CHAPTER 5 PERCENTAGES
EXERCISE 23 Page 40
1. Express 0.0032 as a percentage.
0.0032 = 0.0032 × 100% = 0.32%
2. Express 1.734 as a percentage.
1.734 = 1.734 × 100% = 173.4%
3. Express 0.057 as a percentage.
0.057 = 0.057 × 100% = 5.7%
4. Express 0.374 as a percentage.
0.374 = 0.374 × 100% = 37.4%
5. Express 1.285 as a percentage.
1.285 = 1.285 × 100% = 128.5%
6. Express 20% as a decimal number.
20% =
20
= 0.02
100
7. Express 1.25% as a decimal number.
1.25% =
1.25
= 0.0125
100
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© 2014, John Bird
8. Express
11
as a percentage.
16
11 11
1100
=
×100% =
% = 68.75%
16
16 16
9. Express
5
as a percentage, correct to 3 decimal places.
13
5
5
500
=
% = 38.4615384... = 38.462%, correct to 3 decimal places
×100% =
13 13
13
10. Express as percentages, correct to 3 significant figures:
(a)
7
33
(b)
19
24
(c) 1
11
16
(a)
7
7
700
= ×100% = = 21.21212…% = 21.2%, correct to 3 significant figures
33 33
33
(b)
19 19
1900
= ×100% = =
79.1666...% = 79.2%, correct to 3 significant figures
24 24
24
11
(c)=
1
1.6875
= 1.6875 ×100% = 168.75% = 169%, correct to 3 significant figures
16
11. Place the following in order of size, the smallest first, expressing each as percentages, correct to
1 decimal place:
(a)
12
21
(b)
9
17
(c)
5
9
(d)
6
11
(a)
12 12
1200
= 57.14%
= ×100% =
21 21
21
(b)
9
9
900
= ×100% = = 52.94%
17 17
17
(c)
5 5
500
=
×100% = = 55.56%
9 9
9
(d)
6 6
600
=×100% = = 54.55%
11 11
11
Hence, the order is: (b) 52.94% (d) 54.55%
(c) 55.56% (a) 57.14%
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© 2014, John Bird
12. Express 65% as a fraction in its simplest form.
65% =
65
13
=
100
20
13. Express 31.25% as a fraction in its simplest form.
31.25% =
31.25 31.25 × 4 125 25
5
=
=
= =
100
100 × 4
400 80 16
14. Express 56.25% as a fraction in its simplest form.
56.25% =
56.25 56.25 × 4 225 45
9
=
=
= =
100
100 × 4
400 80 16
15. Evaluate A to J in the following table:
Decimal number
Fraction
0.5
C
E
G
I
0.5 =
Percentage
A
1
4
F
3
5
J
5 1 1
50%
==
×100% =
10 2 2
B
D
30
H
85
Hence, A =
1
2
and
B = 50%
1
=0.25 =0.25 ×100% =25%
4
Hence, C = 0.25 and
D = 25%
30
3
= = 0.3
100 10
Hence, E = 0.30 and
F=
3
=0.60 =0.60 ×100% =60%
5
Hence, G = 0.60 and
30% =
85% =
85 17
= = 0.85
100 20
Hence, I = 0.85 and
76
3
10
H = 60%
J=
17
20
© 2014, John Bird
16. A resistor has a value of 820 Ω ± 5%. Determine the range of resistance values expected.
5% of 820 =
5
× 820 = 41
100
The lowest value expected is 820 – 5% of 820 i.e. 820 – 41 = 779 Ω
The highest value expected is 820 + 5% of 820 i.e. 820 + 41 = 861 Ω
Hence, range of values expected is: 779 Ω to 861 Ω
77
© 2014, John Bird
EXERCISE 24 Page 41
1. Calculate 43.6% of 50 kg.
43.6% of 50 kg =
43.6
× 50 = 21.8 kg
100
2. Determine 36% of 27 m.
36% of 27 m =
36
× 27 = 9.72 m
100
3. Calculate correct to 4 significant figures:
(a) 18% of 2758 tonnes
(a) 18% of 2758 =
(b) 47% of 18.42 grams
(c) 147% of 14.1 seconds
18
× 2758 =×
18 27.58 = 496.44 t = 496.4 t, correct to 4 significant figures
100
(b) 47% of 18.42 =
47
×18.42 =
47 × 0.1842 = 8.6574 g = 8.657 g, correct to 4 significant figures
100
(c) 147% of 14.1 =
147
×14.1 =147 × 0.141 = 20.727 s, = 20.73 s, correct to 4 significant figures
100
4. When 1600 bolts are manufactured, 36 are unsatisfactory. Determine the percentage
that are unsatisfactory.
Percentage unsatisfactory =
36
36
= 2.25%
×100% =
1600
16
5. 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
(a) 140 kg as a percentage of 1 t
140
140
×100% == 14%
1000
10
(b) 47 s as a percentage of 5 min =
47
47
47
×100% = ×100% = = 15.67%
5 × 60
300
3
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© 2014, John Bird
(c) 13.4 cm as a percentage of 2.5 m =
13.4
13.4 134
= 5.36%
×100% =
=
250
2.5
25
6. A block of monel alloy consists of 70% nickel and 30% copper. If it contains 88.2 g of nickel,
determine the mass of copper in the block.
70% nickel represents 88.2 g hence 1% represents
88.2
= 1.26 g
70
Mass of copper = 30 × 1.26 = 37.8 g
7. An athlete runs 5000 m in 15 minutes 20 seconds. With intense training, he is able to reduce
this time by 2.5%. Calculate his new time.
2.5% of 15 min 20 s = 2.5% of (15 × 60 + 20) s =
2.5
× 920 = 23 s
100
Hence, his new time is: 15 min 20 s – 23 s = 14 min 57 s
8. A copper alloy comprises 89% copper, 1.5% iron and the remainder aluminium. Find the
amount of aluminium, in grams, in a 0.8 kg mass of the alloy.
Percentage aluminium = (100 – 89 – 1.5) = 9.5%
Amount of aluminium = 9.5% of 0.8 kg = 9.5% of 800 g
=
9.5
× 800 = 76 g
100
9. A computer is advertised on the internet at £520, exclusive of VAT. If VAT is payable at
17.5%, what is the total cost of the computer?
17.5% of £520 =
17.5
× 520 = £91
100
Total cost of the computer = £520 + £91 = £611
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© 2014, John Bird
10. Express 325 mm as a percentage of 867 mm, correct to 2 decimal places.
325 mm as a percentage of 867 mm =
325
×100%
867
= 37.485582…% = 37.49% correct to 2 decimal places
11. A child sleeps on average 9 hours 25 minutes per day. Express this as a percentage of the
whole day, correct to 1 decimal place.
25
60 ×100% = 39.2%
24
9
Sleep as a percentage of whole day is:
12. Express 408 g as a percentage of 2.40 kg.
408 g as a percentage of 2.40 kg is the same as 408 g as a percentage of 2400 g
i.e.
408
×100% = 17%
2400
13. 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.
% pay increase =
=
new value − original value
×100%
original value
21 500 − 15 500
6000
×100
×100 % =
15 500
15 500
= 38.7%, correct 3 significant figures
14. A metal rod 1.80 m long is heated and its length expands by 48.6 mm. Calculate the percentage
increase in length.
% length increase =
new value − original value
×100%
original value
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© 2014, John Bird
=
48.6 mm
×100 % = 2.7%
1.8 ×1000 mm
15. 12.5% of a length of wood is 70 cm. What is the full length?
12.5% of a length of wood is 12.5%
Hence, 1% of a length of wood is
and 100% of a length of wood is:
70
cm
12.5
70
×100% = 560 cm
12.5
i.e. full length of wood = 5.60 m
16. A metal rod, 1.20 m long, is heated and its length expands by 42 mm. Calculate the percentage
increase in length.
% length increase =
=
new value − original value
×100%
original value
42 mm
×100 % = 3.5%
1.20 ×1000 mm
17. For each of the following resistors, determine the (i) minimum value, (ii) maximum value:
(a) 820 Ω ± 20% (b) 47 kΩ ± 5%
(a) 20% of 820 Ω =
20
× 820 = 164 Ω
100
Hence, (i) minimum value = 820 – 164 = 656 Ω
(ii) maximum value = 820 + 164 = 984 Ω
(b) 5% of 47 kΩ =
5
× 47 = 2.35 kΩ
100
Hence, (i) minimum value = 47 – 2.35 = 44.65 kΩ
(ii) maximum value = 47 + 2.35 = 49.35 kΩ
81
© 2014, John Bird
18. An engine speed is 2400 rev/min. The speed is increased by 8%. Calculate the new speed.
8% of 2400 rev/min =
8
× 2400 = 192 rev/min
100
New speed = 2400 + 192 = 2592 rev/min
82
© 2014, John Bird
EXERCISE 25 Page 43
1.
A machine part has a length of 36 mm. The length is incorrectly measured as 36.9 mm.
Determine the percentage error in the measurement.
% error =
error
36.9 − 36
×100%
=
×100%
correct value
36
=
0.9
×100% = 2.5%
36
The percentage measurement error is 2.5% too high, which may be written as + 2.5% error
2.
When a resistor is removed from an electrical circuit the current flowing increases from 450 µA
to 531 µA. Determine the percentage increase in the current.
% error in current =
error
531 − 450
×100%
=
×100%
correct value
450
=
3.
81
×100% = 18%
450
In a shoe shop sale, everything is advertised as ‘40% off’. If a lady pays £186 for a pair of
Jimmy Choo shoes, what was their original price?
Original price of shoes =
new value
186
×100
=
×100
100 − % change
100 − 40
=
4.
186
×100 = £310
60
Over a four-year period a family home increases in value by 22.5% to £214 375. What was the
value of the house four years ago?
Original cost of house =
new value
214 375
×=
100
×100
100 + % change
100 + 22.5
=
214375
×100 = £175 000
122.5
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© 2014, John Bird
5.
An electrical retailer makes a 35% profit on all its products. What price does the retailer pay for
a dishwasher which is sold for £351?
Original cost of dishwasher =
new value
351
×100
=
×100
100 + % change
100 + 35
=
6.
351
×100 = £260
135
The cost of a sports car is £23 500 inclusive of VAT at 17.5%. What is the cost of the car
without the VAT added?
Original cost of car =
new value
23500
×=
100
×100
100 + % change
100 + 17.5
=
7.
23500
×100 = £20 000
117.5
£8000 is invested in bonds at a building society which is offering a rate of 6.75% per annum.
Calculate the value of the investment after two years.
After one year, value of investment = £8000 + 6.75% of £8000
= £8000 +
6.75
× £8000 = £8000 + £540 = £8540
100
After two years, value of investment = £8540 + 6.75% of £8540
= £8540 +
8.
6.75
× £8540 = £8540 + £576.45 = £9116.45
100
An electrical contractor earning £36 000 per annum receives a pay rise of 2.5%. He pays 22%
of his income as tax and 11% on National Insurance contributions. Calculate the increase he
will actually receive per month.
Pay rise = 2.5% of £36 000 = £900 p.a.
Pay rise per month = £900/12 = £75
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© 2014, John Bird
Total tax = 22% + 11% = 33%, hence he actually keeps 67% of his pay
67% of £75 =
9.
67
× £75 = £50.25 = the increase he will actually receive per month
100
Five mates enjoy a meal out. With drinks, the total bill comes to £176. They add a 12.5% tip
and divide the amount equally between them. How much does each pay?
Total bill = £176 +
12.5
× £176 = £176 + £22 = £198
100
Each mate therefore pays £198 ÷ 5 = £39.60
10. In December a shop raises the cost of a £920, 40-inch LCD TV, by 5%. It does not sell and in
its January sale it reduces the TV by 5%. What is the sale price of the monitor?
Cost of television in December after 5% rise = £920 +
5
× £920 = £920 + £46 = £966
100
In January this price is reduced by 5%, hence cost of television in January
= £966 –
5
× £966 = £966 – £48.30 = £917.70
100
11. A man buys a business and makes a 20% profit when he sells it three years later for £222 000.
What did he pay originally for the business?
Original house cost =
new value
222 000
×100
=
×100
100 + % change
100 + 20
=
222000
×100 = £185 000
120
12. A drilling machine should be set to 250 rev/min. The nearest speed available on the machine is
268 rev/min. Calculate the percentage overspeed.
Percentage over speed =
268 − 250
18
18 180
×100% = ×100% = = = 7.2%
250
250
2.5 25
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© 2014, John Bird
13. Two kilograms of a compound contains 30% of element A, 45% of element B, and 25% of
element C. Determine the masses of the three elements present.
Mass of element A = 30% of 2 kg =
30
60
× 2 kg = = 0.6 kg
100
100
Mass of element B = 45% of 2 kg =
45
90
× 2 kg = = 0.9 kg
100
100
Mass of element C = 25% of 2 kg =
25
50
× 2 kg = = 0.5 kg
100
100
14. A concrete mixture contains seven parts by volume of ballast, four parts by volume of sand and
two parts by volume of cement. Determine the percentage of each of these three constituents
correct to the nearest 1% and the mass of cement in a two-tonne dry mix, correct to 1
significant figure.
Ballast:sand:cement = 7:4:2
Total number of parts = 7 + 4 + 2 = 13
Percentage of ballast =
Percentage of sand =
7
700
×100% == 53.846% = 54%, correct to the nearest 1%
13
13
4
400
×100% == 30.769% = 31%, correct to the nearest 1%
13
13
Percentage of cement =
2
200
×100% == 15.38% = 15%, correct to the nearest 1%
13
13
Mass of cement in two-tonne mix = 15% of 2 = 0.3 t
15. In a sample of iron ore, 18% is iron. How much ore is needed to produce 3600 kg of iron?
Iron ore required =
3600 360 000
= 20 000 kg or 20 Mg or 20 tonnes
=
0.18
18
16. A screw’s dimension is 12.5 ± 8% mm. Calculate the possible maximum and minimum length
of the screw.
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© 2014, John Bird
Maximum length of screw = 12.5 + 8% of 12.5 = 12.5 +
Minimum length of screw = 12.5 – 8% of 12.5 = 12.5 –
8
×12.5 = 12.5 + 1 = 13.5 mm
100
8
×12.5 = 12.5 – 1 = 11.5 mm
100
17. The output power of an engine is 450 kW. If the efficiency of the engine is 75%, determine the
power input.
Efficiency =
from which,
power output
power input
i.e. 75% =
power input =
450
power input
i.e. 0.75 =
450
power input
450 45 000
= 600 kW
=
0.75
75
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© 2014, John Bird