Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.The concentration of a pollutant in a lake, in arbitrary units, follows the recurrence C_{n+1} = 0.75C_n + 40, with C_0 = 500: each week, 25% of the pollutant breaks down naturally, and then a further 40 units enter the lake from run-off. An ecologist classifies the lake as safe once C_n first drops below 300. Find the first whole number of weeks after which the lake is safe.
- 2.Ffion is paid £11.20 per hour. She works 6 hours on Monday and 4.5 hours on Tuesday. Work out her total pay for the two days.
- 3.Write 400 g : 1.5 kg as a ratio in its simplest form.
- 4.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.
- 5.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 6.A shop sells identical pens. The total cost, C pence, of n pens is given by C = 12n. Write the ratio n : C in its simplest form.
- 7.A jug holds 3 litres of a drink that is 60% fruit juice. 1 litre of water is added to the jug. Work out the percentage of the new mixture that is fruit juice.
- 8.A cyclist rides 19.3 km in 47 minutes. Work out her average speed, in km/h, to 1 decimal place.
- 9.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 10.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.
- 11.An acid solution has a concentration of 20%. The solution contains 40 g of acid, and the rest of the solution is water. Work out the mass of water in the solution.
- 12.A cyclist's distance travelled, in metres, is recorded against time, in seconds. From t = 3 to t = 8 the distance increases from 12 m to 32 m. A tangent to the distance–time graph at t = 6 has gradient 3. Work out the average speed of the cyclist over the interval from t = 3 to t = 8.
- 13.A gardener mixes 300 ml of plant feed concentrate with 1.2 litres of water to make a spray. Write the ratio of concentrate to water in its simplest form.
- 14.A company's turnover was £128 000 in 2023 and £176 000 in 2024. Work out the multiplier that changes the 2023 turnover into the 2024 turnover.
- 15.The number of tickets a group can afford is inversely proportional to the price per ticket. At £4 per ticket, the group can afford 12 tickets. Work out how many tickets the group can afford at £6 per ticket.
Answer key
- (a) 4 weeks — Apply the recurrence week by week. C_1 = 0.75 × 500 + 40 = 375 + 40 = 415. C_2 = 0.75 × 415 + 40 = 311.25 + 40 = 351.25. C_3 = 0.75 × 351.25 + 40 = 263.4375 + 40 = 303.4375. C_4 = 0.75 × 303.4375 + 40 = 227.578125 + 40 = 267.578125. C_3 = 303.4375 is still above 300, but C_4 = 267.58 has dropped below it, so the lake first becomes safe after 4 weeks. Taking 25% of the ORIGINAL 500 every week instead of 25% of the current amount, a flat 125 each time, gives 500 − 125 + 40 = 415, then 415 − 125 + 40 = 330, then 330 − 125 + 40 = 245, which crosses 300 a week too early and gives the wrong answer of 3 weeks. Continuing one extra step to C_5 = 0.75 × 267.578125 + 40 = 200.68 + 40 = 240.68 and calling it 5 weeks overshoots, since the concentration had already dropped below 300 at C_4. Forgetting the 40 units of run-off each week and only applying the decay gives C_1 = 0.75 × 500 = 375, then C_2 = 0.75 × 375 = 281.25 — this is already below 300 after only 2 weeks, because without the run-off the concentration falls much faster.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (a) 4 : 15 — Convert to the same unit first: 1.5 kg = 1500 g, since 1 kg = 1000 g. This gives the ratio 400 : 1500. Divide both parts by their highest common factor, 100, to get 4 : 15. Giving 40 : 150 divides by 10 only, which is not the highest common factor, so it is not fully simplified. Giving 15 : 4 swaps the order. Giving 4 : 1.5 has not converted 1.5 kg into grams, so the two parts are not measured in the same unit.
- (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.
- (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.
- (b) 1:12 — C = 12n means that for every 1 pen there are 12 pence of cost, so n : C = 1 : 12, and the highest common factor of 1 and 12 is 1, so this is already in its simplest form. Writing C : n instead of n : C gives 12 : 1, the ratio the wrong way round. Reading C = 12n as '12 more than n' instead of '12 times n', so n = 1 gives C = 13, gives 1 : 13, from adding instead of multiplying. Choosing n = 12, so C = 12 × 12 = 144, gives the correct pair of values n : C = 12 : 144, but the ratio the right way round left unsimplified — 12 and 144 share a common factor of 12, which has not been cancelled.
- (c) 45% — Method: adding water changes the total volume but not the amount of fruit juice, so find the juice, find the new total volume, and write the first as a percentage of the second. Working: 3 × 0.6 = 1.8 litres of fruit juice; the new volume is 3 + 1 = 4 litres; 1.8 ÷ 4 = 0.45, which is 45%. Answer: 45%. The distractors: 60% is the strength before the water goes in, and assumes that adding water leaves the strength unchanged; 15% comes from dividing the 60% by the 4 litres of mixture instead of dividing the 1.8 litres of juice by the 4 litres; 75% is the fraction of the new mixture that came out of the original jug, 3 litres out of 4, which ignores that only 60% of that 3 litres was juice.
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (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.
- (a) 160 g — Method: use the ratio 20:100 to find the mass of the whole solution from the mass of acid, then take the acid away to leave the water. Working: 20:100 = 40:m, and 40 ÷ 20 = 2, so m = 2 × 100 = 200 g of solution; the water is 200 − 40 = 160 g. Answer: 160 g. The distractors: 200 g is the mass of the whole solution, which is the middle step and includes the acid the question asks you to leave out; 8 g comes from working out 20% of 40 g, which treats the 40 g as the whole solution rather than as the 20% inside it; 10 g comes from reading the 40 g as the 80% that is water, giving a solution of 50 g and a difference of 50 − 40.
- (c) 4 m/s — The average rate of change of distance with respect to time over an interval is the change in distance divided by the change in time — the gradient of the chord joining the two endpoints, not the gradient of any tangent inside the interval. From t = 3 to t = 8 the change in time is 8 − 3 = 5 and the change in distance is 32 − 12 = 20, so the average speed is 20 ÷ 5 = 4 m/s. Reporting the change in distance on its own, as 20 m/s, is not a speed: those 20 metres were covered over the whole 5 seconds, not in one second, so the 20 still has to be divided by the 5. The tangent's gradient of 3 m/s is the instantaneous speed at the single moment t = 6, not the average over the whole 5-second interval, so it must not be used here. Adding the change in distance and the change in time instead of dividing gives 20 + 5 = 25, which is not a speed. The average speed of the cyclist over the interval is 4 m/s.
- (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (a) 1.375 — The multiplier is the new value divided by the original value: 176 000 ÷ 128 000 = 1.375. Choosing 0.375 comes from working out the increase as a fraction of the original (48 000 ÷ 128 000 = 0.375) and stopping there, without adding on the 1 whole that represents the original turnover. Choosing 0.727 comes from dividing the 2023 turnover by the 2024 turnover, the wrong way round (128 000 ÷ 176 000, rounded to 3 decimal places). Choosing 137.5 comes from writing the multiplier as a percentage — the 2024 turnover is 137.5% of the 2023 turnover — instead of converting it to the decimal multiplier 1.375.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
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