Notesale: Turn your study into money

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Chapter 6

Ratio and proportion
6
...

Some practical examples include mixing paint, sand
and cement, or screen wash
...

Two quantities are in direct proportion when they
increase or decrease in the same ratio
...
Also, calculating currency exchange rates
and converting imperial to metric units rely on direct
proportion
...
For example, the time taken
to do a job is inversely proportional to the number of
people in a team: double the people, half the time
...

For this chapter you will need to know about decimals
and fractions and to be able to use a calculator
...
2

Ratios

Ratios are generally shown as numbers separated by a
colon (:) so the ratio of 2 and 7 is written as 2 :7 and we
read it as a ratio of ‘two to seven
...
e
...
e
...
e
...
1016/B978-1-85617-697-2
...
The paint mix is
4 parts total, with 3 parts red and 1 part white
...

Problem 1
...
Reduce the ratio to its simplest
form
(i) Both 6 and 27 can be divided by 3
...

6 :27 and 2 :9 are called equivalent ratios
...
In this example, the simplest form is 2 : 9 which
means for every 2 females in the class there are 9 male
students
...
A gear wheel having 128 teeth is in
mesh with a 48-tooth gear
...

(i) 128 and 48 can both be divided by 2, i
...
128 :48
is the same as 64 :24
(ii) 64 and 24 can both be divided by 8, i
...
64 :24 is
the same as 8 :3
(iii) There is no number that divides completely into
both 8 and 3 so 8 :3 is the simplest ratio, i
...
the
gear ratio is 8 : 3

Ratio and proportion
Thus, 128 :48 is equivalent to 64 :24 which is equivalent
to 8 :3 and 8 : 3 is the simplest form
...
A wooden pole is 2
...
Divide
it in the ratio of 7 to 19
(i) Since the ratio is 7 :19, the total number of parts
is 7 + 19 = 26 parts
...
08 m = 208 cm, hence,
208
= 8
...

Hence, 2
...

(Check: 56 + 152 must add up to 208, otherwise an error
would have been made
...
In a competition, prize money of
£828 is to be shared among the first three in the
ratio 5 : 3 : 1
(i) Since the ratio is 5 : 3 :1 the total number of parts
is 5 + 3 + 1 = 9 parts
...

828
= £92, 3 parts cor9
responds to 3 × £92 = £276 and 5 parts corresponds to 5 × £92 = £460
...
(Check: 460 + 276 + 92 must add up to
828, otherwise an error would have been made
...
A map scale is 1 : 30 000
...

Determine the actual distance between the schools,
giving the answer in kilometres
Actual distance between schools
= 6 × 30 000 cm = 180 000 cm
=

180,000
m = 1800 m
100

1800
m = 1
...
6 km, hence the schools are just over 1 mile
apart
...
In a box of 333 paper clips, 9 are defective
...

2
...
Determine the gear ratio in its
simplest form
...
In a box of 2000 nails, 120 are defective
...

4
...
36 m long is to be cut into two
in the ratio 6 to 15
...

5
...
How long will it take to cook a 7 kg
turkey?
6
...
Calculate the
amount each receives
...
A local map has a scale of 1 :22 500
...
7 km
...
Prize money in a lottery totals £3801 and is
shared among three winners in the ratio 4 :2 : 1
...

Problem 6
...
65 in its
simplest form
(i) Changing both quantities to the same units, i
...
to
pence, gives a ratio of 45 :765
(ii) Dividing both quantities by 5 gives
45 : 765 ≡ 9 :153
(iii) Dividing both quantities by 3 gives
9 : 153 ≡ 3 : 51
(iv) Dividing both quantities by 3 again gives
3 : 51 ≡ 1 : 17

42 Basic Engineering Mathematics
Thus, 45 p as a ratio of £7
...


2
...
How much acid is needed to make
266 ml of the mixture?

Problem 7
...
If 45 ml of water is added and
the mixture stirred, what is now the alcohol content?

3
...
If 18 ml of water is added and the
mixture stirred, determine the new percentage
alcoholic content
...

alcohol =
100
(ii) After 45 ml of water is added we have 30 + 45
= 75 ml of fluid, of which alcohol is 12 ml
...
A wooden beam 4 m long weighs 84 kg
...


= 16%
...
20 tonnes of a mixture of sand and
gravel is 30% sand
...

100
(ii) If the mixture has 6 t of sand then amount of
gravel = 20 − 6 = 14 t
...
1% would be
40
14
mixture would be
× 100 t = 35 t
...

(v) We already have 6 t of sand, so amount of sand
to be added to produce a mixture with 40%
gravel = 21 − 6 = 15 t
...
An alloy is made up of metals P and Q in the
ratio 3
...
How much of P has to
be added to 4
...
15 000 kg of a mixture of sand and gravel is
20% sand
...


6
...
For example,
if 12 cans of lager have a mass of 4 kg, then 24 cans of
lager will have a mass of 8 kg; i
...
, if the quantity of cans
doubles then so does the mass
...

In the previous section we had an example of mixing
1 shovel of cement to 4 shovels of sand; i
...
, the ratio
of cement to sand was 1 : 4
...
e
...
This is another example
of direct proportion
...
)
Now try the following Practice Exercise
Practice Exercise 25 Further ratios
(answers on page 342)
1
...
95 kg
...
e
...


(b) Charles’s law states that, for a given mass of gas
at constant pressure, the volume V is directly proportional to its thermodynamic temperature T , i
...

V ∝ T
...
e
...


Ratio and proportion
Here are some worked examples to help us understand
more about direct proportion
...
3 energy saving light bulbs cost
£7
...
Determine the cost of 7 such light bulbs
(i) 3 light bulbs cost £7
...
80
= £2
...
60 = £18
...
If 56 litres of petrol costs £59
...
92
59
...
07
56
Hence, 32 litres cost 32 × 1
...
24
(ii) Therefore, 1 litre of petrol costs

Problem 11
...
When, for mild steel, the stress
is 63 MPa, the strain is 0
...
Determine (a) the
value of strain when the stress is 42 MPa, (b) the
value of stress when the strain is 0
...

(i) When the stress is 63 MPa, the strain is
0
...
0003
strain of
63
(iii) Thus, the value of strain when the stress is
0
...
0002
63
(b) Strain is proportional to stress
...
0003, the stress is
63 MPa
...
0001 corresponds to
63
MPa
...
00072 =
× 7
...
2 MPa
...
Charles’s law states that for a given
mass of gas at constant pressure, the volume is
directly proportional to its thermodynamic
temperature
...
4 litres

43

at 600 K
...
2 litres, (b) the volume at 540 K
(a)

Volume is directly proportional to temperature
...
4 litres, the temperature is 600 K
...

2
...
2 = 800 K
...
2 litres =
2
...

(i) When the temperature is 600 K, the volume
is 2
...

(ii) Hence, a temperature of 1 K corresponds to
2
...

600
(iii) Thus, the volume at a temperature of
2
...
16 litres
...
3 engine parts cost £208
...
Calculate the cost
of 8 such parts
...
If 9 litres of gloss white paint costs £24
...

3
...
6 kg
...

4
...
Determine the effort, in grams, to lift a
load of 5
...

5
...
32 kg, what will
28 cans weigh?
6
...
When, for copper, the stress is
60 MPa, the strain is 0
...
Determine
(a) the strain when the stress is 24 MPa and
(b) the stress when the strain is 0
...
Charles’s law states that volume is directly
proportional to thermodynamic temperature
for a given mass of gas at constant pressure
...
8 litres at 330 K
...
4 litres and (b) the volume when the
temperature is 396 K
...

Problem 13
...
The gutter spans 8
...
How much lower
should the low end be?
(i) The guttering has to decline in the ratio 3 :700 or
3
700
(ii) If d is the vertical drop in 8
...

voltage of
3
(iii) Thus, when the current is 4
...
2 = 126 mV
...
Some approximate imperial to
metric conversions are shown in Table 6
...
Use the
table to determine
(a) the number of millimetres in 12
...
1
length

1 mile = 1
...

(i) When voltage is 90 mV, the current is 3 A
...

90
(iii) Thus, when the voltage is 60 mV, the
3
current = 60 ×
= 2A
...

(i) When current is 3 A, the voltage is 90 mV
...
2 lb = 1 kg
(1 lb = 16 oz)

capacity

i
...
d = 36 mm, which is how much the lower end
should be to allow rainwater to drain
...
Ohm’s law state that the current
flowing in a fixed resistance is directly proportional
to the applied voltage
...

Determine (a) the current when the voltage is
60 mV and (b) the voltage when the current is 4
...
54 cm

1
...
5 inches = 12
...
54 cm = 31
...
73 cm = 31
...
5 mm

(b) 50 m
...
h
...
6 km/h = 80 km/h
300
miles = 186
...
6
20
kg = 9
...
2
(e) 56 kg = 56 × 2
...
2 lb
(c)

300 km =

0
...
2 × 16 oz = 3
...

Thus, 56 kg = 123 lb 3 oz, correct to the nearest
ounce
...
1 litres
192 pints =
1
...
76 pints = 105
...
6
105
...
2 gallons
8
Problem 16
...
2
...
Determine (a) the current flowing
when the p
...
is 5 V and (b) the p
...
when the
current is 10 mA
...


The tourist rate for the Swiss franc is quoted in
a newspaper as £1 = 1
...
How many francs
can be purchased for £326
...


If 1 inch = 2
...


4
...
2 lb = 1 kg and 1lb = 16 oz, determine the
number of pounds and ounces in 38 kg (correct
to the nearest ounce)
...


If 1 litre = 1
...


6
...
When for brass the stress is 21 MPa,
the strain is 0
...
Determine the stress
when the strain is 0
...


7
...
48 cm, find the number of
millimetres in 23 inches
...


The tourist rate for the Canadian dollar is
quoted in a newspaper as £1 = 1
...
How
many Canadian dollars can be purchased for
£550?

(d) the number of American dollars which can be
purchased for £92
...
2

(a)

France

£1 = 1
...
50 kronor

Switzerland

£1 = 1
...
80 dollars

£1 = 1
...
25 euros
= 68
...


(b) £1 = 185 yen, hence £23 = 23 × 185 yen
= 4255 yen
...
50
= £610
...
50 kronor, hence 6405 lira = £

(d) £1 = 1
...
50 = 92
...
80 dollars = $166
...
95 Swiss francs, hence
2925
= £1500
2925 pesetas = £
1
...


Ohm’s law states that current is proportional
to p
...
in an electrical circuit
...
d
...
4

Inverse proportion

Two variables, x and y, are in inverse proportion to one
1
1
k
another if y is proportional to , i
...
y α or y = or
x
x
x
k = x y where k is a constant, called the coefficient of
proportionality
...

For example, the time for a journey is inversely proportional to the speed of travel
...
p
...
a journey
is completed in 20 minutes, then at 60 m
...
h
...
Double the speed,
half the journey time
...
)
In another example, the time needed to dig a hole is
inversely proportional to the number of people digging
...
Half the
men, twice the time
...
)
Here are some worked examples on inverse proportion
...
It is estimated that a team of four
designers would take a year to develop an
engineering process
...
Hence, 3 designers would
4
take years, i
...
1 year 4 months
...
A team of five people can deliver
leaflets to every house in a particular area in four
hours
...
Hence, 3 peo2
20
hours, i
...
6 hours, i
...
6 hours
ple would take
3
3
40 minutes
...
The electrical resistance R of a
piece of wire is inversely proportional to the
cross-sectional area A
...
02 ohms
...
5 × 106 pascals, determine (a) the coefficient of
proportionality and (b) the volume if the pressure is
changed to 4 × 106 pascals
(a)

V ∝

1
k
i
...
V = or k = pV
...
5 × 106)(0
...
12 × 106
(b) Volume,V =

k
0
...
03 m3
=
p
4 × 106

Now try the following Practice Exercise
Practice Exercise 28 Further inverse
proportion (answers on page 342)
1
...
Assuming all eat the same
amount, how long will the potatoes last if there
are only two in the family?
2
...
If y is inversely proportional to x and
y = 15
...
6, determine (a) the
coefficient of proportionality, (b) the value of
y when x is 1
...
2

coefficient of proportionality, k = (7
...
A car travelling at 50 km/h makes a journey in
70 minutes
...
e
...
Hence, when
A
A
R = 7
...
Hence, when R = 4,
R
36
= 9 mm2
the cross sectional area, A =
4

(b) Since k = R A then A =

Problem 20
...
If
a gas occupies a volume of 0
...
Boyle’s law states that, for a gas at constant
temperature, the volume of a fixed mass of
gas is inversely proportional to its absolute
pressure
...
5 m3
at a pressure of 200 × 103 pascals, determine
(a) the constant of proportionality, (b) the
volume when the pressure is 800 × 103 pascals and (c) the pressure when the volume is
1
...


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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