CHAPTER 1 SOME MATHEMATICS REVISION
EXERCISE 1, Page 4
1.
Evaluate 378.37 – 298.651 + 45.64 – 94.562
By calculator, 378.37 – 298.651 + 45.64 – 94.562 = 30.797
2.
Evaluate
17.35  34.27
correct to 3 decimal places
41.53  3.76
By calculator,
17.35  34.27
= 53.83187... = 53.832, correct to 3 decimal places
41.53  3.76
3.
 4.527  3.63  0.468
correct to 5 significant figures
 452.51  34.75
Evaluate
By calculator,
4.
 4.527  3.63  0.468 = 1.0944077.... = 1.0944, correct to 5 significant figures
 452.51  34.75
Evaluate 52.34 
By calculator, 52.34 
5.
Evaluate
By calculator,
6.
 912.5  41.46 
correct to 3 decimal places
 24.6  13.652 
 912.5  41.46 
= 50.329663... = 50.330, correct to 3 decimal places
 24.6  13.652 
52.14  0.347 11.23
correct to 4 significant figures
19.73  3.54
52.14  0.347 11.23
= 36.45494.... = 36.45, correct to 4 significant figures
19.73  3.54
Evaluate 6.852 correct to 3 decimal places
By calculator, 6.852 = 46.9225 = 46.923, correct to 3 decimal places
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1
7.
Evaluate  0.036  in engineering form
2
By calculator,  0.036  = 0.001296 = 1.296  103 in engineering form
2
8.
Evaluate 1.33
By calculator, 1.33 = 2.197
9.
Evaluate  0.38  correct to 4 decimal places
3
By calculator,  0.38  = 0.054872 = 0.0549, correct to 4 decimal places
3
10.
Evaluate  0.018 in engineering form
3
By calculator,  0.018 = 5.832  106 in engineering form
3
11.
Evaluate
1
correct to 1 decimal place
0.00725
By calculator,
1
= 137.93103... = 137.9, correct to 1 decimal place
0.00725
12.
1
1

correct to 4 significant figures
0.065 2.341
Evaluate
By calculator,
13.
1
1

= 14.957447... = 14.96, correct to 4 significant figures
0.065 2.341
Evaluate 2.14
By calculator, 2.14 = 19.4481
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2
14.
Evaluate  0.22  correct to 5 significant figures in engineering form
5
By calculator,  0.22  = 5.153632 104 = 515.36  10 6 , correct to 5 significant figures in
5
engineering form
15.
Evaluate 1.012  correct to 4 decimal places
7
By calculator, 1.012  = 1.087085... = 1.0871, correct to 4 decimal places
7
16.
Evaluate 1.13  2.94  4.42 correct to 4 significant figures
By calculator, 1.13  2.94  4.42 = 52.6991 = 52.70, correct to 4 significant figures
17.
34528 correct to 2 decimal places
Evaluate
By calculator,
34528 = 185.81711... = 185.82, correct to 2 decimal places
18.
3
Evaluate
17 correct to 3 decimal places
By calculator, 3 17 = 2.57128159... = 2.571, correct to 3 decimal places
19.
Evaluate
6
2451  4 46 correct to 3 decimal places
By calculator, 6 2451  4 46 = 1.0676068... = 1.068, correct to 3 decimal places
20.
Evaluate 5 103  7 108 and express in engineering form.
By calculator, 5 103  7 108 = 3.5  106 in engineering form
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21.
Evaluate
6 103 14 104
and express in engineering form.
2 106
By calculator,
6 103 14 104
= 4.2  10 6 in engineering form
6
2 10
22.
56.43 103  3 104
correct to 3 decimal places in engineering form
8.349 103
Evaluate
56.43 103  3 104
By calculator,
= 0.202766798... = 202.767  103 , correct to 3 decimal places
3
8.349 10
in engineering form
23.
Evaluate
99 105  6.7 103
correct to 4 significant figures in engineering form
36.2 104
99 105  6.7 103
By calculator,
= 18323204.42 = 18320000 = 18.32  10 6 correct to 4 significant
4
36.2 10
figures in engineering form
24.
Evaluate
4 1
 as a decimal, correct to 4 decimal places
5 3
By calculator,
4 1
 = 0.466666... = 0.4667, correct to 4 decimal places
5 3
25.
2 1 3
  as a fraction
3 6 7
Evaluate
By calculator,
26.
2 1 3 13
  
3 6 7 14
5 5
Evaluate 2  1 as a decimal, correct to 4 significant figures
6 8
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5 5 107
By calculator, 2  1 
= 4.4583333... = 4.458, correct to 4 significant figures
6 8 24
6
1
27. Evaluate 5  3 as a decimal, correct to 4 significant figures
7
8
6
1 153
By calculator, 5  3 
= 2.7321428... = 2.732, correct to 4 significant figures
7
8 56
28.
Evaluate
By calculator,
29.
3 4 2 4
   as a fraction
4 5 3 9
3 4 2 4
9
   
4 5 3 9
10
8
2
Evaluate 8  2 as a mixed number
9
3
8
2 10
1
3
By calculator, 8  2 
9
3 3
3
30.
1 1
7
Evaluate 3  1  1
as a decimal, correct to 3 decimal places
5 3 10
1 1
7 77
By calculator, 3 1  1 
= 2.566666... = 2.567, correct to 3 decimal places
5 3 10 30
 1 2
 4 1  2
5 3
31. Evaluate 
 as a decimal, correct to 3 significant figures
3 9
 1
3  2 
5
 4
 1 2
 4  1  2 118
5 3
By calculator, 
= 0.07758053... = 0.0776, correct to 3 significant figures
 
3  9 1521
 1
3  2 
5
 4
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32.
Evaluate sin 67 correct to 4 decimal places
By calculator, sin 67 = 0.9205, correct to 4 decimal places
33.
Evaluate tan 71 correct to 4 decimal places
By calculator, tan 71 = 2.9042, correct to 4 decimal places
34.
Evaluate cos 63.74 correct to 4 decimal places
By calculator, cos 63.74 = 0.4424, correct to 4 decimal places
35.
Evaluate tan 39.55 - sin 52.53 correct to 4 decimal places
By calculator, tan 39.55 - sin 52.53 = 0.0321, correct to 4 decimal places
36.
Evaluate sin(0.437 rad) correct to 4 decimal places
By calculator, sin(0.437 rad) = 0.4232, correct to 4 decimal places
37.
Evaluate tan(5.673 rad) correct to 4 decimal places
By calculator, tan(5.673 rad) = - 0.6992, correct to 4 decimal places
38. Evaluate
 sin 42.6 tan 83.2 correct to 4 decimal places
By calculator,
 sin 42.6 tan 83.2 = 5.8452, correct to 4 decimal places
39.
cos 13.8
cos 13.8
Evaluate 1.59 correct to 4 significant figures
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By calculator, 1.59 = 4.995, correct to 4 significant figures
40.
Evaluate 2.7( - 1) correct to 4 significant figures
By calculator, 2.7( - 1) = 5.782, correct to 4 significant figures
41.
Evaluate 2
By calculator, 2
42.
 13 1 correct to 4 significant figures
 13 1 = 25.72, correct to 4 significant figures
Evaluate 8.5e 2.5 correct to 4 significant figures
By calculator, 8.5e 2.5 = 0.6977, correct to 4 significant figures
43.
Evaluate 3e 2 1 correct to 4 significant figures
By calculator, 3e 2 1 = 591.0, correct to 4 significant figures
44.
Evaluate
5.52


 2
 correct to 4 significant figures
 2e  26.73 
By calculator,
5.52


 2
 = 3.520, correct to 4 significant figures
 2e  26.73 
45. Evaluate
 e 2  3  

 correct to 4 significant figures
  8.57 
By calculator,
 e 2  3  

 = 0.3770, correct to 4 significant figures
  8.57 
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EXERCISE 2 Page 5
1. The area A of a rectangle is given by the formula A = lb. Evaluate the area when l = 12.4 cm and
b = 5.37 cm
Area, A = l × b = 12.4 × 5.37 = 66.588 cm 2 = 66.59 cm 2
2. The circumference C of a circle is given by the formula C = 2r. Determine the circumference
given r = 8.40 mm
Circumference, C = 2r = 2 ×  × 8.40 = 52.78 mm
3. A formula used in connection with gases is R =
PV
. Evaluate R when P = 1500, V = 5 and
T
T = 200
R = (PV)/T =
(1500)(5)
= 37.5
200
4. The velocity of a body is given by v = u + at. The initial velocity u is measured when time t is
15 seconds and found to be 12 m/s. If the acceleration a is 9.81 m/s2 calculate the final velocity v.
Velocity, v = u + at = 12 + 9.81 × 15 = 159.15 m/s
5. Calculate the current I in an electrical circuit, when I = V/R amperes when the voltage V is
measured and found to be 7.2 V and the resistance R is 17.7 
Current, I =
V 7.2

= 0.407 A
R 17.7
6. Find the distance s, given that s =
1 2
gt . Time t = 0.032 seconds and acceleration due to gravity
2
g = 9.81 m/s2. Give the answer in millimetres.
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1
1
2
Distance, s = gt 2   9.81 0.032  = 0.00502 m or 5.02 mm (since 1 m = 1000 mm)
2
2
7. The energy stored in a capacitor is given by E =
1
CV2 joules. Determine the energy when
2
capacitance C = 5  10  6 farads and voltage V = 240 V.
Energy, E =
1
1
CV2 =  5 106  2402 = 0.144 J
2
2
8. Find the area A of a triangle, given A =
1
bh, when the base length b is 23.42 m and the height h is
2
53.7 m
Area of triangle, A =
1
1
bh =  23.42  53.7 = 628.8 m 2
2
2
9. Resistance R2 is given by R2 = R1(1 + t). Find R2, correct to 4 significant figures, when R1 = 220,
 = 0.00027 and t = 75.6
Resistance, R 2 = R1 1  t   220 1   0.00027 75.6    220 1  0.020412 
= 220  1.020412
= 224.5, correct to 4 significant figures.
10. Density =
mass
. Find the density when the mass is 2.462 kg and the volume is 173 cm3. Give
volume
the answer in units of kg/m3. Note that 1 cm 3 = 10 6 m3 .
Volume = 173 cm3 = 173  10 6 m 3
Density =
mass
2.462 kg

= 14231.21387 = 14230 kg/ m 3 , correct to 4 significant figures.
volume 173 106 m3
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11. Evaluate resistance RT, given
1
1
1
1



when R1 = 5.5 , R2 = 7.42  and
R T R1 R 2 R 3
R3 = 12.6 
1
1
1
1
1
1
1






 0.181818  0.134771  0.079365 = 0.395954
R T R1 R 2 R 3 5.5 7.42 12.6
RT =
and
1
= 2.526 
0.395954
12. The potential difference, V volts, available at battery terminals is given by V = E - Ir. Evaluate V
when E = 5.62, I = 0.70 and R = 4.30
Potential difference, V = E – Ir = 5.62 – 0.70 × 4.30 = 5.62 – 3.01 = 2.61 V
13. The current I amperes flowing in a number of cells is given by I =
nE
. Evaluate the current
R  nr
when n = 36. E = 2.20, R = 2.80 and r = 0.50
Current, I =
nE
(36)(2.20)
79.2
79.2



= 3.81 A, correct to 3 significant
R  n r 2.80  (36)(0.50) 2.80  18 20.80
figures.
14. Energy, E joules, is given by the formula E =
1 2
LI . Evaluate the energy when L = 5.5 and
2
I = 1.2
1
1
1
2
Energy, E = LI 2   5.5 1.2    5.5 1.44  = 3.96 J
2
2
2
8. The current I amperes in an a.c. circuit is given by I =
V
(R 2  X 2 )
. Evaluate the current when
V = 250, R = 11.0 and X = 16.2
Current, I =
V
R X
2
2

250
11.0  16.2
2
2

250
250

383.44 19.581624...
= 12.77 A, correct to 4 significant figures.
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EXERCISE 3, Page 7
1 1

3 4
1. Evaluate
A common denominator can be obtained by multiplying the two denominators together, i.e. the
common denominator is 3  4 = 12
The two fractions can now be made equivalent, i.e.
1 4
1 3


and
3 12
4 12
so that they can be easily added together, as follows:
1 1
4
3
43
7
 =
+
=
=
3 4 12 12
12
12
1 1

5 4
2. Evaluate
A common denominator can be obtained by multiplying the two denominators together, i.e. the
common denominator is 5  4 = 20
The two fractions can now be made equivalent, i.e.
1 4

5 20
so that they can be easily added together, as follows:
1 1
4
5
9
45
 =
+
=
=
5 4
20
20
20
20
3. Evaluate
and
1 5

4 20
1 1 1
 
6 2 5
1 1 1
5  15  6 14
7
  =

=
6 2 5
30
30 15
4. Use a calculator to evaluate
1 3 8
 
3 4 21
1 3 8
1 1 2
 
=   by cancelling
3 4 21 3 1 7
=
1 2 76
1
 
=
3 7
21
21
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5. Use a calculator to evaluate
3 4 2 4
  
4 5 3 9
3 4 2 4 3 1 2 9
   =   
4 5 3 9 1 5 3 4
=
6. Evaluate
3 1 1 3
3 3 6  15
9
   =  =
=1 5 1 2
5 2
10
10
3 5 1
  as a decimal, correct to 4 decimal places.
8 6 2
3 5 1
9  20  12 17
  =

= 0.7083 correct to 4 decimal places
8 6 2
24
24
8
2
7. Evaluate 8  2 as a mixed number.
9
3
8
2 80 8 80 3 10 1 10
1
8 2       
=3
9
3 9 3 9 8 3 1 3
3
1 1
7
8. Evaluate 3  1  1
as a decimal, correct to 3 decimal places.
5 3 10
1 1
7 16 4 17 64 17
3 1  1    

5 3 10 5 3 10 15 10
=
9. Determine
128  51 77
17

2
= 2.567 correct to 3 decimal places
30
30
30
2 3
 as a single fraction.
x y
2 2y

x xy
and
3 3x

y xy
Hence,
2y  3x
3x  2y
2 3
2y 3x
 =
+
=
or
xy
xy
xy
xy
x y
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12
EXERCISE 4, Page 9
1. Express 0.057 as a percentage
0.057 = 0.057  100% = 5.7%
2. Express 0.374 as a percentage
0.374 = 0.374  100% = 37.4%
3. Express 20% as a decimal number
20% =
4. Express
20
= 0.20
100
11
as a percentage
16
11 11
1100
  100% 
% = 68.75%
16 16
16
5. Express
5
as a percentage, correct to 3 decimal places
13
5
5
500
 100% 
% = 38.461538…. by calculator
13 13
13
= 38.462% correct to 3 decimal places
6. Place the following in order of size, the smallest first, expressing each as percentages, correct to
1 decimal place:
(a)
(a)
12
21
(b)
9
17
12 12
1200
 100% 
% = 57.1%
21 21
21
(c)
5
9
(d)
6
11
(b)
9
9
900
 100% 
% = 52.9%
17 17
17
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(c)
5 5
500
  100% 
% = 55.6%
9 9
9
(d)
6 6
600
 100% 
% = 54.6%
11 11
11
Placing them in order of size, the smallest first, gives: (b), (d), (c) and (a)
7. Express 65% as a fraction in its simplest form
65% =
65
65
13
and by dividing the numerator and denominator by 5 gives: 65% =
=
100
100
20
8. Calculate 43.6% of 50 kg
43.6% of 50 kg =
43.6
 50 kg = 21.8 kg
100
9. Determine 36% of 27 m
36% of 27 m =
36
 27 m = 9.72 m
100
10. Calculate correct to 4 significant figures:
(a) 18% of 2758 tonnes
(a) 18% of 2758 t =
(b) 47% of 18.42 grams
(c) 147% of 14.1 seconds
18
 2758 t= 496.4 t
100
(b) 47% of 18.42 g =
47
 18.42 g = 8.657 g
100
(c) 147% of 14.1 s =
147
 14.1 s = 20.73 s
100
11. Express: (a) 140 kg as a percentage of 1 t
(b) 47 s as a percentage of 5 min
(c) 13.4 cm as a percentage of 2.5 m
It is essential when expressing one quantity as a percentage of another that both quantities are in the
same units.
© John Bird Published by Taylor and Francis
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(a) 1 tonne = 1000 kg, hence 140 kg as a percentage of 1 t =
140
100% = 14%
1000
(b) 5 minutes = 5  60 = 300 s, hence 47 s as a percentage of 5 minutes =
47
 100% = 15.67 %
300
(c) 2.5 m = 2.5  100 = 250 cm, hence 13.4 cm as a percentage of 2.5 m =
13.4
100% = 5.36 %
250
12. A computer is advertised on the internet at £520, exclusive of VAT. If VAT is payable at
20%, what is the total cost of the computer?
VAT = 20% of £520 =
20
 520 = £104
100
Total cost of computer = £520 + £104 = £624
13. Express 325 mm as a percentage of 867 mm, correct to 2 decimal places.
325 mm as a percentage of 867 mm =
325
 100% = 37.49%
867
14. When signing a new contract, a Premiership footballer’s pay increases from £15,500 to
£21,500 per week. Calculate the percentage pay increase, correct 3 significant figures.
Percentage change is given by:
i.e.
% increase =
new value  original value
100%
original value
21500  15500
6000
100% 
100% = 38.7%
15500
15500
15. A metal rod 1.80 m long is heated and its length expands by 48.6 mm. Calculate the percentage
increase in length.
% increase =
48.6
48.6
100% 
100% = 2.7%
1.80 1000
1800
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15
EXERCISE 5 Page 10
1.
In a box of 333 paper clips, 9 are defective. Express the non-defective paper clips as a ratio of
the defective paper clips, in its simplest form.
Non-defective paper clips = 333 – 9 = 324
The ratio of non-defective clips to defective clips is: 324:9
Dividing both by 9 gives the ratio as: 36:1
2.
A gear wheel having 84 teeth is in mesh with a 24 tooth gear. Determine the gear ratio in its
simplest form.
Gear ratio = 84:24
Dividing both by 4 gives: 21:6
Dividing both by 3 gives:
3.
7:2 or 3.5:1
A metal pipe 3.36 m long is to be cut into two in the ratio 6 to 15. Calculate the length of each
piece.
Since the ratio is 6:15, the total number of parts is 6 + 15 = 21 parts
21 parts corresponds to 3.36 m = 336 cm, hence, 1 part corresponds to
336
= 16
21
Thus, 6 parts corresponds to 6  16 = 96 cm,
and 15 parts corresponds to 15  16 = 240 cm
Hence, 3.36 m divides in the ratio of 6:15 as 96 cm to 240 cm
4.
In a will, £6440 is to be divided between three beneficiaries in the ratio 4:2:1. Calculate the
amount each receives.
Since the ratio is 4:2:1 the total number of parts is 4 + 2 + 1 = 7 parts
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7 parts corresponds to £6440
1 part corresponds to
6440
= £920, 2 parts corresponds to 2  £920 = £1840
7
and 4 parts corresponds to 4  £920 = £3680
Hence, £6440 divides in the ratio of 4:2:1 as £3680 to £1840 to £920
5.
A local map has a scale of 1:22,500. The distance between two motorways is 2.7 km. How far
are they apart on the map?
Distance between motorways =
6.
2.7 km 2700 m 270000cm


= 12 cm
22500
2250
22500
Express 130 g as a ratio of 1.95 kg.
Changing both quantities to the same units, i.e. to grams, gives a ratio of: 130:1950
Dividing both quantities by 10 gives:
130:1950  13:195
Dividing both quantities by 13 gives:
13:195  1:15
Thus, 130 g as a ratio of 1.95 kg is:
7.
1:15
In a laboratory, acid and water are mixed in the ratio 2:5. How much acid is needed to make
266 ml of the mixture?
In a ratio of 2:5 there are 2 + 5 = 7 parts
1 part = 266 ml  7 = 38 ml
Amount of acid in the mixture = 2 parts = 2 × 38 = 76 ml
8.
A glass contains 30 ml of gin which is 40% alcohol. If 18 ml of water is added and the mixture
stirred, determine the new percentage alcoholic content.
The 30 ml of gin contains 40% alcohol =
40
 30 = 12 ml
100
© John Bird Published by Taylor and Francis
17
After 18 ml of water is added we have 30 + 18 = 48 ml of fluid of which alcohol is 12 ml
Fraction of alcohol present =
12
48
Percentage of alcohol present =
9.
12
 100% = 25%
48
A wooden beam 4 m long weighs 84 kg. Determine the mass of a similar beam that is 60 cm
long.
4 m of beam weighs 84 kg, hence, 1 m of beam = 84 kg ÷ 4 = 21 kg
60 cm, i.e. 0.6 m, will weigh 0.6 × 21 = 12.6 kg
10.
An alloy is made up of metals P and Q in the ratio 3.25:1 by mass. How much of P has to be
added to 4.4 kg of Q to make the alloy.
For every 1 part of Q there is 3.25 parts of P
For 4.4 kg of Q, P = 3.25 × 4.4 = 14.3 kg
© John Bird Published by Taylor and Francis
18
EXERCISE 6 Page 11
1.
3 engine parts cost £208.50. Calculate the cost of 8 such parts.
If 3 engine parts cost £208.50, then 1 engine part costs £208.50 ÷ 3 = £69.50
Hence, the cost of 8 engine parts = 8 × £69.50 = £556
2.
If 9 litres of gloss white paint costs £24.75, calculate the cost of 24 litres of the same paint.
If 9 litres of paint cost £24.75, then 1 engine part costs £24.75 ÷ 9 = £2.75
Hence, the cost of 24 litres = 24 × £2.75 = £66
3.
The total mass of 120 household bricks is 57.6 kg. Determine the mass of 550 such bricks.
If 120 bricks have a mass of 57.6 kg, then 1 brick has a mass of 57.6 kg ÷ 120 = 0.48 kg
Hence, the mass of 550 bricks = 550 × 0.48 kg = 264 kg
4.
Hooke’s law states that stress is directly proportional to strain within the elastic limit of a
material. When, for copper, the stress is 60 MPa, the strain is 0.000625.
Determine (a) the strain when the stress is 24 MPa, and (b) the stress when the strain is 0.0005
(a) Stress is directly proportional to strain.
When the stress is 60 MPa, the strain is 0.000625,
hence a stress of 1 MPa corresponds to a strain of
0.000625
60
and the value of strain when the stress is 24 MPa =
0.000625
 24 = 0.00025
60
(b) If when the strain is 0.000625, the stress is 60 MPa,
then a strain of 0.0001 corresponds to
60
MPa
6.25
© John Bird Published by Taylor and Francis
19
and the value of stress when the strain is 0.0005 =
5.
60
 5 = 48 MPa
6.25
Charles’s law states that volume is directly proportional to thermodynamic temperature for a
given mass of gas at constant pressure. A gas occupies a volume of 4.8 litres at 330 K.
Determine (a) the temperature when the volume is 6.4 litres, and (b) the volume when the
temperature is 396 K.
(a) Volume is directly proportional to temperature.
When the volume is 4.8 litres, the temperature is 330 K
hence a volume of 1 litre corresponds to a temperature of
and the temperature when the volume is 6.4 litres =
330
K
4.8
330
 6.4 = 440 K
4.8
(b) Temperature is proportional to volume
When the temperature is 330 K, the volume is 4.8 litres
4.8
litre
330
hence a temperature of 1 K corresponds to a volume of
and the volume at a temperature of 396 K =
4.8
 396 = 5.76 litres
330
6. Ohm’s law states that current is proportional to p.d. in an electrical circuit. When a p.d. of
60 mV is applied across a circuit a current of 24 A flows. Determine:
(a) the current flowing when the p.d. is 5 V, and
(b) the p.d. when the current is 10 mA
(a) Current is directly proportional to the voltage.
When p.d. is 60 mV, the current is 24 A,
hence a p.d. of 1 mV corresponds to a current of
24
A
60
and a p.d. of 1 volt corresponds to a current of 1000 ×
24
A
60
© John Bird Published by Taylor and Francis
20
and when the p.d. is 5 V, the current = 5 × 1000 
24
= 2000 A = 2 mA
60
(b) Voltage is directly proportional to the current.
When current is 24 A, the voltage is 60 mV,
hence a current of 1 A corresponds to a voltage of
60  103
= 2.5 kV
24  106
and when the current is 10 mA, the voltage = 10 10 3  2.5 kV = 25 V
7.
If 2.2 lb = 1 kg, and 1 lb = 16 oz, determine the number of pounds and ounces in 38 kg (correct
to the nearest ounce).
38 kg = 38 × 2.2 = 83.6 lb
0.6 lb = 0.6 × 16 oz = 9.6 oz = 10 oz to the nearest ounce
Hence, 38 kg = 83 lb 10 oz
8. If 1 litre = 1.76 pints, and 8 pints = 1 gallon, determine (a) the number of litres in 35 gallons,
and (b) the number of gallons in 75 litres.
(a) 35 gallons = 35 × 8 pints = 280 pints
Number of litres in 35 gallons = 280 ÷ 1.76 = 159.1 litres
(b) 75 litres = 75 × 1.76 pints = 132 pints
132 pints = 132 ÷ 8 gallons = 16.5 gallons
© John Bird Published by Taylor and Francis
21
EXERCISE 7 Page 12
1. A 10 kg bag of potatoes lasts for a week with a family of 7 people. Assuming all eat the same
amount, how long will the potatoes last if there were only two in the family?
If a 10 kg bag of potatoes lasts for a week with a family of 7 people, then it would last for 7 weeks
for 1 person.
For two persons, it would last for 7/2 = 3.5 weeks
2. If 8 men take 5 days to build a wall, how long would it take 2 men?
If 8 men take 5 days to build a wall, then 1 man would take 8 × 5 = 40 days.
Hence, 2 men would take 40/2 = 20 days to build the wall.
3. If y is inversely proportional to x and y = 15.3 when x = 0.6, determine (a) the coefficient of
proportionality, (b) the value of y when x is 1.5, and (c) the value of x when y is 27.2
y
1
x
i.e. y =
k
x
where k is the constant of proportionality.
(a) When y = 15.3, x = 0.6, hence 15.3 =
k
0.6
from which, constant of proportionality, k = (15.3)(0.6) = 9.18
(b) When x = 1.5, y =
k 9.18

= 6.12
x 1.5
(c) When y = 27.2, x =
k 9.18

= 0.3375
y 27.2
4. A car travelling at 50 km/h makes a journey in 70 minutes. How long will the journey take at
70 km/h?
If the car takes 70 minutes at 50 km/h then distance travelled = 50 km/h ×
© John Bird Published by Taylor and Francis
70
h = 175/3 km
60
22
If the speed is 70 km/h then time for journey =
175 / 3km 5
5
 h   60 min = 50 minutes
70 km / h 6
6
5. 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. If a gas occupies a volume of 1.5 m3 at a pressure of
200  103 Pascal’s, determine (a) the constant of proportionality, (b) the volume when the pressure
is 800  103 Pascals and (c) the pressure when the volume is 1.25 m3
Volume 
1
absolute pressure
i.e. V 
1
p
or V =
(a) When V = 1.5 m 3 , p = 200 103 Pa, hence, 1.5 =
k
where k is the constant of proportionality.
p
k
200  103
from which, constant of proportionality, k = (1.5)( 200 103 ) = 300  10 3
(b) When p = 800 103 , V =
(c) When V = 1.25 m 3 , p =
k 300 103
= 0.375 m 3

3
p 800 10
k 300 103

= 240  10 3 Pa
V
1.25
© John Bird Published by Taylor and Francis
23
EXERCISE 8, Page
1. Evaluate 22  2  24
22  2  24  2214  27
by law 1 of indices
= 128
2. Evaluate 35  33  3 in index form
35  33  3  3531 = 3 9
by law 1 of indices
27
23
3. Evaluate
27
 27 3  2 4
23
by law 2 of indices
= 16
4. Evaluate
33
35
33
 335  32
5
3
by law 2 of indices
=
1
32
by law 5 of indices
=
1
9
5. Evaluate 7 0
70 = 1
6. Evaluate
by law 4 of indices
23  2  26
27
© John Bird Published by Taylor and Francis
24
23  2  26 2316 210
 7  7  2107  23 = 8
27
2
2
by laws 1 and 2 of indices
10 106
7. Evaluate
105
10 106
 10165  102 = 100
5
10
by laws 1 and 2 of indices
8. Evaluate 10 4  10
104  10 
104
 1041  103 = 1000
1
10
by law 2 of indices
103 104
9. Evaluate
109
103 104
1
1
 103 49  102  2 
= 0.01
9
10
10 100
by law 2 of indices
10. Evaluate 56  52  57
56  52  57 
56  52
 56 27  51 = 5
57
by laws 1 and 2 of indices
11. Evaluate (72)3 in index form
(72)3 = 7 23 = 7 6 by law 3 of indices
12. Evaluate (33)2
(33)2 = 33 2 = 3 6 = 3  3  3  3  3  3 = 729
© John Bird Published by Taylor and Francis
25
13. Evaluate
37  34
in index form
35
37  34
 37 45 = 3 6
5
3
14. Evaluate
by laws 1 and 2 of indices
(9  32 )3
(3  27)2
2
2
34 

(9  32 )3  3  3 
312



 3128 = 3 4 = 81 by laws 1, 2 and 3 of indices
(3  27) 2  3  33 2  34 2 38
3
15. Evaluate
3
(16  4) 2
(2  8)3
4
2
26 

(16  4) 2  2  2 
212



=1
3
3
12
(2  8)3
 2  23   2 4  2
2
16. Evaluate
2
by laws 1, 2 and 3 of indices
5 2
5 4
52
2  4
 5    52 4  52 = 25
4
5
by law 2 of indices
32  34
17. Evaluate
33
1
32  34
1
 3243  3243  35  5 =
3
243
3
3
18. Evaluate
by laws 1, 2 and 5 of indices
7 2  7 3
7  7 4
72  73 7 23 7 1


 713  713  72 = 49
7  74 714 73
by laws 1 and 2 of indices
© John Bird Published by Taylor and Francis
26
19. Simplify, giving the answer as a power: z 2  z6
z 2  z 6  z 2 6 = z 8
by law 1 of indices
20. Simplify, giving the answer as a power: a  a 2  a 5
a  a 2  a 5  a1 25 = a 8
by law 1 of indices
21. Simplify, giving the answer as a power: n8  n 5
n8  n 5  n85 = n 3
by law 1 of indices
22. Simplify, giving the answer as a power: b 4  b 7
b 4  b7  b 47 = b 11
by law 1 of indices
23. Simplify, giving the answer as a power: b 2  b 5
b 2  b5 
b2
 b25 = b 3 or
5
b
1
b3
by laws 2 and 5 of indices
24. Simplify, giving the answer as a power: c5  c3  c 4
c5  c3  c 4 
c5  c3 c53 c8
 4  4  c84 = c4
c4
c
c
by laws 1 and 2 of indices
m5  m 6
25. Simplify, giving the answer as a power:
m 4  m3
m5  m6 m56 m11
 43  7  m117 = m 4
4
3
m m
m
m
by laws 1 and 2 of indices
© John Bird Published by Taylor and Francis
27
26. Simplify, giving the answer as a power:
(x 2 )(x) x 21 x 3
 6  6 = x 3 or
6
x
x
x
1
x3
by laws 1, 2 and 5 of indices
27. Simplify, giving the answer as a power:
x   x
3 4
34
= x12
2 3
23
1
y6
= y  6 or
t  t   t   t  = t
1 3 2
4 2
8
7 2
72
= c14
y 
2 3
t  t 
3 2
by laws 1 and 3 of indices
30. Simplify, giving the answer as a power:
c   c
3 4
by laws 3 and 5 of indices
29. Simplify, giving the answer as a power:
3 2
x 
by law 3 of indices
28. Simplify, giving the answer as a power:
y   y
(x 2 )(x)
x6
c 
7 2
by law 3 of indices
 a2 
31. Simplify, giving the answer as a power:  5 
a 
3
 a2 
9
2 5 3
3 3
 5    a    a  = a or
a 
1
a9
3
by laws 3 and 5 of indices
© John Bird Published by Taylor and Francis
28
 1 
32. Simplify, giving the answer as a power:  3 
b 
4
 1
12
3 4
or
 3   b  = b
b 
1
b12
4
by laws 3 and 4 of indices
 b2 
33. Simplify, giving the answer as a power:  7 
b 
2
2
 b2 
10
2  7 2
5 2
 7   b   b  = b
b
 
by laws 2 and of indices
34. Simplify, giving the answer as a power:
1
1
 33 = 9 or s 9
s
s  s
1
3 3
1
s 
3 3
by laws 3 and 5 of indices
35. Simplify, giving the answer as a power: p3qr 2  p2q5 r  pqr 2
p3qr 2  p2q5r  pqr 2  p321  q151  r 212  p6  q 7  r 5 = p 6q 7r 5
36. Simplify, giving the answer as a power:
x 3 y2 z
 x 35 y 21z13 = x2 y z 2 or
x 5 y z3
x3 y2z
x 5 y z3
y
x z2
2
© John Bird Published by Taylor and Francis
29
EXERCISE 9, Page 15
1. Evaluate A given A = 3( 2 + 1 + 4)
A = 3( 2 + 1 + 4) = 3(7) = 3  7 = 21
2. Evaluate A given A = 4[5(2 + 1) – 3(6 – 7]
A = 4[5(2 + 1) – 3(6 – 7] = 4[5(3) – 3(- 1)]
= 4[15 + 3] = 4[18]
= 4  18 = 72
3. Expand the brackets: 2(x – 2y + 3)
2(x – 2y + 3) = 2(x) – 2(2y) + 2(3)
= 2x – 4y + 6
4. Expand the brackets: (3x – 4y) + 3(y – z) – (z – 4x)
(3x – 4y) + 3(y – z) – (z – 4x) = 3x – 4y + 3y – 3z – z + 4x
= 3x + 4x – 4y + 3y – 3z – z
= 7x – y – 4z
5. Expand the brackets: 2x + [y – (2x + y)]
2x + [y – (2x + y)] = 2x + [y – 2x – y]
= 2x + [- 2x] = 2x - 2x = 0
6. Expand the brackets: 24a – [2{3(5a – b) – 2(a + 2b)} + 3b]
24a – [2{3(5a – b) – 2(a + 2b)} + 3b] = 24a – [2{15a – 3b – 2a – 4b} + 3b]
= 24a – [2{13a – 7b} + 3b]
© John Bird Published by Taylor and Francis
30
= 24a – [26a – 14b + 3b]
= 24a – [26a – 11b]
= 24a – 26a + 11b
= - 2a + 11b or 11b – 2a
7. Expand the brackets: ab[c + d – e(f – g + h{i + j})]
ab[c + d – e(f – g + h{i + j})] = ab[c + d – e(f – g + hi + hj)]
= ab[c + d – ef + eg – ehi – ehj]
= abc + abd – abef + abeg – abehi - abehj
© John Bird Published by Taylor and Francis
31
EXERCISE 10 Page 17
1. Solve: 2x + 5 = 7
Since 2x + 5 = 7 then 2x = 7 – 5
and
2x = 2 i.e. x =
2
=1
2
2. Solve: 8 - 3t = 2
Since 8 – 3t = 2
then 8 – 2 = 3t
i.e.
6 = 3t
3. Solve:
Since
from which,
t=
6
=2
3
2
c-1=3
3
2
2
c - 1 = 3 then
c=3+1=4
3
3
from which,
c=
3
4 = 6
2
4. Solve: 2x - 1 = 5x + 11
Since 2x – 1 = 5x + 11
then
- 1 – 11 = 5x – 2x
– 12 = 3x
i.e.
from which,
x=
12
=-4
3
5. Solve: 2a + 6 - 5a = 0
Since 2a + 6 – 5a = 0
i.e.
then
6 = 5a – 2a
6 = 3a
from which,
a=
6
=2
3
© John Bird Published by Taylor and Francis
32
6. Solve: 3x - 2 - 5x = 2x - 4
Since 3x - 2 - 5x = 2x – 4 then 4 – 2 = 2x - 3x + 5x
i.e.
2 = 4x from which,
x=
2
1
=
4
2
7. Solve: 20d - 3 + 3d = 11d + 5 - 8
Since 20d – 3 + 3d = 11d + 5 – 8
then
20d + 3d – 11d = 5 – 8 + 3
i.e.
12d = 0
from which, d = 0
8. Solve: 2(x - 1) = 4
Since 2(x - 1) = 4 then 2x – 2 = 4 and 2x = 4 + 2 = 6
from which,
x=
6
=3
2
9. Solve: 16 = 4(t + 2)
Since 16 = 4(t + 2)
then 16 = 4t + 8
16 – 8 = 4t
and
i.e. 8 = 4t
and
t=
8
=2
4
10. Solve: 5(f - 2) - 3(2f + 5) + 15 = 0
Since 5(f – 2) – 3(2f + 5) + 15 = 0
then
5f – 10 – 6f – 15 + 15 = 0
5f – 6f = 10 + 15 – 15
i.e.
and
- f = 10
from which, f = - 10
11. Solve: 2x = 4(x - 3)
© John Bird Published by Taylor and Francis
33
Since 2x = 4(x - 3) then 2x = 4x – 12
i.e.
12 = 4x – 2x = 2x
and
x=
12
=6
2
12. Solve: 6(2 - 3y) - 42 = - 2(y - 1)
Since 6(2 – 3y) – 42 = - 2(y – 1)
12 – 18y – 42 = - 2y + 2
then
-18y + 2y = 2 – 12 + 42
i.e.
and
- 16y = 32
from which,
y=
32
32

=-2
16
16
13. Solve: 2(3g - 5) - 5 = 0
Since 2(3g - 5) - 5 = 0 then 6g – 10 – 5 = 0
i.e.
6g = 10 + 5 = 15
and
g=
15
= 2.5
6
14. Solve: 4(3x + 1) = 7(x + 4) - 2(x + 5)
Since 4(3x + 1) = 7(x + 4) – 2(x + 5)
then
12x + 4 = 7x + 28 – 2x – 10
i.e.
12x – 7x + 2x = 28 – 10 – 4
and
7x = 14
from which,
x=
14
=2
7
15. Solve: 11 + 3(r - 7) = 16 - (r + 2)
Since
11 + 3(r - 7) = 16 - (r + 2)
then
11 + 3r – 21 = 16 – r – 2
i.e.
3r + r = 16 – 2 – 11 + 21
© John Bird Published by Taylor and Francis
34
i.e.
4r = 24
and
r=
24
=6
4
16. Solve: 8 + 4(x - 1) - 5(x - 3) = 2(5 - 2x)
Since
8 + 4(x - 1) - 5(x - 3) = 2(5 - 2x)
then
8 + 4x – 4 – 5x + 15 = 10 – 4x
i.e.
4x – 5x + 4x = 10 – 8 + 4 – 15
i.e.
3x = - 9
from which,
x=
17. Solve:
Since
9
=-3
3
1
d+3=4
5
1
d + 3 = 4 then
5
from which,
1
d=4–3=1
5
d = (5) (1) = 5
18. Solve: 2 +
3
2
5
y=1+ y+
4
3
6
Multiplying each term by 12 (the lowest common denominator of 3, 4 and 6) gives:
3
2
5
(12)(2)  (12) y  (12)(1)  (12) y  (12)
4
3
6
i.e.
24 + 9y = 12 + 8y + 10
and
9y – 8y = 12 + 10 – 24
i.e.
19. Solve:
y=-2
1
1
(2x - 1) + 3 =
4
2
© John Bird Published by Taylor and Francis
35
Multiplying each term by 4 gives:
1
1
(4) (2x  1)  (4)(3)  (4)
4
2
2x – 1 + 12 = 2
i.e.
and
2x = 2 + 1 – 12
i.e.
2x = - 9
20. Solve:
from which, x = 
9
1
=4
2
2
1
1
2
(2f - 3) + (f - 4) +
=0
5
6
15
Multiplying each term by 30 gives:
1
1
2
(30) (2f  3)  (30) (f  4)  (30)  0
5
6
15
i.e.
6(2f – 3) + 5(f – 4) + 4 = 0
and
12f – 18 + 5f – 20 + 4 = 0
12f + 5f = 18 + 20 – 4
i.e.
and
17f = 34
21. Solve:
from which, f =
34
=2
17
1
1
1
(3m - 6) - (5m + 4) + (2m - 9) = - 3
3
4
5
Multiplying each term by 60 gives:
1
1
1
(60) (3m  6)  (60) (5m  4)  (60) (2m  9)  (60)(3)
3
4
5
i.e.
20(3m – 6) – 15(5m + 4) + 12(2m – 9) = -180
i.e.
60m – 120 – 75m – 60 + 24m – 108 = - 180
and
60m – 75m + 24m = - 180 + 120 + 60 + 108
i.e.
9m = 108
from which,
m=
108
= 12
9
© John Bird Published by Taylor and Francis
36
22. Solve:
x x
- =2
3 5
Multiplying each term by 15 gives:
(15)
x
x
- (15) = (15)(2)
3
5
i.e.
5x – 3x = 30
i.e.
2x = 30
from which,
23. Solve:
Since
x=
30
= 15
2
2
3
=
a
8
2
3
=
then by cross-multiplying: (2)(8) = 3a
a
8
i.e.
16 = 3a
from which,
a=
24. Solve:
16
1
=5
3
3
1
1
7
+
=
3n
4n
24
Multiplying each term by 24n gives:
(24n)
1
1
7
 (24n)
 (24n)
3n
4n
24
i.e.
25. Solve:
8 + 6 = 7n
i.e.
14 = 7n
from which,
n=
14
=2
7
x 3 x 3
=
+2
4
5
Multiplying each term by 20 gives:
© John Bird Published by Taylor and Francis
37
(20)
x 3
x 3
 (20)
 (20)(2)
4
5
i.e.
5(x + 3) = 4(x – 3) + 40
i.e.
5x + 15 = 4x – 12 + 40
and
5x – 4x = -12 + 40 – 15
from which,
x = 13
26. Solve:
3t
6t
2t 3
=
+
20
12
15 2
Multiplying each term by 60 gives:
(60)
3t
6t
2t
3
= (60)
+ (60)
- (60)
20
12
15
2
3(3t) = 5(6 – t) + 4(2t) – 30(3)
i.e.
9t = 30 – 5t + 8t – 90
i.e.
9t + 5t – 8t = 30 – 90
and
6t = - 60
from which,
27. Solve:
 60
= - 10
6
t=
y
7
5 y
+
=
5
20
4
Multiplying each term by 20 gives:
y
7
5 y
(20)  (20)
 (20)
5
20
4
i.e.
4y + 7 = 5(5 – y)
i.e.
4y + 7 = 25 – 5y
and
4y + 5y = 25 – 7
i.e.
9y = 18
from which,
y=
18
=2
9
© John Bird Published by Taylor and Francis
38
28. Solve:
v2
1
=
2v  3 3
Cross-multiplying gives:
3(v – 2) = 1(2v – 3)
i.e.
3v – 6 = 2v - 3
and
3v – 2v = 6 – 3
i.e.
v=3
29. Solve:
2
3
=
a  3 2a  1
Multiplying each term by (a – 3)(2a + 1) gives:
(a  3)(2a  1)
2
3
 (a  3)(2a  1)
a 3
2a  1
i.e.
2(2a + 1) = 3(a – 3)
i.e.
4a + 2 = 3a – 9
and
4a – 3a = - 9 - 2
from which,
a = - 11
30. Solve: 3 t = 9
Since
3 t = 9 then
t =
9
=3
3
t =  3 = 9
2
from which,
31. Solve: 2 y = 5
Since
2 y =5
then
y =
5
2
© John Bird Published by Taylor and Francis
39
2
1
25
5
y=   
=6
4
4
2
from which,
x 
32. Solve: 10 = 5   1
2 
Dividing both sides by 5 gives:
2
x
1
2
Squaring both sides gives:
4
x
1
2
4 1 
x
2
i.e.
33. Solve: 16 =
Since
16 =
and
34. Solve:
t2
9
i.e.
5=
x
2
from which, x = (5)(2) = 10
t2
9
then 16 × 9 = t 2
t  16  9
i.e. t = ±12
 y2
1

 =
2
 y2
y2 1
 
y2 2
Squaring both sides gives:
2
i.e.
y2 1

y2 4
Multiplying each term by 4(y – 2) gives:
4(y  2)
y2
1
 4(y  2)
y2
4
i.e.
4(y + 2) = (y – 2)
i.e.
4y + 8 = y – 2
i.e.
4y – y = - 2 – 8
© John Bird Published by Taylor and Francis
40
and
3y = - 10
35. Solve:
Since
from which,
y=
10
1
= 3
3
3
6
2a
=
a
3
6
2a
=
a
3
then (6)(3) = (a)(2a)
i.e.
18 =
2a2
i.e.
a2 =
18
=9
2
Hence,
a=
9 =±3
© John Bird Published by Taylor and Francis
41
EXERCISE 11 Page 19
1. Transpose a + b = c - d – e for d
Since
a + b = c - d – e then d = c – e – a - b
2. Transpose y = 7x
for x
y
7
Since y = 7x then x =
3. Transpose pv = c for v
Since pv = c then v =
c
p
4. Transpose v = u + at for a
Since
v = u + at
then
v – u = at
and
a=
vu
t
5. Transpose V = IR for R
Since V = IR then R =
V
I
6. Transpose x + 3y = t for y
Since x + 3y = t then 3y = t – x
and dividing both sides by 3 gives:
y
tx
1
or y   t  x 
3
3
© John Bird Published by Taylor and Francis
42
7. Transpose c = 2r for r
Dividing both sides of c = 2r by 2 gives:
8. Transpose y = mx + c
Since
c
c
 r or r =
2
2
for x
y = mx + c
then
y – c = mx
and
x=
y c
m
9. Transpose I = PRT for T
Since
I = PRT
then
T=
I
PR
10. Transpose XL  2 f L for L
Since
X L  2 f L
then
L=
11. Transpose I =
E
R
XL
2f
for R
Multiplying both sides of I =
E
by R gives: I R = E
R
and dividing both sides by I gives:
R=
E
I
© John Bird Published by Taylor and Francis
43
12. Transpose y 
x
 3 for x
a
x
3
a
Since
y
then
y–3=
and
x = a(y  3)
13. Transpose F =
Rearranging F =
x
a
9
C + 32
5
for C
9
C  32 gives:
5
Multiplying both sides by
5
 F  32   C
9
1
2 f C
for f
Since
XC 
1
2 f C
then
f=
14. Transpose X C 
or
C
5
 F  32 
9
1
2 C XC
a
for r
1 r
Multiplying both sides of S =
i.e.
9
C
5
5
5
5 9 
gives:
 F  32    
 C 
9
9
 9  5 
i.e.
15. Transpose S =
F – 32 =
a
by (1 – r) gives: S(1 – r) = a
1 r
S – Sr = a
© John Bird Published by Taylor and Francis
44
from which,
S – a = Sr
and dividing both sides by S gives:
Sa
Sa
a
= r i.e. r =
or r = 1 S
S
S
16. Transpose y =
 (x  d)
for x
d
Multiplying both sides of y =
 x  d
by d gives:
d
yd
 x d

Dividing both sides by  gives:
d
and
Alternatively, from the first step,
yd = (x – d)
i.e.
yd = x - d
yd
=x

or
x  d
yd

yd + d = x
and
from which,
17. Transpose A =
Since A =
x=
3(F  f )
L
yd  d d  y   



i.e. x =
d
y  

for f
3(F  f )
then AL = 3(F – f)
L
i.e.
and
AL = 3F – 3f
3f = 3F – AL
from which,
f=
Since
yd = (x – d)
3F  AL
3
3F  AL 3F AL
AL
3F  AL
 
 F
then f =
3
3
3
3
3
may also be written as f = F -
© John Bird Published by Taylor and Francis
AL
3
45
18. Transpose y =
A B2
for D
5 CD
A B2
Since y =
5 CD
then D =
A B2
5Cy
19. Transpose R = R0(1 + t)
for t
by cross-multiplying
Removing the bracket in R  R 0 1  t  gives: R  R 0  R 0  t
from which,
R  R0  R0  t
and
R  R0
=t
R 0
20. Transpose I =
Ee
Rr
or t =
R  R0
R 0
for R
Multiplying both sides by (R + r) gives:
I(R + r) = E – e
i.e.
and
IR+Ir=E–e
IR=E–e–Ir
and dividing both sides by I gives:
R=
E e  Ir
I
or R =
Ee
r
I
21. Transpose y = 4ab2c2 for b
Dividing both sides by 4a c 2 gives:
Taking the square root of both sides gives:
22. Transpose for t = 2
L
g
y
 b2
2
4ac
or
b2 
y
4ac 2
b=
y
4ac2
for L
© John Bird Published by Taylor and Francis
46
Dividing both sides of t = 2
L
by 2 gives:
g
t
L

2
g
2
L
 t 
  
g
 2 
Squaring both sides gives:
L  t 
 
g  2 
or
 t 
L = g

 2 
Multiplying both sides by g gives:
2
2
or L =
g t2
4 2
23. Transpose v2 = u2 + 2as for u
Since v2 = u2 + 2as then v2 – 2as = u2
from which,
24. Transpose N =
u=
v2  2as
ax

 for a
 y 
Squaring both sides of N =
ax
gives:
y
ax
y
N2 y  a  x
Multiplying both sides by y gives:
or
a + x = N2 y
a = N2y - x
from which,
25. Transpose Z =
N2 
R 2  (2fL) 2 for L, and evaluate L when Z = 27.82, R = 11.76 and f = 50.
Squaring both sides of Z =
R 2   2fL 
2
gives:
Z2  R 2   2fL 
from which,
Z2  R 2   2fL 
Taking the square root of both sides gives:
Z2  R 2  2fL
Dividing both sides by 2f gives:
When Z = 27.82, R = 11.76 and f = 50, L =
L=
2
2
or
2fL  Z2  R 2
Z2  R 2
2f
27.822  11.762 25.2122

= 0.080
2(50)
2(50)
© John Bird Published by Taylor and Francis
47
EXERCISE 12, Page 20
1. If 5 apples and 3 bananas cost £1.45 and 4 apples and 6 bananas cost £2.42, determine how
much an apple and a banana each cost.
Let an apple = A and a banana = B, then:
5A + 3B = 145
(1)
4A + 6B = 242
(2)
From equation (1),
5A = 145 – 3B
and
A=
From equation (2),
4A = 242 – 6B
and
A=
145  3B
= 29 – 0.6B
5
(3)
242  6B
= 60.5 – 1.5B
4
Equating (3) and (4) gives:
(4)
29 – 0.6B = 60.5 – 1.5B
i.e.
1.5B – 0.6B = 60.5 – 29
and
0.9B = 31.5
and
B=
31.5
= 35
0.9
A = 29 – 0.6(35) = 29 – 21 = 8
Substituting in (3) gives:
Hence, an apple costs 8p and a banana costs 35p
2. If 7 apples and 4 oranges cost £2.64 and 3 apples and 3 oranges cost £1.35, determine how much
an apple and an orange each cost.
Let an apple = A and an orange = R, then:
Multiplying equation (1) by 3 gives:
7A + 4R = 264
(1)
3A + 3R = 135
(2)
21A + 12R = 792
(3)
Multiplying equation (2) by 4 gives:
12A + 12R = 540
(4)
© John Bird Published by Taylor and Francis
48
Equation (3) – equations (4) gives:
9A = 252
from which,
A=
Substituting in (3) gives:
252
= 28
9
21(28) + 12R = 792
i.e.
12R = 792 – 21(28)
i.e.
12R = 204
and
B=
204
= 17
12
Hence, an apple costs 28p and an orange costs 17p
3. Three new cars and four new vans supplied to a dealer together cost £93000, and five new cars
and two new vans of the same models cost £99000. Find the respective costs of a car and a van.
Let a car = C and a van = V, then working in £1000’s:
3C + 4V = 93
(1)
5C + 2V = 99
(2)
Multiplying equation (2) by 2 gives:
10C + 4V = 198
(3)
Equation (3) – equations (1) gives:
7 C = 105
from which,
C=
Substituting in (1) gives:
105
= 15
7
3(15) + 4V = 93
i.e.
4V = 93 – 3(15)
i.e.
4V = 48
and
V=
48
= 12
4
Hence, a car costs £15000 and a van costs £12000
4. In a system of forces, the relationship between two forces F 1 and F 2 is given by:
5F 1 + 3F 2 = - 6
3F 1 + 5F 2 = - 18
Solve for F 1 and F 2
© John Bird Published by Taylor and Francis
49
5F 1 + 3F 2 = - 6
(1)
3F 1 + 5F 2 = - 18
(2)
Multiplying equation (1) by 5 gives:
25F 1 + 15F 2 = - 30
(3)
Multiplying equation (2) by 3 gives:
9F 1 + 15F 2 = - 54
(4)
Equation (3) – equation (4) gives:
16F 1 = - 30 - - 54 = - 30 + 54 = 24
from which,
F1 =
Substituting in (1) gives:
24
= 1.5
16
5(1.5) + 3F 2 = - 6
i.e.
3F 2 = - 6 - 5(1.5)
i.e.
3F 2 = - 13.5
and
F2 =
13.5
= - 4.5
3
Hence, F 1 = 1.5 and F 2 = - 4.5
5. Solve the simultaneous equations:
a+b=7
a–b=3
a+b=7
(1)
a–b=3
(2)
Adding equations (1) and (2) gives:
2a = 10
from which,
a=
Substituting in (1) gives:
10
=5
2
5+b=7
b=7–5=2
i.e.
Hence, a = 5 and b = 2
6. Solve the simultaneous equations:
8a - 3b = 51
3a + 4b = 14
© John Bird Published by Taylor and Francis
50
8a - 3b = 51
(1)
3a + 4b = 14
(2)
Multiplying equation (1) by 4 gives:
32a – 12b = 204
(3)
Multiplying equation (2) by 3 gives:
9a + 12b = 42
(4)
Equation (3) + equations (4) gives:
41a = 246
from which,
a=
246
=6
41
Substituting in (1) gives:
48 – 3b = 51
i.e.
48 - 51 = 3b
i.e.
– 3 = 3b
and
b=-1
Hence, a = 6 and b = - 1
© John Bird Published by Taylor and Francis
51