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### Basic engineering mathematics e5

1. 1. Basic Engineering Mathematics
2. 2. In memory of Elizabeth
3. 3. Basic Engineering MathematicsFifth editionJohn Bird, BSc(Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollTAMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORDPARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYONewnes is an imprint of Elsevier
5. 5. ContentsPreface ixAcknowledgements xInstructor’s Manual xi1 Basic arithmetic 11.1 Introduction 11.2 Revision of addition and subtraction 11.3 Revision of multiplication and division 31.4 Highest common factors and lowestcommon multiples 51.5 Order of precedence and brackets 62 Fractions 92.1 Introduction 92.2 Adding and subtracting fractions 102.3 Multiplication and division of fractions 122.4 Order of precedence with fractions 13Revision Test 1 153 Decimals 163.1 Introduction 163.2 Converting decimals to fractions andvice-versa 163.3 Signiﬁcant ﬁgures and decimal places 173.4 Adding and subtracting decimal numbers 183.5 Multiplying and dividing decimal numbers 194 Using a calculator 224.1 Introduction 224.2 Adding, subtracting, multiplying anddividing 224.3 Further calculator functions 234.4 Evaluation of formulae 285 Percentages 335.1 Introduction 335.2 Percentage calculations 335.3 Further percentage calculations 355.4 More percentage calculations 36Revision Test 2 396 Ratio and proportion 406.1 Introduction 406.2 Ratios 406.3 Direct proportion 426.4 Inverse proportion 457 Powers, roots and laws of indices 477.1 Introduction 477.2 Powers and roots 477.3 Laws of indices 488 Units, preﬁxes and engineering notation 538.1 Introduction 538.2 SI units 538.3 Common preﬁxes 538.4 Standard form 568.5 Engineering notation 57Revision Test 3 609 Basic algebra 619.1 Introduction 619.2 Basic operations 619.3 Laws of indices 6410 Further algebra 6810.1 Introduction 6810.2 Brackets 6810.3 Factorization 6910.4 Laws of precedence 7111 Solving simple equations 7311.1 Introduction 7311.2 Solving equations 7311.3 Practical problems involving simpleequations 77Revision Test 4 8212 Transposing formulae 8312.1 Introduction 8312.2 Transposing formulae 8312.3 Further transposing of formulae 8512.4 More difﬁcult transposing of formulae 8713 Solving simultaneous equations 9013.1 Introduction 9013.2 Solving simultaneous equations in twounknowns 9013.3 Further solving of simultaneous equations 92
6. 6. vi Contents13.4 Solving more difﬁcult simultaneousequations 9413.5 Practical problems involving simultaneousequations 9613.6 Solving simultaneous equations in threeunknowns 99Revision Test 5 10114 Solving quadratic equations 10214.1 Introduction 10214.2 Solution of quadratic equations byfactorization 10214.3 Solution of quadratic equations by‘completing the square’ 10514.4 Solution of quadratic equations byformula 10614.5 Practical problems involving quadraticequations 10814.6 Solution of linear and quadratic equationssimultaneously 11015 Logarithms 11115.1 Introduction to logarithms 11115.2 Laws of logarithms 11315.3 Indicial equations 11515.4 Graphs of logarithmic functions 11616 Exponential functions 11816.1 Introduction to exponential functions 11816.2 The power series for ex 11916.3 Graphs of exponential functions 12016.4 Napierian logarithms 12216.5 Laws of growth and decay 125Revision Test 6 12917 Straight line graphs 13017.1 Introduction to graphs 13017.2 Axes, scales and co-ordinates 13017.3 Straight line graphs 13217.4 Gradients, intercepts and equationsof graphs 13417.5 Practical problems involving straight linegraphs 14118 Graphs reducing non-linear laws to linear form 14718.1 Introduction 14718.2 Determination of law 14718.3 Revision of laws of logarithms 15018.4 Determination of law involving logarithms 15019 Graphical solution of equations 15519.1 Graphical solution of simultaneousequations 15519.2 Graphical solution of quadratic equations 15619.3 Graphical solution of linear and quadraticequations simultaneously 16019.4 Graphical solution of cubic equations 161Revision Test 7 16320 Angles and triangles 16520.1 Introduction 16520.2 Angular measurement 16520.3 Triangles 17120.4 Congruent triangles 17520.5 Similar triangles 17620.6 Construction of triangles 17921 Introduction to trigonometry 18121.1 Introduction 18121.2 The theorem of Pythagoras 18121.3 Sines, cosines and tangents 18321.4 Evaluating trigonometric ratios of acuteangles 18521.5 Solving right-angled triangles 18821.6 Angles of elevation and depression 191Revision Test 8 19322 Trigonometric waveforms 19522.1 Graphs of trigonometric functions 19522.2 Angles of any magnitude 19622.3 The production of sine and cosine waves 19822.4 Terminology involved with sine andcosine waves 19922.5 Sinusoidal form: Asin(ωt ± α) 20223 Non-right-angled triangles and some practicalapplications 20523.1 The sine and cosine rules 20523.2 Area of any triangle 20523.3 Worked problems on the solution oftriangles and their areas 20623.4 Further worked problems on the solutionof triangles and their areas 20723.5 Practical situations involving trigonometry 20923.6 Further practical situations involvingtrigonometry 21124 Cartesian and polar co-ordinates 21424.1 Introduction 21424.2 Changing from Cartesian to polarco-ordinates 21424.3 Changing from polar to Cartesianco-ordinates 21624.4 Use of Pol/Rec functions on calculators 217
7. 7. Contents viiRevision Test 9 21825 Areas of common shapes 21925.1 Introduction 21925.2 Common shapes 21925.3 Areas of common shapes 22125.4 Areas of similar shapes 22926 The circle 23026.1 Introduction 23026.2 Properties of circles 23026.3 Radians and degrees 23226.4 Arc length and area of circles and sectors 23326.5 The equation of a circle 236Revision Test 10 23827 Volumes of common solids 24027.1 Introduction 24027.2 Volumes and surface areas of commonshapes 24027.3 Summary of volumes and surface areas ofcommon solids 24727.4 More complex volumes and surface areas 24727.5 Volumes and surface areas of frusta ofpyramids and cones 25227.6 Volumes of similar shapes 25628 Irregular areas and volumes, and mean values 25728.1 Areas of irregular ﬁgures 25728.2 Volumes of irregular solids 25928.3 Mean or average values of waveforms 260Revision Test 11 26429 Vectors 26629.1 Introduction 26629.2 Scalars and vectors 26629.3 Drawing a vector 26629.4 Addition of vectors by drawing 26729.5 Resolving vectors into horizontal andvertical components 26929.6 Addition of vectors by calculation 27029.7 Vector subtraction 27429.8 Relative velocity 27629.9 i, j and k notation 27730 Methods of adding alternating waveforms 27830.1 Combining two periodic functions 27830.2 Plotting periodic functions 27830.3 Determining resultant phasors by drawing 28030.4 Determining resultant phasors by the sineand cosine rules 28130.5 Determining resultant phasors byhorizontal and vertical components 283Revision Test 12 28631 Presentation of statistical data 28831.1 Some statistical terminology 28831.2 Presentation of ungrouped data 28931.3 Presentation of grouped data 29232 Mean, median, mode and standard deviation 29932.1 Measures of central tendency 29932.2 Mean, median and mode for discrete data 29932.3 Mean, median and mode for grouped data 30032.4 Standard deviation 30232.5 Quartiles, deciles and percentiles 30333 Probability 30633.1 Introduction to probability 30633.2 Laws of probability 307Revision Test 13 31234 Introduction to differentiation 31334.1 Introduction to calculus 31334.2 Functional notation 31334.3 The gradient of a curve 31434.4 Differentiation from ﬁrst principles 31534.5 Differentiation of y = axn by thegeneral rule 31534.6 Differentiation of sine and cosine functions 31834.7 Differentiation of eax and lnax 32034.8 Summary of standard derivatives 32134.9 Successive differentiation 32234.10 Rates of change 32335 Introduction to integration 32535.1 The process of integration 32535.2 The general solution of integrals of theform axn 32535.3 Standard integrals 32635.4 Deﬁnite integrals 32835.5 The area under a curve 330Revision Test 14 335List of formulae 336Answers to practice exercises 340Index 356
8. 8. viii ContentsWebsite Chapters(Goto:http://www.booksite.elsevier.com/newnes/bird)Preface iv36 Number sequences 136.1 Simple sequences 136.2 The n’th term of a series 136.3 Arithmetic progressions 236.4 Geometric progressions 537 Binary, octal and hexadecimal 937.1 Introduction 937.2 Binary numbers 937.3 Octal numbers 1237.4 Hexadecimal numbers 1538 Inequalities 1938.1 Introduction to inequalities 1938.2 Simple inequalities 1938.3 Inequalities involving a modulus 2038.4 Inequalities involving quotients 2138.5 Inequalities involving square functions 2238.6 Quadratic inequalities 2339 Graphs with logarithmic scales 2539.1 Logarithmic scales and logarithmicgraph paper 2539.2 Graphs of the form y = axn 2539.3 Graphs of the form y = abx 2839.4 Graphs of the form y = aekx 29Revision Test 15 32Answers to practice exercises 33
10. 10. AcknowledgementsThe publisher wishes to thank CASIO Electronic Co.Ltd, London for permission to reproduce the image ofthe Casio fx-83ES calculator on page 23.The publishers also wish to thank the AutomobileAsso-ciation for permission to reproduce a map of Portsmouthon page 131.
11. 11. Instructor’s ManualFull worked solutions and mark scheme for all theAssignments are contained in this Manual which isavailable to lecturers only.To download the Instructor’s Manual visit http://www.booksite.elsevier.com/newnes/bird
13. 13. Chapter 1Basic arithmetic1.1 IntroductionWhole numbers are called integers. +3,+5 and +72are examples of positive integers; −13,−6 and −51are examples of negative integers. Between positiveand negative integers is the number 0 which is neitherpositive nor negative.The four basic arithmetic operators are add (+), subtract(−), multiply (×) and divide (÷).It is assumed that adding, subtracting, multiplying anddividing reasonably small numbers can be achievedwithout a calculator. However, if revision of this areais needed then some worked problems are included inthe following sections.When unlike signs occur together in a calculation, theoverall sign is negative. For example,3 + (−4) = 3 + −4 = 3 − 4 = −1and(+5) × (−2) = −10Like signs together give an overall positive sign. Forexample,3 − (−4) = 3 − −4 = 3 + 4 = 7and(−6) × (−4) = +241.2 Revision of addition andsubtractionYou can probably already add two or more numberstogether and subtract one number from another. How-ever, if you need a revision then the following workedproblems should be helpful.Problem 1. Determine 735 + 167H T U7 3 5+ 1 6 79 0 21 1(i) 5 + 7 = 12. Place 2 in units (U) column. Carry 1in the tens (T) column.(ii) 3 + 6 + 1 (carried) = 10. Place the 0 in the tenscolumn. Carry the 1 in the hundreds (H) column.(iii) 7 + 1 + 1 (carried) = 9. Place the 9 in the hun-dreds column.Hence, 735 + 167 = 902Problem 2. Determine 632 − 369H T U6 3 2− 3 6 92 6 3(i) 2 − 9 is not possible; therefore ‘borrow’ 1 fromthe tens column (leaving 2 in the tens column). Inthe units column, this gives us 12 − 9 = 3.(ii) Place 3 in the units column.(iii) 2 − 6 is not possible; therefore ‘borrow’ 1 fromthe hundreds column (leaving 5 in the hun-dreds column). In the tens column, this gives us12 − 6 = 6.(iv) Place the 6 in the tens column.DOI: 10.1016/B978-1-85617-697-2.00001-6
14. 14. 2 Basic Engineering Mathematics(v) 5 − 3 = 2.(vi) Place the 2 in the hundreds column.Hence, 632 − 369 = 263Problem 3. Add 27,−74,81 and −19This problem is written as 27 − 74 + 81 − 19.Adding the positive integers: 2781Sum of positive integers is 108Adding the negative integers: 7419Sum of negative integers is 93Taking the sum of the negative integersfrom the sum of the positive integers gives 108−9315Thus, 27 − 74 + 81 − 19 = 15Problem 4. Subtract −74 from 377This problem is written as 377 − −74. Like signstogether give an overall positive sign, hence377 − −74 = 377 + 74 3 7 7+ 7 44 5 1Thus, 377 − −74 = 451Problem 5. Subtract 243 from 126The problem is 126 − 243. When the second number islarger than the ﬁrst, take the smaller number from thelarger and make the result negative. Thus,126 − 243 = −(243 − 126) 2 4 3− 1 2 61 1 7Thus, 126 − 243 = −117Problem 6. Subtract 318 from −269The problem is −269 − 318. The sum of the negativeintegers is2 6 9+ 3 1 85 8 7Thus, −269 − 318 = −587Now try the following Practice ExercisePracticeExercise 1 Further problems onaddition and subtraction (answers onpage 340)In Problems 1 to 15, determine the values of theexpressions given, without using a calculator.1. 67kg − 82 kg + 34kg2. 73m − 57m3. 851mm − 372mm4. 124− 273 + 481 − 3985. £927 − £114+ £182 − £183 − £2476. 647 − 8727. 2417 − 487 + 2424− 1778 − 47128. −38419 − 2177 + 2440− 799 + 28349. £2715 − £18250+ £11471 − £1509 +£11327410. 47 + (−74) − (−23)11. 813 − (−674)12. 3151 − (−2763)13. 4872 g− 4683g14. −23148 − 4772415. \$53774− \$3844116. Holes are drilled 35.7mm apart in a metalplate. If a row of 26 holes is drilled, deter-mine the distance, in centimetres, betweenthe centres of the ﬁrst and last holes.17. Calculate the diameter d and dimensions Aand B for the template shown in Figure 1.1.All dimensions are in millimetres.
15. 15. Basic arithmetic 3126050 38120110BAdFigure 1.11.3 Revision of multiplication anddivisionYou can probably already multiply two numberstogether and divide one number by another. However, ifyou need a revision then the followingworked problemsshould be helpful.Problem 7. Determine 86 × 7H T U8 6× 76 0 24(i) 7 × 6 = 42. Place the 2 in the units (U) columnand ‘carry’ the 4 into the tens (T) column.(ii) 7 × 8 = 56;56 + 4 (carried) = 60. Place the 0 inthe tens column and the 6 in the hundreds (H)column.Hence, 86 × 7 = 602A good grasp of multiplication tables is needed whenmultiplying such numbers; a reminder of the multipli-cation table up to 12 × 12 is shown below. Conﬁdencewith handling numbers will be greatly improved if thistable is memorized.Problem 8. Determine 764× 387 6 4× 3 86 1 1 22 2 9 2 02 9 0 3 2Multiplication table× 2 3 4 5 6 7 8 9 10 11 122 4 6 8 10 12 14 16 18 20 22 243 6 9 12 15 18 21 24 27 30 33 364 8 12 16 20 24 28 32 36 40 44 485 10 15 20 25 30 35 40 45 50 55 606 12 18 24 30 36 42 48 54 60 66 727 14 21 28 35 42 49 56 63 70 77 848 16 24 32 40 48 56 64 72 80 88 969 18 27 36 45 54 63 72 81 90 99 10810 20 30 40 50 60 70 80 90 100 110 12011 22 33 44 55 66 77 88 99 110 121 13212 24 36 48 60 72 84 96 108 120 132 144
16. 16. 4 Basic Engineering Mathematics(i) 8 × 4 = 32. Place the 2 in the units column andcarry 3 into the tens column.(ii) 8 × 6 = 48;48 + 3 (carried) = 51. Place the 1 inthe tens column and carry the 5 into the hundredscolumn.(iii) 8 × 7 = 56;56 + 5 (carried) = 61. Place 1 in thehundredscolumn and 6 in thethousandscolumn.(iv) Place 0 in the units column under the 2.(v) 3 × 4 = 12. Place the 2 in the tens column andcarry 1 into the hundreds column.(vi) 3 × 6 = 18;18 + 1 (carried) = 19. Place the 9 inthe hundreds column and carry the 1 into thethousands column.(vii) 3 × 7 = 21;21 + 1 (carried) = 22. Place 2 in thethousands column and 2 in the ten thousandscolumn.(viii) 6112+ 22920 = 29032Hence, 764 × 38 = 29032Again, knowing multiplication tables is ratherimportantwhen multiplying such numbers.It is appreciated, of course, that such a multiplicationcan,and probably will,beperformed using a calculator.However, there are times when a calculator may not beavailable and it is then useful to be able to calculate the‘long way’.Problem 9. Multiply 178 by −46When the numbers have different signs, the result willbe negative. (With this in mind, the problem can nowbe solved by multiplying 178 by 46). Following theprocedure of Problem 8 gives1 7 8× 4 61 0 6 87 1 2 08 1 8 8Thus, 178 × 46 = 8188 and 178 × (−46) = −8188Problem 10. Determine 1834 ÷ 72627 1834(i) 7 into 18 goes 2, remainder 4. Place the 2 abovethe 8 of 1834 and carry the 4 remainder to thenext digit on the right, making it 43.(ii) 7 into 43 goes 6, remainder 1. Place the 6 abovethe 3 of 1834 and carry the 1 remainder to thenext digit on the right, making it 14.(iii) 7 into 14 goes 2, remainder 0. Place 2 above the4 of 1834.Hence, 1834÷ 7 = 1834/7 =18347= 262.The method shown is called short division.Problem 11. Determine 5796 ÷ 1248312 5796489996363600(i) 12 into 5 won’t go. 12 into 57 goes 4; place 4above the 7 of 5796.(ii) 4 × 12 = 48; place the 48 below the 57 of 5796.(iii) 57 − 48 = 9.(iv) Bring down the 9 of 5796 to give 99.(v) 12 into 99 goes 8; place 8 above the 9 of 5796.(vi) 8 × 12 = 96; place 96 below the 99.(vii) 99 − 96 = 3.(viii) Bring down the 6 of 5796 to give 36.(ix) 12 into 36 goes 3 exactly.(x) Place the 3 above the ﬁnal 6.(xi) Place the 36 below the 36.(xii) 36 − 36 = 0.Hence, 5796 ÷ 12 = 5796/12 =579612= 483.The method shown is called long division.
17. 17. Basic arithmetic 5Now try the following Practice ExercisePracticeExercise 2 Further problems onmultiplication and division (answers onpage 340)Determine the values of the expressions given inproblems 1 to 9, without using a calculator.1. (a) 78 × 6 (b) 124 × 72. (a) £261 × 7 (b) £462 × 93. (a) 783kg × 11 (b) 73kg × 84. (a) 27mm × 13 (b) 77mm × 125. (a) 448 × 23 (b) 143 × (−31)6. (a) 288m ÷ 6 (b) 979m ÷ 117. (a)18137(b)896168. (a)2142413(b) 15900 ÷ 159. (a)8873711(b) 46858 ÷ 1410. A screw has a mass of 15grams. Calculate,in kilograms, the mass of 1200 such screws(1kg = 1000g).1.4 Highest common factors andlowest common multiplesWhen two or more numbers are multiplied together, theindividualnumbers are called factors. Thus, a factoris anumber which divides into another number exactly. Thehighest common factor (HCF) is the largest numberwhich divides into two or more numbers exactly.For example, consider the numbers 12 and 15.The factors of 12 are 1, 2, 3, 4, 6 and 12 (i.e. all thenumbers that divide into 12).The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbersthat divide into 15).1 and 3 are the only common factors; i.e., numberswhich are factors of both 12 and 15.Hence, the HCF of 12 and 15 is 3 since 3 is the highestnumber which divides into both 12 and 15.A multiple is a number which contains another numberan exact number of times. The smallest number whichis exactly divisible by each of two or more numbers iscalled the lowest common multiple (LCM).For example, the multiples of 12 are 12, 24, 36, 48,60, 72,... and the multiples of 15 are 15, 30, 45,60, 75,...60 is a common multiple (i.e. a multiple of both 12 and15) and there are no lower common multiples.Hence, the LCM of 12 and 15 is 60 since 60 is thelowest number that both 12 and 15 divide into.Here are some further problems involving the determi-nation of HCFs and LCMs.Problem 12. Determine the HCF of the numbers12, 30 and 42Probably the simplest way of determining an HCF is toexpress each number in terms of its lowest factors. Thisis achieved by repeatedly dividing by the prime numbers2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,12 = 2 × 2 × 330 = 2 × 3 × 542 = 2 × 3 × 7The factors which are common to each of the numbersare 2 in column 1 and 3 in column 3, shown by thebroken lines. Hence, the HCF is 2 × 3; i.e., 6. That is,6 is the largest number which will divide into 12, 30and 42.Problem 13. Determine the HCF of the numbers30, 105, 210 and 1155Using the method shown in Problem 12:30 = 2 × 3 × 5105 = 3 × 5 × 7210 = 2 × 3 × 5 × 71155 = 3 × 5 × 7 × 11The factors which are common to each of the numbersare 3 in column 2 and 5 in column 3. Hence, the HCFis 3 × 5 = 15.Problem 14. Determine the LCM of the numbers12, 42 and 90
18. 18. 6 Basic Engineering MathematicsThe LCM is obtained by ﬁnding the lowest factors ofeach of the numbers, as shown in Problems 12 and 13above, and then selecting the largest group of any of thefactors present. Thus,12 = 2 × 2 × 342 = 2 × 3 × 790 = 2 × 3 × 3 × 5The largest group of any of the factors present is shownby the broken lines and are 2 × 2 in 12, 3 × 3 in 90, 5 in90 and 7 in 42.Hence, the LCM is 2 × 2 × 3 × 3 × 5 × 7 = 1260 andis the smallest number which 12, 42 and 90 will alldivide into exactly.Problem 15. Determine the LCM of the numbers150, 210, 735 and 1365Using the method shown in Problem 14 above:150 = 2 × 3 × 5 × 5210 = 2 × 3 × 5 × 7735 = 3 × 5 × 7 × 71365 = 3 × 5 × 7 × 13Hence, the LCM is 2 × 3 × 5 × 5 × 7 × 7 × 13 =95550.Now try the following Practice ExercisePracticeExercise 3 Further problems onhighest common factors and lowest commonmultiples (answers on page 340)Find (a) the HCF and (b) the LCM of the followinggroups of numbers.1. 8, 12 2. 60, 723. 50, 70 4. 270, 9005. 6, 10, 14 6. 12, 30, 457. 10, 15, 70, 105 8. 90, 105, 3009. 196, 210, 462, 910 10. 196, 350, 7701.5 Order of precedence and brackets1.5.1 Order of precedenceSometimes addition, subtraction, multiplication, divi-sion, powers and brackets may all be involved in acalculation. For example,5 − 3 × 4 + 24 ÷ (3 + 5) − 32This is an extreme example but will demonstrate theorder that is necessary when evaluating.When we read, we read from left to right. However,with mathematics there is a deﬁnite order of precedencewhich we need to adhere to. The order is as follows:BracketsOrder (or pOwer)DivisionMultiplicationAdditionSubtractionNotice that the ﬁrst letters of each word spell BOD-MAS, a handy aide-m´emoire. Order means pOwer. Forexample, 42 = 4 × 4 = 16.5 − 3 × 4 + 24 ÷ (3 + 5) − 32 is evaluated asfollows:5 − 3 × 4 + 24 ÷ (3 + 5) − 32= 5 − 3 × 4 + 24 ÷ 8 − 32(Bracket is removed and3 + 5 replaced with 8)= 5 − 3 × 4 + 24 ÷ 8 − 9 (Order means pOwer; inthis case, 32= 3 × 3 = 9)= 5 − 3 × 4 + 3 − 9 (Division: 24 ÷ 8 = 3)= 5 − 12 + 3 − 9 (Multiplication: − 3 × 4 = −12)= 8 − 12 − 9 (Addition: 5 + 3 = 8)= −13 (Subtraction: 8 − 12 − 9 = −13)In practice, it does not matter if multiplicationis per-formed before divisionor if subtraction is performedbefore addition. What is important is that the pro-cessofmultiplicationanddivisionmustbecompletedbefore addition and subtraction.1.5.2 Brackets and operatorsThe basic laws governing the use of brackets andoperators are shown by the following examples.
19. 19. Basic arithmetic 7(a) 2 + 3 = 3 + 2; i.e., the order of numbers whenadding does not matter.(b) 2 × 3 = 3 × 2; i.e., the order of numbers whenmultiplying does not matter.(c) 2 + (3 + 4) = (2 + 3) + 4; i.e., the use of bracketswhen adding does not affect the result.(d) 2 × (3 × 4) = (2 × 3) × 4; i.e., the use of bracketswhen multiplying does not affect the result.(e) 2 × (3 + 4) = 2(3 + 4) = 2 × 3 + 2 × 4; i.e., anumber placed outside of a bracket indicatesthat the whole contents of the bracket must bemultiplied by that number.(f) (2 + 3)(4 + 5) = (5)(9) = 5 × 9 = 45; i.e., adja-cent brackets indicate multiplication.(g) 2[3 + (4 × 5)] = 2[3 + 20] = 2 × 23 = 46; i.e.,when an expression contains inner and outerbrackets, the inner brackets are removedﬁrst.Here are some further problems in which BODMASneeds to be used.Problem 16. Find the value of 6 + 4 ÷ (5 − 3)The order of precedence of operations is rememberedby the word BODMAS. Thus,6 + 4 ÷ (5 − 3) = 6 + 4 ÷ 2 (Brackets)= 6 + 2 (Division)= 8 (Addition)Problem 17. Determine the value of13 − 2 × 3 + 14 ÷ (2 + 5)13 − 2 × 3 + 14÷ (2 + 5) = 13 − 2 × 3 + 14 ÷ 7 (B)= 13 − 2 × 3 + 2 (D)= 13 − 6 + 2 (M)= 15 − 6 (A)= 9 (S)Problem 18. Evaluate16 ÷(2 + 6) + 18[3 + (4 × 6) − 21]16 ÷ (2 + 6) + 18[3 + (4 × 6) − 21]= 16 ÷ (2 + 6) + 18[3 + 24 − 21] (B: inner bracketis determined ﬁrst)= 16 ÷ 8 + 18 × 6 (B)= 2 + 18 × 6 (D)= 2 + 108 (M)= 110 (A)Note that a number outside of a bracket multiplies allthat is inside the brackets. In this case,18[3 + 24 − 21] = 18[6], which means 18 × 6 = 108Problem 19. Find the value of23 − 4(2 × 7) +(144 ÷ 4)(14 − 8)23 − 4(2 × 7) +(144 ÷ 4)(14 − 8)= 23 − 4 × 14+366(B)= 23 − 4 × 14+ 6 (D)= 23 − 56 + 6 (M)= 29 − 56 (A)= −27 (S)Problem 20. Evaluate3 + 52 − 32 + 231 + (4 × 6) ÷ (3 × 4)+15 ÷ 3 + 2 × 7 − 13 ×√4+ 8 − 32 + 13 + 52 − 32 + 231 + (4 × 6) ÷ (3 × 4)+15 ÷ 3 + 2 × 7 − 13 ×√4 + 8 − 32 + 1=3 + 4 + 81 + 24 ÷ 12+15 ÷ 3 + 2 × 7 − 13 × 2 + 8 − 9 + 1=3 + 4 + 81 + 2+5 + 2 × 7 − 13 × 2 + 8 − 9 + 1=153+5 + 14 − 16 + 8 − 9 + 1= 5 +186= 5 + 3 = 8
20. 20. 8 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 4 Further problems onorder of precedenceand brackets (answerson page 340)Evaluate the following expressions.1. 14+ 3 × 152. 17 − 12 ÷ 43. 86 + 24 ÷ (14 − 2)4. 7(23 − 18) ÷ (12 − 5)5. 63 − 8(14 ÷ 2) + 266.405− 42 ÷ 6 + (3 × 7)7.(50 − 14)3+ 7(16 − 7) − 78.(7 − 3)(1 − 6)4(11 − 6) ÷ (3 − 8)9.(3 + 9 × 6) ÷ 3 − 2 ÷ 23 × 6 + (4 − 9) − 32 + 510.4 × 32 + 24 ÷ 5 + 9 × 32 × 32 − 15 ÷ 3+2 + 27 ÷ 3 + 12 ÷ 2 − 325 + (13 − 2 × 5) − 411.1 +√25 + 3 × 2 − 8 ÷ 23 × 4− 32 + 42 + 1−(4 × 2 + 7 × 2) ÷ 11√9+ 12 ÷ 2 − 23
21. 21. Chapter 2Fractions2.1 IntroductionA mark of 9 out of 14 in an examination may be writ-ten as914or 9/14.914is an example of a fraction. Thenumber above the line, i.e. 9, is called the numera-tor. The number below the line, i.e. 14, is called thedenominator.When the value of the numerator is less than thevalue of the denominator, the fraction is called aproper fraction.914is an example of a properfraction.When thevalueofthenumeratorisgreaterthan thevalueof the denominator, the fraction is called an improperfraction.52is an example of an improper fraction.A mixed number is a combination of a whole numberand a fraction. 212is an example of a mixed number. Infact,52= 212.There are a number of everyday examples in whichfractions are readily referred to. For example, threepeople equally sharing a bar of chocolate would have13each. A supermarket advertises15off a six-pack ofbeer; if the beer normally costs £2 then it will nowcost £1.60.34of the employees of a company arewomen; if the company has 48 employees, then 36 arewomen.Calculators are able to handle calculations with frac-tions. However, to understand a little more about frac-tions we will in this chapter show how to add, subtract,multiply and divide with fractions without the use of acalculator.Problem 1. Change the following improperfractions into mixed numbers:(a)92(b)134(c)285(a)92means 9 halves and92= 9 ÷ 2, and 9 ÷ 2 = 4and 1 half, i.e.92= 412(b)134means 13 quarters and134= 13 ÷ 4, and13 ÷ 4 = 3 and 1 quarter, i.e.134= 314(c)285means 28 ﬁfths and285= 28 ÷ 5, and 28 ÷ 5 =5 and 3 ﬁfths, i.e.285= 535Problem 2. Change the following mixed numbersinto improper fractions:(a) 534(b) 179(c) 237(a) 534means 5 +34. 5 contains 5 × 4 = 20 quarters.Thus, 534contains 20 + 3 = 23 quarters, i.e.534=234DOI: 10.1016/B978-1-85617-697-2.00002-8
22. 22. 10 Basic Engineering MathematicsThe quick way to change 534into an improperfraction is4 × 5 + 34=234.(b) 179=9 × 1 + 79=169.(c) 237=7 × 2 + 37=177.Problem 3. In a school there are 180 students ofwhich 72 are girls. Express this as a fraction in itssimplest formThe fraction of girls is72180.Dividing both the numerator and denominator by thelowest prime number, i.e. 2, gives72180=3690Dividing both the numerator and denominator again by2 gives72180=3690=18452 will not divide into both 18 and 45, so dividing both thenumerator and denominator by the next prime number,i.e. 3, gives72180=3690=1845=615Dividing both the numerator and denominator again by3 gives72180=3690=1845=615=25So72180=25in its simplest form.Thus,25of the students are girls.2.2 Adding and subtracting fractionsWhen the denominators of two (or more) fractions tobe added are the same, the fractions can be added ‘onsight’.For example,29+59=79and38+18=48.In the latter example, dividing both the 4 and the 8 by4 gives48=12, which is the simpliﬁed answer. This iscalled cancelling.Additionand subtraction of fractions is demonstratedin the following worked examples.Problem 4. Simplify13+12(i) Make the denominators the same for each frac-tion. The lowest number that both denominatorsdivideinto is called the lowest commonmultipleor LCM (see Chapter 1, page 5). In this example,the LCM of 3 and 2 is 6.(ii) 3 divides into 6 twice. Multiplying both numera-tor and denominator of13by 2 gives13=26=(iii) 2 dividesinto 6, 3 times. Multiplyingboth numer-ator and denominator of12by 3 gives12=36=(iv) Hence,13+12=26+36=56+ =Problem 5. Simplify34−716(i) Make the denominators the same for each frac-tion. The lowest common multiple (LCM) of 4and 16 is 16.(ii) 4 divides into 16, 4 times. Multiplying bothnumerator and denominator of34by 4 gives34=1216=(iii)716already has a denominator of 16.
23. 23. Fractions 11(iv) Hence,34−716=1216−716=516− =Problem 6. Simplify 423− 116423− 116is the same as 423− 116which is thesame as 4 +23− 1 +16which is the same as4 +23− 1 −16which is the same as 3 +23−16whichis the same as 3 +46−16= 3 +36= 3 +12Thus, 423− 116= 312Problem 7. Evaluate 718− 537718− 537= 7 +18− 5 +37= 7 +18− 5 −37= 2 +18−37= 2 +7 × 1 − 8 × 356= 2 +7 − 2456= 2 +−1756= 2 −1756=11256−1756=112 − 1756=9556= 13956Problem 8. Determine the value of458− 314+ 125458− 314+ 125= (4 − 3 + 1) +58−14+25= 2 +5 × 5 − 10× 1 + 8 × 240= 2 +25 − 10 + 1640= 2 +3140= 23140Now try the following Practice ExercisePracticeExercise 5 Introduction tofractions (answers on page 340)1. Change the improper fraction157into amixed number.2. Change the improper fraction375into amixed number.3. Change the mixed number 249into animproper fraction.4. Change the mixed number 878into animproper fraction.5. A box contains 165 paper clips. 60 clipsare removed from the box. Express this asa fraction in its simplest form.6. Order the following fractions from the small-est to the largest.49,58,37,12,357. A training college has 375 students of which120 are girls. Express this as a fraction in itssimplest form.Evaluate, in fraction form, the expressions given inProblems 8 to 20.8.13+259.56−41510.12+2511.716−1412.27+31113.29−17+2314. 325− 21315.727−23+5916. 5313+ 33417. 458− 32518. 1037− 82319. 314− 445+ 15620. 534− 125− 312
24. 24. 12 Basic Engineering Mathematics2.3 Multiplication and division offractions2.3.1 MultiplicationTo multiply two or more fractions together, the numer-ators are ﬁrst multiplied to give a single numberand this becomes the new numerator of the com-bined fraction. The denominators are then multipliedtogether to give the new denominator of the combinedfraction.For example,23×47=2 × 43 × 7=821Problem 9. Simplify 7 ×257 ×25=71×25=7 × 21 × 5=145= 245Problem 10. Find the value of37×1415Dividing numerator and denominator by 3 gives37×1415=17×145=1 × 147 × 5Dividing numerator and denominator by 7 gives1 × 147 × 5=1 × 21 × 5=25This process of dividingboth the numerator and denom-inator of a fraction by the same factor(s) is calledcancelling.Problem 11. Simplify35×4935×49=15×43by cancelling=415Problem 12. Evaluate 135× 213× 337Mixed numbers must be expressed as improper frac-tions before multiplication can be performed. Thus,135× 213× 337=55+35×63+13×217+37=85×73×247=8 × 1 × 85 × 1 × 1=645= 1245Problem 13. Simplify 315× 123× 234The mixed numbers need to be changed to improperfractions before multiplication can be performed.315× 123× 234=165×53×114=41×13×111by cancelling=4 × 1 × 111 × 3 × 1=443= 14232.3.2 DivisionThe simple rule for division is change the divisionsign into a multiplication sign and invert the secondfraction.For example,23÷34=23×43=89Problem 14. Simplify37÷82137÷821=37×218=31×38by cancelling=3 × 31 × 8=98= 118Problem 15. Find the value of 535÷ 713The mixed numbers must be expressed as improperfractions. Thus,535÷ 713=285÷223=285×322=145×311=4255Problem 16. Simplify 323× 134÷ 234
25. 25. Fractions 13Mixed numbersmust beexpressed as improperfractionsbefore multiplication and division can be performed:323× 134÷ 234=113×74÷114=113×74×411=1 × 7 × 13 × 1 × 1by cancelling=73= 213Now try the following Practice ExercisePracticeExercise 6 Multiplying anddividing fractions (answers on page 340)Evaluate the following.1.25×472. 5 ×493.34×8114.34×595.1735×15686.35×79× 1277.1317× 4711× 34398.14×311× 15399.29÷42710.38÷456411.38÷53212.34÷ 14513. 214× 12314. 113÷ 25915. 245÷71016. 234÷ 32317.19×34× 11318. 314× 135÷2519. A ship’s crew numbers 105, of which17arewomen. Of the men,16are ofﬁcers. Howmany male ofﬁcers are on board?20. If a storage tank is holding 450 litres whenit is three-quarters full, how much will itcontain when it is two-thirds full?21. Three people, P, Q and R, contribute to afund. P provides 3/5 of the total, Q pro-vides 2/3 of the remainder and R provides£8. Determine (a) the total of the fund and(b) the contributions of P and Q.22. A tank contains 24,000 litres of oil. Initially,710of the contents are removed, then35ofthe remainder is removed. How much oil isleft in the tank?2.4 Order of precedence withfractionsAs stated in Chapter 1, sometimes addition, subtraction,multiplication, division, powers and brackets can all beinvolved in a calculation. A deﬁnite order of precedencemust be adhered to. The order is:BracketsOrder (or pOwer)DivisionMultiplicationAdditionSubtractionThis is demonstrated in the followingworked problems.Problem 17. Simplify720−38×45720−38×45=720−3 × 12 × 5by cancelling (M)=720−310(M)=720−620=120(S)Problem 18. Simplify14− 215×58+91014− 215×58+910=14−115×58+910=14−111×18+910by cancelling=14−118+910(M)=1 × 104 × 10−11 × 58 × 5+9 × 410 × 4(since the LCM of 4, 8 and 10 is 40)
26. 26. 14 Basic Engineering Mathematics=1040−5540+3640=10 − 55 + 3640(A/S)= −940Problem 19. Simplify212−25+34÷58×23212−25+34÷58×23=52−2 × 45 × 4+3 × 54 × 5÷58×23(B)=52−820+1520÷58×23(B)=52−2320÷54×13by cancelling (B)=52−2320÷512(B)=52−2320×125(D)=52−235×35by cancelling=52−6925(M)=5 × 252 × 25−69 × 225 × 2(S)=12550−13850(S)= −1350Problem 20. Evaluate13of 512− 334+ 315÷45−1213of 512− 334+ 315÷45−12=13of 134+ 315÷45−12(B)=13×74+165÷45−12(O)(Note that the ‘of ’ is replaced with amultiplication sign.)=13×74+165×54−12(D)=13×74+41×11−12by cancelling=712+41−12(M)=712+4812−612(A/S)=4912= 4112Now try the following Practice ExercisePracticeExercise 7 Order of precedencewith fractions (answers on page 340)Evaluate the following.1. 212−35×20272.13−34×16273.12+35÷915−134.15+ 223÷59−145.45×12−16÷25+236.35−23−12÷56×327.12of 425− 3710+ 313÷23−258.623× 125−13634÷ 1129. 113× 215÷2510.14×25−15÷23+41511.23+ 315× 212+ 113813÷ 31312.113of 2910− 135+ 213÷23−34
27. 27. Revision Test 1 : Basic arithmetic and fractionsThis assignment covers the material contained in Chapters 1 and 2. The marks available are shown in brackets at theend of each question.1. Evaluate1009cm − 356cm − 742cm + 94cm. (3)2. Determine £284 × 9. (3)3. Evaluate(a) −11239 − (−4732) + 9639(b) −164 × −12(c) 367 × −19 (6)4. Calculate (a) \$153 ÷ 9 (b) 1397g÷ 11 (4)5. A small component has a mass of 27 grams.Calculate the mass, in kilograms, of 750 suchcomponents. (3)6. Find (a) the highest common factor and (b) thelowest common multiple of the following num-bers: 15 40 75 120. (7)Evaluate the expressions in questions 7 to 12.7. 7 + 20 ÷ (9 − 5) (3)8. 147 − 21(24 ÷ 3) + 31 (3)9. 40 ÷ (1 + 4) + 7[8 + (3 × 8) − 27] (5)10.(7 − 3)(2 − 5)3(9 − 5) ÷ (2 − 6)(3)11.(7 + 4 × 5) ÷ 3 + 6 ÷ 22 × 4 + (5 − 8) − 22 + 3(5)12.(42 × 5 − 8) ÷ 3 + 9 × 84 × 32 − 20 ÷ 5(5)13. Simplify(a)34−715(b) 158− 213+ 356(8)14. A training college has 480 students of which 150are girls. Express this as a fraction in its simplestform. (2)15. A tank contains 18000litresof oil. Initially,710ofthe contents are removed, then25of the remainderis removed. How much oil is left in the tank? (4)16. Evaluate(a) 179×38× 335(b) 623÷ 113(c) 113× 215÷25(10)17. Calculate(a)14×25−15÷23+415(b)23+ 315× 212+ 113813÷ 313(8)18. Simplify113of 2910− 135+ 213÷23−34(8)
28. 28. Chapter 3Decimals3.1 IntroductionThe decimal system of numbers is based on the digits0 to 9.Thereareanumberofeveryday occurrencesin which weuse decimal numbers. For example, a radio is, say, tunedto 107.5MHz FM; 107.5 is an example of a decimalnumber.In a shop, a pair of trainers cost, say, £57.95; 57.95 isanotherexampleofadecimal number.57.95 isadecimalfraction, where a decimal point separates the integer, i.e.57, from the fractional part, i.e. 0.9557.95 actually means (5 × 10)+ (7 × 1)+ 9 ×110+ 5 ×11003.2 Converting decimals to fractionsand vice-versaConverting decimals to fractions and vice-versa isdemonstrated below with worked examples.Problem 1. Convert 0.375 to a proper fraction inits simplest form(i) 0.375 may be written as0.375 × 10001000i.e.0.375 =3751000(ii) Dividing both numerator and denominator by 5gives3751000=75200(iii) Dividing both numerator and denominator by 5again gives75200=1540(iv) Dividing both numerator and denominator by 5again gives1540=38Since both 3 and 8 are only divisible by 1, we cannot‘cancel’ any further, so38is the ‘simplest form’ of thefraction.Hence, the decimal fraction 0.375 =38as a properfraction.Problem 2. Convert 3.4375 to a mixed number(i) 0.4375 may be written as0.4375 × 1000010000i.e.0.4375 =437510000(ii) Dividing both numerator and denominator by 25gives437510000=175400(iii) Dividing both numerator and denominator by 5gives175400=3580(iv) Dividing both numerator and denominator by 5again gives3580=716Since both 5 and 16 are only divisible by 1, wecannot ‘cancel’ any further, so716is the ‘lowestform’ of the fraction.(v) Hence, 0.4375 =716Thus, the decimal fraction 3.4375=3716as a mixednumber.DOI: 10.1016/B978-1-85617-697-2.00003-X
29. 29. Decimals 17Problem 3. Express78as a decimal fractionTo convert a proper fraction to a decimal fraction, thenumerator is divided by the denominator.0.8 7 58 7.0 0 0(i) 8 into 7 will not go. Place the 0 above the 7.(ii) Place the decimal point above the decimal pointof 7.000(iii) 8 into 70 goes 8, remainder 6. Place the 8 abovethe ﬁrst zero after the decimal point and carry the6 remainder to the next digit on the right, makingit 60.(iv) 8 into 60 goes 7, remainder 4. Place the 7 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 8 into 40 goes 5, remainder 0. Place 5 above thenext zero.Hence, the proper fraction78= 0.875 as a decimalfraction.Problem 4. Express 51316as a decimal fractionFor mixed numbers it is only necessary to convert theproper fraction part of the mixed number to a decimalfraction.0.8 1 2 516 13.0 0 0 0(i) 16 into 13 will not go. Place the 0 above the 3.(ii) Place the decimal point above the decimal pointof 13.0000(iii) 16 into 130 goes 8, remainder 2. Place the 8 abovethe ﬁrst zero after the decimal point and carry the2 remainder to the next digit on the right, makingit 20.(iv) 16 into 20 goes 1, remainder 4. Place the 1 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 16 into 40 goes 2, remainder 8. Place the 2 abovethe next zero and carry the 8 remainder to the nextdigit on the right, making it 80.(vi) 16 into 80 goes 5, remainder 0. Place the 5 abovethe next zero.(vii) Hence,1316= 0.8125Thus, the mixed number 51316= 5.8125 as a decimalfraction.Now try the following Practice ExercisePracticeExercise 8 Converting decimals tofractions and vice-versa (answers onpage 341)1. Convert 0.65 to a proper fraction.2. Convert 0.036 to a proper fraction.3. Convert 0.175 to a proper fraction.4. Convert 0.048 to a proper fraction.5. Convert the following to proper fractions.(a) 0.65 (b) 0.84 (c) 0.0125(d) 0.282 (e) 0.0246. Convert 4.525 to a mixed number.7. Convert 23.44 to a mixed number.8. Convert 10.015 to a mixed number.9. Convert 6.4375 to a mixed number.10. Convert the following to mixed numbers.(a) 1.82 (b) 4.275 (c) 14.125(d) 15.35 (e) 16.212511. Express58as a decimal fraction.12. Express 61116as a decimal fraction.13. Express732as a decimal fraction.14. Express 11316as a decimal fraction.15. Express932as a decimal fraction.3.3 Signiﬁcant ﬁgures and decimalplacesA number which can be expressed exactly as adecimal fraction is called a terminating decimal.
30. 30. 18 Basic Engineering MathematicsFor example,3316= 3.1825 is a terminating decimalA number which cannot be expressed exactly as a deci-mal fraction is called a non-terminating decimal. Forexample,157= 1.7142857... is a non-terminating decimalThe answer to a non-terminating decimal may beexpressed in two ways, depending on the accuracyrequired:(a) correct to a number of signiﬁcant ﬁgures, or(b) correct to a number of decimal places i.e. thenumber of ﬁgures after the decimal point.The last digit in the answer is unaltered if the next digiton the right is in the group of numbers 0, 1, 2, 3 or 4.For example,1.714285... = 1.714 correct to 4 signiﬁcant ﬁgures= 1.714 correct to 3 decimal placessince the next digit on the right in this example is 2.The last digit in the answer is increased by 1 if the nextdigit on the right is in the group of numbers 5, 6, 7, 8 or9. For example,1.7142857... = 1.7143 correct to 5 signiﬁcant ﬁgures= 1.7143 correct to 4 decimal placessince the next digit on the right in this example is 8.Problem 5. Express 15.36815 correct to(a) 2 decimal places, (b) 3 signiﬁcant ﬁgures,(c) 3 decimal places, (d) 6 signiﬁcant ﬁgures(a) 15.36815 = 15.37 correct to 2 decimal places.(b) 15.36815 = 15.4 correct to 3 signiﬁcant ﬁgures.(c) 15.36815 = 15.368 correct to 3 decimal places.(d) 15.36815 = 15.3682 correct to 6 signiﬁcantﬁgures.Problem 6. Express 0.004369 correct to(a) 4 decimal places, (b) 3 signiﬁcant ﬁgures(a) 0.004369 = 0.0044 correct to 4 decimal places.(b) 0.004369 = 0.00437 correct to 3 signiﬁcantﬁgures.Note that the zeros to the right of the decimal point donot count as signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 9 Signiﬁcant ﬁgures anddecimal places (answers on page 341)1. Express 14.1794 correct to 2 decimal places.2. Express 2.7846 correct to 4 signiﬁcant ﬁgures.3. Express 65.3792 correct to 2 decimal places.4. Express 43.2746 correct to 4 signiﬁcantﬁgures.5. Express 1.2973 correct to 3 decimal places.6. Express 0.0005279 correct to 3 signiﬁcantﬁgures.3.4 Adding and subtracting decimalnumbersWhen adding or subtracting decimal numbers, careneeds to be taken to ensure that the decimal pointsare beneath each other. This is demonstrated in thefollowing worked examples.Problem 7. Evaluate 46.8 + 3.06 + 2.4 + 0.09and give the answer correct to 3 signiﬁcant ﬁguresThe decimal points are placed under each other asshown. Each column is added, starting from theright.46.83.062.4+0.0952.3511 1(i) 6 + 9 = 15. Place 5 in the hundredths column.Carry 1 in the tenths column.(ii) 8 + 0 + 4 + 0 + 1 (carried) = 13. Place the 3 inthe tenths column. Carry the 1 into the unitscolumn.(iii) 6 + 3 + 2 + 0 + 1 (carried) = 12. Place the 2 intheunitscolumn.Carry the1 into thetenscolumn.
31. 31. Decimals 19(iv) 4 + 1(carried) = 5. Place the 5 in the hundredscolumn.Hence,46.8 + 3.06 + 2.4 + 0.09 = 52.35= 52.4,correct to 3signiﬁcant ﬁguresProblem 8. Evaluate 64.46 − 28.77 and give theanswer correct to 1 decimal placeAs with addition, the decimal points are placed undereach other as shown.64.46−28.7735.69(i) 6 − 7 is not possible; therefore ‘borrow’ 1 fromthe tenths column. This gives 16 − 7 = 9. Placethe 9 in the hundredths column.(ii) 3 − 7 is not possible; therefore ‘borrow’ 1 fromthe unitscolumn. This gives 13 − 7 = 6. Place the6 in the tenths column.(iii) 3 − 8 is not possible; therefore ‘borrow’ from thehundreds column. This gives 13 − 8 = 5. Placethe 5 in the units column.(iv) 5 − 2 = 3. Place the 3 in the hundreds column.Hence,64.46 − 28.77 = 35.69= 35.7 correct to 1 decimal placeProblem 9. Evaluate 312.64 − 59.826 − 79.66+38.5 and give the answer correct to 4 signiﬁcantﬁguresThe sum of the positive decimal fractions= 312.64 + 38.5 = 351.14.The sum of the negative decimal fractions= 59.826 + 79.66 = 139.486.Taking the sum of the negative decimal fractions fromthe sum of the positive decimal fractions gives351.140−139.486211.654Hence, 351.140 − 139.486 = 211.654 = 211.7, cor-rect to 4 signiﬁcant ﬁgures.Now try the following Practice ExercisePracticeExercise 10 Adding andsubtracting decimal numbers (answers onpage 341)Determine the following without using a calcula-tor.1. Evaluate 37.69 + 42.6, correct to 3 signiﬁ-cant ﬁgures.2. Evaluate 378.1 − 48.85, correct to 1 decimalplace.3. Evaluate 68.92 + 34.84 − 31.223, correct to4 signiﬁcant ﬁgures.4. Evaluate 67.841 − 249.55 + 56.883, correctto 2 decimal places.5. Evaluate 483.24 − 120.44 − 67.49, correctto 4 signiﬁcant ﬁgures.6. Evaluate 738.22−349.38−427.336+56.779,correct to 1 decimal place.7. Determine the dimension marked x in thelength of the shaft shown in Figure 3.1. Thedimensions are in millimetres.82.9227.41 8.32 34.67xFigure 3.13.5 Multiplying and dividing decimalnumbersWhen multiplying decimal fractions:(a) the numbers are multiplied as if they were integers,and(b) the position of the decimal point in the answer issuch that there are as many digits to the right ofit as the sum of the digits to the right of the dec-imal points of the two numbers being multipliedtogether.This is demonstrated in the followingworked examples.
32. 32. 20 Basic Engineering MathematicsProblem 10. Evaluate 37.6 × 5.4376×5415041880020304(i) 376 × 54 = 20304.(ii) As there are 1 + 1 = 2 digits to the right ofthe decimal points of the two numbers beingmultiplied together, 37.6 × 5.4, then37.6 × 5.4 = 203.04Problem 11. Evaluate 44.25 ÷ 1.2, correct to(a) 3 signiﬁcant ﬁgures, (b) 2 decimal places44.25 ÷ 1.2 =44.251.2The denominator is multiplied by 10 to change it into aninteger. The numerator is also multiplied by 10 to keepthe fraction the same. Thus,44.251.2=44.25 × 101.2 × 10=442.512The long division is similar to the long division ofintegers and the steps are as shown.36.87512 442.50036827210596908460600(i) 12 into 44 goes 3; place the 3 above the second4 of 442.500(ii) 3 × 12 = 36; place the 36 below the 44 of442.500(iii) 44 − 36 = 8.(iv) Bring down the 2 to give 82.(v) 12 into 82 goes 6; place the 6 above the 2 of442.500(vi) 6 × 12 = 72; place the 72 below the 82.(vii) 82 − 72 = 10.(viii) Bring down the 5 to give 105.(ix) 12 into 105 goes 8; place the 8 above the 5 of442.500(x) 8 × 12 = 96; place the 96 below the 105.(xi) 105 − 96 = 9.(xii) Bring down the 0 to give 90.(xiii) 12 into 90 goes 7; place the 7 above the ﬁrstzero of 442.500(xiv) 7 × 12 = 84; place the 84 below the 90.(xv) 90 − 84 = 6.(xvi) Bring down the 0 to give 60.(xvii) 12 into 60 gives 5 exactly; place the 5 abovethe second zero of 442.500(xviii) Hence, 44.25 ÷ 1.2 =442.512= 36.875So,(a) 44.25 ÷ 1.2 = 36.9, correct to 3 signiﬁcantﬁgures.(b) 44.25 ÷ 1.2 = 36.88, correct to 2 decimalplaces.Problem 12. Express 723as a decimal fraction,correct to 4 signiﬁcant ﬁguresDividing 2 by 3 gives23= 0.666666...and 723= 7.666666...Hence, 723= 7.667 correct to 4 signiﬁcant ﬁgures.Note that 7.6666... is called 7.6 recurring and iswritten as 7..6Now try the following Practice ExercisePracticeExercise 11 Multiplying anddividing decimal numbers (answers onpage 341)In Problems 1 to 8, evaluate without using acalculator.1. Evaluate 3.57 × 1.42. Evaluate 67.92 × 0.7
33. 33. Decimals 213. Evaluate 167.4 × 2.34. Evaluate 342.6 × 1.75. Evaluate 548.28 ÷ 1.26. Evaluate 478.3 ÷ 1.1, correct to 5 signiﬁcantﬁgures.7. Evaluate 563.48 ÷ 0.9, correct to 4 signiﬁ-cant ﬁgures.8. Evaluate 2387.4 ÷ 1.5In Problems 9 to 14, express as decimal fractionsto the accuracy stated.9.49, correct to 3 signiﬁcant ﬁgures.10.1727, correct to 5 decimal places.11. 1916, correct to 4 signiﬁcant ﬁgures.12. 53511, correct to 3 decimal places.13. 133137, correct to 2 decimal places.14. 8913, correct to 3 signiﬁcant ﬁgures.15. Evaluate 421.8 ÷ 17, (a) correct to 4 signif-icant ﬁgures and (b) correct to 3 decimalplaces.16. Evaluate0.01472.3, (a) correct to 5 decimalplaces and (b) correct to 2 signiﬁcant ﬁgures.17. Evaluate (a)12..61.5(b) 5..2 × 1218. A tank contains 1800 litres of oil. How manytins containing 0.75 litres can be ﬁlled fromthis tank?
34. 34. Chapter 4Using a calculator4.1 IntroductionIn engineering, calculations often need to be performed.For simple numbers it is useful to be able to use men-tal arithmetic. However, when numbers are larger anelectronic calculator needs to be used.There are several calculators on the market, many ofwhich will be satisfactory for our needs. It is essentialto have a scientiﬁc notationcalculator which will haveall the necessary functions needed and more.This chapter assumes you have a CASIO fx-83EScalculator, or similar, as shown in Figure 4.1.Besides straightforward addition, subtraction, multipli-cation and division, which you will already be ableto do, we will check that you can use squares, cubes,powers, reciprocals, roots, fractions and trigonomet-ric functions (the latter in preparation for Chapter 21).There are several other functions on the calculatorwhich we do not need to concern ourselves with at thislevel.4.2 Adding, subtracting, multiplyingand dividingInitially, after switching on, press Mode.Of the three possibilities, use Comp, which is achievedby pressing 1.Next, press Shift followed by Setup and, of the eightpossibilities, use Mth IO, which is achieved by press-ing 1.By all means experiment with the other menu options –refer to your ‘User’s guide’.All calculators have +, −, × and ÷ functions andthese functions will, no doubt, already have been usedin calculations.Problem 1. Evaluate 364.7 ÷ 57.5 correct to 3decimal places(i) Type in 364.7(ii) Press ÷.(iii) Type in 57.5(iv) Press = and the fraction3647575appears.(v) Press the S ⇔ D functionand the decimal answer6.34260869... appears.Alternatively, after step (iii) press Shift and = and thedecimal will appear.Hence, 364.7 ÷ 57.5 = 6.343 correct to 3 decimalplaces.Problem 2. Evaluate12.47 × 31.5970.45 × 0.052correct to 4signiﬁcant ﬁgures(i) Type in 12.47(ii) Press ×.(iii) Type in 31.59(iv) Press ÷.(v) The denominator must have brackets; i.e. press (.(vi) Type in 70.45 × 0.052 and complete the bracket;i.e. ).(vii) Press = and the answer 107.530518... appears.Hence,12.47 × 31.5970.45 × 0.052= 107.5 correct to 4 signiﬁcantﬁgures.DOI: 10.1016/B978-1-85617-697-2.00004-1
35. 35. Using a calculator 23Figure 4.1 A Casio fx-83ES calculatorNow try the following Practice ExercisePracticeExercise 12 Addition, subtraction,multiplication and division using a calculator(answers on page 341)1. Evaluate 378.37 − 298.651 + 45.64 − 94.5622. Evaluate 25.63 × 465.34 correct to 5 signif-icant ﬁgures.3. Evaluate 562.6 ÷ 41.3 correct to 2 decimalplaces.4. Evaluate17.35 × 34.2741.53 ÷ 3.76correct to 3 decimalplaces.5. Evaluate 27.48 + 13.72 × 4.15 correct to4 signiﬁcant ﬁgures.6. Evaluate(4.527 + 3.63)(452.51 ÷ 34.75)+ 0.468 correctto 5 signiﬁcant ﬁgures.7. Evaluate 52.34 −(912.5 ÷ 41.46)(24.6 − 13.652)correct to3 decimal places.8. Evaluate52.14 × 0.347 × 11.2319.73 ÷ 3.54correct to4 signiﬁcant ﬁgures.9. Evaluate451.224.57−363.846.79correct to 4 signiﬁ-cant ﬁgures.10. Evaluate45.6 − 7.35 × 3.614.672 − 3.125correct to 3decimal places.4.3 Further calculator functions4.3.1 Square and cube functionsLocate the x2 and x3 functions on your calculator andthen check the following worked examples.Problem 3. Evaluate 2.42(i) Type in 2.4(ii) Press x2 and 2.42 appears on the screen.(iii) Press = and the answer14425appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal 5.76Alternatively, after step (ii) press Shift and = .Thus, 2.42 = 5.76Problem 4. Evaluate 0.172 in engineering form(i) Type in 0.17(ii) Press x2 and 0.172 appears on the screen.
36. 36. 24 Basic Engineering Mathematics(iii) Press Shift and = and the answer 0.0289 appears.(iv) Press the ENG function and the answer changesto 28.9 × 10−3, which is engineering form.Hence, 0.172 = 28.9×10−3 in engineering form. TheENG function is extremely important in engineeringcalculations.Problem 5. Change 348620 into engineeringform(i) Type in 348620(ii) Press = then ENG.Hence, 348620 = 348.62×103 in engineering form.Problem 6. Change 0.0000538 into engineeringform(i) Type in 0.0000538(ii) Press = then ENG.Hence, 0.0000538 = 53.8×10−6 in engineering form.Problem 7. Evaluate 1.43(i) Type in 1.4(ii) Press x3 and 1.43 appears on the screen.(iii) Press = and the answer343125appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal: 2.744Thus, 1.43 = 2.744.Now try the following Practice ExercisePracticeExercise 13 Square and cubefunctions (answers on page 341)1. Evaluate 3.522. Evaluate 0.1923. Evaluate 6.852 correct to 3 decimal places.4. Evaluate (0.036)2 in engineering form.5. Evaluate 1.5632 correct to 5 signiﬁcantﬁgures.6. Evaluate 1.337. Evaluate 3.143 correct to 4 signiﬁcantﬁgures.8. Evaluate (0.38)3 correct to 4 decimal places.9. Evaluate (6.03)3correct to 2 decimal places.10. Evaluate (0.018)3 in engineering form.4.3.2 Reciprocal and power functionsThe reciprocal of 2 is12, the reciprocal of 9 is19and thereciprocal of x is1x, which from indices may be writtenas x−1. Locate the reciprocal, i.e. x−1 on the calculator.Also, locate the power function, i.e. x , on yourcalculator and then check the following workedexamples.Problem 8. Evaluate13.2(i) Type in 3.2(ii) Press x−1 and 3.2−1 appears on the screen.(iii) Press = and the answer516appears.(iv) Press the S ⇔ D function and the fractionchanges to a decimal: 0.3125Thus,13.2= 0.3125Problem 9. Evaluate 1.55 correct to 4 signiﬁcantﬁgures(i) Type in 1.5(ii) Press x and 1.5 appears on the screen.(iii) Press 5 and 1.55 appears on the screen.(iv) Press Shift and = and the answer 7.59375appears.Thus, 1.55 = 7.594 correct to 4 signiﬁcant ﬁgures.Problem 10. Evaluate 2.46 − 1.94 correct to 3decimal places(i) Type in 2.4(ii) Press x and 2.4 appears on the screen.
37. 37. Using a calculator 25(iii) Press 6 and 2.46appears on the screen.(iv) The cursor now needs to be moved; this isachieved by using the cursor key (the largeblue circular function in the top centre of thecalculator). Press →(v) Press −(vi) Type in 1.9, press x , then press 4.(vii) Press = and the answer 178.07087... appears.Thus, 2.46 − 1.94 = 178.071 correct to 3 decimalplaces.Now try the following Practice ExercisePracticeExercise 14 Reciprocal and powerfunctions (answers on page 341)1. Evaluate11.75correct to 3 decimal places.2. Evaluate10.02503. Evaluate17.43correct to 5 signiﬁcant ﬁgures.4. Evaluate10.00725correct to 1 decimal place.5. Evaluate10.065−12.341correct to 4 signiﬁ-cant ﬁgures.6. Evaluate 2.147. Evaluate (0.22)5correct to 5 signiﬁcantﬁgures in engineering form.8. Evaluate (1.012)7 correct to 4 decimalplaces.9. Evaluate (0.05)6 in engineering form.10. Evaluate 1.13 + 2.94 − 4.42 correct to 4 sig-niﬁcant ﬁgures.4.3.3 Root and ×10x functionsLocate the square root function√and the√function (which is a Shift function located abovethe x function) on your calculator. Also, locate the×10x function and then check the following workedexamples.Problem 11. Evaluate√361(i) Press the√function.(ii) Type in 361 and√361 appears on the screen.(iii) Press = and the answer 19 appears.Thus,√361 = 19.Problem 12. Evaluate 4√81(i) Press the√function.(ii) Type in 4 and 4√appears on the screen.(iii) Press → to move the cursor and then type in 81and 4√81 appears on the screen.(iv) Press = and the answer 3 appears.Thus,4√81 = 3.Problem 13. Evaluate 6 × 105 × 2 × 10−7(i) Type in 6(ii) Press the ×10x function (note, you do not haveto use ×).(iii) Type in 5(iv) Press ×(v) Type in 2(vi) Press the ×10x function.(vii) Type in −7(viii) Press = and the answer325appears.(ix) Press the S ⇔ D function and the fractionchanges to a decimal: 0.12Thus, 6 × 105 × 2 × 10−7 = 0.12Now try the following Practice ExercisePracticeExercise 15 Root and ×10xfunctions (answers on page 341)1. Evaluate√4.76 correct to 3 decimal places.2. Evaluate√123.7 correct to 5 signiﬁcantﬁgures.
38. 38. 26 Basic Engineering Mathematics3. Evaluate√34528correct to 2 decimal places.4. Evaluate√0.69 correct to 4 signiﬁcantﬁgures.5. Evaluate√0.025 correct to 4 decimal places.6. Evaluate 3√17 correct to 3 decimal places.7. Evaluate 4√773 correct to 4 signiﬁcantﬁgures.8. Evaluate 5√3.12 correct to 4 decimal places.9. Evaluate 3√0.028 correct to 5 signiﬁcantﬁgures.10. Evaluate6√2451−4√46 correct to 3 decimalplaces.Express the answers to questions 11 to 15 inengineering form.11. Evaluate 5 × 10−3 × 7 × 10812. Evaluate3 × 10−48 × 10−913. Evaluate6 × 103 × 14× 10−42 × 10614. Evaluate56.43 × 10−3 × 3 × 1048.349 × 103correct to3 decimal places.15. Evaluate99 × 105 × 6.7 × 10−336.2 × 10−4correct to 4signiﬁcant ﬁgures.4.3.4 FractionsLocate the and functions on your calculator(the latter function is a Shift function found abovethe function) and then check the following workedexamples.Problem 14. Evaluate14+23(i) Press the function.(ii) Type in 1(iii) Press ↓ on the cursor key and type in 4(iv)14appears on the screen.(v) Press → on the cursor key and type in +(vi) Press the function.(vii) Type in 2(viii) Press ↓ on the cursor key and type in 3(ix) Press → on the cursor key.(x) Press = and the answer1112appears.(xi) Press the S ⇔ D function and the fractionchanges to a decimal 0.9166666...Thus,14+23=1112= 0.9167 as a decimal, correct to4 decimal places.It is also possible to deal with mixed numbers on thecalculator. Press Shift then the function andappears.Problem 15. Evaluate 515− 334(i) Press Shift then the function and appearson the screen.(ii) Type in 5 then → on the cursor key.(iii) Type in 1 and ↓ on the cursor key.(iv) Type in 5 and 515appears on the screen.(v) Press → on the cursor key.(vi) Typein – and then pressShift then the functionand 515− appears on the screen.(vii) Type in 3 then → on the cursor key.(viii) Type in 3 and ↓ on the cursor key.(ix) Type in 4 and 515− 334appears on the screen.(x) Press = and the answer2920appears.(xi) Press S ⇔ D function and the fraction changes toa decimal 1.45Thus, 515− 334=2920= 1920= 1.45 as a decimal.