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.
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Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (b) 4 — Method: find the length scale factor by taking the cube root of the volume scale factor, then square it to get the area scale factor. Working: 8 = 2³, so the length scale factor is 2, and the area scale factor is 2² = 4. Answer: 4. 8 comes from using the volume scale factor itself as if it were the area scale factor. 64 comes from squaring the volume scale factor, 8² = 64, instead of first taking its cube root. 2 comes from correctly finding the length scale factor but then forgetting to square it.
- (c) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (b) 1.00 litres — Total volume = 350 + 650 = 1000 cm³. Since 1000 cm³ = 1 litre, the smoothie is 1.00 litre. Using only the orange juice's 650 cm³ and converting that gives 0.65 litres, forgetting the mango juice entirely. Using only the mango juice's 350 cm³ gives 0.35 litres, forgetting the orange juice. Adding both volumes correctly to get 1000 cm³ but not converting to litres leaves the answer as 1000.00, which is the volume in the wrong unit.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (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.
- (c) 40 minutes — Method: this is inverse proportion — fewer taps means longer, not shorter — so the number of taps × the time taken stays constant. Working: 6 × 20 = 120, and with 3 taps the time is 120 ÷ 3 = 40 minutes. So 3 taps take 40 minutes. Distractor 10 minutes comes from treating it as direct proportion instead of inverse, working out 20 × 3 ÷ 6. Distractor 30 minutes comes from halving the number of taps and adding half the original time, 20 + 10, instead of doubling the time. Distractor 17 minutes comes from subtracting the number of taps removed, 3, directly from the original time, 20.
- (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.
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
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