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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Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (c) 3 — Method: find the height scale factor, cube it to find the volume (and coverage) scale factor, use it to find one large sack's coverage, then divide the total lawn area by this and round up to a whole number of sacks. Working: height scale factor = 40 ÷ 20 = 2, so coverage scale factor = 2³ = 8, and each large sack covers 30 × 8 = 240 m². 500 ÷ 240 = 2.08…, which rounds UP to 3 whole sacks. Answer: 3. 2 comes from correctly finding that each large sack covers 240 m², but then rounding 500 ÷ 240 down instead of up, which would leave part of the lawn untreated. 5 comes from squaring the height scale factor (2² = 4) instead of cubing it, giving a coverage of only 30 × 4 = 120 m² per sack. 17 comes from forgetting to scale the coverage at all and dividing 500 by the smaller sack's coverage of 30 m².
- (c) Firm B — £3 per mile against Firm A's £2 per mile — Firm A's gradient is (14 − 4) ÷ 5 = 2, so it charges £2 per mile. Firm B's gradient is (21 − 6) ÷ 5 = 3, so it charges £3 per mile. £3 is more than £2, so Firm B charges more per mile. Swapping the two firms' gradients gives the answer with Firm A at £3 and Firm B at £2, which has the labels the wrong way round. Dividing the change in miles by the change in cost, instead of the other way round, gives 5 ÷ 10 = £0.50 for Firm A and 5 ÷ 15 = £0.33 for Firm B and so names Firm A — that is the gradient upside down. And a positive fixed charge does not mean two firms charge the same rate: the rate is found from the gradient, not from whether the intercept is positive.
- (b) 4.5 m² — The height ratio is 40 : 60, which simplifies to 2 : 3, so the larger bookcase is 1.5 times as tall as the smaller one. Areas scale with the square of the length scale factor, so the wood needed scales by 1.5² = 2.25. 2 × 2.25 = 4.5, so the larger bookcase needs 4.5 m² of wood. Giving 3 m² uses the length scale factor, 1.5, without squaring it (2 × 1.5 = 3). Giving 6.75 m² cubes the scale factor, 1.5³ = 3.375, as if wood coverage were a volume (2 × 3.375 = 6.75). Giving 2.25 m² is the squared scale factor on its own, without multiplying by the smaller bookcase's wood area of 2 m².
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (d) 135 m² — Method: for area, the scale factor must be squared. Working: area scale = 300² = 90 000. 15 × 90 000 = 1,350,000 cm². Convert to m² by dividing by 10 000: 1,350,000 ÷ 10 000 = 135 m². Wrong options: 0.45 m² comes from using the linear scale factor (×300) instead of squaring it; 1,350,000 m² comes from forgetting to convert the answer from cm² to m²; 13,500 m² comes from dividing by 100 instead of 10 000 when converting units.
- (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.
- (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.
- (d) 4.7 km — Multiply the map length by the scale factor: 9.4 × 50 000 = 470 000 cm. Convert to kilometres: 470 000 cm = 4700 m = 4.7 km. Converting only to metres and calling the answer 4700 kilometres mistakes metres for kilometres. Misreading the scale as 1 : 5000 instead of 1 : 50 000, 9.4 × 5000 = 47 000 cm = 0.47 km, is ten times too small. Misplacing the decimal point in 9.4 and effectively using 94, 94 × 50 000 = 4 700 000 cm = 47 km, is ten times too big.
- (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.
- (d) £30.00 — The gradient is (130 − 70) ÷ (5 − 2) = 60 ÷ 3 = £20 per hour. Using the point (2, 70): the cost for 2 hours at £20 per hour is 20 × 2 = £40, so the call-out fee is 70 − 40 = £30. Taking the C-value of the first point as the fee without subtracting the hourly cost gives £70.00 — but that point already includes 2 hours of the hourly rate. Using the gradient itself as the fee, £20.00, confuses the rate per hour with the fixed charge. Subtracting 20 × 3 = 60 instead of 20 × 2 = 40 (using the wrong h-value) gives 70 − 60 = £10.00.
- (b) 450.00 kg — 1 m³ = 100 × 100 × 100 = 1,000,000 cm³, so 0.5 m³ = 500,000 cm³. Mass = density × volume = 0.9 × 500,000 = 450,000 g. Converting to kilograms by dividing by 1000 gives 450,000 ÷ 1000 = 450.00 kg. Skipping the m³-to-cm³ conversion and multiplying 0.9 × 0.5 = 0.45 treats the volume as if it were already 0.5 cm³, giving 0.45 kg. Finding the mass correctly in grams, 450,000 g, but not converting to kilograms leaves 450000.00 kg, out by a factor of 1000. Using the area conversion factor of 10,000, as if converting m² to cm², instead of the volume factor of 1,000,000 gives 0.5 × 10,000 = 5,000 'cm³', and a mass of 0.9 × 5,000 = 4,500 g, which is 4.50 kg.
- (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.
- (d) £144 — Method: find the length (perimeter) scale factor by taking the square root of the area ratio, then apply it to the cost. Working: 12 : 27 simplifies to 4 : 9, and the square root of each part gives the length ratio 2 : 3, so the scale factor from the smaller to the larger pond is 3 ÷ 2 = 1.5. Cost = £96 × 1.5 = £144. Answer: £144. £216 comes from using the area ratio itself as the cost ratio, £96 × (27 ÷ 12) = £216, without taking the square root. £64 comes from using the length ratio the wrong way round, £96 × (2 ÷ 3) = £64. £111 comes from simply adding the difference in area, 27 − 12 = 15, onto the original cost, £96 + £15 = £111, instead of scaling proportionally.
- (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.
- (d) The line is straight and passes through the origin. — Direct proportion means y = kx for a constant k. This is a straight line, and when x = 0, y = 0, so it passes through the origin. For a positive k it slopes upward from left to right.
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