Notesale: Turn your study into money

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---

Chapter 1

Basic arithmetic
1
...
+3, +5 and +72
are examples of positive integers; −13, −6 and −51
are examples of negative integers
...

The four basic arithmetic operators are add (+), subtract
(−), multiply (×) and divide (÷)
...
However, if revision of this area
is needed then some worked problems are included in
the following sections
...
For example,
3 + (−4) = 3 + −4 = 3 − 4 = −1

Problem 1
...
Place 2 in units (U) column
...

(ii) 3 + 6 + 1 (carried) = 10
...
Carry the 1 in the hundreds (H) column
...
Place the 9 in the hundreds column
...
For
example,
3 − (−4) = 3 − −4 = 3 + 4 = 7

Problem 2
...
2 Revision of addition and
subtraction
You can probably already add two or more numbers
together and subtract one number from another
...

DOI: 10
...
00001-6

(i) 2 − 9 is not possible; therefore ‘borrow’ 1 from
the tens column (leaving 2 in the tens column)
...

(ii) Place 3 in the units column
...
In the tens column, this gives us
12 − 6 = 6
...


2 Basic Engineering Mathematics
(v) 5 − 3 = 2
...
The sum of the negative
integers is

(vi) Place the 2 in the hundreds column
...


269
+318

Add 27, −74, 81 and −19

This problem is written as 27 − 74 + 81 − 19
...


Subtract −74 from 377

This problem is written as 377 − −74
...


73 m − 57 m

3
...


124 − 273 + 481 − 398

5
...


647 − 872
2417 − 487 + 2424 − 1778 − 4712

8
...


£2715 − £18250 + £11471 − £1509 +
£113274

10
...
When the second number is
larger than the first, take the smaller number from the
larger and make the result negative
...
813 − (−674)
12
...
4872 g − 4683 g
14
...
$53774 − $38441

243
−126
117

Thus, 126 − 243 = −117
Problem 6
...


7
...


377
+ 74

Thus, 377 − −74 = 451
Problem 5
...
Holes are drilled 35
...
If a row of 26 holes is drilled, determine the distance, in centimetres, between
the centres of the first and last holes
...
Calculate the diameter d and dimensions A
and B for the template shown in Figure 1
...

All dimensions are in millimetres
...
Determine 86 × 7

110

HTU
86
×
7

B

12

6 0 2
4

d

(i) 7 × 6 = 42
...


A

(ii) 7 × 8 = 56; 56 + 4 (carried) = 60
...


60

50

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
...


38
120

Figure 1
...
Determine 764 × 38

1
...
However, if
you need a revision then the following worked problems
should be helpful
...
Place the 2 in the units column and
carry 3 into the tens column
...
Place the 1 in
the tens column and carry the 5 into the hundreds
column
...
Place 1 in the
hundreds column and 6 in the thousands column
...

(v) 3 × 4 = 12
...

(vi) 3 × 6 = 18; 18 + 1 (carried) = 19
...

(vii) 3 × 7 = 21; 21 + 1 (carried) = 22
...


 262
7 1834
(i) 7 into 18 goes 2, remainder 4
...

(ii) 7 into 43 goes 6, remainder 1
...

(iii) 7 into 14 goes 2, remainder 0
...

1834
Hence, 1834 ÷ 7 = 1834/7 =
= 262
...


Problem 11
...

It is appreciated, of course, that such a multiplication
can, and probably will, be performed using a calculator
...


Problem 9
...
(With this in mind, the problem can now
be solved by multiplying 178 by 46)
...


Determine 1834 ÷ 7

 483
12 5796
48
99
96
36
36
00
(i) 12 into 5 won’t go
...

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

(iv) Bring down the 9 of 5796 to give 99
...

(vi) 8 × 12 = 96; place 96 below the 99
...

(viii) Bring down the 6 of 5796 to give 36
...

(x) Place the 3 above the final 6
...

(xii) 36 − 36 = 0
...

12
The method shown is called long division
...

1
...


(a) £261 × 7

(b) £462 × 9

3
...


(a) 27 mm × 13

(b) 77 mm × 12

5
...


(a) 288 m ÷ 6

(b) 979 m ÷ 11

7
...


(a)

21424
13

(b) 15900 ÷ 15

9
...
A screw has a mass of 15 grams
...


1
...
Thus, a factor is a
number which divides into another number exactly
...

For example, consider the numbers 12 and 15
...
e
...

The factors of 15 are 1, 3, 5 and 15 (i
...
all the numbers
that divide into 15)
...
e
...

Hence, the HCF of 12 and 15 is 3 since 3 is the highest
number which divides into both 12 and 15
...
The smallest number which

is exactly divisible by each of two or more numbers is
called the lowest common multiple (LCM)
...
and the multiples of 15 are 15, 30, 45,
60, 75,
...
e
...

Hence, the LCM of 12 and 15 is 60 since 60 is the
lowest number that both 12 and 15 divide into
...

Problem 12
...
This
is achieved by repeatedly dividing by the prime numbers
2, 3, 5, 7, 11, 13, … (where possible) in turn
...
Hence, the HCF is 2 × 3; i
...
, 6
...

Problem 13
...
Hence, the HCF
is 3 × 5 = 15
...
Determine the LCM of the numbers
12, 42 and 90

5

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
...
Determine the LCM of the numbers
150, 210, 735 and 1365
Using the method shown in Problem 14 above:

× 7

3 ×5

× 7×7

1365 =

3 ×5

× 7

1
...
1 Order of precedence
Sometimes addition, subtraction, multiplication, division, powers and brackets may all be involved in a
calculation
...

When we read, we read from left to right
...
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
...
For
example, 42 = 4 × 4 = 16
...

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
...
5

× 13

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

1
...
60, 72

3
...
270, 900

5
...
12, 30, 45

7
...
90, 105, 300

9
...
196, 350, 770

= 8 − 12 − 9
= −13

(Division: 24 ÷ 8 = 3)
(Multiplication: − 3 × 4 = −12)
(Addition: 5 + 3 = 8)
(Subtraction: 8 − 12 − 9 = −13)

In practice, it does not matter if multiplication is performed before division or if subtraction is performed
before addition
...


1
...
2 Brackets and operators
The basic laws governing the use of brackets and
operators are shown by the following examples
...
e
...


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


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


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


16 ÷ (2 + 6) + 18[3 + (4 × 6) − 21]
= 16 ÷ (2 + 6) + 18[3 + 24 − 21] (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
...
Find the value of
(144 ÷ 4)
23 − 4(2 × 7) +
(14 − 8)
23 − 4(2 × 7) +

Here are some further problems in which BODMAS
needs to be used
...
Find the value of 6 + 4 ÷ (5 − 3)
The order of precedence of operations is remembered
by the word BODMAS
...
Determine the value of
13 − 2 × 3 + 14 ÷ (2 + 5)

13 − 2 × 3 + 14 ÷ (2 + 5) = 13 − 2 × 3 + 14 ÷ 7
= 13 − 2 × 3 + 2

(D)

= 13 − 6 + 2

(M)

= 15 − 6

(A)

=9

(S)

Problem 18
...
Evaluate


3 + 52 − 32 + 23
15 ÷ 3 + 2 × 7 − 1
√
+
1 + (4 × 6) ÷ (3 × 4) 3 × 4 + 8 − 32 + 1

3+

+

15 ÷ 3 + 2 × 7 − 1
√
3 × 4 + 8 − 32 + 1

=

3+4+8
15 ÷ 3 + 2 × 7 − 1
+
1 + 24 ÷ 12
3×2+8−9+1

=

3+4+8
5+2×7−1
+
1+2
3×2+8−9+1

=

5 + 14 − 1
15
+
3
6+8−9+1

= 5+

18
6

= 5+3 = 8

(B)

7

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
...


14 + 3 × 15

2
...


86 + 24 ÷ (14 − 2)

4
...


63 − 8(14 ÷ 2) + 26

6
...


40
− 42 ÷ 6 + (3 × 7)
5
(50 − 14)
+ 7(16 − 7) − 7
3

8
...


10
...


√
25 + 3 × 2 − 8 ÷ 2

−

3 × 4 − 32 + 42 + 1

1+

(4 × 2 + 7 × 2) ÷ 11
√
9 + 12 ÷ 2 − 23

List of formulae
Laws of indices:

Areas of plane figures:

a m × a n = a m+n
a m/n =

√
n m
a

am
an

= a m−n (a m )n = a mn

a −n =

Quadratic formula:
If ax 2 + bx + c = 0

1
an

Area = l × b

(i) Rectangle

a0 = 1
b

√
−b ± b2 − 4ac
x=
2a

then

Equation of a straight line:

l

(ii) Parallelogram Area = b × h

y = mx + c

Definition of a logarithm:
If y = a x

then

h

x = loga y

Laws of logarithms:
log(A × B) = log A + log B
 
A
= log A − log B
log
B

b

(iii) Trapezium

log An = n × log A

1
Area = (a + b)h
2
a

Exponential series:
ex = 1 + x +

x2 x3
+
+···
2! 3!

h

(valid for all values of x)
b

Theorem of Pythagoras:
b 2 = a 2 + c2

(iv) Triangle

Area =

1
×b×h
2

A

c

B

b
h

a

C
b

List of formulae
Area = πr 2 Circumference = 2πr

(v) Circle

(iii) Pyramid
If area of base = A and
perpendicular height = h then:

r

s

␪

Volume =

1
× A×h
3

r

2π radians = 360 degrees

Radian measure:

h

For a sector of circle:
θ◦
(2πr) = rθ
360

(θ in rad)

1
θ◦
(πr 2 ) = r 2 θ
360
2

(θ in rad)

s=

arc length,

shaded area =

Equation of a circle, centre at origin, radius r:
x 2 + y2 = r 2
Equation of a circle, centre at (a, b), radius r:

Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
1
Volume = πr2 h
3
Curved Surface area = πrl

(x − a)2 + (y − b)2 = r 2

Total Surface area = πrl + πr2

Volumes and surface areas of regular
solids:
l

(i) Rectangular prism (or cuboid)

h

Volume = l × b × h
Surface area = 2(bh + hl + lb)

r

(v) Sphere

l

h

4
Volume = πr3
3
Surface area = 4πr2

b

(ii) Cylinder
Volume = πr2 h
Total surface area = 2πrh + 2πr2
r

r

h

337

338 Basic Engineering Mathematics
Areas of irregular figures by approximate
methods:
Trapezoidal rule


 
width of 1 first + last
Area ≈
interval
2 ordinate

+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule



1 width of
first + last
Area ≈
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

For a general sinusoidal function y = A sin (ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
= frequency, f hertz
2π
2π
= periodic time T seconds
ω
α = angle of lead or lag (compared with
y = A sin ωt )

Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r, θ) then

y
r = x 2 + y 2 and θ = tan−1
x
If co-ordinate (r, θ) = (x, y) then
x = r cosθ and y = r sin θ

Mean or average value of a waveform:
area under curve
length of base
sum of mid-ordinates
=
number of mid-ordinates

mean value, y =

Triangle formulae:
Sine rule:
Cosine rule:

b
c
a
=
=
sin A sin B
sin C
a 2 = b2 + c2 − 2bc cos A

B

b

a

If a = first term and d = common difference, then the
arithmetic progression is: a, a + d, a + 2d,
...

The n’th term is: arn−1
Sum of n terms, Sn =

A
c

Arithmetic progression:

a (1 − r n )
a (r n − 1)
or
(1 − r )
(r − 1)

If − 1 < r < 1, S∞ =

a
(1 − r )

C

Area of any triangle
1
= × base × perpendicular height
2
1
1
1
= ab sin C or
ac sin B or
bc sin A
2
2
2

a +b+c
= [s (s − a) (s − b) (s − c)] where s =
2

Statistics:
Discrete data:

#
mean, x¯ =

x

n
)
(
 #

(x − x¯ )2

standard deviation, σ =
n

List of formulae
Grouped data:

#
fx
mean, x¯ = #
f

)
 #$
%(

f (x − x)
¯ 2

#
standard deviation, σ =
f

Standard integrals
y
axn
cos ax

Standard derivatives
sin ax

y or f(x)

dy
= or f (x)
dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1
x

eax
1
x

&

a

y dx
x n+1
+ c (except when n = −1)
n +1

1
sin ax + c
a
1
− cos ax + c
a
1 ax
e +c
a
ln x + c

339

Answers

Answers to practice exercises
Chapter 1

Chapter 2

Exercise 1 (page 2)
1
...

7
...

13
...

17
...
16 m
3
...
£565
6
...
−36 121
9
...
1487
12
...
−70872
15
...
25 cm
d = 64 mm, A = 136 mm, B = 10 mm

1
...

4
...

7
...


5
...

13
...


Exercise 2 (page 5)
(a) 468 (b) 868
2
...

(a) 259 (b) 56
8
...


2
...

6
...

10
...


(a) 12 (b) 360
(a) 90 (b) 2700
(a) 3 (b) 180
(a) 15 (b) 6300
(a) 14 (b) 53 900

Exercise 4 (page 8)
2
...
68

3
...
5

DOI: 10
...
00040-5

3
...

10
1
14
...
1
21
6
...


16
...


2
6
8
2
...

35
9
11
3
1
7
...

5
13
5
3
2
12
...
3
2
5
12
4
3
1
17
...
13
4
9
(a) £60, P£36, Q£16

7
...

15
...


22
9
8
25
3
16
4
27
17
60

4
...

12
...

20
...
5
9
...
−33
10
...
1
2
12
14
...


71
8
11
15
3
16
51
8
52
17
20

19
...

15
5
...
4
20
...
2880 litres

Exercise 7 (page 14)
1
18
7
6
...
2
1
...
22
11
...

(a) 4 (b) 24
(a) 10 (b) 350
(a) 2 (b) 210
(a) 5 (b) 210
(a) 14 (b) 420 420

2
...

3
...

7
...


1
7
4
11
17
30
43
77
9
1
40

1
...
4

1
9
19
7
...
2
20
2
...
1
8
...
7
3

4
...

15
5
...
0
...
14
...
1
...


13
20

7
40
21
1
141
(b)
(c)
(d)
(e)
25
80
500
11
3
7
...
10
25
200
11
1
7
(b) 4
(c) 14
(d) 15
40
8
20
2
...
4
40
41
10
...
(a)

11
...
625

9
250

3
...


6
125

9
...

4
...

10
...


2
...
11
...
185
...
8307
5
...
1581
6
...
571
5
...
1
...
0
...
068
11
...
5 ×10
12
...
5 ×103
−6
−3
4
...
202
...
18
...
6
...
0
...
11
...
0
...
14
...
1
...
2
...
65
...
0
...
329
...
18
...
43
...
72
...
12
...
−124
...
4
...
0
...
−

9
10

4
...
732
8
...
0
...
0
...
0
...
−0
...
0
...
0
...
5
...
2
...
0
...
0
...
0
...
998 2
...
544
3
...
02 4
...
42
456
...
434
...
626
...
1591
...
444 10
...
62963
11
...
563 12
...
455
13
...
8
...
(a) 24
...
812
(a) 0
...
0064 17
...
4˙ (b) 62
...
4
...
0
...
3
...
13
...
50
...
53
...
36
...
12
...
0
...
46
...
1
...
2
...
2
...
30
...
0
...
219
...
5
...
5
...
52
...
0
...
25
...
591
...
69
...
17
...

4
...

10
...
11927
6
...
0944
10
...
325

2
...
30
...
84
...
10
...
2
...


Exercise 10 (page 19)

1
...

9
...

16
...


2
...
0
...
137
...
19
...
515
...
15
...
52
...
0
...
80
...
295
...
59 cm2
159 m/s
0
...

5
...

11
...
78 mm
0
...
8 m2
281
...

6
...

12
...
5
5
...
5
2
...

4
...

10
...


£589
...
508
...
V = 2
...
5
5
...
81 A 6
...
79 s
E = 3
...
I = 12
...
s = 17
...
184 cm2 11
...
327
(a) 12
...
p
...

(c) 13
...
15 h

342 Basic Engineering Mathematics
Exercise 26 (page 43)

Chapter 5

1
...
£66 3
...
450 g 5
...
56 kg
6
...
00025 (b) 48 MPa 7
...
76 litre

Exercise 21 (page 34)
1
...
32%
2
...
4% 3
...
7% 4
...
4%
5
...
5%
6
...
20
7
...
0125 8
...
6875
9
...
462% 10
...
2% (b) 79
...
(b), (d), (c), (a) 12
...

14
...
A = , B = 50%, C = 0
...
30,
2
17
3
F = , G = 0
...
85, J =
10
20

Exercise 27 (page 45)
1
...
170 fr
3
...
8 mm
4
...
(a) 159
...
5 gallons
6
...
4 MPa
7
...
2 mm 8
...

3
...

7
...

14
...
8 kg 2
...
72 m
(a) 496
...
657 g
(a) 14%
(b) 15
...
49% 11
...
2%
2
...
5
...
73 s 4
...
36% 6
...
76 g
9
...
17% 13
...
3
...
25%
37
...
7%

1
...
5 weeks
2
...
(a) 9
...
12 (c) 0
...
50 minutes
5
...
375 m2 (c) 24 × 103 Pa

Chapter 7
Exercise 29 (page 48)

Exercise 23 (page 38)
1
...

9
...

14
...

16
...
5%
2
...
£310
4
...
£20 000 7
...
45 8
...
25
£39
...
£917
...
£185 000 12
...
2%
A 0
...
9 kg, C 0
...
3 t
20 000 kg (or 20 tonnes)
13
...
5 mm 17
...
27
6
...
128
7
...
100 000
8
...
24
10
...
96
9
...
1 6
...
16 4
...
01 10
...
76
7
...
1000 9
...
36 13
...
34 15
...
25
1
1
1
17
...
49 19
...
5
20
...
128

2
...
36 : 1

2
...
5 : 1 or 7 : 2 3
...
96 cm, 240 cm 5
...
£3680, £1840, £920 7
...
£2172

Exercise 31 (page 52)
1
...
9

Exercise 25 (page 42)
1
...
76 ml
3
...
12
...
14
...
25 000 kg

147
148
17
13
...


2
...
±3
10
...
64

19
56

32
25
1
7
...


11
...
4

1
2

4
...
±
13
1
12
...

3
...

7
...

11
...

15
...

19
...

23
...

27
...

4
...

8
...

12
...

16
...

20
...

24
...

28
...

y
20
...
(x − y)(a + b)
1
...

5
...

9
...

13
...


2x(y − 4z)
2x(1 + 2y)
4x(1 + 2x)
x(1 + 3x + 5x 2 )
r(s + p + t )


2 p q 2 2 p2 − 5q
2x y(y + 3x + 4x 2 )
7y(4 + y + 2x)
2r
18
...

t
21
...
(a − 2b)(2x + 3y)

2
...

6
...

10
...

14
...


1
...
2

6
...
2

11
...
2

4
...
6
1
8
...
6
18
...
−10
17
...
0
14
...
12y 2 − 3y
4
...
1

7
...
6a 2 + 5a −
11
...
9x 2 +

1
...
−2

3
...
15

7
...
5

11
...
2

16
...
6

21
...
−3

1
4
1
3

10
...
10a 2 − 3a + 2

15
...
5

1
2

1
3

4
...
12

9
...
13

13
...
−11

15
...
3

19
...
10

23
...
±4

Exercise 44 (page 79)
1
...

5
...


10−7
2
...
3
...
8  (b) 30 
digital camera battery £9, camcorder battery £14
800 
7
...
12 cm, 240 cm2
4
...
30 kg

2
...
004
5
...
12 m, 8 m

3
...
£312, £240
9
...
5 N

Chapter 12
Exercise 46 (page 84)
1
...
3

Exercise 43 (page 76)

Exercise 41 (page 72)
1
...
4b − 15b2
3
5
...
2

c
p
V
5
...
r =
2π
3
...
a =
t
1
6
...
x =
m
2
...
R =
I
5
13
...
T =

10
...
C =

ω ωL −

12
...
f =

13
...
λ =

R2


5

345

+ , 63
...
r =
or 1 −
S
S
2
...
f =

AL
3F − AL
or f = F −
3
3

4
...
t =

R − R0
R0 α

6
...
b =
9
...
R =

t 2g
4π 2


360 A
12
...
080
14
...
L =

11
...
a =
m −n
3(x + y)
3
...
b = √
1 − a2
a( p2 − q 2 )
7
...
t2 = t1 +
11
...
725



√
v 2 − 2as

M
+ r4
π
mrCR
4
...
r =
x+y



2
...
965
10
...
03L
8
...
5
p = 2, q = −1
x = 3, y = 2
a = 5, b = 2
s = 2, t = 3
m = 2
...
5
x = 2, y = 5

2
...

6
...

10
...

14
...


x = 3, y = 4
x = 4, y = 1
x = 1, y = 2
a = 2, b = 3
x = 1, y = 1
x = 3, y = −2
a = 6, b = −1
c = 2, d = −3

Exercise 50 (page 94)

13
...

3
...

7
...

11
...

15
...

3
...

7
...

4
...

8
...
30, b = 0
...
x = , y =
2
4
1
1
3
...
c = 3, d = 4
3
7
...
a = , b = −
3
2
4
...
r = 3, s =
2
8
...

3
...

6
...


a = 0
...
5, c = 3
α = 0
...
56 
a = 4, b = 10

2
...
47, I2 = 4
...
£15 500, £12 800
7
...
40
9
...
5, F2 = −4
...
x = 2, y = 1, z = 3
3
...
x = 2, y = −2, z = 2
4
...

7
...

10
...


x = 2, y = 4, z = 5 6
...
x = −4, y = 3, z = 2
x = 1
...
5, z = 4
...

4
...

8
...


1
...
0
...
905 A
0
...
38 m
1
...
835 m or 18
...

5
...

9
...
84 cm
0
...
78 cm
7m

Chapter 14
Exercise 58 (page 110)

Exercise 54 (page 104)
1
...
−1
...
5
7
...

13
...


4
−2 or −3
4 or −3
2

19
...
4 or −8
4
5
...
−5
11
...
2 or 7
17
...
−1
...
x = 1, y = 3 and x = −3, y = 7

6
...
x = , y = − and −1 , y = −4
5
5
3
3

9
...

15
...


23
...
5

1
4
25
...

5

1
1
26
...

or −
3
2

1
2 or −1
−4
3 or −3

1
8
1
24
...

8
5
30
...


22
...
x 2 − 4x + 3 = 0
33
...
x 2 − 36 = 0

3
...
3
7
...
−2
9
...

2
3
1
11
...
10 000 13
...
9 15
...
0
...

18
...
4

or −2
3
2

32
...
4x 2 − 8x − 5 = 0
36
...
7x − 1
...
4

3
...
−3

5
...
−3
...
268
3
...
468 or −1
...
2
...
307

1
8

3
...
−3
...
637
4
...
290 or 0
...
−2
...
351

1
...
log 12

2
...
log 500

3
...
log 100

4
...
log 6

9
...
log 1 = 0 11
...
log 243 or log 35 or 5 log3
13
...
log 64 or log26 or 6 log2

Exercise 56 (page 107)
1
...

5
...

9
...

13
...
637 or −3
...
781 or 0
...
608 or −1
...
851 or −2
...
481 or −1
...
167
4
...
438

2
...

6
...

10
...


0
...
792
0
...
693
1
...
232
2
...
086
4
...
676
7
...
641

15
...
5
16
...
5
19
...
x = 2

17
...
5 18
...
a = 6 22
...
1
...
3
...
0
...
6
...
2
...
3
...
2
...
−0
...
316
...

2
...

4
...


(a)
0
...
0988
(a) 4
...
04106
2
...


Exercise 67 (page 134)

0
...
064037
2
...
07482
120
...
446
8
...
08286

Exercise 63 (page 120)
1
...
0601

2
...
389 (b) 0
...

x − 2x 4
3
1
4
...
(a) Horizontal axis: 1 cm = 4 V (or 1 cm = 5 V),
vertical axis: 1 cm = 10 
(b) Horizontal axis: 1 cm = 5 m, vertical axis:
1 cm = 0
...
2 mm
2
...
5 (d) 5
3
...
5
4
...
1 (b) −1
...
The 1010 rev/min reading should be 1070 rev/min;
(a) 1000 rev/min (b) 167 V

1 − 2x 2 −

Exercise 64 (page 122)
1
...
95, 2
...
(a) 28 cm3 (b) 116 min

2
...
65, −1
...
(a) 70◦C (b) 5 minutes

Exercise 65 (page 124)
1
...

3
...

11
...

17
...
55547 (b) 0
...
8941
(a) 2
...
33154 (c) 0
...
4904 4
...
5822 5
...
197
6
...
2
0
...
11
...
1
...
1
...
962 12
...
4
147
...
4
...
3
...
e
...
500 19
...
Missing values: −0
...
25, 0
...
25, 2
...
(a) 4, −2 (b) −1, 0 (c) −3, −4 (d) 0, 4
1
1
1 1
(b) 3, −2 (c) ,
2
2
24 2
4
...
(a) 2,

1
2
2
1
2
5
...
(a) (b) −4 (c) −1
5
6
7
...
(2, 1)

9
...
5, 6)

10
...
(a) 89 cm (b) 11 N (c) 2
...
4 W + 48
12
...
15 W + 3
...
a = −20, b = 412

Exercise 69 (page 144)
1
...
(a) 850 rev/min (b) 77
...
(a) 150◦ C (b) 100
...
(a) 0
...
25L + 12
2
...
21 kPa

3
...
32 volts (b) 71
...
(a) 1
...
293 m 5
...
45 s
6
...
37 N
7
...
04 A (b) 1
...
2
...
£2424

347

9
...
07 A (b) 0
...
5 N (e) 592 N (f) 212 N
4
...
003, 8
...
(a) 22
...
43 s (c) v = 0
...
5
6
...
9L − 0
...
(a) 1
...
89% (c) F = −0
...
21
8
...
00022 (c) 28
...
a = 0
...
3 kPa, 275
...
(a)

Chapter 18

9
...
6, 13
...
6, 0
...
6 or 0
...
x = −1
...
5 (a) −30 (b) 2
...
50
(c) 2
...
8

Exercise 74 (page 161)

Exercise 70 (page 149)
1
...
(a) y (b)

√

x (c) b (d) a

y
1
(c) f (d) e 4
...
(a) (b) 2 (c) a (d) b
x
x
6
...
5, b = 0
...
78 mm2 7
...
15

1
...
5, y = −5
...
(a) x = −1
...
5 (b) x = −1
...
24
(c) x = −1
...
0

3
...
(a) 950 (b) 317 kN
9
...
4, b = 8
...
4 (ii) 11
...
x = −2
...
5 or 1
...
x = −2, 1 or 3, Minimum at (2
...
1),
Maximum at (−0
...
2)
3
...
x = −2
...
4 or 2
...
x = 0
...
5
6
...
3, 1
...
8
7
...
5

Exercise 71 (page 154)
1
...

3
...

5
...

7
...


(a) lg y (b) x (c) lg a (d) lg b
(a) lg y (b) lg x (c) L (d) lg k
(a) ln y (b) x (c) n (d) ln m
I = 0
...
75 candelas
a = 3
...
5
a = 5
...
6, 38
...
0
R0 = 26
...
42
8
...
08e0
...
4 N, μ = 0
...
0 N, 1
...
5, y = 1
...
3, y = −1
...
4, b = 1
...
82◦ 27
2
...
51◦11 4
...
15 44 17 6
...
72
...
27
...
37◦ 57
10
...
reflex 2
...
acute 4
...
(a) 21◦ (b) 62◦ 23 (c) 48◦56 17

Chapter 19

1
...

5
...


Chapter 20

2
...
x = −1, y = 2
6
...
(a) Minimum (0, 0) (b) Minimum (0, −1)
(c) Maximum (0, 3) (d) Maximum (0, −1)
2
...
4 or 0
...
−3
...
9
4
...
1 or 4
...
−1
...
2
6
...
5 or −2, Minimum at (−1
...
1)
7
...
7 or 1
...
(a) ±1
...
3

6
...
(a) 60◦ (b) 110◦ (c) 75◦ (d) 143◦ (e) 140◦
(f ) 20◦ (g) 129
...
Transversal (a) 1 & 3, 2 & 4, 5 & 7, 6 & 8
(b) 1 & 2, 2 & 3, 3 & 4, 4 & 1, 5 & 6, 6 & 7,
7 & 8, 8 & 5, 3 & 8, 1 & 6, 4 & 7 or 2 & 5
(c) 1 & 5, 2 & 6, 4 & 8, 3 & 7 (d) 3 & 5 or 2 & 8
9
...
a = 69◦ , b = 21◦ , c = 82◦ 11
...
1
...
0
...
40◦55

Exercise 78 (page 173)
1
...
a = 40◦ , b = 82◦, c = 66◦,
d = 75◦, e = 30◦ , f = 75◦
3
...
52◦
5
...
5◦
6
...
40◦, 70◦, 70◦, 125◦, isosceles
8
...
a = 103◦, b = 55◦ , c = 77◦ , d = 125◦,
e = 55◦, f = 22◦, g = 103◦, h = 77◦ ,
i = 103◦, j = 77◦, k = 81◦
10
...
A = 37◦, B = 60◦ , E = 83◦

349

4
3
4
3
2
...
sin A = , tan A =
17
15
112
15
, cos X =
4
...
(a)
(b)
(c)
17
17
15
7
24
6
...
(a) 9
...
625

Exercise 79 (page 176)
1
...
proof

Exercise 84 (page 187)
1
...

5
...

13
...
7550 2
...
846
3
...
52
(a) 0
...
1010 (c) 0
...
33◦
6
...
25◦
7
...
78◦
8
...
41 54 11
...
05
12
...
3586 14
...
803

Exercise 80 (page 178)

Exercise 85 (page 189)

1
...
54 mm, y = 4
...
9 cm, 7
...
(a) 2
...
3 m

1
...
22 (b) 5
...
87 (d) 8
...
595 (f ) 5
...
(a) AC = 5
...
04◦ , ∠C = 30
...
928 cm, ∠D = 30◦, ∠F = 60◦
(c) ∠J = 62◦, HJ = 5
...
59 cm
(d) ∠L = 63◦ , LM = 6
...
37 cm
(e) ∠N = 26◦, ON = 9
...
201 cm
(f ) ∠S = 49◦, RS = 4
...
625 cm

Exercise 81 (page 180)
1–5
...


3
...
54 m

4
...
40 mm

Chapter 21
Exercise 82 (page 182)

Exercise 86 (page 192)

1
...

7
...

11
...


1
...
15 m
2
...
249
...
110
...
53
...
9
...
107
...
9
...
56 m
9
...
24 m
3
...
54 mm
20
...
7
...
11
...
11 mm 8
...
20 cm each (b) 45◦
10
...
81 km
3
...
132
...
94 mm
14
...
sin Z = , cos Z = , tan X = , cos X =
41
41
9
41

1
...
78◦ and 137
...
53◦ and 351
...
(a) 29
...
92◦ (b) 123
...
14◦
3
...
21◦ and 224
...
12◦ and 293
...
t = 122◦7 and 237◦53
5
...
θ = 39◦44 and 219◦44

Exercise 88 (page 202)
1
...
180◦
3
...
120◦
◦
◦
◦
5
...
2, 144
7
...
5, 720◦
7
9
...
6, 360◦ 11
...
5 ms
2
13
...
100 μs or 0
...
625 Hz 16
...
leading

Exercise 91 (page 209)
1
...
2 cm, Q = 47
...
65◦,
area = 77
...
p = 6
...
83◦, R = 44
...
938 m2
3
...
33◦, Y = 52
...
05◦,
area = 27
...
Z = 29
...
50◦ , Z = 96
...
(a) 40 (b) 25 Hz (c) 0
...
29 rad (or 16
...

3
...

7
...
(a) 122
...
80◦ , 40
...
54◦
(a) 11
...
55◦
4
...
4 m
BF = 3
...
0 m 6
...
35 m, 5
...
48 A, 14
...
(a) 75 cm (b) 6
...
157 s
(d) 0
...
94◦ ) lagging 75 sin 40t
3
...
01 s or 10 ms
(d) 0
...
61◦) lagging 300 sin 200πt
4
...
43) volts

π

A or
5
...
524)A
6
...
2 sin(100πt + 0
...
(a) 5A, 50 Hz, 20 ms, 24
...
093A (c) 4
...
375 ms (e) 3
...
C = 83◦ , a = 14
...
9 mm,
area = 189 mm2
2
...
568 cm, a = 7
...
65 cm2
3
...
0 cm,
area = 134 cm2
4
...
08 mm,
area = 185
...
J = 44◦29 , L = 99◦31 , l = 5
...
132 cm2 , or, J = 135◦ 31 , L = 8◦29 ,
l = 0
...
917 cm2
6
...
2 mm,
area = 820
...
19 mm, area = 174
...
80
...
38◦, 40
...
(a) 15
...
07◦
3
...
25 cm, 126
...
19
...
36
...
x = 69
...
130◦ 8
...
66 mm

Chapter 24
Exercise 94 (page 215)
1
...

3
...

5
...

7
...


(5
...
04◦) or (5
...
03 rad)
(6
...
82◦) or (6
...
36 rad)
(4
...
57◦) or (4
...
03 rad)
(6
...
58◦) or (6
...
54 rad)
(7
...
20◦) or (7
...
55 rad)
(4
...
31◦) or (4
...
12 rad)
(5
...
04◦) or (5
...
74 rad)
(15
...
75◦) or (15
...
37 rad)

Exercise 95 (page 217)
(1
...
830)
2
...
917, 3
...
362, 4
...
(−2
...
154)
(−9
...
400)
6
...
615, −3
...
750, −1
...
(4
...
233)
(a) 40∠18◦, 40∠90◦, 40∠162◦, 40∠234◦, 40∠306◦
(b) (38
...
36), (0, 40), (−38
...
36),
(−23
...
36), (23
...
36)
10
...
0 mm
1
...

5
...

9
...
p = 105◦ , q = 35◦
3
...
r = 142◦, s = 95◦

Exercise 97 (page 225)
1
...
91 cm
2
...
7 cm2 3
...
27
...
18 cm
6
...
(a) 29 cm2 (b) 650 mm2
8
...
3
...
6750 mm
11
...
30 cm2
12
...

4
...

9
...

13
...

18
...


113 cm2
2
...
1790 mm2
2
2
802 mm
5
...
1269 m2
2
1548 m
8
...
0 cm (b) 783
...
46 m 10
...
80 cm, 74
...
86 mm (b) 197
...
26
...
67 cm, 54
...
82
...
748
(a) 0
...
2 m2
17
...
47 m2
2
(a) 396 mm (b) 42
...
701
...
74 mm

Exercise 104 (page 237)
1
...
Centre at (3, −2), radius 4
3
...
Circle, centre (0, 0), radius 6

Chapter 27
Exercise 98 (page 226)
1
...

3
...

7
...
27 cm2 (b) 706
...
(a) 20
...
41 mm
(a) 53
...
9 mm2
6
...
89 m

Exercise 99 (page 228)
1
...
1624 mm2 3
...
918 ha (b) 456 m

Exercise 105 (page 243)
1
...

5
...

10
...

14
...

17
...


1
...
5 cm3
3
...
15 cm3 , 135 g
7
...
(a) 35
...
3 cm2
1
...
37
...
63 cm
13
...
709 cm, 153
...
99 cm
16
...
099 m2
8
...
22 m
18
...
5 min
4 cm
20
...
08 m3

Exercise 100 (page 229)
1
...
80 m2

3
...
14 ha

Chapter 26
Exercise 101 (page 231)
1
...
24 cm 2
...
5 mm 3
...
629 cm 4
...
68 cm
5
...
73 cm 6
...
97
...
201
...
0 cm2 2
...
68 cm3 , 25
...
113
...
1 cm2 4
...
131 cm 5
...
2681 mm3 7
...
083 cm3
(b) 20 106 mm2 or 201
...
8
...
(a) 512 × 106 km2 (b) 1
...
664

Exercise 102 (page 232)

Exercise 107 (page 251)

5π
5π
π
(b)
(c)
6
12
4
2
...
838 (b) 1
...
054
3
...
(a) 0◦ 43 (b) 154◦8 (c) 414◦ 53

1
...

4
...


1
...
90 cm2
(a) 56
...
82 cm2
3
...
57 kg
5
...
4 m2
29
...
5 cm3 (b) 84
...
4 cm3 (b) 32
...
0 cm3

352 Basic Engineering Mathematics

7
...

11
...


(b) 146 cm2 (vi) (a) 86
...
9 cm (b) 38
...
125 cm3
3
2
10
...
5 m
10
...
220
...
1 cm3 , 1027 cm2
2
(a) 1458 litres (b) 9
...
45

147 cm3 , 164 cm2
10480 m3 , 1852 m2
10
...
14 m

14
...
72◦ to the 5 N force
29
...
04◦ to the horizontal
9
...
70◦
9
...
89 m/s at 159
...
62 N at 26
...
07 knots, E 9
...

3
...

7
...

7
...

10
...


2
...
1707 cm2
6
...
(a) 54
...
16◦ (b) 45
...
66◦
2
...
71 m/s at 121
...
55 m/s at 8
...
83
...
6◦ to the vertical
2
...
22
...
78◦ N

Exercise 109 (page 256)
1
...
137
...
4
...
54
...
63
...
4
...
143 m2

Exercise 111 (page 260)
1
...
59 m3

2
...
20
...
(a) 2 A (b) 50 V (c) 2
...
0
...
1 A
5
...
13 cm2 , 368
...

3
...

7
...


i − j − 4k
−i + 7j − k
−3i + 27j − 8k
i + 7
...
6i + 4
...
9k

2
...

6
...

10
...
5j − 10k
2i + 40j − 43k

Chapter 30
Exercise 118 (page 279)

2
...
5 V (b) 3 A
4
...
83 V (b) 0

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Chapter 29
Exercise 119 (page 281)
Exercise 113 (page 266)
1
...

2
...
scalar
4
...
scalar
6
...
vector 8
...
vector

Exercise 114 (page 273)
1
...

3
...

5
...
35 N at 18
...
62◦ to the 12 m/s velocity
16
...
57◦ to the 13 N force
28
...
30◦ to the 18 N force
32
...
80◦ to the 30 m displacement

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Exercise 120 (page 283)
1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395) 4
...
11 sin(ωt + 0
...
8
...
173)

Answers to practice exercises
Exercise 121 (page 284)
11
...
324)
2
...
73 sin(ωt − 0
...
79 sin(ωt − 0
...
695 sin(ωt + 0
...
38 sin(ωt + 1
...
3 sin(314
...
233) V (b) 50 Hz
(a) 10
...
3t + 0
...
(a) 79
...
352)V (b) 150 Hz
(c) 6
...

3
...

5
...

7
...
(a) continuous (b) continuous (c) discrete
(d) continuous
2
...
If one symbol is used to represent 10 vehicles, working correct to the nearest 5 vehicles, gives 3
...
5,
6, 7, 5 and 4 symbols respectively
...
If one symbol represents 200 components, working
correct to the nearest 100 components gives: Mon 8,
Tues 11, Wed 9, Thurs 12 and Fri 6
...

3
...

4
...

5
...

6
...

7
...

P increases by 20% at the expense of Q and R
...
Four rectangles of equal height, subdivided as follows: week 1: 18%, 7%, 35%, 12%, 28%; week 2:
20%, 8%, 32%, 13%, 27%; week 3: 22%, 10%, 29%,
14%, 25%; week 4: 20%, 9%, 27%, 19%, 25%
...

9
...
5◦ , 22
...
5◦, 167
...


353

10
...

11
...
(a) £16 450 (b) 138

Exercise 124 (page 297)
1
...
3–39
...
5–39
...
7–39
...
9–40
...
1–40
...
3–40
...
5–40
...
7–40
...

2
...
35, 39
...
75, 39
...
and
heights of 1, 5, 9, 17,
...
There is no unique solution, but one solution is:
20
...
9 3; 21
...
4 10; 21
...
9 11;
22
...
4 13; 22
...
9 9; 23
...
4 2
...
There is no unique solution, but one solution is:
1–10 3; 11–19 7; 20–22 12; 23–25 11;
26–28 10; 29–38 5; 39–48 2
...
20
...
45 13; 21
...
45 37; 22
...
45 48
6
...
5, 15, 21, 24, 27, 33
...
5
...
3, 0
...
67, 2
...
5 and 0
...

7
...
95 2), (11
...
95 19), (12
...
95
42), (13
...
A graph of cumulative frequency against upper class
boundary having co-ordinates given in the answer
to problem 7
...
(a) There is no unique solution, but one solution is:
2
...
09 3; 2
...
14 10; 2
...
19 11;
2
...
24 13; 2
...
29 9; 2
...
34 2
...
07, 2
...
and heights of 3, 10,
...
095 3; 2
...
195 24; 2
...
295 46; 2
...


(d) A graph of cumulative frequency against upper
class boundary having the co-ordinates given
in part (c)
...
Mean 7
...
Mean 27
...
−2542 A/s
2
...
16 cd/V (b) 312
...
(a) −1000 V/s (b) −367
...
−1
...
(a)
15
2
3
4
...
(a)

5
...


7
...

9
...
(a) 8 x + 8 x3 +
x +c
5

7x 2
+c
2
3
(b) t 8 + c
8
5 4
x +c
(b)
24
3
(b) 2t − t 4 + c
4
(b)

(b) 4θ + 2θ 2 +

5
(a) θ 2 − 2θ + θ 3 + c
2
3
2
3
(b) x 4 − x 3 + x 2 − 2x + c
4
3
2
4
1
(a) − + c
(b) − 3 + c
3x
4x
4√ 5
1√
4 9
(a)
x +c
(b)
x +c
5
9
15 √
10
5
(b)
x +c
(a) √ + c
7
t

7
(b) − cos 3θ + c
3
1
(b) 18 sin x + c
3
−2
(b)
+c
15e5x

13
...
(a) 4x + c

3
sin 2x + c
2
1
11
...
(a) e + c
8
10
...
(a) 1
...
5

2
...
5

3
...
333

4
...
75 (b) 0
...
(a) 10
...
1667

6
...
(a) 1 (b) 4
...
(a) 0
...
638

9
...
09 (b) 2
...
(a) 0
...
099

Exercise 141 (page 334)
1
...
37
...
1
...
proof
6
...
5
9
...
67

3
...
1
10
...
29

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