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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (b) The ratio C : m is not constant because the formula includes a fixed charge of £3 as well as the charge per mile. — For the ratio C : m to stay constant, C must be directly proportional to m, i.e. C = km with no constant term. Because of the +3 fixed charge, C is not directly proportional to m: for example m = 1 gives C = 5.5 (ratio 5.5 : 1), while m = 10 gives C = 28 (ratio 2.8 : 1) — the ratio has changed.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (c) 32 km/h — Method: to change mph into km/h, multiply by the number of kilometres in a mile. Working: 20 × 1.6 = 32 km/h. So the cyclist's speed is 32 km/h. Distractor 12.5 km/h comes from dividing by 1.6 instead of multiplying. Distractor 21.6 km/h comes from adding 1.6 instead of multiplying by it. Distractor 20 km/h comes from not converting the units at all.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (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.
- (c) 36 cm — A perimeter is built from lengths, so it is multiplied by the length scale factor and not by the square of it. The scale factor from the smaller rectangle to the larger one is 10 ÷ 5 = 2. The larger perimeter is therefore 18 × 2 = 36 cm.
- (c) 4 : 5 — Method: a formula of the form y = kx says that for every 1 of x there are k of y, so the ratio y : x is k : 1, which is then written with whole numbers and cancelled. Working: here k = 0.8, so y : x = 0.8 : 1; multiplying both parts by 10 gives 8 : 10, and dividing both parts by their highest common factor 2 gives 4 : 5. Answer: 4 : 5. The distractors: 5 : 4 comes from writing x : y, reversing the order asked for; 8 : 10 has the decimal cleared but the ratio left unsimplified, and the question asks for the simplest form; 8 : 1 comes from multiplying only the left-hand part by 10 and leaving the 1 unchanged, which changes the relationship instead of keeping it.
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (d) 64 — Since y is directly proportional to √x, y = k√x. Using x = 25, y = 20: √25 = 5, so 20 = k × 5, giving k = 20 ÷ 5 = 4. The equation is y = 4√x. When y = 32: √x = 32 ÷ 4 = 8, and x = 8² = 64. Stopping at √x = 8 without squaring leaves the square root of x, not x itself. Treating the relationship as if y were proportional to x itself gives k = 20 ÷ 25 = 0.8 and then x = 32 ÷ 0.8 = 40, which is a different relationship entirely. Multiplying instead of dividing when isolating √x gives √x = 32 × 4 = 128, far too large to be a square root here. When y = 32, x = 64.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (d) 13/10 — First find the selling price: £150 + £45 = £195. Put the selling price over the cost price: 195/150. Divide both numbers by their highest common factor, 15: 195÷15 = 13, 150÷15 = 10, giving 13/10. (10/13 comes from writing the prices the wrong way round. 3/10 is just the profit written as a fraction of the cost price, 45/150, not the selling price. 13/23 comes from comparing the selling price to the combined total of the cost price and the selling price, 195/345.)
- (b) They are in inverse proportion, because x × y = 60 for both pairs. — Testing inverse proportion means checking that x × y is the same for every pair: 4 × 15 = 60 and 6 × 10 = 60, so the quantities are in inverse proportion. Saying they are not in inverse proportion because x + y differs uses addition, which is not the correct test. Saying they are not in inverse proportion because y ÷ x differs uses the test for direct proportion, and finding that it differs tells us nothing about inverse proportion. Saying x × y = 40 for both pairs is an arithmetic slip: 4 × 15 = 60, not 40.
- (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.
- (c) 14 litres — Method: find the amount of fuel used per km first, then use it to find the fuel needed for 175 km. Working: 24 ÷ 300 = 0.08 litres per km, and 0.08 × 175 = 14 litres. So 14 litres are needed. Distractor 24 litres comes from assuming the same amount of fuel is used no matter the distance, without scaling. Distractor 21 litres comes from misreading the original distance as 200 km instead of 300 km. Distractor 1.4 litres comes from a decimal-point slip, giving an answer ten times too small.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
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