Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.A spring's extension is directly proportional to the force applied to it. A force of 5 N produces an extension of 12 mm. Work out the extension produced by a force of 20 N.
- 2.A map has a scale of 1 : 1000. A road is 300 m long in real life. Work out the length of the road on the map, in centimetres.
- 3.x × y is used to test whether two quantities are in inverse proportion. For the pairs x = 4, y = 15 and x = 6, y = 10, which statement is correct?
- 4.A recipe for 8 muffins needs 200 g of sugar. Sam wants to make 20 muffins for a bake sale, and he already has 350 g of sugar. Work out how many more grams of sugar he needs to buy.
- 5.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 6.Two mathematically similar logos are printed on a poster. They have areas 20 cm² and 45 cm². Nadia says that the length scale factor from the smaller logo to the larger logo is 45 ÷ 20 = 2.25. Give a reason why Nadia is incorrect, and work out the correct length scale factor.
- 7.A film runs for 2.25 hours. Work out the length of the film in minutes.
- 8.A charity collects donations from adults and children in the ratio 5:2. Altogether, £238 is collected. Work out how much more the adults donate than the children.
- 9.Priya is paid £58.50 for working 7.5 hours on a Saturday. Work out her rate of pay, in £ per hour.
- 10.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
- 11.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 12.A water butt is being filled from a hosepipe at a constant rate while a small leak drains water out at a constant rate, giving a constant net rate of change. The volume of water in the butt, V litres, is shown on a straight-line graph against time, t minutes. The line passes through the points (5, 20) and (15, 60). The butt is empty at t = 0. Work out how many minutes it takes to reach a volume of 100 litres.
- 13.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 14.A scale model of a shipping container is built at a scale of 1 : 30, using material with the same density as the real container. The model has a mass of 400 g. Work out the mass of the real container, giving your answer in kilograms.
- 15.A tap fills a tank at a rate of 15 litres per minute. Given that 1 litre = 1000 cm³, work out the rate at which the tank fills in cm³ per second.
Answer key
- (c) 48 mm — Method: extension = k × force, where k = extension ÷ force. Working: k = 12 ÷ 5 = 2.4 mm per N. At 20 N: extension = 2.4 × 20 = 48 mm. Wrong options: 32 mm comes from adding the extension and force numbers instead of scaling (12 + 20); 3 mm comes from treating the relationship as inverse proportion (12 × 5 ÷ 20); 36 mm comes from using an incorrect scale factor of 3 between the forces instead of the correct factor of 4 (20 ÷ 5).
- (d) 30 cm — Convert 300 m to centimetres: 300 m = 30 000 cm. The scale 1 : 1000 means the map length is the real length divided by 1000: 30 000 ÷ 1000 = 30, giving 30 cm. Dividing the unconverted 300 (metres, not centimetres) by 1000 gives 0.3 cm, far too small to be a real map measurement. Dividing by 10 000 instead of 1 000 gives 3 cm, ten times too small. Multiplying instead of dividing, 300 × 1000 = 300 000, gives 300 000 cm — using the scale the wrong way round, as if the map were bigger than real life.
- (b) They are in inverse proportion, because x × y = 60 for both pairs. — Testing inverse proportion means checking that x × y is the same for every pair: 4 × 15 = 60 and 6 × 10 = 60, so the quantities are in inverse proportion. Saying they are not in inverse proportion because x + y differs uses addition, which is not the correct test. Saying they are not in inverse proportion because y ÷ x differs uses the test for direct proportion, and finding that it differs tells us nothing about inverse proportion. Saying x × y = 40 for both pairs is an arithmetic slip: 4 × 15 = 60, not 40.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (d) 1.5 — Method: the length scale factor is the square root of the area scale factor, not the area scale factor itself. Working: the area scale factor is 45 ÷ 20 = 2.25, and the square root of 2.25 is 1.5. Answer: 1.5. Nadia's answer, 2.25, is the AREA scale factor — she never took the square root to get back to the length scale factor. 4.5 comes from doubling the area scale factor instead of taking its square root. 0.67 comes from taking the square root in the wrong direction, finding the scale factor from the larger rug to the smaller rug instead of the other way round.
- (a) 135 minutes — Method: to change hours into minutes, multiply by 60. Working: 2.25 × 60 = 135 minutes. So the film runs for 135 minutes. Distractor 145 minutes comes from reading the '.25' as 25 minutes instead of a quarter of an hour, giving 2 hours 25 minutes. Distractor 225 minutes comes from multiplying by 100 instead of 60. Distractor 150 minutes comes from rounding 2.25 hours to 2.5 hours before converting.
- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (c) 10,800 kg — Method: since the model and the real container are similar and made of the same material, mass scales with volume, so the mass scale factor is the length scale factor cubed. Working: 30³ = 27,000, so the real container's mass is 400 × 27,000 = 10,800,000 g, which is 10,800,000 ÷ 1,000 = 10,800 kg. Answer: 10,800 kg. 12 kg comes from using the length scale factor directly, 400 × 30 = 12,000 g, without cubing it. 360 kg comes from squaring the length scale factor instead of cubing it, 400 × 30² = 360,000 g. 10,800,000 kg comes from correctly cubing the scale factor but then forgetting to convert the mass from grams into kilograms.
- (d) 250 cm³/s — Method: first change litres per minute into cm³ per minute, then change per minute into per second. Working: 15 × 1000 = 15000 cm³ per minute, then 15000 ÷ 60 = 250 cm³ per second. So the tank fills at 250 cm³ per second. Distractor 15000 cm³/s comes from stopping after the first step and forgetting to change minutes into seconds. Distractor 900000 cm³/s comes from multiplying by 60 instead of dividing. Distractor 2500 cm³/s comes from dividing by 6 instead of 60.
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