The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

engineeringmathematics-131205224415-phpapp01

Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by siti yasmin, 2019-11-28 21:30:55

ENGINEERING MATH

engineeringmathematics-131205224415-phpapp01

Engineering Mathematics

In memory of Elizabeth

Engineering Mathematics

Fourth Edition
JOHN BIRD, BSc(Hons) CMath, FIMA, CEng, MIEE, FCollP, FIIE

Newnes

OXFORD AMSTERDAM BOSTON LONDON NEW YORK PARIS
SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO

Newnes
An imprint of Elsevier Science
Linacre House, Jordan Hill, Oxford OX2 8DP
200 Wheeler Road, Burlington MA 01803

First published 1989
Second edition 1996
Reprinted 1998 (twice), 1999
Third edition 2001
Fourth edition 2003
Copyright  2001, 2003, John Bird. All rights reserved

The right of John Bird to be identified as the author of this work
has been asserted in accordance with the Copyright, Designs and
Patents Act 1988
No part of this publication may be reproduced in any material
form (including photocopying or storing in any medium by
electronic means and whether or not transiently or incidentally to some
other use of this publication) without the written permission of the
copyright holder except in accordance with the provisions of the Copyright,
Designs and Patents Act 1988 or under the terms of a licence issued by the
Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London,
England W1T 4LP. Applications for the copyright holder’s written
permission to reproduce any part of this publication should be
addressed to the publisher

Permissions may be sought directly from Elsevier’s Science and Technology Rights
Department in Oxford, UK: phone: (+44) (0) 1865 843830; fax: (+44) (0) 1865
853333; e-mail: [email protected] You may also complete your request
on-line via the Elsevier Science homepage (http://www.elsevier.com), by selecting
‘Customer Support’ and then ‘Obtaining Permissions’

British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library

ISBN 0 7506 5776 6

For information on all Newnes publications visit our website at www.Newnespress.com

Typeset by Laserwords Private Limited, Chennai, India
Printed and bound in Great Britain

Contents

Preface xi 7 Partial fractions 51
7.1 Introduction to partial fractions 51
Part 1 Number and Algebra 1 7.2 Worked problems on partial fractions
with linear factors 51
1 Revision of fractions, decimals and 7.3 Worked problems on partial fractions
percentages 1 with repeated linear factors 54
1.1 Fractions 1 7.4 Worked problems on partial fractions
1.2 Ratio and proportion 3 with quadratic factors 55
1.3 Decimals 4
1.4 Percentages 7 8 Simple equations 57
8.1 Expressions, equations and
2 Indices and standard form 9 identities 57
2.1 Indices 9 8.2 Worked problems on simple
2.2 Worked problems on indices 9 equations 57
2.3 Further worked problems on 8.3 Further worked problems on simple
indices 11 equations 59
2.4 Standard form 13 8.4 Practical problems involving simple
2.5 Worked problems on standard equations 61
form 13 8.5 Further practical problems involving
2.6 Further worked problems on standard simple equations 62
form 14
Assignment 2 64
3 Computer numbering systems 16
3.1 Binary numbers 16 9 Simultaneous equations 65
3.2 Conversion of binary to decimal 16 9.1 Introduction to simultaneous
3.3 Conversion of decimal to binary 17 equations 65
3.4 Conversion of decimal to binary via 9.2 Worked problems on simultaneous
octal 18 equations in two unknowns 65
3.5 Hexadecimal numbers 20 9.3 Further worked problems on
simultaneous equations 67
4 Calculations and evaluation of 9.4 More difficult worked problems on
formulae 24 simultaneous equations 69
4.1 Errors and approximations 24 9.5 Practical problems involving
4.2 Use of calculator 26 simultaneous equations 70
4.3 Conversion tables and charts 28
4.4 Evaluation of formulae 30 10 Transposition of formulae 74
10.1 Introduction to transposition of
Assignment 1 33 formulae 74
10.2 Worked problems on transposition of
5 Algebra 34 formulae 74
5.1 Basic operations 34 10.3 Further worked problems on
5.2 Laws of Indices 36 transposition of formulae 75
5.3 Brackets and factorisation 38 10.4 Harder worked problems on
5.4 Fundamental laws and precedence 40 transposition of formulae 77
5.5 Direct and inverse proportionality 42
11 Quadratic equations 80
6 Further algebra 44 11.1 Introduction to quadratic equations 80
6.1 Polynomial division 44 11.2 Solution of quadratic equations by
6.2 The factor theorem 46 factorisation 80
6.3 The remainder theorem 48

vi CONTENTS Multiple choice questions on chapters 1 to
16 127
11.3 Solution of quadratic equations by
‘completing the square’ 82 Part 2 Mensuration 131

11.4 Solution of quadratic equations by 17 Areas of plane figures 131
formula 84 17.1 Mensuration 131
17.2 Properties of quadrilaterals 131
11.5 Practical problems involving quadratic 17.3 Worked problems on areas of plane
equations 85 figures 132
17.4 Further worked problems on areas of
11.6 The solution of linear and quadratic plane figures 135
equations simultaneously 87 17.5 Worked problems on areas of
composite figures 137
12 Logarithms 89 17.6 Areas of similar shapes 138
12.1 Introduction to logarithms 89
12.2 Laws of logarithms 89 18 The circle and its properties 139
12.3 Indicial equations 92 18.1 Introduction 139
12.4 Graphs of logarithmic functions 93 18.2 Properties of circles 139
18.3 Arc length and area of a sector 140
Assignment 3 94 18.4 Worked problems on arc length and
sector of a circle 141
13 Exponential functions 95 18.5 The equation of a circle 143
13.1 The exponential function 95
13.2 Evaluating exponential functions 95 19 Volumes and surface areas of
13.3 The power series for ex 96 common solids 145
13.4 Graphs of exponential functions 98 19.1 Volumes and surface areas of
13.5 Napierian logarithms 100 regular solids 145
13.6 Evaluating Napierian logarithms 100 19.2 Worked problems on volumes and
13.7 Laws of growth and decay 102 surface areas of regular solids 145
19.3 Further worked problems on volumes
14 Number sequences 106 and surface areas of regular
14.1 Arithmetic progressions 106 solids 147
14.2 Worked problems on arithmetic 19.4 Volumes and surface areas of frusta of
progression 106 pyramids and cones 151
14.3 Further worked problems on arithmetic 19.5 The frustum and zone of a sphere 155
progressions 107 19.6 Prismoidal rule 157
14.4 Geometric progressions 109 19.7 Volumes of similar shapes 159
14.5 Worked problems on geometric
progressions 110 20 Irregular areas and volumes and mean
14.6 Further worked problems on geometric values of waveforms 161
progressions 111 20.1 Areas of irregular figures 161
14.7 Combinations and permutations 112 20.2 Volumes of irregular solids 163
20.3 The mean or average value of a
15 The binomial series 114 waveform 164
15.1 Pascal’s triangle 114
15.2 The binomial series 115 Assignment 5 168
15.3 Worked problems on the binomial
series 115 Part 3 Trigonometry 171
15.4 Further worked problems on the
binomial series 117 21 Introduction to trigonometry 171
15.5 Practical problems involving the 21.1 Trigonometry 171
binomial theorem 120 21.2 The theorem of Pythagoras 171
21.3 Trigonometric ratios of acute
16 Solving equations by iterative angles 172
methods 123
16.1 Introduction to iterative methods 123
16.2 The Newton–Raphson method 123
16.3 Worked problems on the
Newton–Raphson method 123

Assignment 4 126

21.4 Fractional and surd forms of CONTENTS vii
trigonometric ratios 174
25.7 Worked problems (iv) on trigonometric
21.5 Solution of right-angled triangles 175 equations 212
21.6 Angles of elevation and
26 Compound angles 214
depression 176 26.1 Compound angle formulae 214
21.7 Evaluating trigonometric ratios of any 26.2 Conversion of a sin ωt C b cos ωt into
R sin ωt C ˛) 216
angles 178 26.3 Double angles 220
21.8 Trigonometric approximations for small 26.4 Changing products of sines and cosines
into sums or differences 221
angles 181 26.5 Changing sums or differences of sines
and cosines into products 222
22 Trigonometric waveforms 182
22.1 Graphs of trigonometric functions 182 Assignment 7 224
22.2 Angles of any magnitude 182
22.3 The production of a sine and cosine Multiple choice questions on chapters 17
wave 185 to 26 225
22.4 Sine and cosine curves 185
22.5 Sinusoidal form A sin ωt š ˛ 189 Part 4 Graphs 231
22.6 Waveform harmonics 192
27 Straight line graphs 231
23 Cartesian and polar co-ordinates 194 27.1 Introduction to graphs 231
23.1 Introduction 194 27.2 The straight line graph 231
23.2 Changing from Cartesian into polar 27.3 Practical problems involving straight
co-ordinates 194 line graphs 237
23.3 Changing from polar into Cartesian
co-ordinates 196 28 Reduction of non-linear laws to linear
23.4 Use of R ! P and P ! R functions on form 243
calculators 197 28.1 Determination of law 243
28.2 Determination of law involving
Assignment 6 198 logarithms 246

24 Triangles and some practical 29 Graphs with logarithmic scales 251
applications 199 29.1 Logarithmic scales 251
24.1 Sine and cosine rules 199 29.2 Graphs of the form y D axn 251
24.2 Area of any triangle 199 29.3 Graphs of the form y D abx 254
24.3 Worked problems on the solution of 29.4 Graphs of the form y D aekx 255
triangles and their areas 199
24.4 Further worked problems on the 30 Graphical solution of equations 258
solution of triangles and their 30.1 Graphical solution of simultaneous
areas 201 equations 258
24.5 Practical situations involving 30.2 Graphical solution of quadratic
trigonometry 203 equations 259
24.6 Further practical situations involving 30.3 Graphical solution of linear and
trigonometry 205 quadratic equations simultaneously
263
25 Trigonometric identities and 30.4 Graphical solution of cubic equations
equations 208 264
25.1 Trigonometric identities 208
25.2 Worked problems on trigonometric 31 Functions and their curves 266
identities 208 31.1 Standard curves 266
25.3 Trigonometric equations 209 31.2 Simple transformations 268
25.4 Worked problems (i) on trigonometric 31.3 Periodic functions 273
equations 210 31.4 Continuous and discontinuous
25.5 Worked problems (ii) on trigonometric functions 273
equations 211 31.5 Even and odd functions 273
25.6 Worked problems (iii) on trigonometric 31.6 Inverse functions 275
equations 212
Assignment 8 279

viii CONTENTS 38.3 Worked problems on probability 327
38.4 Further worked problems on
Part 5 Vectors 281
probability 329
32 Vectors 281 38.5 Permutations and combinations 331
32.1 Introduction 281
32.2 Vector addition 281 39 The binomial and Poisson distribution 333
32.3 Resolution of vectors 283 39.1 The binomial distribution 333
32.4 Vector subtraction 284 39.2 The Poisson distribution 336

33 Combination of waveforms 287 Assignment 10 339
33.1 Combination of two periodic
functions 287 40 The normal distribution 340
33.2 Plotting periodic functions 287 40.1 Introduction to the normal distribution
33.3 Determining resultant phasors by 340
calculation 288 40.2 Testing for a normal distribution 344

Part 6 Complex Numbers 291 41 Linear correlation 347
41.1 Introduction to linear correlation 347
34 Complex numbers 291 41.2 The product-moment formula for
34.1 Cartesian complex numbers 291 determining the linear correlation
34.2 The Argand diagram 292 coefficient 347
34.3 Addition and subtraction of complex 41.3 The significance of a coefficient of
numbers 292 correlation 348
34.4 Multiplication and division of complex 41.4 Worked problems on linear
numbers 293 correlation 348
34.5 Complex equations 295
34.6 The polar form of a complex 42 Linear regression 351
number 296 42.1 Introduction to linear regression 351
34.7 Multiplication and division in polar 42.2 The least-squares regression lines 351
form 298 42.3 Worked problems on linear
34.8 Applications of complex numbers 299 regression 352

35 De Moivre’s theorem 303 43 Sampling and estimation theories 356
35.1 Introduction 303 43.1 Introduction 356
35.2 Powers of complex numbers 303 43.2 Sampling distributions 356
35.3 Roots of complex numbers 304 43.3 The sampling distribution of the
means 356
Assignment 9 306 43.4 The estimation of population
parameters based on a large sample
Part 7 Statistics 307 size 359
43.5 Estimating the mean of a population
36 Presentation of statistical data 307 based on a small sample size 364
36.1 Some statistical terminology 307
36.2 Presentation of ungrouped data 308 Assignment 11 368
36.3 Presentation of grouped data 312
Multiple choice questions on chapters 27
37 Measures of central tendency and to 43 369
dispersion 319
37.1 Measures of central tendency 319 Part 8 Differential Calculus 375
37.2 Mean, median and mode for discrete
data 319 44 Introduction to differentiation 375
37.3 Mean, median and mode for grouped 44.1 Introduction to calculus 375
data 320 44.2 Functional notation 375
37.4 Standard deviation 322 44.3 The gradient of a curve 376
37.5 Quartiles, deciles and percentiles 324 44.4 Differentiation from first
principles 377
38 Probability 326
38.1 Introduction to probability 326
38.2 Laws of probability 326

44.5 Differentiation of y D axn by the CONTENTS ix
general rule 379
49.6 Worked problems on integration using
44.6 Differentiation of sine and cosine the tan  substitution 424
functions 380
Assignment 13 425
44.7 Differentiation of eax and ln ax 382
50 Integration using partial fractions 426
45 Methods of differentiation 384 50.1 Introduction 426
45.1 Differentiation of common functions 50.2 Worked problems on integration using
384 partial fractions with linear
45.2 Differentiation of a product 386 factors 426
45.3 Differentiation of a quotient 387 50.3 Worked problems on integration using
45.4 Function of a function 389 partial fractions with repeated linear
45.5 Successive differentiation 390 factors 427
50.4 Worked problems on integration using
46 Some applications of differentiation 392 partial fractions with quadratic
46.1 Rates of change 392 factors 428
46.2 Velocity and acceleration 393
46.3 Turning points 396 q
46.4 Practical problems involving maximum 51 The t = substitution 430
and minimum values 399
46.5 Tangents and normals 403 2
46.6 Small changes 404 51.1 Introduction 430

Assignment 12 406 Â
51.2 Worked problems on the t D tan
Part 9 Integral Calculus 407
2
47 Standard integration 407 substitution 430
47.1 The process of integration 407 51.3 Further worked problems on the
47.2 The general solution of integrals of the
form axn 407 Â
47.3 Standard integrals 408 t D tan substitution 432
47.4 Definite integrals 411
2
48 Integration using algebraic substitutions
414 52 Integration by parts 434
48.1 Introduction 414 52.1 Introduction 434
48.2 Algebraic substitutions 414 52.2 Worked problems on integration by
48.3 Worked problems on integration using parts 434
algebraic substitutions 414 52.3 Further worked problems on integration
48.4 Further worked problems on integration by parts 436
using algebraic substitutions 416
48.5 Change of limits 416 53 Numerical integration 439
53.1 Introduction 439
49 Integration using trigonometric 53.2 The trapezoidal rule 439
substitutions 418 53.3 The mid-ordinate rule 441
49.1 Introduction 418 53.4 Simpson’s rule 443
49.2 Worked problems on integration of
sin2 x, cos2 x, tan2 x and cot2 x 418 Assignment 14 447
49.3 Worked problems on powers of sines
and cosines 420 54 Areas under and between curves 448
49.4 Worked problems on integration of 54.1 Area under a curve 448
products of sines and cosines 421 54.2 Worked problems on the area under a
49.5 Worked problems on integration using curve 449
the sin  substitution 422 54.3 Further worked problems on the area
under a curve 452
54.4 The area between curves 454

55 Mean and root mean square values 457
55.1 Mean or average values 457
55.2 Root mean square values 459

56 Volumes of solids of revolution 461
56.1 Introduction 461
56.2 Worked problems on volumes of solids
of revolution 461

x CONTENTS 59.4 De Morgan’s laws 490
59.5 Karnaugh maps 491
56.3 Further worked problems on volumes 59.6 Logic circuits 495
of solids of revolution 463 59.7 Universal logic circuits 500

57 Centroids of simple shapes 466 60 The theory of matrices and determinants
57.1 Centroids 466 504
57.2 The first moment of area 466 60.1 Matrix notation 504
57.3 Centroid of area between a curve and 60.2 Addition, subtraction and multiplication
the x-axis 466 of matrices 504
57.4 Centroid of area between a curve and 60.3 The unit matrix 508
the y-axis 467 60.4 The determinant of a 2 by 2 matrix
57.5 Worked problems on centroids of 508
simple shapes 467 60.5 The inverse or reciprocal of a 2 by 2
57.6 Further worked problems on centroids matrix 509
of simple shapes 468 60.6 The determinant of a 3 by 3 matrix
57.7 Theorem of Pappus 471 510
60.7 The inverse or reciprocal of a 3 by 3
58 Second moments of area 475 matrix 511
58.1 Second moments of area and radius of
gyration 475 61 The solution of simultaneous equations by
58.2 Second moment of area of regular matrices and determinants 514
sections 475 61.1 Solution of simultaneous equations by
58.3 Parallel axis theorem 475 matrices 514
58.4 Perpendicular axis theorem 476 61.2 Solution of simultaneous equations by
58.5 Summary of derived results 476 determinants 516
58.6 Worked problems on second moments 61.3 Solution of simultaneous equations
of area of regular sections 476 using Cramers rule 520
58.7 Worked problems on second moments
of areas of composite areas 480 Assignment 16 521

Assignment 15 482 Multiple choice questions on chapters 44–61
522
Part 10 Further Number and Algebra 483
Answers to multiple choice questions 526
59 Boolean algebra and logic circuits 483
59.1 Boolean algebra and switching circuits Index 527
483
59.2 Simplifying Boolean expressions 488
59.3 Laws and rules of Boolean algebra
488

Preface

This fourth edition of ‘Engineering Mathematics’ 1. Algebraic techniques: 10, 14, 15,
covers a wide range of syllabus requirements. In 28–30, 34, 59–61
particular, the book is most suitable for the latest
National Certificate and Diploma courses and 2. Trigonometry: 22–24, 26
Vocational Certificate of Education syllabuses in 3. Calculus: 44–49, 52–58
Engineering. 4. Statistical and probability: 36–43

This text will provide a foundation in mathematical (iii) Applied Mathematics in Engineering, the
principles, which will enable students to solve mathe- compulsory unit for Advanced VCE (for-
matical, scientific and associated engineering princi- merly Advanced GNVQ), to include all or
ples. In addition, the material will provide engineer- part of the following chapters:
ing applications and mathematical principles neces-
sary for advancement onto a range of Incorporated 1. Number and units: 1, 2, 4
Engineer degree profiles. It is widely recognised that 2. Mensuration: 17–20
a students’ ability to use mathematics is a key element 3. Algebra: 5, 8–11
in determining subsequent success. First year under- 4. Functions and graphs: 22, 23, 27
graduates who need some remedial mathematics will 5. Trigonometry: 21, 24
also find this book meets their needs.
(iv) Further Mathematics for Engineering, the
In Engineering Mathematics 4th Edition, theory optional unit for Advanced VCE (formerly
is introduced in each chapter by a simple outline of Advanced GNVQ), to include all or part of
essential definitions, formulae, laws and procedures. the following chapters:
The theory is kept to a minimum, for problem solv-
ing is extensively used to establish and exemplify 1. Algebra and trigonometry: 5, 6,
the theory. It is intended that readers will gain real 12–15, 21, 25
understanding through seeing problems solved and
then through solving similar problems themselves. 2. Graphical and numerical techniques:
20, 22, 26–31
For clarity, the text is divided into ten topic
areas, these being: number and algebra, mensura- 3. Differential and integral calculus:
tion, trigonometry, graphs, vectors, complex num- 44–47, 54
bers, statistics, differential calculus, integral calculus
and further number and algebra. (v) The Mathematics content of Applied Sci-
(vi) ence and Mathematics for Engineering,
This new edition will cover the following syl- (vii) for Intermediate GNVQ
labuses: (viii) Mathematics for Engineering, for Founda-
tion and Intermediate GNVQ
(i) Mathematics for Technicians, the core unit Mathematics 2 and Mathematics 3 for City
for National Certificate/Diploma courses in & Guilds Technician Diploma in Telecom-
Engineering, to include all or part of the munications and Electronic Engineering
following chapters: Any introductory/access/foundation co-
urse involving Engineering Mathematics at
1. Algebra: 2, 4, 5, 8–13, 17, 19, 27, 30 University, Colleges of Further and Higher
2. Trigonometry: 18, 21, 22, 24 education and in schools.
3. Statistics: 36, 37
4. Calculus: 44, 46, 47, 54 Each topic considered in the text is presented in
a way that assumes in the reader little previous
(ii) Further Mathematics for Technicians, knowledge of that topic.
the optional unit for National Certifi-
cate/Diploma courses in Engineering, to
include all or part of the following chapters:

xii ENGINEERING MATHEMATICS lecturers could set the Assignments for students to
attempt as part of their course structure. Lecturers’
‘Engineering Mathematics 4th Edition’ provides may obtain a complimentary set of solutions of the
a follow-up to ‘Basic Engineering Mathematics’ Assignments in an Instructor’s Manual available
and a lead into ‘Higher Engineering Mathemat- from the publishers via the internet — full worked
ics’. solutions and mark scheme for all the Assignments
are contained in this Manual, which is available to
This textbook contains over 900 worked lecturers only. To obtain a password please e-mail
problems, followed by some 1700 further [email protected] with the following details:
problems (all with answers). The further problems course title, number of students, your job title and
are contained within some 208 Exercises; each work postal address.
Exercise follows on directly from the relevant
section of work, every two or three pages. In To download the Instructor’s Manual visit
addition, the text contains 234 multiple-choice http://www.newnespress.com and enter the book
questions. Where at all possible, the problems title in the search box, or use the following direct
mirror practical situations found in engineering URL: http://www.bh.com/manuals/0750657766/
and science. 500 line diagrams enhance the
understanding of the theory. ‘Learning by Example’ is at the heart of ‘Engi-
neering Mathematics 4th Edition’.
At regular intervals throughout the text are some
16 Assignments to check understanding. For exam- John Bird
ple, Assignment 1 covers material contained in
Chapters 1 to 4, Assignment 2 covers the material University of Portsmouth
in Chapters 5 to 8, and so on. These Assignments
do not have answers given since it is envisaged that

Part 1 Number and Algebra

1

Revision of fractions, decimals
and percentages

1.1 Fractions Alternatively:

Step (2) Step (3)

When 2 is divided by 3, it may be written as 2 or ##
3
2 1 2 7ð1 C 3ð2
2/3. 3 is called a fraction. The number above the CD
37 21
line, i.e. 2, is called the numerator and the number
"
below the line, i.e. 3, is called the denominator.
Step (1)
When the value of the numerator is less than

the value of the denominator, the fraction is called Step 1: the LCM of the two denominators;
Step 2:
a proper fraction; thus 2 is a proper fraction. 1
3 Step 3: 3
When the value of the numerator is greater than for the fraction , 3 into 21 goes 7 times,

the denominator, the fraction is called an improper 7 ð the numerator is 7 ð 1;

fraction. Thus 7 is an improper fraction and can also for the fraction 2 , 7 into 21 goes 3 times,
3 7
be expressed as a mixed number, that is, an integer
3 ð the numerator is 3 ð 2.
7
and a proper fraction. Thus the improper fraction 3

is equal to the mixed number 2 1 . 1 2 7 C 6 13
3 Thus C D D as obtained previously.
When a fraction is simplified by dividing the 3 7 21 21

numerator and denominator by the same number,

the process is called cancelling. Cancelling by 0 is 21
Problem 2. Find the value of 3 2
not permissible.
36

12 One method is to split the mixed numbers into
Problem 1. Simplify C integers and their fractional parts. Then

37 21 2 1
3 2 D 3C 2C
The lowest common multiple (i.e. LCM) of the two 36 3
denominators is 3 ð 7, i.e. 21 6

Expressing each fraction so that their denomina- 2 1
tors are 21, gives: D3C 2

12 1723 7 6 3 6
CDðCðD C 4 131
D1C
3 7 3 7 7 3 21 21 6 D1 D1
7 C 6 13 662

DD Another method is to express the mixed numbers as
21 21 improper fractions.

2 ENGINEERING MATHEMATICS 8 1 7 24 8 8 ð 1 ð 8
D ðð D
9 2 9 2 11
Since 3 D , then 3 D C D 5 13 71 5ð1ð1
64 4
3 3 33 3 D D 12
1 12 1 13 55
Similarly, 2 D C D
6 66 6 3 12
2 1 11 13 22 13 9 1 Problem 6. Simplify ł
Thus 3 2 D D D D 1
3 63 6 6 66 2 7 21
as obtained previously.

Problem 3. Determine the value of 3
512
3 ł 12 D 7
4 3 C1 7 21 12
845
21

512 512 Multiplying both numerator and denominator by the
4 3 C1 D 4 3C1 C C reciprocal of the denominator gives:
845 845

5 ð 5 10 ð 1 C 8 ð 2 3 1 3 21 3 3
D2C ð
D 17 12 4 D 4 D 3
40 7 1 12 21 1 1 4
25 10 C 16 12
D2C
21 ð
40 1 21 12 1
31 31
D2C D2 This method can be remembered by the rule: invert
40 40 the second fraction and change the operation from
division to multiplication. Thus:

3 14 3 12 1 3 21 3 3
Problem 4. Find the value of ð ł D ð D as obtained previously.

7 15 7 21 1 7 12 4 4

Dividing numerator and denominator by 3 gives: 31
Problem 7. Find the value of 5 ł 7
1 3 14 1 14 1 ð 14
ð Dð D 53

7 15 5 7 5 7 ð 5 The mixed numbers must be expressed as improper
Dividing numerator and denominator by 7 gives: fractions. Thus,

1 ð 14 2 1 ð 2 2 3 1 28 22 14 28 3 42
DD 5 ł7 D ł D ð D
5 3 5 3 5 22 11 55
17ð5 1ð5 5
This process of dividing both the numerator and
denominator of a fraction by the same factor(s) is
called cancelling.

Problem 8. Simplify

313 1 21 31
Problem 5. Evaluate 1 ð 2 ð 3 Cłð
3 54 83
537

Mixed numbers must be expressed as improper The order of precedence of operations for problems
containing fractions is the same as that for inte-
fractions before multiplication can be performed. gers, i.e. remembered by BODMAS (Brackets, Of,
Division, Multiplication, Addition and Subtraction).
Thus, Thus,

313
1 ð2 ð3

537

53 61 21 3 1 21 31
DCðCð C Cłð
55 33 77 3 54 83

REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 3

1 4ð2C5ð1 31 23 2 12
Dł (B) 2. (a) C (b) C
(D) 7 11 9
3 20 24 8 (M) 73
1 13 8 2 (S) 43 47
Dð (b)
3 5 20 1 (a)
1 26 77 63
D
35 3 2 145
5 ð 1 3 ð 26 3. (a) 10 8 (b) 3 4 C 1
D
7 3 456
15 16 17
73 13
D D −4 (a) 1 (b)
15 15 21 60

35 17 15
4. (a) ð (b) ð
49 35 119

Problem 9. Determine the value of 53
(a) (b)

12 49

7 11 131 37 2 13 7 4
of 3 2 C 5 ł
6 24 8 16 2 5. (a) ð ð 1 (b) ð 4 ð 3
59 7 17 11 39

7 11 13 1 3
of 3 2 C 5 ł 2 (a) (b) 11

5

6 24 8 16 3 45 15

7 1 41 3 1 6. (a) ł (b) 1 ł 2
D of 1 C ł 8 64 39
(B)
6 4 8 16 2 (O) 8 12
7 5 41 3 1 (D) (a) (b)
DðC ł (M)
6 4 8 16 2 (A) 15 23
(A)
7 5 41 16 2 1 (S) 13 8 1 7
DðC ð 2 7. C ł 1

6 4 18 3 2 5 15 3 24
35 82 1
DC 7 5 3 15 4
24 3 2 8. of 15 ð C ł 5
35 C 656 1
D 15 7 4 16 5

24 2 12132 13
691 1 9. ð łC 126
D 43357
24 2
691 12 21 21 3 28
D 10. ð 1 ł C C 1 2
34 34 5 55
24
679 7 1.2 Ratio and proportion
D D 28
24 24 The ratio of one quantity to another is a fraction, and
is the number of times one quantity is contained in
Now try the following exercise another quantity of the same kind. If one quantity is
directly proportional to another, then as one quan-
Exercise 1 Further problems on fractions tity doubles, the other quantity also doubles. When a
Evaluate the following: quantity is inversely proportional to another, then
as one quantity doubles, the other quantity is halved.
12 7 1
1. (a) C (b) 4 Problem 10. A piece of timber 273 cm
25 16 long is cut into three pieces in the ratio of 3
9 to 7 to 11. Determine the lengths of the three
(a) 3 pieces
(b)
10
16

4 ENGINEERING MATHEMATICS 1 person takes three times as long, i.e.
4 ð 3 D 12 hours,
The total number of parts is 3 C 7 C 11, that is, 21.
Hence 21 parts correspond to 273 cm 5 people can do it in one fifth of the time that
12
273
1 part corresponds to D 13 cm one person takes, that is hours or 2 hours
5
21
3 parts correspond to 3 ð 13 D 39 cm 24 minutes.
7 parts correspond to 7 ð 13 D 91 cm
11 parts correspond to 11 ð 13 D 143 cm Now try the following exercise

i.e. the lengths of the three pieces are 39 cm, Exercise 5 Further problems on ratio and
91 cm and 143 cm. proportion

(Check: 39 C 91 C 143 D 273) 1. Divide 621 cm in the ratio of 3 to 7 to 13.
[81 cm to 189 cm to 351 cm]
Problem 11. A gear wheel having 80 teeth
is in mesh with a 25 tooth gear. What is the 2. When mixing a quantity of paints, dyes of
gear ratio?
four different colours are used in the ratio
80 16
Gear ratio D 80:25 D D D 3.2 of 7:3:19:5. If the mass of the first dye

25 5 used is 3 1 g, determine the total mass of
i.e. gear ratio D 16 : 5 or 3.2 : 1 2
the dyes used. [17 g]
Problem 12. An alloy is made up of
metals A and B in the ratio 2.5 : 1 by mass. 3. Determine how much copper and how
How much of A has to be added to 6 kg of much zinc is needed to make a 99 kg
B to make the alloy? brass ingot if they have to be in the
proportions copper : zinc: :8 : 3 by mass.
Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1)
A 2.5 [72 kg : 27 kg]

or D D 2.5 4. It takes 21 hours for 12 men to resurface
B1
A a stretch of road. Find how many men

When B D 6 kg, D 2.5 from which, it takes to resurface a similar stretch of
6
road in 50 hours 24 minutes, assuming
A D 6 ð 2.5 D 15 kg
the work rate remains constant. [5]
Problem 13. If 3 people can complete a
task in 4 hours, how long will it take 5 5. It takes 3 hours 15 minutes to fly from
people to complete the same task, assuming city A to city B at a constant speed. Find
the rate of work remains constant how long the journey takes if

(a) the speed is 1 1 times that of the
2
original speed and

(b) if the speed is three-quarters of the
original speed.

[(a) 2 h 10 min (b) 4 h 20 min]

1.3 Decimals

The more the number of people, the more quickly The decimal system of numbers is based on the
the task is done, hence inverse proportion exists. digits 0 to 9. A number such as 53.17 is called
a decimal fraction, a decimal point separating the
3 people complete the task in 4 hours, integer part, i.e. 53, from the fractional part, i.e. 0.17

A number which can be expressed exactly as REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 5

a decimal fraction is called a terminating deci- 87.23
81.70
mal and those which cannot be expressed exactly 5.53

as a decimal fraction are called non-terminating Thus 87.23 − 81.70 = 5.53

decimals. Thus, 3 D 1.5 is a terminating decimal, Problem 16. Find the value of
2 23.4 17.83 57.6 C 32.68
4
but 3 D 1.33333. . . is a non-terminating decimal. The sum of the positive decimal fractions is
23.4 C 32.68 D 56.08
1.33333. . . can be written as 1.3P, called ‘one point-
The sum of the negative decimal fractions is
three recurring’. 17.83 C 57.6 D 75.43

The answer to a non-terminating decimal may be Taking the sum of the negative decimal fractions
from the sum of the positive decimal fractions gives:
expressed in two ways, depending on the accuracy
56.08 75.43
required: i.e. 75.43 56.08 D −19.35

(i) correct to a number of significant figures, that Problem 17. Determine the value of
is, figures which signify something, and 74.3 ð 3.8

(ii) correct to a number of decimal places, that is,
the number of figures after the decimal point.

The last digit in the answer is unaltered if the next
digit on the right is in the group of numbers 0, 1,
2, 3 or 4, but is increased by 1 if the next digit
on the right is in the group of numbers 5, 6, 7, 8
or 9. Thus the non-terminating decimal 7.6183. . .
becomes 7.62, correct to 3 significant figures, since
the next digit on the right is 8, which is in the group
of numbers 5, 6, 7, 8 or 9. Also 7.6183. . . becomes
7.618, correct to 3 decimal places, since the next
digit on the right is 3, which is in the group of
numbers 0, 1, 2, 3 or 4.

Problem 14. Evaluate When multiplying decimal fractions: (i) the numbers
42.7 C 3.04 C 8.7 C 0.06 are multiplied as if they are integers, and (ii) the
position of the decimal point in the answer is such
The numbers are written so that the decimal points that there are as many digits to the right of it as the
are under each other. Each column is added, starting sum of the digits to the right of the decimal points
from the right. of the two numbers being multiplied together. Thus

42.7 (i) 743
3.04 38
8.7
0.06 5 944
22 290
54.50
28 234
Thus 42.7 Y 3.04 Y 8.7 Y 0.06 = 54.50
(ii) As there are 1 C 1 D 2 digits to the right of
Problem 15. Take 81.70 from 87.23 the decimal points of the two numbers being
multiplied together, (74.3 ð 3.8), then
The numbers are written with the decimal points
under each other. 74.3 × 3.8 = 282.34

Problem 18. Evaluate 37.81 ł 1.7, correct
to (i) 4 significant figures and (ii) 4 decimal
places

6 ENGINEERING MATHEMATICS

37.81 Problem 20. Express as decimal fractions:
37.81 ł 1.7 D 97

1.7 (a) and (b) 5
16 8
The denominator is changed into an integer by
multiplying by 10. The numerator is also multiplied
by 10 to keep the fraction the same. Thus

37.81 ł 1.7 D 37.81 ð 10 378.1 (a) To convert a proper fraction to a decimal frac-
D tion, the numerator is divided by the denomi-
1.7 ð 10 17 nator. Division by 16 can be done by the long
division method, or, more simply, by dividing
The long division is similar to the long division of by 2 and then 8:
integers and the first four steps are as shown:

22.24117.. 4.50 0.5625
2 9.00 8 4.5000
17 378.100000
34 9
Thus, = 0.5625
38
34 16

41 (b) For mixed numbers, it is only necessary to
34
convert the proper fraction part of the mixed
70
68 number to a decimal fraction. Thus, dealing

20 with the 7 gives:
8

0.875 7
8 7.000 i.e. D 0.875

8

(i) 37.81 ÷ 1.7 = 22.24, correct to 4 significant 7
figures, and Thus 5 = 5.875

8

(ii) 37.81 ÷ 1.7 = 22.2412, correct to 4 decimal Now try the following exercise
places.
Exercise 3 Further problems on decimals
Problem 19. Convert (a) 0.4375 to a proper In Problems 1 to 6, determine the values of
fraction and (b) 4.285 to a mixed number the expressions given:

0.4375 ð 10 000 1. 23.6 C 14.71 18.9 7.421 [11.989]
(a) 0.4375 can be written as
2. 73.84 113.247 C 8.21 0.068
10 000 [ 31.265]
without changing its value,
3. 3.8 ð 4.1 ð 0.7 [10.906]
4375
i.e. 0.4375 D 4. 374.1 ð 0.006 [2.2446]

10 000 5. 421.8 ł 17, (a) correct to 4 significant
By cancelling figures and (b) correct to 3 decimal
places.
4375 875 175 35 7
D D DD [(a) 24.81 (b) 24.812]

10 000 2000 400 80 16 0.0147
7 6. , (a) correct to 5 decimal places

i.e. 0.4375 = 2.3
16 and (b) correct to 2 significant figures.
285 57
[(a) 0.00639 (b) 0.0064]
(b) Similarly, 4.285 D 4 D 4
1000 200

7. Convert to proper fractions: REVISION OF FRACTIONS, DECIMALS AND PERCENTAGES 7

(a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282 (a) 1.875 corresponds to 1.875 ð 100%, i.e.
and (e) 0.024 187.5%

13 21 1 141 3 (b) 0.0125 corresponds to 0.0125 ð 100%, i.e.
(a) (b) (c) (d) (e) 1.25%

20 25 80 500 125 Problem 22. Express as percentages:
52
8. Convert to mixed numbers:
(a) and (b) 1
(a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35 16 5

and (e) 16.2125

 41 11 1 To convert fractions to percentages, they are (i) con-
(a) 1 (b) 4 (c) 14 8  verted to decimal fractions and (ii) multiplied by 100
 50
7 40 55
(d) 15 17 (a) By division, D 0.3125, hence corre-
20 (e) 16
80 16 16
sponds to 0.3125 ð 100%, i.e. 31.25%
In Problems 9 to 12, express as decimal frac-
tions to the accuracy stated: 2
(b) Similarly, 1 D 1.4 when expressed as a
4
9. , correct to 5 significant figures. 5
decimal fraction.
9
[0.44444] 2
Hence 1 D 1.4 ð 100% D 140%
17
10. , correct to 5 decimal place. 5

27 Problem 23. It takes 50 minutes to machine
[0.62963] a certain part. Using a new type of tool, the
time can be reduced by 15%. Calculate the
9 new time taken
11. 1 , correct to 4 significant figures.

16
[1.563]

31 15% of 50 minutes 15 750
12. 13 , correct to 2 decimal places. D ð 50 D

37 100 100
[13.84] D 7.5 minutes.

hence the new time taken is

50 7.5 D 42.5 minutes.

1.4 Percentages Alternatively, if the time is reduced by 15%, then

Percentages are used to give a common standard it now takes 85% of the original time, i.e. 85% of
and are fractions having the number 100 as their
85 4250 D 42.5 minutes, as above.
25 50 D ð 50 D
denominators. For example, 25 per cent means 100 100

100 Problem 24. Find 12.5% of £378
1
i.e. and is written 25%. 12.5% of £378 means 12.5 ð 378, since per cent
4
means ‘per hundred’. 100
Problem 21. Express as percentages:
(a) 1.875 and (b) 0.0125 12.51 1
Hence 12.5% of £378 D ð 378 D ð 378 D
A decimal fraction is converted to a percentage by 100 8 8
multiplying by 100. Thus,
378 D £47.25

8

8 ENGINEERING MATHEMATICS

Problem 25. Express 25 minutes as a 2. Express as percentages, correct to 3
percentage of 2 hours, correct to the significant figures:
nearest 1%
7 19 11
(a) (b) (c) 1
33 24 16
Working in minute units, 2 hours D 120 minutes.
25 [(a) 21.2% (b) 79.2% (c) 169%]

Hence 25 minutes is ths of 2 hours. By can- 3. Calculate correct to 4 significant figures:
120 (a) 18% of 2758 tonnes (b) 47% of

25 5 18.42 grams (c) 147% of 14.1 seconds
celling, D
[(a) 496.4 t (b) 8.657 g (c) 20.73 s]
120 24
4. When 1600 bolts are manufactured, 36
Expressing 5 as a decimal fraction gives 0.2083P
24 are unsatisfactory. Determine the percent-

Multiplying by 100 to convert the decimal fraction age unsatisfactory. [2.25%]
to a percentage gives:
5. Express: (a) 140 kg as a percentage of
0.2083P ð 100 D 20.83P% 1 t (b) 47 s as a percentage of 5 min
(c) 13.4 cm as a percentage of 2.5 m
Thus 25 minutes is 21% of 2 hours, correct to the
nearest 1%. [(a) 14% (b) 15.67% (c) 5.36%]

Problem 26. A German silver alloy consists 6. A block of monel alloy consists of 70%
of 60% copper, 25% zinc and 15% nickel.
Determine the masses of the copper, zinc and
nickel in a 3.74 kilogram block of the alloy

nickel and 30% copper. If it contains

By direct proportion: 88.2 g of nickel, determine the mass of

copper in the block. [37.8 g]

100% corresponds to 3.74 kg 7. A drilling machine should be set to

1% corresponds to 3.74 D 0.0374 kg 250 rev/min. The nearest speed available

100 on the machine is 268 rev/min. Calculate

60% corresponds to 60 ð 0.0374 D 2.244 kg the percentage over speed. [7.2%]

25% corresponds to 25 ð 0.0374 D 0.935 kg 8. Two kilograms of a compound contains
30% of element A, 45% of element B and
15% corresponds to 15 ð 0.0374 D 0.561 kg 25% of element C. Determine the masses
of the three elements present.
Thus, the masses of the copper, zinc and nickel are
2.244 kg, 0.935 kg and 0.561 kg, respectively. [A 0.6 kg, B 0.9 kg, C 0.5 kg]

(Check: 2.244 C 0.935 C 0.561 D 3.74) 9. A concrete mixture contains seven parts
by volume of ballast, four parts by vol-
Now try the following exercise ume of sand and two parts by volume of
cement. Determine the percentage of each
Exercise 4 Further problems percentages of these three constituents correct to the
1. Convert to percentages: nearest 1% and the mass of cement in a
two tonne dry mix, correct to 1 significant
(a) 0.057 (b) 0.374 (c) 1.285 figure.
[(a) 5.7% (b) 37.4% (c) 128.5%]
[54%, 31%, 15%, 0.3 t]

2

Indices and standard form

2.1 Indices (i) When multiplying two or more numbers hav-
ing the same base, the indices are added. Thus
The lowest factors of 2000 are 2ð2ð2ð2ð5ð5ð5.
These factors are written as 24 ð 53, where 2 and 5 32 ð 34 D 32C4 D 36
are called bases and the numbers 4 and 3 are called
indices. (ii) When a number is divided by a number having
the same base, the indices are subtracted. Thus
When an index is an integer it is called a power.
Thus, 24 is called ‘two to the power of four’, and 35 D 35 2 D 33
has a base of 2 and an index of 4. Similarly, 53 is 32
called ‘five to the power of 3’ and has a base of 5
and an index of 3. (iii) When a number which is raised to a power
is raised to a further power, the indices are
Special names may be used when the indices are multiplied. Thus
2 and 3, these being called ‘squared’ and ‘cubed’,
respectively. Thus 72 is called ‘seven squared’ and 35 2 D 35ð2 D 310
93 is called ‘nine cubed’. When no index is shown,
the power is 1, i.e. 2 means 21.

Reciprocal (iv) When a number has an index of 0, its value
is 1. Thus 30 D 1

The reciprocal of a number is when the index is (v) A number raised to a negative power is the

1 and its value is given by 1, divided by the base. reciprocal of that number raised to a positive

Thus the reciprocal of 2 is 2 1 and its value is 1 power. Thus 3 4D 1 Similarly, 1 D 23
2 34 23
or 0.5. Similarly, the reciprocal of 5 is 5 1 which

means 1 or 0.2 (vi) When a number is raised to a fractional power
5 the denominator of the fraction is the root of
the number and the numerator is the power.
Square root
p
The square root of a number is when the indpex is 1 , Thus 82/3 D 3 82 D 2 2 D 4
2 and
pp
and the square root of 2 is written as 21/2 or 2. The 251/2 D 2 251 D 251 D š5

value of a square root is the value of the base which pp
(Note that Á 2 )
when multipliedpby itself gives the number. Since
3ðp3 D 9, then 9 D 3. However, 3 ð 3 D 9,
so 9 D 3. There are always two answers when

finding the square root of a number and this is shown 2.2 Worked problems on indices

by putting both a C and a sign in frpont of the Problem 1. Evaluate: (a) 52 ð 53,
answer to apsquare root problem. Thus 9 D š3 (b) 32 ð 34 ð 3 and (c) 2 ð 22 ð 25
and 41/2 D 4 D š2, and so on.
From law (i):
Laws of indices (a) 52ð53 D 5 2C3 D 55 D 5ð5ð5ð5ð5 D 3125

When simplifying calculations involving indices,
certain basic rules or laws can be applied, called
the laws of indices. These are given below.

10 ENGINEERING MATHEMATICS Problem 6. Find the value of

(b) 32 ð 34 ð 3 D 3 2C4C1 D 37 23 ð 24 32 3
D 3 ð 3 ð Ð Ð Ð to 7 terms (a) 27 ð 25 and (b) 3 ð 39
D 2187

(c) 2 ð 22 ð 25 D 2 1C2C5 D 28 D 256

Problem 2. Find the value of: From the laws of indices:
75 57
(a) 23 ð 24 2 3C4 D 27 D 27 12 D 2 5
(a) 73 and (b) 54 27 ð 25 D 2 7C5 212

From law (ii): 11
DD

25 32

(b) 32 3 32ð3 36 D 36 10 D 3 4
3 ð 39 D D
(a) 75 D75 3 D 72 D 49
73 31C9 310

57 11
54 D 34 D 81

(b) D57 4 D 53 D 125

Now try the following exercise

Problem 3. Evaluate: (a) 52 ð 53 ł 54 and Exercise 5 Further problems on indices
(b) 3 ð 35 ł 32 ð 33
In Problems 1 to 10, simplify the expressions
From laws (i) and (ii): given, expressing the answers in index form
and with positive indices:
52 ð 53 5 2C3
(a) 52 ð 53 ł 54 D 54 D 54 1. (a) 33 ð 34 (b) 42 ð 43 ð 44
[(a) 37 (b) 49]
D 55 D55 4 D 51 D 5
54 2. (a) 23 ð 2 ð 22 (b) 72 ð 74 ð 7 ð 73
[(a) 26 (b) 710]
(b) 3 ð 35 ł 32 ð 33 3 ð 35 3 1C5
DD
24 37
32 ð 33 3 2C3 3. (a) 23 (b) 32

D 36 D 36 5 D 31 D 3 [(a) 2 (b) 35]
35

Problem 4. Simplify: (a) 23 4 (b) 32 5, 4. (a) 56 ł 53 (b) 713/710
expressing the answers in index form. [(a) 53 (b) 73]

5. (a) 72 3 (b) 33 2 [(a) 76 (b) 36]

From law (iii): 22 ð 23 37 ð 34
(a) 23 4 D 23ð4 D 212 (b) 32 5 D 32ð5 D 310 6. (a) 24 (b) 35

[(a) 2 (b) 36]

102 3 57 135
Problem 5. Evaluate: 104 ð 102 7. (a) 52 ð 53 (b) 13 ð 132

[(a) 52 (b) 132]

From the laws of indices: 9 ð 32 3 16 ð 4 2

102 3 10 2ð3 106 8. (a) 3 ð 27 2 (b) 2 ð 8 3
104 ð 102 D 10 4C2 D 106
[(a) 34 (b) 1]
D 106 6 D 100 D 1

INDICES AND STANDARD FORM 11

5 2 32 ð 3 4 (Note that it does not matter whether the 4th root
9. (a) 5 4 (b) 33 of 16 is found first or whether 16 cubed is found
first — the same answer will result).
(a) 52 1
(b) 35 p
(c) 272/3 D 3 272 D 3 2 D 9

72 ð 7 3 23 ð 2 4 ð 25 (d) 9 1/2 D 1 D 1 D 1 1
10. (a) 7 ð 7 4 (b) 2 ð 2 2 ð 26 91/2 p š3 D±

9 3

(a) 72 1 41.5 ð 81/3
(b) Problem 10. Evaluate: 22 ð 32 2/5
2

2.3 Further worked problems on p
indices 41.5 D 43/2 D 43 D 23 D 8,

p
81/3 D 3 8 D 2, 22 D 4

and 32 2/5 D 1 1 1 D 1
322/5 Dp D 22 4
33 ð 57
Problem 7. Evaluate: 53 ð 34 5 322

Hence 41.5 ð 81/3 8 ð 2 16 D 16
Alternatively, D D
22 ð 32 1 1
2/5 4ð

The laws of indices only apply to terms having the 4
same base. Grouping terms having the same base,
and then applying the laws of indices to each of the 41.5 ð 81/3 [ 2 2]3/2 ð 23 1/3 23 ð 21
groups independently gives: 22 ð 32 2/5 D 22 ð 25 2/5 D 22 ð 2 2

33 ð 57 D 33 57 D33 4 ð57 3 D 23C1 2 2 D 24 D 16
53 ð 34 ð

34 53

54 625 1 Problem 11. Evaluate: 32 ð 55 C 33 ð 53
31 3 D 208
D3 1 ð 54 D D 34 ð 54
3

Problem 8. Find the value of Dividing each term by the HCF (i.e. highest com-
23 ð 35 ð 72 2 mon factor) of the three terms, i.e. 32 ð 53, gives:

74 ð 24 ð 33 32 ð 55 33 ð 53
32 ð 53 C 32 ð 53
32 ð 55 C 33 ð 53 D 34 ð 54
34 ð 54

23 ð 35 ð 72 2 D 23 4 ð 35 3 ð 72ð2 4 32 ð 53
74 ð 24 ð 33
3 2 2 ð 5 5 3 C 3 3 2 ð 50
D 2 1 ð 32 ð 70 D

34 2 ð54 3

D 1 ð 32 ð 1 D 9 D 1 30 ð 52 C 31 ð 50
4 D 32 ð 51
2 22

1 ð 25 C 3 ð 1 28
DD
Problem 9. Evaluate: 9ð5 45
(a) 41/2 (b) 163/4 (c) 272/3 (d) 9 1/2
Problem 12. Find the value of
p 32 ð 55
(a) 41/2 D 4 D ±2
34 ð 54 C 33 ð 53
p
(b) 163/4 D 4 163 D š2 3 D ±8

12 ENGINEERING MATHEMATICS

To simplify the arithmetic, each term is divided by Now try the following exercise
the HCF of all the terms, i.e. 32 ð 53. Thus

32 ð 55 Exercise 6 Further problems on indices

34 ð 54 C 33 ð 53 In Problems 1 and 2, simplify the expressions
given, expressing the answers in index form
32 ð 55 and with positive indices:

D 32 ð 53
34 ð 54 33 ð 53
C
32 ð 53 32 ð 53 33 ð 52 7 2ð3 2
1. (a) 54 ð 34 (b) 35 ð 74 ð 7 3
32 2 ð55 3
D 34 2 ð54 3 C33 2 ð53 3 11
(a) 3 ð 52 (b) 73 ð 37
30 ð 52 25 25
D DD
32 ð 51 C 31 ð 50 45 C 3 48
42 ð 93 8 2 ð 52 ð 3 4
2. (a) 83 ð 34 (b) 252 ð 24 ð 9 2
43 2
ð 3
Problem 13. Simplify: 32 1
3 5 (a) 25 (b) 210 ð 52
2
3

5 3. Evaluate a 11 b 810.25
giving the answer with positive indices 32 4 1/2

A fraction raised to a power means that both the c 16 1/4 d
9
numerator and the denominator of the fraction are
4 3 43
raised to that power, i.e. D 33 12
3 (a) 9 (b) š3 (c) š (d) š

23

A fraction raised to a negative power has the In Problems 4 to 8, evaluate the expressions
same value as the inverse of the fraction raised to a given.
positive power.

32 1 1 52 52 92 ð 74 147
Thus, 5 D 3 2 D 32 D 1 ð 32 D 32 4. 34 ð 74 C 33 ð 72 148

5 52 24 2 3 2 ð 44 1
5. 23 ð 162 9
23 5 3 53
Similarly, D D 23 13 22
5 2
65
43 3 2 43 52 6. 23 5
ð 32
ð 72
35 53
Thus, 23 D 33 32 5

5 23 44
43 52 23
D 33 ð 32 ð 53 7. 3 [64]
22

22 3 ð 23 9
D
32 3/2 ð 81/3 2 1
3 3C2 ð 5 3 2 8. 3 2 ð 43 1/2 ð 9 1/2 4

29 2
D 35 × 5

INDICES AND STANDARD FORM 13

2.4 Standard form Similarly,

A number written with one digit to the left of the 6 ð 104 6 104 2 D 4 ð 102
decimal point and multiplied by 10 raised to some 1.5 ð 102 D 1.5 ð
power is said to be written in standard form. Thus:
5837 is written as 5.837 ð 103 in standard form, 2.5 Worked problems on standard
and 0.0415 is written as 4.15 ð 10 2 in standard form
form.
Problem 14. Express in standard form:
When a number is written in standard form, the (a) 38.71 (b) 3746 (c) 0.0124
first factor is called the mantissa and the second
factor is called the exponent. Thus the number For a number to be in standard form, it is expressed
5.8 ð 103 has a mantissa of 5.8 and an exponent with only one digit to the left of the decimal point.
of 103. Thus:

(i) Numbers having the same exponent can be (a) 38.71 must be divided by 10 to achieve one
added or subtracted in standard form by adding digit to the left of the decimal point and it
or subtracting the mantissae and keeping the must also be multiplied by 10 to maintain the
exponent the same. Thus: equality, i.e.

2.3 ð 104 C 3.7 ð 104 38.71 D 38.71 ð 10 D 3.871 × 10 in standard
D 2.3 C 3.7 ð 104 D 6.0 ð 104
form 10
and 5.9 ð 10 2 4.6 ð 10 2
D 5.9 4.6 ð 10 2 D 1.3 ð 10 2 (b) 3746 D 3746 ð 1000 D 3.746 × 103 in stan-
1000
When the numbers have different exponents,
one way of adding or subtracting the numbers dard form
is to express one of the numbers in non-
standard form, so that both numbers have the 100 1.24
same exponent. Thus: (c) 0.0124 D 0.0124 ð D

2.3 ð 104 C 3.7 ð 103 100 100
D 2.3 ð 104 C 0.37 ð 104 D 1.24 × 10−2 in standard form
D 2.3 C 0.37 ð 104 D 2.67 ð 104
Problem 15. Express the following
Alternatively, numbers, which are in standard form, as
decimal numbers: (a) 1.725 ð 10 2
2.3 ð 104 C 3.7 ð 103 (b) 5.491 ð 104 (c) 9.84 ð 100

D 23 000 C 3700 D 26 700 (a) 1.725 ð 10 2 D 1.725 D 0.01725
D 2.67 ð 104 100

(ii) The laws of indices are used when multiplying (b) 5.491 ð 104 D 5.491 ð 10 000 D 54 910
or dividing numbers given in standard form. (c) 9.84 ð 100 D 9.84 ð 1 D 9.84 (since 100 D 1)
For example,
Problem 16. Express in standard form,
2.5 ð 103 ð 5 ð 102 correct to 3 significant figures:
D 2.5 ð 5 ð 103C2
D 12.5 ð 105 or 1.25 ð 106 32 9
(a) (b) 19 (c) 741
83 16

14 ENGINEERING MATHEMATICS

3 6. (a) 3.89 ð 10 2 (b) 6.741 ð 10 1
(a) D 0.375, and expressing it in standard form (c) 8 ð 10 3
[(a) 0.0389 (b) 0.6741 (c) 0.008]
8
gives: 0.375 D 3.75 × 10−1 2.6 Further worked problems on
standard form
(b) 2 D 19.6P D 1.97 × 10 in standard form,
19
3
correct to 3 significant figures

(c) 9 D 741.5625 D 7.42 × 102 in standard
741
16
form, correct to 3 significant figures

Problem 17. Express the following Problem 18. Find the value of:

numbers, given in standard form, as fractions (a) 7.9 ð 10 2 5.4 ð 10 2
or mixed numbers: (a) 2.5 ð 10 1 (b) 8.3 ð 103 C 5.415 ð 103 and
(b) 6.25 ð 10 2 (c) 1.354 ð 102 (c) 9.293 ð 102 C 1.3 ð 103 expressing the

(a) 2.5 ð 10 1 D 2.5 D 25 1 answers in standard form.
D
10 100 4 Numbers having the same exponent can be added
or subtracted by adding or subtracting the mantissae
(b) 6.25 ð 10 2 D 6.25 D 625 1 and keeping the exponent the same. Thus:
D
100 10 000 16 (a) 7.9 ð 10 2 5.4 ð 10 2
D 7.9 5.4 ð 10 2 D 2.5 × 10−2
(c) 1.354 ð 102 D 135.4 D 135 4 2
D 135 (b) 8.3 ð 103 C 5.415 ð 103
10 5 D 8.3 C 5.415 ð 103 D 13.715 ð 103
D 1.3715 × 104 in standard form
Now try the following exercise
(c) Since only numbers having the same exponents
Exercise 7 Further problems on standard can be added by straight addition of the man-
form tissae, the numbers are converted to this form
before adding. Thus:
In Problems 1 to 4, express in standard form: 9.293 ð 102 C 1.3 ð 103
D 9.293 ð 102 C 13 ð 102
1. (a) 73.9 (b) 28.4 (c) 197.72 D 9.293 C 13 ð 102
D 22.293 ð 102 D 2.2293 × 103
(a) 7.39 ð 10 (b) 2.84 ð 10
(c) 1.9762 ð 102 in standard form.
Alternatively, the numbers can be expressed as
2. (a) 2748 (b) 33170 (c) 274218 decimal fractions, giving:
(a) 2.748 ð 103 (b) 3.317 ð 104
(c) 2.74218 ð 105 9.293 ð 102 C 1.3 ð 103

3. (a) 0.2401 (b) 0.0174 (c) 0.00923 D 929.3 C 1300 D 2229.3
(a) 2.401 ð 10 1 (b) 1.74 ð 10 2 D 2.2293 × 103
(c) 9.23 ð 10 3 in standard form as obtained previously. This
method is often the ‘safest’ way of doing this
17 31 type of problem.
4. (a) (b) 11 (c) 130 (d)
28 5 32

(a) 5 ð 10 1 (b) 1.1875 ð 10
(c) 1.306 ð 102 (d) 3.125 ð 10 2

In Problems 5 and 6, express the numbers
given as integers or decimal fractions:

5. (a) 1.01 ð 103 (b) 9.327 ð 102
(c) 5.41 ð 104 (d) 7 ð 100

[(a) 1010 (b) 932.7 (c) 54 100 (d) 7]

INDICES AND STANDARD FORM 15

Problem 19. Evaluate 3.5 ð 105 3. (a) 4.5 ð 10 2 3 ð 103
and (b) 7 ð 102 (b) 2 ð 5.5 ð 104
(a) 3.75 ð 103 6 ð 104
[(a) 1.35 ð 102 (b) 1.1 ð 105]

expressing answers in standard form 6 ð 10 3 2.4 ð 103 3 ð 10 2

4. (a) 3 ð 10 5 (b) 4.8 ð 104

(a) 3.75 ð 103 6 ð 104 D 3.75 ð 6 103C4 [(a) 2 ð 102 (b) 1.5 ð 10 3]

D 22.50 ð 107 5. Write the following statements in stan-
dard form:
D 2.25 × 108

(b) 3.5 ð 105 D 3.5 ð 105 2 (a) The density of aluminium is
7 ð 102 7 2710 kg m 3

D 0.5 ð 103 D 5 × 102 [2.71 ð 103 kg m 3]

Now try the following exercise (b) Poisson’s ratio for gold is 0.44
[4.4 ð 10 1]

Exercise 8 Further problems on standard (c) The impedance of free space is
form
376.73 [3.7673 ð 102 ]
In Problems 1 to 4, find values of the expres-
sions given, stating the answers in standard (d) The electron rest energy is
form: 0.511 MeV [5.11 ð 10 1 MeV]

1. (a) 3.7 ð 102 C 9.81 ð 102 (e) Proton charge-mass ratio is
(b) 1.431 ð 10 1 C 7.3 ð 10 1 9 5 789 700 C kg 1
[(a) 1.351 ð 103 (b) 8.731 ð 10 1]
[9.57897 ð 107 C kg 1]
2. (a) 4.831 ð 102 C 1.24 ð 103
(b) 3.24 ð 10 3 1.11 ð 10 4 (f) The normal volume of a perfect gas
[(a) 1.7231 ð 103 (b) 3.129 ð 10 3] is 0.02241 m3 mol 1

[2.241 ð 10 2 m3 mol 1]

3

Computer numbering systems

3.1 Binary numbers Problem 2. Convert 0.10112 to a decimal
fraction
The system of numbers in everyday use is the
denary or decimal system of numbers, using 0.10112 D 1 ð 2 1 C 0 ð 2 2 C 1 ð 2 3
the digits 0 to 9. It has ten different digits C1ð2 4
(0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) and is said to have a 11 1
radix or base of 10.
D 1 ð 2 C 0 ð 22 C 1 ð 23
The binary system of numbers has a radix of 2 1
and uses only the digits 0 and 1.
C 1 ð 24
3.2 Conversion of binary to decimal 11 1
DCC
The decimal number 234.5 is equivalent to 2 8 16
D 0.5 C 0.125 C 0.0625
2 ð 102 C 3 ð 101 C 4 ð 100 C 5 ð 10 1 D 0.687510

i.e. is the sum of terms comprising: (a digit) multi- Problem 3. Convert 101.01012 to a decimal
plied by (the base raised to some power). number

In the binary system of numbers, the base is 2, so 101.01012 D 1 ð 22 C 0 ð 21 C 1 ð 20
1101.1 is equivalent to: C0ð2 1C1ð2 2
C0ð2 3C1ð2 4
1 ð 23 C 1 ð 22 C 0 ð 21 C 1 ð 20 C 1 ð 2 1
D 4 C 0 C 1 C 0 C 0.25
Thus the decimal number equivalent to the binary C 0 C 0.0625
number 1101.1 is
D 5.312510
1
8 C 4 C 0 C 1 C , that is 13.5 Now try the following exercise

2 Exercise 9 Further problems on conver-
sion of binary to decimal num-
i.e. 1101.12 = 13.510, the suffixes 2 and 10 denot- bers
ing binary and decimal systems of numbers respec-
tively.

Problem 1. Convert 110112 to a decimal
number

From above: 110112 D 1 ð 24 C 1 ð 23 C 0 ð 22 In Problems 1 to 4, convert the binary num-
C 1 ð 21 C 1 ð 20 bers given to decimal numbers.

D 16 C 8 C 0 C 2 C 1 1. (a) 110 (b) 1011 (c) 1110 (d) 1001
D 2710 [(a) 610 (b) 1110 (c) 1410 (d) 910]

2. (a) 10101 (b) 11001 (c) 101101 COMPUTER NUMBERING SYSTEMS 17
(d) 110011
[(a) 2110 (b) 2510 (c) 4510 (d) 5110] For fractions, the most significant bit of the result
is the top bit obtained from the integer part of
3. (a) 0.1101 (b) 0.11001 (c) 0.00111 multiplication by 2. The least significant bit of the
(d) 0.01011 result is the bottom bit obtained from the integer
part of multiplication by 2.
(a) 0.812510 (b) 0.7812510
(c) 0.2187510 (d) 0.3437510 Thus 0.62510 = 0.1012
4. (a) 11010.11 (b) 10111.011
(c) 110101.0111 (d) 11010101.10111 Problem 4. Convert 4710 to a binary
(a) 26.7510 (b) 23.37510 number
(c) 53.437510 (d) 213.7187510
From above, repeatedly dividing by 2 and noting the
remainder gives:

2 47 Remainder

2 23 1

2 11 1

3.3 Conversion of decimal to binary 25 1

An integer decimal number can be converted to a 22 1
corresponding binary number by repeatedly dividing
by 2 and noting the remainder at each stage, as 21 0
shown below for 3910
01

2 39 Remainder 1 0 1 11 1
2 19 1
29 1 Thus 4710 = 1011112
24 1
22 0 Problem 5. Convert 0.4062510 to a binary
21 0 number
1
0

(most → 1 0 0 1 1 1 ← (least From above, repeatedly multiplying by 2 gives:

significant bit) significant bit) 0.40625 × 2 = 0. 8125

The result is obtained by writing the top digit of 0.8125 × 2 = 1. 625
the remainder as the least significant bit, (a bit is a
binary digit and the least significant bit is the one 0.625 × 2 = 1. 25
on the right). The bottom bit of the remainder is the
most significant bit, i.e. the bit on the left.

Thus 3910 = 1001112 0.25 × 2 = 0. 5

The fractional part of a decimal number can be con- 0.5 × 2 = 1. 0
verted to a binary number by repeatedly multiplying
by 2, as shown below for the fraction 0.625

0.625 × 2 = 1. 250 .0 1 1 0 1

0.250 × 2 = 0. 500 i.e. 0.4062510 = 0.011012

0.500 × 2 = 1. 000 Problem 6. Convert 58.312510 to a binary
number

(most significant bit) .1 0 1 (least significant bit)

18 ENGINEERING MATHEMATICS 3.4 Conversion of decimal to binary
via octal
The integer part is repeatedly divided by 2, giving:
For decimal integers containing several digits, repe-
2 58 Remainder atedly dividing by 2 can be a lengthy process. In
2 29 0 this case, it is usually easier to convert a decimal
2 14 1 number to a binary number via the octal system of
27 0 numbers. This system has a radix of 8, using the
23 1 digits 0, 1, 2, 3, 4, 5, 6 and 7. The denary number
21 1 equivalent to the octal number 43178 is

01 4 ð 83 C 3 ð 82 C 1 ð 81 C 7 ð 80

1 1 1 01 0 i.e. 4 ð 512 C 3 ð 64 C 1 ð 8 C 7 ð 1 or 225510

The fractional part is repeatedly multiplied by 2 An integer decimal number can be converted to a
giving: corresponding octal number by repeatedly dividing
by 8 and noting the remainder at each stage, as
0.3125 × 2 = 0.625 shown below for 49310
0.625 × 2 = 1.25
0.25 × 2 = 8 493 Remainder
0.5 × 2 = 0.5
8 61 5
.0 1 0 1 1.0
87 5
Thus 58.312510 = 111010.01012
07
Now try the following exercise

Exercise 10 Further problems on conver- 755
sion of decimal to binary
numbers Thus 49310 = 7558
The fractional part of a decimal number can be con-
In Problems 1 to 4, convert the decimal verted to an octal number by repeatedly multiplying
numbers given to binary numbers. by 8, as shown below for the fraction 0.437510

1. (a) 5 (b) 15 (c) 19 (d) 29 0.4375 × 8 = 3 . 5

(a) 1012 (b) 11112 0.5 × 8 = 4 . 0
(c) 100112 (d) 111012
.3 4
2. (a) 31 (b) 42 (c) 57 (d) 63
For fractions, the most significant bit is the top
(a) 111112 (b) 1010102 integer obtained by multiplication of the decimal
(c) 1110012 (d) 1111112 fraction by 8, thus

3. (a) 0.25 (b) 0.21875 (c) 0.28125 0.437510 D 0.348
(d) 0.59375
The natural binary code for digits 0 to 7 is shown
(a) 0.012 (b) 0.001112 in Table 3.1, and an octal number can be converted
to a binary number by writing down the three bits
(c) 0.010012 (d) 0.100112 corresponding to the octal digit.
Thus 4378 D 100 011 1112
4. (a) 47.40625 (b) 30.8125 and 26.358 D 010 110.011 1012
(c) 53.90625 (d) 61.65625

(a) 101111.011012 (b) 11110.11012
(c) 110101.111012 (d) 111101.101012

COMPUTER NUMBERING SYSTEMS 19

Table 3.1 Natural Problem 9. Convert 5613.9062510 to a
binary number binary number, via octal
Octal digit
000 The integer part is repeatedly divided by 8, noting
0 001 the remainder, giving:
1 010
2 011 8 5613 Remainder
3 100
4 101 8 701 5
5 110
6 111 8 87 5
7
8 10 7

81 2

The ‘0’ on the extreme left does not signify any- 01
thing, thus 26.358 D 10 110.011 1012
12755
Conversion of decimal to binary via octal is demon-
strated in the following worked problems. This octal number is converted to a binary number,
(see Table 3.1)
Problem 7. Convert 371410 to a binary
number, via octal 127558 D 001 010 111 101 1012
i.e. 561310 D 1 010 111 101 1012
Dividing repeatedly by 8, and noting the remainder
gives: The fractional part is repeatedly multiplied by 8, and
noting the integer part, giving:

8 3714 Remainder 0.90625 × 8 = 7.25
0.25 × 8 = 2.00
8 464 2

8 58 0 .7 2

87 2

07 This octal fraction is converted to a binary number,
(see Table 3.1)
7202
0.728 D 0.111 0102
From Table 3.1, 72028 D 111 010 000 0102 i.e. 0.9062510 D 0.111 012
i.e. 371410 = 111 010 000 0102
Thus, 5613.9062510 = 1 010 111 101 101.111 012

Problem 8. Convert 0.5937510 to a binary Problem 10. Convert 11 110 011.100 012
number, via octal to a decimal number via octal

Multiplying repeatedly by 8, and noting the integer Grouping the binary number in three’s from the
values, gives: binary point gives: 011 110 011.100 0102

0.59375 × 8 = 4.75 Using Table 3.1 to convert this binary number to
0.75 × 8 = 6.00 an octal number gives: 363.428 and

.4 6 363.428 D 3 ð 82 C 6 ð 81 C 3 ð 80
C4ð8 1C2ð8 2
Thus 0.5937510 D 0.468
From Table 3.1, 0.468 D 0.100 1102 D 192 C 48 C 3 C 0.5 C 0.03125
i.e. 0.5937510 = 0.100 112
D 243.5312510

20 ENGINEERING MATHEMATICS

Now try the following exercise To convert from hexadecimal to decimal:

Exercise 11 Further problems on con- For example
version between decimal and 1A16 D 1 ð 161 C A ð 160
binary numbers via octal D 1 ð 161 C 10 ð 1 D 16 C 10 D 26

In Problems 1 to 3, convert the decimal i.e. 1A16 D 2610
numbers given to binary numbers, via octal. Similarly,

1. (a) 343 (b) 572 (c) 1265 2E16 D 2 ð 161 C E ð 160
D 2 ð 161 C 14 ð 160 D 32 C 14 D 4610
(a) 1010101112 (b) 10001111002
(c) 100111100012 and 1BF16 D 1 ð 162 C B ð 161 C F ð 160
D 1 ð 162 C 11 ð 161 C 15 ð 160
2. (a) 0.46875 (b) 0.6875 (c) 0.71875 D 256 C 176 C 15 D 44710

(a) 0.011112 (b) 0.10112 Table 3.2 compares decimal, binary, octal and hex-
(c) 0.101112 adecimal numbers and shows, for example, that

3. (a) 247.09375 (b) 514.4375 2310 D 101112 D 278 D 1716
(c) 1716.78125
 (a) 11110111.000112  Problem 11. Convert the following
 (b) 1000000010.01112  hexadecimal numbers into their decimal
(c) 11010110100.110012 equivalents: (a) 7A16 (b) 3F16

4. Convert the following binary numbers to (a) 7A16 D 7 ð 161 C A ð 160 D 7 ð 16 C 10 ð 1
decimal numbers via octal: D 112 C 10 D 122

(a) 111.011 1 (b) 101 001.01 Thus 7A16 = 12210
(c) 1 110 011 011 010.001 1 (b) 3F16 D 3 ð 161 C F ð 160 D 3 ð 16 C 15 ð 1

(a) 7.437510 (b) 41.2510 D 48 C 15 D 63
Thus, 3F16 = 6310
(c) 7386.187510
Problem 12. Convert the following
3.5 Hexadecimal numbers hexadecimal numbers into their decimal
equivalents: (a) C916 (b) BD16
The complexity of computers requires higher order
numbering systems such as octal (base 8) and hex- (a) C916 D C ð 161 C 9 ð 160 D 12 ð 16 C 9 ð 1
adecimal (base 16), which are merely extensions D 192 C 9 D 201
of the binary system. A hexadecimal numbering
system has a radix of 16 and uses the following 16 Thus C916 = 20110
distinct digits: (b) BD16 D B ð 161 C D ð 160 D 11 ð 16 C 13 ð 1

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F D 176 C 13 D 189
Thus BD16 = 18910
‘A’ corresponds to 10 in the denary system, B to
11, C to 12, and so on. Problem 13. Convert 1A4E16 into a denary
number

COMPUTER NUMBERING SYSTEMS 21

Table 3.2 Hence 2610 = 1A16

Decimal Binary Octal Hexadecimal Similarly, for 44710

0 0000 0 0 16 447 Remainder
1 0001 1 1
2 0010 2 2 16 27 15 ≡ F16
3 0011 3 3 16 1
4 0100 4 4 11 ≡ B16
5 0101 5 5 0 1 ≡ 116
6 0110 6 6
7 0111 7 7 1 BF
8 1000 10 8
9 1001 11 9 Thus 44710 = 1BF16
10 1010 12 A
11 1011 13 B Problem 14. Convert the following decimal
12 1100 14 C numbers into their hexadecimal equivalents:
13 1101 15 D (a) 3710 (b) 10810
14 1110 16 E
15 1111 17 F (a) 16 37 Remainder
16 10000 20 10 16 2 5 = 516
17 10001 21 11 0 2 = 216
18 10010 22 12
19 10011 23 13 most significant bit → 2 5 ← least significant bit
20 10100 24 14
21 10101 25 15 Hence 3710 = 2516
22 10110 26 16
23 10111 27 17 (b) 16 108 Remainder
24 11000 30 18
25 11001 31 19 16 6 12 = C16
26 11010 32 1A 0 6 = 616
27 11011 33 1B
28 11100 34 1C 6C
29 11101 35 1D
30 11110 36 1E Hence 10810 = 6C16
31 11111 37 1F
32 100000 40 20 Problem 15. Convert the following decimal
numbers into their hexadecimal equivalents:
(a) 16210 (b) 23910

1A4E16 (a) 16 162 Remainder
D 1 ð 163 C A ð 162 C 4 ð 161 C E ð 160 16 10 2 = 216
D 1 ð 163 C 10 ð 162 C 4 ð 161 C 14 ð 160 0 10 = A16
A2
D 1 ð 4096 C 10 ð 256 C 4 ð 16 C 14 ð 1
Hence 16210 = A216
D 4096 C 2560 C 64 C 14 D 6734 (b) 16 239 Remainder

Thus, 1A4E16 = 673410 16 14 15 = F16
0 14 = E16
To convert from decimal to hexadecimal: EF

This is achieved by repeatedly dividing by 16 and Hence 23910 = EF16
noting the remainder at each stage, as shown below
for 2610 To convert from binary to hexadecimal:

16 26 Remainder The binary bits are arranged in groups of four,
16 1 starting from right to left, and a hexadecimal symbol
10 ≡ A16
0 1 ≡ 116

most significant bit → 1 A ← least significant bit

22 ENGINEERING MATHEMATICS

is assigned to each group. For example, the binary (b) Grouping bits in fours from
number 1110011110101001 is initially grouped in
the right gives: 0001 1001 1110

fours as: 1110 0111 1010 1001 and assigning hexadecimal

and a hexadecimal symbol symbols to each group gives: 1 9 E
from Table 3.2
assigned to each group as E 7 A 9
from Table 3.2 Thus, 1100111102 = 19E16

Hence 11100111101010012 = E7A916.

To convert from hexadecimal to binary: Problem 18. Convert the following
hexadecimal numbers into their binary
The above procedure is reversed, thus, for example, equivalents: (a) 3F16 (b) A616
6CF316 D 0110 1100 1111 0011
from Table 3.2 (a) Spacing out hexadecimal

i.e. 6CF316= 1101100111100112 digits gives: 3F

and converting each into

Problem 16. Convert the following binary binary gives: 0011 1111
numbers into their hexadecimal equivalents: from Table 3.2

(a) 110101102 (b) 11001112 Thus, 3F16 = 1111112
(b) Spacing out hexadecimal digits
(a) Grouping bits in fours from the
gives: A6
right gives: 1101 0110
and converting each into binary
and assigning hexadecimal symbols
gives: 1010 0110
to each group gives: D6 from Table 3.2
from Table 3.2
Thus, A616 = 101001102
Thus, 110101102 = D616
(b) Grouping bits in fours from the Problem 19. Convert the following
hexadecimal numbers into their binary
right gives: 0110 0111 equivalents: (a) 7B16 (b) 17D16

and assigning hexadecimal symbols

to each group gives: 67 (a) Spacing out hexadecimal
Thus, 11001112 = 6716 from Table 3.2

digits gives: 7B

Problem 17. Convert the following binary and converting each into
numbers into their hexadecimal equivalents:
binary gives: 0111 1011
(a) 110011112 (b) 1100111102 from Table 3.2

Thus, 7B16 = 11110112

(a) Grouping bits in fours from the (b) Spacing out hexadecimal

right gives: 1100 1111 digits gives: 17D

and assigning hexadecimal and converting each into

symbols to each group gives: C F binary gives: 0001 0111 1101
from Table 3.2 from Table 3.2

Thus, 110011112 = CF16 Thus, 17D16 = 1011111012

COMPUTER NUMBERING SYSTEMS 23

Now try the following exercise 9. 110101112 [D716]
10. 111010102 [EA16]
Exercise 12 Further problems on hexa- 11. 100010112 [8B16]
decimal numbers 12. 101001012 [A516]

In Problems 1 to 4, convert the given hexadec- In Problems 13 to 16, convert the given hex-
imal numbers into their decimal equivalents. adecimal numbers into their binary equiva-
lents.
1. E716 [23110] 2. 2C16 [4410]
13. 3716 [1101112]
3. 9816 [15210] 4. 2F116 [75310]
In Problems 5 to 8, convert the given decimal 14. ED16 [111011012]
numbers into their hexadecimal equivalents.
15. 9F16 [100111112]
5. 5410 [3616] 6. 20010 [C816]
16. A2116 [1010001000012]
7. 9110 [5B16] 8. 23810 [EE16]
In Problems 9 to 12, convert the given binary
numbers into their hexadecimal equivalents.

4

Calculations and evaluation of
formulae

4.1 Errors and approximations 55 could therefore be expected. Certainly
an answer around 500 or 5 would not be
(i) In all problems in which the measurement of expected. Actually, by calculator
distance, time, mass or other quantities occurs, 49.1 ð 18.4 ð 122.1
an exact answer cannot be given; only an
answer which is correct to a stated degree of D 47.31, correct to
accuracy can be given. To take account of this 61.2 ð 38.1
an error due to measurement is said to exist. 4 significant figures.

(ii) To take account of measurement errors it Problem 1. The area A of a triangle is
is usual to limit answers so that the result 1
given is not more than one significant figure
greater than the least accurate number given by A D bh. The base b when
given in the data. 2

(iii) Rounding-off errors can exist with decimal measured is found to be 3.26 cm, and the
fractions. For example, to state that D perpendicular height h is 7.5 cm. Determine
3.142 is not strictly correct, but ‘ D 3.142 the area of the triangle.
correct to 4 significant figures’ is a true state-
ment. (Actually, D 3.14159265 . . .) 11
Area of triangle D bh D ð 3.26 ð 7.5 D
(iv) It is possible, through an incorrect procedure,
to obtain the wrong answer to a calculation. 22
This type of error is known as a blunder. 12.225 cm2 (by calculator).

(v) An order of magnitude error is said to exist The approximate value is 1 ð 3 ð 8 D 12 cm2, so
if incorrect positioning of the decimal point 2
occurs after a calculation has been completed.
there are no obvious blunder or magnitude errors.
(vi) Blunders and order of magnitude errors can However, it is not usual in a measurement type
be reduced by determining approximate val- problem to state the answer to an accuracy greater
ues of calculations. Answers which do not than 1 significant figure more than the least accurate
seem feasible must be checked and the cal- number in the data: this is 7.5 cm, so the result
culation must be repeated as necessary. should not have more than 3 significant figures

An engineer will often need to make a Thus, area of triangle = 12.2 cm2

quick mental approximation for a calcula- Problem 2. State which type of error has
been made in the following statements:
tion. For example, 49.1 ð 18.4 ð 122.1 may
(a) 72 ð 31.429 D 2262.9
61.2 ð 38.1
(b) 16 ð 0.08 ð 7 D 89.6
be approximated to 50 ð 20 ð 120 and then,
(c) 11.714 ð 0.0088 D 0.3247 correct to
60 ð 40 4 decimal places.

by cancelling, 50 ð1 20 ð 120 2 1 29.74 ð 0.0512
D 50. An (d) D 0.12, correct to
1 60 ð 40 2 1
11.89
2 significant figures.

accurate answer somewhere between 45 and

CALCULATIONS AND EVALUATION OF FORMULAE 25

(a) 72 ð 31.429 D 2262.888 (by calculator), 2.19 ð 203.6 ð 17.91
hence a rounding-off error has occurred. The i.e. ³ 80
answer should have stated:
12.1 ð 8.76

(By calculator, 2.19 ð 203.6 ð 17.91 D 75.3,

72 ð 31.429 D 2262.9, correct to 5 significant 12.1 ð 8.76
figures or 2262.9, correct to 1 decimal place.
correct to 3 significant figures.)
8 32 ð 7
(b) 16 ð 0.08 ð 7 D 16 ð ð 7 D Now try the following exercise

100 25 Exercise 13 Further problems on errors
224 24
D D 8 D 8.96
25 25

Hence an order of magnitude error has In Problems 1 to 5 state which type of error,
occurred. or errors, have been made:

(c) 11.714 ð 0.0088 is approximately equal to 1. 25 ð 0.06 ð 1.4 D 0.21
12 ð 9 ð 10 3, i.e. about 108 ð 10 3 or 0.108. [order of magnitude error]

Thus a blunder has been made.

(d) 29.74 ð 0.0512 30 ð 5 ð 10 2 2. 137 ð 6.842 D 937.4
³
11.89 12 Rounding-off error–should add ‘correct
to 4 significant figures’ or ‘correct to
150 15 1 1 decimal place’
D 12 ð 102 D 120 D 8 or 0.125
24 ð 0.008 [Blunder]
3. D 10.42
hence no order of magnitude error has
12.6

occurred. However, 29.74 ð 0.0512 D 0.128 4. For a gas pV D c. When pressure
p D 1 03 400 Pa and V D 0.54 m3 then
11.89 c D 55 836 Pa m3.
correct to 3 significant figures, which equals
Measured values, hence
0.13 correct to 2 significant figures. c D 55 800 Pa m3

Hence a rounding-off error has occurred.

Problem 3. Without using a calculator, 4.6 ð 0.07
determine an approximate value of: 5. D 0.225

52.3 ð 0.274
 Order of magnitude error and rounding- 

11.7 ð 19.1 2.19 ð 203.6 ð 17.91  off error–should be 0.0225, correct to 
3 significant figures or 0.0225,
(a) (b) 12.1 ð 8.76
9.3 ð 5.7
correct to 4 decimal places

11.7 ð 19.1 In Problems 6 to 8, evaluate the expressions
(a) is approximately equal to approximately, without using a calculator.

9.3 ð 5.7 6. 4.7 ð 6.3 [³30 (29.61, by calculator)]
10 ð 20
2.87 ð 4.07
, i.e. about 4 7.
10 ð 5
6.12 ð 0.96
(By calculator, 11.7 ð 19.1 D 4.22, correct to
³2 (1.988, correct to 4 s.f., by
9.3 ð 5.7 calculator)

3 significant figures.)

2.19 ð 203.6 ð 17.91 2 ð 20 200 ð 20 2 72.1 ð 1.96 ð 48.6
(b) ³ 8.
12.1 ð 8.76 1 10 ð 10 1
139.3 ð 5.2
D 2 ð 20 ð 2 after cancelling,
³10 (9.481, correct to 4 s.f., by
calculator)

26 ENGINEERING MATHEMATICS 1
(a) D 0.01896453 . . . D 0.019, correct to 3
4.2 Use of calculator
52.73
The most modern aid to calculations is the pocket- decimal places
sized electronic calculator. With one of these, cal-
culations can be quickly and accurately performed, 1
correct to about 9 significant figures. The scientific (b) D 36.3636363 . . . D 36.364, correct to
type of calculator has made the use of tables and
logarithms largely redundant. 0.0275
3 decimal places
To help you to become competent at using your
calculator check that you agree with the answers to 11
the following problems: (c) C D 0.71086624 . . . D 0.711, cor-

4.92 1.97
rect to 3 decimal places

Problem 4. Evaluate the following, correct Problem 7. Evaluate the following,
to 4 significant figures: expressing the answers in standard form,
correct to 4 significant figures.
(a) 4.7826 C 0.02713 (b) 17.6941 11.8762
(c) 21.93 ð 0.012981 (a) 0.00451 2 (b) 631.7 6.21 C 2.95 2

(a) 4.7826 C 0.02713 D 4.80973 D 4.810, correct (c) 46.272 31.792
to 4 significant figures
(a) 0.00451 2 D 2.03401ð10 5 D 2.034 × 10−5,
(b) 17.6941 11.8762 D 5.8179 D 5.818, correct correct to 4 significant figures
to 4 significant figures
(b) 631.7 6.21 C 2.95 2 D 547.7944 D
(c) 21.93 ð 0.012981 D 0.2846733 . . . D 0.2847, 5.477944 ð 102 D 5.478 × 102, correct to 4
correct to 4 significant figures significant figures

Problem 5. Evaluate the following, correct (c) 46.272 31.792 D 1130.3088 D 1.130 × 103,
to 4 decimal places: correct to 4 significant figures

(a) 46.32 ð 97.17 ð 0.01258 4.621 Problem 8. Evaluate the following, correct
1 (b) to 3 decimal places:

(c) 62.49 ð 0.0172 23.76
2
2.37 2 3.60 2 5.40 2
(a) (b) C
0.0526 1.92 2.45
(a) 46.32 ð 97.17 ð 0.01258 D 56.6215031 . . . D
56.6215, correct to 4 decimal places 15
4.621 (c) 7.62 4.82

(b) D 0.19448653 . . . D 0.1945, correct to 2.37 2
23.76 (a) D 106.785171 . . . D 106.785, correct
4 decimal places
1 0.0526
to 3 decimal places
(c) 62.49 ð 0.0172 D 0.537414 D 0.5374,
2 3.60 2 5.40 2
correct to 4 decimal places

(b) C D 8.37360084 . . . D
1.92 2.45

Problem 6. Evaluate the following, correct 8.374, correct to 3 decimal places
to 3 decimal places:

1 1 11 15
(a) (b) (c) C (c) 7.62 4.82 D 0.43202764 . . . D 0.432, cor-
52.73 0.0275 4.92 1.97
rect to 3 decimal places
























Click to View FlipBook Version