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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (c) 2 hours — Method: time = distance ÷ speed, and the units of the speed fix the unit of the time. Working: 120 ÷ 60 = 2, and because the speed is in kilometres per hour the time is in hours. Answer: 2 hours. The distractors: 0.5 hours comes from dividing the speed by the distance, 60 ÷ 120, the division the wrong way round; 2 minutes comes from the correct division but the wrong unit, reading kilometres per hour as kilometres per minute; 180 minutes comes from adding 120 and 60 instead of dividing.
- (a) 5 — Method: substitute the value of y into the equation and undo the multiplication by dividing both sides by the constant. Working: 35 = 7x, so x = 35 ÷ 7 = 5. Answer: 5. The distractors: 245 comes from multiplying 35 by 7 instead of dividing, which undoes nothing; 28 comes from working out 35 − 7, reading y = 7x as y = x + 7; 7 is the constant itself, read straight off the equation and given as the value of x.
- (c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
- (d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
- (c) Yes, correct, because 24 ÷ 8 × 3 = 9. — Method: divide the count you know by its ratio part to find the value of one part, then multiply by the part you want. Working: one part = 24 ÷ 8 = 3 counters, so green = 3 × 3 = 9 counters, and the student is correct. Dividing 24 by 3 instead of 8 gives 24 ÷ 3 × 8 = 64, the ratio parts the wrong way round. Stopping after 24 ÷ 8 = 3 gives 3, forgetting to multiply by the green ratio part. Adding 24 + 3 = 27 confuses adding a ratio part with scaling by it.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (a) 0.8 — The gradient of a line through the origin is the y-coordinate of a point divided by its x-coordinate: 20 ÷ 25 = 0.8. Choosing 1.25 comes from dividing the wrong way round, 25 ÷ 20. Choosing 20 comes from reading off the cost at the point instead of dividing it by the number of miles. Choosing 5 comes from subtracting the two coordinates (25 − 20) instead of dividing them.
- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (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.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (a) 3/8 — The whole prize fund is split into 3 + 5 = 8 equal parts. The smaller winner receives 3 of these parts, so their share is 3/8 of the whole fund. Taking the larger number of parts, 5, as the numerator instead gives 5/8, the larger winner's share. Writing the ratio numbers directly as a fraction without adding them, 3/5, treats the ratio as a fraction of the OTHER share rather than of the whole. Inverting the ratio gives 5/3, which is not even a valid fraction of a whole, since it is greater than 1.
- (a) 1 : 2 000 000 — Method: a scale is a ratio between two lengths written in the same unit, reduced so that the map distance is 1. Working: 120 km = 120 × 1000 × 100 = 12 000 000 cm, so the ratio is 6 : 12 000 000, and dividing both parts by 6 gives 1 : 2 000 000. Answer: 1 : 2 000 000. The distractors: 1 : 12 000 000 comes from writing the real length in centimetres without dividing by the 6 cm on the map; 1 : 200 000 comes from taking 120 km as 1 200 000 cm, one conversion step short, before dividing by 6; 1 : 20 000 comes from converting 120 km to 120 000 m and treating those metres as centimetres.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (c) 10 g — Method: mass = density × volume. Working: 2 × 5 = 10, in grams because the density is in grams per cubic centimetre. Answer: 10 g. The distractors: 2.5 g comes from dividing the volume by the density, 5 ÷ 2, instead of multiplying; 0.4 g comes from dividing the density by the volume, 2 ÷ 5; 7 g comes from adding the density and the volume.
- (d) 2:5 — Multiply both parts by 10 to clear the decimals: 6 : 15. Divide both parts by their common factor 3: 2 : 5.
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