EXTRA EXERCISE FROM JOHN BIRDS BOOK

Engineering Mathematics

In memory of Elizabeth

Engineering Mathematics

Fifth edition

John Bird BSc(Hons), CEng, CSci, CMath, FIET, MIEE,
FIIE, FIMA, FCollT

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Fourth edition 2003
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7 8 9 10 11 11 10 9 8 7 6 5 4 3 2 1

Contents

Preface xii 6 Further algebra 48
48
Section 1 Number and Algebra 6.1 Polynominal division 50
52
1 6.2 The factor theorem
6.3 The remainder theorem

1 Revision of fractions, decimals 3 7 Partial fractions 54
and percentages 3 7.1 Introduction to partial 54
1.1 Fractions 5 fractions 54
1.2 Ratio and proportion 6 7.2 Worked problems on partial 57
1.3 Decimals 9 fractions with linear factors 58
1.4 Percentages 7.3 Worked problems on partial 60
11 fractions with repeated linear factors 60
2 Indices, standard form and engineering 11 7.4 Worked problems on partial 60
notation 12 fractions with quadratic factors 62
2.1 Indices 13 64
2.2 Worked problems on indices 15 8 Simple equations 65
2.3 Further worked problems on indices 15 8.1 Expressions, equations and
2.4 Standard form identities 67
2.5 Worked problems on standard form 16 8.2 Worked problems on simple
2.6 Further worked problems on equations 68
standard form 17 8.3 Further worked problems on 68
2.7 Engineering notation and common simple equations
prefixes 19 8.4 Practical problems involving 68
19 simple equations 70
3 Computer numbering systems 19 8.5 Further practical problems
3.1 Binary numbers 20 involving simple equations 72
3.2 Conversion of binary to decimal 73
3.3 Conversion of decimal to binary 21 Revision Test 2 77
3.4 Conversion of decimal to 23 77
binary via octal 9 Simultaneous equations
3.5 Hexadecimal numbers 27 9.1 Introduction to simultaneous
27 equations
4 Calculations and evaluation of formulae 29 9.2 Worked problems on
4.1 Errors and approximations 31 simultaneous equations
4.2 Use of calculator 32 in two unknowns
4.3 Conversion tables and charts 9.3 Further worked problems on
4.4 Evaluation of formulae 37 simultaneous equations
9.4 More difficult worked
Revision Test 1 38 problems on simultaneous
38 equations
5 Algebra 40 9.5 Practical problems involving
5.1 Basic operations 42 simultaneous equations
5.2 Laws of Indices
5.3 Brackets and factorisation 44 10 Transposition of formulae
5.4 Fundamental laws and 10.1 Introduction to transposition
precedence 46 of formulae
5.5 Direct and inverse
proportionality

vi Contents 77 15.3 Further worked problems on 115
arithmetic progressions 117
10.2 Worked problems on 78 118
transposition of formulae 15.4 Geometric progressions 119
80 15.5 Worked problems on 120
10.3 Further worked problems on
transposition of formulae 83 geometric progressions 122
15.6 Further worked problems on 122
10.4 Harder worked problems on 83 123
transposition of formulae geometric progressions 123
83 15.7 Combinations and 125
11 Quadratic equations 127
11.1 Introduction to quadratic 85 permutations
equations 130
11.2 Solution of quadratic 87 16 The binomial series 130
equations by factorisation 16.1 Pascal’s triangle 130
11.3 Solution of quadratic 88 16.2 The binomial series 131
equations by ‘completing 16.3 Worked problems on the
the square’ 90 binomial series 133
11.4 Solution of quadratic 16.4 Further worked problems on
equations by formula 91 the binomial series 134
11.5 Practical problems involving 91 16.5 Practical problems involving
quadratic equations 91 the binomial theorem 139
11.6 The solution of linear and 92
quadratic equations 93 17 Solving equations by iterative methods 141
simultaneously 17.1 Introduction to iterative methods 141
94 17.2 The Newton–Raphson method 141
12 Inequalities 95 17.3 Worked problems on the 142
12.1 Introduction in inequalities Newton–Raphson method 145
12.2 Simple inequalities 97 147
12.3 Inequalities involving a modulus 97 Revision Test 4 148
12.4 Inequalities involving quotients 97
12.5 Inequalities involving square 100 Multiple choice questions on 150
functions 101 Chapters 1–17 150
12.6 Quadratic inequalities 150
102 Section 2 Mensuration 152
13 Logarithms 153
13.1 Introduction to logarithms 103 18 Areas of plane figures 155
13.2 Laws of logarithms 103 18.1 Mensuration
13.3 Indicial equations 103 18.2 Properties of quadrilaterals
13.4 Graphs of logarithmic functions 104 18.3 Worked problems on areas of
106 plane figures
Revision Test 3 108 18.4 Further worked problems on
108 areas of plane figures
14 Exponential functions 110 18.5 Worked problems on areas of
14.1 The exponential function composite figures
14.2 Evaluating exponential functions 114 18.6 Areas of similar shapes
14.3 The power series for ex 114
14.4 Graphs of exponential functions 19 The circle and its properties
14.5 Napierian logarithms 114 19.1 Introduction
14.6 Evaluating Napierian logarithms 19.2 Properties of circles
14.7 Laws of growth and decay 19.3 Arc length and area of a sector
19.4 Worked problems on arc
15 Number sequences length and sector of a circle
15.1 Arithmetic progressions 19.5 The equation of a circle
15.2 Worked problems on
arithmetic progressions

20 Volumes and surface areas of 157 24.3 Changing from polar into Contents vii
common solids 157 Cartesian co-ordinates
20.1 Volumes and surface areas of 157 213
regular solids 24.4 Use of R → P and P → R 214
20.2 Worked problems on volumes 160 functions on calculators
and surface areas of regular solids 164
20.3 Further worked problems on 167 Revision Test 6 215
volumes and surface areas of 170
regular solids 172 25 Triangles and some practical 216
20.4 Volumes and surface areas of applications 216
frusta of pyramids and cones 174 25.1 Sine and cosine rules 216
20.5 The frustum and zone of 174 25.2 Area of any triangle 216
a sphere 176 25.3 Worked problems on the solution
20.6 Prismoidal rule 177 of triangles and their areas 218
20.7 Volumes of similar shapes 25.4 Further worked problems on 220
the solution of triangles and 222
21 Irregular areas and volumes and their areas
mean values of waveforms 25.5 Practical situations involving 225
21.1 Area of irregular figures trigonometry 225
21.2 Volumes of irregular solids 25.6 Further practical situations 225
21.3 The mean or average value of involving trigonometry 226
a waveform 227
26 Trigonometric identities and equations 228
Revision Test 5 182 26.1 Trigonometric identities 229
26.2 Worked problems on 229
Section 3 Trigonometry 185 trigonometric identities
26.3 Trigonometric equations 231
22 Introduction to trigonometry 187 26.4 Worked problems (i) on 231
22.1 Trigonometry 187 trigonometric equations 233
22.2 The theorem of Pythagoras 187 26.5 Worked problems (ii) on 236
22.3 Trigonometric ratios of acute angles 188 trigonometric equations
22.4 Fractional and surd forms of 26.6 Worked problems (iii) on 238
trigonometric ratios 190 trigonometric equations
22.5 Solution of right-angled triangles 191 26.7 Worked problems (iv) on 239
22.6 Angle of elevation and depression 193 trigonometric equations
22.7 Evaluating trigonometric
ratios of any angles 195 27 Compound angles
22.8 Trigonometric approximations 27.1 Compound angle formulae
for small angles 197 27.2 Conversion of a sin ωt + b cos ωt
into R sin(ωt + α)
23 Trigonometric waveforms 199 27.3 Double angles
23.1 Graphs of trigonometric functions 199 27.4 Changing products of sines
23.2 Angles of any magnitude 199 and cosines into sums or
23.3 The production of a sine and differences
cosine wave 202 27.5 Changing sums or differences
23.4 Sine and cosine curves 202 of sines and cosines into
23.5 Sinusoidal form A sin(ωt ± α) 206 products
23.6 Waveform harmonics 209
Revision Test 7 241
24 Cartesian and polar co-ordinates 211
24.1 Introduction 211 Multiple choice questions on 242
24.2 Changing from Cartesian into Chapters 18–27
polar co-ordinates 211

viii Contents 247 34.3 Determining resultant 308
phasors by calculation
Section 4 Graphs 249 311
249 Section 6 Complex Numbers
28 Straight line graphs 249 313
28.1 Introduction to graphs 35 Complex numbers 313
28.2 The straight line graph 255 35.1 Cartesian complex numbers 314
28.3 Practical problems involving 35.2 The Argand diagram
straight line graphs 261 35.3 Addition and subtraction of 314
261 complex numbers
29 Reduction of non-linear laws to 35.4 Multiplication and division of 315
linear form 264 complex numbers 317
29.1 Determination of law 35.5 Complex equations
29.2 Determination of law 269 35.6 The polar form of a complex 318
involving logarithms 269 number
269 35.7 Multiplication and division in 320
30 Graphs with logarithmic scales 272 polar form
30.1 Logarithmic scales 273 35.8 Applications of complex 321
30.2 Graphs of the form y = axn numbers
30.3 Graphs of the form y = abx 276 325
30.4 Graphs of the form y = aekx 36 De Moivre’s theorem 325
276 36.1 Introduction 325
31 Graphical solution of equations 36.2 Powers of complex numbers 326
31.1 Graphical solution of 277 36.3 Roots of complex numbers
simultaneous equations 329
31.2 Graphical solution of 281 Revision Test 9
quadratic equations 331
31.3 Graphical solution of linear 282 Section 7 Statistics
and quadratic equations 333
simultaneously 284 37 Presentation of statistical data 333
31.4 Graphical solution of cubic 284 37.1 Some statistical terminology 334
equations 286 37.2 Presentation of ungrouped data 338
291 37.3 Presentation of grouped data
32 Functions and their curves 345
32.1 Standard curves 291 38 Measures of central tendency and 345
32.2 Simple transformations 291 dispersion
32.3 Periodic functions 293 38.1 Measures of central tendency 345
32.4 Continuous and 38.2 Mean, median and mode for
discontinuous functions 297 discrete data 346
32.5 Even and odd functions 38.3 Mean, median and mode for 348
32.6 Inverse functions 299 grouped data
38.4 Standard deviation 350
Revision Test 8 301 38.5 Quartiles, deciles and
301 percentiles 352
Section 5 Vectors 301 352
302 39 Probability 353
33 Vectors 305 39.1 Introduction to probability
33.1 Introduction 39.2 Laws of probability 353
33.2 Vector addition 307 39.3 Worked problems on
33.3 Resolution of vectors probability 355
33.4 Vector subtraction 307 39.4 Further worked problems on
307 probability
34 Combination of waveforms
34.1 Combination of two periodic
functions
34.2 Plotting periodic functions

Contents ix

39.5 Permutations and 44.5 Tangents and normals 411
combinations 357 44.6 Small changes 412

Revision Test 10 359 Revision Test 12 415
360
40 The binomial and Poisson distribution 360 45 Differentiation of parametric 416
40.1 The binomial distribution 363 equations 416
40.2 The Poisson distribution 45.1 Introduction to parametric 416
366 equations 417
41 The normal distribution 366 45.2 Some common parametric
41.1 Introduction to the normal 371 equations 418
distribution 45.3 Differentiation in parameters
41.2 Testing for a normal 374 45.4 Further worked problems on 421
distribution differentiation of parametric 421
375 equations 421
Revision Test 11
381 46 Differentiation of implicit functions 422
Multiple choice questions on 46.1 Implicit functions 423
Chapters 28–41 383 46.2 Differentiating implicit
383 functions 426
Section 8 Differential Calculus 383 46.3 Differentiating implicit 426
384 functions containing 426
42 Introduction to differentiation 385 products and quotients 426
42.1 Introduction to calculus 387 46.4 Further implicit 429
42.2 Functional notation 388 differentiation
42.3 The gradient of a curve 390 431
42.4 Differentiation from first 47 Logarithmic differentiation
principles 392 47.1 Introduction to logarithmic 433
42.5 Differentiation of y = axn by 392 differentiation
the general rule 394 47.2 Laws of logarithms 435
42.6 Differentiation of sine and 395 47.3 Differentiation of logarithmic 435
cosine functions 397 functions 435
42.7 Differentiation of eax and ln ax 398 47.4 Differentiation of [f (x)]x 436
439
43 Methods of differentiation 400 Revision Test 13
43.1 Differentiation of common 400 442
functions 401 Section 9 Integral Calculus 442
43.2 Differentiation of a product 404 442
43.3 Differentiation of a quotient 48 Standard integration
43.4 Function of a function 408 48.1 The process of integration 442
43.5 Successive differentiation 48.2 The general solution of
integrals of the form axn
44 Some applications of differentiation 48.3 Standard integrals
44.1 Rates of change 48.4 Definite integrals
44.2 Velocity and acceleration
44.3 Turning points 49 Integration using algebraic
44.4 Practical problems involving substitutions
maximum and minimum 49.1 Introduction
values 49.2 Algebraic substitutions
49.3 Worked problems on
integration using algebraic
substitutions

x Contents

49.4 Further worked problems on 444 54.3 The mid-ordinate rule 471
integration using algebraic 444 54.4 Simpson’s rule 473
substitutions
447 Revision Test 15 477
49.5 Change of limits 447
55 Areas under and between curves 478
50 Integration using trigonometric 447 55.1 Area under a curve 478
substitutions 449 55.2 Worked problems on the area 479
50.1 Introduction 450 under a curve 482
50.2 Worked problems on 451 55.3 Further worked problems on 484
integration of sin2 x, cos2 x, 453 the area under a curve
tan2 x and cot2 x 55.4 The area between curves 487
50.3 Worked problems on powers 487
of sines and cosines 56 Mean and root mean square values 489
50.4 Worked problems on integration of 56.1 Mean or average values
products of sines and cosines 56.2 Root mean square values 491
50.5 Worked problems on integration 491
using the sin θ substitution 57 Volumes of solids of revolution 492
50.6 Worked problems on integration 57.1 Introduction
using the tan θ substitution 57.2 Worked problems on volumes 493
of solids of revolution
Revision Test 14 454 57.3 Further worked problems on 496
volumes of solids of 496
51 Integration using partial fractions 455 revolution 496
51.1 Introduction 455 496
51.2 Worked problems on 58 Centroids of simple shapes 497
integration using partial 455 58.1 Centroids 497
fractions with linear factors 58.2 The first moment of area 498
51.3 Worked problems on integration 456 58.3 Centroid of area between a 501
using partial fractions with curve and the x-axis
repeated linear factors 457 58.4 Centroid of area between a 505
51.4 Worked problems on integration curve and the y-axis 505
using partial fractions with 460 58.5 Worked problems on 505
quadratic factors 460 centroids of simple shapes 506
460 58.6 Further worked problems 506
θ on centroids of simple shapes 506
52 The t = tan substitution 462 58.7 Theorem of Pappus
507
2 464 59 Second moments of area
52.1 Introduction 464 59.1 Second moments of area and 510
52.2 Worked problems on the 464 radius of gyration
466 59.2 Second moment of area of
t = tan θ substitution 469 regular sections
2 469 59.3 Parallel axis theorem
469 59.4 Perpendicular axis theorem
52.3 Further worked problems on 59.5 Summary of derived results
the t = tan θ substitution 59.6 Worked problems on second
2 moments of area of regular
sections
53 Integration by parts 59.7 Worked problems on second
53.1 Introduction moments of area of
53.2 Worked problems on composite areas
integration by parts
53.3 Further worked problems on
integration by parts

54 Numerical integration
54.1 Introduction
54.2 The trapezoidal rule

Revision Test 16 512 62.2 Solution of simultaneous
equations by determinants
Section 10 Further Number and 513 548
Algebra 62.3 Solution of simultaneous 552
equations using Cramers rule 553
60 Boolean algebra and logic circuits 515
60.1 Boolean algebra and Revision Test 17 555
switching circuits 515
60.2 Simplifying Boolean Section 11 Differential Equations 557
expressions 520 557
60.3 Laws and rules of Boolean 63 Introduction to differential 558
algebra 520 equations 558
60.4 De Morgan’s laws 522 63.1 Family of curves 560
60.5 Karnaugh maps 523 63.2 Differential equations 562
60.6 Logic circuits 528 63.3 The solution of equations of
60.7 Universal logic gates 532 the form dy = f (x) 565
dx
61 The theory of matrices and 536 63.4 The solution of equations of 566
determinants 536 the form dy = f (y) 570
61.1 Matrix notation dx
61.2 Addition, subtraction and 536 63.5 The solution of equations of 571
multiplication of matrices 540 the form dy = f (x) · f (y)
61.3 The unit matrix 540 dx
61.4 The determinant of a 2 by 2 matrix
61.5 The inverse or reciprocal of a 541 Revision Test 18
2 by 2 matrix 542
61.6 The determinant of a 3 by 3 matrix Multiple choice questions on
61.7 The inverse or reciprocal of a 544 Chapters 42–63
3 by 3 matrix
546 Answers to multiple choice questions
62 The solution of simultaneous
equations by matrices and 546 Index
determinants
62.1 Solution of simultaneous
equations by matrices

Preface

Engineering Mathematics 5th Edition covers a wide statistics, differential calculus, integral calculus, further
range of syllabus requirements. In particular, the book number and algebra and differential equations.
is most suitable for the latest National Certificate and
Diploma courses and City & Guilds syllabuses in This new edition covers, in particular, the following
Engineering. syllabuses:

This text will provide a foundation in mathemat- (i) Mathematics for Technicians, the core unit for
ical principles, which will enable students to solve National Certificate/Diploma courses in Engi-
mathematical, scientific and associated engineering neering, to include all or part of the following
principles. In addition, the material will provide chapters:
engineering applications and mathematical principles 1. Algebraic methods: 2, 5, 11, 13, 14, 28, 30
necessary for advancement onto a range of Incorporated (1, 4, 8, 9 and 10 for revision)
Engineer degree profiles. It is widely recognised that a 2. Trigonometric methods and areas and vol-
students’ ability to use mathematics is a key element in umes: 18–20, 22–25, 33, 34
determining subsequent success. First year undergrad- 3. Statistical methods: 37, 38
uates who need some remedial mathematics will also 4. Elementary calculus: 42, 48, 55
find this book meets their needs.
(ii) Further Mathematics for Technicians, the
In Engineering Mathematics 5th Edition, new optional unit for National Certificate/Diploma
material is included on inequalities, differentiation of courses in Engineering, to include all or part of
parametric equations, implicit and logarithmic func- the following chapters:
tions and an introduction to differential equations. 1. Advanced graphical techniques: 29–31
Because of restraints on extent, chapters on lin- 2. Algebraic techniques: 15, 35, 38
ear correlation, linear regression and sampling and 2. Trigonometry: 23, 26, 27, 34
estimation theories have been removed. However, 3. Calculus: 42–44, 48, 55–56
these three chapters are available to all via the
internet. (iii) The mathematical contents of Electrical and
Electronic Principles units of the City & Guilds
A new feature of this fifth edition is that a free Inter- Level 3 Certificate in Engineering (2800).
net download is available of a sample of solutions
(some 1250) of the 1750 further problems contained in (iv) Any introductory/access/foundation course
the book – see below. involving Engineering Mathematics at
University, Colleges of Further and Higher
Another new feature is a free Internet download education and in schools.
(available for lecturers only) of all 500 illustrations
contained in the text – see below. Each topic considered in the text is presented in a way
that assumes in the reader little previous knowledge of
Throughout the text theory is introduced in each that topic.
chapter by a simple outline of essential definitions,
formulae, laws and procedures. The theory is kept to Engineering Mathematics 5th Edition provides
a minimum, for problem solving is extensively used a follow-up to Basic Engineering Mathematics and
to establish and exemplify the theory. It is intended a lead into Higher Engineering Mathematics 5th
that readers will gain real understanding through see- Edition.
ing problems solved and then through solving similar
problems themselves. This textbook contains over 1000 worked problems,
followed by some 1750 further problems (all with
For clarity, the text is divided into eleven topic
areas, these being: number and algebra, mensura-
tion, trigonometry, graphs, vectors, complex numbers,

answers). The further problems are contained within structure. Lecturers’ may obtain a complimentary set
some 220 Exercises; each Exercise follows on directly of solutions of the Revision Tests in an Instructor’s
from the relevant section of work, every two or three Manual available from the publishers via the internet –
pages. In addition, the text contains 238 multiple- see below.
choice questions. Where at all possible, the problems
mirror practical situations found in engineering and sci- A list of Essential Formulae is included in
ence. 500 line diagrams enhance the understanding of the Instructor’s Manual for convenience of reference.
the theory. Learning by Example is at the heart of Engineering
Mathematics 5th Edition.
At regular intervals throughout the text are some 18
Revision tests to check understanding. For example, JOHN BIRD
Revision test 1 covers material contained in Chapters Royal Naval School of Marine Engineering,
1 to 4, Revision test 2 covers the material in Chapters
5 to 8, and so on. These Revision Tests do not have HMS Sultan,
answers given since it is envisaged that lecturers could formerly University of Portsmouth and
set the tests for students to attempt as part of their course
Highbury College,
Portsmouth

Free web downloads

Additional material on statistics

Chapters on Linear correlation, Linear regression and
Sampling and estimation theories are available for free
to students and lecturers at http://books.elsevier.com/
companions/9780750685559

In addition, a suite of support material is available to
lecturers only from Elsevier’s textbook website.

Solutions manual

Within the text are some 1750 further problems arranged
within 220 Exercises. A sample of over 1250 worked
solutions has been prepared for lecturers.

Instructor’s manual

This manual provides full worked solutions and mark
scheme for all 18 Revision Tests in this book.

Illustrations

Lecturers can also download electronic files for all
illustrations in this fifth edition.

To access the lecturer support material, please go to
http://textbooks.elsevier.com and search for the book.
On the book web page, you will see a link to the Instruc-
tor Manual on the right. If you do not have an account
for the textbook website already, you will need to reg-
ister and request access to the book’s subject area. If
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the right subject area, please follow the ’Request Access
to this Subject Area’ link at the top of the subject area
homepage.

Section 1

Number and Algebra

This page intentionally left blank

Chapter 1

Revision of fractions,
decimals and percentages

1.1 Fractions Alternatively:

When 2 is divided by 3, it may be written as 2 or 2/3 or Step (2) Step (3)
3
2 ↓↓
3
2/3. is called a fraction. The number above the line, 1 + 2 = (7 × 1) + (3 × 2)

i.e. 2, is called the numerator and the number below 37 21

the line, i.e. 3, is called the denominator. ↑

When the value of the numerator is less than the Step (1)

value of the denominator, the fraction is called a proper

fraction; thus 2 is a proper fraction. When the value Step 1: the LCM of the two denominators;
3
of the numerator is greater than the denominator, the
1
fraction is called an improper fraction. Thus 7 is Step 2: for the fraction 3 , 3 into 21 goes 7 times,
3
an improper fraction and can also be expressed as a 7 × the numerator is 7 × 1;

mixed number, that is, an integer and a proper frac- Step 3: for the fraction 2 , 7 into 21 goes 3 times,
7
tion. Thus the improper fraction 7 is equal to the mixed 3 × the numerator is 3 × 2.
3
1
number 2 3 . Thus 1 + 2 = 7 + 6 = 13 as obtained previously.
3 7 21 21
When a fraction is simplified by dividing the numer-

ator and denominator by the same number, the pro-

cess is called cancelling. Cancelling by 0 is not 2 1
3
permissible. Problem 2. Find the value of − 2
36

Problem 1. Simplify 1 + 2 One method is to split the mixed numbers into integers
37 and their fractional parts. Then

The lowest common multiple (i.e. LCM) of the two 2 − 1 = 3+ 2 − 2+ 1
denominators is 3 × 7, i.e. 21. 3 2
36 3 6
Expressing each fraction so that their denominators
are 21, gives: =3+ 2 −2− 1
36
1+2=1×7+2×3= 7 + 6
3 7 3 7 7 3 21 21 = 1 + 4 − 1 = 3 = 1
1 1
= 7 + 6 = 13 66 6 2
21 21
Another method is to express the mixed numbers as
improper fractions.

4 Engineering Mathematics

Since 3 = 9 , then 2 = 9 + 2 = 11 = 8 × 17¡ × &2&4 8 = 8 × 1 × 8
3 5 13¡ 7¡1 5 × 1 × 1
Section 1 3 333 3

Similarly, 2 1 = 12 + 1 = 13 = 64 = 4
6666 12
55

Thus 3 2 − 2 1 = 11 − 13 = 22 − 13 = 9 = 1 1 Problem 6. Simplify 3 ÷ 12
3 63 6 6 66 2 7 21

as obtained previously.

Problem 3. Determine the value of 3

5 − 1 + 2 3 ÷ 12 = 7
4 3 1 7 21 12
845
21

5 1 2 5−1+2 Multiplying both numerator and denominator by the
3 1 reciprocal of the denominator gives:

4 − + = (4 − 3 + 1) +
845 845 1¡3 &2&1 3
3 1¡7 &1&2 4 3
= 2 + 5 × 5 − 10 × 1 + 8 × 2 × 3
40 7 4 4
12 = 1&1&2 &2&1 1 = 1 =
= 2 + 25 − 10 + 16 1 &2&1 &1&2 1
40 21 ×

=2+ 31 = 31 This method can be remembered by the rule: invert the
2 second fraction and change the operation from division
40 40 to multiplication. Thus:

Problem 4. Find the value of 3 × 14 3 ÷ 12 = 13¡ × &2&1 3 = 3 as obtained previously.
7 15 7 21 1¡7 &1&2 4 4

Dividing numerator and denominator by 3 gives: Problem 7. Find the value of 3 ÷ 71
5
53
1¡3
7 14 = 1 × 14 = 1 × 14
× &1&55 7 5 7×5
The mixed numbers must be expressed as improper
Dividing numerator and denominator by 7 gives: fractions. Thus,

1 × &1&4 2 = 1×2 = 2 3 ÷ 71 = 28 ÷ 22 = 14 &2&8 × 3 = 42
17¡ × 5 1×5 5 5 3 5 3 5 &2&2 11 55

5

This process of dividing both the numerator and denom-

inator of a fraction by the same factor(s) is called Problem 8. Simplify

cancelling.

1− 2+1 ÷ 3×1
3 1 3 3 54 83
Problem 5. Evaluate 1 × 2 × 3
537

Mixed numbers must be expressed as improper frac- The order of precedence of operations for problems
tions before multiplication can be performed. Thus, containing fractions is the same as that for integers,
i.e. remembered by BODMAS (Brackets, Of, Division,
313 Multiplication, Addition and Subtraction). Thus,
1 ×2 ×3

537

= 5 + 3 × 6 + 1 × 21 + 3 1− 2+1 ÷ 3×1
55 33 77 3 54 83

Revision of fractions,decimals and percentages 5

= 1 − 4 × 2+5 × 1 ÷ 3¡ 1 (B) 2. (a) 2 + 3 (b) 2 − 1 + 2 Section 1
3 20 &2&4 8 (D)
(M) 7 11 973
¡8 2 (S)
= 1 − 13 × 1 43 47
3 5 &2&0 (B) (a) (b)
(O)
(D) 77 63
(M)
1 26 (A) 3. (a) 3 − 2 (b) 1 − 4 4 + 5
=− (A) 10 8 3 1
(S) 73 456
35

= (5 × 1) − (3 × 26) 16 17
15 (a) 1 (b)

= −73 = −4 13 21 60
15 15
4. (a) 3 × 5 (b) 17 × 15
49 35 119

Problem 9. Determine the value of 53
(a) (b)
7 11 131
12 49

of 3 − 2 + 5 ÷ − 5. (a) 3 × 7 × 1 2 (b) 13 × 4 7 × 3 4
6 24 8 16 2

59 7 17 11 39

7 1 − 1 + 1 ÷ 3 −1 3
of 3 2 5 (a) (b) 11
6 24 8 16 2
5

= 7 of 1 + 41 ÷ 3 −1 6. (a) 3 ÷ 45 (b) 1 ÷ 5
1 1 2
6 4 8 16 2 8 64 39

= 7 × 5 + 41 ÷ 3 − 1 8 12
6 4 8 16 2 (a) (b)

= 7 × 5 + 41 × &1&6 2 − 1 15 23
6 4 18¡ 3 2
7. 1 + 3 ÷ 8 − 1 7
= 35 + 82 − 1 2 5 15 3 1
24 3 2
24
= 35 + 656 − 1
24 2 8. 7 15 × 5 + 3 ÷ 15 4
of 5
= 691 − 1 15 7 4 16
24 2 5

= 691 − 12 9. 1 × 2 − 1 ÷ 3 + 2 − 13
24 43357 126

= 679 = 28 7 10. 2 × 1 ÷ 2+1 + 13 28
24 24 1 2
34 34 5
55

11. If a storage tank is holding 450 litres when it is
three-quarters full, how much will it contain
when it is two-thirds full?
[400 litres]

12. Three people, P, Q and R contribute to a fund.

Now try the following exercise P provides 3/5 of the total, Q provides 2/3 of

the remainder, and R provides £8. Determine

(a) the total of the fund, (b) the contributions

Exercise 1 Further problems on fractions of P and Q. [(a) £60 (b) £36, £16]
Evaluate the following:

12 71 1.2 Ratio and proportion
1. (a) + (b) −
25 16 4 The ratio of one quantity to another is a fraction, and
93 is the number of times one quantity is contained in
(a) (b) another quantity of the same kind. If one quantity is

10 16

6 Engineering Mathematics

Section 1 directly proportional to another, then as one quantity 1 person takes three times as long, i.e.
doubles, the other quantity also doubles. When a quan-
tity is inversely proportional to another, then as one 4 × 3 = 12 hours,
quantity doubles, the other quantity is halved.
5 people can do it in one fifth of the time that one
Problem 10. A piece of timber 273 cm long is 12
cut into three pieces in the ratio of 3 to 7 to 11.
Determine the lengths of the three pieces person takes, that is hours or 2 hours 24 minutes.
5
The total number of parts is 3 + 7 + 11, that is, 21.
Hence 21 parts correspond to 273 cm Now try the following exercise

1 part corresponds to 273 = 13 cm Exercise 2 Further problems on ratio and
21 proportion

3 parts correspond to 3 × 13 = 39 cm 1. Divide 621 cm in the ratio of 3 to 7 to 13.
7 parts correspond to 7 × 13 = 91 cm [81 cm to 189 cm to 351 cm]
11 parts correspond to 11 × 13 = 143 cm
2. When mixing a quantity of paints, dyes of
i.e. the lengths of the three pieces are 39 cm, 91 cm
and 143 cm. four different colours are used in the ratio of
(Check: 39 + 91 + 143 = 273)
7 : 3 : 19 : 5. If the mass of the first dye used
Problem 11. A gear wheel having 80 teeth is in
mesh with a 25 tooth gear. What is the gear ratio? is 3 1 g, determine the total mass of the dyes
2
Gear ratio = 80 : 25 = 80 = 16 = 3.2 used. [17 g]
25 5
3. Determine how much copper and how much
i.e. gear ratio = 16 : 5 or 3.2 : 1
zinc is needed to make a 99 kg brass ingot if
Problem 12. An alloy is made up of metals A and
B in the ratio 2.5 : 1 by mass. How much of A has they have to be in the proportions copper :
to be added to 6 kg of B to make the alloy?
zinc: :8 : 3 by mass. [72 kg : 27 kg]
Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1) or
A = 2.5 = 2.5 4. It takes 21 hours for 12 men to resurface a
B1
stretch of road. Find how many men it takes to
When B = 6 kg, A = 2.5 from which,
6 resurface a similar stretch of road in 50 hours

A = 6 × 2.5 = 15 kg 24 minutes, assuming the work rate remains

Problem 13. If 3 people can complete a task in constant. [5]
4 hours, how long will it take 5 people to complete
the same task, assuming the rate of work remains 5. It takes 3 hours 15 minutes to fly from city A
constant to city B at a constant speed. Find how long
the journey takes if
The more the number of people, the more quickly the
task is done, hence inverse proportion exists. (a) the speed is 1 1 times that of the original
2
3 people complete the task in 4 hours. speed and

(b) if the speed is three-quarters of the orig-
inal speed.
[(a) 2 h 10 min (b) 4 h 20 min]

1.3 Decimals

The decimal system of numbers is based on the digits
0 to 9. A number such as 53.17 is called a decimal
fraction, a decimal point separating the integer part,
i.e. 53, from the fractional part, i.e. 0.17.

A number which can be expressed exactly as a deci-
mal fraction is called a terminating decimal and those
which cannot be expressed exactly as a decimal fraction

Revision of fractions,decimals and percentages 7

are called non-terminating decimals. Thus, 3 = 1.5 The sum of the positive decimal fractions is Section 1
2
4 23.4 + 32.68 = 56.08
is a terminating decimal, but 3 = 1.33333. . . is a non-
The sum of the negative decimal fractions is
terminating decimal. 1.33333. . . can be written as 1.3,
17.83 + 57.6 = 75.43
called ‘one point-three recurring’.
Taking the sum of the negative decimal fractions from
The answer to a non-terminating decimal may be the sum of the positive decimal fractions gives:

expressed in two ways, depending on the accuracy 56.08 − 75.43
i.e. −(75.43 − 56.08) = −19.35
required:
Problem 17. Determine the value of 74.3 × 3.8
(i) correct to a number of significant figures, that is,
figures which signify something, and When multiplying decimal fractions: (i) the numbers are
multiplied as if they are integers, and (ii) the position of
(ii) correct to a number of decimal places, that is, the the decimal point in the answer is such that there are as
number of figures after the decimal point. many digits to the right of it as the sum of the digits to
the right of the decimal points of the two numbers being
The last digit in the answer is unaltered if the next multiplied together. Thus
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 (i) 743
is in the group of numbers 5, 6, 7, 8 or 9. Thus the non- 38
terminating decimal 7.6183. . . becomes 7.62, correct to
3 significant figures, since the next digit on the right is 5 944
8, which is in the group of numbers 5, 6, 7, 8 or 9. Also 22 290
7.6183. . . becomes 7.618, correct to 3 decimal places, 28 234
since the next digit on the right is 3, which is in the
group of numbers 0, 1, 2, 3 or 4. (ii) As there are (1 + 1) = 2 digits to the right of
the decimal points of the two numbers being
Problem 14. Evaluate 42.7 + 3.04 + 8.7 + 0.06 multiplied together, (74.3 × 3.8), then

The numbers are written so that the decimal points are 74.3 × 3.8 = 282.34
under each other. Each column is added, starting from
the right. Problem 18. Evaluate 37.81 ÷ 1.7, correct to (i) 4
significant figures and (ii) 4 decimal places
42.7
3.04 37.81 ÷ 1.7 = 37.81
8.7 1.7
0.06
The denominator is changed into an integer by multi-
54.50 plying by 10. The numerator is also multiplied by 10 to
keep the fraction the same. Thus
Thus 42.7 + 3.04 + 8.7 + 0.06 = 54.50
37.81 ÷ 1.7 = 37.81 × 10
Problem 15. Take 81.70 from 87.23 1.7 × 10

The numbers are written with the decimal points under 378.1
each other. =

87.23 17
−81.70

5.53

Thus 87.23 − 81.70 = 5.53

Problem 16. Find the value of
23.4 − 17.83 − 57.6 + 32.68

8 Engineering Mathematics

Section 1 The long division is similar to the long division of (b) For mixed numbers, it is only necessary to convert
integers and the first four steps are as shown:
the proper fraction part of the mixed number to a
22.24117..
decimal fraction. Thus, dealing with the 7 gives:
17 378.100000 8
3_4_
38 0.875 i.e. 7 = 0.875
_34_ 8 7.000 8
41
3__4 Thus 5 7 = 5.875
70 8
_68_
20 Now try the following exercise

(i) 37.81 ÷ 1.7 = 22.24, correct to 4 significant Exercise 3 Further problems on decimals
figures, and
In Problems 1 to 6, determine the values of the
(ii) 37.81 ÷ 1.7 = 22.2412, correct to 4 decimal expressions given:
places.
1. 23.6 + 14.71 − 18.9 − 7.421 [11.989]

Problem 19. Convert (a) 0.4375 to a proper 2. 73.84 − 113.247 + 8.21 − 0.068
fraction and (b) 4.285 to a mixed number [−31.265]

3. 3.8 × 4.1 × 0.7 [10.906]

4. 374.1 × 0.006 [2.2446]

(a) 0.4375 can be written as 0.4375 × 10 000 without 5. 421.8 ÷ 17, (a) correct to 4 significant figures
and (b) correct to 3 decimal places.
changing its value, 10 000 [(a) 24.81 (b) 24.812]

i.e. 0.4375 = 4375 0.0147
10 000 6. , (a) correct to 5 decimal places and

By cancelling 2.3
(b) correct to 2 significant figures.
4375 = 875 = 175 = 35 = 7
10 000 2000 400 80 16 [(a) 0.00639 (b) 0.0064]
i.e. 0.4375 = 7
7. Convert to proper fractions:
16 (a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282 and
(e) 0.024

(b) Similarly, 4.285 = 4 285 = 4 57 13 21 1 141 3
1000 200 (a) (b) (c) (d) (e)
20 25 80 500 125

Problem 20. Express as decimal fractions: 8. Convert to mixed numbers:
97
(a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35 and
(a) and (b) 5
16 8 (e) 16.2125

⎡ 41 11 1⎤
(a) 1 (b) 4 (c) 14
⎢⎣ 50 40 8 ⎦⎥
7 17
(a) To convert a proper fraction to a decimal fraction, (d) 15 (e) 16
the numerator is divided by the denominator. Divi- 20 80
sion by 16 can be done by the long division method,
or, more simply, by dividing by 2 and then 8: In Problems 9 to 12, express as decimal fractions
to the accuracy stated:

4.50 0.5625 4
2 9.00 8 4.5000 9. , correct to 5 significant figures.

9 9
Thus = 0.5625 [0.44444]

16

Revision of fractions,decimals and percentages 9

17 To convert fractions to percentages, they are (i) con- Section 1
10. , correct to 5 decimal places. verted to decimal fractions and (ii) multiplied by 100

27 (a) By division, 5 = 0.3125, hence 5 corresponds
[0.62963] 16 16

9 to 0.3125 × 100%, i.e. 31.25%
11. 1 , correct to 4 significant figures.
(b) Similarly, 2 = 1.4 when expressed as a decimal
16 1
[1.563] 5
fraction.
31
12. 13 , correct to 2 decimal places. Hence 1 2 = 1.4 × 100% = 140%
5
37
[13.84] Problem 23. It takes 50 minutes to machine a
certain part, Using a new type of tool, the time can
13. Determine the dimension marked x in be reduced by 15%. Calculate the new time taken
the length of shaft shown in Figure 1.1.
The dimensions are in millimetres.
[12.52 mm]

82.92

27.41 8.32 x 34.67

15% of 50 minutes = 15 × 50 = 750
100 100
= 7.5 minutes.

hence the new time taken is

Figure 1.1 50 − 7.5 = 42.5 minutes.

14. A tank contains 1800 litres of oil. How many Alternatively, if the time is reduced by 15%, then
it now takes 85% of the original time, i.e. 85% of
tins containing 0.75 litres can be filled from 50 = 85 × 50 = 4250 = 42.5 minutes, as above.

this tank? [2400] 100 100

1.4 Percentages Problem 24. Find 12.5% of £378

Percentages are used to give a common standard and 12.5% of £378 means 12.5 × 378, since per cent means
are fractions having the number 100 as their denomina- 100

25 1 ‘per hundred’.
tors. For example, 25 per cent means i.e. and is
$12$.51 1
100 4 Hence 12.5% of £378 = ¨10¨08 × 378 = 8 × 378 =
written 25%. 378
= £47.25
Problem 21. Express as percentages: 8
(a) 1.875 and (b) 0.0125
Problem 25. Express 25 minutes as a percentage
A decimal fraction is converted to a percentage by of 2 hours, correct to the nearest 1%
multiplying by 100. Thus,
Working in minute units, 2 hours = 120 minutes.
(a) 1.875 corresponds to 1.875 × 100%, i.e. 187.5% 25
(b) 0.0125 corresponds to 0.0125 × 100%, i.e. 1.25%
Hence 25 minutes is ths of 2 hours. By cancelling,
Problem 22. Express as percentages: 120
52
25 = 5
(a) and (b) 1 120 24
16 5
Expressing 5 as a decimal fraction gives 0.2083˙
24

Section 1 10 Engineering Mathematics 4. When 1600 bolts are manufactured, 36

Multiplying by 100 to convert the decimal fraction to a are unsatisfactory. Determine the percentage
percentage gives:
unsatisfactory. [2.25%]
0.2083˙ × 100 = 20.83%
5. Express: (a) 140 kg as a percentage of 1 t
Thus 25 minutes is 21% of 2 hours, correct to the (b) 47 s as a percentage of 5 min (c) 13.4 cm
nearest 1%. as a percentage of 2.5 m
[(a) 14% (b) 15.67% (c) 5.36%]
Problem 26. A German silver alloy consists of
60% copper, 25% zinc and 15% nickel. Determine 6. A block of monel alloy consists of 70% nickel
the masses of the copper, zinc and nickel in a 3.74
kilogram block of the alloy and 30% copper. If it contains 88.2 g of

By direct proportion: nickel, determine the mass of copper in the

100% corresponds to 3.74 kg block. [37.8 g]
1% corresponds to 3.74 = 0.0374 kg
100 7. A drilling machine should be set to

60% corresponds to 60 × 0.0374 = 2.244 kg 250 rev/min. The nearest speed available on
25% corresponds to 25 × 0.0374 = 0.935 kg
15% corresponds to 15 × 0.0374 = 0.561 kg the machine is 268 rev/min. Calculate the

Thus, the masses of the copper, zinc and nickel are percentage over speed. [7.2%]
2.244 kg, 0.935 kg and 0.561 kg, respectively.
(Check: 2.244 + 0.935 + 0.561 = 3.74) 8. Two kilograms of a compound contains 30%
of element A, 45% of element B and 25% of
element C. Determine the masses of the three
elements present.
[A 0.6 kg, B 0.9 kg, C 0.5 kg]

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

10. In a sample of iron ore, 18% is iron. How

much ore is needed to produce 3600 kg of

2. Express as percentages, correct to 3 signifi- iron? [20 000 kg]
cant figures:
11. A screws’ dimension is 12.5 ± 8% mm. Cal-
7 19 11 culate the possible maximum and minimum
(a) (b) (c) 1 length of the screw.
33 24 16 [13.5 mm, 11.5 mm]

[(a) 21.2% (b) 79.2% (c) 169%]

3. Calculate correct to 4 significant figures: 12. The output power of an engine is 450 kW. If
(a) 18% of 2758 tonnes (b) 47% of 18.42
grams (c) 147% of 14.1 seconds the efficiency of the engine is 75%, determine
[(a) 496.4 t (b) 8.657 g (c) 20.73 s]
the power input. [600 kW]

Chapter 2

Indices,standard form and
engineering notation

2.1 Indices Laws of indices

The lowest factors of 2000 are 2 × 2 × 2 × 2 × 5 × 5 × 5. When simplifying calculations involving indices, cer-
These factors are written as 24 × 53, where 2 and 5 are tain basic rules or laws can be applied, called the laws
called bases and the numbers 4 and 5 are called indices. of indices. These are given below.

When an index is an integer it is called a power. Thus, (i) When multiplying two or more numbers having
24 is called ‘two to the power of four’, and has a base of the same base, the indices are added. Thus
2 and an index of 4. Similarly, 53 is called ‘five to the
power of 3’ and has a base of 5 and an index of 3. 32 × 34 = 32+4 = 36

Special names may be used when the indices are 2 and (ii) When a number is divided by a number having the
3, these being called ‘squared’and ‘cubed’, respectively. same base, the indices are subtracted. Thus
Thus 72 is called ‘seven squared’ and 93 is called ‘nine
cubed’. When no index is shown, the power is 1, i.e. 2 35 = 35−2 = 33
means 21. 32

Reciprocal (iii) When a number which is raised to a power is raised
to a further power, the indices are multiplied. Thus
The reciprocal of a number is when the index is −1

and its value is given by 1, divided by the base. Thus the

reciprocal of 2 is 2−1 and its value is 1 or 0.5. Similarly, (35)2 = 35×2 = 310
2
5−1 1
the reciprocal of 5 is which means 5 or 0.2.

Square root (iv) When a number has an index of 0, its value is 1.
Thus 30 = 1

1 (v) A number raised to a negative power is the recip-
2
The square root of a number is when the √index is , and rocal of that number raised to a positive power.
the square root of 2 is written as 21/2 or
2. The value Thus 3−4 = 1 Similarly, 1 = 23
34 2−3
(vi) When a number is raised to a fractional power
of a square root is the value of the base which when

multi√plied by itself gives the number. Since √3 × 3 = 9, the denominator of the fraction is the root of the
then 9 = 3. However, (−3) × (−3) = 9, so 9 = −3.
number and the numerator is the power.

There are always two answers when finding the square Thus 82/3 = √3 82 = (2)2 = 4

root of a number and this is shown by putting both √ √
2 251 251
aro+otanpdroabl−emsi.gnThinusfr√on9t=of±t3heanadns4w1e/r2 to√a square and 251/2 = = = ±5
= 4 = ±2,
(Note that √ ≡ √ )
and so on.
2

12 Engineering Mathematics

Section 1 2.2 Worked problems on indices (102)3
Problem 5. Evaluate: 104 × 102

Problem 1. Evaluate (a) 52 × 53, (b) 32 × 34 × 3 From the laws of indices:
and (c) 2 × 22 × 25
(102)3 10(2×3) 106
From law (i): 104 × 102 = 10(4+2) = 106
(a) 52 × 53 = 5(2+3) = 55 = 5 × 5 × 5 × 5 × 5 = 3125
(b) 32 × 34 × 3 = 3(2+4+1) = 37 = 106−6 = 100 = 1

= 3 × 3 × · · · to 7 terms Problem 6. Find the value of
= 2187
(c) 2 × 22 × 25 = 2(1+2+5) = 28 = 256 23 × 24 (32)3
(a) 27 × 25 and (b) 3 × 39

Problem 2. Find the value of: From the laws of indices:
75 57
(a) 23 × 24 = 2(3+4) = 27 = 27−12 = 2−5
(a) 73 and (b) 54 27 × 25 2(7+5) 212

From law (ii): = 1 = 1
25 32

(a) 75 = 7(5−3) = 72 = 49 (b) (32)3 = 32×3 = 36 = 36−10 = 3−4
73 3 × 39 31+9 310

(b) 57 = 5(7−4) = 53 = 125 = 1 = 1
54 34 81

Problem 3. Evaluate: (a) 52 × 53 ÷ 54 and Now try the following exercise
(b) (3 × 35) ÷ (32 × 33)

From laws (i) and (ii): Exercise 5 Further problems on indices

(a) 52 × 53 ÷ 54 = 52 × 53 = 5(2+3) In Problems 1 to 10, simplify the expressions
54 54 given, expressing the answers in index form and
with positive indices:
55
= 54 = 5(5−4) = 51 = 5 1. (a) 33 × 34 (b) 42 × 43 ×44
[(a) 37 (b) 49]
(b) (3 × 35) ÷ (32 × 33) = 3 × 35 = 3(1+5)
32 × 33 3(2+3)
2. (a) 23 × 2 × 22 (b) 72 × 74 × 7 × 73
36
= 35 = 3(6−5) = 31 =3 [(a) 26 (b) 710]

Problem 4. Simplify: (a) (23)4 (b) (32)5, 24 37 [(a) 2 (b) 35]
expressing the answers in index form. 3. (a) 23 (b) 32

From law (iii): 4. (a) 56 ÷ 53 (b) 713/710
(a) (23)4 = 23×4 = 212 (b) (32)5 = 32×5 = 310 [(a) 53 (b) 73]

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

Indices,standard form and engineering notation 13

22 × 23 37 × 34 Problem 9. Evaluate: Section 1
6. (a) 24 (b) 35 (a) 41/2 (b) 163/4 (c) 272/3 (d) 9−1/2

57 [(a) 2 (b) 36] √
7. (a) 52 × 53 (a) 41/2 = 4 = ±2
135
(b) 13 × 132 √
(b) 163/4 = 4 163 = ( ± 2)3 = ±8
[(a) 52 (b) 132]
(Note that it does not matter whether the 4th root
(9 × 32)3 (16 × 4)2 of 16 is found first or whether 16 cubed is found
8. (a) (3 × 27)2 (b) (2 × 8)3 first — the same answer will result).

[(a) 34 (b) 1] √
(c) 272/3 = 3 272 = (3)2 = 9
5−2 32 × 3−4
9. (a) 5−4 (b) 33 (d) 9−1/2 = 1 = √1 = 1 = ±1
91/2 9 ±3 3
1
(a) 52 (b) 35

72 × 7−3 23 × 2−4 × 25 41.5 × 81/3
10. (a) 7 × 7−4 (b) 2 × 2−2 × 26 Problem 10. Evaluate: 22 × 32−2/5

(a) 72 1 √
(b) 41.5 = 43/2 = 43 = 23 = 8
2
√
81/3 = 3 8 = 2, 22 = 4

and 32−2/5 = 1 = 1 = 1 = 1
322/5 √5 322 22 4
2.3 Further worked problems on
indices 41.5 × 81/3 8 × 2 16 = 16
Hence = =
22 × 32−2/5 4× 1 1
4

33 × 57 Alternatively,
Problem 7. Evaluate: 53 × 34
41.5 × 81/3 [(2)2]3/2 × (23)1/3 23 × 21
The laws of indices only apply to terms having the 22 × 32−2/5 = 22 × (25)−2/5 = 22 × 2−2
same base. Grouping terms having the same base, and
then applying the laws of indices to each of the groups = 23+1−2−(−2) = 24 = 16
independently gives:
Problem 11. Evaluate: 32 × 55 + 33 × 53

33 × 57 33 57 = 3(3−4) × 5(7−3) 34 × 54
53 × 34 34 53
= =

= 3−1 × 54 = 54 = 625 = 1 Dividing each term by the HCF (i.e. highest common
31 3 208 factor) of the three terms, i.e. 32 × 53, gives:

3

Problem 8. Find the value of 32 × 55 + 33 × 53 32 × 55 + 33 × 53
23 × 35 × (72)2 34 × 54 32 × 53 32 × 53
74 × 24 × 33 =
34 × 54

32 × 53

= 3(2−2) × 5(5−3) + 3(3−2) × 50
3(4−2) × 5(4−3)
23 × 35 × (72)2 = 23−4 × 35−3 × 72×2−4
74 × 24 × 33 30 × 52 + 31 × 50
32 × 51
= 2−1 × 32 × 70 =

= 1 × 32 × 1 = 9 = 1 = 1 × 25 + 3 × 1 = 28
4
2 22 9×5 45

14 Engineering Mathematics

Section 1 Problem 12. Find the value of = 43 × 52 × 23
32 × 55 33 32 53

34 × 54 + 33 × 53 (22)3 × 23
= 3(3+2) × 5(3−2)

To simplify the arithmetic, each term is divided by the 29
HCF of all the terms, i.e. 32 × 53. Thus = 35 × 5

32 × 55 Now try the following exercise
34 × 54 + 33 × 53

32 × 55 Exercise 6 Further problems on indices

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

= 3(4−2) 3(2−2) × 5(5−3) × 5(3−3) 33 × 52 7−2 × 3−2
× 5(4−3) + 3(3−2) 1. (a) 54 × 34 (b) 35 × 74 × 7−3

= 32 30 × 52 × 50 = 25 = 25 11
× 51 + 31 45 + 3 48 (a) 3 × 52 (b) 73 × 37

4 3 3 −2 42 × 93 8−2 × 52 × 3−4
2. (a) 83 × 34 (b) 252 × 24 × 9−2
×
Problem 13. Simplify: 32 1
35 (a) 25 (b) 210 × 52
2 −3

5
giving the answer with positive indices

3. Evaluate (a) 1 −1 (b) 810.25
32
A fraction raised to a power means that both the numer-
(c) 16(−1/4) (d) 4 1/2
ator and the denominator of the fraction are raised to 9
3 43
4 33
that power, i.e. 3 = (a) 9 (b) ± 3 (c) ± 1 (d) ± 2
23
A fraction raised to a negative power has the same

value as the inverse of the fraction raised to a positive

power. In Problems 4 to 8, evaluate the expressions given.

Thus, 3 −2 1 = 1 =1 × 52 = 52
3 32 32 32
5 = 2 92 × 74 147
4. 34 × 74 + 33 × 72 148
5 52

2 −3 5 3 53 (24)2 − 3−2 × 44 1
2 23 5. 23 × 162 9
Similarly, = =
5

1 3 2 −2
3 −2 43 52 −
4 3 23 −5 65
3 × 33 × 32 32 72
5 53 6.

Thus, 2 −3 =

5 23 5

Indices,standard form and engineering notation 15

44 [64] (ii) The laws of indices are used when multiplying Section 1
3 1 or dividing numbers given in standard form. For
7. 2 2 4 example,
9 2
(2.5 × 103) × (5 × 102)
(32)3/2 × (81/3)2 = (2.5 × 5) × (103+2)
8. (3)2 × (43)1/2 × (9)−1/2 = 12.5 × 105 or 1.25 × 106

Similarly,

2.4 Standard form 6 × 104 = 6 × (104−2) = 4 × 102
1.5 × 102 1.5
A number written with one digit to the left of the decimal
point and multiplied by 10 raised to some power is said 2.5 Worked problems on standard
to be written in standard form. Thus: 5837 is written form
as 5.837 × 103 in standard form, and 0.0415 is written
as 4.15 × 10−2 in standard form. Problem 14. Express in standard form:
(a) 38.71 (b) 3746 (c) 0.0124
When a number is written in standard form, the first
factor is called the mantissa and the second factor is For a number to be in standard form, it is expressed with
called the exponent. Thus the number 5.8 × 103 has a only one digit to the left of the decimal point. Thus:
mantissa of 5.8 and an exponent of 103.
(a) 38.71 must be divided by 10 to achieve one digit
(i) Numbers having the same exponent can be added to the left of the decimal point and it must also be
or subtracted in standard form by adding or sub- multiplied by 10 to maintain the equality, i.e.
tracting the mantissae and keeping the exponent
the same. Thus: 38.71 = 38.71 ×10 = 3.871 × 10 in standard form
10
2.3 × 104 + 3.7 × 104
= (2.3 + 3.7) × 104 = 6.0 × 104 (b) 3746 = 3746 × 1000 = 3.746 × 103 in standard
1000
and 5.9 × 10−2 − 4.6 × 10−2
= (5.9 − 4.6) × 10−2 = 1.3 × 10−2 form
(c) 0.0124 = 0.0124 × 100 = 1.24
When the numbers have different exponents,
one way of adding or subtracting the numbers is 100 100
to express one of the numbers in non-standard = 1.24 × 10−2 in standard form
form, so that both numbers have the same expo-
nent. Thus: Problem 15. Express the following numbers,
which are in standard form, as decimal numbers:
2.3 × 104 + 3.7 × 103 (a) 1.725 × 10−2 (b) 5.491 × 104 (c) 9.84 × 100
= 2.3 × 104 + 0.37 × 104
= (2.3 + 0.37) × 104 = 2.67 × 104 (a) 1.725 × 10−2 = 1.725 = 0.01725
100
Alternatively,
(b) 5.491 × 104 = 5.491 × 10 000 = 54 910
2.3 × 104 + 3.7 × 103 (c) 9.84 × 100 = 9.84 × 1 = 9.84 (since 100 = 1)
= 23 000 + 3700 = 26 700
= 2.67 × 104

16 Engineering Mathematics

Section 1 Problem 16. Express in standard form, correct to 17 31

3 significant figures: 4. (a) (b) 11 (c) 130 (d)
28 5 32
32 9
(a) (b) 19 (c) 741 (a) 5 × 10−1 (b) 1.1875 × 10
83 16 (c) 1.306 × 102 (d) 3.125 × 10−2

3 In Problems 5 and 6, express the numbers given
(a) = 0.375, and expressing it in standard form as integers or decimal fractions:

8 5. (a) 1.01 × 103 (b) 9.327 × 102
gives: 0.375 = 3.75 × 10−1 (c) 5.41 × 104 (d) 7 × 100
[(a) 1010 (b) 932.7 (c) 54 100 (d) 7]
(b) 19 2 = 19.6˙ = 1.97 × 10 in standard form, correct
3 6. (a) 3.89 × 10−2 (b) 6.741 × 10−1
(c) 8 × 10−3
to 3 significant figures [(a) 0.0389 (b) 0.6741 (c) 0.008]

(c) 741 9 = 741.5625 = 7.42 × 102 in standard form,
16

correct to 3 significant figures

Problem 17. Express the following numbers, 2.6 Further worked problems on
standard form
given in standard form, as fractions or mixed
numbers: (a) 2.5 × 10−1 (b) 6.25 × 10−2 (c) Problem 18. Find the value of:
1.354 × 102
(a) 7.9 × 10−2 − 5.4 × 10−2
(a) 2.5 × 10−1 = 2.5 = 25 = 1 (b) 8.3 × 103 + 5.415 × 103 and
10 100 4 (c) 9.293 × 102 + 1.3 × 103

(b) 6.25 × 10−2 = 6.25 = 625 1 expressing the answers in standard form.
=
100 10 000 16 Numbers having the same exponent can be added or
subtracted by adding or subtracting the mantissae and
(c) 1.354 × 102 = 135.4 = 135 4 = 135 2 keeping the exponent the same. Thus:
10 5
(a) 7.9 × 10−2 − 5.4 × 10−2
Now try the following exercise = (7.9 − 5.4) × 10−2 = 2.5 × 10−2

Exercise 7 Further problems on standard (b) 8.3 × 103 + 5.415 × 103
form
= (8.3 + 5.415) × 103 = 13.715 × 103
In Problems 1 to 4, express in standard form: = 1.3715 × 104 in standard form
(c) Since only numbers having the same exponents can
1. (a) 73.9 (b) 28.4 (c) 197.72 be added by straight addition of the mantissae, the
(a) 7.39 × 10 (b) 2.84 × 10 numbers are converted to this form before adding.
(c) 1.9772 × 102 Thus:

2. (a) 2748 (b) 33 170 (c) 274 218 9.293 × 102 + 1.3 × 103
(a) 2.748 × 103 (b) 3.317 × 104
(c) 2.74218 × 105 = 9.293 × 102 + 13 × 102

3. (a) 0.2401 (b) 0.0174 (c) 0.00923
(a) 2.401 × 10−1 (b) 1.74 × 10−2
(c) 9.23 × 10−3

Indices,standard form and engineering notation 17

= (9.293 + 13) × 102 6 × 10−3 (2.4 × 103)(3 × 10−2) Section 1
= 22.293 × 102 = 2.2293 × 103 4. (a) 3 × 10−5 (b)
in standard form. (4.8 × 104)

Alternatively, the numbers can be expressed as [(a) 2 × 102 (b) 1.5 × 10−3]
decimal fractions, giving:
5. Write the following statements in standard
9.293 × 102 + 1.3 × 103 form:
= 929.3 + 1300 = 2229.3
= 2.2293 × 103 (a) The density of aluminium is 2710 kg m−3
[2.71 × 103 kg m−3]
in standard form as obtained previously. This
method is often the ‘safest’ way of doing this type (b) Poisson’s ratio for gold is 0.44
of problem. [4.4 × 10−1]

Problem 19. Evaluate (a) (3.75 × 103)(6 × 104) (c) The impedance of free space is 376.73
3.5 × 105 [3.7673 × 102 ]

and (b) 7 × 102 (d) The electron rest energy is 0.511 MeV
expressing answers in standard form [5.11 × 10−1 MeV]

(a) (3.75 × 103)(6 × 104) = (3.75 × 6)(103+4) (e) Proton charge-mass ratio is
9 5 789 700 C kg−1
[9.57897 × 107 C kg−1]

(f) The normal volume of a perfect gas is
0.02241 m3 mol−1
[2.241 × 10−2 m3 mol−1]

= 22.50 × 107

= 2.25 × 108

(b) 3.5 × 105 = 3.5 × 105−2 2.7 Engineering notation and
7 × 102 7 common prefixes

= 0.5 × 103 = 5 × 102 Engineering notation is similar to scientific notation
except that the power of ten is always a multiple of 3.
Now try the following exercise
For example, 0.00035 = 3.5 × 10−4 in scientific
Exercise 8 Further problems on standard notation,
form
but 0.00035 = 0.35 × 10−3 or 350 × 10−6 in
In Problems 1 to 4, find values of the expressions engineering notation.
given, stating the answers in standard form:
Units used in engineering and science may be made
1. (a) 3.7 × 102 + 9.81 × 102 larger or smaller by using prefixes that denote multi-
(b) 1.431 × 10−1 + 7.3 × 10−1 plication or division by a particular amount. The eight
[(a) 1.351 × 103 (b) 8.731 × 10−1] most common multiples, with their meaning, are listed
in Table 2.1, where it is noticed that the prefixes involve
2. (a) 4.831 × 102 + 1.24 × 103 powers of ten which are all multiples of 3:
(b) 3.24 × 10−3 − 1.11 × 10−4 For example,
[(a) 1.7231 × 103 (b) 3.129 × 10−3]
5 MV means 5 × 1 000 000 = 5 × 106
3. (a) (4.5 × 10−2)(3 × 103) (b) 1.1 × 105]
(b) 2 × (5.5 × 104) = 5 000 000 volts
[(a) 1.35 × 102
3.6 k means 3.6 × 1000 = 3.6 × 103

= 3600 ohms

18 Engineering Mathematics

Section 1 Table 2.1 Meaning (i.e. × 1012)
Prefix Name multiply by 1 000 000 000 000 (i.e. × 109)
T tera multiply by 1 000 000 000 (i.e. × 106)
G giga multiply by 1 000 000 (i.e. × 103)
M mega multiply by 1000 (i.e. × 10−3)
k kilo divide by 1000 (i.e. × 10−6)
m milli divide by 1 000 000 (i.e. × 10−9)
μ micro divide by 1 000 000 000 (i.e. × 10−12)
n nano divide by 1 000 000 000 000
p pico

7.5 μC means 7.5 ÷ 1 000 000 = 7.5 or Now try the following exercise
106

7.5 × 10−6 = 0.0000075 coulombs

and 4 mA means 4 × 10−3 or = 4 Exercise 9 Further problems on
Similarly, 103 engineering notation and
common prefixes
= 4 = 0.004 amperes
1000 1. Express the following in engineering notation

and in prefix form:

0.00006 J = 0.06 mJ or 60 μJ (a) 100 000 W (b) 0.00054 A

5 620 000 N = 5620 kN or 5.62 MN (c) 15 × 105 (d) 225 × 10−4 V

47 × 104 = 470 000 = 470 k or 0.47 M (e) 35 000 000 000 Hz (f) 1.5 × 10−11 F

(g) 0.000017 A (h) 46200

and 12 × 10−5A = 0.00012 A = 0.12 mA or 120 μA [(a) 100 kW (b) 0.54 mA or 540 μA

(c) 1.5 M (d) 22.5 mV (e) 35 GHz (f ) 15 pF

A calculator is needed for many engineering calcula- (g) 17 μA (h) 46.2 k ]

tions, and having a calculator which has an ‘EXP’ and

‘ENG’ function is most helpful. 2. Rewrite the following as indicated:
For example, to calculate: 3 × 104 × 0.5 × 10−6
(a) 0.025 mA = …….μA
volts, input your calculator in the following order: (b) 1000 pF = …..nF
(c) 62 × 104 V = …….kV
(a) Enter ‘3’ (b) Press ‘EXP’ (or ×10x) (c) (d) 1 250 000 = …..M

Enter ‘4’(d) Press ‘×’(e) Enter ‘0.5’(f) Press ‘EXP’ [(a) 25 μA (b) 1 nF (c) 620 kV (d) 1.25 M ]

(or ×10x) (g) Enter ‘−6’ (h) Press ‘=’

The answer is 0.015 V or 7 Now press the ‘ENG’

200 3. Use a calculator to evaluate the following in

button, and the answer changes to 15 × 10−3 V. engineering notation:
(a) 4.5 × 10−7 × 3 × 104
The ‘ENG’ or ‘Engineering’ button ensures that the
(1.6 × 10−5)(25 × 103)
value is stated to a power of 10 that is a multiple of (b) (100 × 106)

3, enabling you, in this example, to express the answer [(a) 13.5 × 10−3 (b) 4 × 103]

as 15 mV.

Chapter 3

Computer numbering
systems

3.1 Binary numbers From above: 110112 = 1 × 24 + 1 × 23 + 0 × 22
+ 1 × 21 + 1 × 20
The system of numbers in everyday use is the denary
or decimal system of numbers, using the digits 0 to 9. = 16 + 8 + 0 + 2 + 1
It has ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) = 2710
and is said to have a radix or base of 10.
Problem 2. Convert 0.10112 to a decimal fraction
The binary system of numbers has a radix of 2 and
uses only the digits 0 and 1.

0.10112 = 1 × 2−1 + 0 × 2−2 + 1 × 2−3

3.2 Conversion of binary to decimal + 1 × 2−4

= 1 × 1 + 0 × 1 + 1 × 1
2 22 23
The decimal number 234.5 is equivalent to
2 × 102 + 3 × 101 + 4 × 100 + 5 × 10−1 1
+ 1 × 24

i.e. is the sum of term comprising: (a digit) multiplied =1+1+ 1
by (the base raised to some power). 2 8 16

In the binary system of numbers, the base is 2, so = 0.5 + 0.125 + 0.0625
1101.1 is equivalent to:
= 0.687510
1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2−1
Problem 3. Convert 101.01012 to a decimal
number

Thus the decimal number equivalent to the binary 101.01012 = 1 × 22 + 0 × 21 + 1 × 20
number 1101.1 is + 0 × 2−1 + 1 × 2−2
+ 0 × 2−3 + 1 × 2−4
8 + 4 + 0 + 1 + 1 that is 13.5
, = 4 + 0 + 1 + 0 + 0.25
2
+ 0 + 0.0625
i.e. 1101.12 = 13.510, the suffixes 2 and 10 denoting
binary and decimal systems of number respectively. = 5.312510

Problem 1. Convert 110112 to a decimal number

Section 1 20 Engineering Mathematics

Now try the following exercise For fractions, the most significant bit of the result is the
top bit obtained from the integer part of multiplication
Exercise 10 Further problems on by 2. The least significant bit of the result is the bottom
conversion of binary to bit obtained from the integer part of multiplication by 2.
decimal numbers Thus 0.62510 = 0.1012

In Problems 1 to 4, convert the binary number given Problem 4. Convert 4710 to a binary number
to decimal numbers.
From above, repeatedly dividing by 2 and noting the
1. (a) 110 (b) 1011 (c) 1110 (d) 1001 remainder gives:
[(a) 610 (b) 1110 (c) 1410 (d) 910]

2. (a) 10101 (b) 11001 (c) 101101 (d) 110011
[(a) 2110 (b) 2510 (c) 4510 (d) 5110]

3. (a) 0.1101 (b) 0.11001 (c) 0.00111 (d) 0.01011
(a) 0.812510 (b) 0.7812510
(c) 0.2187510 (d) 0.3437510

4. (a) 11010.11 (b) 10111.011 (c) 110101.0111
(d) 11010101.10111
(a) 26.7510 (b) 23.37510
(c) 53.437510 (d) 213.7187510

3.3 Conversion of decimal to binary Thus 4710 = 1011112

An integer decimal number can be converted to a cor- Problem 5. Convert 0.4062510 to a binary
responding binary number by repeatedly dividing by 2 number
and noting the remainder at each stage, as shown below
for 3910 From above, repeatedly multiplying by 2 gives:

The result is obtained by writing the top digit of the i.e. 04062510 = 0.011012
remainder as the least significant bit, (a bit is a binary
digit and the least significant bit is the one on the right).
The bottom bit of the remainder is the most significant
bit, i.e. the bit on the left.

Thus 3910 = 1001112

The fractional part of a decimal number can be converted
to a binary number by repeatedly multiplying by 2, as
shown below for the fraction 0.625

Computer numbering systems 21

Problem 6. Convert 58.312510 to a binary 3.4 Conversion of decimal to binary Section 1
number via octal
The integer part is repeatedly divided by 2, giving:
For decimal integers containing several digits, repeat-
The fractional part is repeatedly multiplied by 2 giving: edly dividing by 2 can be a lengthy process. In this
case, it is usually easier to convert a decimal number to
a binary number via the octal system of numbers. This
system has a radix of 8, using the digits 0, 1, 2, 3, 4,
5, 6 and 7. The denary number equivalent to the octal
number 43178 is

4 × 83 + 3 × 82 + 1 × 81 + 7 × 80

i.e. 4 × 512 + 3 × 64 + 1 × 8 + 7 × 1 or 225510

An integer decimal number can be converted to a cor-
responding octal number by repeatedly dividing by 8
and noting the remainder at each stage, as shown below
for 49310

Thus 58.312510 = 111010.01012

Now try the following exercise

Exercise 11 Further problems on Thus 49310 = 7558
conversion of decimal The fractional part of a decimal number can be con-
to binary numbers verted to an octal number by repeatedly multiplying by
8, as shown below for the fraction 0.437510
In Problem 1 to 4, convert the decimal numbers
given to binary numbers. For fractions, the most significant bit is the top integer
obtained by multiplication of the decimal fraction by 8,
1. (a) 5 (b) 15 (c) 19 (d) 29 thus

(a) 1012 (b) 11112 0.437510 = 0.348
(c) 100112 (d) 111012 The natural binary code for digits 0 to 7 is shown
in Table 3.1, and an octal number can be converted
2. (a) 31 (b) 42 (c) 57 (d) 63 to a binary number by writing down the three bits
corresponding to the octal digit.
(a) 111112 (b) 1010102 Thus 4378 = 100 011 1112
(c) 1110012 (d) 1111112 and 26.358 = 010 110.011 1012

3. (a) 0.25 (b) 0.21875 (c) 0.28125
(d) 0.59375

(a) 0.012 (b) 0.001112
(c) 0.010012 (d) 0.100112

4. (a) 47.40625 (b) 30.8125
(c) 53.90625 (d) 61.65625

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

22 Engineering Mathematics

Section 1 Table 3.1 Thus 0.5937510 = 0.468
Octal digit
Natural 0.468 = 0.100 1102
binary number From Table 3.1.

0 000 i.e. 0.5937510 = 0.100 112

1 001 Problem 9. Convert 5613.9062510 to a binary
number, via octal
2 010
The integer part is repeatedly divided by 8, noting the
3 011 remainder, giving:

4 100

5 101

6 110

7 111

The ‘0’ on the extreme left does not signify anything, This octal number is converted to a binary number, (see
thus 26.358 = 10 110.011 1012 Table 3.1)

Conversion of decimal to binary via octal is demon- 127558 = 001 010 111 101 1012
strated in the following worked problems. i.e. 561310 = 1 010 111 101 1012

Problem 7. Convert 371410 to a binary number, The fractional part is repeatedly multiplied by 8, and
via octal noting the integer part, giving:

Dividing repeatedly by 8, and noting the remainder
gives:

From Table 3.1, 72028 = 111 010 000 0102 This octal fraction is converted to a binary number, (see
i.e. 371410 = 111 010 000 0102 Table 3.1)

Problem 8. Convert 0.5937510 to a binary 0.728 = 0.111 0102
number, via octal i.e. 0.9062510 = 0.111 012

Multiplying repeatedly by 8, and noting the integer Thus, 5613.9062510 = 1 010 111 101 101.111 012
values, gives:
Problem 10. Convert 11 110 011.100 012 to a
decimal number via octal

Grouping the binary number in three’s from the binary
point gives: 011 110 011.100 0102

Computer numbering systems 23

Using Table 3.1 to convert this binary number to an 3.5 Hexadecimal numbers Section 1
octal number gives: 363.428 and
The complexity of computers requires higher order
363.428 = 3 × 82 + 6 × 81 + 3 × 80 numbering systems such as octal (base 8) and hexadeci-
+ 4 × 8−1 + 2 × 8−2 mal (base 16), which are merely extensions of the binary
system. A hexadecimal numbering system has a radix
= 192 + 48 + 3 + 0.5 + 0.03125 of 16 and uses the following 16 distinct digits:
= 243.5312510
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F
Now try the following exercise
‘A’ corresponds to 10 in the denary system, B to 11, C
to 12, and so on.

Exercise 12 Further problems on To convert from hexadecimal to decimal:
conversion between decimal
and binary numbers via octal For example
1A16 = 1 × 161 + A × 160
In Problems 1 to 3, convert the decimal numbers = 1 × 161 + 10 × 1 = 16 + 10 = 26
given to binary numbers, via octal.
i.e. 1A16 = 2610
1. (a) 343 (b) 572 (c) 1265
⎡⎤ Similarly,
⎣ (a) 1010101112 (b) 10001111002 ⎦ 2E16 = 2 × 161 + E × 160
(c) 100111100012 = 2 × 161 + 14 × 160 = 32 + 14 = 4610

2. (a) 0.46875 (b) 0.6875 (c) 0.71875 and 1BF16 = 1 × 162 + B × 161 + F × 160
= 1 × 162 + 11 × 161 + 15 × 160
⎡⎤ = 256 + 176 + 15 = 44710
⎣ (a) 0.011112 (b) 0.10112 ⎦
Table 3.2 compares decimal, binary, octal and hexadec-
(c) 0.101112 imal numbers and shows, for example, that

3. (a) 247.09375 (b) 514.4375 2310 = 101112 = 278 = 1716

(c) 1716.78125 Problem 11. Convert the following hexadecimal
numbers into their decimal equivalents: (a) 7A16
⎡ ⎤ (b) 3F16
⎥⎥⎦
⎢⎢⎣ (a) 11110111.000112 (a) 7A16 = 7 × 161 + A × 160 = 7 × 16 + 10 × 1
(b) 1000000010.01112 = 112 + 10 = 122

(c) 11010110100.11.0012 Thus 7A16 = 12210
(b) 3F16 = 3 × 161 + F × 160 = 3 × 16 + 15 × 1
4. Convert the following binary numbers to dec-
imal numbers via octal: = 48 + 15 = 63
Thus, 3F16 = 6310
(a) 111.011 1 (b) 101 001.01
Problem 12. Convert the following hexadecimal
(c) 1 110 011 011 010.001 1 numbers into their decimal equivalents: (a) C916
(b) BD16
⎡⎤
⎣ (a) 7.437510 (b) 41.2510 ⎦

(c) 7386.187510

24 Engineering Mathematics

Section 1 Table 3.2 Binary Octal Hexadecimal (a) C916 = C × 161 + 9 × 160 = 12 × 16 + 9 × 1
0000 0 0 = 192 + 9 = 201
Decimal 0001 1 1
0 0010 2 2 Thus C916 = 20110
1 0011 3 3 (b) BD16 = B × 161 + D × 160 = 11 × 16 + 13 × 1
2 0100 4 4
3 0101 5 5 = 176 + 13 = 189
4 0110 6 6 Thus, BD16 = 18910
5 0111 7 7
6 1000 8 Problem 13. Convert 1A4E16 into a denary
7 1001 10 9 number
8 1010 11 A 1A4E16
9 1011 12 B
10 1100 13 C = 1 × 163 + A × 162 + 4 × 161 + E × 160
11 1101 14 D = 1 × 163 + 10 × 162 + 4 × 161 + 14 × 160
12 1110 15 E = 1 × 4096 + 10 × 256 + 4 × 16 + 14 × 1
13 1111 16 F = 4096 + 2560 + 64 + 14 = 6734
14 10000 17 10 Thus, 1A4E16 = 673410
15 10001 20 11
16 10010 21 12 To convert from decimal to hexadecimal:
17 10011 22 13 This is achieved by repeatedly dividing by 16 and noting
18 10100 23 14 the remainder at each stage, as shown below for 2610.
19 10101 24 15
20 10110 25 16 Hence 2610 = 1A16
21 10111 26 17 Similarly, for 44710
22 11000 27 18
23 11001 30 19 Thus 44710 = 1BF16
24 11010 31 1A
25 11011 32 1B Problem 14. Convert the following decimal
26 11100 33 1C numbers into their hexadecimal equivalents:
27 11101 34 1D (a) 3710 (b) 10810
28 11110 35 1E
29 11111 36 1F
30 37 20
31 100000 40
32

(a) Computer numbering systems 25 Section 1

Hence 3710 = 2516 6CF316 = 0110 1100 1111 0011
(b) from Table 3.2

Hence 10810 = 6C16 i.e. 6CF316 = 1101100111100112
Problem 15. Convert the following decimal
numbers into their hexadecimal equivalents: Problem 16. Convert the following binary
(a) 16210 (b) 23910 numbers into their hexadecimal equivalents:
(a) (a) 110101102 (b) 11001112

Hence 16210 = A216 (a) Grouping bits in fours from the
(b)
right gives: 1101 0110

and assigning hexadecimal symbols

to each group gives: D6

from Table 3.2

Thus, 110101102 = D616

(b) Grouping bits in fours from the

right gives: 0110 0111

and assigning hexadecimal symbols

to each group gives: 67

from Table 3.2

Thus, 11001112 = 6716

Problem 17. Convert the following binary
numbers into their hexadecimal equivalents:
(a) 110011112 (b) 1100111102

Hence 23910 = EF16 (a) Grouping bits in fours from the 1100 1111
right gives:
To convert from binary to hexadecimal: and assigning hexadecimal CF
symbols to each group gives: from Table 3.2
The binary bits are arranged in groups of four, start-
ing from right to left, and a hexadecimal symbol is Thus, 110011112 = CF16
assigned to each group. For example, the binary number
1110011110101001 is initially grouped in (b) Grouping bits in fours from
the right gives:
fours as: 1110 0111 1010 1001 and assigning hexadecimal 0001 1001 1110
symbols to each group gives:
and a hexadecimal symbol 19E
Thus, 1100111102 = 19E16 from Table 3.2
assigned to each group as E 7 A 9
from Table 3.2

Hence 11100111101010012 = E7A916

To convert from hexadecimal to binary: Problem 18. Convert the following hexadecimal
The above procedure is reversed, thus, for example,
numbers into their binary equivalents: (a) 3F16
(b) A616

26 Engineering Mathematics

Section 1 (a) Spacing out hexadecimal digits gives: 3 F Now try the following exercise

and converting each into

binary gives: 0011 1111 Exercise 13 Further problems on
from Table 3.2 hexadecimal numbers

Thus, 3F16 = 1111112 In Problems 1 to 4, convert the given hexadecimal
numbers into their decimal equivalents.
(b) Spacing out hexadecimal digits

gives: A6 1. E716 [23110] 2. 2C16 [4410]

and converting each into binary 3. 9816 [15210] 4. 2F116 [75310]

gives: 1010 0110 In Problems 5 to 8, convert the given decimal
from Table 3.2 numbers into their hexadecimal equivalents.

Thus, A616 = 101001102 5. 5410 [3616] 6. 20010 [C816]

Problem 19. Convert the following hexadecimal 7. 9110 [5B16] 8. 23810 [EE16]

numbers into their binary equivalents: (a) 7B16 In Problems 9 to 12, convert the given binary
(b) 17D16 numbers into their hexadecimal equivalents.

9. 110101112 [D716]

(a) Spacing out hexadecimal 7B 10. 111010102 [EA16]
digits gives:
and converting each into 0111 1011 11. 100010112 [8B16]
binary gives: from Table 3.2
12. 101001012 [A516]
Thus, 7B16 = 11110112 1 7D
(b) Spacing out hexadecimal In Problems 13 to 16, convert the given hexadeci-
0001 0111 1101 mal numbers into their binary equivalents.
digits gives: from Table 3.2
and converting each into 13. 3716 [1101112]
binary gives:
14. ED16 [111011012]
Thus, 17D16 = 1011111012
15. 9F16 [100111112]

16. A2116 [1010001000012]

Chapter 4

Calculations and evaluation
of formulae

4.1 Errors and approximations 50 × 20 × 120
to and then, by cancelling,
(i) In all problems in which the measurement of dis-
tance, time, mass or other quantities occurs, an 60 × 40
exact answer cannot be given; only an answer
which is correct to a stated degree of accuracy 50 × 1&2&0 ר12¨02£1 = 50. An accurate answer
can be given. To take account of this an error 1&6&0 ×&4&02£1
due to measurement is said to exist. somewhere between 45 and 55 could therefore be

(ii) To take account of measurement errors it is usual expected. Certainly an answer around 500 or 5
to limit answers so that the result given is not
more than one significant figure greater than would not be expected. Actually, by calculator
the least accurate number given in the data.
49.1 × 18.4 × 122.1 = 47.31, correct to 4 sig-
(iii) Rounding-off errors can exist with decimal frac- 61.2 × 38.1
tions. For example, to state that π = 3.142 is not nificant figures.
strictly correct, but ‘π = 3.142 correct to 4 sig-
nificant figures’ is a true statement. (Actually, Problem 1. The area A of a triangle is given by
π = 3.14159265. . .) A = 1 bh. The base b when measured is found to be

(iv) It is possible, through an incorrect procedure, to 2
obtain the wrong answer to a calculation. This 3.26 cm, and the perpendicular height h is 7.5 cm.
type of error is known as a blunder.
Determine the area of the triangle.
(v) An order of magnitude error is said to exist if
incorrect positioning of the decimal point occurs 11
after a calculation has been completed. Area of triangle = bh = × 3.26 × 7.5

(vi) Blunders and order of magnitude errors can be 22
reduced by determining approximate values of = 12.225 cm2 (by calculator).
calculations. Answers which do not seem feasi- The approximate values is 1 × 3 × 8 = 12 cm2, so
ble must be checked and the calculation must be
repeated as necessary. 2
there are no obvious blunder or magnitude errors. How-
An engineer will often need to make a quick men- ever, it is not usual in a measurement type problem to
tal approximation for a calculation. For exam- state the answer to an accuracy greater than 1 significant
figure more than the least accurate number in the data:
49.1 × 18.4 × 122.1 this is 7.5 cm, so the result should not have more than
ple, 61.2 × 38.1 may be approximated 3 significant figures
Thus, area of triangle = 12.2 cm2

Problem 2. State which type of error has been
made in the following statements:

(a) 72 × 31.429 = 2262.9
(b) 16 × 0.08 × 7 = 89.6

28 Engineering Mathematics

(c) 11.714 × 0.0088 = 0.3247 correct to 4 decimal i.e. 2.19 × 203.6 × 17.91 ≈ 80
places. 12.1 × 8.76
Section 1
(d) 29.74 × 0.0512 = 0.12, correct to 2 significant (By calculator, 2.19 × 203.6 × 17.91 ≈ 75.3,
11.89 12.1 × 8.76
correct to 3 significant figures.)
figures.

(a) 72 × 31.429 = 2262.888 (by calculator), hence a Now try the following exercise
rounding-off error has occurred. The answer
should have stated: Exercise 14 Further problems on errors

72 × 31.429 = 2262.9, correct to 5 significant fig-
ures or 2262.9, correct to 1 decimal place.

(b) 16 × 0.08 × 7 = 16 × 8 × 7 = 32 × 7 In Problems 1 to 5 state which type of error, or
100 25 errors, have been made:

= 224 = 8 24 = 8.96 1. 25 × 0.06 × 1.4 = 0.21
25 25 [order of magnitude error]

Hence an order of magnitude error has occurred. 2. 13⎡7 × 6.842 = 937.4 ⎤

(c) 11.714 × 0.0088 is approximately equal to Rounding-off error–should add ‘correct
12 × 9 × 10−3, i.e. about 108 × 10−3 or 0.108.
⎣ to 4 significant figures’ or ‘correct to ⎦
Thus a blunder has been made.
1 decimal place’

29.74 × 0.0512 30 × 5 × 10−2 3. 24 × 0.008 = 10.42
(d) ≈ 12.6 [Blunder]
11.89 12

150 15 1 4. For a gas pV = c. When pressure p = 1 03 400
= 12 × 102 = 120 = 8 or 0.125 Pa and V = 0.54 m3 then c = 55 836 Pa m3.

hence no order of magnitude error has occurred. Measured values, hence
However, 29.74 × 0.0512 = 0.128 correct to 3 c = 55 800 Pa m3

11.89 5. 4.6 × 0.07 = 0.225
significant figures, which equals 0.13 correct to 52⎡.3 × 0.274
⎤
2 significant figures.

Hence a rounding-off error has occurred. ⎣⎢⎢ Order of magnitude error and rounding- ⎦⎥⎥

off error–should be 0.0225, correct to 3
significant figures or 0.0225,
Problem 3. Without using a calculator, determine
an approximate value of: correct to 4 decimal places

11.7 × 19.1 2.19 × 203.6 × 17.91 In Problems 6 to 8, evaluate the expressions
(a) (b) approximately, without using a calculator.
9.3 × 5.7 12.1 × 8.76

11.7 × 19.1 10 × 20 6. 4.7 × 6.3 [≈30 (29.61, by calculator)]

(a) 9.3 × 5.7 is approximately equal to 10 × 5 2.87 × 4.07
7.
i.e. about 4
6.12 × 0.96
(By calculator, 11.7 × 19.1 = 4.22, correct to 3 ≈2 (1.988, correct to 4 s.f., by
9.3 × 5.7 calculator)

significant figures.) 72.1 × 1.96 × 48.6
8.
(b) 2.19 × 203.6 × 17.91 ≈ 2 × 20¨20¨0 ×&2&02
12.1 × 8.76 1&1&0 ×&1&01 139.3 × 5.2
≈10 (9.481, correct to 4 s.f., by
= 2 × 20 × 2 after cancelling, calculator)

Calculations and evaluation of formulae 29

4.2 Use of calculator (a) 1 = 0.01896453 . . . = 0.019, correct to 3 Section 1
52.73
The most modern aid to calculations is the pocket-sized decimal places
electronic calculator. With one of these, calculations
can be quickly and accurately performed, correct to (b) 1 = 36.3636363 . . . = 36.364, correct to 3
about 9 significant figures. The scientific type of calcu- 0.0275
lator has made the use of tables and logarithms largely decimal places
redundant.
(c) 1 + 1 = 0.71086624 . . . = 0.711, correct to
To help you to become competent at using your cal- 4.92 1.97
culator check that you agree with the answers to the 3 decimal places
following problems:
Problem 7. Evaluate the following, expressing
Problem 4. Evaluate the following, correct to 4 the answers in standard form, correct to 4
significant figures: significant figures.

(a) 4.7826 + 0.02713 (b) 17.6941 − 11.8762 (a) (0.00451)2 (b) 631.7 − (6.21 + 2.95)2
(c) 21.93 × 0.012981
(c) 46.272 − 31.792
(a) 4.7826 + 0.02713 = 4.80973 = 4.810, correct to
4 significant figures (a) (0.00451)2 = 2.03401 × 10−5 = 2.034 × 10−5,
correct to 4 significant figures
(b) 17.6941 − 11.8762 = 5.8179 = 5.818, correct to
4 significant figures (b) 631.7 − (6.21 + 2.95)2 = 547.7944
= 5.477944 × 102 = 5.478 × 102, correct to 4
(c) 21.93 × 0.012981 = 0.2846733. . . = 0.2847, cor- significant figures
rect to 4 significant figures
(c) 46.272 − 31.792 = 1130.3088 = 1.130 × 103,
correct to 4 significant figures

Problem 5. Evaluate the following, correct to 4 Problem 8. Evaluate the following, correct to 3
decimal places: decimal places:

(a) 46.32 × 97.17 × 0.01258 4.621 (2.37)2 (b) 3.60 2 5.40 2
(b) (a)
+
23.76 0.0526 1.92 2.45

(c) 1 × 0.0172) 15
(62.49 (c) 7.62 − 4.82
2

(a) 46.32 × 97.17 × 0.01258 = 56.6215031. . . (a) (2.37)2 = 106.785171. . . = 106.785, correct to 3
= 56.6215, correct to 4 decimal places 0.0526
decimal places
(b) 4.621 = 0.19448653. . . = 0.1945, correct to 4
23.76 (b) 3.60 2 5.40 2
decimal places
+ = 8.37360084. . . = 8.374,
(c) 1 (62.49 × 0.0172) = 0.537414 = 0.5374, correct 1.92 2.45
2
to 4 decimal places correct to 3 decimal places

15
(c) 7.62 − 4.82 = 0.43202764. . . = 0.432, correct to

3 decimal places

Problem 6. Evaluate the following, correct to 3 Problem 9. Evaluate the following, correct to 4
decimal places: significant figures:

11 (c) 1 + 1 √√√
(a) (b) (a) 5.462 (b) 54.62 (c) 546.2
52.73 0.0275 4.92 1.97

30 Engineering Mathematics

Section 1 √ Problem 13. Evaluate the following, expressing
(a) 5.462 = 2.3370922. . . = 2.337, correct to 4 sig- the answers in standard form, correct to 4 decimal
places:
nificant figures
√ (a) (5.176 × 10−3)2
(b) 54.62 = 7.39053448. . . = 7.391, correct to 4
significant figures 1.974 × 101 × 8.61 × 10−2 4
√ (b)
(c) 546.2 = 23.370922 . . . = 23.37, correct to 4
significant figures 3.462
√
Problem 10. Evaluate the following, correct to 3 (c) 1.792 × 10−4

decimal places: √√
√ (b) 52.91 − 31.76

(a) 0.007328 (a) (5.176 × 10−3)2 = 2.679097. . . × 10−5
√ = 2.6791 × 10−5, correct to 4 decimal places

(c) 1.6291 × 104

√ 1.974 × 101 × 8.61 × 10−2 4
(a) 0.007328 = 0.08560373 = 0.086, correct to 3 (b) = 0.05808887. . .

decimal places 3.462
√√
(b) 52.91 − 31.76 = 1.63832491. . . = 1.638, = 5.8089 × 10−2, correct to 4 decimal places
correct to 3 decimal places √
√√ (c) 1.792 × 10−4 = 0.0133865. . . = 1.3387 × 10−2,
(c) 1.6291 × 104 = 16291 = 127.636201. . .
correct to 4 decimal places
= 127.636, correct to 3 decimal places

Problem 11. Evaluate the following, correct to 4 Now try the following exercise
significant figures:
Exercise 15 Further problems on the use of
√ a calculator
(a) 4.723 (b) (0.8316)4 (c) 76.212 − 29.102

(a) 4.723 = 105.15404. . . = 105.2, correct to 4 signif- In Problems 1 to 9, use a calculator to evaluate the
icant figures quantities shown correct to 4 significant figures:

(b) (0.8316)4 = 0.47825324. . . = 0.4783, correct to 4 1. (a) 3.2492 (b) 73.782 (c) 311.42 (d) 0.06392
significant figures
√ (a) 10.56 (b) 5443 (c) 96970
(d) 0.004083
(c) 76.212 − 29.102 = 70.4354605. . . = 70.44, cor- √√√
rect to 4 significant figures 2. (a) √4.735 (b) 35.46 (c) 73 280
(d) 0.0256
Problem 12. Evaluate the following, correct to 3
significant figures: (a) 2.176 (b) 5.955 (c) 270.7
(d) 0.1600
(a) 6.092√ √
25.2 × 7 (b) 3 47.291 11 1

3. (a) (b) (c)
√ 7.768 48.46 0.0816
(c) 7.2132 + 6.4183 + 3.2914 1
(d)
1.118

6.092 (a) 0.1287 (b) 0.02064
√
(a) = 0.74583457. . . = 0.746, correct (c) 12.25 (d) 0.8945

25.2 × 7 4. (a) 127.8 × 0.0431 × 19.8

to 3 significant figures (b) 15.76 ÷ 4.329
√ [(a) 109.1 (b) 3.641]
(b) 3 47.291 = 3.61625876. . . = 3.62, correct to 3

significant figures 5. (a) 137.6 (b) 11.82 × 1.736
√
(c) 7.2132 + 6.4183 + 3.2914 = 20.8252991. . ., 552.9 0.041

= 20.8 correct to 3 significant figures [(a) 0.2489 (b) 500.5]

Calculations and evaluation of formulae 31

6. (a) 13.63 (b) 3.4764 (c) 0.1245 Calculate: Section 1

[(a) 2515 (b) 146.0 (c) 0.00002932] (a) how many French euros £27.80 will buy

24.68 × 0.0532 3 (b) the number of Japanese yen which can be
7. (a) bought for £23

7.412 (c) the pounds sterling which can be exchanged
0.2681 × 41.22 4 for 7114.80 Norwegian kronor
(b)
32.6 × 11.89 (d) the number of American dollars which can be
purchased for £90, and
[(a) 0.005559 (b) 1.900]
(e) the pounds sterling which can be exchanged
14.323 4.8213 for 2990 Swiss francs
8. (a) 21.682 (b) 17.332 − 15.86 × 11.6

[(a) 6.248 (b) 0.9630]

(15.62)2 (a) £1 = 1.46 euros, hence
9. (a) √ £27.80 = 27.80 × 1.46 euros = 40.59 euros
29.21 × 10.52
(b) £1 = 220 yen, hence
√ £23 = 23 × 220 yen = 5060 yen
(b) 6.9212 + 4.8163 − 2.1614
(c) £1 = 12.10 kronor, hence
[(a) 1.605 (b) 11.74] 7114.80 kronor = £ 7114.80 = £588
12.10
10. Evaluate the following, expressing the
answers in standard form, correct to 3 dec- (d) £1 = 1.95 dollars, hence
imal places: (a) (√8.291 × 10−2)2 £90 = 90 × 1.95 dollars = $175.50
(b) 7.623 × 10−3
(e) £1 = 2.30 Swiss francs, hence
[(a) 6.874 × 10−3 (b) 8.731 × 10−2] 2990 franc = £ 2990 = £1300
2.30

4.3 Conversion tables and charts Problem 15. Some approximate imperial to
metric conversions are shown in Table 4.2
It is often necessary to make calculations from var-
ious conversion tables and charts. Examples include Table 4.2
currency exchange rates, imperial to metric unit con-
versions, train or bus timetables, production schedules length 1 inch = 2.54 cm
and so on. 1 mile = 1.61 km

Problem 14. Currency exchange rates for five weight 2.2 lb = 1 kg
countries are shown in Table 4.1 (1 lb = 16 oz)

capacity 1.76 pints = 1 litre
(8 pints = 1 gallon)

Table 4.1 £1 = 1.46 euros Use the table to determine:
France £1 = 220 yen
Japan £1 = 12.10 kronor (a) the number of millimetres in 9.5 inches,
Norway £1 = 2.30 francs (b) a speed of 50 miles per hour in kilometres
Switzerland £1 = 1.95 dollars ($)
U.S.A. per hour,
(c) the number of miles in 300 km,
(d) the number of kilograms in 30 pounds weight,
(e) the number of pounds and ounces in 42

kilograms (correct to the nearest ounce),
(f) the number of litres in 15 gallons, and
(g) the number of gallons in 40 litres.

32 Engineering Mathematics

Section 1 (a) 9.5 inches = 9.5 × 2.54 cm = 24.13 cm 2. Below is a list of some metric to imperial
conversions.
24.13 cm = 24.13 × 10 mm = 241.3 mm
Length 2.54 cm = 1 inch
(b) 50 m.p.h. = 50 × 1.61 km/h = 80.5 km/h 1.61 km = 1 mile
(c) 300 km = 300 miles = 186.3 miles Weight
Capacity 1 kg = 2.2 lb (1 lb = 16 ounces)
1.61
(d) 30 lb = 30 kg = 13.64 kg 1 litre = 1.76 pints
(8 pints = 1 gallon)
2.2
(e) 42 kg = 42 × 2.2 lb = 92.4 lb Use the list to determine (a) the number

0.4 lb = 0.4 × 16 oz = 6.4 oz = 6 oz, correct to the of millimetres in 15 inches, (b) a speed of
nearest ounce
35 mph in km/h, (c) the number of kilome-
Thus 42 kg = 92 lb 6 oz, correct to the nearest
ounce. tres in 235 miles, (d) the number of pounds

(f) 15 gallons = 15 × 8 pints = 120 pints and ounces in 24 kg (correct to the near-
120 pints = 120 litres = 68.18 litres
1.76 est ounce), (e) the number of kilograms in

(g) 40 litres = 40 × 1.76 pints = 70.4 pints 15 lb, (f) the number of litres in 12 gallons
70.4 pints = 70.4 gallons = 8.8 gallons
8 and (g) the number of gallons in 25 litres.
⎡⎤
(a) 381 mm (b) 56.35 km/h
⎢⎢⎣ ⎦⎥⎥
(c) 378.35 km (d) 52 lb 13 oz
(e) 6.82 kg (f) 54.55 litre

(g) 5.5 gallons

3. Deduce the following information from the
train timetable shown in Table 4.3:

Now try the following exercise (a) At what time should a man catch a train
at Mossley Hill to enable him to be in
Manchester Piccadilly by 8.15 a.m.?

Exercise 16 Further problems conversion (b) A girl leaves Hunts Cross at 8.17 a.m. and
tables and charts travels to Manchester Oxford Road. How
long does the journey take? What is the
1. Currency exchange rates listed in a newspaper average speed of the journey?
included the following:
(c) A man living at Edge Hill has to be at
Italy £1 = 1.48 euro
Japan £1 = 225 yen work at Trafford Park by 8.45 a.m. It takes
Australia £1 = 2.50 dollars
Canada £1 = $2.20 him 10 minutes to walk to his work from
Sweden £1 = 13.25 kronor
Trafford Park station. What time train
Calculate (a) how many Italian euros £32.50
will buy, (b) the number of Canadian dol- should he catch from Edge Hill?
lars that can be purchased for £74.80, (c) the ⎡⎤
pounds sterling which can be exchanged for (a) 7.09 a.m.
14 040 yen, (d) the pounds sterling which can ⎣ (b) 52 minutes, 31.15 m.p.h. ⎦
be exchanged for 1754.30 Swedish kronor, and
(e) the Australian dollars which can be bought (c) 7.04 a.m.
for £55
4.4 Evaluation of formulae
[(a) 48.10 euros (b) $164.56
(c) £62.40 (d) £132.40 The statement v = u + at is said to be a formula for v
(e) 137.50 dollars] in terms of u, a and t.
v, u, a and t are called symbols.
The single term on the left-hand side of the equation, v,
is called the subject of the formulae.

Table 4.3 Liverpool, Hunt’s Cross and Warrington → Manchester

MX MO SX SO SX

BHX BHX BHX

♦♦ ♦

AC C

Miles $$

II I

0 Liverpool Lime Street 82, 99 d 05 25 05 37 06 03 06 23 06 30
06 34
1 1 Edge Hill 82, 99 d 06 39
2 82 d 06 41
82 d 06 43
3 1 Mossley Hill 82 d 06 73
2

4 1 West Allerton
2

5 1 Allerton
2

– Liverpool Central 101 d

– Garston (Merseyside) 101 d 06 26

7 1 Hunt’s Cross d 05u38 05u50 06 17 06 47
2

8 1 Halewood d 06 20 06 50
2

10 1 Hough Green d 06 24 06 54
2

12 1 Widnes d 06 27 06 57
2

16 Sankey for Penketh d 00 02 06 32 07 02

18 1 Warrington Central a 00 07 05 50 06 02 06 37 06 46 07 07
2

– d 05 51 06 03 06 30 06 46 07 00

20 1 Padgate d 06 33 07 03
2

21 1 Birchwood d 05 56 06 08 06 36 06 51 07 06
2

24 1 Glazebrook d 06 41 07 11
2

25 1 Iriam d 06 02 06 44 07 14
2

28 Flixton d 06 06 06 48 07 18

28 1 Chassen Road d 06 08 06 50 07 20
2

29 Urmston d 00 03 06 10 06 52 07 22

30 1 Humphrey Park d 00 13 06 13 06 55 07 25
2

31 Trafford Park d 06 15 06 57 07 27

34 Deansgate 81 d 00 23 07 03 07 33
00 27 06 22 07 05 07 08 07 35
34 1 Manchester Oxford Road 81 a 00 27 06 23 06 22 07 09
2 06 23

–d

35 Manchester Piccadilly 81 a 00 34 06 25 06 25 07 11

– Stockport 81, 90 a 06 34 06 34 07 32

– Shetfield 90 a 07 30 07 30

BHX SX BHX BHX SX BHX Ch04-H8555.tex 31/7/2007 9: 53 page 33
♦ BHX ♦ BHX
D E 08 30
♦ ♦ 08 34
C 08 39
08 41
I II 08 43
08 75
06 54 07 00 07 17 07 30 07 52 08 00 08 23 08 26
08 47
07 04 07 34 08 04 08 50
08 54
07 09 07 39 08 09 08 57
09 02
07 11 07 41 08 11 09 07
09 00
07 13 07 43 08 13 09 03
09 06
06 45 07 15 07 45 09 11
09 14
06 56 07 26 07 56 09 18
09 20
07u07 07 17 07 47 08u05 08 17 09 22
09 25
07 20 07 50 08 20 09 27
09 33
07 24 07 54 08 24 09 35

07 27 07 35 07 57 08 12 08 27 (Continued)

07 32 08 02 08 32

07 19 07 37 07 43 08 07 08 20 08 37 08 46

07 20 07 30 07 43 08 00 08 20 08 30 08 46

07 33 08 03 08 33

07 25 07 36 07 48 08 06 08 25 08 34 08 51

07 41 08 11 08 41

07 44 07 54 08 14 08 34 08 44

07 48 08 15 08 38 08 48 Calculations and evaluation of formulae 33

07 50 08 20 08 40 08 50

07 52 08 22 08 42 08 52

07 55 08 25 08 45 08 55

07 57 08 27 08 47 08 57

08 03 08 33 08 52 09 03

07 40 08 05 08 08 08 35 08 40 08 54 09 05 09 08

07 41 08 09 08 37 08 41 08 55 09 09

07 43 08 11 08 39 08 43 08 57 09 11

07 54 08 32 08 54 09 19 09 32

08 42 08 42

Section 1

Table 4.3 Continued

BHX
♦ ♦ ♦♦
I I II

Liverpool Lime Street 82, 99 d 08 54 09 00 09 23 09 30 09 56 10 00 10 23

Edge Hill 82, 99 d 09 04 09 34 10 04

Mossley Hill 82 d 09 09 09 39 10 09

West Allerton 82 d 09 11 09 41 10 11

Allerton 82 d 09 13 09 43 10 13

Liverpool Central 101 d 09 45 09 15 09 45

Garston (Merseyside) 101 d 09 56 09 26 09 56

Hunt’s Cross d 09u09 09 17 09 47 10u09 10 17

Halewood d 09 20 09 50 10 20

Hough Green d 09 24 09 54 10 24

Widnes d 09 27 09 57 10 27

Sankey for Penketh d 09 32 10 02 10 32

Warrington Central a 09 21 09 37 09 46 10 07 10 21 10 37 10 46

d 09 22 09 30 09 46 10 00 10 22 10 46 11
11
Padgate d 09 33 10 03 11
11
Birchwood d 09 36 09 51 10 06 10 51 11
11
Glazebrook d 09 41 10 11 11
11
Iriam d 09 44 10 14 11
11
Flexton d 09 48 10 18 11
11
Chassen Road d 09 50 10 20

Urmston d 09 52 10 22

Humphrey Park d 09 55 10 25

Trafford Park d 09 57 10 27

Deansgate 81 d 10 03 10 33

Manchester Oxford Road 81 a 09 40 10 05 10 08 10 35 10 40 11 08

d 09 41 10 06 10 09 10 41 11 09

Manchester Piccadilly 81 a 09 43 10 08 10 11 10 43 11 11

Stockport 81, 90 a 09 54 10 25 10 32 10 54 11 32

Sheffield 90 a 10 42 11 42

Reproduced with permission of British Rail