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## Basic Engg. Mathematics

Basic Engineering Mathematics

Basic arithmetic

1.1 Introduction

Whole numbers are called integers. +3,+5 and +72 are examples of positive integers; 13,6 and 51 are examples of negative integers. Between positive and negative integers is the number 0 which is neither positive nor negative. The four basic arithmetic operators are add (+), subtract (), multiply (×) and divide (÷). It is assumed that adding, subtracting, multiplying and dividing reasonably small numbers can be achieved without a calculator. However, if revision of this area is needed then some worked problems are included in the following sections. When unlike signs occur together in a calculation, the overall sign is negative. For example, 3+(4) = 3+4 = 34 = 1 and (+5)×(2) = 10 Like signs together give an overall positive sign.

For example, 3(4) = 3−−4 = 3+4 = 7 and (6)×(4) = +24

You can probably already add two or more numbers together and subtract one number from another. However, if you need a revision then the following worked problems should be helpful.

Problem 1. Determine 735+167

HTU

7 3 5

+ 1 6 7

9 0 2

1 1

(i) 5+7 = 12. Place 2 in units (U) column. Carry 1 in the tens (T) column.

(ii) 3+6+1 (carried) = 10. Place the 0 in the tens column. Carry the 1 in the hundreds (H) column.

(iii) 7+1+1 (carried) = 9. Place the 9 in the hundreds column.

Hence, 735+167 = 902

Problem 2. Determine 632369

HTU

6 3 2

3 6 9

2 6 3

(i) 29 is not possible; therefore ‘borrow’ 1 from the tens column (leaving 2 in the tens column). In the units column, this gives us 129 = 3.

(ii) Place 3 in the units column.

(iii) 26 is not possible; therefore ‘borrow’ 1 from the hundreds column (leaving 5 in the hundreds column). In the tens column, this gives us 126 = 6.

(iv) Place the 6 in the tens column. 2 Basic Engineering Mathematics

(v) 53 = 2.

(vi) Place the 2 in the hundreds column. Hence, 632369 = 263 Problem 3. Add 27,74, 81 and 19

This problem is written as 2774+8119. Adding the positive integers: 27 81 Sum of positive integers is 108 Adding the negative integers: 74

19 Sum of negative integers is 93 Taking the sum of the negative integers from the sum of the positive integers gives 108 −93 15

Thus, 2774+8119 = 15

Problem 4. Subtract 74 from 377

This problem is written as 377−−74. Like signs together give an overall positive sign, hence 377−−74 = 377+74 3 7 7

+ 7 4 4 5 1

Thus, 377−−74 = 451

Problem 5. Subtract 243 from 126 The problem is 126243.When the second number is larger than the first, take the smaller number from the larger and make the result negative. Thus,

126243=(243126) 2 4 3

1 2 6

1 1 7

Thus, 126243=117

Problem 6. Subtract 318 from 269

The problem is 269318. The sum of the negative integers is

2 6 9

+3 1 8

5 8 7

Thus, 269318=587 Now try the following Practice Exercise

Practice Exercise 1 Further problems on addition and subtraction (answers on page 340)

In Problems 1 to 15, determine the values of the expressions given, without using a calculator.

1. 67kg82 kg+34kg

2. 73m57m

3. 851mm372mm

4. 124273+481398

5. £927£114+£182£183£247

6. 647872

7. 2417487+242417784712

8. 384192177+2440799+2834

9. £2715£18250+£11471£1509+£113274

10. 47+(74)(23)

11. 813(674)

12. 3151(2763)

13. 4872 g4683g

14. 2314847724

15. 774441

16. Holes are drilled 35.7mm apart in a metal plate. If a row of 26 holes is drilled, determine the distance, in centimetres, between the centres of the first and last holes.

17. Calculate the diameter d and dimensions A and B for the template shown in Figure 1.1. All dimensions are in millimetres.3 Revision of multiplication and You can probably already multiply two numbers together and divide one number by another. However, if you need a revision then the following worked problems should be helpful. Problem 7. Determine 86×7

HTU

8 6

× 7

6 0 2

4

(i) 7×6 = 42. Place the 2 in the units (U) column and ‘carry’ the 4 into the tens (T) column.

(ii) 7×8 = 56;56+4 (carried) = 60. Place the 0 in the tens column and the 6 in the hundreds (H) column. Hence, 86×7 = 602

A good grasp of multiplication tables is needed when multiplying such numbers; a reminder of the multiplication table up to 12×12 is shown below. Confidence with handling numbers will be greatly improved if this table is memorized.

Problem 8. Determine 764×38

7 6 4

× 3 8

6 1 1 2

2 2 9 2 0

2 9 0 3 2

Multiplication table

× 2 3 4 5 6 7 8 9 10 11 12

2 4 6 8 10 12 14 16 18 20 22 24

3 6 9 12 15 18 21 24 27 30 33 36

4 8 12 16 20 24 28 32 36 40 44 48

5 10 15 20 25 30 35 40 45 50 55 60

6 12 18 24 30 36 42 48 54 60 66 72

7 14 21 28 35 42 49 56 63 70 77 84

8 16 24 32 40 48 56 64 72 80 88 96

9 18 27 36 45 54 63 72 81 90 99 108

10 20 30 40 50 60 70 80 90 100 110 120

11 22 33 44 55 66 77 88 99 110 121 132

12 24 36 48 60 72 84 96 108 120 132 144

4 Basic Engineering Mathematics

(i) 8×4 = 32. Place the 2 in the units column and carry 3 into the tens column.

(ii) 8×6 = 48;48+3 (carried) = 51. Place the 1 in the tens column and carry the 5 into the hundreds column.

(iii) 8×7 = 56;56+5 (carried) = 61. Place 1 in the hundreds column and 6 in the thousands column.

(iv) Place 0 in the units column under the 2.

(v) 3×4 = 12. Place the 2 in the tens column and carry 1 into the hundreds column.

(vi) 3×6 = 18;18+1 (carried) = 19. Place the 9 in the hundreds column and carry the 1 into the thousands column.

(vii) 3×7 = 21;21+1 (carried) = 22. Place 2 in the thousands column and 2 in the ten thousands column.

(viii) 6112+22920 = 29032 Hence, 764×38 = 29032

Again, knowing multiplication tables is rather important when multiplying such numbers. It is appreciated, of course, that such a multiplication can, and probably will, be performed using a calculator. However, there are times when a calculator may not be available and it is then useful to be able to calculate the ‘long way’.

Problem 9. Multiply 178 by 46 When the numbers have different signs, the result will be negative. (With this in mind, the problem can now be solved by multiplying 178 by 46). Following the procedure of Problem 8 gives

1 7 8

× 4 6

1 0 6 8

7 1 2 0

8 1 8 8

Thus, 178×46 = 8188 and 178×(46)=8188

Problem 10. Determine 1834÷7

262

7

_

1834

(i) 7 into 18 goes 2, remainder 4. Place the 2 above the 8 of 1834 and carry the 4 remainder to the next digit on the right, making it 43. (ii) 7 into 43 goes 6, remainder 1. Place the 6 above the 3 of 1834 and carry the 1 remainder to the next digit on the right, making it 14.

(iii) 7 into 14 goes 2, remainder 0. Place 2 above the 4 of 1834.

Hence, 1834÷7 = 1834/7 = 1834

7 = 262.

The method shown is called short division.

Problem 11. Determine 5796÷12

483

12

_

5796

48

99

96

36

36

00

(i) 12 into 5 won’t go. 12 into 57 goes 4; place 4 above the 7 of 5796.

(ii) 4×12 = 48; place the 48 below the 57 of 5796.

(iii) 5748 = 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) 9996 = 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) 3636 = 0. Hence, 5796 ÷ 12 = 5796/12 = 5796 12 = 483.

The method shown is called long division. Basic arithmetic 5 Now try the following Practice Exercise Practice Exercise 2 Further problems on multiplication and division (answers on page 340)

Determine the values of the expressions given in problems 1 to 9, without using a calculator.

1. (a) 78 × 6 (b) 124 × 7

2. (a) £261 × 7 (b) £462 × 9

3. (a) 783kg × 11 (b) 73kg × 8

4. (a) 27mm × 13 (b) 77mm × 12

5. (a) 448 × 23 (b) 143×(31)

6. (a) 288m ÷ 6 (b) 979m ÷ 11

7. (a) 1813 7(b)896 16

8. (a) 21424 13 (b) 15900 ÷ 15

9. (a) 88737 11 (b) 46858 ÷ 14

10. 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 multiples

When two or more numbers are multiplied together, the individual numbers are called factors. Thus, a factor is a number which divides into another number exactly. The highest common factor (HCF) is the largest number which 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 the numbers that divide into 12). The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbers that divide into 15). 1 and 3 are the only common factors; i.e., numbers which are factors of both 12 and 15. Hence, the HCF of 12 and 15 is 3 since 3 is the highest number which divides into both 12 and 15. A multiple is a number which contains another number an exact number of times. The smallest number which is exactly divisible by each of two or more numbers is called 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 and 15) and there are no lower common multiples. Hence, the LCM of 12 and 15 is 60 since 60 is the lowest number that both 12 and 15 divide into. Here are some further problems involving the determination of HCFs and LCMs.

Problem 12. Determine the HCF of the numbers 12, 30 and 42 Probably the simplest way of determining an HCF is to express each number in terms of its lowest factors. This is achieved by repeatedly dividing by the prime numbers 2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,

12 = 2 × 2 × 3

30 = 2 × 3 × 5

42 = 2 × 3 × 7

The factors which are common to each of the numbers are 2 in column 1 and 3 in column 3, shown by the broken lines. Hence, the HCF is 2 × 3; i.e., 6. That is, 6 is the largest number which will divide into 12, 30 and 42.

Problem 13. Determine the HCF of the numbers 30, 105, 210 and 1155 Using the method shown in Problem 12:

30 = 2 × 3 × 5

105 = 3 × 5 × 7

210 = 2 × 3 × 5 × 7

1155 = 3 × 5 × 7 × 11

The factors which are common to each of the numbers are 3 in column 2 and 5 in column 3. Hence, the HCF is 3 × 5 = 15.

Problem 14. Determine the LCM of the numbers 12, 42 and 90 6 Basic Engineering Mathematics The LCM is obtained by finding the lowest factors of each of the numbers, as shown in Problems 12 and 13 above, and then selecting the largest group of any of the factors present. Thus, 12 = 2×2 × 3 42 = 2 × 3 × 7 90 = 2 × 3×3 × 5 The largest group of any of the factors present is shown by the broken lines and are 2×2 in 12, 3×3 in 90, 5 in 90 and 7 in 42.

Hence, the LCM is 2 ×2×3×3×5×7 = 1260 and is the smallest number which 12, 42 and 90 will all divide into exactly. Problem 15. Determine the LCM of the numbers 150, 210, 735 and 1365 Using the method shown in Problem 14 above:

150 = 2 × 3 × 5×5

210 = 2 × 3 × 5 × 7

735 = 3 × 5 × 7×7

1365 = 3 × 5 × 7 × 13

Hence, the LCM is 2 × 3 × 5 × 5 × 7 × 7 × 13 = 95550.

Now try the following Practice Exercise Practice Exercise 3 Further problems on highest common factors and lowest common

multiples (answers on page 340) Find (a) the HCF and (b) the LCMof the following groups of numbers.

1. 8, 12 2. 60, 72

3. 50, 70 4. 270, 900

5. 6, 10, 14 6. 12, 30, 45

7. 10, 15, 70, 105 8. 90, 105, 300

9. 196, 210, 462, 910 10. 196, 350, 770

1.5 Order of precedence and brackets

1.5.1 Order of precedence

Sometimes addition, subtraction, multiplication, division, powers and brackets may all be involved in a calculation. For example, 53×4+24÷(3+5)32 This is an extreme example but will demonstrate the order that is necessary when evaluating. When we read, we read from left to right. However, with mathematics there is a definite order of precedence which we need to adhere to. The order is as follows: Brackets Order (or pOwer)

Division Multiplication Addition Subtraction Notice that the first letters of each word spell BODMAS, a handy aide-m´emoire. Order means pOwer. For example, 42 = 4×4 = 16.

5 3 × 4 + 24 ÷ (3 + 5) 32 is evaluated as follows: 53×4+24÷(3+5)32

= 53×4+24÷832 (Bracket is removed and 3+5 replaced with 8) = 53×4+24÷89 (Order means pOwer; in

this case, 32 = 3×3 = 9) = 53×4+39 (Division: 24÷8 = 3)

= 512+39 (Multiplication: 3×4=12)

= 8129 (Addition: 5+3 = 8)

=13 (Subtraction: 8129=13)

In practice, it does not matter if multiplication is performed before division or if subtraction is performed before addition. What is important is that the process of multiplication and division must be completed before addition and subtraction.

1.5.2 Brackets and operators

The basic laws governing the use of brackets and operators are shown by the following examples. Basic arithmetic 7

(a) 2+3 = 3+2; i.e., the order of numbers when adding does not matter.

(b) 2×3 = 3×2; i.e., the order of numbers when multiplying does not matter.

(c) 2+(3+4) = (2+3)+4; i.e., the use of brackets when adding does not affect the result.

(d) 2×(3×4) = (2×3)×4; i.e., the use of brackets when multiplying does not affect the result.

(e) 2×(3+4) = 2(3+4) = 2×3+2×4; i.e., a number placed outside of a bracket indicates that the whole contents of the bracket must be multiplied by that number.

(f ) (2+3)(4 +5) = (5)(9) = 5×9 = 45; i.e., adjacent brackets indicate multiplication.

(g) 2[3+(4×5)] = 2[3+20] = 2×23 = 46; i.e., when an expression contains inner and outer brackets, the inner brackets are removed first.

Here are some further problems in which BODMAS needs to be used. Problem 16. Find the value of 6+4÷(53)

The order of precedence of operations is remembered by the word BODMAS. Thus,

6+4÷(53) = 6+4÷2 (Brackets)

= 6+2 (Division)

Problem 17. Determine the value of

132×3+14÷(2+5)

132×3+14÷(2+5) = 132×3+14÷7 (B)

= 132×3+2 (D)

= 136+2 (M)

= 156 (A)

= 9 (S)

Problem 18. Evaluate

16 ÷(2 +6)+18[3+(4×6)21]

16÷(2 +6)+18[3+(4 ×6)21]

= 16÷(2 +6)+18[3+2421] (B: inner bracket

is 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 all that is inside the brackets. In this case, 18[3+2421] = 18, which means 18×6 = 108

Problem 19. Find the value of 234(2×7)+ (144÷4) (148) 234(2×7)+ (144÷4) (148)

= 234×14+ 36 6

(B) = 234×14+6 (D)

= 2356+6 (M)

= 2956 (A)

= 27 (S)

Problem 20. Evaluate 3+ __ 52 32_ +23 1+(4×6)÷(3×4) + 15÷3+2×71 3× √ 4+832 +1 3+ __ 52 32_ +23 1+(4×6)÷(3×4) + 15÷3+2×71 3×  √4+832 +1 = 3+4+8 1+24÷12 + 15÷3+2×71 3×2+89+1 = 3+4+8 1+2 + 5+2×71 3×2+89+1 = 153 + 5+141 6+89+1 = 5+ 18 6 = 5+3 = 8

8 Basic Engineering Mathematics

Now try the following Practice Exercise Practice Exercise 4 Further problems on order of precedence and brackets (answers

on page 340)

Evaluate the following expressions.

1. 14+3×15

2. 1712÷4

3. 86+24÷(14 2)

4. 7(2318)÷(125)

5. 638(14÷2)+26

6. 40 5 −42÷6+(3×7)

7. (5014) 3 +7(167)7

8. (73)(16) 4(116)÷(38)

9. (3+9×6)÷32÷2 3×6+(49)32 +5

10. _4×32 +24_÷5+9×32×32 15÷3+2+27÷3+12÷2325+(132×5)4

11.1+√25+3×28÷23×4−__32 +42_+1−(4×2+7×2)÷11√9+12÷2 23apter 2Fractions

2.1 Introduction

A mark of 9 out of 14 in an examination may be written as 9 14 or 9/14. 9  14 is an example of a fraction. The number above the line, i.e. 9, is called the numerator. The number below the line, i.e. 14, is called the denominator. When the value of the numerator is less than the value of the denominator, the fraction is called a proper fraction. 9  14 is an example of a proper fraction. When the value of the numerator is greater than the value of the denominator, the fraction is called an improper

fraction.

5 2 is an example of an improper fraction. A mixed number is a combination of a whole number and a fraction. 2 1 2

is an example of a mixed number. In fact, 5     = 2  12.

There are a number of everyday examples in which fractions are readily referred to. For example, three people equally sharing a bar of chocolate would have 1 3 each. A supermarket advertises 1 5 off a six-pack of beer; if the beer normally costs £2 then it will now cost £1.60. 3 4 of the employees of a company are women; if the company has 48 employees, then 36 are women. Calculators are able to handle calculations with  fractions. However, to understand a little more about fractions we will in this chapter show how to add, subtract, multiply and divide with fractions without the use of a calculator.

Now try the following Practice Exercise

Practice Exercise 5 Introduction to fractions (answers on page 340)

1. Change the improper fraction 15 7 into a mixed number

2. Change the improper fraction 37 5 into a mixed number.

3. Change the mixed number 2 49 into an improper fraction.

4. Change the mixed number 8 7 8 into an improper fraction.

5. A box contains 165 paper clips.

6. clips are removed from the box. Express this as a fraction in its simplest form.

7. A training college has 375 students of which 120 are girls. Express this as a fraction in its simplest form. Evaluate, in fraction form, the expressions given in Problems 8 to 20.   12 Basic Engineering Mathematics

2.3 Multiplication and division of fractions

2.3.1 Multiplication

To multiply two or more fractions together, the numerators are first multiplied to give a single number and this becomes the new numerator of the combined fraction. The denominators are then multiplied together to give the new denominator of the combined fraction. For example,

This process of dividingboth the numerator and denominator of a fraction by the same factor(s) is called cancelling. Mixed numbers must be expressed as improper fractions before multiplication can be performed.

2.3.2 Division

The simple rule for division is change the division sign into a multiplication sign and invert the second fraction

Fractions 13 Mixed numbers must be expressed as improper fractions before multiplication and division can be performed.

A ship’s crew numbers 105, of which17 are women. Of the men,16 are officers. How many male officers are on board? 20. 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? 21. Three people, P, Q and R, contribute to a fund. 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 and (b) the contributions of P and

Q. 22. A tank contains 24,000 litres of oil. Initially,710 of the contents are removed, then 35 of the remainder is removed. How much oil is left in the tank?4 Order of precedence with fractions As stated in Chapter 1, sometimes addition, subtraction, multiplication, division, powers and brackets can all be involved in a calculation. A definite order of precedence must be adhered to. The order is:

Revision Test 1 : Basic arithmetic and fractions

This assignment covers the material contained in Chapters 1 and 2. The marks available are shown in brackets at the end of each question.

1. Evaluate 1009cm356cm742cm+94cm. (3)

2. Determine £284×9. (3)

3. Evaluate

(a) 11239(4732)+9639

(b) 164×−12

(c) 367×−19 (6)

4. Calculate (a) 3÷9 (b) 1397g÷11 (4)

5. A small component has a mass of 27 grams.

Calculate the mass, in kilograms, of 750 such components. (3)

6. Find (a) the highest common factor and (b) the lowest common multiple of the following numbers:

15 40 75 120. (7) Evaluate the expressions in questions 7 to 12.

7. 7+20÷(95) (3)

8. 14721(24÷3)+31 (3)

9. 40÷(1+4)+7[8+(3×8)27] (5)

10.(73)(2 5) 3(95)÷(2 6) (3)

11. (7+4×5)÷3+6÷2 2×4+(58)22 +3 (5)

12.(42 ×58)÷3+9×8 4×32 20÷5 (5)

14. A training college has 480 students of which 150 are girls. Express this as a fraction in its simplest form. (2)

15. A tank contains 18000litres of oil. Initially,710 of the contents are removed, then 25 of the remainder is removed. How much oil is left in the tank? (4) Chapter 3 Decimals

3.1 Introduction

The decimal system of numbers is based on the digits 0 to 9.

There are a number of everyday occurrences in which we use decimal numbers. For example, a radio is, say, tuned to 107.5MHz FM; 107.5 is an example of a decimal number. In a shop, a pair of trainers cost, say, £57.95; 57.95 is another example of a decimal number. 57.95 is a decimal fraction, where a decimal point separates the integer, i.e. 57, from the fractional part, i.e. 0.95

.2 Converting decimals to fractions and vice-versa

Converting decimals to fractions and vice-versa is demonstrated below with worked examples.

Problem 1. Convert 0.375 to a proper fraction in its simplest form

(i) 0.375 may be written as 0.375×1000 1000

i.e. 0.375 = 375 1000

(ii) Dividing both numerator and denominator by 5 gives 375 1000 = 75200

(iii) Dividing both numerator and denominator by 5 again gives 75 200 = 15 40

(iv) Dividing both numerator and denominator by 5 again gives 15 40 = 3 8 Since both 3 and 8 are only divisible by 1, we cannot ‘cancel’ any further, so 3 8 is the ‘simplest form’ of the fraction. Hence, the decimal fraction 0.375 = 3 8 as a proper fraction.

Problem 2. Convert 3.4375 to a mixed number

(i) 0.4375 may be written as 0.4375×10000  10000  i.e.  0.4375 = 4375  10000

(ii) Dividing both numerator and denominator by 25 gives 4375  10000 =  175  400

(iii) Dividing both numerator and denominator by 5 gives 175 400 = 35  80

(iv) Dividing both numerator and denominator by 5 again gives 35 80 = 7 16 Since both 5 and 16 are only divisible by 1, we cannot ‘cancel’ any further, so 7 16 is the ‘lowest form’ of the fraction.

(v) Hence, 0.4375 = 7 16 Thus, the decimal fraction 3.4375=3 7 16 as a mixed number. DOI: 10.1016/B978-1-85617-697-2.00003-X

Decimals 17

Problem 3. Express 7 8 as a decimal fraction To convert a proper fraction to a decimal fraction, the numerator is divided by the denominator.

0.8 7 5 8 _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 point of 7.000

(iii) 8 into 70 goes 8, remainder 6. Place the 8 above the first zero after the decimal point and carry the 6 remainder to the next digit on the right,making it 60. (iv) 8 into 60 goes 7, remainder 4. Place the 7 above the next zero and carry the 4 remainder to the next digit on the right, making it 40.

(v) 8 into 40 goes 5, remainder 0. Place 5 above the next zero. Hence, the proper fraction 78 = 0.875 as a decimal fraction. Problem 4. Express 5 13 16 as a decimal fraction For mixed numbers it is only necessary to convert the proper fraction part of the mixed number to a decimal fraction. 0.8 1 2 5 16 _ 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 point of 13.0000

(iii) 16 into 130 goes 8, remainder 2. Place the 8 above the first zero after the decimal point and carry the 2 remainder to the next digit on the right,making it 20.

(iv) 16 into 20 goes 1, remainder 4. Place the 1 above the next zero and carry the 4 remainder to the next digit on the right, making it 40.

(v) 16 into 40 goes 2, remainder 8. Place the 2 above the next zero and carry the 8 remainder to the next digit on the right, making it 80.

(vi) 16 into 80 goes 5, remainder 0. Place the 5 above the next zero.

(vii) Hence, 13 16 = 0.8125 Thus, the mixed number 5 13 16 = 5.8125 as a decimal fraction. Now try the following Practice Exercise

Practice Exercise 8 Converting decimals to fractions and vice-versa (answers on page 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) 0.024

6. 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) 16.2125

11. Express 58 as a decimal fraction.  places A number which can be expressed exactly as a decimal fraction is called a terminating decimal.

18 Basic Engineering Mathematics For example, 3 3 16 = 3.1825 is a terminating decimal A number which cannot be expressed exactly as a decimal fraction is called a non-terminating decimal. For example, 1 5 7 = 1.7142857. . . is a non-terminating decimal The answer to a non-terminating decimal may be expressed in two ways, depending on the accuracy required:

(a) correct to a number of significant figures, or

(b) correct to a number of decimal places i.e. the number of figures after the decimal point.

The last digit in the answer is unaltered if the next digit on the right is in the group of numbers 0, 1, 2, 3 or 4.

For example, 1.714285. . .= 1.714 correct to 4 significant figures = 1.714 correct to 3 decimal places since the next digit on the right in this example is 2. The last digit in the answer is increased by 1 if the next digit on the right is in the group of numbers 5, 6, 7, 8 or

9. For example, 1.7142857. . .= 1.7143 correct to 5 significant figures = 1.7143 correct to 4 decimal places since 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 significant figures.

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 significant figures.

Note that the zeros to the right of the decimal point do not count as significant figures.

Now try the following Practice Exercise

Practice Exercise 9 Significant figures and

decimal 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 significant figures.

5. Express 1.2973 correct to 3 decimal places.

6. Express 0.0005279 correct to 3 significant figures.

When adding or subtracting decimal numbers, care needs to be taken to ensure that the decimal points are beneath each other. This is demonstrated in the following worked examples.

Problem 7. Evaluate 46.8+3.06+2.4+0.09 and give the answer correct to 3 significant figures The decimal points are placed under each other as shown. Each column is added, starting from the right.

(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 in the tenths column. Carry the 1 into the units column.

(iii) 6+3+2+0+1 (carried) = 12. Place the 2 in the units column.Carry the 1 into the tens column. Decimals 19

(iv) 4+1(carried) = 5. Place the 5 in the hundreds column. Hence, 46.8+3.06+2.4 +0.09 = 52.35 = 52.4, correct to 3 significant figures

Problem 8. Evaluate 64.4628.77 and give the answer correct to 1 decimal place As with addition, the decimal points are placed under each other as shown. 64.46 −28.77 35.69

(i) 67 is not possible; therefore ‘borrow’ 1 from the tenths column. This gives 167 = 9. Place the 9 in the hundredths column.

(ii) 37 is not possible; therefore ‘borrow’ 1 from the units column. This gives 137 = 6. Place the 6 in the tenths column.

(iii) 38 is not possible; therefore ‘borrow’ from the hundreds column. This gives 138 = 5. Place the 5 in the units column. (iv) 52 = 3. Place the 3 in the hundreds column. Hence, 64.4628.77 = 35.69 = 35.7 correct to 1 decimal place

Problem 9. Evaluate 312.6459.82679.66+ 38.5 and give the answer correct to 4 significant Figures The 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 from the sum of the positive decimal fractions gives 351.140 −139.486 211.654

Hence, 351.140139.486 = 211.654 = 211.7, correct to 4 significant figures. Now try the following Practice Exercise

Practice Exercise 10 Adding and subtracting decimal numbers (answers on page 341)

Determine the following without using a calculator.

1. Evaluate 37.69+42.6, correct to 3 significant figures.

2. Evaluate 378.148.85, correct to 1 decimal place.

3. Evaluate 68.92+34.8431.223, correct to 4 significant figures.

4. Evaluate 67.841249.55+56.883, correct to 2 decimal places.

5. Evaluate 483.24120.4467.49, correct to 4 significant figures.

6. Evaluate 738.22349.38427.336+56.779, correct to 1 decimal place.

7. Determine the dimension marked x in the length of the shaft shown in Figure 3.1. The dimensions are in millimetres. 82.92 27.41 8.32 x 34.67

Figure 3.1

3.5 Multiplying and dividing decimal numbers

When multiplying decimal fractions:

(a) the numbers are multiplied as if they were integers, and

(b) the position of the decimal point in the answer is such that there are as 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 multiplied together. This is demonstrated in the followingworked examples. 20 Basic Engineering Mathematics

Problem 10. Evaluate 37.6×5.4 376 ×54 1504 18800 20304

(i) 376×54= 20304.

(ii) As there are 1+1 = 2 digits to the right of the decimal points of the two numbers being multiplied together, 37.6×5.4, then 37.6×5.4 = 203.04

Problem 11. Evaluate 44.25÷1.2, correct to (a) 3 significant figures, (b) 2 decimal places 44.25÷1.2 = 44.25 1.2 The denominator is multiplied by 10 to change it into an integer. The numerator is also multiplied by 10 to keep the fraction the same. Thus, 44.25 1.2 = 44.25×10 1.2×10 = 442.5 12

The long division is similar to the long division of integers and the steps are as shown. 36.875 12 _ 442.50036

(i) 12 into 44 goes 3; place the 3 above the second 4 of 442.500

(ii) 3×12 = 36; place the 36 below the 44 of 442.500

(iii) 4436 = 8.

(iv) Bring down the 2 to give 82.

(v) 12 into 82 goes 6; place the 6 above the 2 of

442.500

(vi) 6×12 = 72; place the 72 below the 82.

(vii) 8272 = 10.

(viii) Bring down the 5 to give 105.

(ix) 12 into 105 goes 8; place the 8 above the 5 of 442.500

(x) 8×12 = 96; place the 96 below the 105.

(xi) 10596 = 9.

(xii) Bring down the 0 to give 90.

(xiii) 12 into 90 goes 7; place the 7 above the first zero of 442.500

(xiv) 7×12 = 84; place the 84 below the 90.

(xv) 9084 = 6.

(xvi) Bring down the 0 to give 60.

(xvii) 12 into 60 gives 5 exactly; place the 5 above the second zero of 442.500

(xviii) Hence, 44.25÷1.2 = 442.5 12 = 36.875 So,

(a) 44.25÷1.2 = 36.9, correct to 3 significant figures.

(b) 44.25÷1.2 = 36.88, correct to 2 decimal places. Now try the following Practice Exercise Practice Exercise 11 Multiplying and dividing decimal numbers In Problems 1 to 8, evaluate without using a calculator.

1. Evaluate 3.57×1.4

2. Evaluate 67.92×0.7 Decimals 21

3. Evaluate 167.4×2.3

4. Evaluate 342.6×1.7

5. Evaluate 548.28÷1.2

6. Evaluate 478.3÷1.1, correct to 5 significant figures.

7. Evaluate 563.48÷0.9, correct to 4 significant figures.

8. Evaluate 2387.4÷1.5 In Problems 9 to 14, express as decimal fractions to the accuracy stated.

9. 49, correct to 3 significant figures.

10. 17 27 , correct to 5 decimal places.

4.1 Introduction

In engineering, calculations often need to be performed. For simple numbers it is useful to be able to use mental arithmetic. However, when numbers are larger an electronic calculator needs to be used. There are several calculators on the market, many of which will be satisfactory for our needs. It is essential to have a scientific notation calculator which will have all 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, multiplication and division, which you will already be able to do, we will check that you can use squares, cubes, powers, reciprocals, roots, fractions and trigonometric functions (the latter in preparation for Chapter 21). There are several other functions on the calculator which we do not need to concern ourselves with at this level.

4.2 Adding, subtracting, multiplying and dividing

Initially, after switching on, press Mode. Of the three possibilities, use Comp, which is achieved by pressing 1. Next, press Shift followed by Setup and, of the eight possibilities, use Mth IO, which is achieved by pressing 1.     By all means experiment with the other menu options – refer to your ‘User’s guide’. All calculators have +, , × and ÷ functions and these functions will, no doubt, already have been used in calculations.

Problem 1. Evaluate 364.7 ÷ 57.5 correct to 3 decimal places

(i) Type in 364.7

(ii) Press ÷.

(iii) Type in 57.5

(iv) Press = and the fraction 3647 575 appears.

(v) Press the S D function and the decimal answer 6.34260869. . . appears. Alternatively, after step (iii) press Shift and = and the decimal will appear. Hence, 364.7÷57.5 = 6.343 correct to 3 decimal places.

Problem 2. Evaluate 12.47×31.59 70.45×0.052 correct to 4 significant 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, 12.47×31.59 70.45×0.052 = 107.5 correct to 4 significant figures.

DOI: 10.1016/B978-1-85617-697-2.00004-1 Using a calculator 23 Figure 4.1 A Casio fx-83ES calculator

Now try the following Practice Exercise Practice Exercise 12 Addition, subtraction, multiplication and division using a calculator

1. Evaluate 378.37298.651+45.6494.562

2. Evaluate 25.63 × 465.34 correct to 5 significant figures.

3. Evaluate 562.6 ÷ 41.3 correct to 2 decimal places.

4. Evaluate 17.35×34.27 41.53÷3.76 correct to 3 decimal places.

5. Evaluate 27.48+13.72×4.15 correct to 4 significant figures.

6. Evaluate (4.527+3.63) (452.51÷34.75) +0.468 correct to 5 significant figures.

7. Evaluate 52.34(912.5÷41.46) (24.613.652) correct to 3 decimal places.

8. Evaluate 52.14×0.347×11.23 19.73÷3.54 correct to 4 significant figures.

9. Evaluate 451.2 24.57 − 363.8 46.79 correct to 4 significant figures.

10. Evaluate 45.67.35×3.61 4.6723.125 correct to 3 decimal places.

4.3 Further calculator functions

4.3.1 Square and cube functions

Locate the x2 and x3 functions on your calculator and then 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 answer 144 25 appears.

(iv) Press the S D function and the fraction changes to a decimal 5.76 Alternatively, after step (ii) press Shift and = .Thus, 2.42 = 5.76

Problem 4. Evaluate 0.172 in engineering form

(i) Type in 0.17

(ii) Press x2 and 0.172 appears on the screen. 24 Basic Engineering Mathematics

(iii) Press Shift and = and the answer 0.0289 appears.

(iv) Press the ENG function and the answer changes to 28.9×103, which is engineering form.

Hence, 0.172 = 28.9×103 in engineering form. he ENG function is extremely important in engineering calculations.

Problem 5. Change 348620 into engineering form

(i) Type in 348620

(ii) Press = then ENG.

Hence, 348620 = 348.62×103 in engineering form.

Problem 6. Change 0.0000538 into engineering form

(i) Type in 0.0000538

(ii) Press = then ENG.

Hence, 0.0000538 = 53.8×106 in engineering form.

The AWL is the most recent and widely referred word list for teaching and learning academic vocabulary. The AWL was developed by Averil Coxhead from the Victoria University of Wellington,