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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- (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.
- (d) 400 — The area scale factor is the length scale factor squared: 20² = 400, so the real car's surface area is 400 times the model's. 20 comes from using the length scale factor itself, without squaring it. 8000 comes from cubing the length scale factor (20³), instead of squaring it — cubing is the rule for volume, not area. 40 comes from doubling the length scale factor (2 × 20), instead of squaring it.
- (d) 40 — Substitute x = 5 and y = 8 into y = k ÷ x to get 8 = k ÷ 5, so k = 8 × 5 = 40. Getting 13 comes from adding the two numbers (5 + 8) instead of multiplying. Getting 1.6 comes from dividing 8 by 5 instead of multiplying. Getting 3 comes from subtracting the two numbers (8 − 5) instead of multiplying.
- (d) 600 g — Find the flour used for one person: 240 ÷ 6 = 40 g per person. Then scale up to 15 people: 40 × 15 = 600 g. Dividing 240 by 15 and then multiplying by 6 gives 96 g — that uses the ratio the wrong way round, scaling down instead of up. Stopping at 40 g only gives the amount for one person, not for 15. Adding 15 − 6 = 9 to 240 gives 249 g, which treats the number of people and the grams of flour as if they were the same kind of quantity — proportion means scaling, not adding. The recipe needs 600 g of flour.
- (a) 2 km — Method: multiply the map distance by the scale to get the real distance in centimetres, then convert centimetres to kilometres using 100 cm = 1 m and 1000 m = 1 km. Working: 4 × 50 000 = 200 000 cm; 200 000 ÷ 100 = 2000 m; 2000 ÷ 1000 = 2. Answer: 2 km. The distractors: 200 km comes from dividing the 200 000 cm by 1000 in a single step, as if a kilometre were 1000 cm rather than the 100 000 cm it is; 20 km comes from converting to metres correctly, 200 000 ÷ 100 = 2000 m, and then dividing those metres by 100 instead of by 1000; 0.2 km comes from dividing by 1000 to reach metres, as if a metre were 1000 cm, and then dividing by 1000 again, so 200 000 is divided by 1 000 000 altogether.
- (c) 3 : 7 — Method: a ratio 'of A to B' is written with A first, and it is simplified by dividing both parts by their highest common factor. Working: vans are named first, so the ratio starts as 15 : 35; the highest common factor of 15 and 35 is 5, and 15 ÷ 5 = 3 while 35 ÷ 5 = 7. Answer: 3 : 7. The distractors: 7 : 3 comes from writing the two numbers in the order they appear in the question, cars before vans, instead of the order asked for; 3 : 10 comes from comparing the vans with the total number of vehicles, 15 : 50, a part-to-whole ratio in place of the part-to-part ratio asked for; 15 : 35 is the right comparison left unsimplified, and the question asks for the simplest form.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (b) 1 : 2.25 — Method: to write a ratio in the form 1 : n, divide both parts by the first part. Working: 4 ÷ 4 = 1 and 9 ÷ 4 = 2.25, so 4 : 9 = 1 : 2.25. Working out 9 ÷ 4 = 2.25 correctly but then writing it as the first part gives 2.25 : 1, the two parts the wrong way round. Subtracting 9 − 4 = 5 gives 1 : 5, confusing the difference between the parts with the ratio. Multiplying 4 × 9 = 36 gives 1 : 36, confusing the product of the parts with the ratio.
- (d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (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.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (b) 12 — Method: work out what one part of the ratio is worth, then multiply that by the number of parts in the other share. Working: the smaller number matches the 2 parts, so 8 ÷ 2 = 4 for one part, and 3 × 4 = 12. Answer: 12. The distractors: 16 comes from multiplying 8 by 2, the ratio part that belongs to the smaller number; 24 comes from multiplying 8 by 3 without first finding the value of one part; 9 comes from adding the difference between the ratio parts, 3 − 2 = 1, to 8.
- (b) £50 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then multiply by the parts in the largest share. Working: 3 + 4 + 5 = 12 parts, £120 ÷ 12 = £10 for one part, and the largest share is 5 parts, so 5 × £10 = £50. Answer: £50. The distractors: £10 is the value of one part only; £30 is the 3-part share, which is the smallest one; £40 is the 4-part share, the middle one.
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