy(x1, y3) …

Pxy(x1, yn)

Px(x1)

X2

Pxy(x2, y1)

Pxy(x2, y2)

Pxy(x3, y3) …

Pxy(x3, yn)

Px(x2)

X3 . . . .

Pxy(x3, y1) . . . .

Pxy(x3, y2) . . . .

Pxy(x3, y3) … . . . .

Pxy(x3, yn) . . . .

Px(x3) . . . .

Xm

Pxy(xm, y1)

Pxy(xm, y2)

Pxy(xm, y3) …

Pxy(xm, yn)

Px(xm)

Py(y1)

Py(y2)

Py(y3) …

Py(yn)

j

m

∑ P (x , y ) xy

i =1

i

j

From the above table, it can be easily observed that the marginal probability mass functions of x and y nomely Px(x) and Py(y) respectively can be obtained from the joint probability mass function Pxy(x, y) as Px (xi) = And Py(yj) =

n

∑ P ( x , y ) , for i = 1, 2, … m xy

j =1

i

m

∑ P (x , y ) j =1

xy

i

j

j

Pxy(xi, yj) for j = 1, 2, 3, … n

Pxy (xi, yj) ≥ 0 ∀ i, j m

∑ i =1

functions respectively. Then a function fxy(x, y) is called the joint probability density function of the two dimensional random variable (X, Y) if the probability that the point (x, y) will lie in the infinitesimal rectangular region of area dx dy is fxy(x, y) dx dy 1 1 1 1 ⎞ ⎛ i.e. P ⎜ x − dx ≤ X ≤ x + dx, y − dy ≤ Y ≤ y + dy ⎟ 2 2 2 2 ⎠ ⎝ = fXY (x, y) dx dy ∞ ∞

n

∑ Pxy ( xi y j ) = 1

∫∫

j =1

The cumulative joint distribution function of the two dimensional random variable (x, y) is given by fxy (x, y) = P(X ≤ x, Y ≤ y).

Joint Probability Density Function Let X and Y be two continuous random variables on the same sample space S with fx(x) and fy(y) as the probability density

f xy ( x, y )dxdy = 1

−∞ −∞

The marginal probability density functions fx(x) and fy(y) of the two continuous random variables x and y are given by fx(x) =

∞

∫

−∞

f xy ( x, y )dy and fy(y) =

∞

∫

f xy ( x, y )dx

−∞

The cumulative joint distribution function fxy(x, y) of the two dimensional random variable (x, y) (where x and y are any

Chapter 2 Probability | 2.257 two continuous random variables defined on the same sample space) is given by fxy(x, y) =

∞ ∞

∫∫

f xy ( x, y )dxdy

−∞ −∞

Discrete Series Example: x : Number of children in a family f : Number of families Total number of families = 50

Conditional Probability Functions of Random Variables Let x and y be two discrete (continuous) random variables defined on the same sample space with joint probability mass (density) function fxy(x, y) then (i) the conditional probability mass (density) function fx/y(x/y)of x, given Y = y is defined as f xy ( x, y ) , where fy(y) ≠ 0 and (ii) the confx/y (x/y) = f y ( y) ditional probability mass (density) function fy/x(y/x) of y, f xy ( x, y ) where fx(x) ≠ 0 given X = x is defined as fy/x (y/x) = f x ( x)

Pxy(x, y) = Px(x) Py(y) Where Px(x) and Py(y) are the marginal probability mass (density) functions of the random variables X and Y respectively.

0

1

2

3

4

f

8

10

19

8

5

Continuous Series Example: Total number of students = 50

Independent Random Variables Two discrete (continuous) random variables X and Y defined on the same sample space with joint probability mass (density) function PXY(x, y) are said to be independent, if and only if

x

Class Interval (CI)

Frequency (f)

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

8 12 13 10 7

In order to analyse and get insight into the data some mathematical constants are devised. These constants concisely describe any given series of data. Basically we deal with two of these constants

1. Averages or measures of central tendencies 2. Measures of spread or dispersion.

Measures of Central Tendencies

Note: If the random variables X and Y are independent then Pxy (a ≤ X ≤ b, c ≤ Y ≤ d) = Px (a ≤ X ≤ b) Py(c ≤ Y ≤ d).

These tell us about how the data is clustered or concentrated. They give the central idea about the data. The measures are

Statistics

Statistics is basically the study of numeric data. It includes methods of collection, classification, presentation, analysis and inference of data. Data as such is qualitative or quantitative in nature. If one speaks of honesty, beauty, colour etc, the data is qualitative while height, weight, distance, marks etc are quantitative. The present course aims to systematically study statistics of quantitative data. The quantitative data can be divided into three categories

1. Individual series 2. Discrete series and 3. Continuous series.

Individual Series Examples: (a) Heights of 8 students 5.0, 4.9, 4.5, 5.1, 5.3, 4.8, 5.1, 5.3 (in feet) (b) The weight of 10 students 46, 48, 52, 53.4, 47, 56.8, 52, 59, 55, 52 (in kgs)

1. Arithmetic mean or mean 2. Geometric mean 3. Harmonic mean 4. Median 5. Mode

The first three are mathematical averages and the last two are averages of position.

Measures of Dispersion It is possible that two sets of data may have the same central value, yet they may differ in spread. So there is a need to study about the spread of the data. The measures we deal with are,

1. Range 2. Quartile deviation or semi interquartile range 3. Mean deviation 4. Standard deviation (including variance)

The formulae for each of the above mentioned measures is listed for each of the series in what follows.

2.258 | Engineering Mathematics

Measures of Central Tendencies 1. Arithmetic mean (A.M. or x) (a) Individual series: x + x + … + xn ∑ xi = x= 1 2 n n (b) Discrete series: f x + f 2 x2 + … + f n xn ∑ fi xi = x= 1 1 f1 + f 2 + … + f n ∑ fi

where x1, x2, … xn are n distinct values with frequencies f1, f2, f3, …, fn respectively. (c) Continuous series: f m + f 2 m2 + … + f n mn ∑ fi mi x= 1 1 = f1 + f 2 + … + f n ∑ fi where f1, f2, f3, … fn are the frequencies of the classes whose mid-values are m1, m2, … mn respectively. Some Important Results Based on A.M. 1. The algebraic sum of deviations taken about mean is zero. 2. Its value is based on all items. 3. Mean of first n natural numbers is (n + 1)/2 4. Arithmetic mean of two numbers a and b is (a + b)/2 5. If b is A.M. of a and c then a, b, c are in arithmetic progression. Combined Mean: If x1 and x2 are the arithmetic means of two series with n1 and n2 observations respectively, the combined mean, xc =

n1 x1 + n2 x2 n1 + n2

2. Median: If for a value the total frequency above (or below) it is half of the overall total frequency the value is termed as Median. Median is the middlemost item. Individual series: If x1, x2, … xn are arranged in ascending order of magnitude then the median is the th ⎛ n + 1⎞ size of ⎜ item. ⎝ 2 ⎟⎠ Some Results Based on Median 1. Median does not take into consideration all the items. 2. The sum of absolute deviations taken about median is least. 3. Median is the abscissa of the point of intersection of the cumulative frequency curves. 4. Median is the best suited measure for open end classes.

Mode The most frequently found item is called mode. Being so, it is easy and straight forward to find for individual and discrete series.

Empirical Formula • For moderately symmetrical distribution, • Mode = 3 median – 2 mean, • For a symmetric distribution, Mode = Mean = Median. This formula is to be applied in the absence of sufficient data. Given any two, of the mean, median or mode the third can be found.

Measures of Dispersion 1. Range: The range of a distribution is the difference between the greatest and the least values observed. Some Important Results Based on Range (a) Range is a crude measure of dispersion as it is based only on the value of extreme observations. (b) It is also very easy to calculate. (c) It does not depend on the frequency of items. Q3 − Q1 2 (a) Individual series: The numbers are first arranged in ascending or descending order, then we find the quartiles Q1 and Q3 as Q1 → size of (n + 1)/4th item Q3 → size of 3(n + 1)/4th item The first quartile (or the lower quartile) Q1 is that value of the variable, which is such that onequarter of the observations lies below it. The third quartile Q3 is that value of the variable, which is such that three-quarters of the observations lie below it. 2. Quartile Deviation (Q.D.): Q.D. =

3. Mean Deviation (M.D.): It is defined as the arithmetic mean of the deviation from origin, which may be either mean or median or mode. (a) Individual Series: M.D. =

x1 − A + x2 − A + … + xn − A

n where x1, x2 … xn are the n observations and A is the mean or median or mode. Some Results Based on M.D.: (a) Mean deviation depends on all items. (b) By default, mean deviation is to be computed about mean. (c) Mean deviation about the median is the least. a−b (d) Mean deviation of two numbers a and b is . 2 4. Standard Deviation (S.D.): Standard deviation is referred to as root mean squared deviation about the mean. (a) Individual series: ( x1 − x ) 2 + ( x2 − x ) 2 + … + ( xn − x ) 2 n where x1, x2, … xn are n observations with mean as x S.D. (s) =

Chapter 2 Probability | 2.259

∑x

⎛ ∑ xi −⎜ Alternatively σ = ⎜ n n ⎝ formula for computational purpose i

2

2

⎞ ⎟⎟ is an useful ⎠

Some Results Based on S.D. (a) The square of standard deviation is termed as variance. (b) S.D. is the least mean square deviation. (c) If each item is increased by a fixed constant the S.D. does not alter or S.D. is independent of change of origin. (d) Standard deviation depends on each and every data item. (e) For a discrete series in the form a, a + d, a + 2d, … (A.P.), the standard deviation is given by n2 −1 , where n is number of terms in S.D. = d 12 the series. S.D. ×100 Coefficient of variation (C.V.) is defined as, C.V. = AM This is a relative measure, which helps in measuring the consistency. Smaller the co-efficient of variation, greater is the consistency. Example 17: For the individual series, compute the mean, median and mode 8, 11, 14, 17, 20, 23, 26, 29

∑ xi = 8 + 11 + … + 29 = 18.5

n 8 Median: As the numbers are in ascending order and the number 17 and 20 being middle terms. Median =

Mode = 3 median – 2 mean = 3(18.5) – 2(18.5) = 18.5

Example 18: The arithmetic mean of 8, 14, x, 20 and 24 is 16; then find x? 8 + 14 + x + 20 + 24 = 16 5 x = 80 – 66 = 14

Solution: x = ⇒

Example 19: Calculate standard deviation of first five prime numbers. Solution: Given set of observations {2, 3, 5, 7, 11}

Sx2/n = 208/5

Sx/n = 28/5

\ S.D. =

Co-efficient of Variation (C.V.)

Solution: Mean = x =

Mode: As no term can be regarded as ‘most often found’, mode is not-defined. However using empirical formula,

∑x n

2

⎛∑x⎞ −⎜ ⎜ n ⎟⎟ ⎝ ⎠ 2

2

=

208 ⎛ 28 ⎞ − = 3.2 5 ⎜⎝ 5 ⎟⎠

Example 20: In a series of observations, coefficient of variation is 25 and mean is 50. Find the variance. Solution: Coefficient of variation: C.V. = ⇒ S.D. =

C.V. ⋅x 100

S.D. ×100 x

= 50 × 25/100 = 12.5

variance = (12.5)2 = 156.25

17 + 20 37 = = 18.5 2 2

Exercises Practice Problems I Directions for questions 1 to 55: Select correct alternative from the given choices. 1. If eight unbiased coins are tossed together, then the probability that the number of heads exceeds the number of tails is 31 1 93 57 (A) (B) (C) (D) 128 2 256 256 2. If A and B are two mutually exclusive and exhaustive events and the probability that the non-occurrence of A is 3/4, then the probability of occurrence of B is 1 1 3 1 (A) (B) (C) (D) 4 2 4 16 3. If four fair dice are rolled together, then the probability that the total score on the four dice is less than 22 is

26 3 427 83 (A) (B) (C) (D) 27 432 432 108 4. A bag contains five red balls, three black balls and a white ball. If three balls are drawn from the bag, the probability that the three balls are of different colours is (A) 23/28 (B) 5/28 (C) 3/28 (D) None of these 5. From a box containing 18 bulbs, of which exactly 1/3rd are defective, five bulbs are chosen at random to fit into the five bulb holders in a room. The probability that the room gets lighted is 6

(A) 1 − 18

6 C5 C5 (B) 18 C5 C5

12 12 C C5 1 − 18 5 (C) (D) 18 C5 C5

2.260 | Engineering Mathematics 6. On a biased dice, any even number appears four times as frequently as any odd number. If the dice is rolled thrice what is the probability that the sum of the scores on them is more than 16? 26 112 26 112 (A) (B) (C) (D) 375 375 3375 3375 7. A five digit number is formed using the digits 0, 1, 2, 3, 4 and 5 at random but without repetition. The probability that the number so formed is divisible by 5 is 1 2 4 9 (A) (B) (C) (D) 5 5 25 25 8. If ten students are to be seated in a row, then the probability that two particular students never sit together is 2 4 1 3 (A) (B) (C) (D) 5 5 5 5 9. If six people sit around a circular table, the probability that two specified persons always sit side by side is 14 11 2 4 (A) (B) (C) (D) 15 15 5 15 10. Eight letters are to be placed in eight addressed envelopes. If the letters are placed at random into the envelopes, the probability that exactly one letter is placed in a wrong addressed envelopes is 1 1 (A) (B) 6 8! 1 (C) (D) None of these 7! 11. A puzzle in logic was given to three students A, B and C 1 3 1 and respecwhose chances of solving it are , 2 4 4 tively. The probability that the problem being solved is 29 31 1 7 (A) (B) (C) (D) 32 32 8 8 12. If A and B are two events of an experiment such that 3 7 P(A ∪ B) = , P(A) = , then find P(B) given that 4 20 (i) A and B are mutually exclusive 1 1 3 2 (A) (B) (C) (D) 4 5 5 5 (ii) A and B are equally likely 7 3 2 13 (A) (B) (C) (D) 20 4 5 20 (iii) A and B are independent events 7 8 6 2 (A) (B) (C) (D) 13 13 13 5 13. The probability that a square selected at random from a 8 × 8 chessboard is of size 3 × 3 is 8 14 3 25 (A) (B) (C) (D) 51 17 17 204

14. A letter is taken randomly from the word SISTER, and another letter is taken randomly from the word ‘RESIST’ the probability that the two letters are same letters is 2 3 7 2 (A) (B) (C) (D) 3 5 36 9 15. I had to type a 6 character password. The probability that I make a mistake in typing a character is 0.3. The password that I typed turned out to be wrong. Find the probability that only the last character that I entered is wrong.

( 0.7 ) × (0.3) ( 0.7 ) (B) 6 (A) 6 1 − ( 0.7 ) 1 − ( 0.7 ) 5 5 ( 0.7 ) ( 0.7 )( 0.3) (C) (D) 6 6 1 − ( 0.7 ) 1 − ( 0.7 ) 5

6

16. There are two groups X and Y. Each group contains three different types of people. In group X, there are 10 Indians, 8 Americans and 7 Japanese. In group Y, there are 8 Indians, 6 Japanese and 6 Americans. If two people are selected at random from the two groups, then what is the probability that both selected are Indians from group X? 56 57 (A) (B) 113 103 57 57 (C) (D) 113 123 17. A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a spade. The person who picks a spade is declared the winner. If A begins the game, then the probability that B wins the game is 5 4 3 4 (A) (B) (C) (D) 9 9 7 7 18. If three unbiased dice are rolled simultaneously in a random experiment, the sum of the numbers showing up on them is 14. What is the probability of an event of showing up 6 on any one of the dice? 7 11 4 3 (A) (B) (C) (D) 15 15 5 5 19. At the Wimbledon, the probability that Federer qualifies for the final is 0.7, and the probability that Nadal qualifies for the semifinal is 0.5. The probability that Federer qualifies for the final or Nadal qualifies for the semifinal is 0.8. Given that Nadal qualifies for the semifinal, find the probability that Federer qualifies for the final. (A) 0.2 (B) 0.8 (C) 0.6 (D) 0.9 1 ⎡ P( A ∪ B) ∩ A ⎤ = and ⎣ ⎦ 3 1 P( A ∩ B) + P( A ∩ B) = , then what is the value of 2 P(A/B)?

20. P(A

∪

B)

=

2/3,

1 1 1 1 (A) (B) (C) (D) 3 6 5 2

Chapter 2 Probability | 2.261 21. If P(A) = ⎛ Ac ⎞ P⎜ c ⎟ . ⎝B ⎠

6 1 3 , P(Bc) = and P(A ∩ B) = , then find 4 7 5

17 19 29 71 (A) (B) (C) (D) 60 60 60 120 22. If two events A and B are such that P( A ) = 0.4, P(B) = ⎛ B ⎞ 0.7 and P(A ∩ B) = 0.2, then P ⎜ is ⎝ A ∪ B ⎟⎠ 3 2 1 4 (C) (D) (A) (B) 5 5 4 5 23. A random variable X has mean 2 and E(X2) = 6. Then the standard deviation of X is (A) 6 (B) 2 (C) 2 (D) 4 24. The standard deviation of 3x + 2 is 4; then the variance of x is (A) 12 (B) 4/3 (C) 16/9 (D) Cannot be determined Common data for questions 25 and 26: Probability mass function of a variate x is as follows. x

0

1

2

3

4

P(X = x)

k

2k

3k

4k

5k

2 5. k = 2 1 3 4 (A) (B) (C) (D) 15 15 15 15 26. P(x ≥ 3) = 1 4 3 5 (A) (B) (C) (D) 3 15 5 7 Common data for questions 27 and 28: A variate x has the probability distribution as x

4

8

12

P(X = x)

1 3

3 5

1 15

27. Values of E(x) and E(x2) respectively are 102 150 104 160 (A) , , (B) 15 3 15 3 21 160 104 151 (C) , (D) , 3 5 15 3 28. The value of E[(3x + 2)2] is ________. (A) 675.2 (B) 560.2 (C) 134.56 (D) 567.2 29. For a random variable x, the p.d.f. ⎧kx 2 0 ≤ x ≤ 1 f(x) = ⎨ . Find E(x). ⎩ 0 otherwise 1 1 3 3 (A) (B) (C) (D) 8 4 8 4

30. In the random experiment of drawing a card from 15 cards numbered 1 to 15, if x is the random variable defined by the number appeared on the card, then the expectation of x is (A) 8 (B) 7 (C) 6 (D) 5 31. For a binominal distribution, mean is 6 and variance is 2. The number of Bernouli trials is (A) 8 (B) 9 (C) 10 (D) 11 32. If X(n, p) follows a binominal distribution with n = 6 such that 9P [X = 4] = P[X = 2], then p = 1 1 1 (A) (B) (C) 1 (D) 3 2 4 33. The distribution of the number of male children in a family of 5 children follows which of the following distributions? (A) Normal (B) Poisson (C) Binomial (D) Negative binomial Common data for questions 34 and 35: Let ABC be a bulb manufacturing company. The probability that a bulb manufactured by ABC defective is 0.25. 34. What is the probability that in a sample of 8 bulbs at most one will be defective? 8

8

7

11 ⎛ 3 ⎞ 11 ⎛ 3 ⎞ 1 ⎛3⎞ (A) ⎜ ⎟ (B) 11⎜ ⎟ (C) ⎜ ⎟ (D) 8 3 ⎝4⎠ 4 3 4 4 ⎝ ⎠ ⎝ ⎠ 35. Atleast 3 will be defective? 5

⎛ 3 ⎞ ⎛ 61 ⎞ (B) 1 – ⎜ ⎟ ⎜ ⎟ ⎝4⎠ ⎝ 9 ⎠

8

⎛3⎞ ⎛ 1 ⎞ (D) 1 – ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 48 ⎠

⎛ 3 ⎞ ⎛ 25 ⎞ (A) 1 – ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 48 ⎠

⎛ 3 ⎞ ⎛ 24 ⎞ (C) 1 – ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 48 ⎠

8

6

36. A random variable X follows a Poisson distribution such that P[X = 1] = P[X = 2]. Its mean and variance are, respectively, (A) 1, 1 (B) 2, 2 (C) 3 , 2 (D) 2 , 2 37. The probability that a person hits a target is 0.003. What is the probability of hitting the target with 2 or more bullets if the number of shots is 2000? (A) 1 – e–6 (B) 1 – e6 (C) 1 – 7e6 (D) 1 – 7e–6 1 38. If X is a continuous random variable with p.d.f f(x) = 4 if - 2 ≤ x ≤ 2 and f(x) = 0 elsewhere, the mean of X is _____ (A) 1 (B) 1.5 (C) 2 (D) 0 39. If x is a uniformly distributed random variable in [1, 4] 3⎞ ⎛ then P ⎜ x > ⎟ is 2⎠ ⎝ 1 1 5 1 (A) (B) (C) (D) 6 2 6 4

2.262 | Engineering Mathematics 40. If x is a uniformly distributed random variable in [2, 5] then E(x2) is (A) 2 (B) 8 (C) 13 (D) 15 41. If the life time of bulbs (in months) is exponential with mean 5 months, then the probability that the bulb lasts for atleast 7 months is (A) 0.2466 (B) 0.7534 (C) 0.4932 (D) 0.5068 42. The median of a normal variate X with p.d.f 1

f ( x) =

2π σ

e −( x − μ )

2

/ 2σ 2

is,

μ (A) 0 (B) s (C) m (D) σ 43. x is a normal variate with mean 35 and variance 25 probability of 31 ≤ x < 45 is (– 0.8 ≤ z < 0 = 0.2881)

45. If X and Y are two independent random variables with expectations 3 and 4 respectively. Then the expectation of X Y is (A) 1 (B) 7 (C) 12 (D) 16 46. If X and Y are two independent random variables that are uniformly distributed over the same interval [2, 5] 11 11 ⎞ ⎛ then P ⎜ X ≤ , Y ≥ ⎟ is 4 3⎠ ⎝ 1 2 1 4 (C) (D) (A) (B) 9 9 3 7 47. Two fair dice are rolled simultaneously. Let X denote the number on the first die and Y denote the number on the second die. Then the value of P(X + Y ≤ 7/Y ≥ 5) is 1 1 1 1 (A) (B) (D) (C) 3 6 2 4

0.4772

48. The mean of cubes of first 10 natural numbers is (A) 305 (B) 300 (C) 302.5 (D) 310 0

2

(A) 0.6735 (B) 0.7563 (C) 0.7653 (D) 0.5736 4 4. Let X1 and Y1 be two discrete random variables with joint probability mass function as given below Y X

2

3

P(X1 = xi)

1

1 15

2 15

1 5

4 15 1 3

8 15 2 3

4 5

4 P(Y1 = yi)

–1 3 P(Y2 = yi)

50. If the mean of a set of 12 observations is 10 and another set of 8 observations is 12, then the mean of combined set is (A) 12.6 (B) 10.8 (C) 12.8 (D) 10.6 51. The mode of a distribution of 13 and its mean is 4 then its median is (A) 7 (B) 9 (C) 8 (D) 11

Let X2 and Y2 be two discrete random variables with joint probability mass function as given below Y2 X2

49. The mean of 25 observations was found to be 38. It was later discovered that 23 and 38 were misread as 25 and 36, then the mean is (A) 32 (B) 36 (C) 38 (D) None of these

0

4

7

P(X2 = xi)

1 7 4 21 1 3

3 14 2 7 1 2

1 14 2 21 1 6

3 7 4 7

Which of the following statements is TRUE about the random variables X1, X2, Y1 and Y2? (A) Only X1 and Y1 are independent (B) Only X2 and Y2 are independent (C) X1 and Y1 are independent as well as X2 and Y2 are independent (D) Neither X1 and Y1 are independent nor X2 and Y2 are independent

52. Consider the non-decreasing series of the numbers, 1, 8, 8, 13, 14, 14, x, y, 18, 20, 31, 34, 38 and 40. If the median of the series is 15, then the mode of the series is (A) 14 (B) 16 (C) 18 (D) Cannot be determined 53. The standard deviation of 5, 5, 5, 5, 5, 5, 5, 13 is (A) 2 2 (B) 6

(C) 5

(D) 7

54. If the standard deviation of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 is M, then the standard deviation of 101, 102, 103, 104, ….. and 111 is (A) M (B) 100 + M (C) 100 - M (D) M - 100 55. The arithmetic mean of five observations is 6.4 and the variance is 8.24. If three of the observations are 3, 4, 8, then find the other two observations. (A) 6, 11 (B) 10, 7 (C) 8, 9 (D) 5, 12

Chapter 2 Probability | 2.263

Previous Years’ Questions 7. A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. 3 1 5 3 (A) (B) (C) (D) The probability that two such randomly generated 8 2 8 4 strings are not identical is:[2005] 2. An examination paper has 150 multiple choice ques1 1 tions of one mark each, with each question having (A) n (B) 1 four choices. Each incorrect answer fetches -0.25 n 2 mark. Suppose 1000 students choose all their answers 1 1 randomly with uniform probability. The sum total of (C) (D) 1 - n n! the expected marks obtained by all these students is 2 [2004] 8. We are given a set X = {x , …, x } where x = 2i. A sam1 n i (A) 0 (B) 2550 (C) 7525 (D) 9375 ple S ⊆ X is drawn by selecting each x independently 1. If a fair coin is tossed four times. What is the probability that two heads and two tails will result? [2004]

i

3. Two n bit binary strings, S1 and S2 are chosen randomly with uniform probability. The probability that Hamming distance between these strings (The number of bit positions where the two strings differ) is equal to d is [2004]

1 . The expected value of the 2 smallest number in sample S is:[2006] 1 (A) (B) 2 (C) n (D) n n

n n C Cd (A) nd (B) 2 2d

9. For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is: [2006] ⎛ 2n⎞ ⎛ 2n⎞ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ (A) n (B) 4 2n 1 ⎛ 2n⎞ 1 (C) ⎜⎝ n ⎟⎠ (D) 2

d 1 (C) n (D) 2 2d 4. A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0, 0), (1, 0), (1, 2) and (0, 2). If p is the length of the position vector of the point, the expected value of p2 is [2004] 2 (A) (B) 1 3 4 5 (C) (D) 3 3 5. Let f(x) be the continuous probability density function of a random variable X. The probability that a < X ≤ b is:[2005] (A) f (b - a) (B) f (B) - f (A) b

b

a

a

(C) f ∫ ( x)dx (D) ∫ xf ( x)dx 6. Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q 1 2 are and , respectively. Given that a ball selected 3 3 in the above process is a red ball, the probability that it came from the box P is [2005]

with probability pi =

10. Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday? [2008] (A) 0.24 (B) 0.36 (C) 0.4 (D) 0.6 11. Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown. If P(X ≤ –1) = P(Y ≥ 2), the standard deviation of Y is [2008] (A) 3 (B) 2 (C) 2 (D) 1

12. Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3, ….., 20. What is the probability that 2 appears at an earlier 4 5 2 19 (A) (B) (C) (D) position than any other even number in the selected 19 19 9 30 permutation? [2008]

2.264 | Engineering Mathematics 1 1 (A) (B) 2 10 9! (C) 20 !

(D) None of these

13. An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even, given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3? [2009] (A) 0.453 (B) 0.468 (C) 0.485 (D) 0.492 14. Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. [2010] What is the probability of a computer being declared faulty? (A) pq + (1 − p) (1 − q) (B) (1 − q)p (C) (1 − p) q (D) pq 15. What is the probability that divisor of 10 is a multiple of 1096? [2010] 1 4 12 16 (A) (B) (C) (D) 625 625 625 625 99

16. If the difference between the expectation of the square of a random variable (E [x2]) and the square of the expectation of the random variable (E[X])2 is denoted by R then [2011] (A) R = 0 (B) R < 0 (C) R ≥ 0 (D) R > 0

(A) Index position of mode of X in X is the same as the index position of mode of Y in Y. (B) Index position of median of X in X is the same as the index position of median of Y in Y. mY = amx + b (C) (D) sy = asx + b 19. A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?[2011] 1 4 1 2 (A) (B) (C) (D) 5 25 4 5 20. Consider a random variable X that takes values +1 and –1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = –1 and +1 are [2012] (A) 0 and 0.5 (B) 0 and 1 (C) 0.5 and 1 (D) 0.25 and 0.75 21. Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2 or 3, the die is rolled second time. What is the probability that the sum total of values that turn up is at least 6? [2012] (A) 10/21 (B) 5/12 (C) 2/3 (D) 1/6 22. Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? [2013] (A) 8/(2e3) (B) 9/(2e3) (C) 17/(2e3) (D) 26/(2e3) 23. Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ______. [2014]

24. 17. If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads? [2011] 25. 1 1 1 2 (A) (B) (C) (D) 3 4 2 3 18. Consider a finite sequence of random values X = [x1, x2, …, xn ]. Let mx be the mean and σ x be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi = a*xi + b, where a and b are positive constants. Let my be the mean and σ y be the standard deviation of this sequence. Which one of the following statements is INCORRECT? [2011]

Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is X/1296. The value of X is ______. [2014] The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by P. Then 100 P = ______. [2014]

26. Each of the nine words in the sentence ‘The quick brown fox jumps over the lazy dog’ is written on a

Chapter 2 Probability | 2.265 separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is ______. (The answer should be rounded to one decimal place).[2014] 27. The probability that a given integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _____. [2014] 28. Let S be a sample space and two mutually exclusive events A and B be such that A ∪ B = S. If P(.) denotes the probability of the event, the maximum value of P(A) P(B) is ______. [2014] 29. Suppose ℒ = {p, q, r, s, t} is a lattice represented by the following Hasse diagram: t

q

r

s

p

For any x, y ∈ℒ, not necessarily distinct, x ∨ y and x ∧ y are join and meet of x, y, respectively. Let ℒ3 = {(x, y, z): x, y, z ∈ℒ} be the set of all ordered triplets of the elements of ℒ. Let pr be the probability that an element (x, y, z) ∈ℒ3 chosen equiprobably satisfies x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). Then [2015]

(A) pr = 0

(B) pr = 1 1 1 (C) < pr < 1 0 < pr ≤ (D) 5 5 30. Let X and Y denote the sets containing 2 and 20 distinct objects respectively and F denote the set of all possible functions be from X to Y. Let f be randomly chosen from f . The probability of f being one-to-one is ______. [2015] 31. A probability density function on the interval [a,1] is given by 1/x2 and outside this interval the value of the function is zero. The value of a is _____ . [2016] 32. Consider the following experiment. Step1. Flip a fair coin twice. Step2. If the outcomes are (TAILS, HEADS) then output Y and stop. Step3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop. Step 4 If the outcomes are (TAILS, TAILS), then go to Step 1. The probability that the output of the experiment is Y is (up to two decimal places) ______. [2016] 33. Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _______. [2016]

Answer Keys

Exercises Practice Problems 1 1. C 11. A 19. B 29. D 39. C 49. C

2. C 3. C 4. B 12. (i) D (ii) A (iii) B 20. B 21. B 22. B 30. A 31. B 32. D 40. C 41. A 42. C 50. B 51. A 52. D

5. A 13. C 23. B 33. C 43. C 53. D

6. D 14. D 24. C 34. A 44. C 54. A

7. D 15. A 25. B 35. B 45. C 55. A

8. B 16. C 26. C 36. B 46. A

9. C 17. C 27. A 37. D 47. C

10. D 18. C 28. D 38. D 48. C

Previous Years’ Questions 1. A 2. D 11. A 12. B 21. B 22. C 27. 0.259 to 0.261

3. A 4. D 13. B 14. A 23. 0.24 to 0.27 28. 0.25 to 0.25

5. C 6. A 7. D 8. D 9. A 10. C 15. A 16. C 17. A 18. D 19. A 20. C 24. 10 to 10 25. 11.85 to 11.95 26. 3.8 to 3.9 29. D 30. 0.95 31. 0.5 32. 0.33 to 0.34 33. 0.55

Chapter 3 Set Theory and Algebra LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • • •

Sets Some deﬁnitions Basic set operations Sets in number system Functions Algebra Binary operation

• • • • • • •

Permutations Even and odd permutations Fundamental theorem of homomorphism Partial ordering and lattices Lattice Boolean algebra Standard forms

sets

Finite and Infinite Sets

A set is a well defined collection of objects. The objects of the set are called its elements. Sets are usually denoted by capital letters and the elements of the set are denoted by lower case letters. If an element x belongs to set A, it is denoted by x ∈ A. If x is not an element of A, it is denoted by x ∉ A. A set, in general is represented in two forms:

A set ‘A’ is said to be ﬁnite if it is either an empty set or contains finite number of elements, otherwise, it is said to be inﬁnite. Examples: 1. The set of vowels in English alphabet is finite. 2. The set of natural numbers is infinite.

1. Roster Form: In this form a set is described by actually listing out the elements. For example, the set of all even natural numbers less than 12 is represented by {2, 4, 6, 8, 10}. 2. Set Builder Form: In this form a set is described by a characterising property. For example, the set of all even natural numbers less than 12 is represented by {x ∈ N | x < 12 and x is even}. The symbol | is read as ‘such that.’

sOme deFinitiOns

Cardinality of a Set The number of distinct elements in a set is called the cardinality (or order) of the set. If a finite set A has n distinct elements, the cardinality of the set is n and is denoted by O(A) or n(A). The cardinality of empty set is 0. Example: Cardinality of A = {x, y, z, t} is 4.

Singleton Set

Null Set

A set consisting of a single element is called a singleton set, i.e. a set whose cardinality is 1 is a singleton set. Examples: (i) {3} (ii) {a}

A set is said to be a null set if it has no elements. It is also called an empty set or a void set and is denoted by f.

Equal Sets

Examples: 1. {x | x is an integer, 1< x < 2} 2. {x | x is a real number and x2 < 0}

Two sets A and B are said to be equal if they have the same elements, i.e., if every member of A is a member of B and every member of B is a member of A.

Chapter 3 Set Theory and Algebra | 2.267

Subsets and Supersets Let A and B be two sets. If every element of A belongs to B, then A is said to be a subset of B and is denoted by A ⊆ B. If A is a subset of B we say that B contains A or B is a superset of A and is denoted by B ⊂ A. If A is not a subset of B then this is denoted by A ⊄ B. Note: 1. Every set is a subset of itself. 2. Empty set is a subset of every set. 3. If A is a finite set of cardinality n, then the total number of subsets of A are 2n.

Power Set If A is any set, then the set of all subsets of A is called the power set of A and is denoted by P(A), i.e. P(A) = {S|S ⊆ A} Example: If A = {1, 2, 3}, then P(A) = {f, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} Note: 1. A, f ∈ P(A). 2. If A is a finite set having n elements, then the cardinality of P(A) is 2n.

Universal Set The set that contains all the objects in a given context is called Universal Set. It is denoted by m or U. Examples: 1. In Plane Geometry, the set of all points in the plane is the universal set. 2. In the context of divisibility tests, natural numbers will be the universal set.

Basic Set Operations Union of Sets If A and B are two sets, the union of A and B is the set of all those elements which belong to either A or to B or to both A and B and is denoted by A ∪ B. A ∪ B = {x |x ∈ A or x ∈ B} Note: 1. If A ⊂ B, then A ∪ B = B 2. A ∪ f = A 3. A ∪ m = m

Intersection of Sets Let A and B, be two sets. The intersection of A and B is the set of all those elements that belong to A and B. It is denoted by A ∩ B. A ∩ B = {x | x ∈ A and x ∈ B}

Note: 1. A ⊂ B, then A ∩ B = A. 2. A ∩ f = f. 3. A ∩ m = A.

Disjoint Sets Two sets A and B are said to be disjoint if they have no element in common. Note: A ∩ B = f.

Difference of Sets Let A and B be two sets. The difference of A and B, denoted by A - B, is the set of all those elements of A which do not belong to B. A - B = {x | x ∈ A and x ∉ B} = {x ∈ A| x ∉ B} Similarly, B - A = { x ∈ B| x ∉ A}. Example: A = {1, 2, 3, 4}, B = {3, 4, 8, 10} A - B = {1, 2} B - A = {8, 10}

Complement of a Set The complement of set A is the set of all those elements that do not belong to set A, i.e. the complement of a set A is the difference of the universal set and set A and is denoted by A’ or Ac. Example: If m is the set of natural numbers, A is the set of even natural numbers, then A’ is the set of odd natural numbers.

Symmetric Difference of Two Sets Let A and B be two sets. The symmetric difference of the sets A and B is the set (A - B) ∪ (B - A) and is denoted by A D B. Example: A = {1, 2, 3, 4}, B = {3, 4, 5, 6} then A D B = {1, 2, 5, 6}

Some Results 1. A ∪ A = A; A ∩ A = A 2. A ∪ B = B ∪ A 3. A ∩ B = B ∩ A 4. A ∪ (B ∪ C) = (A ∪ B) ∪ C 5. A ∩ (B ∩ C) = (A ∩ B) ∩ C 6. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 7. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 8. C - (C - A) = C ∩ A 9. C - (A ∪ B) = (C - A) ∩ (C - B) 10. C - (A ∩ B) = (C - A) ∪ (C - B) 11. (Ac)c = A 12. (A ∪ B)c = Ac ∩ Bc 13. (A ∩ B)c = Ac ∪ Bc

2.268 | Engineering Mathematics

Sets in Number system Natural numbers : Integers : Whole numbers : Rational numbers : Real numbers : Complex numbers:

N = {1, 2, 3, 4, .........} Z = {0, ±1, ±2, ±3, .......} W = { 0, 1, 2, 3, .........} Q = {x|x = p/q; p, q ∈ Z, q ≠ 0} R = {x|x is a rational or x is an irrational number} C = {x|x = a + ib; a, b ∈ R, i = −1 }

R-1 = {(b, a)| (a, b) ∈ R} We note that Domain of R = Range of R-1 and Range of R = Domain of R-1.

Identity Relation Let A be any set and let R = {(a, a)| a ∈ A}. Then R is called an Identity Relation or Equality Relation or Diagonal Relation in A.

Intervals

Example: A = {1, 2, 3}. Then R = {(1, 1) (2, 2) (3, 3)} is an identity relation.

(a, b) = a < x < b, x ∈ R [a, b] = a ≤ x ≤ b, x ∈ R

Reflexive Relation

Example: [0, 1] means all real numbers between 0 and 1 (both 0 and 1 included) (0, 1) means all real numbers between 0 and 1(0 and 1 excluded)

Cartesian Product of Two Sets Let A and B be any two sets. Then the Cartesian Product of A and B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B. The product is denoted by A × B. A × B = {(a, b)| a ∈ A, b ∈ B} Example: A = {1, 2, 3} B = {a, b} then A × B = {(1, a), (2, a), (3, a), (1, b), (2, b), (3, b)} B × A = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)} Note: 1. If A and B are two sets such that O(A) = m and O(B) = n, then the number of ordered pairs in A × B is mn 2. A × B ≠ B × A 3. n(A × B) = n(B × A)

Relations

A relation R on a set A is said to be reflexive if for every x ∈ A, (x, x) ∈ R. Example: Let A = {1, 2, 3}. R = { (1, 1) (1, 2) (1, 3) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (3, 3)} is a reflexive relation in A. Note: If ‘A’ has n elements then a reflexive relation must have at least ‘n’ ordered pairs.

Symmetric Relation A relation R on a set is said to be symmetric if x R y ⇒ y R x, i.e., (x, y) ∈ R ⇒ (y, x) ∈ R Example: Let A = {1, 2, 3}. R = {(3, 3), (1, 1) (1, 2) (2, 1)} is a symmetric relation in A. Note: The necessary and sufficient condition for a relation R to be symmetric is R = R-1.

Transitive Relation A relation R on a set A is said to be transitive if x R y and y R z ⇒ x R z, i.e., (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R.

Any subset of A × B is a relation from A to B. If a ∈ A is paired with b ∈ B in a relation R then (x, y) ∈ R and x R y are the notations used. A subset of A × A is called a relation on the set A.

Equivalence Relation

Example: A = {1, 2, 3}, B = {a, b} then R = {(1, a) (3, b)} is a possible relation.

A relation which is reflexive, symmetric and transitive is called an equivalence relation.

Domain and Range of a Relation The set of all first coordinates of the members of a relation R is called the Domain of R and the set of all second coordinates of members of R is called the range of R, i.e. if R is a relation from A to B then Domain of R = Dom R = {a |(a, b) ∈ R} Range of R = Ran R = {b|(a, b) ∈ R}

Inverse of a Relation Let A and B be two sets and let R be a relation from A to B. Then the inverse of R denoted by R-1 is a relation from B to A and is defined by

Example: In the set of natural numbers, the relation ‘is a factor of’ is a transitive relation.

Note: 1. The smallest equivalence relation on a set A is the identity relation. 2. The largest equivalence relation on a set A is the cartesian product A × A.

Antisymmetric Relation A relation R on a set A is said to be antisymmetric, if a R b and b R a if and only if a = b i.e. R is antisymmetric if (a, b) ∈ R, (b, a) ∈ R ⇔ a = b. Example: The Relation ‘≤ ’ on the set R of real numbers is antisymmetric.

Chapter 3 Set Theory and Algebra | 2.269

Partially Ordered Relation Let R be a relation on a set A, then R is said to be a partial order relation if it is reflexive, antisymmetric and transitive. A set on which a partial order is defined is said to be a Partially Ordered Set (poset). Solved Examples

3. As a ≤ b and b ≤ c ⇒ a ≤ c. \ R is transitive. \ R satisfies the required reflexive, Anti-symmetric and transitive properties. \ R is a partially ordered relation.

Functions

Example 1: If A, B, C are three sets, then (A – B) ∩ (B – C)

Definition

Solution:

A relation f, which associates to each element of a set A exactly one element of set B is called a function from A to B and is denoted as f: A → B (read as f maps A into B). If (a, b) ∈ f then ‘b’ is called the image of ‘a’ under f and is written as b = f (a) and ‘a’ is called the pre-image of ‘b’. If f: A → B then A is called the domain of f and B is called the co-domain of f and the set R = {f (a)/a ∈ A} is called the range of f. The range of f is also called the image of f and is denoted by Im(f ) or f (A).

A

B

C

From the figure A – B = and B – C =

\ (A – B) ∩ (B – C) = f

Example 2: Classify the relation ‘is parallel to’ in the set of all lines in a plane. Solution: (i) R is reflexive because any line is parallel to itself. (ii) R is symmetric because if a line 1, is parallel to another line 2, then 2 is parallel to 1. (iii) R is transitive If 1, is parallel to 2, 2 is parallel to line 3, then 1 is parallel to 3. (iv) Not anti-symmetric. The line 1 is parallel to 2 and 2 is parallel to 1 does not imply 1 = 2. Note: Since ‘is parallel to’ satisfies reflexive, symmetric and transitive properties. Therefore ‘is parallel to’ is an equivalence relation. Example 3: If A = {1, 2, 3, 4}, then show that a reflexive relation in A has at least 4 elements. Solution: Let the relation in A be {(1, 1), (2, 2), (3, 3), (4, 4)}. It consists 4 elements and it is reflexive. A reflexive relation in A should have at least all possible pairs of the form (x, x) ∈ A × A. Example 4: Show that the relation R = {(x, y) / x ≤ y} on the set of natural numbers is partially ordered. Solution: 1. Every natural number is equal to itself. \ R satisfies reflexive property 2. As a ≤ b and b ≤ a ⇔ a = b; a; b ∈ N. \ R is anti-symmetric.

Note: 1. Range ⊆ Co-domain. 2. f ⊆ A × B. 3. Every element of A has a unique f-image in B. 4. Two or more elements of A can have the same f-image in B. 5. There may be elements in B which are not f-images of any element of A. 6. The number of functions from a set A containing m elements to another set B containing n elements is nm.

One-one Function (Injection) A function f: A → B is called as one-to-one function if distinct elements of A have distinct images in B. That is, f: A → B is one-one if x1 ≠ x2 ⇒ f (x1) ≠ f (x2) equivalently f (x1) = f (x2) ⇒ x1 = x2. Note: 1. If n(A) = m, n(B) = n, then the number of one to one functions is nPm (n ≥ m). 2. A one-one function is possible from A to B if n(A) ≤ n(B).

Many-one Function A function which is not one-one is called many-one function.

Onto Function (Surjection) A function f : A → B is called an onto function if every element of B is an image of at least one element of A i.e. f : A → B is onto, if for each y ∈ B, there exists x ∈ A such that f(x) = y. Note: 1. If f is onto, Range of f = co-domain of f. 2. If n(A) = m, n(B) = n then number of onto functions are nm - nC1 (n - 1)m + nC2 (n - 2)m ....+…(n ≤ m)

2.270 | Engineering Mathematics 3. An onto function is possible from A to B if n(A) ≥ n(B).

Into Function A function which is not onto is called into function, i.e., f : A → B is into if Range of f is a proper subset of B.

Bijection If a function is both one-one and onto, then it is called a bijective function or bijection. Note: 1. Number of bijections on a set with n elements is n! 2. A bijection from A to B is possible if n(A) = n(B).

Constant Function A function f : A → B is said to be a constant function if f(x) = k, for all x ∈ A, where k is a fixed element of B. Note: Range of f = {k}

Identity Function The function f : A → A defined by f(x) = x is called the identity function, denoted by IA. Note: 1. For the identity function, Range = Domain. Symbolically, IA(x) = x for x ∈ A. 2. Identity function is a bijection.

Inverse Function

4. If f: A → B is bijective then f -1 of = IA and fof -1 = IB. 5. If f : A → B and g: B → A are two functions such that gof = IA and fog = IB then g = f -1. 6. If h: A → B, g: B → C and f : C → D be any three functions then fo(goh) = (fog)oh. 7. (fof -1) (x) = x or fof-1 = I. 8. If gof is one-one then ‘f ’ must be one-one but ‘g’ need not be. 9. If gof is surjection then ‘g’ must be onto but ‘f ’ need not be. 10. If gof is bijection then ‘f ’ must be one-one and g’ must be onto.

Real Function If A is a non-empty subset of R, then a function f : A → R is called a real valued function.

Operations on Real-valued Functions If f : D1 → R and g: D2 → R and D = D1 ∩ D2 then we can define (a) f ± g : D → R such that for all x ∈ D, ( f ± g)(x) = f(x) ± g(x) (b) f × g : D → R such that for all x ∈ D, ( fg)(x) = f(x) × g(x) (c) f/g: D → R such that for all x ∈ D, f/g(x) = f(x)/g(x) provided g(x) ≠ 0 (d) for some constant c, (cf ) : D1 → R such that for all x ∈ D1, (cf )(x) = cf (x). Note: Domain of f ± g, fg, f/g, is D1 ∩ D2 where D1 is the domain of f and D2 is domain of g.

If f : A → B be a bijective function then the function f -1: B → A, where f -1 = {(b, a)/(a, b) ∈ f } is called the inverse of the function f.

Even and Odd Functions

Note: 1. If f : A → B is bijective, f -1 : B → A is also bijective 2. If f (a) = b, then a = f -1 (b)

Examples: (i) x2 + 5 (ii) cosx

Composition of Functions If f : A → B and g : B → C are two functions, then gof is a function from A to C such that gof (a) = g( f (a)) for every a ∈ A and is called the composition of f and g, read as ‘g composition f ’. Note: 1. If f: A → B and g : B → C are two functions then, (a) f and g are injective ⇒ gof is injective. (b) f and g are surjective ⇒ gof is surjective. (c) f and g are bijective ⇒ gof is bijective. 2. If f : A → A and is injective. If A is finite then f is bijective. But if A is infinite this is not true. 3. If f : A → B, g: B → C are two bijective functions then (gof )-1 = f -1og -1.

A function f(x) is said to be an even function if f(-x) = f(x) for all x in its domain.

A function f is said to be odd if f(-x) = - f(x) for every x in its domain. Examples: (iv) x3 (ii) sinx Note: If f and g are two functions then 1. f is even and g is even ⇒ fog is an even function 2. f is odd and g is odd ⇒ fog is an odd function 3. f is even and g is odd ⇒ fog is an even function 4. f is odd and g is even ⇒ fog is an even function

Periodic Function A function f (x) is said to be periodic if there exists a nonnegative real number T such that f(x + T) = f (x) ∀ x ∈ R. If T is the smallest positive real number such that f (x + T) = f (x) for all x ∈ R, then T is called the fundamental period of f (x).

Chapter 3 Set Theory and Algebra | 2.271 Note: 1. If f (x) is a periodic function with period T, then the function f (ax + b), where a > 0 is periodic with period T/a. The periods of both sin x and cos x is 2 p and the period of tan x is p. 2. By period, one means fundamental period. Example 5: Which of the following relations is a function on {1, 2, 3}? 1. f1 = {(2, 1), (2, 2), (3, 3)} 2. f2 = {(1, 2), (2, 1), (3, 3)} 3. f3 = {(1, 1), (1, 3), (2, 2)} 4. f4 = {(3, 1), (3, 2)} Solution: Clearly f2 is a function while in f1, ‘2’ has more than one image. In f3, 1 has more than one image and in f4, 3 has more than one image.

\ f1, f3, and f4 are not functions.

Example 6: Let f and g be two real valued functions defined on the domain {-2, -1, 0, 1}. If f = {(-2, -1), (-1, 0), (0, 1), (1, 2)}and g = {(-2, 0), (-1, 1), (0, 2), (1, 3)} then, find 2 f + 3g, f 2 and fg. Solution: (i) 2f (-2) + 3g(-2) = 2(-1) + 3(0) = -2 2f (-1) + 3g(-1) = 2(0) + 3(1) = 3 2f (0) + 3g(0) = 2(1) + 3(2) = 8 2f (1) + 3g(1) = 2(2) + 3(3) = 13 \ 2f + 3g = {(-2, -2), (-1, 3), (0, 8), (1, 13)} 2 (ii) f (x) = {f(x)}2 = {(-2, 1), (-1, 0), (0, 1), (1, 4)} (iii) fg(x) = f(x) ⋅ g(x) f (-2) ⋅ g(-2) = (-1)(0) = 0, f(-1)g(-1) = 0, f (0)g(0) = 1 ⋅ 2 = 2 and f (1)g(1) = (2) (3) = 6 \ fg = {(-2, 0), (-1, 0), (0, 2), (1, 6)} Example 7: Let f (x) = 3x2 - 7x + 9. For what values of x is 3f (x) = f (3x)? Solution: Given 3f(x) = f(3x) ⇒ 3(3x2 - 7x + 9) = 3(3x)2 - 7(3x) + 9 ⇒ 9x2 - 21x + 27 = 27x2 - 21x + 9 ⇒ 18x2 = 18 ⇒ x = ±1 Example 8: Find the domain for which the functions f(x) = 2x3 - 5x + 2 and g(x) = x3 -4x + 2 are equal. Solution: Given, f(x) and g(x) are equal ⇒ f(x) = g(x) ⇒ 2x3 - 5x + 2 = x3 -4x + 2 ⇒ x3 - x = 0 ⇒ x(x2 - 1) = 0 ⇒ x = 0 and x2 - 1 = 0 ⇒ x = 0 or x = ± 1 \ Required domain set is {-1, 0, 1} Example 9: If f(x) = function or not.

cosec x sec x , then verify f(x) is even tan x + cot x

cosec( − x ) sec ( − x ) −cosec x sec x = tan( − x ) + cot ( − x ) −(tan x + cot x ) = f(x) \ f(x) is even.

Solution: f(-x) =

Example 10: Find domain of the function 1 f(x) = . | x | −x Solution: Given f(x) =

1 | x | −x

is defined only for |x| – x

> 0 i.e. |x| > x i.e. x < 0 \ Domain is (- ∞, 0).

Algebra Binary Operation A binary operation ‘*’ on a set is a rule that assigns to each ordered pair of elements of the set, a unique element of the set. Formally, a binary operation on a set S is just a function of two variables on S, with values in S, or it is a map from S × S to S. i.e., * : S × S → S defined by * (a, b) = a * b where (a, b) ∈ S × S and a * b ∈S. Thus if * is a binary operation, then a * b ∈ S for all a, b ∈ S or equivalently we say that S is closed under the operation ‘*’. Note: The function notation is not very convenient. So various symbols are used to denote different binary operations in a set. We can use symbols +, o, *, (×) etc. However the most important notations are the additive and multiplicative notations. In terms of additive notation, the element of S associated with the element (a, b) of S × S is denoted by a + b. In terms of multiplication notation, the element of S associated to the element (a, b) of S × S is denoted by ab. Throughout this chapter, the following standard notation is used. Z → the set of all integers. Z+ → the set of all positive integers. Ze → the set of all even integers. R → the set of all real numbers. R+ → the set of all positive real numbers. R* → the set of all non-zero real numbers. Q → the set of all rational numbers. Q+ → the set of all positive rational numbers. N → the set of all natural numbers. C → the set of complex numbers. Examples: 1. Addition is a binary operation on N. i.e., for all a, b ∈ N, a + b ∈ N. We say N is closed with respect to addition. Similarly we have addition is a binary operation on Z, Q, R and C.

2.272 | Engineering Mathematics 2. Multiplication is a binary operation on N. i.e., for all a, b ∈ N, a × b ∈ N. Similarly we have multiplication is a binary operation on Z, Q, R and C. 3. Let S be a set and P(S) the power set of S. The operations union and intersection are binary operations on P (S). 4. Let S be the set of all m × n matrices. The operation of addition of matrices is a binary operation. 5. The composition of functions is a binary operation on the set of all functions on a non-empty set. Note: Subtraction is not a binary operation in N, and division is not a binary operation in N, Z, R and C.

Types of Binary Operations Associative Operation

Examples: 1. For the binary operation addition in R, 0 is the identity and for any a ∈ R, its negative (– a) is its additive inverse since, (– a) + a = 0 = a + (– a) for all a ∈ R. 2. For the binary operation multiplication in Q, 1 is the identity and for any non-zero rational number a/b, b/a is its ⎛b⎞ ⎛a⎞ ⎛a⎞ ⎛b⎞ multiplicative inverse, since ⎜ ⎟ ⎜ ⎟ = 1 = ⎜ ⎟ ⎜ ⎟ . ⎝b⎠ ⎝a⎠ ⎝a⎠ ⎝b⎠

Algebraic Structure A non-empty set G equipped with a binary operation * is called an algebraic structure denoted by (G, *)

Semi-Group

A binary operation * on a set S is associative if (a * b) * c = a * (b * c) for all a, b, c ∈ S.

An algebraic structure (G, *) is called a semi-group if * is associative in G, i.e., ∀ a, b, c ∈ G we have (a * b) * c = a * (b * c).

Commutative Operation

Monoid

A binary operation * on a set S is called commutative if a * b = b * a for all a, b ∈ S.

A semi-group in which the identity element exists is called a monoid.

Examples: 1. Addition and multiplication are both associative and commutative operations in R. 2. Matrix multiplication is associative but not commutative. 3. Subtraction and division operations are neither associative nor commutative in R - {0}. Note: 1. The number of binary operations on a set G of order n 2 is ( n) n . 2. The number of commutative binary operations on G of order n is nn(n+1)/2 .

Operation with Identity Elements The operation * on a set S is said to be an operation with identity element, if there exists an element e ∈ S such that a * e = a = e * a, for all a ∈ S. The element e if exists, is called the identity element of S. Examples: 1. For the binary operation ‘addition’ in Q, the identity element is 0; since, a + 0 = a = 0 + a for all a ∈ Q. 2. For the binary operation ‘multiplication’ in Q the identity element is 1; since, a * 1 = a = 1 × a for all a ∈ Q.

Invertible Elements An element a ∈ S is said to be invertible under a binary operation * defined on S, if there exists an element b ∈ S such that a * b = b * a = e, where e is the identity element of the set S.

Group A group is a non empty set together with a binary operation * defined on it, such that the following axioms are satisfied: 1. Closure: a, b ∈ G ⇒ a * b ∈ G for all a, b ∈ G This axiom reaffirms that ‘*’ is a binary operation. 2. Associativity: a * (b * c) = (a * b) * c for all a, b, c ∈ G. 3. Existence of Identity: There is an element e ∈ G such that a * e = a = e * a for all a ∈ G. Then e is called the identity element in G. 4. Existence of Inverse: For each element a ∈ G there exists an element b ∈ G such that a * b = e = b * a. The element b is called an inverse of a with respect to * and is denoted by a–1.

Abelian (Commutative) Group A group (G, *) is said to be abelian or commutative if its binary operation * is commutative in G. i.e., If a * b = b * a for all a, b ∈ G, then (G, *) is an abelian group. Examples: 1. (N, +) is a semi group where + is the usual addition. 2. (N, ×) is a monoid where × is the usual multiplication. 3. The set of all 2 × 2 matrices ⎛ p q⎞ ⎜ ⎟, ⎝ r s⎠

Chapter 3 Set Theory and Algebra | 2.273 where p, q, r, s ∈ R is a group under matrix addition since, closure property and associativity hold. Clearly ⎛0 0⎞ ⎜ ⎟ ⎝0 0⎠ is the identity element and ⎛ − p −q ⎞ ⎜ ⎟ ⎝ −r − s ⎠ is the additive inverse of ⎛ p q⎞ ⎜ ⎟. ⎝ r s⎠ 4. The set S of all 2 × 2 non-singular matrices ⎛ p q⎞ ⎜ ⎟ ⎝ r s⎠ where p, q, r, s ∈ R is a group under matrix multiplication, since: Let ⎛ p q⎞ ⎛w x⎞ A= ⎜ ⎟ and B = ⎜ ⎟ r s ⎝ ⎠ ⎝ y z⎠ where p, q, r, s, w, x, y, z ∈ R. Also |A| ≠ 0 and |B| ≠ 0. Then A, B ∈ S

⎛ pw + qy px + qz ⎞ ⇒ AB = ⎜ ⎟ ∈ S, |A B| ≠ 0 ⎝ rw + sy rx + sz ⎠ [∵|AB| = |A||B| ≠ 0] \ Closure property holds. Matrix multiplication is clearly associative.

⎛1 0⎞ ⎜ ⎟ is the identity element in S. ⎝0 1⎠ ⎛ p q⎞ 1 ⎛ s −q ⎞ The inverse of A = ⎜ ⎟ is ⎜ ⎟ ∈S since | A| ⎝ −r p ⎠ ⎝ r s⎠ | A | ≠ 0. Note: Since matrix multiplication is not commutative, it is not an abelian group. 5. Under usual addition Z, Q, R and C are groups. 6. Z under the usual multiplication is not a group even though identity element 1 exists. In fact any element other than 1 and – 1 do not have an inverse in Z. 7. Under usual multiplication the set Q* consisting nonzero rational numbers is an infinite abelian group. We note that Closure property and associativity hold. Clearly, 1 ∈ Q* and 1. x = x .1 = x for all x ∈ Q*. \ 1 is the identity element. 1 1 1 If x ∈ Q* then ∈ Q* and x. = .x=1 x x x \ Each element of Q* has an inverse in Q*; Also xy = yx for all x, y ∈ Q* . Since there are an infinite number of elements in Q*, Q* is an infinite group.

8. Under usual multiplication R* consisting non-zero real numbers forms an infinite abelian group. This can be proved similarly as above. 9. Trivial group is a group with just one element e, which is the identity element.

Order of a Group A group (G, *) is called a finite group, if G contains a finite number of distinct elements. Otherwise, G is called an infinite group. The number of elements in a finite group is called the order of the group. An infinite group is said to be of infinite order. The order of a group G is denoted by the symbol O(G).

Integral Powers of an Element Let G be a group and let a ∈ G. Let n be any positive integer. Then a * a * a * a * … * a (n times) is denoted by an. Note: 1. a0 = e 2. a -n = a-1 * a-1 *… a-1 (n times) = (an)-1 Note: For the sake of convenience and brevity, we often refer a group (G, *) as simply G and for a, b ∈ G, a * b is written as ab.

Some Results If (G, *) is a group, then we have 1. The identity element of G (w.r.t ‘.’) is unique. 2. Every element in G has a unique inverse in G. 3. For each a ∈ G, (a-1)-1 = a. 4. For all a, b ∈ G, (ab)-1 = b-1a-1. This is called the reversal law which can be extended. i.e., a1, a2, ……..an ∈ G, (a1a2a3……an)-1 = an-1 an - 1-1 an - 2-1……..a1-1 5. Let a, b ∈ G. Then for m, n ∈ N, we have (i) aman = am + n (ii) (am)n = amn (iii) en = e 6. If a, b, c ∈ G, then ab = ac ⇒ b = c (Left cancellation Law) ba = ca ⇒ b = c (Right cancellation Law) 7. If a, b are any two elements of the group G, then the equations ax = b and ya = b have unique solutions in G. 8. A semi-group (G, *) is a group if the equations ax = b, ya = b have unique solutions in G for a, b ∈ G. 9. A finite semi-group (G, *) satisfying the cancellation laws is a group. Note: To prove that an algebraic structure (G, *) is a group, it is enough to prove the existence of left identity in G and the existence of left inverse for each element in G w.r.t. * (alternatively, existence of right identity and right inverse can be checked for).

Results on Abelian Group 1. If G is a group such that for all a, b ∈ G (ab)2 = a2b2, then G is abelian.

2.274 | Engineering Mathematics 2. If G is a group in which ∀a, b ∈G, (ab)i = aibi for three consecutive integers i, then G is abelian. 3. Let G be a finite group whose order is not divisible by 3. If ∀a, b ∈ G, (ab)3 = a3b3, then G must be abelian. 4. If every element of the group G is its own inverse, then G is abelian. 5. Every group of order less than or equal to 5 is abelian. 6. If G is an abelian group, then ∀ a, b ∈ G and all integers n, (ab)n = an bn.

Order of an Element Let G be a group and a ∈ G. If there exists a least positive integer n such that an = e, then n is called the order of the element a. The order of ‘a’ is denoted by O(a). If no such positive integer n exists, then a is of infinite order. Example 1: Consider the multiplicative group G = {1, – 1, i, – i}. The order of each element of this group is: O (1) = 1 O (– 1) = 2 (∵ (- 1)2 = 1) O (i) = 4 (∵ (i)4 = 1) O (– i) = 4 (∵ (- i)4 = 1) Example 2: Consider (Q*, ×) where × is the usual multiplication and Q* is the set of all non-zero rational numbers. Here O(1) = 1, O(- 1) = 2 and every other element of Q* has an infinite order.

Some Results Suppose G is a group with identity element e, 1. O(e) = 1. 2. O(a) = O(a-1) for every a ∈ G. 3. If am = e for a ∈ G, then O(a) ≤ m. Further O(a) divides m. 4. If a ∈ G with order n and p be a number relatively prime to n, then O(ap) = n. 5. If G is a finite group and a ∈ G, then O(a) divides O(G). 6. If G is a finite group and a ∈ G, then ao(G) = e. 7. If every element of a group G except the identity element is of order 2, then G is abelian. 8. Suppose G is an abelian group and a, b ∈ G with O(a) = m and O(b) = n such that (m, n) = 1, then O(ab) = mn. 9. Let (G, *) be a group and a ∈ G such that O(a) = n. If H = {a, a2, a3, …, an}, then (H, *) is also a group. 10. The order of any positive integral power of an element in a group G cannot exceed the order of the element, i.e., for any a ∈ G, O(am) ≤ O(a), where m is a positive integer.

Cyclic Groups A group (G, ×) is said to be cyclic if there exists a ∈ G such that every element of G is expressible as some integral

power of a. If G = {an/n ∈ Z}, ‘a’ is called the generator of G and we write G = < a >. Examples: 1. G = {1, - 1, i, - i} is a cyclic group generated by i or - i. 2. (Z, +) is a cyclic group generated by 1 (can also be generated by -1) (It is an infinite cyclic group). 3. (G, +4), where G = {0, 1, 2, 3} and +4 is addition modulo 4, is a cyclic group generated by 1 or 3. 4. The nth roots of unity forms a cyclic group w.r.t. multiplication, w being generator. Results: 1. Every cyclic group is abelian but not conversely. 2. Every subgroup of a cyclic group is cyclic. 3. If a is a generator of a cyclic group G, then a-1 is also a generator of G. 4. A group of prime order is cyclic. 5. The order of a cyclic group is the same as the order of its generator. 6. Every infinite cyclic group has two and only two generators which are inverses of each other. 7. A cyclic group of order n has f (n) (Eulers f function) number of generators. 8. A finite group of order n containing an element of order n must be cyclic. Example 3: If G is a finite group with four elements, then prove that G must be abelian. Solution: Let G = {e, x, y, z}, where e is the identity element. So e = e–1. Since G has four elements, G has at least one element, say x, such that x2 = e. Two cases arise now. Case 1: y–1 = y and z–1 = z. In this case G is clearly abelian. Considering other elements in the table we have, x * y = y or x * y = z x*y=y⇒x=e But x ≠ e \x*y=z Similarly y * z = x; z * x = y. The composition table is * E X Y

E E X Y

x x e z

y y z e

z z y x

Z

Z

y

x

e

This group is called the Klein-4 group.

Chapter 3 Set Theory and Algebra | 2.275 Case 2: y–1 = z and z–1 = y The composition table is

*

E

x

y

z

E

E

x

y

z

X

x

E

z

y

y

y

z

x

e

z

z

y

e

x

G is clearly abelian. Example 4: Suppose x y = x in a group G. Then (a) y = 0 (b) y = e (c) x = 0 (d) x = e Solution: xy = x ⇒ x–1 xy = x-1 x ⇒ y = e Example 5: Under usual addition, is the set ⎫⎪ ⎪⎧ a b 2 : a, b, c, d ∈Q * ⎬ G =⎨ + d ⎪⎩ c ⎪⎭ where cd ≠ 0, a group? Solution: Let x1 = x2 =

a1 b1 + 2; c1 d1 a2 b2 + c2 d2

2

Where a1, b1, c1 , d1, a2, b2, c2, d2 ∈ Q* and c1 d1 ≠ 0 and c2 d2 ≠ 0

x1 + x2 =

=

a1 a2 ⎛ b1 b2 ⎞ + +⎜ + ⎟ 2 c1 c2 ⎝ d1 d2 ⎠

a1c2 + a2 c1 c1c2

⎛bd +b d ⎞ +⎜ 1 2 2 1 ⎟ 2 ∈ G ⎝ d1d2 ⎠

\ Closure property holds. Associativity holds in G, since 0 ∉ Q*. The identity element does not exists in G. \ G is not a group. Example 6: Let G = {1, 8}. Show that under multiplication modulo 9, G is a group? Solution: G = {1, 8} 1 x9 1 = 1; 1 x9 8 = 8; 8 x9 8 = 1 x9

1

8

1

1

8

8

8

1

From the table it follows that G is a group. Example 7: Let G be a group with identity e. Then, what is the value of x if a b x = c? Solution: a b x = c ⇒ (a b x)–1 = c–1, (taking inverse) ⇒ x–1 b–1 a–1 = c–1 ⇒ x x–1 b–1 a–1 = x c–1, (left multiply by x)

⇒ b–1 a–1 = x c–1 ⇒ e b–1 a–1 = x c–1 –1 –1 –1 ⇒ b a c = x c c, (right multiply by c) ⇒ b–1 a–1 c = x e = x (by definition of inverse).

ab a+b when a + b ≠ 0, ab ≠ 0. Show that Q* along with this operation is a semi–group. ab Solution: a * b = where ab ≠ 0, a + b ≠ 0. Closure a+b property exists ⎛ ab ⎞ (a * b) * c = ⎜ ⎟*c ⎝a+b⎠ ab c abc = a+b = ab ab + ac + bc +c a + b abc abc bc ⎞ ⎛ a *(b * c) = a * ⎜ = b+c = ⎟ bc ab + ac + bc ⎝ b+c⎠ a+ c \ (a * b) * c = a * (bb*+c) \ * is associative. Hence Q* along with this operation is a semi-group. Example 8: On Q* define the operation * by a * b =

Example 9: In R* define the operation # by a # b ab = 2 (a + b) − . 2 Show that (R*, #) is not a group. ab Solution: a # b = 2 (a + b) − 2 a, b ∈ R* ⇒ a # b ∈ R* Closure property is satisfied ab ⎞ ⎛ (a # b) # c = ⎜ 2a + 2b − ⎟ # c 2 ⎠ ⎝ c ⎛ ab ⎞ ab ⎞ ⎛ = 2 ⎜ 2a + 2b − ⎟ + 2c – ⎜ 2a + 2b − ⎟ 2 2 ⎠ 2 ⎝ ⎠ ⎝ abc = 4a + 4b + 2c – ab – ac – bc + 4 bc ⎞ ⎛ a # (b # c) = a # ⎜ 2b + 2c − ⎟ 2 ⎠ ⎝

bc ⎞ a ⎛ = 2a + 2 ⎜ 2b + 2c − ⎟ − 2 ⎠ 2 ⎝

= 2a + 4b + 4c – bc – ab – ac +

(a # b) # c ≠ a # (b # c) \ # is not associative. \ (R*, #) is not a group.

bc ⎞ ⎛ ⎜ 2b + 2c − 2 ⎟ ⎝ ⎠ abc 4

Example 10: Let G be the set of all real 2 × 2 matrices ⎛a b⎞ ⎜ ⎟ ⎝ c 0⎠ where bc ≠ 0. Is G a group under matrix multiplication?

2.276 | Engineering Mathematics 1. For all p ∈ P, p ≤ p (reflexivity). 2. For all p, q ∈ P, if p ≤ q and q ≤ p, then p = q (antisymmetry). 3. For all p, q, r ∈ P, if p ≤ q and q ≤ r, then p ≤ r (transitivity).

⎫⎪ ⎪⎧⎛ a b ⎞ Solution: G = ⎨⎜ ⎟ : a, b, c ∈ R, bc ≠ 0 ⎬ ⎩⎪⎝ c 0 ⎠ ⎭⎪

⎛d ⎛a b⎞ Let A = ⎜ ⎟ and B = ⎜ ⎝ c 0⎠ ⎝f

⎛a b⎞ ⎛ d AB = ⎜ ⎟⎜ ⎝ c 0⎠ ⎝ f

e⎞ ⎟. 0⎠

e ⎞ ⎛ ad + bf ⎟ =⎜ 0 ⎠ ⎝ cd

ae ⎞ ⎟∉ G ce ⎠

\ Closure axiom not satisfied. \ G is not a group. Example 11: With the help of Cayley’s table check whether the set {0, 1, 2,…, 6} is a group under ‘addition modulo 7’ as operation. Solution: G = {0, 1, 2,…., 6} The Cayley’s table is as follows: +7

0

1

2

3

4

5

6

0 1 2 3 4 5 6

0 1 2 3 4 5 6

1 2 3 4 5 6 0

2 3 4 5 6 0 1

3 4 5 6 0 1 2

4 5 6 0 1 2 3

5 6 0 1 2 3 4

6 0 1 2 3 4 5

It follows clearly from the table that G is a group under addition modulo 7. Example 12: With respect to multiplication modulo 5, is the set {1, 2, 3, 4} a group. Is this set a group with respect to addition modulo 5? Solution: G = {1, 2, 3, 4} x5

1

2

3

4

1 2 3 4

1 2 3 4

2 4 1 3

3 1 4 2

4 3 2 1

The table contains only elements of G. Associativity follows from associativity of integer multiplication. 1 is the identity element. Every row and column contains the identity element \ every element has an inverse. \ G is a group with respect to multiplication modulo 5. But G with respect to addition modulo 5 is not a group, since the identity element 0 ∉ G.

Partial Ordering and Lattices A relation R on a non-empty set P is said to be a Partial ordering or Partial order if R is reflexive anti-symmetric and transitive. A set P with Partial ordering R is said to be a Partially ordered set or Poset, and it is represented as (P, R). But it is common to use ‘≤ ’ to represent an arbitrary Partial order on P. So, symbolically it is represented as (P, ≤) and its characteristic properties are as given below.

Examples of Posets

1. ‘R’ – set of real numbers and ‘≤ ’ less than or equal to is a poset. We have, (i) x ≤ x for all x ∈ R For x, y ∈ R (ii) x ≤ y and y ≤ x ⇒ x = y (iii) for x, y, z ∈ R , x ≤ y and y ≤ z ⇒ x ≤ z. Thus ‘≤ ’ partially ordered. Note: Just as ‘less than or equal to’ is a partially ordered relation in R, ‘≥ –greater than or equal’ to would also be a partially ordered relation in R. We say ‘ ≥’ is dual of ‘≤ ’. 2. Let S be a non empty set and P(S) be its power set, the relation set inclusion ⊆ induces a partially ordered relation. Clearly, ‘ ⊇ – is a super set of ’ will be the dual. (P(s), ⊆) is a poset.

Comparability and Incomparability If such that for any a, b ∈ P, if either a ≤ b or b ≤ a, then we say the elements, a and b are comparable. If for some a, b ∈ P, neither a ≤ b nor b ≤ a, then a, b are said to be incomparable. The whole idea of partial ordering is to comment on comparability of 2 elements in the poset. Total order or Linear order If (P, ≤) is poset and for every a, b ∈ P, if either a ≤ b, or b ≤ a, then we say P is totally ordered. In a nut shell, any 2 elements must be comparable. Examples: 1. (R, ≤) is totally ordered because for any pair of real numbers a, b either a ≤ b, or b ≤ a holds. 2. Let D16 = {1, 2, 4, 8,16}, (factors of 16). Consider the relation ‘divides’ on D16. We know that ‘divides’ is a partially ordered relation in D16. We note that for any 2 elements a, b ∈ D16 either a divides b or b divides a. Note: Consider D6 = {1, 2, 3, 6} (factors of 6). The relation ‘divides’ is partially ordered. However it is not totally ordered. Consider 2, 3 ∈ D6. We observe that neither ‘2 divides 3’ nor 3 divides 2.

Hasse Diagrams b

a

Chapter 3 Set Theory and Algebra | 2.277 The pictoral representation of posets are done using Hasse diagrams. The construction of Hasse diagram is as follows. Let (P, ≤) be a poset. The elements of P are denoted as points or nodes. If a, b ∈ P, are such that a ≤ b then the point representing ‘a’ is placed below, the point which represents ‘b’. The vertices (nodes) implying transitive property could be avoided for instance if a ≤ b and b ≤ c we do not need a path connecting a and c. The dotted line in the diagram is avoided. c

Greatest Element and Least Element in a Poset If (P, ≤) is a poset, the element ‘l’ is said to be the least element in P. if l ≤ x for all x ∈ P, while the element ‘g’ is said to the greatest element in P, if x ≤ g for all x ∈ P. Note: l ≤ x ≤ g for all x ∈ P. Examples: 1. In D16, 1 is the least element, while 16 is the greatest element. 2. In D6, 6 is the greatest element and 1 is the least element. Note: A poset may or may not have the greatest or least element.

b a

d

e

Example: The Hasse diagrams of

a b

16

6

8 3

4

2

c

In the above Hasse diagram, the greatest and least elements does not exist. We notice that e, d, are maximal elements while b, c are minimal elements. c

d

2 1

1

b

D6

D16

D6 = {1, 2, 3, 6}

D16 = {1, 2, 4, 8, 16}

Now on, we describe the posets by using Hasse diagrams.

a

In the above Hasse diagram, a is the least element. The greatest element doesn’t exist. c and d are maximal elements.

Least Upper Bound or Supremum (lub)

Minimal Element in a Poset Let (P, ≤) be a poset. a ∈ P is said to be minimal element if there does not exist c ∈ P such that c ≤ a.

Maximal Element in a Poset

Let A be a non empty subset of a poset, (P, ≤) . ∈ P, is said to be the least upper bound (supremum of A) (i) If x ≤ l ∀ x ∈ A (ii) If 1 ∈ P is an upper bound of A, then l ≤ l1.

Let (P ≤) be a poset. b ∈ P is said to be a maximal element in P, if there does not exist c ∈ P, such that b ≤ c.

The first condition says l is an upper bound, the second condition says, no other upper bound can be less than (leastness).

Example: Consider the Hasse diagram for P = {a, b, c, d, e, f, g, h, i, j}

Greatest Lower Bound or Infimum (glb)

g

h

e

(i) If g ≤ x ∀ x ∈ A (ii) If g1 ∈ P is any lower bound of A, then g1 ≤ g.

f d b

c

j a

Let A be a non empty subset of a poset (P, ≤); g ∈ P is said to be the greatest lower bound of A

i

a, i, j are minimal elements whereas e, g, h are maximal elements of P. Note: There could be multiple maximal and multiple minimal elements in a poset.

The first condition says g is the lower bound of A. The second condition says, every lower bound of A is less than or equal to g (greatness). Examples: 1. Let D24 = {1, 2, 3, 4, 6, 8, 12, 24} (the factors of 24) with ‘divides’ as a partially ordered relation be considered as a poset.

2.278 | Engineering Mathematics Let A = {2, 4, 8, 12, 24} The Hasse diagram of A and D4 are as follows: • 24

• 24 12 •

12 •

•8

6•

•4

•8

•2

3•

•4

•1

2

Dual of a Statement

2 is the g l b and 24 is the l u b of A. We note that 1, is a lower bound of A though it is not a member of A.

Lattice Lattice is a partially ordered set in which every pair of the elements has l u b and g l b. Join (∨): If L is a lattice and for a, b ∈ L the join of a, b is defined as lub (a, b). It is denoted by a ∨ b. Meet (∧): If L is a lattice and for a, b ∈ L, the meet of a, b is defined as glb (a, b). It is denoted by a ∧ b. Examples: 1. D6 = {1, 2, 3, 6} We could form 4C2 = 6 pairs of elements out of D6 (non-identical pairs) {1, 2}, {1, 3}, {1, 6}…{3. 6}. Pair 1, 2 1, 3 1, 6 2, 3 2, 6 3, 6

lub 2 3 6 6 6 6

glb 1 1 1 1 2 3

Note: For identical pairs the element itself will be LUB as well as GLB. 2. Let A = {1, 2, 3}. The poset (P(A), ⊆) is a lattice. P(A) is power set of A, under set inclusion. The Hasse Diagram is a follows {1, 2, 3} {1, 3}

Properties of join and meet: 1. Commutative law: for a, b ∈ L a∧b=b∧a A ∨ b = b ∨ 2. Associative law: for a, b, c, ∈ P a ∧ (b ∧ c) = (a ∧ b) ∧c a ∨ (b ∨ c) = (a ∨ b)∨c 3. Absorption law: for a, b, ∈ L a ∧ (a ∨ b) = a a ∨ (a ∧ b) = a

{1, 2}

The dual of any statement in lattice (L, ∧, ∨) is defined to be the statement obtained by interchanging ∨ and ∧. Example: For the statement a ∨ (b ∧ a) = a, we note that a ∧ (b ∨ a) = a is the dual statement.

Idempotent Law In a lattice (L, ∨, ∧) for a ∈ L, a ∨ a, = a and a ∧ a = a.

Lattice and Order if (L, ∧ , ∨) is a lattice, let ‘≤ ’ be a relation defined on L as a ≤ b if a ∧ b = a for a, b, ∈ L. Note: We could also define as a ≤ b if a ∨ b = b

Bounded Lattice A lattice L is said to have an upper bound 1 and a lower bound 0 if for all x ∈ L. 0 ≤ x and x ≤ 1 1 is the upper bound and 0 is the lower bound of L. Note: (i) 0 ∨ x = x (ii) 0 ∧ x = 0 (iii) 1 ∨ x = 1 (iv) 1 ∧ x = x (v) Some lattices may be bounded above but not be below Example: positive integers under the usual greater than are equal to (z+, ≥) is bounded above but not below. (vi) Some lattices may be bounded below but not above, Example: Positive integers under usual less than or equal to’ (z+, ≤) is bounded below but above.

Distributive Lattice {2, 3}

A lattice L is said to be distributive if for any a, b, c ∈ L, a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) 1

{1}

{3}

{2}

a

b

φ o

Chapter 3 Set Theory and Algebra | 2.279 Example 1: The Hasse diagram shown below represents a distribute lattice

Example 2: 1

Important results Examples of a lattice that is non distributive:

b•

a•

• o

1 •

a•

ND1

ND2

Standard result: A lattice L is non-distributive if and only if it contains a sublattice isomorphic to ND1 or ND2.

a ∨ (b ∧ c) = a∨ (o) = a

Complemented lattice: A bounded lattice with lower bound 0 and upper bound 1 is said to be a complemented lattice if for each x ∈ L, there exists a unique x′ ∈ L such that x ∨ x′ = 1 and x ∧ x′ = 0. x’ is said to be complement of x.

(a ∨ b) ∧ (a ∨ c) = b ∧ 1 = b Thus

c

a ∨ (b ∧ c) = a ∨ (0) = a (a ∨ b) ∧ (a ∨ c) = 1 ∧ 1 Thus a ∨ (b ∧ c) ≠ (a ∨ b) ∧ (a ∨ c)

c

o

•b

a ∨ (b ∧ c) ≠ (a ∨ b) ∧ (a ∨ b)

Exercises Practice Problems 1 Directions for questions 1 to 74: Select the correct alternative from the given choices. 1. If A = {2, 4, 5, 6, 7}, B = {3, 4, 6, 7, 8} and C = {2, 3, 4, 5, 6}, then (A ∪ B) ∩ C = (A) {2, 4, 5, 6, 7} (B) {1, 2, 3, 4, 5} (C) {3, 4, 5, 6, 7} (D) {2, 3, 4, 5, 6} 2. If A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}, then (B – A) ∪ B = (A) A (B) B (C) A∪B (D) A ∩ B 3. If A and B are two sets, then (A – B) ∩ A = (A) A (B) B (C) A – B (D) A ∪ B 4. For non empty sets A and B, which of the following is a non empty set? (A) A – Ac (B) (A – B) ∩ (A ∩ B) (C) m – (A ∪ Ac) (D) A ∩ Ac 5. (A ∪ B) – (A ∩ B) = (A) A ∩ B (C) (A –B) ∩ (B – A)

(B) A ∪ B (D) (A – B) ∪ (B – A)

6. If A – B = X and B – A = Y, then Y – X = (A) Y (B) X (C) X ∩ Y (D) X ∪ Y 7. If A, B and C are any three sets, then A – (B ∪ C) = (A) (A – B) ∪ (A – C) (B) (A – B) ∪ (B – C) (C) (A – B) ∩ (A – C) (D) (A – B) ∩ (B – C)

8. Which of the following is false? (A) A – (B ∩ C) = (A – B) ∪ (A – C) (B) (A ∪ B)c = Ac∩ Bc (C) (A ∩ B)c = Ac ∩ Bc (D) A ∩ Ac = f 9. If A = {(x, y)/y = x2 where x, y ∈ Z} and B = {(x, y)/y = x where x, y ∈ Z}, then (A) n(A ∩ B) = 0 (B) n(A ∩ B) = 1 (C) n(A ∩ B) = 2 (D) n(A ∩ B) = 3 1 10. If A = {(x, y) / y = x+ , x ∈ Z and x ≠ 0} and B = {(x, x y)/y = 2x x∈ Z}, then (A) n(A ∩ B) = 0 (B) n(A ∩ B) = 2. (C) n(A ∩ B) = 3 . (D) n(A ∩ B) = 1 11. If B ⊇ A and C ⊆ A, then which of the following is necessarily true? (A) B ⊆ C (B) B = C (C) A = B ≠ C (D) C ⊆ B 12. P, Q and R are three sets, such that P ⊆ Q, Q ⊆ R and R ⊆ P implies (A) P = Q (B) Q = R (C) P = Q = R (D) P = Q = f 13. If A – B = C – B and A ∩ B = C ∩ B, then which of the following is true? (A) A ⊆ B (B) A ∩ C = B (C) A = C (D) A = B 14. Which of the following is not true? (A) If A and B are two sets, then B – A = B ∩ A (B) If A and B are two sets then A ∩ B = A ∪ B

2.280 | Engineering Mathematics

(C) If A, B and C are three sets, then A – (B – C) = (A – B) – (A – C) (D) If A, B and C are three sets then A ∩ (B ∩ C) ⊆ B ∩ C. 15. If the cardinality of a power set P(A) is 512, then the cardinality of the set A is (A) 8 (B) 9 (C) 10 (D) 7 16. Twenty members have attended a party, 3/5 of them belong to group A, 2/5 of them belong to group B and 1/5 of them belong to both the groups. How many members do not belong to neither of the groups? (A) 0 (B) 4 (C) 2 (D) 8 17. In a class of 125 students 72 can speak Tamil, 60 can speak Malayalam and 45 can speak Kanada. 48 speak Tamil and Malayalam, 37 speak Kanada and Tamil, 36 speak kanada and Malayalam and 30 speak all the three languages. How many speak none of the languages? (A) 31 (B) 39 (C) 53 (D) 48 Common data for questions 18 to 20: In a group of 20 students, 10 play cricket, 12 play volleyball and 3 play neither of the games. 18. How many students play both cricket and volleyball? (A) 2 (B) 3 (C) 4 (D) 5 19. How many students play exactly one game? (A) 12 (B) 10 (C) 5 (D) 7 20. How many students play at least one game? (A) 12 (B) 17 (C) 20 (D) 7 Common data for questions 21 to 24: In a survey of 250 families in a locality, it was found that 140 families subscribe for newspaper Deccan Chronicle, 110 families subscribe for Times of India and 110 families subscribe for The Hindu, 60 subscribe for both Deccan Chronicle and Times of India, 40 for both Deccan Chronicle and The Hindu, 50 for both Times of India and The Hindu and 20 are not subscribers of any of these newspapers 21. Find the number of families subscribing for all the three news papers. (A) 10 (B) 30 (C) 40 (D) 20 22. How many families subscribe for only one news paper? (A) 100 (B) 110 (C) 120 (D) 140 23. How many families subscribe for exactly two newspapers? (A) 90 (B) 100 (C) 110 (D) 150 24. How many families subscribe at least two newspapers? (A) 90 (B) 100 (C) 110 (D) 150 25. If A = {3, 4, 5, 6} and B = {a, b}, then the number of relations defined from A to B is (B) 28 (C) 12 (D) 8 (A) 26 26. If a set has 4 elements, then the number of relations on A is (A) 46 (B) 28 (C) 416 (D) 216

27. If n (A) = 6, then the number of elements in the identity relation defined on A is (A) 36 (B) 6 (C) 10 (D) 5 28. If A = {1, 3, 5, 7}, then the maximum number of elements in any relation defined on A is (A) 4 (B) 8 (C) 16 (D) 20 29. Let R = {(1, 1), (2, 2), (3, 3)} be a relation in the set A = {1, 2, 3}. Then R is ____. (A) Symmetric (B) Anti-symmetric (C) Both (A) and (B) (D) Neither (A) nor (B) 30. The relation ⊇, defined on the power set of a non empty set A, does not satisfies _____ property. (A) Reflexive (B) Symmetric (C) Transitive (D) Anti-symmetric 31. The relation R = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 1), (2, 1)} in the set X = {1, 2, 3} is (A) Reflexive and symmetric (B) Reflexive and transitive (C) Symmetric and transitive (D) Neither symmetric nor transitive 32. A relation R, defined as x + y = 10 in the set of real numbers, satisfies ______ property. (A) Reflexive (B) Symmetric (C) Transitive (D) Equivalence 33. If ‘R’ is a relation in a set X such that R-1 = R, then the relation R is (A) Transitive (B) Anti-symmetric (C) Symmetric (D) None of these 34. Let R be a reflexive relation in a finite set A having x-elements and let there be y ordered pairs in R. Then which of the following is true? (A) y ≥ x. (B) y ≤ x. (C) y = x. (D) None of these. 35. R1 and R2 are two relations which are equivalence in a set A. Then (A) R1 ∩ R2 is equivalence on A (B) R1 ∪ R2 is equivalence on A (C) R1 – R2 is equivalence on A (D) None of these 36. If A = {1, 3, 5, 7} and B = {2, 4, 5, 6, 7}, then which of the following set of ordered pairs represents a function from A to B? (A) {(1, 2), (5, 6), (3, 4)} (B) {(1, 2), (3, 4), (1, 6), (5, 7), (7, 6)} (C) {(1, 2), (3, 4), (5, 6), (7, 7)} (D) {(1, 2), (3, 4), (5, 6), (6, 7)} 37. If A = {2, 3, 4}, then which of the following relations represents a function in A to A? (A) f1 = {(x, y) / y = x - 1} (B) f2 = {(x, y) / y = 6 - x} (C) f3 = {(x, y) / y = x + 2} (D) f4 = {(x, y) / y = x2}

Chapter 3 Set Theory and Algebra | 2.281 38. If A and B are two sets such that n(A) = 3 and n(B) = 5, then the number of mappings from A to B are (B) 35 (C) 28 (D) 3 × 5 (A) 53 39. The number of one-one mappings from set A to set B when A = {1, 2, 3, 4} and B = {2, 3, 4, 5, 6} is (A) 5 (B) 45 (C) 120 (D) 0 40. If n(A) = 4 and n(B) = 3, then the number of onto functions from set A to set B is (A) 43 (B) 34 (C) 36 (D) None of these 41. If n(A) = 3 and n(B) = 4, then the number of onto mappings defined from set A to set B is (A) 1 (B) 34 (C) 43 (D) 0 42. The number of bijections defined from set A to A, where A = {1, 2, 3, 4} is (A) 6 (B) 24 (C) 44 (D) 24 2 43. f : R → R and f(x) = , then f is 1+ x2 (A) One-one but not onto. (B) Onto, but not one one (C) Bijective (D) None of these 44. If f(x) = 2x + 2–x and g(x) = 2x - 2–x, then the value of f(x).g(y) + f(y) g(x) = (A) g(x + y) (B) f(x + y) (C) 2g(x + y) (D) 2f(x + y) 45. The inverse of the function f(x) =

2x −1 is 3x + 2

1 + 2x 2x −1 (A) (B) 3x − 2 2 − 3x 2x + 1 2 − 3x (C) (D) 1 + 2x x −3 46. If f(x) =

3x + 2 3 , for x ≠ , then f –1of (0) is 2x − 3 2

(A) 0 (B) 1 (C) -1 (D) 2 2 47. If f(x) = x + 1 and g(x) = 2x -3, then the value of (gof )–1(5) can be (A) 0 (B) 1 (C) 2 (D) -2 48. Let f(x) be a function. Then f(x) + f(-x) is always (A) An odd function. (B) An even function. (C) A constant function. (D) One that cannot be said. 49. On the set of integers Z, which of the following is not a binary operation? (A) Addition (B) Subtraction (C) Multiplication (D) Division 50. Which of the following binary operations is not commutative? (A) G = {(a, b) : a, b ∈ R, a ≠ 0} with (a, b) * (c, d) = (ac, bc + d)

(B) G = {(a, b) : a, b ∈ Z, b ≠ 0} with (a, b) * (c, d) = (ac, bd) (C) G = {(a, b) : a, b ∈ Z, b ≠ 0} with (a, b) * (c, d) = (ad + bc, bd) (D) G = {(a, b) : a, b ∈ Q} with (a, b) * (c, d) = (a + c, b + d) 51. Let a binary operation * be defined in R by a * b = 6 ab. Then identity e = (A) 1/6 (B) 1/4 (C) 1/3 (D) 1/2 52. A set G consists of ordered pairs (a, b) such that a, b ∈ R, a ≠ 0. A binary operation * is defined on G as (a, b) * (c, d) = (a c, b + d). Then the inverse of (a, b) ∈ G is (A) (- a, - b) (B) (1/a, b) (C) (- 1/a, b) (D) (1/a, - b) 53. Which of the following algebraic structure does not form a group? (A) (Z, +) (Integers) (B) (R, +) (Real numbers) (C) (R+, ×) (positive real numbers) (D) (N, ×) (natural numbers) 54. Which of the following sets is not a group under multiplication? (A) G = {1, w, w2}, where w is a complex cube root of unity (B) G = {1, –1, i, –i} (C) G = C {C is the set of complex numbers} (D) G = R* (non-zero real numbers) 55. Which of the following is an example of an infinite non-abelian group? (A) Set of all non-singular n×n matrices with entries from Q under multiplication. (B) Q under addition (C) R under addition (D) C* under multiplication. (non-zero complex numbers) 56. The set G = {1, 2, 3, 4, 5} under multiplication modulo 6 is (A) An algebraic structure (B) A non-abelian group (C) An abelian group (D) None of the above 57. In a group G = {0, 1, 2, 3, 4} under addition modulo 5, the inverse of 4 is (A) 0 (B) 1 (C) 2 (D) 3 58. The identity element of the group of residue classes of integers modulo 6 with respect to addition of residue classes is (A) [1] (B) [0] (C) [2] (D) [3] 59. Order of the element 5 in the group (R*, o) where o denotes multiplication is (A) 0 (B) 1 (C) 2 (D) ∞ 60. The order of the element –1 in the multiplication group Q* is (A) 1 (B) 2 (C) 0 (D) Not defined

2.282 | Engineering Mathematics 61. Find the index of the subgroup H = {0, ± 3, ± 6, ± 9,……} in the additive group of integers. (A) 3 (B) 2 (C) 4 (D) not defined + 62. Let G = (R, +) and G′ = (R , 0) denote groups of real numbers under addition and positive real numbers under multiplication respectively. Which of the following is a group homomorphism? ⎧log x, x >1, (A) f: G → G′, f(x) = ⎨ ⎩ 1; x ≤ 1 (B) f: G → G′, f(x) = x + 3 (C) f: G′ → G f(x) = ex (D) f: G′ → G f(x) = logx 63. If R0 is the multiplicative group of non-zero real numbers and f: R0 → R0 is a map defined as f(x) = ax for a, x ∈ R0. Which of the following statement/s is/are true? I. f is a homomorphism for a = 1 II. f is homomorphism for a = – 1 III. f is not a homomorphism for a ≠ 1, – 1 (A) I, II (B) III alone (C) I, III (D) All of these 64. Which of the following is not a poset? (R – set of real numbers) (A) (R, ≤) (B) (R, ≥) (C) (R, =) (D) (R, ≠) 65. If Dn = {x : x is a divisor of n}, for n ∈ N, then which of the following poset is totally ordered with respect to the relation a R b if a is a multiple of b? (A) D36 (B) D42 (C) D625 (D) D90 66. Which of the following statements is true? P: All totally ordered sets have least elements Q: The Hasse diagram of a totally ordered set is a line. (A) P alone (B) Q alone (C) Both P, Q (D) Neither P nor Q Common data for questions 67 to 71: b

a

h

These questions are based on the following Hasse diagrams L = {a, b, c, d e, f, g, h, i}. 67. Which of the following is the least element of L? (A) d (B) f (C) e (D) Does not exist 68. How many maximal elements does L have? (A) 1 (B) 2 (C) 3 (D) 4 69. How many minimal elements does L have? (A) 1 (B) 2 (C) 3 (D) 4 70. Which of the following subset of L has the least element? (A) {a, b, c, d, h} (B) {b, c, d, f } (C) {a, b, c, e} (D) {c, g} 71. Which of the following pair of elements are comparable ? (A) {a, b} (B) {b, e} (C) {a, i} (D) {c, g} Common data for questions 72 and 73: Let A = {p, q, r}. The power set of A with usual union and intersection as join and meet is considered as a lattice (L, ∨ , ∧) 72. The lattice is (A) Complemented (C) Bounded

73. The complement of {p, q} is (A) {p, q} (B) A (C) f

g

d

f

(D) {r}

74. Which of the following Hasse Diagram represents a non distributive lattice? (A) (B) 1 1 •

•b

a•

a

b•

c•

•d

• o

e

c

(B) Distributive (D) All the above

o

(C) •I

(D) None of these

•O

i

Previous Years’ Questions 1. Consider the binary relation: S = {(x, y)|y = x + 1 and x, y ∈ {0, 1, 2}} The reflexive transitive closure of S is (A) {(x, y)|y > x and x, y ∈ {0, 1, 2}} (B) {(x, y)|y ≥ x and x, y ∈ {0, 1, 2}} (C) {(x, y)|y < x and x, y ∈ {0, 1, 2}} (D) {(x, y)|y ≤ x and x, y ∈ {0, 1, 2}}

2. The following is the incomplete operation table of a 4-element group. [2004]

* e a b c

e e a

a a b

b b c

c c e

Chapter 3 Set Theory and Algebra | 2.283 The last row of the table is [2004] (A) c a e b (B) cbae (C) c b e a (D) ceab 3. The inclusion of which of the following sets into S = {{1, 2}, {1, 2, 3}, {1, 3, 5}, {1, 2, 4}, {1, 2, 3, 4, 5}} is necessary and sufficient to make S a complete lattice under the partial order defined by set containment? [2004] (A) {1} (B) {1}, {2, 3} (C) {1}, {1, 3} (D) {1}, {1, 3}, {1, 2, 3, 4}, {1, 2, 3, 5} 4. Let A, B and C be non-empty sets and let X = (A - B) - C and Y = (A - C) - (B - C) Which one of the following is TRUE? [2005] (A) X = Y (B) X ⊂ Y (C) Y ⊂ X (D) None of these 5. The following is the Hasse diagram of the poset [{a, b, c, d, e}, ≤ ] a

b

d

c

e

The poset is: [2005] (A) Not a lattice (B) A lattice but not a distributive lattice (C) A distributive lattice but not a Boolean algebra (D) A Boolean algebra 6. The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively:[2005] (A) 3 and 13 (B) 2 and 11 (C) 4 and 13 (D) 8 and 14 7. Let R and S be any two equivalence relations on a nonempty set A. Which one of the following s tatements is TRUE?[2005] (A) R ∪ S, R ∩ S are both equivalence relations (B) R ∪ S is an equivalence relation (C) R ∩ S is an equivalence relation (D) Neither R ∪ S nor R ∩ S is an equivalence relation 8. Let f: B → C and g: A →B be two functions and let h = fog. Given that h is an onto function. Which one of the following is TRUE? [2005] (A) f and g should both be onto functions (B) f should be onto but g need not be onto

(C) g should be onto but f need not be onto (D) both f and g need not be onto 9. Consider the set H of all 3 × 3 matrices of the type ⎡a f ⎢ ⎢0 b ⎢⎣0 0

e⎤ d ⎥⎥ c ⎥⎦

where a, b, c, d, e and f are real numbers and abc ≠ 0. Under the matrix multiplication operation, the set H is: [2005] (A) A group (B) A monoid but not a group (C) A semi group but not a monoid (D) Neither a group nor a semi-group 10. Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X × Y and E be the set of all subsets of W. The number of functions from Z to E is: [2006] (A) z (B) z × 2xy (C) z2 (D) 2xyz 11. The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four pausible reasons. Which one of them is false? [2006] (A) It is not closed (B) 2 does not have an inverse (C) 3 does not have an inverse (D) 8 does not have an inverse 12. A relation R is defined on ordered pairs of integers as follows: (x, y) R (u, v) if x < u and y > v. Then R is: [2006] (A) Neither a Partial Order nor an Equivalence Relation (B) A Partial Order but not a Total order (C) A Total order (D) An Equivalence Relation 13. Let E, F and G be finite sets. Let X = (E ∩ F) - (F ∩ G) and Y = (E - (E ∩ G)) - (E -F) Which one of the following is true? [2006] (A) X ⊂ Y (B) X ⊂ Y (C) X = Y (D) X - Y ≠ f and Y - X ≠ f 14. Given a set of elements N = {1, 2,...., n} and two arbitrary subsets A ⊆ N and B ⊆ N, how many of the n! permutations p from N to N satisfy min (p (A)) = min (p(B)), where min (S) is the smallest integer in the set of integers S, and p (S) is the set of integers obtained by applying permutation p to each element of S? [2006]

2.284 | Engineering Mathematics

(A) (n - |A ∪ B|)|A||B| (B) (|A|2 + |B|2) n2

p2

(D)

p1

p3

| A∩B| (C) n! | A ∪ B | ⎛ n ⎞ (D) |A ∩ B|2/ ⎜ ⎟ ⎝| A ∪ B |⎠ 1 5. Let S = {1, 2, 3,..., m}, m > 3. Let X1.....Xn be subsets of S each of size 3. Define a function f from S to the set of natural numbers as, f (i) is the number of sets Xj that contain the element i. That is f (i) = |{j|i ∈ Xj}|.

m

Then

∑ f (i )

is:[2006]

i =1

(A) 3m (B) 3n (C) 2m + 1 (D) 2n + 1 16. Let S be a set of n elements. The number of ordered pairs in the largest and the smallest equivalence relations on S are:[2007] (A) n and n (B) n2 and n 2 (C) n and 0 (D) n and 1 17. What is the maximum number of different Boolean functions involving n Boolean variables? [2007] n

2

2 2n (A) n2 (B) 2n (C) 2 (D) 18. How many different non-isomorphic Abelian groups of order 4 are there?[2007] (A) 2 (B) 3 (C) 4 (D) 5 19. Consider the set S = {a, b, c, d}. Consider the following 4 partitions p1, p2, p3, p4 on S : p1 = {a b c d} , p2 =

{ab, cd} , p3 = {abc, d} ,

p4 = {a, b, c, d} . Let p be the

partial order on the set of partitions S/ = {p1, p2, p3, p4} defined as follows: pi p pj if and only if pi refines pj. The poset diagram for (S/, p) is:[2007] p1 (A) p1 (B)

p2

p2

p4

p3 p3

p4

(C)

p1

p2

p3

p4

p4

20. If P, Q, R are subsets of the universal set U, then (P ∩ Q ∩ R) ∪ (Pc ∩ Q ∩ R) ∪ Qc ∪ Rc is [2008] (A) Qc ∪ Rc (B) P ∪ Qc ∪ Rc (C) Pc ∪ Qc ∪ Rc (D) U 21. Which one of the following is NOT necessarily a property of a Group?[2009] (A) Commutativity (B) Associativity (C) Existence of inverse for every element (D) Existence of identity 22. Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE? [2009] (A) R is symmetric but NOT antisymmetric (B) R is NOT symmetric but antisymmetric (C) R is both symmetric and antisymmetric (D) R is neither symmetric nor antisymmetric 23. For the composition table of a cyclic group shown below * a b c d

a a b c d

b b a d c

c c d b a

d d c a b

Which one of the following choices is correct? [2009] (A) a, b are generators (B) b, c are generators (C) c, d are generators (D) d, a are generators 24. What is the possible number of reflexive relations on a set of 5 elements?[2010] (A) 210 (B) 215 (C) 220 (D) 225 25. Consider the set S = {1, w, w2}, where w and w2 are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) forms [2010] (A) A group (B) A ring (C) An integral domain (D) A field 26. How many onto (or surjective) functions are there from an n element (n ≥ 2) set to a 2-element set? [2012] (A) 2n (B) 2n – 1 (C) 2n – 2 (D) 2(2n – 2) 27. A binary operation ⊕ on a set of integers is defined as x ⊕ y = x2 + y2. Which one of the following statements is TRUE about ⊕?[2013]

Chapter 3 Set Theory and Algebra | 2.285 2 8.

(A) Commutative but not associative (B) Both commutative and associative (C) Associative but not commutative (D) Neither commutative nor associative Let S denote the set of all functions f: (0, 1)4 → {0, 1}. Denote by N the number of functions from S to the set {0, 1}. The value of log2 log2 N is ______. [2014]

29. Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U < V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements: S1: There is a subset of S that is larger than every other subset. S2: There is a subset of S that is smaller than every other subset. Which one of the following is CORRECT? [2014] (A) Both S1 and S2 are true. (B) S1 is true and S2 is false. (C) S2 is true and S1 is false. (D) Neither S1 nor S2 is true. 30. Let X and Y be finite sets and f: X → Y be a function. Which one of the following statements is TRUE? [2014] (A) For any subsets A and B of X, |f(A ∪ B)| = |f(A)| + |f(B)| (B) For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B) (C) For any subsets A and B of X, |f(A ∩ B)| = min{|f(A)|, |f(B)|} (D) For any subsets S and T of Y, f –1(S ∩ T) = f –1(S) ∩ f –1(T) 31. Let G be a group with 15 elements. Let L be a subgroup of G. It is known that L ≠ G and that the size of L is at least 4. The size of L is ––––. [2014] 32. Consider the set of all functions f :{0,1, ………,2014} → {0, 1 …..,2014} such that f(f(i)) = i, for all 0 ≤ i ≤ 2014. Consider the following statements: P. For each such function it must be the case that for every i, f(i) = i. Q: For each such function it must be the case that for some i, f(i) = i. R. Each such function must be onto. Which one of the following is CORRECT? [2014] (A) P, Q and R are true (B) Only Q and R are true (C) Only P and Q are true (D) Only R is true

33. There are two elements x, y in a group (G, *) such that every element in the group can be written as a product of some number of x’s and y’s in some order. It is known that x*x=y*y =x*y*x*y=y*x*y*x=e Where e is the identify element. The maximum number of elements in such a group is ––––––. [2014] x g ( h( x )) is , then x −1 h( g ( x )) [2015]

34. If g(x) = 1 – x and h(x) =

h( x ) −1 (B) g( x) x x g( x) (C) (D) (1 − x ) 2 h( x ) (A)

35. For a set A, the power set of A is denoted by 2A. If A = {5, {6}, {7}}, which of the following options are TRUE?[2015] I. ∅ ∈ 2A II. ∅ ⊆ 2A A III. {5, {6}} ∈ 2 IV. {5, {6}} ⊆ 2A (A) I and III only (B) II and III only (C) I, II and III only (D) I, II and IV only 36. The cardinality of the power set of {0, 1, 2,…, 10} is _____[2015] 37. Let R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1. Which one of the following statements about R is true?[2015] (A) R is symmetric and reflexive but not transitive (B) R is reflexive but not symmetric and not transitive (C) R is transitive but not reflexive and not symmetric (D) R is symmetric but not reflexive and not transitive 38. The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = (a, b, c} is ______. [2015] 39. Suppose U is the power set of the set S = {1,2,3,4,5,6}. For any T ∈ U, let |T | denote the number of elements in T and T 1 denote the complement of T. For any T, R ∈ U, let T \R be the set of all elements in T which are not in R. Which one of the following is true?[2015] (A) ∀X ∈ U (|X | = |X 1|) (B) $X ∈ U $ Y ∈ U (|X | = 5, |Y | = 5 and X ∩ Y = f) (C) ∀X ∈ U ∀Y ∈ U(|X| = 2, |Y| = 3 and X \Y = f) (D) ∀X ∈ U ∀Y ∈ U (X \ Y = Y 1\ X 1) 40. Let R be a relation on the set of ordered pairs of positive integers such that ((p, q),(r, s)) ∈ R if and only if p – s = q – r. Which one of the following is true about R?[2015]

2.286 | Engineering Mathematics (A) Both reflexive and symmetric (B) Reflexive but not symmetric (C) Not reflexive but symmetric (D) Neither reflexive nor symmetric 41. A function f: N+→N+, defined on the set of positive integers N+, satisfies the following properties: f (n) = f (n/2) if n is even f (n) = f (n+5) if n is odd let R = {i|∃ j: f (j) = i} be the set of distinct values that f takes. The maximum possible size of R is ____ . [2016]

42. A binary relation R on N × N is defined as follows: (a, b) R (c, d) if a ≤ c or b ≤ d. consider the following propositions:[2016] P : R is reflexive Q : R is transitive Which one of the following statements is TRUE? (A) Both P and Q are true. (B) P is true and Q is false (C) P is false and Q is true (D) Both P and Q are false.

Answer Keys

Exercises Practice Problems 1 1. D 11. D 21. D 31. B 41. D 51. A 61. A 71. C

2. B 12. C 22. C 32. B 42. B 52. D 62. D 72. D

3. C 13. C 23. A 33. C 43. D 53. D 63. C 73. D

4. A 14. C 24. C 34. A 44. C 54. C 64. D 74. B

5. D 15. B 25. B 35. A 45. A 55. A 65. C

6. A 16. B 26. D 36. C 46. A 56. D 66. B

7. C 17. B 27. B 37. B 47. B 57. B 67. D

8. C 18. D 28. C 38. A 48. B 58. B 68. C

9. C 19. A 29. C 39. C 49. D 59. D 69. B

10. B 20. B 30. B 40. C 50. A 60. B 70. A

4. A 5. B 14. C 15. B 24. C 25. A 33. 4 to 4 34. A

6. C 16. B 26. C 35. C

7. C 8. B 9. A 17. C 18. A 19. C 27. A 28. 16 to 16 36. 2048 37. D 38. 36

10. D 20. D 29. A 39. D

Previous Years’ Questions 1. B 2. D 3. A 11. C 12. A 13. C 21. A 22. D 23. C 30. D 31. 5 to 5 32. B 40. C 41. 2 42. B

Chapter 4 Combinatorics LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • •

Permutations Combinations Generating permutations Lexicographic order

introDuction Permutations and Combinations is one of the important areas in many exams because of two reasons. The first is that solving questions in this area is a measure of students’ reasoning ability. Secondly, solving problems in areas like Probability requires thorough knowledge of Permutations and Combinations. Before discussing Permutations and Combinations, let us look at what is called as the ‘fundamental rule’. ‘If one operation can be performed in ‘m’ ways and (when, it has been performed in any one of these ways), a second operation then can be performed in ‘n’ ways, the number of ways of performing the two operations will be m × n’. This can be extended to any number of operations. If there are three cities A, B and C such that there are 3 roads connecting A and B and 4 roads connecting B and C, then the number of ways one can travel from A to C is 3 × 4, i.e., 12. This is a very important principle and we will be using it extensively in Permutations and Combinations. Because we use it very extensively, we do not explicitly state every time that the result is obtained by the fundamental rule but directly write down the result.

Permutations Each of the arrangements which can be made by taking some or all of a number of items is called a Permutation. Permutation implies ‘arrangement’ or that ‘order of the items’ is important. The permutations of three items a, b and c taken two at a time are ab, ba, ac, ca, cb and bc. Since the order in which the items are taken is important, ab and ba are counted as two different

• • • •

Recurrence relations Homogenous linear recurrence relation Generating functions Useful generating functions

permutations. The words ‘permutation’ and ‘arrangement’ are synonymous and can be used interchangeably. The number of permutations of n things taking r at a time is denoted by nPr (and read as ‘nPr’).

comBinations Each of the groups or selections which can be made by taking some or all of a number of items is called a Combination. In combinations, the order in which the items are taken is not considered as long as the specific things are included. The combination of three items a, b and c taken two at a time are ab, bc and ca. Here, ab and ba are not considered separately because the order in which a and b are taken is not important but it is only required that a combination including a and b is what is to be counted. The words ‘combination’ and ‘selection’ are synonymous. The number of combinations of n things taking r at a time is denoted by nCr (and read as ‘nCr’). When a problem is read, it should first be clear to you as to whether it is a permutation or combination that is being discussed. Some times the problem specifically states whether it is the number of permutations (or arrangements) or the number of combinations (or selections) that you should find out. The questions can be as follows: For permutations, ‘Find the number of permutations that can be made .....’ OR ‘Find the number of arrangements that can be made.....’ OR ‘Find the number of ways in which you can arrange.....’

2.288 | Engineering Mathematics For combinations, ‘Find the number of combinations that can be made .....’ OR ‘Find the number of selections that can be made.....’ OR ‘Find the number of ways in which you can select.....’ Some times, the problem may not explicitly state whether what you have to find out is a permutation or a combination but the nature of what is to be found out will dictate whether it is the number of permutations or the number of combinations that you have to find out. Let us look at the following two examples to clarify this. ‘How many four digit numbers can be made from the digits 1, 2, 3 and 4 using each digit once?’ Here, since we are talking of numbers, the order of the digits matters and hence what we have to find out is permutations. ‘Out of a group of five friends that I have, I have to invite two for dinner. In how many different ways can I do this?’ Here, if the five friends are A, B, C, D and E, whether the two friends that I call for dinner on a particular day are A and B or B and A, it does not make any difference, i.e., here the order of the ‘items’ does not play any role and hence it is the number of combinations that we have to find out. Now we will find out the number of permutations and combinations that can be made from a group of given items. Initially, we impose two constraints (conditions) while looking at the number of permutations. They are 1. All the n items are distinct or dissimilar (or no two items are of the same type) 2. Each item is used at most once (i.e., no item is repeated in any arrangement)

Number of Linear Permutations of ‘n’ Dissimilar Items Taken ‘r’ at a Time Without Repetition (nPr) Consider r boxes each of which can hold one item. When all the r boxes are filled, what we have is an arrangement of r items taken from the given n items. So, each time we fill up the r boxes with items taken from the given n items, we have an arrangement of r items taken from the given n items without repetition. Hence the number of ways in which we can fill up the r boxes by taking things from the given n things is equal to the number of permutations of n things taking r at a time. Boxes

� � 1 2

� � .......................... 3 4

� r

The first box can be filled in n ways (because any one of the n items can be used to fill this box). Having filled the first box, to fill the second box we now have only (n – 1) items; any one of these items can be used to fill the second box and hence the second box can be filled in (n – 1) ways;

similarly, the third box in (n – 2) ways and so on the rth box can be filled in {n – (r – 1)} ways, i.e. [n – r + 1] ways. Hence, from the Fundamental Rule, all the r boxes together be filled up in

n × (n – 1) × (n – 2) … (n – r + 1) ways

So,

n

Pr = n × (n – 1) × (n – 2) … (n – r + 1)

This can be simplified by multiplying and dividing the right hand side by (n – r) (n – r – 1) … 3.2.1 giving us nPr = n(n – 1) (n – 2) … [n – (r – 1)] ( n − 1) ( n − 2)…[n − ( r − 1)].( n − r )… 3.2.1 = ( n − r )… 3.2.1 n! = ( n − r )! The number of permutations of n distinct items taking r items at a time is n! n Pr = ( n − 4)!

If we take n items at a time, then we get nPn. From a discussion similar to that we had for filling the r boxes above, we can find that nPn is equal to n! The first box can be filled in n ways, the second one in (n – 1) ways, the third one in (n – 2) ways and so on, then the nth box in 1 way; hence, all the n boxes can be filled in n(n – 1) (n – 2) … 3.2.1 ways, i.e., n! ways. Hence, n

Pr = n !

But if we substitute r = n in the formula for nPr, then we n! get nPn = ; since we already found that nPn = n!, we can 0! conclude that 0! = 1. Number of combinations of n dissimilar things taken r at a time. Let the number of combinations nCr be x. Consider one of these x combinations. Since this is a combination, the order of the r items is not important. If we now impose the condition that order is required for these r items, we can get r! arrangements from this one combination. So each combination can give rise to r! permutations. x combinations will thus give rise to x . r! permutations. But since these are all permutations of n things taken r at a time, this must be equal to nPr. So, x.r! = nPr = ⇒

n! ( n − r )!

n! Cr = r !.( n − r )!

n

The number of combinations of n dissimilar things taken all at a time is 1.

Chapter 4 Combinatorics | 2.289 Out of n things lying on a table, if we select r things and remove them from the table, we are left with (n – r) things on the table – that is, whenever r things are selected out of n things, we automatically have another selection of the (n – r) things. Hence, the number of ways of making combinations taking r out of n things is the same as selecting (n – r) things out of n given things, i.e., n

Cr = nCn–r

When we looked at nPr, we imposed two constraints which we will now release one by one and see how to find out the number of permutations.

Number of Arrangements of n Items of which p are of One Type, q are of a Second Type and the Rest are Distinct When the items are all not distinct, then we cannot talk of a general formula for nPr for any r but we can talk of only n Pn (which is given below). If we want to find out nPr for a specific value of r in a given problem, we have to work on a case to case basis (this has been explained in one of the solved examples). The number of ways in which n things may be arranged taking them all at a time, when p of the things are exactly alike of one kind, q of them exactly alike of another kind, r of them exactly alike of a third kind, and the rest all distinct is n! . p! q! r !

Number of Arrangements of n Distinct Items where each Item can be used any Number of T imes (i.e., Repetition Allowed) You are advised to apply the basic reasoning given while deriving the formula for nPr to arrive at this result also. The first box can be filled up in n ways; the second box can be filled in again in n ways (even though the first box is filled with one item, the same item can be used for filling the second box also because repetition is allowed); the third box can also be filled in n ways and so on ... the rth box can be filled in n ways. Now all the r boxes together can be filled in {n.n.n.n ... r times} ways, i.e., nr ways. The number of permutations of n things, taken r at a time when each item may be repeated once, twice, … up to r times in any arrangement is nr. What is important is not this formula by itself but the reasoning involved. So, even while solving problems of this type, you will be better off if you go from the basic reasoning and not just apply this formula.

Total Number of Combinations Out of n given things, the number of ways of selecting one or more things is where we can select 1 or 2 or 3 ... and so

on n things at a time; hence the number of ways is nC1 + nC2 + nC3 + … nCn This is called ‘the total number of combinations’ and is equal to 2n – 1 where n is the number of things. The same can be reasoned out in the following manner also. There are n items to select from. Let each of these be represented by a box. 1 2 3 4 n No. of ways of dealing � � � � .. ..... � with the boxes 2 2 2 2 2 The first box can be dealt with in two ways. In any combination that we consider, this box is either included or not included. These are the two ways of dealing with the first box. Similarly, the second box can be dealt with in two ways, the third one in two ways and so on, the n th box in two ways. By the Fundamental Rule, the number of ways of dealing with all the boxes together is 2 × 2 × 2 × … ways, i.e., in n times 2n ways. But out of these, there is one combination where we ‘do not include the first box, do not include the second box, do not include the third box and so on, do not include the n th box.’ That means, no box is included. But this is not allowed because we have to select one or more of the items (i.e., at least one item). Hence this combination of no box being included is to be subtracted from the 2n ways to give the result of Number of ways of selecting one or more items from n given items is 2n – 1 Number of integral solutions of the equation x1 + x2 + … + xn = s Consider the equation x1 + x2 + x3 = 10 If we consider all possible integral solutions of this equation, there are infinitely many. But the number of positive (or non-negative) integral solutions is finite. We would like the number of positive integral solutions of this equation, i.e., values of (x1, x2, x3) such that each xi > 0. We imagine 10 identical objects arranged on a line. There are 9 gaps between these 10 objects. If we choose any two of these gaps, we are effectively splitting the 10 identical objects into 3 parts of distinct identity. Conversely, every split of these 10 objects corresponds to a selection of 2 gaps out of the 9 gaps. Therefore, the number of positive integral solutions is 9 C2. In general, if x1 + x2 + … + xn = s where s ≥ n, the number of positive integral solutions is s–1Cn–1. If we need the number of non negative integral solutions, we proceed as follows. Let a1, a2, … be a non–negative integral solution. Than a1 + 1, a2 + 1, … an + 1 is a positive integral solution of the equation x1 + x2 + … + xn = s + n. Therefore, the number of non-negative integral solutions of

2.290 | Engineering Mathematics the given equation is equal to the number of positive integral solutions of x1 + x2 + … + xn = s + n, which is s+n–1Cn–1. For x1 + x2 + x3 + … + xn = s where s ≥ 0, the number of positive integral solutions (when s ≥ n) is s–1Cn–1 and the number of non-negative integral solutions is n+s–1 C n–1

Solved Examples Example 1: Find the number of 3 digit numbers that can be formed using the digits 0, 2, 5, 7, 8 and 9 no digit occurring more than once in a number. Solution: Note that we cannot give the answer 6P3 in this case since such 3 digit numbers include numbers with 0 in the 100th place (and this is not a 3 digit number). We tackle this problem from first principles i.e., the number of 3 digit numbers that can be formed equals the number of ways of filling 3 spaces in a row using the 6 digits. The 100th place in the number can be filled in 5 ways (as 0 cannot be used). We are left with 5 digits (including 0) and therefore, the 10th place can be filled in 5 ways and the unit place can then be filled in four ways. The number of 3 digit numbers that can be formed using the given digits is given by 5 × 5 × 4 = 100. Example 2: How many even numbers of 4 digits can be formed with the digits 2, 4, 5, 6, 7, no digit being used more than once in each number? Solution: The unit place can be filled by using 2, 4 or 6 only, since the number is to be even. That is, there are 3 ways of filling the unit place. We are left with 4 digits and the number of ways of filling the 10th, 100th and 1000th places is equal to 4P3 = 24. By using the fundamental counting principle, the number of 4 digit numbers which are even is equal to 3 × 24 = 72

2. The boys can be arranged in 5P5 or 5! ways and since no two girls are to be together the girls can be seated at the beginning or at the end of the row or at the spaces between two boys. Hence the 3 girls can be placed in 6 positions in 6P3 ways. Therefore, the total number of arrangements where no two girls are together is 5! × 6P3 = 120 × 120 = 14400 3. Proceeding as in (1) the 5 boys together may be considered as one block. The 3 girls and the set consisting of 5 boys can be arranged in 4! ways. The boys among themselves can be arranged in 5! ways. The total number of arrangements is therefore 4! × 5! = 24 × 120 = 2880. Example 4: How many different numbers can be formed with the digits 4, 4, 4, 4, 5, 5, 1, 3, 3, 6, 7, 3 using all the digits each time? Solution: Note that all the digits are not distinct. 4 is repeated 4 times, 5 is repeated twice and 3 is repeated thrice. The total number of numbers that can be formed is 12 ! therefore . 4 ! 2! 3! Example 5: How many different numbers each of 4 digits can be formed with digits 2, 4, 5, 6, 8, 9 if no digit should occur more than thrice in each number? Solution: The number of numbers possessing the given properties = the number of 4 digit numbers with the given digits when each digit can occur zero, one, two, three or four times minus the number of 4 digit numbers in which each digit occurs four times = 64 - 6 = 1290. Example 6: Find the number of combinations of 15 things taken 10 at a time if 10 of the 15 things are alike and the rest unlike.

Example 3: Find the number of ways in which 5 boys and 3 girls may be arranged in a row so that

Solution: 10 alike + 5 distinct from each other. We can have selections (or combinations) in the following ways.

1. 10 alike 2. 9 alike + 1 from the group of 5 distinct things 3. 8 alike + 2 from the group of 5 4. 7 alike + 3 from the group of 5 5. 6 alike + 4 from the group of 5 and 6. 5 alike + all the 5 from the group of 5. The number of selections = 1 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 = 32.

1. The girls may be together 2. No two girls may be together 3. The boys may be together

Solution: Let the boys be B1, B2, B3, B4, B5 and the girls be G1, G2, G3. 1. Since G1, G2, G3 are always to be together, consider them as one block, say G. In all arrangements with the boys, wherever G is, the three girls are together. The 5 boys and the set of girls G can be arranged in 6P6 or 6! ways. In each of these arrangements, the girls can be arranged among themselves in 3P3 or 3! ways. Hence the total number of arrangements in which girls are together is 6! × 3! = 4320.

Example 7: There are 15 points in a plane. By joining them in all possible ways, how many different straight lines can be obtained if

1. No three of the points are collinear. 2. 5 of them are in the same line and no three of the rest are collinear.

Chapter 4 Combinatorics | 2.291 Solution: 1. A straight line can be obtained by joining any two points. Since there are 15 points and no three points are collinear, the number of straight lines that can be formed is equal to the number of combinations of 15 15 × 14 = 105. things taken 2 at a time which is 15C2 = 2 2. In this case, the number of straight lines that can be formed using the 5 collinear points is only 1. The number of straight lines when 5 of the 15 points are collinear is given by 15C2- 5C2 + 1 = 105 - 10 + 1 = 96. Example 8: 1. In how many ways can a pack of 52 cards be divided into four sets of 13 each? 2. In how many ways can 52 cards be divided among four players so that each may have 13? Solution: 1. In this case, pack of 52 cards is to be divided into four equal sets. Therefore, the number of ways in this case 52 ! is (13!) 4 × 4 ! 2. In the first case the cards are already divided into equal sets of 13 each. We have to now distribute the four sets to four players. Therefore, the number of 52 ! 52 ! . ways is × 4! = (13!) 4 × 4 ! (13!) 4 Example 9: There are 18 guests at a dinner party. They are to sit 9 on each side of a long table; three particular persons desire to sit on one side and two others on the other side. In how many ways can the guests be arranged? Solution: Since three persons desire to sit on one side (say side A) and 2 others on the other side (say side B), we have to select 6 persons from the remaining 13 persons for side A. This can be done in 13C6 ways. The 7 persons will automatically go to side B. Now, after the selections, the 9 persons on side A can be arranged among themselves in 9! ways. Similarly, the 9 persons on side B can be arranged among themselves in 9! ways. Total number of arrangements is equal to 13C6 × (9!)2. Example 10: There are 29 intermediate stations b etween Trivandrum and Mangalore. How many different kinds of second class tickets will have to be stocked in the Trivandrum – Mangalore section for journey within the section? Solution: T

29 stations between T and M

M

T = Trivandrum M = Mangalore

The number of different kinds of tickets to be stocked for the onward journey from T to M = 30C2 Similarly, for the journey from M to T, the number of different kind of tickets to be stocked = 30C2 Therefore, total number of tickets to be stocked = 2 2 × 30 × 29 30 × C2 = = 870 2

Example 11: In how many ways can a committee of 3 ladies and 4 gentlemen be appointed from a group of 8 ladies and 7 gentlemen? What will be the number of ways if Mrs. L refuses to serve in a committee if Mr. G is member of the committee? Solution: In the first case, number of ways = 8C3 × 7C4 = 1960. The committee formed by any one of the 1960 ways above will be in one of the forms below. 1. Both G and L are in 2. Both G and L are out 3. G is in, L is out 4. L is in, G is out Since Mrs. L refuses to serve in the committee if Mr. G is there, the number of ways of forming the committee = 1960 – number of ways the committee is formed as (i). = 1960 – 7C2 × 6C3 = 1540. Example 12: 5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. In how many different ways can we place the balls so that no box remains empty? Solution: It is clear that each box must contain at least one ball since no box remains empty. There are 2 cases. Case I: No. of balls:

Box I 1

Box II Box III 1 3

This can be done in 5C1 × 4C1 × 1 or 20 ways. But the box containing 3 balls can be chosen in 3 ways. Required number = 20 × 3 = 60 Case II: No. of balls:

Box I 1

Box II Box III 2 2

This can be done in 5C1 × 4C2 = 30 ways. But the box containing 1 ball can be chosen in 3 ways. Required number = 30 × 3 = 90 Thus the total number of ways = 60 + 90 = 150. Example 13: In how many ways 20 half-rupee coins can be distributed among 4 children? [or find the number of non negative integer solutions of the equation x1 + x2 + x3 + x4 = 20] Solution: We solve the problem by adding three identical one rupee coins and then permuting the 23 coins in all possible ways. 23! The required number of ways = 20 ! 3! 21.22.23 = = 1771 1.2.3 Explanation: In any arrangement, the number of half-rupee coins to the left of the first 1` coin gives the value of x1, the number of half-rupee coins between the first and second 1` coins in the arrangement gives the value of x2 and so on.

2.292 | Engineering Mathematics Example 14: How many diagonals are there in a polygon of 15 sides? Solution: If we select any 2 of the 15 vertices, by joining them we get a side or a diagonal. Two vertices can be chosen from the 15 in 15 C2 = 105 ways. But the number of sides = 15. Therefore the number of diagonals in a polygon = 105 – 15. The number of diagonals in a polygon of n sides = nC2 – n = n( n − 3) . 2 Example 15: There is a question paper consisting of 5 questions – Each question has an internal choice of two questions. In how many ways a student can attempt one or more questions of the given 5 questions in the paper? Solution: For each question, the student has 3 options. Attempt choice 1 or attempt choice 2 or leave the question. Since there are 5 questions, the total number of options = 3 × 3 × 3 × 3 × 3 = 243. Number of ways in which the student can leave all the questions = 1(the student should exercise 3rd option of leaving the question in all the 5 cases and there is only one way in which this can be done). \ The number of ways a student can attempt one or more questions is equal to = 243 - 1 = 242.

Generating Permutations Lexicographic Order Let set A contain n elements {a1, a2, a3, …, an}. The number of permutations formed with these elements is n!. The order we get by arranging these permutations in the increasing order of magnitude is known as Lexicographic order i.e., let a1, a2, …, an – 1, an be the one permutation the next permutation in lexicographic order can be found as follows. Let us consider the last two positions i.e., (n – 1)th position and nth position. If an–1 < an, then we interchange the two positions we get the next permutation. If an–1 > an, then we consider the last three positions i.e., (n – 2)th, (n – 1)th and nth position. If an–2 < an–1 then Case (i): When an–2 < an Case (ii): When an–2 > an, then put an–1, an, and an–2 in the (x – 2)th, (n – 1)th and nth places respectively. Put an, an–2, an–1 in the (n – 2)th, (n – 1)th and nth places respectively. If an–2 > an–1, then we take the last four digits and continue in the similar manner. Example 1: Consider the permutation 123456 Here a5 < a6, we interchange the two positions we get the next permutation. \ The next permutation is 123465.

Example 2: Consider the permutation 423561 Here a5 > a6 So we take last thee places. \ Put the value of a5 and a4 and write the remaining two values in increasing order. \ The next permutation in lexicographic order is 423615.

Recurrence Relations A recurrence relation for the sequence {an} is an equation that express an in terms of one or more of the previous terms of the sequence. Examples: 1. an = 3an–1 2. an = an – 1 + an –2 3. an = 3an – 1 + 4an–2 Example 16: The number of bacteria in a colony is tribled every hour. Write the recurrence relation for the number of bacteria after n hours. Solution: Let initially the number of bacteria in the colony is a0. Then, after one hour the number bacteria is 3a0. i.e., a1 = 3a0. After 2nd hour the number of bacteria is a2 = 3a1 after 3rd hour the number of bacteria is a3 = 3a2 similarly the number of bacteria after the nth hour is an = 3an – 1. Example 17: A person invests some amount in a bank which pays an interest of 10% p.a. compounded annually. Write a recurrence relation for the amount that the person gets after n years. Suppose he invest `500, how much amount does he get after ten years? Solution: Let Pn be the amount in the account after n years. This is equal to the amount in the account after (n – 1) years plus the interest for the nth year. \

⎛ 10 ⎞ Pn = Pn −1 + ⎜ ⎟ Pn −1 ⎝ 100 ⎠

Pn = (1.1) Pn – 1(1)

put n = 1, 2, 3, … in the equation (1), we get

P1 = (1.1)P0, P2 = (1.1)P1, = (1.1). (1.1) P0

P3 = (1.1)P2 = (1.1) (1.1) (1.1)P0 = (1.1)3 P0

\

Pn = (1.1)nP0

Given n = 10, P0 = 500

P10 = (1.1)10 500 = `1296.87

Example 18: Find a recurrence relation and give the initial conditions for the number of strings of length n that do not have two consecutive x’s. Given that the string has only x’s and y’s. Solution: Let an be the number of strings of length n, satisfying the given condition.

Chapter 4 Combinatorics | 2.293 If there are two digits, the length of string is 2. If there are three digits, that the length of the string is 3. Similarly, if it has n digits, then the length of the string is ‘n’. If n = 1, the strings is either x or y. i.e, a1 = 2 If n = 2, the strings are xy, yx, yy, i.e a2 = 3 If n ≥ 3; the number of strings are found as follows. Case 1: The string of length n ending with y that do not have two consecutive x’s are the strings of length (n – 1) with no two consecutive x’s with y at end. So, there are an – 1 such strings. Case 2: If the string of length n is ending with x, then (n –1) place must be y. Other wise they would be end with pair of x’s. The string of length n ending with x that does not have two consecutive x’s are the strings of length (n – 2) with no two consecutive x’s with yx added at the end. There are an – 2 such strings. \ The recurrence relation is

an = an – 1 + an–2 ∀ n ≥ 3

If r1 and r2 are distinct, then the solution of (4) is of the form an = c1 rn1 + c2 rn2 . If r1 = r2 \ the roots are equal then the solution of (4) is of the form an = c1 r1n + c2 nrn 2. Let the characteristic equation be r5 – c1 r 4 + c2 r3 – c3 r2 – c4r – c5 = 0 have three equal roots of one kind, two equal roots of other kind namely, r1 and r2. Then the solution of recurrence relation is of the form. an = (a0 + na1 + n2a2)r1n + (b0 + nb1)r2n

Linear Non Homogeneous Recurrence Relation with Constant Coefficients The recurrence relation of the form

an = c1 an–1 + c2 an – 2 + … + cK an – K + F(n)(6)

is called linear non-homogenous recurrence relation where F(n) is a function that depends on ‘n’. The recurrence relation

\ initial conditions are a1 = 2, a2 = 3

Homogenous Linear Recurrence Relation

is called the associated homogenous recurrence relation of (6).

Let c1, c2, c3 …, ck be constants. The relation an = c1an – 1 + c2 an – 2 + … + ck an – k, ck ≠ 0 is called linear homogenous recurrence relation of degree k. Examples: 1. Pn = (1.1)Pn– 1 is called a recurrence relation of degree 1. 2. an = 3an – 4 is called a recurrence relation of degree 4.

Solution of Linear Homogenous Recurrence Relation with Constant Coefficients Consider the kth degree linear homogenous equation

an = c1 an – 1 + c2 an – 2 + … + ck an – k(2)

Let we assume an = rn be the solution of the above equation. n r = c1 r n – 1 + c2 r n – 2 + …+ ck r n – k

⇒

r k = c1 r k – 1 + c2 r k – 2 + … + ck(3)

(multiplying with rk–n) The above equation is called characteristic equation of (2). The solution of the equation (3) are called the characteristic roots of the recurrence relation. The solution of the (2) depends on the roots of (3).

1. Consider the 2nd degree recurrence relation an = c1 an – 1 + c2 an – 2(4)

The characteristic equation is

r2 – c1 r – c2 = 0,

Let the roots of (5) are r1 and r2.

(5)

an = c1 an – 1 + c2 an – 2 + … + cK an –K

(7)

Examples: 1. an = an–1 + 2n 2. an = 3an – 1 + 4n The solution (6) can be expressed as the sum of the solution of (7) and the solution of F(n). The solution of F(n) in know as particular solution. 1. If F(n) is linear equation in n, then the particular solution is taken in the form an + b. 2. If F(n) is exponential form i.e dn (d is constant) d is not the root of the characteristic equation (7) then the particular solution is taken in the form p.dn. 3. If F(n) is exponential form i.e dn and d is the root of the characteristic equation (7) and it is repeated q times the particular solution is taken in the form nq . dn (where is constant). 4. Let F(n) is in the form (j0 + j1n + j2n2 + … + ji ni) dn. If d is not root of the characteristic equation (7) then particular solution is taken in the form (0 + 1n + 2n2 + … + ini) dn. If d is the root of the characteristic equation (7) and it is repeated q times the particular solution is taken in the form nq (J0 + J1n + J2n2 + … + Ji ni) dn. Example 19: Solve the following recurrences relations 1. an = an – 1 + 2 an – 2 2. an = 4 an–1 – 4an–2 with initial conditions a0 = 3, a1 = 5. 3. an = 4 an –1 + 4n

2.294 | Engineering Mathematics Solution: 1. an = an – 1 + 2 an – 2(8) The characteristic equation of (8) is r2 – r – 2 = 0 The roots are 2 and – 1 The solution of (8) is an = a1 2n + a2 (–1)n 2. an = 4an – 1 – 4an – 2 The characteristic equation of the above equation is r2 – 4r + 4 = 0 the root of the above equation is r = 2, only one root repeated twice. \The solution is of the from an = a1 2n + a2 n 2n given conditions are a0 = 3, a1 = 5 Put n = 0, 3 = a1 Put n = 1, 5 = 3 . 2 + a2 2 ⇒ –1 = 2a2 ⇒ a = –1/2 2 n⎞ ⎛ \ The solution is an = ⎜ 3 − ⎟ 2n 2⎠ ⎝ 3. an = 3an – 1 + 4n.(9) The above equation is non homogenous equation. The associated homogenous equation is an = 3an – 1. \ The solution of the equation is an = a 3n. Since 4n is the first degree polynomial the particular solution is of the form cn + d. substitute these solution in (9) cn + d = 3[c(n – 1) + d)] + 4n Compare n and constant terms c = 3c + 4 ⇒ c = –2 d = –3c + 3d 2d = 3c

⇒ d = –3 \ The particular solution is –2n – 3 \ General solution is an = a 3n – 3 – 2n

Generating Functions

Useful Generating Functions 1. (1 + x)n = C0 + C1x + C2x2 + … + Cr xr + … + Cnxn =

G(x) = b0 + b1x + b2x2 … bkxk + … ∞

bk x k = ∑ k =0 For example, the generating function of the sequence 1, 2,

3, 4, 5, … is given by 16, … is given by

∞

∞

∑ kx k =0

∑ 2k xk k =0

k

and for the sequence 1, 2, 4, 8,

r

r

r =0

2. (1 - x)n = C0 - C1x + C2x2 - + … + (-1)r Crxr + … + (-1)nCn xn =

n

∑ (−1) C x r

r

r

r =0

⎛ n⎞ where Cr stands for ⎜ ⎟ or nCr. ⎝r⎠ -1 2 3. (1 + x) = 1 - x + x - x3 + - … + ∞

∞

= ∑ ( −1) k x k k =0

4. (1 - x)-1 = 1 + x + x2 + x3 + … + ∞ =

5. (1 + x)-2 = 1- 2x + 3x2 - 4x3 + … + ∞

∞

∑x

k

k =0

∞

= ∑ ( −1) k ( k + 1) x k k =0

6. (1 - x)-2 = 1 + 2x + 3x2 + 4x3 + … + ∞

∞

= ∑ ( k + 1) x k k =0

7. (1 - x)-k = 1+ kx +

k ( k + 1) 2 x + 2!

k ( k + 1) ( k + 2) 3 x +…+ ∞ 3!

∞

= ∑ ( k + r −1) r =0

Cr

xr

∞

xk x2 x3 =1+x + + +… 2 ! 3! k =1 k ! 9. ln(1 + x)

8. ex = ∑

x2 x3 x 4 ( −1) k +1 k x =x − + − +… k 2 3 4 k =0 ∞

=∑

Example 20: What is the generating function of 1 + 8x + 24x2 + … n terms? Solution: We know that the generating function of n( n − 1) 2 ( ax ) + ... is (1 + ax)n 1.2 \ Equating coefficients of x and x2 1 + n(ax) +

The generating function for the sequence b1, b2, b3, …, bk … is an infinite series given by

n

∑C x

na = 8 and

n ( n − 1) 2

a 2 = 24

Substituting the value of na in the above equation. n2 a2 – na.a = 48

64 – 8a = 48

⇒ 8a = 16 ⇒

a = 2. \ The Generating function is (1 + 2x)n.

Chapter 4 Combinatorics | 2.295

Exercises Practice Problems 1 Directions for questions 1 to 40: Select the correct alternative from the given choices. 1. In how many ways can the letters of the word ‘DOUBLE’ be arranged such that no two vowels are together? (A) 6! - 4! (B) 6! - 4! 3! (C) 2.3! 4! (D) 3! 4! 2. Among the arrangements that can be made by using all the letters of the word ‘EXAMINATION’, in how many arrangements A’s come together? 11! (A) 11! (B) 2! 10 ! (C) (D) 11! - 10! 2! 2! 2! 3. How many five-digit numbers can be formed using the digits 0 to 8 if no digit is to occur more than once in any number? (A) 1680 (B) 6720 (C) 13440 (D) 1344 4. How many numbers between 60000 and 80000 can be formed using the digits 3 to 7, when any digit can be used any number of times? (A) 250 (B) 625 (C) 1250 (D) 725 5. There are 5 copies of a Maths book, 4 copies of a Physics book and 3 copies of a Chemistry book. The number of ways in which one or more books can be given away is (A) 89 (B) 119 (C) 60 (D) 59 6. An advertisement board is to be designed with six vertical stripes using some or all of the colours: red, green, blue, black and orange. In how many ways can the board be designed such that no two adjacent stripes have the same colour? (A) 1204 (B) 1024 (C) 5120 (D) 2150 7. Let k be an integer such that the sum of the digits of k is 4 and 105 < k < 106. How many values can ‘k’ have? (A) 36 (B) 46 (C) 45 (D) 56 8. If all the letters of the word RATE are taken and permuted and arranged in alphabetical order as in a dictionary, then what is the rank of the word TEAR? (A) 20 (B) 23 (C) 22 (D) 21 9. If all possible four-digit numbers are formed using the digits 3, 5, 6, 9 without repetition and arranged in ascending order of magnitude, then the position of the number 6953 is (A) 20 (B) 16 (C) 18 (D) 15 10. In how many ways can 8 letters be posted into 5 letter boxes? 8 (A) P5 (B) 58 (C) 85 (D) 56

11. In how many ways can 4 prizes each having 1st, 2nd and 3rd positions be given to 3 boys, if each boy is eligible to receive one prize for each event? 12 (A) P3 (B) 64 3 12 (C) 4 (D) C4 x 3! 12. There are 15 points in a plane of which 8 of them are on a straight line. Then how many (i) straight lines can be formed? (A) 105 (B) 21 (C) 78 (D) 288 (ii) triangles can be formed? (A) 399 (B) 400 (C) 234 (D) 72 13. In how many ways can a representation of 12 students consisting of 8 boys and 4 girls be selected from 15 boys and 10 girls, if a particular boy A and a particular girl B are never together in the representation? 14 (A) C8 x 9C4 15 (B) C8 x 10C4 – 14C8 x 9C4 15 (C) C8 x 10C4 – 14C7 x 9C3 14 (D) C7 x 9C3 Common data for question 14: A man has 8 friends whom he wants to invite for dinner. The number of ways in which he can invite 14. at least one of them is (A) 256 (B) 255 (C) 8! (D) 8! - 1 15. Nine villages in a district are divided into three zones with three villages per zone. The telephone department of the district intends to connect the villages with telephone lines such that every two villages in the same zone are connected with four direct lines and every two villages belonging to different zones are connected with two direct lines. How many direct lines are required? (A) 85 (B) 90 (C) 57 (D) 36 16. A question paper consists of 10 sections with each section having two questions. In how many ways can a candidate attempt one or more questions choosing not more than one question per section? (A) 310 (B) 310 –1 (C) 210 – 1 (D) 210 17. In how many ways can the crew of a ten oared boat be arranged, when of the 10 persons available, 2 of whom can row only on the bow side and 3 of whom can row only on the stroke side? 10 ! 10 ! (A) (B) 2 ! 3! 8! 7 ! 5! (5!)3 (C) (D) 3! 2 ! 3! 2 ! 18. A double decked bus can accommodate 110 passengers, 50 in the upper deck and 60 in the lower deck. In how many ways can the passengers be accommodated

2.296 | Engineering Mathematics if 15 refuse to be in the upper deck while 10 others refuse to be in the lower deck? 85! 50 ! 60 ! 85! (A) (B) 40 ! 45! 40 ! 45! 110 ! 110 ! 50 ! 60 ! (C) (D) 50 ! 60 ! 40 ! 45! 19. How many numbers can be formed using all the digits 5, 4, 7, 6, 1, 4, 5, 4, 1 such that the even digits always occupy the even places? 9! 9! (A) (B) 2! 2! 2! 4! 4! 4 ! 5! 4 ! 5! (C) (D) 3! 2 ! 2! 2! 2! 20. Find the number of selections that can be made by taking 4 letters from the word ‘ENTRANCE’. (A) 70 (B) 36 (C) 35 (D) 72 21. In the above word, the number of arrangements by taking 4 letters are (A) 1680 (B) 606 (C) 625 (D) 1250 22. Seven boxes numbered 1 to 7 are arranged in a row. Each is to be filled by either a black or blue coloured balls such that no two adjacent boxes contain blue coloured balls. In how many ways can the boxes be filled with the balls? (A) 23 (B) 42 (C) 34 (D) 32 23. The number of non-negative integer solutions for the equation x + y + z + t = 5 is (A) 65 (B) 56 (C) 73 (D) 84 24. The number of positive integral solutions of the equation x1 + x2 + x3 + x4 + x5 = 10 is (A) 150 (B) 130 (C) 126 (D) 90 25. The number of terms in the expansion (a + b + c + d)20 is ____. (A) 1671 (B) 1717 (C) 1771 (D) 7171 26. What is the generating function of 1, 1 ,1 ,1 ,1 ,1, 1? x6 − 1 x7 −1 (A) (B) x −1 x −1 5 7 (C) x – 1 (D) x + 1 1 1 1 27. The generating function of 1, 1, , , ,… is 2 ! 3! 4 ! (A) ex (B) ln (1 + x) (C) e2x (D) ln (1 – x) 28. If x lies between –1 and 1, the coefficient of x6 in the expansion of (x2 + x + 1)–3 is (A) 6 (B) 9 (C) 3 (D) 4 29. Number of integral solutions of the equation x1 + x2 + x3 = 25 where 5 ≤ xi ≤ 10, i = 1, 2, 3 is (A) 12 (B) 21 (C) 18 (D) 81

30. The number of positive integral solutions of 2x + 3y + 5z = 20, is (A) 1 (B) 2 (C) 3 (D) 4 31. The first three terms of the series defined by an = 7an – 1; a0 = 3 is (A) 3, 21, 147 (B) 5, 6, 127 (C) 3, 28, 153 (D) 3, 21, 153 Linked Answer Question 32: 32. The recurrence relation for the number of bit strings of length n that do not have two consecutive 1’s is (the string contain only 0’s and 1’s). (A) an = 5an – 1 + an – 2 (B) an = an – 1 – 3an – 2 (C) an = an – 1 + an –2 (D) an = 3an – 1 + an – 2 Common data for questions 33 and 34: A factory manufactures two wheelers. In the first month only one vehicle is produced, in the second month three vehicles are produced, in the third month five vehicles are produced and so on. 33. The recurrence relation for the number of vehicles produced in the first n months (an) is (B) an – 1 + 2n– 3 (A) an –1 + 2n –1 (C) 2an – 1 + 1 (D) 2an – 1 – 1 34. The explicit formula for number vertices in first n months is (A) an = 3n2 (B) an = n2 + n 2 (C) an = 3n + n (D) an = n2 Common data for question 35: A person deposits `1500 in a bank that yields an interest of 5% p.a compounded yearly 35. The solution of the recurrence relation for the amount (in `) he would got at the end of n years is (A) (1.05)n –2 1500 (B) (1.05)n –1 1500 n (C) (1.05) 1500 (D) (1.05) 1500 36. The solution of the recurrence relation an = 3an –1 – 2an – 2 with initial conditions ao = 5 and a1 = 8 is (A) an = 2(1)n – 3(2)n (B) an = 2(1)n + 3(2)n n n (C) an = 3 + 3(2) (D) an = 2(1)n – 5(2)n 37. If the roots of the characteristic equation of a linear homogenous recurrence relation are 3, 3, 3 then the general form of the solution is an = (A) ( α 0 + α1 n + α 2 n2 )3n (B) α 0 n + α1, n2 + α 2 n3 .3n (C) a 33n (D) [(a0 + na1)3n]2

(

)

38. If r2 – c1r – c2 = 0 has only one root r0, then the general solution of the recurrence relation an = c1 an – 1 + c2 an – 2 is (A) an = a1 r0 – a2 n (r0) (B) an = a1 (r0) + a2 (r0n)

Chapter 4 Combinatorics | 2.297 (C) an = a1(r0n) + a2 n (r0n)

(D) None of these

39. What form does a particular solution of the linear non homogenous recurrence relation an = 10an – 1 – 25an – 2 + 5n? (A) n + k5n (B) kn 5n (C) kn25n (D) n2 + k.5n

Linked Answer Question 40: 40. Find all solutions of the recurrence relation an = 4an–1 + 3n. 4 (A) an = α 4 n −1 + n + 3 4 n (B) an = α 4 − n − 3 4 n −1 (C) a =α4 − n+ 3 4 n (D) an = α . 4 + n − 3

Previous Years’ Questions 1. The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power (2, x) is same as 2x)[2004] (A) power (2, n) (B) power (2, n2) (C) power (2, (n2 + n)/2) (D) power (2, (n2 - n)/2) 2. Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?[2004] (A) 9 (B) 8 (C) 7 (D) 6 3. In an M × N matrix, all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is[2004] (A) ≤ a + b (B) ≤ max (a, b) (C) ≤ min (M - a, N - b) (D) ≤ min (a, b) 4. The recurrence equation T(1) = 1 T(n) = 2T (n - 1) + n, n ≥ 2 evaluates to (A) 2n+1 - n - 2 (B) 2n - n n+1 (C) 2 - 2n - 2 (D) 2n + n 5. Let G(x) = g(i)?

[2004]

∞ 1 = g (i ) x i , where |x| < 1. What is ∑ (1 − x ) 2 i = 0 [2005]

Common data for questions 6 and 7: Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i, j) then it can move to either (i + 1, j) or (i, j +1)

6. How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0, 0)?[2007] ⎛ 20⎞ (B) 220 (A) ⎜⎝ 10 ⎟⎠ (C) 210 (D) None of the above 7. Suppose that the robot is not allowed to traverse the line segment from (4, 4) to (5, 4). With this constraint, how many distinct paths are there for the robot to reach (10, 10) starting from (0, 0)? [2007] 8. Let P =

∑

1≤ i ≤ 2 k i odd

i and Q =

∑

i , where k is a positive

1≤ i ≤ 2 k i even

integer. Then [2008] (A) P = Q – k (B) P=Q+k (C) P = Q (D) P = Q + 2k Common data for questions 9 and 10: Let xn denote the number of binary strings of length n that contain no consecutive 0’s. 9. Which of the following recurrences does xn satisfy? [2008] (A) xn = 2xn–1 (B) xn = x⌊n/2⌋ + 1 (C) xn = x⌊n/2⌋ + n (D) xn = xn–1 + xn–2 10. The value of x5 is [2008] (A) 5 (B) 7 (C) 8 (D) 16 (A) 29 (B) 219 ⎛ 20⎞ ⎛ 8⎞ ⎛11⎞ ⎛ 8⎞ ⎛11⎞ (C) ⎜⎝ 10 ⎟⎠ − ⎜⎝ 4⎟⎠ × ⎜⎝ 5 ⎟⎠ ⎜⎝ 4⎟⎠ × ⎜⎝ 5 ⎟⎠ (D) (A) i (B) i+1 (C) 2i (D) 2i 11. A pennant is a sequence of numbers, each number being 1 or 2. An n-pennant is a sequence of numbers with sum equal to n. for example (1, 1, 2) is a

2.298 | Engineering Mathematics 4-pennant. The set of all possible 1-pennants is {(1)}, the set of all possible 2-pennants, is {(2), (1, 1)} and the set of all 3-pennants is {(2, 1), (1, 1, 1), (1, 2)}. Note that the pennant (1, 2) is not the same as the pennant (2, 1). The number of 10-pennats is –––––. [2014] 1 2. The number of distinct positive integral factors of 2014 is ––––––––. [2014] 99

13.

1

∑ x( x + 1) = ________.

[2015]

x =1

14. Let an represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for an? [2015] (A) an-2 + an-1 + 2n-2 (B) an-2 + 2an-1 + 2n-2 (C) 2an-2 + an-1 + 2n-2 (D) 2an-2 +2an-1 + 2n-2 15. The number of divisors of 2100 is ______ [2015]

16. The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is _____ [2015] 17. Let an be the number of n - bit strings that do NOT contain two consecutive 1s. which one of the following is the recurrence relation for an? [2016] (A) an = an-1+2an-2 (B) an = an-1+ an-2 (C) an = 2an-1+an-2 (D) an = 2an-1+2an-2 18. The coefficient of x12 in [2016] 3 4 5 6 3 (x + x + x + x +…) is ____ . 19. Consider the recurrence relation a1=8, an = 6n2 + 2n + an-1. Let a99 = K × 104. The value of K is _____. [2016] 20. The value of the expression 1399 (mod 17), in the range 0 to 16, is ______ . [2016]

Answer Keys

Exercises Practice Problems 1 1. D 11. B 21. B 31. A

2. C 12. A 22. C 32. C

3. C 13. C 23. B 33. A

4. C 14. B 24. C 34. D

5. B 15. B 25. C 35. C

6. C 16. B 26. B 36. B

7. D 17. D 27. A 37. A

8. D 18. B 28. C 38. C

9. C 19. C 29. B 39. C

7. D 14. A

8. A 15. 36

9. D 16. 15

Previous Years’ Questions 1. C 2. D 3. D 4. A 5. B 6. A 10. No choice 11. 89 to 89 12. 8 to 8 13. 0.99 17. B or C 18. 10 19. 197.9 : 198.1 20. 4

10. B 20. B 30. D 40. B

Chapter 5 Graph Theory LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • • • •

Graph Finite and inﬁnite graphs Non-isomorphic graphs Subgraph Circuit Unicursal graph Complete graph Cyclic graph

• • • • • • • •

GraPh

Wheel graph Tree Spanning tress Rooted trees Binary tree Cut-sets and cut-vertices Planar graphs Euler’s formula

Self-loop

A graph G = (V, E) consists of V, a non empty set of vertices (or nodes) and E, a set of edges. Each edge is associated with either one or two vertices called end points and an edge is said to connect its end points. Other names for vertex are node, junction, point, o-cell or o-simplex Other names for edge are branch, line, element 1-cell, arc or 1-simplex. Other names for graph are linear complex, 1-complex or one dimensional complex. Example:

An edge having the same vertex as both its end vertices. Ex: e1 (in Figure 1)

Parallel Edges More than one edge associated with a given pair of vertices. Example: e6, e7 (in Figure 1)

Simple Graph A graph that has neither self-loops nor parallel edges.

General Graph e1 v1

A graph in which parallel edges and self-loops are allowed.

v2

e2

Note: e3

e8

s v3

v5

e6 e7

e4

e5

a4

a5 a6

v4 v6 Figure 1

a1

a3 Figure 2

a2

2.300 | Engineering Mathematics In Figure 2, edges a5 and a6 seem to intersect at a point that does not represent a vertex. Such edges should be thought of as being in different planes and thus having no common point.

In Figure 4, edges e5 and e6 are adjacent edges. v3 and v5 are adjacent vertices but v2 and v5 are not. e9 and e3 are not adjacent edges. Degree of v1, d(v1) = 4, d(v2) = 3, d(v3) = 3, d(v4) = 1, d(v5) = 4 and so on.

Finite and Infinite Graphs

Two Important Theorems

A graph with a finite number of vertices and edges. Otherwise, it is an infinite graph. Example: Finite graph v1 e6

e3

e4

Infinite graph

e5

v2

v3

e1

Regular Graph A graph in which all vertices have equal degree is called a regular graph.

e2 v4

1. The sum of the degrees of all vertices in a graph G is twice the number of edges in G. 2. The number of vertices of odd degree in any graph is always even. In Figure 4, the total number of edges = 9 Sum of the degrees of all vertices = 18 \18 = 2 (9) and the number of vertices of odd degree = 4 (even)

Figure 3(a)

Figure 3(b)

Example:

Note: We confine ourselves to the study of finite graphs and unless otherwise stated the term ‘graph’ will always mean a finite graph.

v1

v3

v2

Incidence When a vertex vi is an end point of some edge ek, vi and ek are said to be incident with (on or to) each other.

v5

v4

Adjacent Edges

Figure 5

Two non-parallel edges incident on a common vertex are called adjacent edges.

Adjacent Vertices Two vertices are said to be adjacent if they are the end points of the same edge.

Degree of a Vertex or Valency The number of edges incident on a vertex vi, with self-loops counted twice is called the degree of the vertex vi and is represented as d(vi) Example:

e1

d (v1) = d (v2) = d (v3) = d (v4) = d (v5) = d (v6) = 3

Isolated Vertex A vertex having no incident edge is called an isolated vertex and its degree is zero.

Pendant or End Vertex A vertex of degree one is called a pendant vertex or end vertex.

Series Two adjacent edges are said to be in series if their common vertex is of degree two. Example:

v1 v2

e9

e8

e3

e6

e1 v6

e7

e4 v5 v4 Figure 4

e8

v4

e5 v7

v8

e5 v3

v1

v6

e2 v5

e7

v6

e6

v2

e2 e4 v3 Figure 6

e3

Chapter 5 Graph Theory | 2.301

Subgraph

1. Vertices v7 and v8 are isolated vertices. 2. Vertex v6 is a pendant vertex \ d (v6) = 1 3. e2 and e4 are said to be in series d (v3) = 2

A graph ‘g’ is said to be a subgraph of G if all the vertices and edges of g are in G and each edge of g has the same end vertices in g as in G and is denoted by g ⊂ G.

Null Graph In a graph G(V, E) if E is empty then it is a null graph (or) A graph having only isolated vertices is called a null graph

Example: f

Example: v1 •

9

v2 •

•

•

v3

v4

e

2 6

6

b

5

7 8

a

a

1

d

3

d

2 5 b

4

4 c Figure 9(a)

c Figure 9(b)

3

Isomorphism

Two graphs G and G′ are said to be isomorphic if there is a one-to-one correspondence between their vertices and between their edges such that the incidence relationship is preserved.

The graph in Figure 9(b) is a subgraph of the graph in Figure 9(a).

Example: p

u

6

e1

1

t

q

4 3

5

2

e4

e6

v

Edge-disjoint and Vertex-disjoint Subgraphs

e2

Two (or more) subgraphs g1 and g2 of G are said to be edge disjoint if they do not have edges in common. If they do not have vertices in common they are said to be vertex-disjoint.

e5

s

e3

y

r

x

Figure 7(a)

Notes: 1. Every graph is its own subgraph. 2. A single vertex and a single edge with its end vertices is also a subgraph of G.

w Figure 7(b)

Since vertices p, q, r, s and t corresponds to y, v, u, w and x and the edges 1, 2, 3, 4, 5, 6 corresponds to e3, e1, e2, e4, e5 and e6.

Walk (or) Chain (or) an Edge Train It is defined as a finite alternating sequence of vertices and edges beginning and ending with vertices. No edge appears more than once in a walk. For example: In Figure 10.

Non-isomorphic Graphs

p

v1

Example: x 6

3

5 s

e1

Figure 8(a)

r

q

e6

2

y

e4

q

v2

w

1 4

t

t

s

p

v4

r

e2

v3

u Figure 10

e5

e3

v

u Figure 8(b)

d(v) = 4 in Figure 8(b) is not corresponding to any vertex in Figure 8(a). Notes: If two graphs G and G1 are isomorphic, then 1. They have the same number of vertices. 2. Same number of edges. 3. An equal number of vertices with a given degree.

v1sv2uv2rv3tv1pv4qv3 is a walk Notes: 1. The end vertices of a given walk are called terminal vertices. 2. If the terminal vertices are same, then it is called a closed walk, otherwise it is an open walk.

Path An open walk in which no vertex appears more than once is called a path.

2.302 | Engineering Mathematics Notes: 1. A path does not intersect itself. 2. The number of edges in a path is called the length of a path. 3. A self-loop can be there in a walk but not in a path. 4. The terminal vertices of a path are of degree one and the rest of the vertices (called intermediate vertices) are of degree two.

Circuit A closed walk in which no vertices (except the initial and final) appears more than once is called a circuit. It is a closed non-intersecting walk. Examples:

Theorem A graph G is disconnected iff its vertex V can be partitioned into two non-empty disjoint subsets V1 and V2 such that there exists no edge in G whose end vertex is in subset V1 and the other in subset V2.

Two Important Results Regarding Connectedness Theorem 1: If a graph (connected or disconnected) has exactly two vertices of odd degree, there must be a path joining these two vertices. Theorem 2: A simple graph with n vertices and k compo( n − k )( n − k +1) nents can have at most edges. 2

Euler Graphs and Euler line

Figure 11(a)

Figure 11(b)

Notes: 1. Every vertex in a circuit is of degree two. 2. A circuit is also called as a cycle, elementary cycle, circular path, polygon or loop.

Connected Graphs and Disconnected Graphs A graph G is said to be connected if there is at least one path between every pair of vertices in G, otherwise G is disconnected. i.e. in a connected graph we can reach any vertex from any other vertex by travelling along the edges. Example: a

b

e

c

Connected graph

Figure 12(a)

Notes: 1. Since a walk is always connected, an Euler graph is always connected except for any isolated vertices the graph may have. 2. A given connected graph G is an Euler graph if and only if all the vertices of G are of even degree.

Unicursal Line or an Open Euler Line An open walk that includes all edges of a graph without retracing any edges is called a unicursal line or an open Euler line.

Unicursal Graph

a

d

In a graph G, a closed walk running through every edge of G exactly once is called an Euler line and the graph that consists of an Euler line is called an Euler graph.

f

b

A connected graph that has a unicursal line is called a unicursal graph.

e

c

Theorem

d

Disconnected graph

In a connected graph G, with exactly 2k odd vertices, there exists ‘k’ edge-disjoint subgraphs such that they together contain all edges of G and that each is a unicursal graph.

Figure 12(b)

Note: A null graph of more than one vertex is disconnected.

Component The connected subgraphs of a disconnected graph are components.

Figure 13 A disconnected graph with two components

Operations on Graphs If G1 = (V1, E1) and G2 = (V2, E2) are any two graphs, then 1. Union of G1 and G2 is G3 = G1 ∪ G2 = (V3, E3) where V3 = V1 ∪ V2 and E3 = E1 ∪ E2 2. Intersection of G1 and G2 is G4 = G1 ∩ G2 = (V4, E4) where V4 = V1 ∩ V2 and E4 = E1 ∩ E2 3. The ring sum of two graphs G1 and G2 is G5 = G1 ⊕ G2 = (G1 ∪ G2) - (G1 ∩ G2)

Chapter 5 Graph Theory | 2.303 Example:

(Graph by deleting the edge ej from G i.e G – ej)

v1 v1

a

v2

g

c

v4 f

e G1

g

v4

c

i

d f

e

G2

V5

v6

v1

A simple graph in which there exists an edge between every pair of vertices is called a complete graph. The complete graph with ‘n’ vertices is denoted by kn. Example:

v3

d

v3

Complete Graph

v2

a

b

Figure 15(c)

v4 e

v5 (G1 ∪ G2)

v5

Figure 14(c)

(G1 ∩ G2)

k b

l

i

v4

K2

Figure 16(a)

Figure 16(b)

v2 c

K1

Figure 14(d)

v1

g

V4

Figure 14(b) k

v2

a

V6

v6 V7

v4

Figure 14(a) v1

v3

V3

V2

d

v5

V1 k

e

v5

v3

d

v2

i

b

a

v6

K5

K4

K3

Figure 16(c)

v3

Figure 16(e)

Cyclic Graph

f v5

The cyclic graph denoted by Cn (n ≥ 3) is the graph with n vertices v1, v2, …, vn and edges {v1, v2}, {v2, v3}, …,{vn - 1, vn}, {vn, v1} Example:

(G1 ⊕ G2)

Figure 14(e)

Vertex Deletion and Edge Deletion

a

Deletion of vertex always means the deletion of all edges incident on that vertex. Deletion of the edge ei of G imply the subgraph G - ei. It does not imply the deletion of its end vertices. \ G - ei = G ⊕ ei Example:

a

a

c

b C3

c C4

b

V1

Figure 17(b)

(Graph obtained by deleting vertex V7 from G)

(G)

b

d

Figure 17(a)

a

V3 V6

ej

V2

V4

V5

Figure 15(a)

V4

V5 Figure 15(b)

f

c C5 Figure 17(c)

e C6

V6 V7

b

d f

V3

e

c

V1 V2

Figure 16(d)

Figure 17(d)

Wheel Graph The wheel graph denoted by Wn (n ≥ 3) is the graph obtained from Cn by adding a vertex and edges from this vertex to the original vertices in Cn

2.304 | Engineering Mathematics a a

b

a

a b

e

f c

b

c

d

(a)

c

d

(b)

b g d

e

(c)

c

(d)

Figure 18

deg- (V3) = 2, deg+ (V3) = 3 deg- (V4) = 1, deg+ (V4) = 1 deg- (V5) = 3, deg+ (V5) = 1 Note: The sum of the in degrees and the sum of the out degrees of all vertices in a graph with directed edges are the same and is eqaul to the number of edges in the graph. i.e.,

∑ deg

−

v∈V

Qn (n-cube), n ≥ 1 The graph that has the 2n bit strings of length n as its vertices and edges connecting every pair of bit strings that differ by exactly one bit.

Directed Graph Consists of set of vertices, a set of edges and a function that maps every edge onto some order pair say (a, b).

(V ) = ∑ deg + (V ) = E v∈V

Bipartite Graphs A simple graph G is called bipartite if its vertex set V can be partitioned into two disjoint non-empty sets V and V ′ such that every edge in the graph connects a vertex in V and a vertex in V ′ Example: V1

Example: a

V2

b

a

V3

g

f

V4

c V5 e

V6

d V7

Figure 19

For the edge (a, b) in Figure 21 ‘a’ is called the initial vertex of (a, b) and ‘b’ is called the terminal vertex of (a, b).

V8 Figure 21

Note: The initial and terminal vertex of a loop are the same.

In-degree and Out Degree of a Vertex ‘V’ in a Directed Graph

V = {V1, V3, V5, V7}, V ′ = {V2, V4, V6, V8}

Complete Bipartite Graphs

The in-degree of a vertex V, denoted by deg- (V), is the number of edges with ‘V’ as their terminal vertex. The out degree of V, denoted by deg+ (V) is the number of edges with V as their initial vertex.

The complete bipartite graph, denoted by Km,n is the graph that has its vertex set partitioned into two subsets of m and n vertices respectively, such that two vertices are connected by an edge iff one is in the first subset and the other is in the second subset.

Note: A loop at a vertex contributes ‘1’ to both the in degree and the out degree of the vertex.

Example: K3, 4

Example: V1

V2

V4 Figure 22(a) V3

K2, 4

V5

K2, 2

Figure 20

Solution: deg- (V1) = 2, deg+ (V1) = 3 deg- (V2) = 2, deg+ (V2) = 2

Figure 22(b)

Figure 22(c)

Chapter 5 Graph Theory | 2.305

Some Miscellaneous Points 1. A complete graph with m vertices consists of a total m( m −1) of number of edges. 2 2. The minimum number of edges in a connected graph with m vertices is m - 1

3. The maximum number of edges in a connected simple m( m −1) graph with m vertices is 2 4. The minimum number of edges in a simple graph (not necessarily connected) with m vertices and p components is m - p.

Cut-sets and Cut-vertices

In Figure 33, a spanning tree T (Heavy line) and fundamental cut-sets w. r. t. T (dotted lines) are {e4, e5, e6}, {e1, e3, e4}, {e4, e5, e7} etc.

Edge Connectivity The number of edges in the smallest cut set (i.e cutset with fewest number of edges) is defined as the edge connectivity of G.

Planar Graphs Planar Graph A graph that can be drawn in a plane without a cross over between its edges is called planar. Otherwise it is called non-planar.

Cut-set or Co-cycle

Embedding

It is the set of edges in a connected graph G, whose removal from G leaves G disconnected, provided removal of no proper subset of these edges disconnects G.

A drawing of a geometric representation of a graph on any surface such that no edges intersect is called embedding.

Example: e1

a1

e4

e7 a6

e9 a5

A plane representation of a graph divides the plane into regions (or windows or faces or meshes) as shown in Figure 34.

a2

e6

Region

e2

e8

1

2

4

a3

e3

3 e5 Figure 25 a4

Euler’s Formula

Figure 23

In Figure 32, the set of edges {e1, e7, e8, e3} is a cut-set. Even edge {e5} alone is a cut set.

Fundamental Cut-set (or) Basic Cut-set A cut-set S containing exactly one branch of a spanning tree T is called a fundamental cut-set w.r.t. T. Example: V1 e2

e1

V5

V4

V2

e5 e4

e7

Representing Graphs To represent graphs we use two types of matrices, one is based on the adjacency of vertices, and the other is based on incidence of vertices and edges.

Adjacency Matrix A (AG)

e6

e3

Though the planar graph may have different plane representations, the number of regions resulting from each embedding is the same. The Euler’s formula gives the number of regions in any planar graph. i.e. if ‘G’ is a connected planar graph with n vertices and e edges, the total number of regions in G is (e - n + 2).

Let G = (V, E) be a simple graph with V = {V1, V2, …, Vn} The adjacency matrix A(AG) of G is n × n zero – one matrix with 1 as its (i, j)th entry when Vi, Vj are adjacent and 0 as its (i, j)th entry if Vi, Vj are not adjacent i.e., if A = [aij], then aij = 1, if {Vi, Vj} is an edge of G

V3 Figure 24

0, other wise.

2.306 | Engineering Mathematics Solved Examples Example 1: Use the adjacency matrix to represent the figure given below. q

p

r

Incidence Matrix Let G = (V, E) be a simple undirected graph such that V = {V1, V2, …, Vn} and E = {e1, e2, …, em}. The incidence matrix with respect to this ordering of V and E is the n × m matrix M = [mij], where mij = 1, where edge ej is incident with Vi, 0, other wise Note: Loops are represented with an entry equal to ‘1’

t

s

Example 3: Represent the pseudo graph shown below using an incidence matrix.

Figure 26

Solution: We order the vertices p, q, r, s, t. The matrix representing this graph is

⎡0 ⎢1 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣1

1 0 0 1 1

0 0 0 1 1

1⎤ 1 ⎥⎥ 1⎥ ⎥ 0⎥ 0 ⎥⎦

0 1 1 0 0

e3

e11

e6 V4

V3

e7

Figure 28

Example 2: Use an adjacency matrix to represent the pseudo graph given below.

Solution: The incidence matrix is

⎡e1 V1 ⎢⎢ 1 V2 ⎢ 0 ⎢ V3 ⎢ 0 V4 ⎢⎣ 1

e2 1 0 0 1

e3 0 1 1 0

e4 0 1 1 0

e5 1 1 0 0

e6 0 0 1 1

e7 0 0 1 0

e8 ⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 1 ⎥⎦

Matrix representation for directed graph: For a directed graph G = (V, E) the matrix representation has a ‘1’ in its (i, j)th position if there is an edge from vi to vj where v1, v2, …, vn is an arbitary listing of the vertices of the directed graph i.e., A = [aij] where aij = 1, if (vi, vj) is an edge of G,

b

0, other wise.

Example 4: Represent the directed graph given below using adjacency matrix.

c d

e

e4

e2

e8

Notes: 1. The adjacency matrix of a simple undirected graph is symmetric. 2. A loop at the vertex ai is represented by a ‘1’ at the (i, i)th position of the adjacency matrix. 3. When multiple edges are present the adjacency matrix is no longer a zero – one matrix, since the (i, j)th entry equals the number of edges that are associated to {ai, aj}. 4. The adjacency matrix for a directed graph does not have to be symmetric.

a

V2

e55

V1

a

b

d

c

f

Figure 27

Solution: The adjacency matrix using the ordering of the vertices a, b, c, d, e, f is

⎡1 ⎢3 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣0

3 0 1 0 1 0

0 1 0 1 1 0

0 0 1 0 3 0

0 1 1 3 0 1

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 1⎥ ⎥ 1 ⎥⎦

Figure 29

Solution: The adjacency matrix using the ordering of vertices a, b, c, d is ⎡0 ⎢1 ⎢ ⎢1 ⎢ ⎣0

0 0 0 0

0 1 0 1

1⎤ 0 ⎥⎥ 1⎥ ⎥ 0⎦

Chapter 5 Graph Theory | 2.307

Diameter of a Graph

Example: a

The diameter of a graph is the maximum distance between two distinct vertices.

e

b

Complementary Graph G The complementary graph G of a simple graph G has the same vertices as G. Two vertices are adjacent in G if and only if they are not adjacent in G.

Elementary Sub Division If a graph is planar, so will be any graph obtained by removing an edge {u, v} and adding a new vertex ‘w’ together with edges {u, w] and {w, v}. Such an operation is called an elementary sub division.

f d c

g

Here {(b, d, f ), (a, c, e, g)} is a chromatic partition of the given graph.

Matching A matching in a graph is a subset of edges in which no two edges are adjacent. Actually matching is a problem of assignment of one set of vertices in to another Note: A single edge in a graph is a matching

Chromatic Number

Maximal Matching

If a graph ‘G’ can be coloured using a minimum number of n colours, then n is called the chromatic number of G and is denoted by X(G) and if X(G) = k then we say that the graph G is k – chromatic. Here we shall restrict our selves to simple graphs.

It is a matching to which no edge in the graph can be added without making the edge adjacent to one of the edges of the matching.

Notes: 1. Chromatic number of a graph ‘G’ is always less than or equal to number of vertices. 2. The chromatic number of a complete graph of n vertices is n. 3. If a sub graph of G requires ‘n’ colours then X(G) ≥ n 4. If ‘d ’ is the degree of a vertex v then at most d colours are required to colour the adjacent vertices of v. 5. Every k – chromatic graph has at least k vertices Vi such that deg (Vi) ≥ k – 1 k – Critical graph: A graph G is a said to be k – critical if X(G) = k and X(G – v) < X(G) for each vertex v of G. Notes: If G is a k – critical graph, then (i) G is connected (ii) The degree of each vertex of G is at least k – 1, i.e d(G) ≥ k – 1, d(G) is the minimum degree of any vertex G. 6. For any graph G, X(G) ≤ 1 + D(G), D(G) is the largest degree of all vertices. 7. If d(G) is the minimum degree of any vertex of G, then X(G) ≥ V ( V − δ ( G ) ) where V is the number of vertices of G.

Figure 30

Chromatic Partitioning Given a simple connected graph G, partition of all vertices of G into the smallest possible number of disjoint, independent sets is known as the chromatic partitions of graphs.

In Figure 42, the edges shown with heavy lines is a maximal matching. A graph can have many different maximal matching with different sizes. Among these matching the maximal matching with the largest number of edges is called the largest maximal matching. The number of edges in a largest maximal matching is called the matching number of the graph. Matching is mostly studied in the context of bipartite graph.

Complete Matching This is a bipartite graph having a vertex partition, V1 and V2, in which every vertex in V1 is matched against some vertex in V2. Notes: 1. A complete matching if it exists is a largest maximal matching and the converse of this need not be true. 2. A complete matching of V1 into V2 in a bipartite graph exists if and only if every subset of r vertices in V1 is collectively adjacent to r or more vertices in V2 for all values of r. 3. In a bipartite graph a complete matching of V1 into V2 exists if there is a positive integer m for which degree of every vertex in V1 ≥ m ≥ degree of every vertex in V2. 4. A set of n1 vertices in V1 is collectively incident on, say n2 vertices of V2. The maximum value of the number (n1 – n2) taken over all values of n1 = 1, 2, … and all subsets of V1 is called the deficiency, d(G), of the bipartite graph G. The maximal number of vertices in set V1 that can be matched into V2 is equal to number of vertices in V1 – d(G)

2.308 | Engineering Mathematics

Exercises Practice Problems I

7. Find the diameter of the following graph b

Directions for questions 1 to 49: Select the correct alternative from the given choices. 1. Which of the following statements are true? (i) Any graph which contains some parallel edges is called a multigraph. (ii) An edge of a graph which joins a node to itself is called a self-loop. (iii) If there is no more than one edge between a pair of nodes, then such a graph is called simple-graph. (iv) In a digraph, every edge is directed with an ordered pair. (A) Only (i), (ii) and (iv) are correct (B) Only (iii) and (iv) are correct (C) Only (i) and (iii) are correct (D) All are correct 2. Which among the following are true regarding graphs? (i) All pseudographs are multiple graphs (ii) A graph in which weights are assigned to every edge is called a weighted graph (iii) A null graph contains only isolated nodes (iv) All simple graphs are regular (A) All are correct (B) Only (ii), (iii) and (iv) are correct (C) Only (i), (iii) and (iv) are correct (D) Only (i), (ii) and (iii) are correct 3. A graph is drawn with d(A) = 6, d(B) = 4, d(C) = 3 and d(D) = 2, then the graph is a (A) Pseudo graph (B) Multigraph (C) Regular graph (D) None of these 4. How many edges are there in a graph with ‘v’ vertices each of degree ‘d’? vd (A) vd (B) 2 v+d (C) (D) None of these 2 5. In a simple graph with (p + 1) vertices, the maximum degree of any vertex is (A) p + 1 (B) p (C) p - 1 (D) p - 2 6. Which of the following degree sequences cannot represent an undirected graph? (i) {3, 4, 2, 2} (ii) {3, 1, 2, 2} (iii) {1, 4, 2, 2, 3, 5} (iv) {5, 5, 4, 4} (A) (iv) (B) (i) and (iii) (C) (iii) (D) (ii) and (iv)

g

k

h

l

f

c a

n

j i d

m

e

(A) 10 (C) 6

(B) 8 (D) None of these

8. Consider the following graph, v2

a

v1

b

g

v5

e

v3 d

c

f

h

v4

Which of the following is true? (A) v1 a v2 b v3d v4 is a path (B) v2 b v3 d v4 e v2 is a circuit (C) v1 a v2 b v3 c v3 d v4 e v2 f v5 is a walk (D) All the above. 9.

Which of the following is not true about graphs? (A) Every path is a walk. (B) Every cycle is a path. (C) Every trail is a cycle. (D) Every circuit is a trail.

10. Which among the pair of graphs is isomorphic? (i)

a1

a2

a3

(ii)

a1

b4 b2

b1

b8

b7

a3

a5

b3

b4

a4 a7

a8

a2

b6

b5

a7 a8

a1

b2

b3

a2

a6

(iii)

b1

a4

a3

a4

a5

b8

a6

b1

b2

b3

b7

b4

b5

b6

Chapter 5 Graph Theory | 2.309 (iv)

b6

a2

a6

a3

b2

(A)

(B) Only (i) and (iii) (D) All of these

v4

v6

v1

(I)

v6

v5

v2

11. Which among the following pairs are isomorphic?

v3

v7

(B)

v1 v3

v2

v5

v7

a4

(A) Only (ii) and (iv) (C) Only (i) and (iv)

v3

v2

b3

b1

v4

v1

b4 b7

a7

a5

13. The isomorphic graph of the graph G given below is

b5

a1

v4

(II)

v5

v7

v6

(C) v3

v6 v5 v4

v1 v2 v7

(D)

(III)

v2 v1

v7 v3

v6

v4

v5

(A) I only (C) II and III only

14. Which among the following graph is homeomorphic of K3, 3

(B) I and II only (D) All I, II and III

(A)

c b

a

12. The complement of the following graph G is isomorphic to

d

v1 v2

v3

v4

v5

(B) (A) v5

(B)

v1 v2 v7

v1

v3

(D)

v1

v5 • v2

b g

h f

v3

v2 •

h a

v1 • •v3

d • v4

c

b i

g

v4

h

c

(C) v2

v5

a

e

v4

v6

(C)

d

v5

v4

v3

g

f

e

(D) None of these

f

2.310 | Engineering Mathematics 15. Which among the following is true about the nature of the given graph?

(A) Complement of graph G is connected (B) Graph G is a cycle (C) Graph G is not complete (D) Graph G is a pseudo graph 22. Which of the following is the correct match? List-I

(A) A disconnected graph with two components (B) A disconnected graph with only one component (C) A connected graph (D) A connected graph with two components 16. If the number of edges in a simple graph G having 10 vertices is 37, then the graph is definitely (A) a disconnected graph (B) a connected graph (C) a closed graph (D) All the above 17. The minimum number of edges in a connected graph having 19 vertices is (A) 19 (B) 20 (C) 17 (D) 18 18. The minimum number of edges to be removed to disconnect the following graph is v1

v2 v11

v3

v10

v9 v8

v4

v13 v14

v5

(A) 4 (B) 2 (C) 8 (D) None of these 19. The maximum number of edges in a simple graph with ‘m’ vertices and ‘s’ components is ( m + s)( m + s −1) ( m − s)( m − s +1) (A) (B) 2 2 ( m − s)( m + s −1) (C) 2

(D) None of these

20. Find the minimum number of connected components of a simple graph with 10 vertices and 7 edges. (A) 3 (B) 4 (C) 10 (D) 7 21. Which of the following about the given graph G is true? v1

v2

v3

G

v4

v5

(1) m + n vertices, mn edges

(b) Cn

(2) 2n vertces, n.2n-1 edges

(c) W

(3) n vertices, n edges

(d) Km,n

(4) n vertices,

(e) Qn

n(n − 1) edges 2 (5) n + 1 vertices, 2n edges

(A) a-3, b- 4, c-5, d-1, e-2 (B) a-4, b-5, c-3, d-1, e-2 (C) a-4, b-3, c-5, d-1, e-2 (D) None of these 23. The maximum number of edges possible in a simple bipartite graph with 18 vertices is (A) 49 (B) 64 (C) 81 (D) None of these 24. If the simple graph G has m vertices and n edges. How many edges does G have?

m ( m + 1) m ( m − 1) (A) − n (B) − n 2 2 m ( m − 1) n( n −1) (C) + n (D) − m 2 2

v12

v7 v6

List-II

(a) Kn

25. The complement of the star K1,n has (A) One cut vertex (B) One bridge (C) One isolated vertex (D) Exactly n edges 26. The minimum number of vertices for a simple 3 -regular graph is (A) 3 (B) 4 (C) 5 (D) 6 27. A sales man travels from one city to another, A, B, C and D with an equal distance of 15 kms, and returns to the starting position, the graph indicating his route is (A) A multigraph (B) A pseudo graph (C) A connected graph (D) None of these 28. The statement, which is true about a regular graph, is (A) A regular graph with one vertex is 1-regular graph (B) Every complete graph need not be regular (C) Every regular graph need not be complete (D) All the above

Chapter 5 Graph Theory | 2.311 29. The number of cut vertices in the graph below is v1 v10

v9

v2

v11

v8

v12

v7

v3 v4

v5 v6

(A) 5 (B) 11 (C) 1 (D) 3 30. Which among the following can be represented as plane graphs? a (I)

e

c

d

(II)

List-I ()

(a) There exists a path between every distinct pair of vertices

(2) Connected graph

()

(b) A path that contains every edge of a graph exactly once

(3) Euler path

()

(c) A graph that can be drawn in a plane with no crossings

(4) Planar graph

()

(d) A path that begins and ends at the same vertex

b

a

f

c

e

(III)

⎡1 ⎢2 ⎢ ⎢0 ⎢ ⎣1

2 0 3 0

(A) a

c

d

b

(D) None of these

e

(A) I only (B) I and II only (C) II and III only (D) None of these 31. The number of regions in a graph G with 20 vertices, each of degree 4 is (A) 22 (B) 21 (C) 20 (D) None of these 32. If a polyhedron has 4 vertices and 4 faces the number of edges is (A) 12 (B) 30 (C) 16 (D) 6

d

c

36. Represent the adjacency matrix of the graph a

c

c

a

c

a

d

1⎤ 0 ⎥⎥ 1⎥ ⎥ 0⎦

d (B)

b

(C) a

0 3 1 1

with order a, b, c, d.

d

b

List-II

(1) Circuit

(A) (1)–(d), (2)–(a), (3)–(b), (4)–(c) (B) (1)–(d), (2)–(b), (3)–(a), (4)–(c) (C) (1)–(b), (2)–(c), (3)–(b), (4)–(a) (D) None of these 35. Draw an undirected graph represented by the adjacency matrix

b

f

33. Which of the following statement is/are true for the graphs G and its dual G*? (A) A pendent edge in G* yields an edge forming a self loop in G (B) A dual of graph G is unique (C) Parallel edges in G produce edges in series in G* (D) All the above 34. Which of the following is a correct match?

b

d

b

2.312 | Engineering Mathematics ⎡1 ⎢0 ⎢ (A) ⎢0 ⎢ ⎣1

1 0 1 0

0 0 0 0

⎡1 0 1⎤ ⎢0 1 ⎥ 0⎥ ⎢ (B) ⎢0 0 1⎥ ⎢ ⎥ 1⎦ ⎣0 1

⎡0 ⎢1 ⎢ (C) ⎢1 ⎢ ⎣1

1 0 1 0

0 0 0 1

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎦

0 1 0 1

39. A graph (without self loops) with one or more edges is at least (A) 1 – chromatic (B) bichromatic (C) 3 – chromatic (D) None of these

1⎤ 0 ⎥⎥ 0⎥ ⎥ 1⎦

40. If 4 is the chromatic number of a graph G, then which of the following is false for the number of colours that can be used to properly colour G? (A) 5 (B) 4 (C) 3 (D) 6

(D) None of these

37. Find the adjacency matrix of the directed multi graph with order a, b, c and d. a

b

42. A wheel graph with eleven vertices has a chromatic number of (A) 3 (B) 4 (C) 2 (D) None of these 43. The chromatic number of a wheel graph with six vertices is ____ (A) 2 (B) 3 (C) 4 (D) 5

c

⎡1 ⎢1 ⎢ (A) ⎢1 ⎢ ⎣0

1 0 0 2

2 0 1 1

1⎤ ⎡1 1 ⎥ ⎢2 0 2⎥ ⎢ (B) ⎢1 1 1⎥ ⎥ ⎢ 0⎦ ⎣0 1

⎡1 ⎢2 ⎢ (C) ⎢2 ⎢ ⎣0

0 1 1 0

0 0 1 1

1⎤ 1 ⎥⎥ 0⎥ ⎥ 2⎦

38. The directed multi matrix ⎡0 0 ⎢0 0 ⎢ ⎢1 1 ⎢ ⎣1 1

1 0 0 2

2⎤ 1 ⎥⎥ 1⎥ ⎥ 0⎦

44. A tree with two or more vertices is (A) 3-chromatic (B) 2-chromatic (C) 1-chromatic (D) None of these 45. In the given graph which of the following denotes the chromatic partitions.

(D) None of these

r p

graph with the given adjacency

1⎤ 0 ⎥⎥ 1⎥ ⎥ 0⎦

1 1 0 1

with order a, b, c, d is ______ b (A)

a

(C) a

41. The chromatic number of a complete graph of five vertices is (A) 5 (B) 4 (C) 6 (D) 3

d

d (B)

c b

a

(D) None of these

b

c

u t

q s

v

(A) {(r, s, u), (q, n) (p, t)} (B) {(r, s, n), (q, u), (p, t)} (C) {(p, r, s, u), (q, n), (t)} (D) All the above

46. Which of the following is false? (A) Every pendent edge in a graph is included in every covering of the graph. (B) A covering of 5 vertices graphs will have at least 3 edges. (C) every covering of a graph contain a minimal covering. (D) A spanning tree in a connected graph is not a covering of the graph. 47. For the graph given the maximal independent set is/are r p

d

c

(A) {p, r, s, u} (C) {q, u}

u t

q s

v

(B) {p, r, s, v} (D) All the above

Chapter 5 Graph Theory | 2.313 48. What is the independence number of Kn? (A) 1 (B) 3 (C) 2 (D) None 49. The maximal number of vertices in set V1 that can be matched into set V2 is given by d(G) where d(G) represents the deficiency of a bipartite graph G

(A) Number of vertices in V2 – d(G)

(B) Number of vertices in V2 – number of vertices in V1

(C) Number of vertices in V1 – d(G)

(D) number of vertices in V1 + d(G)

Previous Years’ Questions 1. The minimum number of colors required to color the following graph, such that no two adjacent vertices are assigned the same color, is [2004]

6. Which one of the following graphs is NOT planar? [2005] G2:

G1:

G3:

(A) 2 (B) 3 (C) 4 (D) 5 2. How many graphs on n labeled vertices exist which

(n have atleast

2

− 3n

)

edges. 2 [2004] ( (A)

n ^ 2 − n) / 2

( n ^ 2 −3n) / 2

C( n ^ 2 −3n) / 2 (B) ∑

( n ^ 2 − n)

k =0

n

Ck

( n ^ 2 − n) / 2 (C) ∑ ( ) Ck C (D) n

n^ 2−n / 2

k =0

3. An undirected graph G has n nodes. Its adjacency matrix is given by an n × n square matrix whose (i) diagonal elements are 0’s and (ii) non-diagonal elements are 1’s. Which one of the following is TRUE? [2005] (A) Graph G has no minimum spanning tree (MST) (B) Graph G has a unique MST of cost n - 1 (C) Graph G has a multiple distinct MTSs, each of cost n - 1. (D) Graph G has multiple spanning trees of different costs. 4. Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is: [2005] (A) 6 (B) 8 (C) 9 (D) 13 5. Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is [2005] (A) 12 (B) 8 (C) Less than 8 (D) More than 12

G4:

(A) G1 (B) G2 (C) G3 (D) G4 7. The number of vertices of degree zero in G is: [2006] (A) 1 (B) n (C) n + 1 (D) 2n 8. The maximum degree of vertex in G is: ⎛n⎞ n ⎜ 2 ⎟ 22 (B) 2n - 2 (A) ⎜⎜ ⎟⎟ ⎝2 ⎠

(C) 2n - 3 × 3

[2006]

(D) 2n - 1

9. The number of connected components in G is: [2006] (A) n (B) n+2 n 2n (C) 2 2 (D) n 10. Let G be the non-planar graph with the minimum possible number of edges. Then G has [2007] (A) 9 edges and 5 vertices (B) 9 edges and 6 vertices (C) 10 edges and 5 vertices (D) 10 edges and 6 vertices

11. Which of the following graphs has an Eulerian circuit? [2007] (A) Any k-regular graph where k is an even number. (B) A complete graph on 90 vertices.

2.314 | Engineering Mathematics

(C) The complement of a cycle on 25 vertices. (D) None of the above

12. Which of the following statements is true for every planar graph on n vertices?[2008] (A) The graph is connected (B) The graph is Eulerian (C) The graph has a vertex-cover of size at most 3n/4 (D) The graph has an independent set of size atleast n/3 13. G is a graph on n vertices and 2n – 2 edges. The edges of G can be partitioned into two edge-disjoint spanning trees. Which of the following is NOT true for G? [2008] (A) For every subset of k vertices, the induced subgraph has atmost 2k – 2 edges (B) The minimum cut in G has atleast two edges (C) There are two edge-disjoint paths between every pair of vertices (D) There are two vertex-disjoint paths between every pair of vertices 14. What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? Assume n ≥ 2. [2009] (A) 2 (B) 3 (C) n ≥ 1 (D) n 15. Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices? [2009] (A) No two vertices have the same degree. (B) Atleast two vertices have the same degree. (C) Atleast three vertices have the same degree. (D) All vertices have the same degree. 16. Let G = (V, E) be a graph. Define ξ ( G ) = ∑ id × d ,

18. K4 and Q3 are graphs with the following structures. [2011]

Which one of the following statements is TRUE in relation to these graphs? (A) K4 is planar while Q3 is not (B) Both K4 and Q3 are planar (C) Q3 is planar while K3 is not (D) Neither K4 nor Q3 is planar 19. Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to [2012] (A) 3 (B) 4 (C) 5 (D) 6 20. Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to [2012] (A) 15 (B) 30 (C) 90 (D) 360 21. Which of the following graphs is isomorphic to [2012]

(A)

(B)

(C)

(D)

d

where id is the number of vertices of degree d in G. If S and T are two different trees with ξ (S) = ξ (T), then [2010] (A) |S | = 2|T | (B) |S | = |T | − 1 (C) |S | = |T | (D) |S | = |T | + 1 17. The degree sequence of a simple graph is the sequence of the degrees of the nodes in the graph in decreasing order. Which of the following sequences can not be the degree sequence of any graph?[2010] I. 7, 6, 5, 4, 4, 3, 2, 1 II. 6, 6, 6, 6, 3, 3, 2, 2 III. 7, 6, 6, 4, 4, 3, 2, 2 IV. 8, 7, 7, 6, 4, 2, 1, 1 (A) I and II (B) III and IV (C) IV only (D) II and IV

22. Which of the following statements is/are TRUE for undirected graphs? P: Number of odd degree vertices is even. Q: Sum of degrees of all vertices is even. [2013] (A) P only (B) Q only (C) Both P and Q (D) Neither P nor Q

Chapter 5 Graph Theory | 2.315 23. Consider an undirected random graph of eight vertices. The probability that there is an edge between a 1 pair of vertices is . What is the expected number of 2 unordered cycles of length three? [2013] 1 (A) (B) 1 (C) 7 (D) 8 8 24. The line graph L(G) of a simple graph G is defined as follows: • There is exactly one vertex v(e) in L(G) for each edge e in G. • For any two edges e and e′ in G, L(G) has an edge between v(e) and v(e′), if and only if e and e′ are incident with the same vertex in G. Which of the following statements is/are TRUE? [2013] (P) The line graph of a cycle is a cycle. (Q) The line graph of a clique is a clique. (R) The line graph of a planar graph is planar. (S) The line graph of a tree is a tree. (A) P only (B) P and R only (C) R only (D) P, Q and S only 25. Let G = (V, E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G? [2014] (A) G1 = (V, E1), where E1 = {(u, n) [(u, n) ∉E} (B) G2 = (V, E2), where E2 = {(u, n)| (n, u) ∈ E} (C) G3 = (V, E3), where E3 = {(u, n)| there is a path of length ≤ 2 from u to n in E} (D) G4 = (V4, E), where V4 is the set of vertices in G which are not isolated. 26. Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j): 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a, b) and (c, d) if |a - c| ≤ 1 and |b - d| ≤ 1. The number of edges in this graph is –––––. [2014] 27. An ordered n - tuple (d1, d2, …, dn), with d1 ≥ d2 ≥ …≥ dn is called graphic if there exists a simple undirected graph with n vertices having degrees d1, d2,……dn respectively, which of the following 6-tuples is NOT graphic?[2014] (A) (1, 1, 1, 1, 1, 1) (B) (2, 2, 2, 2, 2, 2) (C) (3, 3, 3, 1, 0, 0) (D) (3, 2, 1, 1, 1, 0) 28. The maximum number of edges in a bipartite graph on 12 vertices is ––––––.[2014] 29. A cycle on n vertices is isomorphic to its complement. The value of n is –––––––.[2014]

30. The number of distinct minimum spanning trees for the weighted graph below is –––––. [2014] 2 2

2

2 1

1

2

1

1

2

1 2

2

31. If G is a forest with n vertices and k connected components, how many edges does G have? [2014] (A) n/k (B) n/k (C) n – k (D) n–k+1 32. Let d denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with d ≥ 3, which one of the following is TRUE? [2014] (A) In any planar embedding, the number of faces is n at least + 2 2 (B) In any planar embedding, the number of faces is n less than + 2 2 (C) There is a planar embedding in which the numn ber of faces is less than + 2 2 (D) There is a planar embedding in which the numn ber of faces is at most δ +1 33. Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three then the number of edges in G is ______. [2015] 34. Let G = (V, E) be a simple undirected graph, and s be a particular vertex in it called the source. For x ∈ V, let d(x) denote the shortest distance in G from s to x. A breadth first search (BFS) is performed starting at s. Let T be the resultant BFS tree. If (u, v) is an edge of G that is not in T, then which one of the following CANNOT be the value of d(u) – d(v)? [2015] (A) -1 (B) 0 (C) 1 (D) 2 35. A graph is self-complementary if it is isomorphic to its complement. For all self-complementary graphs on n vertices, n is [2015] (A) A multiple of 4 (B) Even (C) Odd (D) Congruent to 0 mod 4, or, 1 mod 4 36. In a connected graph, a bridge is an edge whose removal disconnects a graph. Which one of the following statements is true? [2015]

2.316 | Engineering Mathematics

(A) A tree has no bridges (B) A bridge cannot be part of a simple cycle (C) Every edge of a clique with size ≥ 3 is a bridge (A clique is any complete subgraph of a graph) (D) A graph with bridges cannot have a cycle

37. Let G be a connected undirected graph of 100 vertices and 300 edges. The weight of a minimum spanning tree of G is 500. When the weight of each edge of G is increased by five, the weight of a minimum spanning tree becomes______ [2015] 38. Consider the weighted undirected graph with 4 vertices, where the weight of edge {i, j} is given by the entry Wij in the matrix W. 0 2 W = 8 5

2 0 5 8

The largest possible integer value of x, which at least one shortest path between some pair of vertices will contain the edge with weight x is ____. [2016] 39. The minimum number of colours that is sufficient to vertex-colour any planar graph is _____. [2016] 40. Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements: [2016]

8 5 5 8 0 x x 0

I. Each compound in U\S reacts with an odd of compounds. II. At least one compound in U\S reacts with an odd number of compounds. III. Each compound in U\S reacts with an even number of compounds.

Which one of the above statements is ALWAYS TRUE? (A) Only I (B) Only II (C) Only III (D) None

Answer Keys

Exercises Practice Problems 1 1. D 11. D 21. C 31. A 41. A

2. D 12. B 22. C 32. D 42. A

3. D 13. C 23. C 33. D 43. C

4. B 14. C 24. B 34. A 44. B

5. B 15. C 25. C 35. C 45. D

6. B 16. B 26. B 36. A 46. D

7. C 17. D 27. D 37. A 47. D

8. D 18. B 28. C 38. B 48. A

9. C 19. B 29. C 39. B 49. C

10. A 20. A 30. C 40. C

Previous Years’ Questions 1. C 2. D 3. C 11. A 12. C 13. D 21. B 22. C 23. C 29. 5 to 5 30. 6 to 6 31. C 39. 4 40. B

4. B 5. A 6. A 7. C 8. C 9. B 10. B 14. A 15. B 16. C 17. D 18. B 19. D 20. A 24. A 25. B 26. 506 to 506 27. C 28. 36 to 36 32. A 33. 24 34. D 35. D 36. B 37. 995 38. 12

Chapter 6 Linear Algebra LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • • • •

Matrix Type of matrices Determinants Minor and cofactor of a matrix Singular and non-singular matrices Adjoint of a matrix Evaluating inverse of a square matrix Elementary operations (or) elementary transformations

• • • • • • •

Systems of linear equations Cramers method Inverse method Gauss–Jordan method Gauss elimination method The characteristic equation of a matrix Caley–Hemilton theorem

introDuction

Column Matrix

A set of ‘mn’ elements arranged in the form of rectangular array having ‘m’ rows and ‘n’ columns is called a m × n matrix (read as ‘m by n matrix’) and is denoted by A = [aij] where 1 ≤ i ≤ m; 1 ≤ j ≤ n or

A matrix which has only one column.

⎛ a11 ⎜ a A = ⎜ 21 ⎜ : ⎜⎜ ⎝ am1

a12 a22 : am 2

a13 ⋅⋅⋅⋅⋅ a1n ⎞ ⎟ a23 ⋅⋅⋅⋅⋅ a2 n ⎟ : ⋅⋅⋅⋅⋅⋅ : ⎟ ⎟ am 3 ⋅⋅⋅⋅⋅ amn ⎟⎠

The element aij lies in the i-th row and j-th column.

tYPE oF MatricEs Square Matrix A matrix A = [aij]m×n is said to be a square matrix, if m = n (i.e., Number of rows of A = No. of columns of A) The elements a11, a22, a33 …, ann are called ‘DIAGONAL ELEMENTS’. The line containing the diagonal elements is the ‘PRINCIPAL DIAGONAL’. The sum of the diagonal elements of ‘A’ is the ‘TRACE’ of A.

Row Matrix A matrix A = [aij]m×n is said to be row matrix, if m = 1 (i.e., the matrix has only one row) General form is A = [a1, a2 …, an] or [aij]1 × n

⎛ a1 ⎞ ⎜ ⎟ a A =⎜ 2 ⎟ ⎜: ⎟ ⎜⎜ ⎟⎟ ⎝ an ⎠

or

[aij]n×1

Diagonal Matrix A Square Matrix is said to be a Diagonal Matrix if all its elements except those in the principal diagonal are zeros. i.e. if 1. m = n (A is a square Matrix) and 2. aij = 0 if i ≠ j (The non-diagonal elements are zeros) A diagonal matrix of order ‘n’ with diagonal elements d1, d2, …, dn is donated by Diag [d1 d2 … dn].

Scalar Matrix A Diagonal matrix whose diagonal elements are all equal is called a Scalar Matrix. i.e. if 1. m = n 2. aij = 0 if i ≠ j 3. aij = k if i = j for some constant ‘k’.

2.318 | Engineering Mathematics

Unit or Identity Matrix

Equal Matrices

A Scalar Matrix of order ‘n’ in which each diagonal elements is ‘1’ (unity) is called a Unit Matrix or Identity Matrix of order ‘n’ and is denoted by In. i.e.

Two comparable matrices are said be ‘equal’, if the corresponding elements are equal, i.e., A = [aij]m×n and B = [bij]p×q are equal if

1. m=n 2. aij = 0 if i ≠ j 3. aij = 1 if i = j

1. m = p; n = q (i.e. they are of the same order) 2. aij = bij ∀ i, j (i.e. the corresponding elements are equal)

⎛1 0 ⎞ Example: I1 = [1], I2 = ⎜ ⎟ , I3 = ⎝ 0 1⎠

⎛1 0 0 ⎞ ⎜ ⎟ ⎜0 1 0⎟ ⎜ 0 0 1⎟ ⎝ ⎠

Null Matrix or Zero Matrix A matrix is a ‘Null Matrix’ or Zero Matrix if all its elements are zeros.

Transpose of a Matrix The matrix obtained by interchanging the rows and the columns of a given matrix ‘A’ is called the ‘Transpose’ of A and is denoted by AT or A′. If A is an (m × n) matrix, AT will be an (n × m) matrix. Thus if A = [aij]m × n then AT = [uij]n×m where uij = aji.

Upper Triangular Matrix

Properties of T ranspose

A Square Matrix is said to be an upper triangular Matrix, if each element below the Principal Diagonal is zero. i.e.

T - 1 : (A′)′ = A for any matrix A T - 2 : (A + B)′ = A′ + B′ for any two matrices A, B of same order T - 3 : (KA)′ = KA′ for any matrix A T - 4 : (AB)′ = B′A′ for any matrices A, B such that number of columns of A = number of rows of B (REVERSAL LAW) T - 5 : (An)′ = (A′)n for any square matrix A

1. m=n 2. aij = 0 if i > j

⎛1 0 Example: ⎜ ⎜0 ⎝0

4 −1 0 0

3 6 3 0

2 1 2 9

⎞ ⎟ ⎟ ⎠4× 4

Lower Triangular Matrix A Square Matrix is said to be a Lower Triangular Matrix, if each element above the Principal Diagonal is zero, i.e. if 1. m=n 2. aij = 0 if i < j ⎛1 ⎜ −2 Example: ⎜ ⎜0 ⎜5 ⎝

Trace of a Matrix Let ‘A’ be a Square Matrix. The trace of A is defined as the sum of elements of ‘A’ lying in the principal diagonal. Thus if A = [aij]n × n then trace of ‘A’ denoted by trA = a11 + a22 + … + ann

Properties of T race of a Matrix 0 1 7 4

0 0 8 2

0⎞ 0⎟ ⎟ 0⎟ 1 ⎟⎠

Horizontal Matrix If the number of rows of a matrix is less than the number of columns, i.e. m < n, then the matrix is called Horizontal Matrix.

Vertical Matrix If the number of columns in a matrix is less than the number of rows, i.e. if m > n, then the matrix is called a Vertical Matrix.

Comparable Matrices Two matrices A = [aij]m×n and B = [bij]p×q are said to be comparable, if they are of same order, i.e. m = p; n = q.

Let A and B be any two square matrices and K any Scalar then, 1. tr(A + B) = trA + trB 2. tr(KA) = KtrA 3. tr(AB) = tr(BA)

Conjugate of a Matrix A matrix obtained by replacing each element of a matrix ‘A’ by its complex conjugate is called the ‘Conjugate Matrix’ of A and is denoted by A. If A = [aij]m x n, then A = ⎡⎣ aij ⎤⎦ where aij is the conjugate of ‘aij’.

Properties of Conjugate of a Matrix C - 1

(( A )) = A for any matrix ‘A’ (

)

C - 2 A + B =A + B for any matrices A, B of same order. C - 3 (KA) = K A for any matrix ‘A’ and any Scalar K.

Chapter 6 Linear Algebra | 2.319

( ) ( )

C - 4 AB = A ⋅ B for any matrices A and B with the condition that number of columns of A = number of rows of B. C - 5

( A ) = ( A ) for any square matrix ‘A’. n

n

Tranjugate or Transposed Conjugate of a Matrix Tranjugate of a matrix ‘A’ is obtained by transposing the

( )

conjugate of A and is denoted by Aq .Thus Aq = A

T

.

Properties of T ranjugate of a Matrix TC - 1 (Aq)q = A for any matrix A. TC - 2 (A + B)q = Aq + Bq for any matrices A, B of the same order. TC - 3 (KA)q = KAq for any matrix A and any scalar K. TC - 4 (BA)q = BqAq for any matrix A, B with the condition that number of columns of A = number of rows of B. TC - 5 (An)q = (Aq)n for any square matrix ‘A’.

Symmetric Matrix A Matrix A is said to be Symmetric, if AT = A (i.e. Transpose of A = A). Note: A Symmetric matrix must be a square matrix.

Skew-symmetric Matrix A matrix ‘A’ is said to be Skew-Symmetric Matrix, if AT = (-A) i.e. A = [aij]m x n is Skew Symmetric if 1. m=n 2. ajI = -aij ∀ i, j Note: In a Skew-Symmetric Matrix, all the elements of the principal diagonal are zero.

Orthogonal Matrix A Square Matrix ‘A’ of order n × n is said to be an Orthogonal Matrix, if AAT = ATA = In.

Involutory Matrix A square matrix A is said to be Involutory Matrix, if A² = I (where I is Identity Matrix).

Idempotent Matrix A square matrix ‘A’ is said to be an Idempotent Matrix, if A² = A.

Nilpotent Matrix A Square Matrix ‘A’ is said to be Nilpotent Matrix, if there exists a natural number ‘n’ such that An = O. If ‘n’ is the least natural number such that An = O, then ‘n’ is called the index of the nilpotent matrix ‘A’. (Where O is the null matrix)

Unitary Matrix A Square Matrix ‘A’ is said to be a Unitary Matrix if, AAq = AqA = I. (Where Aq is the transposed Conjugate of A)

Hermitian Matrix A Matrix ‘A’ is said to be a Hermitian Matrix, if Aq = A, i.e., A = [aij]m×n is Hermitian if 1. m=n 2. aij = a ij ∀ i, j Note: The diagonal elements in a Hermitian Matrix are real numbers.

Skew-Hermitian Matrix A Matrix ‘A’ is said to be a Skew-Hermitian Matrix, if Aq = -A.

Operations on Matrices Scalar multiplication of matrices: If A is a matrix of order m × n and ‘K’ be any scalar (a real or complex number), then KA is defined to be a m × n matrix whose elements are obtained by multiplying each element of ‘A’ by K, i.e. if A = [aij]m×n then KA = [Kaij]m×n in particular if K = -1; then KA = -A is called the negative of A and is such that A + (-A) = [aij] + [-aij] = [aij - aij] = [0] = O (Zero Matrix) (-A) + A = [-aij] + [aij] = [-aij + aij] = [0] = O i.e., A + (-A) = (-A) + A = O

Properties of Scalar Multiplication Let A, B be two matrices of same order and a, b be any scalars, then S - 1: a(A + B) = aA + aB S - 2: (a + b)A = aA + bA S - 3: a(bA) = (ab)A S - 4: 1. A = A

Addition of Matrices If A and B are two matrices of the same order, then they are ‘conformable’ for addition and their sum ‘A + B’ is obtained by adding the corresponding elements of A and B i.e., if A = [aij]m×n; B = [bij]m×n, then A + B = [aij + bij]m×n

Properties of Addition Let A, B and C be three matrices of same order say m × n, then A - 1: A + B is also a m × n matrix (CLOSURE) A - 2: (A + B) + C = A + (B + C) (ASSOCIATIVITY) A - 3: If ‘O’ is the m × n Zero (Null) matrix, then A + O = O + A = A (‘O’ is the ADDITIVE IDENTITY) A - 4: A + (-A) = (-A) + A = O (-A is the ADDITIVE INVERSE) A - 5: A + B = B + A (COMMUTATIVITY)

2.320 | Engineering Mathematics Note: The set of matrices of same order form an ‘Abelian Group’ under Addition. Multiplication of matrices: Let A and B be two m atrices. A and B are conformable for multiplication, only if the number of columns of A is equal to the number of rows of B. Let A = [aij] be an m × n matrix, B = [bjk] be an n × p matrix. Then the product ‘AB’ is defined as the matrix C = [cik] of order mxp where cik = ai1b1k + ai2b2k + … + ainbnk =

n

∑a b j =1

ij

jk

cij calculated for i = 1, 2, …, m and k = 1, 2, …, m will give all the elements of the matrix C.

value is obtained by taking the sum of the products of the elements of any row (or column) by their corresponding cofactors. Thus for A, D = a1A1 + b1B1 + c1C1

b c a = a1 2 2 − b1 2 b3 c3 a3

D = a1A1 + a2A2 + a3A3

b = a1 2 b3

c2 b − a2 1 c3 b3

c2 a + c1 2 c3 a3

c1 b + a3 1 c3 b2

b2 or also b3

c1 c2

Properties of Multiplication M - 1: If A, B, C be m × n, n × p, p × q matrices respectively, then (AB)C = A(BC) (ASSOCIATIVITY) M - 2: If A is a m × n matrix, then AIn = A and Im A = A and if A is a square matrix i.e., m = n, then AI = IA = A (I is the MULTIPLICATIVE IDENTITY) M - 3: If A, B, C be m × n, n × p, p × q matrices respectively, then A(B + C) = AB + AC (DISTRIBUTIVE LAW) M - 4: Matrix multiplication is NOT COMMUTATIVE in general. M - 5: The INVERSE of a given matrix may not always exist.

Determinants Let A = [aij] be a Square Matrix of order ‘n’. Then the determinant of order ‘n’ associated with ‘A’ is denoted by |A| or |aij| or Det(A) or D. Notes: 1. Determinant of a matrix exists, only if it is a square matrix. 2. The value of a determinant is a single number. Then the minor of an element aij of ‘A’ is the determinant of the 2 × 2 matrix obtained after deleting the ith row and jth column of A and is denoted by Mij. The cofactor of aij small size is denoted by Aij and is defined as (–1)i + j Mij i.e., Aij = (-1)i+j Mij

Determinant of Order 3 (Third Order Determinant) If A is ⎛ a1 b1 ⎜ ⎜ a2 b2 ⎜a b 3 ⎝ 3

a Square Matrix of order ‘3’ given by A = c1 ⎞ ⎟ c2 ⎟ . Then the determinant of ‘A’ given by D = c3 ⎟⎠

a1 Det A = a2 a3

b1 b2 b3

c1 c2 c3

Properties of Determinant 1. If two rows (or columns) of a determinant are interchanged, the value of the determinant is multiplied by (–1). 2. If the rows and columns of a determinant are interchanged, the value of the determinant remains unchanged, i.e. Det(A) = Det(AT). 3. If all the elements of a row (or column) of a determinant are multiplied by a scalar (say ‘K’), the value of the new determinant is equal to ‘K’ times the value of the original determinant. 4. If two rows (or columns) of a determinant are identical, then the value of the determinant is zero. 5. If the elements of a row (or a column) in a determinant are proportional to the elements of any other row (or column), then the determinant is ‘0’. 6. If every element of any row (or column) is zero, then determinant is ‘0’. 7. If each element in a row (or column) of a determinant is the sum of two terms, then its determinant can be expressed as the sum of two determinants of the same order. 8. (The Theorem of ‘false cofactor’) The sum of products of elements of a row (or column) with the cofactors of any other row (or column) is zero. ⎛ a1 b1 c1 ⎞ ⎜ ⎟ Thus in A = ⎜ a2 b2 c2 ⎟ ⎜a b c ⎟ 3 3 ⎠ ⎝ 3 a1A2 + b1B2 + c1C2 = 0 a2A1 + b2B1 + c2C1 = 0 and so on in general arAs + brBs + crCs = 0 if r ≠ s 9. If the elements of a determinant are polynomials in x and the determinant vanishes for x = a, then x – a is a factor of the determinant.

Singular and Non-singular Matrices is a determinant of order 3 and the

A square matrix ‘A’ is said to be singular, if Det(A) = 0 and is non-singular, if Det(A) ≠ 0.

Chapter 6 Linear Algebra | 2.321 Notes: 1. A unit matrix is non-singular (since its Det = 1) 2. If A and B are non-singular matrices of the same ‘type’, then AB is non-singular of the same ‘type’.

Inverse of a Matrix Let ‘A’ be a square matrix. A matrix ‘B’ is said to be an inverse of ‘A’, if AB = BA = I Note: If B is the inverse of ‘A’, then ‘A’ is the inverse of ‘B’.

Some Results of Inverse 1. Inverse of a square matrix, when it exists, is unique. 2. The inverse of a square matrix exists, if and only if it is non-singular. 3. If ‘A’ and ‘B’ are square matrices of the same order, then ‘AB’ is invertible (i.e., inverse of AB exists) if ‘A’ and ‘B’ are both invertible. 4. If ‘A’ and ‘B’ are invertible matrices of the same order, then (AB)-1 = B-1 A-1 5. If A is invertible, then so is AT and (AT)- 1 = (A-1)T. 6. If A is invertible, then so is Aq and (Aq)-1 = (A-1)q.

Adjoint of a Matrix The adjoint of a square matrix ‘A’ is the Transpose of the matrix obtained by replacing the elements of ‘A’ by their corresponding cofactors. Note: The ces and the ⎛ a1 ⎜ b If A = ⎜ 1 ⎜ ..... ⎜⎜ ⎝ 1

adjoint is defined only for square matriadjoint of a matrix ‘A’ is denoted by Adj(A). a2 .... an ⎞ ⎟ b2 .... bn ⎟ ........ ...... ⎟ ⎟ 2 n ⎟⎠ T

⎛ A1 A2 ....... An ⎞ ⎜ ⎟ ⎛ A1 B1 ........L1 ⎞ ⎜ B1 B2 ........ Bn ⎟ ⎜ A B .......L ⎟ 2 2⎟ Adj A = ⎜ ...................... ⎟ = ⎜ 2 ⎜ ⎟ ⎜ ...................... ⎟ ⎟⎟ ⎜ ....................... ⎟ ⎜⎜ ⎜ L L .......L ⎟ ⎝ An Bn ........Ln ⎠ 2 n ⎝ 1 ⎠ Results: 1. If ‘A’ is of order 3 × 3 and K is any number, then Adj(KA) = K²(Adj A) 2. A(Adj A) = (Adj A)A = |A| I for any square matrix ‘A’ 3. Adj I = I; Adj O = O where I is the identity matrix and O is the null matrix. 4. Adj(AB) = (Adj B) (Adj A) if A, B are non-singular and are of same type. 5. If A = An×n, then Det(Adj A) = (Det A)n–1. Adj(Adj A) = (Det A)n–2(A). 2 |Adj (Adj A)| = (det A)( n−1)

Evaluating Inverse of a Square Matrix If A is a square matrix, then A-1 =

1 (Adj A) A

Notes: 1. The inverse of an identity matrix is itself. 1 2. (Adj A)-1 = A A 3. If A is a non-singular square matrix (say of order 3) and K is any non-zero number, then 1 -1 A (KA)–1 = K

Rank and Nullity of a Matrix 1. Rank of a matrix: The Matrix ‘A’ is said to be of rank ‘r’, if and only if it has at least one non-singular square submatrix of order ‘r’ and all square sub-matrices of order (r + 1) and higher orders are singular. The rank of a matrix A is denoted by rank (A) or r(A). 2. Nullity of a matrix: If A is a square matrix of order ‘n’, then n - r(A) i.e., n - rank (A) is defined as nullity of matrix ‘A’ and is denoted by N(A). Remark 1: If there is a non-singular square submatrix of order ‘K’, then r(A) ≥ K. Remark 2: If there is no non-singular square submatrix of order ‘K’, then r(A) < K. Remark 3: If A ′ is the transpose of A, then r(A) = r(A ′) Remark 4: The rank of a null matrix is ‘0’. Remark 5: The rank of a non-singular square matrix of order ‘n’ is ‘n’ and its Nullity is ‘0’. Remark 6: Elementary operations do not change the rank of a matrix. Remark 7: If the product of two matrices A and B is defined, then r(AB) ≤ r(A) and r(AB) ≤ r(B). i.e., the rank of product of two matrices cannot exceed the rank of either of them.

Elementary Operations (or) Elementary Transformations 1. Elementary Row Operations (i) Ri ↔ Rj: Interchanging of ith and jth rows (ii) Ri → kRi: Multiplication of every element of ith row with a non zero scalar k (iii) Ri → Ti + kRj: Addition of k times the elements of jth row to the corresponding elements of ith row 2. Elementary column operations (i) Ci ↔ Cj: Interchanging of ith and jth columns (ii) Ci → kCi: Multiplication of every element of ith column with a non zero scalar k (iii) Ci → Ci + kCj: Addition of k times the elements of jth column to the corresponding elemnt of ith column

2.322 | Engineering Mathematics ⎡ 2 3 −4 1 ⎤ Consider the matrix A = ⎢⎢ 3 0 1 5 ⎥⎥ R2 → 2R2 ⎢⎣ 4 7 1 2 ⎥⎦ ⎡ 2 3 −4 1 ⎤ ~ ⎢6 0 2 10 ⎥ ⎢ ⎥ ⎢⎣ 4 7 1 2 ⎥⎦ C2 ↔ C3 ⎡ 2 −4 3 1 ⎤ ⎢ ~ ⎢ 3 1 0 5 ⎥⎥ ⎢⎣ 4 1 7 2 ⎥⎦ C1 → C1 – 2C4 ⎡ 0 −4 3 1 ⎤ ⎢ ~ ⎢ −7 1 0 5 ⎥⎥ ⎢⎣ 0 1 7 2 ⎥⎦ Note: The rank of a matrix is invariant under elementary operations

Row and Column Equivalence Matrices Row Equivalence Matrix If B is a matrix obtained by applying a finite number of elementary row operations successively on matrix A, then matrix B is said to be row equivalent to A. (or a row equivalent matrix of A)

Column Equivalence Matrix If B is obtained by applying a finite number of elementary column operations successively on matrix A, then matrix B is said to be column equivalent to A. (or a column equivalent matrix of A)

Row Reduced Matrix A matrix A of order m × n is said to be row reduced if 1. The first non zero element of a non zero row is 1. 2. Every other element in the column in which such 1’s occur is 0. ⎛1 0 2⎞ ⎜ ⎟ A = ⎜ 0 1 3 ⎟ is a row reduced matrix. ⎜0 0 0⎟ ⎝ ⎠ ⎛1 0 4⎞ ⎜ ⎟ B = ⎜ 0 5 0 ⎟ is not a row reduced matrix. ⎜0 0 0⎟ ⎝ ⎠

Row Reduced Echelon Matrix A matrix ‘X’ is said to be row reduced echelon matrix if 1. X is row reduced. 2. There exists integer P (0 ≤ p ≤ m) such that first ‘p’ rows of X are non-zero and all the remaining rows are zero rows. 3. For the ith non-zero row, if the first non-zero element of the row (i.e., 1) occurs in the jth column then, j1 < j2 < j3 < … < jp. ⎛1 ⎜ 0 Example: P = ⎜ ⎜0 ⎜ ⎝0

⎡1 3 2 ⎤ Example: B = ⎢⎢3 4 −4 ⎥⎥ C2 – 3C1, ⎢⎣1 1 6 ⎥⎦ 1⎤ ⎡1 0 C3 ⎢ → ⎢3 −5 −2 ⎥⎥ = C (say ) 2 ⎢⎣1 − 2 6 ⎥⎦ C is a column equivalent to B.

0 0 1 0

2⎞ ⎟ 3⎟ ;Q= 4⎟ ⎟ 0⎠

⎛0 1 2 0⎞ ⎜ ⎟ ⎜0 0 0 1⎟ ⎜0 0 0 0⎟ ⎝ ⎠

echelon matrices. The number of non-zero rows (i.e. value of P) are 3 and 2 respectively. The value of i and j are tabulated below P:

⎡1 3 4 ⎤ ⎢ Example: A = ⎢ 2 5 −2 ⎥⎥ ⎢⎣1 4 −3⎥⎦ 4 ⎤ ⎡1 3 R 2 − 2R1 ⎢⎢0 −1 −10 ⎥⎥ = B (say ) ⎢⎣0 1 −7 ⎥⎦ B is a row equivalent matrix of A.

0 1 0 0

Q:

i

1

2

3

j

1

2

3

i

1

2

j

1

4

Normal Form of a Matrix By means of Elementary transformations, every matrix ‘A’ of order m × n and rank r(> 0) can be reduced to one of the following forms. ⎛ Ir 0 ⎞ 1. ⎜ ⎟ ⎝ 0 0⎠ 2. [Ir/0] 3. [Ir] and these are called the normal forms. Ir is the unit matrix of order ‘r’. Note: If a m × n matrix ‘A’ has been reduced to the normal ⎛ I 0⎞ form say ⎜ r ⎟ then ‘r’ is the rank of A. ⎝ 0 0⎠

Chapter 6 Linear Algebra | 2.323

Systems of Linear Equations Let a11x1 + a12x2 + … + a1nxn = b1 a12x1 + a22x2 + … + a2nxn = b2 ............................................... .............................................. an1x1 + an2x2 + … + annxn = bn

(1)

be a system of ‘n’ linear equations in ‘n’ variables x1, x2, …, xn. The above system of equations can be written as ⎛ a11 a12 ........ a1n ⎞ ⎛ x1 ⎞ ⎛ b1 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ a21 a22 .........a2 n ⎟ ⎜ x2 ⎟ ⎜ b2 ⎟ ⎜ .......................... ⎟ ⎜ ... ⎟ = ⎜ ... ⎟ or AX = B ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ......................... ⎟ ⎜ ... ⎟ ⎜ ... ⎟ ⎜a ⎟⎜ ⎟ ⎜ ⎟ ⎝ n1 a2 n ..........an n ⎠ ⎝ xn ⎠ ⎝ bn ⎠ where ⎛ a11 a12 ......... a1n ⎞ ⎛ x1 ⎞ ⎛ b1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ a21 a22 ......... a2 n ⎟ ⎜ x2 ⎟ ⎜ b2 ⎟ ⎜ ⎟ ⎜ ⎟ A = .......................... , X = ... , B = ⎜ ... ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ......................... ⎟ ⎜ ... ⎟ ⎜ ... ⎟ ⎜a ⎟ ⎜x ⎟ ⎜b ⎟ a ....... a mn ⎠ ⎝ n⎠ ⎝ m⎠ ⎝ m1 m 2

Any set of values of x1, x2, x3 … which simultaneously satisfy these equations is called a solution of the system. When the system of equations has one or more solutions, the equations are said to be CONSISTENT and the system of equations are said to be INCONSISTENT if it does not admit any solution. The system of equation (1) is said to be

1. HOMOGENEOUS, if B = 0

2. NON-HOMOGENEOUS, if B ≠ 0

⎛ a11 a12 ..........a1n b1 ⎞ ⎜ ⎟ ⎜ a21 a22 .........a2 n b2 ⎟ The matrix ⎜ .................................. ⎟ is called the aug⎜ ⎟ ⎜ ................................. ⎟ ⎜a ⎟ ⎝ m1 am 2 .........amn bm ⎠ mented matrix of the system of equations and is denoted by [A : B]. Let AX = B represent ‘m’ linear equations with ‘n’ variables. Let Rank of A = r and Rank (A, B) = r1 [where (A, B) is an augmented matrix] If r1 ≠ r, then the system of equations are inconsistent.

A is called the coefficient matrix.

If r1 = r, the table follows: m=n

m>n

m 0, clearly f(c) = 0.

If lim f(x) = f(a). Then f is continuous, otherwise it is

Note: 1. If f(x) is continuous in [a, b], then f takes all values between m and M at least once as x moves from a to b, where M = Supremum of f on [a, b] and m = infimum of f on [a, b].

x→a

x→a

not continuous at x = a. Problems on continuous functions can be grouped into the following categories.

Chapter 7 Calculus | 2.335

2. If f(x) is continuous in [a, b], then | f | is also continuous on [a, b], where | f | (x) = | f(x)| x ∈ [a, b]. 3. Converse may not be true ⎧1; 0 < x ≤ 3 For instance, f(x) = ⎨ ⎩1; 3 < x ≤ 5

is not continuous at x = 3, but | f |(x) = 1 x ∈[0, 5], being a constant function is continuous [0, 5].

Inverse-function Theorem If f is a continuous one-to-one function on [a, b] then f –1 is also continuous on [a, b].

Uniform Continuity A function f defined on an interval I is said to be uniformly continuous on I if given ∈ > 0 there exists a d > 0 such that if x, y are in I and |x – y | < d then | f(x) – f(y) | < ∈. Note: Continuity on [a, b] implies uniform continuity whereas continuity on (a, b) does not mean uniform continuity.

Types of Discontinuity If f is a function defined on an interval I, it is said to have (TD1) a removable discontinuity at p ∈ I, if lim f(x) x→ p exists, but is not equal to f(p). (TD2) a discontinuity of first kind from the left at p if lim− f(x) exists but is not equal to f(p). x→ p

(TD3) a discontinuity of first kind from the right at p if lim+ f(x) exists but is not equal to f(p). x→ p

(TD4) a discontinuity of first kind at p If lim− f(x) and x→ p lim+ f(x) exists but they are unequal. x→ p

(TD5) a discontinuity of second kind from the left at p if lim− f(x) does not exist as x approaches p from left. x→ p

(TD6) a discontinuity of second kind from the right at p if lim+ f(x) does not exist as x approaches p from right. x→ p

(TD7) a discontinuity of second kind at p if neither lim− x→ p f(x) nor lim+ f(x) exist. x→ p

Examples for each type is presented in the following table: Type

Example

TD1

f(x) =

TD2

Point of discontinuity

x2 −1 ,x≠1 x −1 f(1) = 3

x=1

f(x) = x + 3 for 0 < x < 1 f(x) = 5 for x ≥ 1

x=1

(Continued)

Point of discontinuity

Type

Example

TD3

f(x) = x + 3, for x > 2 f(x) = 8 for x ≤ 2

x=2

TD4

x + 3; x > 2 f (x ) = 7; x =2 x − 3; x < 2

x=2

TD5

f(x) = tan x for x < p/2 f(x) = 1, for x ≥ p/2

x=

TD6

f(x) = 1, for x ≤ p/2 f(x) = tan x for x > p/2

x = p/2

TD7

f(x) = 1/x at x ≠ 0 f(0) = 3 at x = 0

x=0

π 2

Note: 1. Every differentiable function is continuous, but the converse is not true. The example of a function which is continuous but not differentiable at a point: f(x) = |x - 3| for x ∈ R is continuous at x = 3, but it is not differentiable at x = 3. 2. The function may have a derivative at a point, but the derivative may not be continuous. For example the function 1 ⎧ 3 ⎪ x sin ; x ≠ 0 has the derivative function as f(x) = ⎨ x ⎪⎩ 0; x =0 1 1 ⎧ 2 ⎪3 x sin − x cos ; x ≠ 0 f ′(x) = ⎨ x x ⎪⎩ x =0 0; However lim f ’( x ) doesn’t exist. x →0

Solved Examples Example 1: Discuss the continuity of the function at x = 1 where f(x) is defined by 3x − 2 for 0 < x ≤ 1 f (x) = x sin( x − 1) = for x > 1 ( x − 1) Solution: Consider the left and right handed limits 3x − 2 = 1 Lt f (x) = Lt x→1 x→1− x sin( x − 1) Lt f (x) = Lt x→1 x→1+ x −1 sin( x − 1) = Lt = 1 ( x −1) → 0 ( x − 1) and \

3(1) − 2 =1 1 Lt− f (x ) = Lt+ f (x) = f (1) f (1) =

x→1

x→1

\ f is continuous at x = 1.

2.336 | Engineering Mathematics

(2

Example 2: If f (x) =

x

)

−1

2

find the values of a and b.

for x ≠ 0 and f(x) = 2 sin 2 x log(1 + x ) log 2 for x = 0, discuss the continuity at x = 0. Solution: Lt f(x) = Lt x→0

x→0

( 2 x − 1) 2 sin 2 x log(1 + x ) 2

⎛ 2x − 1 ⎞ ⎜ ⎟ ⎝ x ⎠ = Lt x→0 sin 2 x log(1 + x ) ( 2) 2x x 2 x ⎛ 2 −1 ⎞ ⎜ ⎟ ⎝ x ⎠ = Lt x→0 1 ⎛ sin 2 x ⎞ log(1 + x ) x 2⎜ ⎟ ⎝ 2x ⎠

2

⎛ 2 −1 ⎞ Lt ⎜ ⎟ ⎝ x ⎠ 1 = 1 sin 2 x 2 Lt log Lt (1 + x ) x 2 x →0 0 x → 2x 1 = (log2)2. 2 x

x →0

But given f (x) = 2 log 2 at x = 0 \

Lt f(x) ≠ f (0)

x→0

x−4 +a x →4 x →4 x − 4 x−4+a = – 1 + a = Lt + x→ 4 − ( x − 4) x−4 Lt f (x) = Lt + +b x→ 4+ x→ 4 x−4 x−4 = Lt + + b = 1 + b x→ 4 ( x − 4) Since given f (x) is continuous at x = 4 Lt − f (x) = f (4) = Lt − f (x) Solution: Lt − f ( x ) = Lt −

x→ 4

⇒

x→ 4

–1 + a = a + b = 1 + b

⇒

a = 1, b = – 1

Example 5: Examine the continuity of the given function at origin where ⎧ xe1/ x ;x≠0 ⎪ f ( x ) = ⎨ 1 + e1/ x ⎪ 0 ;x = 0 ⎩ 1/ x xe Solution: lim f ( x ) = lim =0 1/ x x → 0− x → 0− 1 + e x lim f ( x ) = lim −1/ x =0 + + e +1 x →0 x →0 Then

lim f(x) = lim+ f(x) = lim f(x) = 0

x → 0−

x →0

x →0

\ f(x) is not continuous at x = 0.

Thus the function is continuous at the origin.

Example 3: Find the value of k if

Derivatives

f (x) =

2 x 3 − 5 x 2 + 4 x + 11 , for x ≠ –1 x +1

and f (–1) = k is continuous at x = –1. Solution: Given f (x) is continuous at x = –1 ⇒

Lt f (x) = f (–1) = k.

x→−1

⎡ 2 x 3 − 5 x 2 + 4 x + 11 ⎤ Lt f (x) = Lt ⎢ ⎥ x→−1 x→−1 x +1 ⎣ ⎦ ( x + 1)( 2 x 2 − 7 x + 11) x +1

= Lt

= 2(–1)2 – 7 (–1) + 11

= 2 + 7 + 11 = 20

\

x→−1

k = 20

Example 4: If f (x) = =

x−4 + a, for x < 4, = a + b at x = 4, x−4

x−4 + b, for x > 4 and f (x) is continuous at x = 4, then x−4

In this section we will look at the simplistic form of the definition of a derivative, the derivatives of certain standard functions and application of derivatives. For a function f(x), the ratio [ f(a + h) - f (a)]/h is the rate of change of f (x) in the interval [a, (a + h)]. The limit of this ratio as h tends to zero is called the derivative of f (x). This is represented as f ′(x) i.e., f ( a + h) − f ( a) = f ′(x) h d{ f ( x )} or The derivative f ′(x) is also represented as dx d {f(x)}. dx Hence, if y = f (x), i.e., y is a function of x, then dy/dx is the derivative of y with respect to x. lim h→0

Note: 1. dy/dx is the rate of change of y with respect to x. 2. If the function y can be represented as a general curve, and a tangent is drawn at any point where the tangent makes an angle q with the horizontal (as shown in the figure), then dy/dx = tanq. In other words, derivative of a function at a given point is the slope of the curve at that point, i.e., tan of the angle, the tangent drawn

Chapter 7 Calculus | 2.337 to the curve at that point, makes with the horizontal. y

y = f(x)

If f is differentiable function of x and the derivative f ′ is also a differentiable function of x, then f ′′ is called the second derivative of f. Similarly 3rd, 4th … nth derivative of f may be defined and are denoted by f ′′′, f ′′′′, … f n or y3, y4 … yn.

q

O

Successive Differentiation

x

Application of Derivatives

Standard Results If f(x) and g(x) are two functions of x and k is a constant, then d 1. (c) = 0 (c is a constant) dx d d f(x) (k is a constant) 2. k · f(x) = k dx dx d d d 3. ( f(x) ± g(x)) = f(x) ± g(x) dx dx dx

Product Rule 4.

d { f(x) · g(x)} = f ′(x) · g(x) + f(x) · g′(x) dx

Quotient Rule 5.

d g ( x ) ⋅ f ’( x ) − f ( x ) ⋅ g ’( x ) {f(x)/g(x)} = dx ( g ( x )) 2

Chain Rule 6. If y = f(u) and u = g(x) be two functions, then dy/dx = (dy/du) × (du/dx)

Derivatives of Some Important Functions 1. (a) d/dx (xn) = n · xn-1 (b) d/dx [1/xn] = –n/xn+1 1 (c) d/dx (√x) = ;x≠0 2 x 2. d/dx [axn + b] = an · xn-1 3. d/dx [ax + b]n = n a(ax + b)n-1 4. d/dx [eax] = a · eax 5. d/dx [logx] = 1/x; x > 0 6. d/dx [ax] = ax loga; a > 0 7. (a) d/dx [sinx] = cosx (b) d/dx [cosx] = –sinx (c) d/dx [tanx] = sec2x (d) d/dx [cotx] = –cosec2x (e) d/dx [secx] = secx ·tanx (f) d/dx [cosecx] = –cosecx·cotx

Inverse Rule If y = f(x) and its inverse x = f –1(y) is also defined, then dy 1 = . dx dx / dy

1. Maxima and Minima: A function takes a maximum value or a minimum value when the slope of the tangent of the curve at that point is zero i.e., when the first derivative of the function is zero. If y = f(x), then y is maximum or minimum at the point x = x1 if ⎛ dy ⎞ = 0. ⎜ dx ⎟ ⎝ ⎠ x = x1

Thus we can find the value of x1 by equating dy/dx = 0. As mentioned above that y can have a maximum or a minimum value at x = x1. Whether y is a maximum value or minimum is governed by the sign of the second derivative. The function y has a minimum value if the second derivative is positive. In other words, y is maximum at x = x1 if d2y/dx2 < 0 at x = x1. y is minimum at x = x1 if d2y/dx2 > 0 ⎛ dy ⎞ at x = x1. ⎜ ⎟ = 0. in both the cases discussed above. ⎝ dx ⎠ x = x1 (a) If f ′(c) = 0 and f ′″(c) is negative, then f(x) is maximum for x = c (b) If f ′(c) = 0 and f ′″(c) is positive, then f(x) is minimum for x = c (c) If f ′(c) = f ′′(c) = … = f r - 1(c) = 0 and f r(c) ≠ 0, then (i) If r is even, then f(x) is maximum or minimum for x = c according as f r(c) is negative or positive. (ii) If r is odd, then there is neither maximum nor a minimum for f(x) at x = c.

Rolle’s Theorem Let f be a function defined on [a, b] such that 1. f is continuous on [a, b]; 2. f is differentiable on (a, b) and 3. f (a) = f(b), then there exists c ∈ (a, b) such that f ′(c) = 0

Lagrange’s Mean Value Theorem Let f be a function defined on [a, b] such that 1. f is continuous on [a, b], 2. f is differentiable on (a, b) then there exists c ∈ (a, b) f ( b) − f ( a) such that f ′(c) = . b−a Another form: If f is defined on [a, a + h] such that 1. f is continuous on [a, a + h]. 2. f is differentiable on (a, a + h) then there exists atleast one q ∈ (0, 1) such that f(a + h) = f(a) + hf ′(a + q h).

2.338 | Engineering Mathematics

Cauchy’s Mean Value Theorem

Total Differential Coefficient

Let f and g be two functions defined on [a, b] such that

If u be a continuous function of x and y and if x and y receive small increments Δx and Δy, u will receive, in turn, a small increment Δu. This Δu is called total increment of u.

1. f and g are continuous on [a, b] 2. f and g are differentiable on (a, b) 3. g′(x) ≠ 0 for any x ∈ (a, b) then there exists atleast one real number c ∈ (a, b) such that f (b) − f ( a) f ’(c) = . g (b) − g ( a) g ’(c)

Series Expansions of Some Standard Functions x 2 x3 xn (a) ex = 1 + x + + +…+ +… 2 ! 3! n! (b) sin x = x -

x3 x5 ( −1) n x 2 n +1 + -…+ +… 3! 5! ( 2n + 1)!

(c) cos x = 1 –

x2 x4 ( −1) n x 2 n + −… + +… 2! 4 ! ( 2n)!

x3 x5 x 2n +1 (d) sinh x = x + + +…+ +… 3! 5! ( 2n +1)! (e) cosh x = 1 +

x2 x4 x 2n + +…+ +… 2! 4 ! ( 2n)!

(f) log (1 + x) = x -

x 2 x3 x 4 ( −1) n −1 x n + − −… + 2 3 4 n

(g) (1 + x)–1 = 1 - x + x2 - x3 + … (h) (1 - x)–1 = 1 + x + x2 + x3 + … (i) (1 + x)–2 = 1 - 2x + 3x2 - 4x3 + … x 1⋅ 3 2 1 ⋅ 3 ⋅ 5 3 (j) (1 - x)-1/2 = 1 + + x + ⋅ x + ... 2 2 ⋅3 2⋅ 4 ⋅6 (k) tan–1x = x -

x3 x5 ( −1) n −1 2n- 1 + − ... + x + ... 3 5 ( 2n − 1)

(l) sin–1 x = x +

1 x3 1⋅ 3 x5 ⋅ + ⋅ + ...... 2 3 2⋅4 5

Partial Differentiation Let u be a function of two variables x and y. Let us assume the functional relation as u = f(x, y). Here x alone or y alone or both x and y simultaneously may be varied and in each case a change in the value of u will result. Generally the change in the value of u will be different in each of these three cases. Since x and y are independent, x may be supposed to vary when y remains constant or the reverse. The derivative of u w.r.t. x when x varies and y remains constant is called the partial derivative of u w.r.t. x and is denoted by ∂u/∂x ∂ 2 u ∂ ⎛ ∂u ⎞ ∂ 2 u ∂ ⎛ ∂u ⎞ = ⎜ ⎟, = ⎜ ⎟. 2 ∂x ∂x ⎝ ∂x ⎠ ∂x∂y ∂x ⎝ ∂y ⎠

Δu = f(x + Δx, y + Δy) - f(x, y) In the differential form, this can be written as du = ∂u ∂u dx + dy . ∂x ∂y du is called the total differential of u. If u = f(x, y, z) then

du ∂u ∂x ∂u ∂y ∂u ∂z = ⋅ + ⋅ +⋅ ⋅ dt ∂x ∂t ∂y ∂t ∂z ∂t

Implicit Function If the relation between x and y be given in the form f(x, y) = c where c is a constant, then the total differential coefficient w.r.t. x is zero.

Homogeneous Functions Let us consider the function f(x, y) = a0xn + a1xn-1y + a2xn2 2 y + … + any n. In this expression the sum of the indices of the variable x and y in each term is n. Such an expression is called a homogeneous function of degree n.

Euler’s Theorem If f(x, y) is a homogeneous function of degree n, then x ∂f ∂f +y = nf. ∂x ∂y This is known as Euler’s theorem on homogeneous function. The nth derivatives of some special functions: dn (a) n xn = n! dx dn m! (b) n xm = xm-n s(m being a positive integer ( m − n)! dx more than n) dn (c) n eax = an eax dx d n ⎛ 1 ⎞ ( −1) n ⋅ n ! (d) n ⎜ ;x≠-a ⎟= dx ⎝ x + a ⎠ ( x + a) n +1 dn ( −1) n −1 ( n − 1)! ; (x + a) > 0 (e) n log (x + a) = dx ( x + a) n dn ⎞ ⎛ nπ (f) n sin (ax + b) = an sin ⎜ + ax + b ⎟ dx ⎝ 2 ⎠ dn ⎞ ⎛ nπ (g) n cos (ax + b) = an cos ⎜ + ax + b ⎟ dx ⎝ 2 ⎠ d n ax (h) n (e sin bx) = (a2 + b2)n/2 eax sin (bx + n tan-1 b/a) dx dn (i) n (eax cos bx) = (a2 + b2)n/2 eax cos (bx + n tan-1 b/a) dx

Chapter 7 Calculus | 2.339 d n ⎛ 1 ⎞ ( −1) n n ! (j) n ⎜ 2 2 ⎟ = n + 2 sinn+1q sin (n + 1)q where dx ⎝ x + a ⎠ a -1 q = tan (x/a) dn (k) n (tan-1x) = (-1)n-1 (n - 1)! sinnq . sin nq where q = dx cot-1x

2. Similarly, the area enclosed by the curve x = g(y), the d

lines y = c and y = d and the y-axis is A = ∫ g ( y ) dy c

3. when f(x) ≥ 0 for a ≤ x ≤ c and f(x) ≤ 0 for c ≤ x ≤ b, then the area enclosed by the curve y = f(x), the lines x = a

Indefinite Integrals If f(x) and g(x) are two functions of x such that g′(x) = f(x), then the integral of f(x) is g(x). Further, g(x) is called the antiderivative of f(x). The process of computing an integral of a function is called Integration and the function to be integrated is called Integrand. An integral of a function is not unique. If g(x) is any one integral of f(x), then g(x) + c is also its integral, where C is any constant termed as constant of integration.

Definite Integrals

c

b

a

c

an x = b and the x-axis is A = ∫ f ( x ) dx − ∫ f ( x ) dx y = f(x)

y

A o

(c, 0)

x=b

x=a

x

4. The area enclosed by the curves y = f(x) and y = g(x) and the lines x = a and x = b is given by y

The difference in the values of an integral of a function f(x) for two assigned values say a, b of the independent variable x, is called the Definite Integral of f(x) over the interval [a,b]

f(x) A

b

and is denoted by ∫ f(x) dx

g(x)

a

The number ‘a’ is called the lower limit and the number ‘b’ is the upper limit of integration.

o

x=a

y

x=b

x

x=b

x

f(x)

Fundamental Theorem of Integral Calculus: If f(x) is a function of x continuous in [a, b], then

A

∫ f ( x) dx = g(b) -g(a) where g(x) is a function such that

g(x)

b

a

o

d g(x) = f(x) dx

x=a

b

A = ∫ f ( x ) − g ( x ) dx = a

Applications of Integration

b

Area as a Definite Integral

∫ ( f ( x ) − g ( x )) dx, if f ( x ) ≥ g ( x ) , a ≤ x ≤ b

1. The area enclosed by a curve y = f(x), the lines x = a and x = b and the x-axis is given by

∫ ( g ( x ) − f ( x )) dx, if f ( x ) ≤ g ( x ) ; a ≤ x ≤ b

b

A= ∫ a

a

⎧b ⎪ ∫ f ( x ) dx, if f ( x ) ≥ 0, a ≤ x ≤ b ⎪a f ( x ) dx = ⎨ b ⎪− f x dx, if f x ≤ 0, a ≤ x ≤ b ( ) ⎪ ∫ ( ) ⎩ a

y

Example 6: Find the area enclosed by the curve y = x3, the line y = 2 and the y-axis in first quadrant?

y

y = f(x)

x=b

x=a

x

o A

A

Solution: The area bounded by y = x3, y = 2 and the y-axis is the area OAB as shown in the figure. So, the region OAB is bounded by the curve x = y1/3, the lines y = 0 and y = 2 and the y-axis and x = y1/3 ≥ 0, y ∈ [0, 2] \ The required area 2

⎤ 3 = ∫ y dy = y 4 / 3 ⎥ 4 ⎥⎦ y=0 2

1/ 3

0

o

x=a

x=b

x

y = f(x)

=

3 × 24 / 3 4

2.340 | Engineering Mathematics points with y-coordinates ‘c’ and ‘d’ is given by S =

3 22 / 3 3 = 3 4 =

d

∫

1 + ( dx / dy ) 2 dy provided dx/dy is continuous on

c

[c, d] y

y=2 A

2. Parametric equations: Let x = f(t) and y = g(t) be parametric functions of ‘t’. The length of the arc between the points{f(t1), g(t1)} and {f(t2), g(t2)} t2

⎡ dx dy ⎤ ⎢ dt + dt ⎥ dt provided dx/dt and ⎣ ⎦ t1 dy/dt are both continuous on [t1, t2]. is given by

o x

y = x3

Example 7: Find the area enclosed by the curve y = x2 and line y = 4? Solution: The area enclosed by the curve y = x and the line y = 4 is the region OAB. \ The region OAB is bounded by line y = 4 and the curve y = x2 from x = –2 to x = 2 and 4 ≥ x2 for all x ∈ [–2, 2] 2

∫

3. Polar equations: Let r = f(q) be a function of q, the length of the arc between the points {f(q1), q1} and {f(q2),q2} is given by S =

θ1

(r1, f(r1)), (r2, f(r2)) is given by S =

r2

∫

1 + r 2 ( dθ / dr ) 2

dr provided dq/dr is continuous along the arc. B (2, 4)

y=4

Improper Integrals

A y = x2

Consider definite integral

o

b

∫ f ( x ) dx (8)

x

O

2

∫

( 4 − x 2 )dx

x = −2 2

= 2∫ ( 4 − x 2 )dx (∵ 4 – x2 as even) 0

2

⎡ x3 ⎤ = 2 ⎢4 x − ⎥ 3 ⎦0 ⎣

=

a

If f(x) is a function defined in a finite interval [a, b] and f(x) is continuous for all x which belongs to [a, b] Then (8) is called Proper integral. If f(x) is violated, at least one of these conditions then the integral is known as improper integral. These improper integrals are classified into three kinds.

Improper Integral of the First Kind In a definite integral if one or both limits of integration are infinite then it is an improper integral of first kind.

32 3

∞

b

1. ∫ f ( x ) dx = Lim ∫ f ( x ) dx. b→∞

a

Rectification

a

(Singularity at upper limit)

The process of determining the length of arcs of plane curves is called Rectification. The length of the arc can be calculated by any one of the methods given.

2. ∫ f ( x ) dx = Lt

1. Cartesian equations: Let y = f(x) be a function of x. The length of arc between the points with

Lt 3. ∫ f ( x ) dx = b →∞

x-coordinates ‘a’ and ‘b’ is given by S = ∫ g ( x ) dx

4. ∫ f ( x ) dx = Lt

r 2 + ( dr / dθ ) 2 dq

r1

(–2, 4) A

∫

provided dr/dq is continuous along the arc. If the equation of the curve is given in the form q = f(r), then the length of the arc between the points

y

\ The required area =

θ2

∞

a

dx, provided dy/dx is continuous on [a, b]. Note: If the equation of the curve is given in the form x = f(y), then the length of the arc between the

b

b

∫

a→−∞

−∞ ∞

−∞

f ( x ) dx.

a

b

∫

f ( x ) dx. or

a →−∞ a

∞

a→− ∞

−∞

0

∫ a

r

or = Lt

r→ ∞

∫

−r

f ( x ) dx.

b

f ( x ) dx + Lt

b→∞

∫ 0

f ( x ) dx.

Chapter 7 Calculus | 2.341

Converges when the limits of the above integral exists or finite then the integral is said to be convergent.

Improper Integral of Third Kind

Diverges If the limits does not exists then they are said to be Divergent. Note 1: Geometrically for f(x) ≥ 0, the improper integral ∞

∫

f ( x ) dx denotes the area of an unbounded region lying

a

between the curve y = f (x) the ordinate x = a and x-axis. Note 2: Let f(x) and g(x) be non-negative functions and 0 ≤ f(x) ≤ g(x) for x ≥ a. If ∞

∞

∫ g ( x ) dx converges then a

is also converges and ∫ f ( x ) dx ≤ a

∞

∫

If these limits exists then it is convergent otherwise it is divergent.

Similarly let 0 ≤ g(x) ≤ f(x) if

∞

∞

∫ f ( x ) dx a

c

Note:

1

∫ ( x −c)

p

dx converges for p < 1 and diverges for

a

∫ g ( x ) dx.

p ≥ 1. This is used for convergence or divergence of an improper integral of second kind.

a

∞

∫ g ( x ) dx diverges then a

f ( x ) dx also diverges.

If the limits of the integral are infinite or f(x) may be discontinuous or both then the improper integral is known as third kind. ∞ 1 Note: ∫ p dx is convergent when p > 1 and it is divergent x 1 p ≤ 1. This result is used in comparison test for testing the convergence or divergence of the integral of first kind.

∞

Example 8: Examine

dx

∫x

p

for convergence/divergence.

1

a

i.e., The convergent or divergent of an improper integral by comparing it with a simple integral.

Solution: Consider

Improper Integral of the Second Kind

if p = 1

Consider

⇒

b

∫ f ( x ) dx (9) a

∫ f ( x ) dx =

c −∈

a

Lt

∈→ 0

∫

[log x ]1

k

k

If both the limits of (1) are finite f(x) is undefined or discontinuous at a point in between a and b, then (9) is known as Improper integral of second kind. This can be evaluted as follows. Let f(x) be undefined at a point c which belongs to (a, b) then b

k

k ⎡ x − p +1 ⎤ dx −p = x dx = ⎢ ⎥ if p ≠ 1 and ∫1 x p ∫1 ⎣ − p + 1 ⎦1 k

b

f ( x ) dx + Lt

∈→ 0

a

∫

f ( x ) dx.

dx = logk – log1 = logk → ∞ when x 1 k → ∞ it does not tend to a finite limit. \ it is divergent.

Case I: if p = 1, ∫

k

Case II: If p ≠ 1 diverges if p ≤ 1.

c +∈

dx

∫x

p

1

=

1 [k 1 − p ] it converges if p >1 and 1− p

Exercises Practice Problems I Directions for questions 1 to 47: Select the correct alternative from the given choices. x 2 −1 , x ≠ −1 x → −1 x + 1 (A) 2 (C) −2

1. Lt

(B) 0 (D) Does not exist

2. Lt {3 x − 9 x 2 − x } = ______. x →∞

1 (A) 6 (C) 6

(B) 3 (D) None of these

⎡ 24 cos x − 24 + 12 x 2 − x 4 ⎤ 3. Lt ⎢ ⎥ = x →0 24 x 6 ⎣ ⎦ 1 −1 (A) (B) 720 120 1 −1 (C) (D) 120 720

(

)

1/ n

= 4. If 0 < x < y, Lt y n + x n n →∞ (A) e (B) x (C) y (D) nxn–1

2.342 | Engineering Mathematics 5. Evaluate Lt (x − [x]), (where [x] is the greatest intex→2.7

ger less than or equal to x) (A) −0.3 (B) 0.7 (C) 4.7 (D) 2

1 6. Evaluate Lt 189 . x →0 x (A) 0 (C) −∞

(B) ∞ (D) None of these

1/ x

⎛ 2 x + 3x ⎞ 7. Lt ⎜ ⎟ = x →0 ⎝ 2 ⎠ (A) 1 (B) 3 (C) 6 (D) 2 8. Evaluate Lt (3x + 4 x )1 x , (x ≥ 1) x →∞

(A) 4 (B) 3 9. Lt |x – 2| + [ x – 2] =

(C) 2

(D) 1

x→2

(A) 0 (B) Only left limit exists (C) Only right limit exists (D) Limit does not exist 10. Let the function f(x) = [x]. where [x] is the greatest integer less than or equal to x. Which of the following is/are true? (A) f(x) has jump discontinuity at all x ∈ Z (B) f(x) has removable discontinuity at all x ∈ Z (C) f(x) is continuous at all irrational values. (D) Both (A) and (C) ⎧ 5x − 4 0 < x ≤ 1 at x = 1 11. f(x) = ⎨ 2 ⎩4 x − 3x 1 < x < 2 (A) Left hand continuous at x = 1 (B) Right hand continuous at x = 1 (C) Continuous at x = 1 (D) None of these 12. Check the continuity of the following function ⎧ sin 2 ax , when x ≠ 0 ⎪ f(x) = ⎨ x 2 at x = 0 ⎪ a 2 when x = 0 ⎩

(A) (B) (C) (D)

Continuous at x = 0 Discontinuous at x = 0 Discontinuous of first kind None of these

x < 5, ⎧7 ⎪ 13. If f(x) = ⎨ax + b 5 < x < 7, ⎪11 x > 7 is continuous on R, ⎩ then the values of a and b are (A) a = 2, b = 3 (B) a = −2, b = 3 (C) a = 3, b = −2 (D) a = 2, b = −3

14. Find values of a, b, c so that ⎧ a cot −1 ( x − 3), ⎪ bx, ⎪⎪ f(x) = ⎨ 1 ⎞ −1 ⎛ ⎪ c tan ⎜ x − 3 ⎟ , ⎝ ⎠ ⎪ ⎪⎩ cos −1 ( 4 − x ) + aπ ,

0≤ x 0, attains its maximum and minimum at x = p and x = q respectively such that p2 = q, then the value of ‘a’ is 1 1 (A) 2 (B) (C) (D) 4 4 8 Common data for questions 21 and 22: The sum of the hypotenuse and one side of a right angled triangle is given as a units. 21. When the area is maximum the ratio of the side and the hypotenuse is ______. (A) 2 : 1 (B) 1 : 3 (C) 1 : 2 (D) 2 : 3

Chapter 7 Calculus | 2.343 22. When the area is maximum, find the angle between the hypotenuse and the other side is ______.

(A) 60° (C) 45°

(B) 30° (D) None of these

23. Consider f(x) = |x2 - 3|, 0 ≤ x ≤ 6 and g(x) = ⎧ 3x , 0 ≤ x ≤ 1, . Then Rolle’s theorem can be applied ⎨ ⎩4 − x, 1 < x ≤ 3 in the respective intervals

(A) (B) (C) (D)

To both f(x) and g(x) Only to f(x) Only to g(x) Neither to f(x) nor to g(x)

25. In the interval [1, 2] the value of q to the function f(x) = 2x3 − 5x + 3 is equal to 7 7 2 4 (A) (B) (C) −1 (D) 3 3 3 5 26. Let f(x) = eax and g(x) = e−ax be two functions defined in [p, q], If the functions satisfies Cauchy mean value theorem then the value of ‘c’ is ______. p+q (A) p + q (B) 2 (C) 2(p + q) (D) None of these

(A) 1

(C) 2y

28. The function f(x, y, z) =

(D) – 2y

x y − 2 y z + 5 x z − 3 xyz 2 2

2

2 2

2 2

x 4 + y 2 z 2 + xy 2 z is a homogeneous function of order _______.

(A) 1

(B) 2

(C) 4

(D) 0

3 4.

(

30. If u = log x +

(A) 0

x + y (B) 1

2

)

∂f ∂f ∂f +y +z = ∂x ∂y ∂z

(A) tan 2 f (C) cos 2 f

(B) cot 2 f (D) sin 2 f

For the function f(x, y) = 2x2 + 4y2 + 4xy + 2x + 10y + 7 (A) Local maximum exists, but no local minimum (B) Local minimum exists, but no local maximum (C) Neither local minimum nor local maximum exists (D) Both local minimum and local maximum exists

,

37. Which of the following function/s is/are integrable but not continuous on (0, 10)? (A) f(x) = [x] (ceiling function) (B) f(x) = |x - 3| (C) f(x) = |x - 5| + |x - 2| (D) f(x) = x2 + 5x + 9 π /2

∫ sin

4

x cos6 xdx = ______.

0

∂u ∂u , then e − y = ∂y ∂x u

(C) –1

(B) tan f (D) sin2 f

35. For the function xyz, if x + y + z = 3, then the local maximum occurs for xyz at the point _____ ⎛ 1 1⎞ (B) (5, –1, –1) (A) ⎜ 4, 2 , 2 ⎟ ⎝ ⎠ (C) (1, 1, 1) (D) (7, –3, –1)

38.

u 4 (A) (B) 2 u (C) 4u (D) 6u 2

(A) 2 tan f (C) 2sin2 f

33. The stationary points of the function f(x, y) = x3 + y4 − 27x + 32y + 100 is/are (A) (3, 2), (3, −2) (B) (−3, 2), (−3, −2) (C) (3, 2), (−3, −2) (D) (3, −2), (−3, −2)

6

⎛4x + 4 y⎞ ⎟ , then x ∂u + y ∂u = 2 9. If u = ⎜ ⎜ 6 x+6 y ⎟ ∂x ∂y ⎝ ⎠

∂u ∂u ∂u +y +z is equal to ∂x ∂y ∂z

36. The ratio of the dimensions of a rectangular box of volume 64 cubic units and open at the top that requires least material for its construction is (A) 2 : 2 : 1 (B) 2 : 4 : 5 (C) 2 : 3 : 4 (D) 1 : 2 : 3

∂z = ∂y

(B) y

1 1 2 2 (A) (B) (C) (D) 3 3 2 3 3

31. The value of x

32. x

24. If the function f(x) = px2 + qx2 + rx + s on [0, 1], satisfies the mean value theorem, then the value of c in the interval (0, 1) is

27. If x = cos (z + y2), then

Common data for questions 31 and 42: x 5/ 2 + y5/ 2 + z 5/ 2 Let u = tan f = 1/ 2 x + y1/ 2 + z1/ 2

(D) u

3π 2π 3π 3π (A) (B) (C) (D) 128 425 2560 512 39. Area bounded by the curve y2 = x and the line x = 3 is ________ sq units

(A) 2 3

(B) 4 3

(C) 6 3

(D) 8 3

40. Area bounded by the curve y = –3x2, x = 2 and the two coordinate axes is ______ sq units. (A) 2 (B) 3 (C) 6 (D) 8

2.344 | Engineering Mathematics 41. The volume of the solid obtained by revolving the area bounded by the parabola y2 = x – 4, x-axis and the lines x = 4 and x = 7, about x-axis is _______ cubic unit 9 11 13 15 (A) p (B) p (C) p (D) p 2 2 2 2 42. The length of arc of the curve y = ln (cos x) from x = 0 π to x = is _______ . 4

( ) (C) ln ( 2 + 3 )

(A) ln 1 + 2

∞

43.

∫x

1 1.0001

( 2 − 1) (D) ln ( 2 − 3 ) (B) In

dx = ______ .

1

(A) 1000 (C) 10000 3

1

(B) 100000 (D) 1000000

∫

−1

∫ ( x −2)

(A) 5 – 21/5

(B) 5 + 21/5

(C) 5(1–2)1/5

(D) 5 ⎡⎣ 1 + 2 1/5 ⎤⎦

dx = ______ .

0

1 dx = ______ x 2/5

(A) 5

(B)

10 5 5 (C) (D) 2 3 3

46. Which of the following converges? ∞

I. ∫ 0

(A) (B) (C) (D) ∞

1 x +1 4

∞

dx

II.

∫ 0

dx 4

x 5 +1

Only (I) Only (II) Both (I) and (II) Neither (I) nor (II)

4 x+7 dx is ______ . x 6 + 10

47.

∫

(A) Convergent (B) Divergent (C) Cannot be determined (D) None of these

3

44.

4/5

1

45.

Previous Years’ Questions 1. Consider the following two statements about the function f(x) = | x| P : f(x) is continuous for all real values of x Q : f(x) is differentiable for all real values of x Which of the following is TRUE? [2007] (A) P is true and Q is false. (B) P is false and Q is true. (C) Both P and Q are true. (D) Both P and Q are false. x − sin x equals [2008] x + cos x (A) 1 (B) –1 (C) ∞ (D) –∞ 3. A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve 3x4 − 16x3 + 24x2 + 37 is [2008] (A) 0 (B) 1 (C) 2 (D) 3 2. lim

x →∞

π /4

4.

∫ (1 − tan x)/(1 + tan x)dx

evaluates to[2009]

0

(A) 0 (C) ln 2

(B) 1 (D) ½ ln 2 2n

⎛ 1⎞ 5. What is the value of lim ⎜1 − ⎟ ?[2010] n →∞ ⎝ n⎠ (A) 0 (C) e–1/2

(B) e–2 (D) 1

6. Given i =

−1 , what will be the evaluation of the π 2 cos x + i sin x definite integral ∫ dx?[2011] cos x − i sin x 0

(A) 0

(B) 2

(C) -i (D) i

7. Consider the function f(x) = sin(x) in the interval x ∈[p/4, 7p/4]. The number and location(s) of the local minima of this function are [2012] (A) One, at p/2 (B) One, at 3p/2 (C) Two, at p/2 and 3p/2 (D) Two, at p/4 and 3p/2 8. Which one of the following functions is continuous at x = 3? [2013] ⎧ ⎪ 2, if x = 3 ⎪ (A) f ( x ) = ⎨ x − 1, if x > 3 ⎪x +3 ⎪ , if x < 3 ⎩ 3 ⎧4, if x = 3 f ( x) = ⎨ (B) ⎩8 − x, if x ≠ 3 ⎧ x + 3, if x ≤ 3 (C) f ( x) = ⎨ ⎩ x − 4, if x > 3 1 (D) f ( x) = 3 , if x ≠ 3 x − 27

Chapter 7 Calculus | 2.345 9. Let the function sin θ cos θ tan θ f(q) = sin(π / 6) cos(π / 6) tan(π / 6) sin(π / 3) cos(π / 3) tan(π / 3)

1 5. limx→∞ x1/x is (A) ∞ (C) 1

⎡π π ⎤ where θ ∈ ⎢ , ⎥ and f 1(q) denote the derivative of ⎣6 3⎦ f with respect to q. Which of the following statements is/are TRUE? ⎡π π ⎤ (I) There exists θ ∈ ⎢ , ⎥ such that f 1(q) = 0 ⎣6 3⎦ 10.

⎡π π ⎤ (II) There exists θ ∈ ⎢ , ⎥ such that f 1(q) ≠ 0 ⎣6 3⎦ [2014]

(A) I only (B) II only (C) Both I and II (D) Neither I nor II The function f(x) = xsinx satisfies the following equation: f ″(x) + f(x) + t cos x = 0. The value of t is –––– ––. [2014] 11. A function f(x) is continuous in the interval [0, 2]. It is known that f(0) = f(2) = –1 and f(1) = 1. Which one of the following statements must be true?[2014] (A) There exists a y in the interval (0, 1) such that f(y) = f(y + 1) (B) For every y in the interval (0, 1), f(y) = f(2 – y) (C) The maximum value of the function in the interval (0, 2) is 1 (D) There exists a y in the interval (0, 1) such that f(y) = –f(2 – y) 12. A non – zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE? [2014] (A) f(0) f(4) < 0 (B) f(0) f(4) > 0 (C) f(0) + f(4) > 0 (D) f(0) + f(4) < 0 2π

13. If

∫

[2014]

to ––––

∫x

2

cos xdx

x

[2014] (A) –2p (B) p (C) –p (D) 2p

2 /π

1/ π

1 cos x dx = _____ . x2

[2015]

17. Let f (x) = x–(1/3) and A denote the area of the region bounded by f (x) and the X-axis, when x varies from –1 to 1. Which of the following statements is/are TRUE? [2015] I) f is continuous in [–1, 1] II) f is not bounded in [–1, 1] III) A is nonzero and finite (A) II only (B) III only (C) II and III only (D) I, II and III −x

18. The value of lim(1 + x 2 )e is x →∞

(A) 0

(C) 1

[2015]

1 (B) 2 (D) ∞

1 1 19. If for non-zero x, af(x) + bf = - 25 where a ≠ x x 2 b then ∫1 f ( x )dx is [2015] (A)

1 a2 − b2

47b a(ln 2 − 25) + 2

(B)

1 a − b2

47b a( 2 ln 2 − 25) − 2

(C)

1 a2 − b2

47b a( 2 ln 2 − 25) + 2

(D)

1 a − b2

47b a(ln 2 − 25) − 2

π

14. The value of the integral given below is

∫

16.

x sin x dx = kp, then the value of k is equal

0

[2015] (B) 0 (D) Not defined

2

2

20. lim sin( × − 4) = _______ x→4

×−4

[2016]

21. Let f (x) be a polynomial and g (x) = f1(x) be its derivative. If the degree of (f (x) + f (-x) is 10, then the degree of (g(x) – g (-x)) is ____. [2016]

2.346 | Engineering Mathematics

Answer Keys

Exercises Practice Problems I 1. C 11. C 21. C 31. A 41. A

2. A 12. A 22. B 32. D 42. A

3. D 13. D 23. D 33. D 43. C

4. C 14. B 24. A 34. B 44. D

5. B 15. B 25. C 35. C 45. B

6. D 16. B 26. B 36. A 46. C

7. C 17. A 27. D 37. A 47. A

8. A 18. D 28. B 38. D

9. D 19. D 29. A 39. B

10. D 20. A 30. A 40. D

4. D 12. A

5. B 6. D 13. 4 to 4 14. A

7. B 15. C

8. A 16. –1

9. C 17. C

18. C

Previous Years’ Questions 1. A 2. A 10. –2 to –2 19. A 20. 1

3. B 11. A 21. 9

Test Engineering Mathematics

Time: 60 min.

Directions for questions 1 to 25: Select the correct alternative from the given choices. ⎡k 2 3 ⎤ 1. If the eigen values of the matrix ⎢⎢ 0 16 5 ⎥⎥ are in geometric ⎢⎣ 0 0 4 k ⎥⎦ progression, then the value of k can be

10. Four dice are thrown. One of the dice shows six. Find the probability that only one die shows six. 125 250 (A) (B) 216 671 125 500 (C) (D) 671 671

11. Find the standard deviation of the first seven natural numbers. (A) 2 (B) 7 (C) 2 (D) 0 12. “Every engineering student wrote GATE or GRE” Convert into symbolic form using S(x), G(x), R(x) where S(x) is x is an engineering student G(x) → x wrote GATE R(x) → x wrote GRE (A) ∀ x (S(x)) → (G(x) ∧ R(x)) (B) ∀ x [S(x) → (G(x) ∨ R(x))] (C) ∀ x (S(x) → [G(x) → R(x)]) (D) ∀ x (S(x) ∧ (G(x) ∨ R(x)) 13. What is the solution of the recurrence relation an = 12an –1 – 36 an–2 with initial conditions a0 = 2 and a1 = 24? (A) an = 2.6n – n. 6n (B) an. = 6n + n.6n–1 (C) an = 2.6n – n.6n–1 (D) None of these 14. A binary operator * is defined on set of real numbers as follows a * b = a + b + ab where ∀ a, b ∈ R – {–1}. If G is a group under operator *, then find the inverse element of 3 (where 3 ∈ G). 4 −3 −4 3 (A) (B) (C) (D) 3 4 3 4

(A) 8 (C) 1

(B) –8 (D) All the above

⎡ 2 1⎤ 2. One of the eigen vector of the matrix ⎢ ⎥ is ⎣ 2 3⎦ ⎡ 2 ⎤ ⎡1⎤ ⎡ 2⎤ ⎡2⎤ (A) ⎢ ⎥ ⎢1⎥ (B) ⎢1 ⎥ (C) ⎢ −1⎥ (D) ⎣⎦ ⎣ ⎦ ⎣ ⎦ ⎢⎣ − 2 ⎥⎦ 3. Find the value of k given that following system of linear equations will have no solution. x + 3y + z = 2 2x + y + 5z = 5 x + ky – 2z = 1 (A) 4 (B) – 4 (C) 8 (D) – 8 ⎡1 7 ⎤ 15 11 3 4. If A = ⎢ ⎥ , find the value of A – 2A + 3A . 0 − 1 ⎣ ⎦ (A) A (B) 2A (C) 3A (D) O 5. The perimeter of a rectangle is 16 cm. Find the maximum area of the rectangle. (A) 16 cm2 (B) 64 cm2 (C) 48 cm2 (D) 7 cm2 2

6. ∫ ⎡⎣ x 2 ⎤⎦ dx =. Where [x] is greatest integer less than or 1

equal to x.

(A) 5 − 2 − 3 (B) 5+ 2 − 3

5 + 2 + 3 (D) 5− 2 + 3 (C) 7. If A, B and C are three events such that P(A) = 0.3 P(B) = 0.4, P(C) = 0.5, P(A ∩ B ∩ C) = 0.3. Find the value of P(A ∪ B ∪ C) + P(A ∩ B) + P(B ∩ C) + P(A ∩ C). (A) 1 (B) 1.2 (C) 1.3 (D) 1.5 8. A number is randomly selected from 1 to 21. Find the probability that it is a multiple of five if it is known that it is an even number. 4 1 2 5 (A) (B) (C) (D) 21 5 5 21 9. A four letter word is formed from the letters A, B, C, D, E, F. Find the probability that it ends with D, if it is known that it starts with A. 1 2 3 4 (A) (B) (C) (D) 5 5 5 5

15. Find the number of solutions of x1 + x2 + x3 = 15 where x1, x2 and x3 are non negative integers satisfying 2 ≤ x1 ≤ 6; 3 ≤ x2 ≤ 7; 4 ≤ x3 ≤ 8. (A) 18 (B) 19 (C) 20 (D) 16 16. F(x) = (x – 2) (x – 3)2, which of the following holds true? ⎛ 7⎞ ⎛ 8⎞ (A) F′ ⎜ ⎟ = 0 (B) F′ ⎜ ⎟ = 0 ⎝ 3⎠ ⎝ 3⎠ ⎛ 9⎞ (C) F′ ⎜ ⎟ = 0 ⎝ 4⎠

⎛ 5⎞ (D) F′ ⎜ ⎟ = 0 ⎝ 2⎠

17. Write the negation of the following statement ∀ x ∃ y (P(x, y) ∨ ∃ z R(x, y, z)). (A) ∃ x ∀ y (~P (x, y) ∧ ∀ z ~ R(x, y, z)) (B) ∃ x ∀ y (P(x, y) ∨ ∀ z ~ R(x, y, z)) (C) ∀ x ∃ y (~ P (x, y) ∧ ∃ z ~ R(x, y, z)) (D) None of these 18. Find the reciprocal of 21 using Newton Raphson method. (Take initial approximation as 0.047 second approximation) (A) 0.047611 (B) 0.047619 (C) 0.046611 (D) 0.047615

2.348 | Engineering Mathematics 6

19. Evaluate

∫

1 + x 2 dx taking h = 1 using Simpson’s

0

Three eight Rule. (A) 18.5091 (B) 19.5091 (C) 17.5091 (D) 19.9501 20. How many moves are needed to solve the Tower of Hanoi problem with 10 disks? (A) 511 (B) 4095 (C) 1025 (D) 1023 21. Find the values of p and q such that the following system of linear equations have infinite number of solutions. x + py + z = 1 x + y + 3z = 2 5x + 5y + 11z = q (A) p = 8, q = 1 (B) p = 1, q = 8 (C) p = 4, q = 1 (D) p = 1, q = 4 22. Find the minimum value of (A) 0 (B) 6 2 3. Consider a given tree graph

How many perfect matching can be formed for the graph? (A) 0 (B) 1 (C) 2 (D) 3 24. Which of the following is a component of a given graph? a f

b

c

e

d (A) efd (B) bcd (C) fab (D) aec 25. What is the sum of degrees of Regions of the given graph?

x4 − x 3 + x 2 + 6. 4 (C) –1 (D) – 6

a

b

e

(A) 12 (C) 15

f

c

d

g

(B) 14 (D) 16

Answer Keys 1. D 11. C 21. B

2. D 12. B 22. B

3. C 13. D 23. A

4. B 14. B 24. D

5. A 15. B 25. D

6. A 16. A

7. D 17. A

8. B 18. B

9. A 19. B

10. D 20. D

Computer Science and Information Technology UNIT 1: Digital Logic 3.351 UNIT II: Computer Organization and Architecture 3.433 UNIT III: Programming and Data Structures3.507 UNIT IV: Databases3.657 UNIT V: Theory of Computation

3.753

UNIT VI: Compiler Design

3.803

UNIT VII: Operating System

3.873

UNIT VIII: Networks, Information Systems, Software Engineering and Web Technology3.969

p a r t III

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Digital Logic

Chapter 1: Number Systems

3.353

Chapter 2: Boolean Algebra and Minimization of Functions

3.364

Chapter 3: Combinational Circuits

3.384

Chapter 4: Sequential Circuits

3.405

U n i t I

This page is intentionally left blank

Chapter 1 Number Systems LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • •

Digital circuits Number system with different base Conversion of number systems Complements Subtraction with complement Numeric codes

DiGital CirCuits Computers work with binary numbers, which use only the digits ‘0’ and ‘1’. Since all the digital components are based on binary operations, it is convenient to use binary numbers when analyzing or designing digital circuits.

Number Systems with Different Base Decimal number system Decimal numbers are usual numbers which we use in our day-today life. The base of the decimal number system is 10. There are ten numbers 0 to 9. The value of the nth digit of the number from the right side = nth digit × (base)n–1 Example 1: (99)10 → 9 × 101 + 9 × 100 = 90 + 9 = 99 Example 2: (332)10 → 3 × 102 + 3 × 101 + 2 × 100 = 300 + 30 + 2 Example 3: (1024)10 → 1 × 103 + 0 × 102 + 2 × 101 + × 100 = 1000 + 0 + 20 + 4 = 1024

Binary number system In binary number system, there are only two digits ‘0’ and ‘1’. Since there are only two numbers, its base is 2. Example 4: (1101)2 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 8 + 4 + 1 = (13)10

Octal number system Octal number system has eight numbers 0 to 7. The base of the number system is 8. The number (8)10 is represented by (10)8.

• • • •

Weighted and non-weighted codes Error detection and correction code Sequential, reﬂective and cyclic codes Self complementing code

Example 5: (658)8 = 6 × 82 + 5 × 81 + 8 × 80 = 384 + 40 + 8 = (432)10

Hexadecimal number system In hexadecimal number system, there are 16 numbers 0 to 9, and digits from 10 to 15 are represented by A to F, respectively. The base of hexadecimal number system is 16. Example 6: (1A5C)16 = 1 × 163 + A × 162 + 5 × 161 + C × 160 = 1 × 4096 + 10 × 256 + 5 × 16 + 12 × 1 = 4096 + 2560 + 80 + 12 = (6748)10 . Table 1 Different number systems Decimal

Binary

Octal

0

000

0

Hexadecimal 0

1

001

1

1

2

010

2

2

3

011

3

3

4

100

4

4

5

101

5

5

6

110

6

6

7

111

7

7

8

1000

10

8

9

1001

11

9

10

1010

12

A

11

1011

13

B

12

1100

14

C

13

1101

15

D

14

1110

16

E

15

1111

17

F (Continued )

3.354 | Digital Logic Table 1 (Continued ) Decimal

Binary

Octal

Hexadecimal

16

10000

20

10

17

10001

21

11

18

10010

22

12

19

10011

23

13

20

10100

24

14

1. For a number system with base n, the number of different symbols in the number system will be n. Example: octal number system will have total of 8 numbers from 0 to 7. 2. The number ‘n’ in the number system with base ‘n’ is represented as (10)n. 3. The equivalent of number (a3a2a1a0 · a–1a–2)n in decimal is a3 × n3 + a2 × n2 + a1 × n1 + a0 × n0 + a–1 × n–1 + a–2 × n–2.

Conversion of Number Systems The conversion of decimal to any other number system involves successive division by the radix until the dividend reaches 0. At each division, the remainder gives a digit of converted number; and the last one is most significant digit, the remainder of the first division is least significant digit. The conversion of other number system to decimal involves multiplying each digit of number system with the weight of the position (in the power of radix) and sum the products calculated, the total is the equivalent value in decimal.

Decimal to binary conversion Example 7: (66)10 2 2 2 2 2

66 33 0 16 1 8 0 4 0 2 0

Example 9: (105.75)10 2 2 2 2 2 2

105 52 26 13 6 3

1 0 0 1 0

1

1

(105)10 = (1101001)2 (0.75)10 Multiply 0.75 by 2 = 1.50 Multiply 0.50 by 2 = 1.00 Reading integers from top to bottom 0.75 = (0.11)2 \ (105.75)10 = (1101001.11)2

Binary to decimal conversion Example 10: (10100011)2 = 1 × 27 + 0 × 26 + 1 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20 = 128 + 0 + 32 + 0 + 0 + 0 + 2 +1 = (163)10 Example 11: (11010011.101)2 = 1 × 27 + 1 × 26 + 0 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20 + (1 × 2–1) + (0 × 2–2) + (1 × 2–3) = 128 + 64 + 0 + 16 + 0 + 0 + 2 + 1 + 0.5 + 0 + 0.125 = (211.625)10

Decimal to octal conversion Reading remainders from bottom to top

Example 12: (16)10 8 16 0 2 Remainder from bottom to top = (20)8

1 0 = (1000010)2

Example 8: (928)10 2 2 2 2 2 2 2 2 2

928 464 232 116 58 29 14 7 3

0 0 0 0 0 1 0 1

1

1

= (1110100000)2

Example 13: (347.93)10 (.93)10 0.93 × 8 = 7.44 0.44 × 8 = 3.52 0.52 × 8 = 4.16 0.16 × 8 = 1.28 …….. Read the integers of octal point from top to bottom. \ (0.93)10 = (0.7341)8 (347)10 8 347 3 8 43 3 5 \ (347)10= (533)8 Ans: (533.7341)8

Chapter 1 Number Systems | 3.355

Octal to decimal conversion Example 14: (33)8 3 × 81 + 3 × 80 = 24 + 3 (27)10 Example 15: (1023.06)8 1 × 83 + 0 × 82 + 2 × 81 + 3 × 80+ 0 × 8–1 + 6 × 8–2 = 512 + 0 + 16 + 3 + 0 + 0.0937 = (2095.0937)10

Octal to binary conversion To convert octal to binary, replace each octal digit with their equivalent 3-bit binary representation. Example 16: (7777)8 Convert each octal digit to binary 7 7 7 7 = 111 111 111 111 = (111 111 111 111)2 Example 17: (567.62)8 5 6 7 . 6 101 110 111 . 110 = (101110111.110010)2

2 010

Binary to octal conversion To convert a binary number to an octal number, starting from the binary point, make groups of 3-bits each on either side of the binary point, and replace each 3-bit binary group by the equivalent octal digit. Example 18: (010011101)2 010 011 101 = (235)8 2 3 5 Example 19: (10010111011.1011)2 010 010 111 011 101 100 ⋅ = (2273.54)8 2 2 7 3 5 4

Decimal to hexadecimal conversion Example 20: (527)10

16 527 16 32 15 2 0

Decimal Hexa 2 → 2 0 → 0 15 → F = (20F)16 Example 21: (18.675)10 (18)10 16 18 1 2 Decimal Hexa 1 – 1 2 – 2 (18)10= (12)16 (0.675)10

0.675 × 16 10.8 0.800 × 16 12.8 0.800 × 16 12.8 0.800 × 16 12.8 Decimal Hexa 10 A 12 C 12 C 12 C = (0.ACCC)16 \ Hexadecimal equivalent is = (12.AC CC)16

Hexadecimal to decimal conversion Example 22: (A3F)16 Decimal Hexa A – 10 3 – 3 F – 15 → 10 × 162 + 3 × 161 + 15 × 160 → 2560 + 48 + 15 → (2623)10 Example 23: (1F63.0EB)16 1 1 F 15 6 6 3 3 0 0 E 14 B 11 3 → 1 × 16 + 15 × 162 + 6 × 161 + 3 × 160 × (0 × 16–1) + (14 × 16–2) + (11 × 16–3) → 4096 + 3840 + 96 + 3 + 0 + 0.0546 + 0.0026 → (8035.0572)10

Hexadecimal to binary number system To represent hexadecimal in binary, represent each HEX number with its 4-bit binary equivalent. Example 24: (34F)16 Hexa Decimal Binary 3 3 0011 4 4 0100 F 15 1111 = (001101001111)2 Example 25: (AFBC . BED) 16

Hexa Decimal Binary A 10 1010 F 15 1111 B 11 1011 C 12 1100 B 11 1011 E 14 1110 D 13 1101 = (1010111110111100.101111101101)2

3.356 | Digital Logic

Binary to hexadecimal number system To convert binary number to a hexadecimal number, starting from the binary point, make groups of 4-bits each on either side of the binary point and replace each 4-bit group by the equivalent hexadecimal digit. Example 26: (11001001)2 1100 1001 → 9 12 → (C9)16 Example 27: (1011011011.01111)2 0010 1101 1011 0111 1000 ⋅ = (2 DB.78)16 2 D B 7 8

Hexadecimal to octal number system The simplest way to convert hexadecimal to octal is, first convert the given hexadecimal number to binary and the Binary number to Octal. Example 28: (C3AF)16 → 001100001110101111 → (141657)8 Example 29: (C6.AE)16 → 0011000110.10101110 → (306.534)8

Octal to hexadecimal number system The simplest way to convert octal to hexadecimal is first convert the given octal number to binary and then the binary number to hexadecimal. Example 30: (775)8 → (000111111101)2 → (1FD)16 Example 31: (34.7)8 → (00011100.1110)2 → (1C.E)16

Complements Complements are used in digital computers to simplify the subtraction operation and for logical manipulation. There are two types of complements for each base r-system. 1. Radix complement (or) r’s complement: the r’s complement of an m digit number N in base r is r m – N for N ≠ 0. For example, N = 0, r’s complement is 0. 2. Diminished radix complement: (or) (r -1)’s complement: Given a number N in base r having m digits, then (r -1)’s complement is (r m - 1) - N. For example, decimal number system will have 10’s complement and 9’s complement. Similarly, binary number system will have 2’s complement and 1’s complement.

Example 32: 10’s complement of (2657)10 is (10)4 – 2657 10000 2657 7343 Example 33: 9’s complement of (2657)10 is (104 -1) - 2657 10000 – 1 9999 2657 7342 •• r’s complement can be obtained by adding 1 to (r – 1)’s complement. r m – N = {(r m - 1) – N} + 1 Example 34: 2’s complement of (101101)2 is = (2)6 – 101101 (26)10 = (100000)2 2’s complement is 100000 –101101 010011 Example 35: 1’s complement of (101101)2 is 26 – 1 = 1000000 – 1 111111 101101 1’s complement –010010 The one’s complement of a binary number is formed by changing 1’s to 0’s and 0’s to 1’s, The 2’s complement can be formed by leaving all least significant 0’s and the first 1 unchanged, and replacing 1’s with zeros and zeros with 1’s in all other bits. If the number M contains radix point, the point should be removed temporarily in order to form r’s/ (r - 1)’s complement. The radix point is then restored to the complemented number in the same relative position. Example 36: What is 1’s complement of (1001.011)2? → Consider without radix point 1001011 Take 1’s complement 0110100 Place radix point again (0110.100)2 Example 37: What is 2’s complement of (1001.011)2? Consider without radix point 1001011 Take 2’s complement 0110101 Place radix point again (0110.101)2 Complement of a complement is equal to the original number r m – (r m – M) = M

Subtraction with Complements Subtraction of two n digit unsigned numbers A - B in base r can be done as follows by r’s complement method. Add A to the r’s complement of B. Mathematically A + (r n - B) = A - B + r n If A ≥ B the sum will produce an end carry r n; which can be discarded. (Discarding carry is equivalent to subtracting r n from result). What is left is the result A – B?

Chapter 1 Number Systems | 3.357 A = 1100 → 1100 B = 1010 (2’s complement) + 0110 Sum: 10010 discard carry (–r n) – 10000 A – B: 0010 If A < B, the sum does not produce an end carry and result is r n - (B - A). Then take r’s complement of the sum, and place a negative sign in front. If A = 1010 B = 1100 A - B can be done as A → 1010 B → 2’s complaint + 0100 Sum: 1110 Here, no carry generated, so result is a negative number. 2’s complement of result → 0010 = 2 result = -2 Subtraction of unsigned numbers by using (r – 1)’s complement can be done in similar way. However, (r – 1)’s complement is one less than the r’s complement. Because of this, the sum produced is one less than the correct difference when an end carry occurs. So end carry will be added to the sum. Removing the end carry and adding 1 to the sum is referred to as an end-around-carry. Consider A = 1100, B = 1010 For A - B A → 1100 B → (1’s complement) + 0101 Sum: 10001 End around carry

+ 1 A - B = 0010

For B - A B → 1010 A → (1’s complement) + 0011 Sum: 1101 There is no end carry, for there result is – (B – A) = –(1’s complement of 1101) = –0010 = –2

Signed Binary Numbers Positive integers can be represented as unsigned numbers; but to represent negative integer, we need a notation for negative values in binary. It is customary to represent the sign with a bit placed in the left most position of the number. The convention is to make the sign bit 0 for positive and 1 for negative. This representation of signed numbers is referred to as sign-magnitude convention S Magnitude +24 is 0 11000 sign magnitude –24 is 1 11000 sign magnitude

Other notation for representation of signed numbers is signed complement system. This is convenient to use in a computer for arithmetic operations. In this system, a negative number is indicated by its complement (i.e., complement of corresponding positive number) whereas the sign-magnitude system negates a number by changing its sign bit, the signed-complement system negates a number by taking its complement. Positive numbers use same notation in sign-magnitude as well as sign-complement systems. The signed-complement system can be used either as the 1’s complement or the 2’s complement. But 2’s complement is the most common. +24 in 1’s/2’s complement representation is 011000 -24 in 1’s complement representation 100111 -24 in 2’s complement representation 101000 Table 2 Signed binary numbers – (4-bits) Decimal

SignedMagnitude

Signed 1’s Complement

Signed 2’s Complement

+7

0111

0111

0111

+6

0110

0110

0110

+5

0101

0101

0101

+4

0100

0100

0100

+3

0011

0011

0011

+2

0010

0010

0010

+1

0001

0001

0001

+0

0000

0000

0000

–0

1000

1111

–

–1

1001

1110

1111

–2

1010

1101

1110

–3

1011

1100

1101

–4

1100

1011

1100

–5

1101

1010

1011

–6

1110

1001

1010

–7

1111

1000

1001

–8

–

–

1000

The ranges of signed binary numbers with n-bits Signed-magnitude: −2n-1 + 1 to +2n-1 − 1 1’s complement representation: −2n-1 + 1 to +2n-1 − 1 2’s complement representation: −2n-1 to +2n-1 − 1 Signed 2’s complement representation can be directly used for arithmetic operations. The carryout of the sign bit position is discarded. In order to obtain a correct answer, we must ensure that the result has a sufficient number of bits to accommodate the sum/product. Example 38: X = 00110, Y = 11100 are represented in 5-bit signed 2’s complement system Then their sum X + Y in 6-bit signed 2’s complemented representation is?

3.358 | Digital Logic Solution: X = 00110 Y = 11100 are 5-bit numbers But result needs to be in 6-bit format. Operands X and Y also should be in 6-bit format X = 000110 Y= 111100 X + Y = (1) 000010 The carry out of sign bit position is discarded result is 000010. Example 39: (36x 70)10 is 10’s complement of (yzyz0)10 Then values of x, y, z are? (A) 4, 5, 2 (B) 4, 6, 3 (C) 3, 6, 3 (D) 3, 5, 4 Solution: (C) (36x70)10 is 10’s complement of (yzyz0)10. 10’s complement of (yzyz0)10 is 105 - yzyz0 = 36 × 70 So 36x70 + yzyz0 = 100000 36x70 +yzyz0 100000 so 7 + z = 10, 1 + x + y = 10 z = 3 1 + 6 + z = 10 y = 6 1 + 3 + y = 10, →x=3 Example 40: The 10’s complement of (843)11 is? (A) (157)11 (B) (267)11 (C) (156)11 (D) (268)11 Solution: (B) Given (843)11 is base 11 number system and the number in the number system range from 0 to 9 & A (A = 10) 10’s complement for (843)11 means (r - 1)’s complement. So (r n - 1) - N = [(11)n - 1] - N (11)n - 1 ⇒ 1000 – 1 AAA – 843 267 10’s complement is (267)11 Example 41: Consider the signed binary number to be 10111011. What is the decimal equivalent of this number if it is in Sign-Magnitude form, or 1’s complement form, or 2’s complement form? Solution: Given binary number = 10111011. As sign bit is 1, it is a negative number. If it is in sign-magnitude format, then MSB is sign bit, and remaining bits represent the magnitude, (0111011)2 = 32 + 16 + 8 + 2 + 1 = 59. So if the given number is in sign-magnitude format, then the number is –59.

If it is in 1’s complement/2’s complement form, then the magnitude of negative number can be obtained by taking 1’s complement/2’s complement for the number, respectively. 10111011 ⇒ 1’s complement ⇒ 01000100 = (68)10. In 1’s complement format, the number is –68. 10111011 ⇒ 2’s complement ⇒ 01000101 = (69)10. In 2’s complement format, the number is –69. Example: Find (–9.625)10 in signed 2’s complement repre sentation. Signed binary fraction can be represented in the same way of signed integer. 2 9 2 4 −1 2 2−0 1− 0 0.625 × 2 = 1.25 0.25 × 2 = 0.5 0.5 × 2 = 1.0 = 0.101 +(9.625) = 01001.101 –9.625 = 10110.011 (by taking 2’s complement)

Binary Multipliers Multiplication of binary number is done in the same way as multiplication of decimal. The multiplicand (m) is multiplied by each bit of the multiplier (N), starting from the LSB. Let M = B3 B2 B1 B0 N = A3 A2 A1 A0 If M × N = P

A3B3 P7P6

A2B3 A3B2 P5

A1B3 A2B2 A3B1 P4

A0B3 A1B2 A2B1 A3B0 P3

A0B2 A1B1 A2B0

A0B1 A1B0

A0B0

P2

P1

P0

=P

Example: Let M = 1 0 1 1 N = 1 1 0 0 M × N = P 1 0 1 1 × 1100 0000 0000 1011 1011 111 _ 10000100=P

Binary Codes Binary codes can be classified as numeric codes and alphanumeric codes. The figure below shows the classification of codes.

Chapter 1 Number Systems | 3.359 Codes

Numeric

Nonweighted

Weighted

Excess 3

Binary

Gray

2421

Selfcomplementing

Five bit BCD codes

2421

BCD

8421

Alphanumeric

3321

8421

5211

4221

Sequential

5311

Cyclic

Reflecting

Gray

Gray ASCII

Excess 3

Parity

Excess 3

5221

Error detecting and correcting

5421

631-1

EBCDIC

Hollerith

Hamming

7421

74-2-1 84-2-1

Numeric Codes

Reflective codes

Numeric Codes are the codes which represent numericals in binary, i.e., only numbers as a series of 0s and 1s.

Binary code in which the n least significant bits for code words 2n through 2n + 1 – 1 are the mirror images of than for 0 through 2n – 1 is called reflective codes. Example: Gray Code

Weighted and non-weighted codes •• The weighted codes are those which obey the positionweighting principle. Each position of a number represents a specific weight. Example: BCD, Binary, 84-2-1, 2421, •• Non-weighted codes are codes which are not assigned fixed values. Example: Excess-3, Gray code 2421, 5211, 84 – 2 – 1 are examples of weighted codes, in which weight is assigned to each position in the number. (27)10 in 2421 code → 0010 1101 (45)10 in 5211 code → 0111 1000 (36)10 in 84 – 2 – 1 code → 0101 1010 Any digit in decimal will be represented by the weights represented by the code.

Error-detecting and correcting codes Codes which allow only error detection are error-detecting codes. Example: Parity Codes which allow error detection as well as correction are called error correcting codes. Example: Hamming codes

Self-complementing codes A code is said to be self-complementing, if the code word of the 9’s complement of number ‘N’, i.e., of “9-N” can be obtained from the code word of ‘N’ by interchanging all the zeros and ones, i.e., by taking 1’s complement. In other words, logical complement of number code is equivalent to representation of its arithmetic complement. Example: 84-2-1, 2421, XS -3. All weighted BCD codes are self-complementing codes.

Binary-coded decimal (BCD) In BCD, each decimal digit 0 to 9 is coded by a 4-bit binary number. BCD codes are convenient to convert to/or from decimal number system.

Decimal BCD Digit

0 0000

1 0001 2 0010 3 0011

Sequential codes

4 0100

A sequential code is one in which each succeeding code word is one binary number greater than the preceding code word. Example: XS–3, BCD

5 0101

Cyclic codes (unit distance codes)

8 1000

Cyclic codes are those in which each successive code word differs from the preceding one in only one bit position. Example: Gray code

6 0110 7 0111 9 1001

Example 42: (628)10 = (0110 0010 1000) BCD

3.360 | Digital Logic BCD addition •• BCD addition is performed by individually adding the corresponding digits of the decimal number expressed in 4-bit binary groups starting from the LSB. •• If there is no carry and the sum term is not an illegal code, no correction is needed. •• If there is a carry out of one group to the next group or if the sum term is an illegal code, the (6)10 is added to the sum term of that group, and the resulting carry is added to the next group.

Excess-3 (XS-3) code

Example 43: 44 + 12 0100 0100 (44 in BCD) 0001 0010 (12 in BCD) 0101 0110 (56 in BCD)

Each gray code number differs from the preceding number by a single bit.

Example 44: 76.9+ 56.6 0111 0110 . 1001 0101 0110 . 0110 1100 1100 . 1111 (all are illegal 0110 0110 . 0110 codes) 0010 0010 . 0101 +1 +1 +1 (propagate carry) 0001 0011 0011 . 0101 1 3 3 . 5 BCD subtraction BCD subtraction is performed by subtracting the digits of each 4-bit group of the subtrahend from the corresponding 4-bit group of the minuend in binary starting from the LSB. Example 45: 42 0100 0010 ( 42 in BCD) −12 −0001 0010 (12 IN BCD) 30 0011 0000 (No borrow, so this is the correct difference) Example 46:

(Borrow 247.7 0010 0100 0111 . 0111 are −156.9 0001 0101 0110 . 1001 present, 90.8 0000 0111 ⋅0000 . 1110 subtract 0110) −01001 − 0110 1001 000 ⋅ 1000 Corrected difference (90.8)

Excess-3 code is a non-weighted BCD code, where each digit binary code word is the corresponding 8421 code word plus 0011. Find the XS-3 code of Example 47: (3)10 → (0011)BCD = (0110)xS3 Example 48: (16)10 → (0001 0110)BCD → (0100 1001)xS3

Gray code Decimal

Gray Code

0 1 2 3 4 5

0000 0001 0011 0010 0110 0111

Binary to gray conversion Step I: Shift the binary number one position to the right, LSB of the shifted number is discarded. Step II: Exclusive or the bits of the binary number with those of the binary number shifted. Example 49: Convert (1001)2 to gray code Binary → 1010 Shifted Binary → 101 ⊕ Gray → 1111 Gray to binary conversion (a) Take the MSB of the binary number is same as MSB of gray code number. (b) X-OR the MSB of the binary to the next significant bit of the gray code. (c) X-OR the 2nd bit of binary to the 3rd bit of Gray code to get 3rd bit binary and so on. (d) Continue this till all the gray bits are exhausted. Example 50: Convert, gray code 1010 to binary 1 0 1 0 Gray 1010 ⇓ ⊕ || ⊕ || ⊕ || 1100 = (1100)2

1

1

0

0

Exercises Practice Problems 1 Directions for questions 1 to 15: Select the correct alternative from the given choices. 1. Assuming all the numbers are in 2’s complement representation, which of the following is divisible by 11110110? (A) 11101010 (B) 11100010 (C) 11111010 (D) 11100111

2. If (84)x (in base x number system) is equal to (64)y (in base y number system), then possible values of x and y are (A) 12, 9 (B) 6, 8 (C) 9, 12 (D) 12, 18 3. Let A = 1111 1011 and B = 0000 1011 be two 8-bit signed 2’s complement numbers. Their product in 2’s complement representation is

Chapter 1 Number Systems | 3.361

(A) 11001001 (C) 11010101

(B) 10011100 (D) 10101101

4. Let r denotes number system’s radix. The only value(s) of r that satisfy the equation 3 (1331) r = (11) r is/are (A) 10 (B) 11 (C) 10 and 11 (D) any r > 3 5. X is 16-bit signed number. The 2’s complement representation of X is (F76A)16. The 2’s complement representation of 8 × X is (A) (1460)16 (B) (D643)16 (C) (4460)16 (D) (BB50)16 6. The HEX number (CD.EF)16 in octal number system is (A) (315.736)8 (B) (513.637)8 (C) (135.673)8 (D) (531.367)8 7. 8-bit 2’s complement representation a decimal number is 10000000. The number in decimal is (A) +256 (B) 0 (C) -128 (D) -256 8. The range of signed decimal numbers that can be represented by 7-bit 1’s complement representation is (A) -64 to + 63 (B) -63 to + 63 (C) -127 to + 128 (D) -128 to +127 9. Decimal 54 in hexadecimal and BCD number system is respectively (A) 63, 10000111 (B) 36,01010100 (C) 66, 01010100 (D) 36, 00110110 10. A new binary-coded hextary (BCH) number system is proposed in which every digit of a base -6 number system is represented by its corresponding 3-bit binary

Practice Problems 2 Directions for questions 1 to 20: Select the correct alternative from the given choices. 1. The hexadecimal representation of (567)8 is (A) 1AF (B) D77 (C) 177 (D) 133 2. (2326)8 is equivalent to (A) (14D6)16 (C) (1283)10

(B) (103112)4 (D) (09AC)16

3. (0.46)8 equivalent in decimal is? (A) 0.59375 (B) 0.3534 (C) 0.57395 (D) 0.3435 4. The 15’s complement of (CAFA)16 is (A) (2051)16 (B) (2050)16 (C) (3506)16 (D) (3505)16 5. 53 in 2’s complement from is? (A) 1001011 (B) 001010 (C) 0110101 (D) 001011 6. Signed 2’s complement representation of (–15)10 is

code. For example, the base -6 number 35 will be represented by its BCH code 011101. In this numbering system, the BCH code 001001101011 corresponds to the following number in base -6 system. (A) 4651 (B) 4562 (C) 1153 (D) 1353 11. The signed 2’s complement representation of (-589)10 in Hexadecimal number system is (A) (F24D)16 (B) (FDB3)16 (C) (F42D)16 (D) (F3BD)16 12. The base of the number system for which the following 66 operation is to be correct = 13 5 (A) 6 (B) 7 (C) 8 (D) 9 13. The solution to the quadratic equation x2 - 11x + 13 = 0 (in number system with radix r) are x = 2 and x = 4. Then base of the number system is (r) = (A) 7 (B) 6 (C) 5 (D) 4 14. The 16’s complement of BADA is (A) 4525 (B) 4526 (C) ADAB (D) 2141 15. (11A1B)8 = (12CD)16, in the above expression A and B represent positive digits in octal number system and C and D have their original meaning in Hexadecimal, the values of A and B are? (A) 2, 5 (B) 2, 3 (C) 3, 2 (D) 3, 5

(A) 11111 (C) 01111

(B) 10001 (D) 10000

7. (0.25)10 in binary number system is? (A) (0.01) (B) (0.11) (C) 0.001 (D) 0.101 8. The equivalent of (25)6 in number system with base 7 is? (A) 22 (B) 23 (C) 24 (D) 26 9. The operation 35 + 26 = 63 is true in number system with radix (A) 7 (B) 8 (C) 9 (D) 11 10. The hexadecimal equivalent of largest binary number with 14-bits is? (A) 2FFF (B) 3FFFF (C) FFFF (D) 1FFFF 11. If x is radix of number system, (211)x = (152)8, then x is (A) 6 (B) 7 (C) 9 (D) 5

3.362 | Digital Logic 12. The value of r for which ( 224) r = (13) r is valid expression, in number system with radix r is? (A) 5 (B) 6 (C) 7 (D) 8 13. Which of the representation in binary arithmetic has a unique zero? (A) sign-magnitude (B) 1’s compliment (C) 2’s complement (D) All of these 14. For the binary number 101101111 the equivalent hexadecimal number is (A) 14E (B) 9E (C) B78 (D) 16F 15. Subtract 1001 from 1110 (A) 0010 (C) 1011

(B) 0101 (D) 1010

16. Which of the following is a positively weighted code? (A) 8421 (B) 84-2-1 (C) EXS-3 (D) 74-2-1

17. Match the items correctly Column 1

Column 2

(P) 8421

(1) Cyclic code

(Q) 2421

(2) self-complementing

(R) 5212

(3) sequential code

(S) Gray code

(4) non-sequential code

(A) P–2, Q–4, R–3, S–1 (B) P–1, Q–4, R–3, S–2 (C) P–3, Q–2, R–4, S–1 (D) P–2, Q–4, R–1, S–2 1 8. Perform the subtraction in XS-3 code 57.6 - 27.8 (A) 0101 1100.1011 (B) 0010 1001.1100 (C) 00011101.1100 (D) 1010 1110.1011 19. The 2’s complement representation of -17 is (A) 101110 (B) 111110 (C) 101111 (D) 110001 20. The decimal 398 is represented in 2421 code by (A) 110000001000 (B) 001110011000 (C) 001111111110 (D) 010110110010

Previous Years’ Questions 1. (1217)8 is equivalent to [2009] (A) (1217)16 (B) (028F)16 (C) (2297)10 (D) (0B17)16 2. P is a 16-bit signed integer. The 2’s complement representation of P is (F87B)16. The 2’s complement representation of 8*P is [2010] (A) (C3D8)16 (B) (187B)16 (C) (F878)16 (D) (987B)16 3. The smallest integer that can be represented by an 8-bit number in 2’s complement form is [2013] (A) –256 (B) –128 (C) –127 (D) 0 4. The base (or radix) of the number system such that the 312 following equation holds is ––––– = 13.1 [2014] 20 5. Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is –––––––. [2014] 6. Consider the equation (43)x = (y3)8 where x and y are unknown. The number of possible solutions is _______ [2015] 7. Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability

mass functions are Pr[Xi = 0] = Pr[Xi = 1] = ½ for i = 1, 2, 3. Define another random variable Y = X1 X2 ⊕ X3, where ⊕ denotes XOR. Then Pr[Y = 0|X3 = 0] = ______

[2015]

8. The 16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is ___ . [2016] 9. Consider an eight - bit ripple - carry adder for computing the sum of A and B, where A and B are integers represented in 2’s complement form. If the decimal value of A is one, the decimal value of B that leads to the longest latency for the sum to stabilize is _____ . [2016] 10. Let x1 ⊕x2 ⊕x3 ⊕x4 = 0 where x1 ,x2 ,x3 ,x4 are Boolean variables, and ⊕ is the XOR operator. Which one of the following must always be TRUE?[2016]

(A) x1 x2 x3 x4 = 0

(B) x1 x3 + x2 = 0 (C) x1 ⊕ x3 = x3 ⊕ x4

(D) x1 + x2 + x3 + x4 = 0

Chapter 1 Number Systems | 3.363

Answer Keys

Exercises Practice Problems 1 1. B 11. B

2. C 12. D

3. A 13. C

4. D 14. B

5. D 15. D

6. A

7. C

8. B

9. B

10. C

4. D 14. D

5. D 15. B

6. B 16. A

7. A 17. C

8. B 18. A

9. B 19. C

10. B 20. C

4. 5

5. 3

6. 5

Practice Problems 2 1. C 11. B

2. B 12. A

3. A 13. C

Previous Years’ Questions 1. B

2. A

3. B

7. 0.75

8. –11

9. –1.0

10. C

Chapter 2 Boolean Algebra and Minimization of Functions LEARNING OBJECTIVES After reading this chapter, you will be able to understand: • • • • • •

Logic gates Boolean algebra AXIOMS and Laws of Boolean algebra Properties of Boolean algebra Conversion from Min term to Max term Minimization of Boolean function

• • • •

K-map method Prime implicant Implementation of function by using NAND-NOR Gates EX-OR, EX-NOR GATE

logic gate s

Table 2 Truth Table Input

1. Inverter or NOT gate (7404 IC): The inverter performs a basic logic operation called inversion or complementation. The purpose of the inverter is to change one logic level to the opposite level. In terms of digital circuits, it converts 1 to 0 and 0 to 1. A

Y Symbol

Output

A

B

Y

0

0

0

0

1

0

1

0

0

1

1

1

3. OR gate (logical adder 7432 IC): The OR gate performs logical, addition commonly known as OR function.

Y=A

Table 1 Truth Table

A B

Y

Input

Output

A

Y

0

1

Symbol Y=A+B

1

0

Figure 2 2 input OR gate

2. AND gate (logical multiplier 7408 IC): The AND gate performs logical multiplication more commonly know as AND function. The AND gate is composed of 2 or more inputs and a single output 2 i/p AND Gate A·B

A B Y = A·B

Figure 1 2 input AND gate

Table 3 Truth Table Input

Output

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

1

Chapter 2 Boolean Algebra and Minimization of Functions | 3.365 4. NAND gate (7400 IC): The NAND gate’s function is basically AND + NOT function. A B

if both inputs are unequal, then the output will be low. Other name for X-NOR gate is equivalent gate. A B

Y Y = A·B

Figure 3 2 input NAND gate

Y Symbol Y = A B = AB + A B

Figure 6 2 input X-NOR Gate

Table 4 Truth Table Input

A

B

A⋅B

A + B (Y)

0

0

0

1

0

1

0

1

1

0

0

1

1

1

1

0

5. NOR gate (7402 IC): The NOR gate is basically OR + NOT function. A B Y=A+B Figure 4 2 input NOR gate Table 5 Truth Table Input

Output

A

B

A+B

A + B (Y)

0

0

0

1

0

1

1

0

1

0

1

0

1

1

1

0

6. Exclusive OR gate X-OR (7486 IC): X-OR is a gate in which unequal inputs create a high logic level output and if both inputs are equal, the output will be low. Other name for EX-OR gate is unequivalent gate. 2 input X-OR Gate Y

X-NOR Gate is complement of X-OR Gate.

Boole an Alge b ra

AXIOMS and Laws of Boolean Algebra 1. AXIOMS (a) AND operation (1) 0 ⋅ 0 = 0 (2) 0 ⋅ 1 = 0 (3) 1 ⋅ 0 = 0 (4) 1 ⋅ 1 = 1 (b) OR operation (5) 0 + 0 = 0 (6) 0 + 1 = 1 (7) 1 + 0 = 1 (8) 1 + 1 = 1 (c) NOT operation (9) 1 = 0 (10) 0 = 1

Symbol Y = A ⊕ B = AB AB Figure 5 2 input X-OR Gate

2. Laws (a) Complementation law (1) 0 = 1

Table 6 Truth Table

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

0

Y 1 0 0 1

Boolean algebra is a system of mathematical logic. It is an algebraic system consisting of the set of elements (0, 1), two binary operators OR and AND and one unary operator NOT. The Boolean algebra is governed by certain well-developed rules and laws.

Y

A B

Table 7 Truth Table A B 0 0 0 1 1 0 1 1

Output

7. Exclusive NOR gate (X-NOR): X-NOR is a gate in which equal inputs create a high logic level output; and

(2) 1 = 0 (3) If A = 0, then A = 1 (4) If A = 1, then A = 0 (5) A = A

(b) AND laws (1) A ⋅ 0 = 0 (NULL Law) (4) A ⋅ 1 = A (Identity Law) (3) A ⋅ A = A (4) A ⋅ A = 0

3.366 | Digital Logic (c) OR laws (1) A + 0 = A (NULL Law) (2) A + 1 = 1 (Identity Law) (3) A + A = A (4) A + A = 1 (d) Commutative laws (1) A + B = B + A (2) A ⋅ B = B ⋅ A (e) Associative laws (1) (A + B) + C = A + (B + C) (2) (A ⋅ B)C = A( B ⋅ C) (f) Distributive laws (1) A(B + C) = AB + AC (2) A + BC = (A + B) (A + C) (g) Redundant literal rule (RLR) (1) A + AB = A + B (2) A( A + B ) = AB (h) Idempotence laws (1) A ⋅ A = A (2) A + A = A (i) Absorption laws (1) A + A ⋅ B = A (2) A (A + B) = A 3. Theorems (a) Consensus theorem Theorem 1: AB + AC + BC = AB + AC Proof: LHS = AB + AC + BC = AB + AC + BC ( A + A) = AB + AC + BCA + BCA = AB(1 + C ) + AC (1 + B ) = AB(1) + AC (1) = AB + AC = RHS. Theorem 2: ( A + B) ( A + C ) ( B + C ) = ( A + B) ( A + C ) Proof: LHS = ( A + B) ( A + C ) ( B + C ) = ( AA + AC + BA + BC ) ( B + C ) = ( AC + BC + AB ) ( B + C )

= 0 + AC + AB + BC = AC + AB + BC ( A + A) = AB + ABC + AC + ABC

= AB + AC = LHS

(c) De Morgan’s theorem

Law 1: A + B = A ⋅ B This law states that the complement of a sum of variable is equal to the product of their individual complements.

Law 2: AB = A + B This law states that the complement of the product of variables is equal to the sum of their individual complements. Example 1: Simplify the Boolean function Y = A( A + B ) Y = A ⋅ A + A⋅ B Solution: Y = A + AB = A(1 + B ) =A Example 2: Simplify the Boolean function Y = A + AB Solution: Y = A ⋅ ( B + 1) + A ⋅ B = A ⋅ B + A + AB = B ( A + A) + A = A+ B Example 3: Simplify the Boolean function Y = A( A + B) + B( A + B) Solution: Y = A ⋅ A + A ⋅ B + B ⋅ A + B ⋅ B = A + B ( A + A) + B = A + B ⋅1 + B = A+ B + B = A+ B Example 4: Simplify the Boolean function Y = ABC + ABC + ABC + ABC Solution: Y = AC ( B + B) + AC ( B + B )

= ABC + BC + AB + AC + BC + ABC

= AC + AC

= AC + BC + AB

= C ( A + A)

RHS = ( A + B) ( A + C ) = AA + AC + BC + AB = AC + BC + AB = LHS. (b) Transposition theorem AB + AC = ( A + C ) ( A + B) Proof: RHS = ( A + C ) ( A + B) = AA + CA + AB + CB

=C Example 5: Simplify the Boolean function Y = ABC + ABC + ABC Solution: = AC ( B + B ) + ABC = AC + ABC = A(C + BC ) = A(C + B) = A+C B

Chapter 2 Boolean Algebra and Minimization of Functions | 3.367 Example 6: Simplify the Boolean function Y = AB + C B + CA + ABD

So the expression inside the parentheses must be evaluated before all the operations. The next operation to be performed is the complement and then follows AND and finally the OR.

Solution: Y = AB(1 + D ) + C B + CA = AB + CB + CA = AB + CB

Prop e rtie s of Boole an Alge b ra n

With n variables, maximum possible distinct functions = 22 . Duality consider the distributive law 1. x (y + z) = xy + xz 2. x + yz = (x + y) (x + z) Second one can be obtained from the first law if the binary operators and the identity elements are interchanged. This important property of Boolean algebra is called the duality principle. The dual of an algebraic expression can be written by interchanging OR and AND operators, 1s by 0, and 0s by 1s. Dual Example 7: x + x ′ = 1 ← → x ⋅ x′ = 0 Dual Solution: xy + xy ′ = x ← → ( x + y ) ( x + y ′) = x Dual x + x ′y = x + y ← → x( x ′ + y ) = xy

Example 8: The dual of F = xy + xz + yz is? Solution: Dual of F = (x + y) (x +z) (y + z) = (x + xz + xy + yz) (y + z) = xy + yz + xz So dual of xy + xz + yz is same as the function itself; For N variables maximum possible self-dual functions n−1 n = 2 2 = 2( 2 / 2 ) Example 9: Which of the following statement/s is/are true S1: The dual of NAND function is NOR S2: The dual of X-OR function is X-NOR (A) S1 and S2 are true (B) S1 is true (C) S2 is true (D) None of these Solution: (A) NAND = (xy)′ = x′ + y′ Dual of NAND = (x + y)′ = x′ y′ X-OR = xy′ + x′y Dual of X-OR = (x + y′) (x′ + y) = xy + x′y′ = X-NOR Both S1 and S2 are true Operator precedence The operator precedence for evaluating Boolean expression is 1. Parentheses 2. NOT 3. AND 4. OR

Complement of function The complement of a function F is F′ is obtained from an interchange of 0s for 1s and 1s for 0s in the value of F. The complements of a function may be derived algebraically through De Morgan’s theorems. (x1 . x2 . x3… xn)′ = x1′ + x′2 + x3′ + … + xn′ (x1 + x2 + x3+ … +xn)′ = x′1 . x2′ . x′3 . x′4 … x′n Example 10: The complement of function F = a(b′c + bc′) is? Solution: (F)′ = [a(b′c + bc′)]′ = a′ + (b′c + bc′)′ = a′ + (b′c)′ ⋅ (bc′)′ = a′ + (b + c)′ (b′ + c) F′ = a′ + bc + b′c′ x y

xy

x y x y x y x y x y

x y

x·y = x + y x+y

x·y x y

x+y

x y

x + y = xy x + y = xy x+y=x+y

x+y x y x y x y

xy xy x+y

Figure 7 Gates with inverted inputs

Boole an Functions, Min Te rm and Max Te rm s

s

The starting point for designing most logic circuits is the truth table, which can be derived from the statement of problem. The truth table is then converted into a Boolean expression and finally create the assembly of logic gates accordingly. Let us consider the example of majority circuit. This circuit takes three inputs (A, B, C) and have one output (Y) which will give the majority of the inputs, i.e., if A, B, C are having more number of zeros. Y = 0 else if A, B, C are having more number of 1s, Y = 1. So from the statements we can derive the truth table as follows: A

B

C

Majority circuit

Y

As we are using three Boolean variables A, B, C, total number of combinations in truth table are 23 = 8.

3.368 | Digital Logic Similarly for n variables, the truth table will have total of 2n combinations, for a Boolean function. Input

Output

Sl. no.

A

B

C

Y

1

0

0

0

0

2

0

0

1

0

3

0

1

0

0

4

0

1

1

1

5

1

0

0

0

6

1

0

1

1

7

1

1

0

1

8

1

1

1

1

→ →

Y = 0, If inputs are having more zeros. Y = 1, If inputs are having more 1’s

For some combinations, output Y = 1, and for others Y = 0. The input combinations for which output Y = 1 are called as min terms. Similarly the input combinations for which output Y = 0 are called as max terms. Min terms are expressed as product terms, Similarly, max terms are expressed as sum terms. The output Y = 1, only in rows 4, 6, 7, 8. So the min terms combinations are 011, 101, 110, 111 in Boolean Algebra, 1 input will be written as A, B, C and 0 input will be written as A, B , C in complement form, we express these min terms as product terms, ABC , ABC , ABC , ABC . To express Y as Boolean expression, we can write it as sum of the min terms. Y = AB C + AB C + ABC + ABC We know that AND operation is a product while OR is sum. So the above equation is a sum of the products (SOP), (or) min terms expression. The other way of expressing Y is Y = Sm (3, 5, 6, 7). Y = m3 + m5 + m6 + m7. The min term numbers are the decimal equivalent of input binary combinations. Similar to SOP we can have product of sums (POS) Boolean expression. The output Y = 0 for the input combinations 000, 001, 010, 100. For max terms 1 input will be indicated as A, B , C in complement form, 0 input will be indicated as A, B, C and max terms are expressed as sum terms. A + B + C, A + B + C , A + B + C , A + B + C Any function can be expressed as product of max terms. So Y = ( A + B + C ) ( A + B + C ) ( A + B + C ) ( A + B + C ) The above equation is a product of sum expression (POS) or max terms expression. In other way Y = pM (0, 1, 2, 4) = M0 ⋅ M1 ⋅ M2 ⋅ M4 The max term numbers are decimal equivalents of corresponding input binary combinations.

Min Term and Max Term All the Boolean expressions can be expressed in a standard sum of product (SOP) form or in a standard product of sum (POS) form. •• A standard SOP form is one in which a number of product terms, each contains all the variables of the function either in complement or non-complement form are summed together. •• A standard POS form is one in which a number of sum terms, each one of which contain all the variable of he function either in complemented or non-complemented form are multiplied together. •• Each of the product term in standard SOP form is called a min term. •• Each of the sum term in the standard POS form is called a max term.

Conversion from min terms to max terms representation Y = ABC + ABC + ABC + ABC Y ′ = ( ABC + ABC + ABC + ABC )′ = ( ABC )′( ABC )′( ABC )′( ABC )′ (Y ′)′ = [( A + B + C )( A + B + C )( A + B + C )( A + B + C )]′ = [p(3, 5, 6, 7)]1 = p(0, 1, 2 ,4) Y = ( A + B + C )( A + B + C )( A + B + C )( A + B + C ) or Y = S(3, 5, 6, 7) = p(0, 1, 2, 4)

Conversion from normal SOP/POS form to canonical SOP/POS Let us consider f ( A, B, C ) = A + BC + AC The above function is in normal (minimized) SOP from, to convert this function to standard SOP(or) canonical SOP form, include missing variable in each and every term, to make it complete. First term A, Missing literals are B, C. Consider A X X, so possible combinations are ABC , ABC , ABC , ABC or we can write A = A = A( B + B)(C + C ) = ABC + ABC + ABC + ABC Second term BC-missing literal is A. Consider XBC ⇒ So possible combinations are ABC, ABC or we can write BC = ( A + A) BC = ABC + ABC

Third term AC = missing literal is B. Consider AXC → so possible combinations are ABC , A, BC or we can write

AC = A( B + B )C = ABC + ABC

Chapter 2 Boolean Algebra and Minimization of Functions | 3.369 m0 m1 m3 m2 m4 m5 m7 m6

Now, f ( A, B, C ) = ABC + ABC + ABC + ABC + ABC , after removing the redundant terms. Now consider

yz 01 00 11 10 x 0 x ′y ′z x ′y ′z x ′yz x ′y z ′ 1 xy ′z ′ xy ′z xyz xyz ′

f ( A, B, C ) = ( A + B ) ( A + C ) To convert this expression to canonical form or standard POS form we can write

3-variable K-map

f ( A, B, C ) = ( A + B + C ⋅ C ) ( A + B ⋅ B + C ) Here the C variables is absent from first term and B from second term. So add C ⋅ C = (0) to first, and B ⋅ B to second, and using distributive law arrive at the result.

Four-variable K-maps The K-map for four variables is shown here, 16 min terms are assigned to 16 squares. yz 00 wx 00 0

f ( A, B, C ) = ( A + B + C ) ( A + B + C ) ( A + B + C )( A + B + C )

Minim

ization ofBoole an Functions

Karnaugh Map (K-map) Method Karnaugh map method is a systematic method of simplifying the Boolean expression. K-map is a chart or a graph composed of an arrangement of adjacent cell, each representing a particular combination of variable in sum or product form. (i.e., min term or max term).

Two-variable K-map x y 0 0 1 1

0 1 0 1

F m0 m1 m2 m3

m0 m1 m2 m3

x 0 1

y 0

1

x ′y ′ x ′y xy ′ xy

Three-variable K-map A three-variable map will have eight min terms (for three variables 23 = 8) represented by 8 squares

11

10

1

3

2

4

5

7

6

11 12

13

15

14

10

9

11

10

01

Simplification Procedure •• Obtain truth table, and write output in canonical form or standard form •• Generate K-map! •• Determine Prime implicants. •• Find minimal set of prime implicants.

01

8

The map is considered to lie on a surface with the top and bottom edges as well as the right and left edge touching each other to form adjacent squares. •• One square → a min term of four literals •• Two adjacent square → a term of three literals •• Four adjacent square → a term of two literals •• Eight adjacent square → a term of one literal •• Sixteen adjacent square → The constant one

Don’t-care Combinations It can often occur that for certain input combinations, the value of the output is unspecified either because the input combination are invalid or because the precise value of the output is of no consequence. The combination for which the values of the expression are not specified are called don’t-care combinations. During the process of design using an SOP, K-map, each don’t-care is treated as a 1 if it is helpful in Map Reduction, otherwise it is treated as a 0 and left alone. During the process of design using a POS K-map, each Don’t-care is treated as a 0 if it is useful in Map Reduction, otherwise it is treated as a 1 and left alone. Example 11: Find the Minimal expression Sm (1, 5, 6, 12, 13, 14) + d (2, 4)

x

y

z

F

0

0

0

m0

0

0

1

m1

0

1

0

m2

0

1

1

m3

1

0

0

m4

1

0

1

m5

10

1

1

0

m6

1

1

1

m7

Solution:

CD 00 AB 00

01 1

10 ×

01

×

1

1

11

1

1

1

11

∴ F = BC + BD + ACD

3.370 | Digital Logic

Pairs, Quads and Octets

Octet

BC 00

A 0

01

11

1

1

The group of eight cells will form one octet.

10

1

The map contains a pair of 1s those are horizontally adjacent. Those cells represent A BC , ABC . For these two min terms, there is change in the form of variable B. By combining these two cells we can form a pair, which is equal to A BC + ABC = AC ( B + B) = AC . If more than one pair exists on K-map, OR the simplified products to get the Boolean function. CD 00 AB 00 1

BC 00

A 0

01

11

1

1

1

10

01

1

11 10

1

F = AC + AC

01 1

11 1

1

01

1

1

1

1

11

1

1

1

1

F=Y Other variable X, Z, W changes their form in octet. Octet can eliminate three variables and their complements.

1

01

11

10 1

01

1

1

1

11

1

1

1

10

1

1

F=D

F = AC D + A BD + AB C + AC D

Other variable A, B, C are vanished.

Eliminating Redundant Groups

BC

BC

1

10

1

Quad is a group of four 1s those are horizontally or vertically adjacent.

0

11

CD 00 AB 00 1

Quad

00

01

10

10

So Pair eliminates one variable by minimization.

A

ZW 00 XY 00

01

11

1

1

1

10

00

01

11

0

1

1

1

1

1

A

1

10

F = C F = AC + AC = ( A + A)C = C By considering two pairs also it will be simplified to C. Quad eliminates two variables from the function CD 00 AB 00 01

1

11

1

10

01 1

11 1

10

1

11

10 1

01

1

1

10

00

A 0

1

1

01

11

1

1

10

1

1

1

AB + AC + BC AB + AC Here BC is redundant pair, which covers already covered min terms of AB, AC . RS 00 PQ 00 11

1

01

11

10

1

1

1

1

1

1

10

1

RS 00 PQ 00 01 11 10

11 10

1

11

This K-map gives fourpairs and one quad.

Corner min terms can from a Quad 01

0

01

1

F = BD + BD

RS 00 PQ 00 1

BC 00

01 1

1

BC A

1

1

F = QS

1

01

11

10

1

1

1

1

1

1

1

But only four pairs are enough to cover all the min times, Quad is not necessary. P RS + PQR + PQR + PRS is minimized function.

Chapter 2 Boolean Algebra and Minimization of Functions | 3.371

Prime Implicant

A

B

C

The group of adjacent min terms is called a Prime Implicant, i.e., all possible pairs, quads, octets, etc. BC 00

01

0

1

1

1

1

A

11

10 1 1

F

Prime implicants are B C , BC , C , A B. Minimized function is C + A B

Essential Prime Implicant The prime implicant which contains at least one min term which cannot be covered by any other prime implicant is called Essential prime implicant. BC 00

A

1

0

01

10

1

1

1 1

1

11

Figure 8 Logic diagram

Example 13: Find the minimal expression for Sm (2, 3, 6, 7, 8, 10, 11, 13, 14) Solution: K-map is: CD AB

Min term 0, 6 can be grouped with only one pair each. The total possible prime implicants are A B, BC , AC , AB but min term 0, 6 can be covered with A B, AB . So we call them as essential prime implicants. Min term 5 can be paired with any of 1 or 7 min term. ZW 00 XY 00 1

01

11

1

1

1

01 11

00 01

1 1

10

The essential prime Implicants are XZW , X Y Z

10

The prime implicant whose min terms are already covered by at least one min term is called redundant prime implicants.

f2

0

00

01

1

1 1

1

11

10

1

Here prime implicants are A B , AC , BC . But BC is already covered by other min terms So BC is redundant prime implicant. Example 12: Find the minimal expression for Sm(1, 2, 4, 6, 7) and implement it using Basic gates. Solution: K-map is BC 00

A

11

1

10 1

1

0 1

01

1

1

F = AC + AB + BC + BC + ABC

4

5 1

8

1 1

13 9

10 3 7

15

1

11

1 1 1 1

2 6 14 10

Example 14: Three Boolean functions are defined as below f1 = Sm(0, 1, 3, 5, 6), f2 = Sm(4, 6, 7) f3 = Sm(1, 4, 5, 7), then find f. f1

BC

1

11 1

∴ F ( A, B, C , D ) = ABCD + ABD + AC + BC + CD