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) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (a) 12/5 — If A is 5/12 of B, then B is the reciprocal of that fraction times A: flip 5/12 to get 12/5, so B is 12/5 of A. 5/12 comes from keeping the same fraction without flipping it, treating the relationship as if it works the same way in both directions. 7/12 comes from computing 1 − 5/12 = 7/12, which is not how a fraction reverses. 12/7 comes from subtracting 5 from 12 to get 7, and writing 12 over that, instead of swapping the numerator and denominator of 5/12.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
- (c) Falling at £950 per year — The gradient of a tangent on a value-age graph is a rate, in pounds per year, so −950 means the van's value is falling at £950 per year at that instant. Writing this as 950% per year mistakes a rate measured in pounds per year for a percentage — the units of a gradient come from the units on the two axes, £ and years, not from a percentage. Saying the value 'falls by £950 over the next year' treats the instantaneous rate at age 2 as if it stayed constant for a whole year, which finds an average future change, not the instantaneous rate at age 2 itself. Reading the sign the wrong way round gives 'rising at £950 per year', which would mean the van is gaining value. Always match the units of a gradient to the units on the two axes of the graph.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (a) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (a) 160 g — Method: use the ratio 20:100 to find the mass of the whole solution from the mass of acid, then take the acid away to leave the water. Working: 20:100 = 40:m, and 40 ÷ 20 = 2, so m = 2 × 100 = 200 g of solution; the water is 200 − 40 = 160 g. Answer: 160 g. The distractors: 200 g is the mass of the whole solution, which is the middle step and includes the acid the question asks you to leave out; 8 g comes from working out 20% of 40 g, which treats the 40 g as the whole solution rather than as the 20% inside it; 10 g comes from reading the 40 g as the 80% that is water, giving a solution of 50 g and a difference of 50 − 40.
- (b) 3/10 — Work out the empty space: 500 − 350 = 150 litres. Form the fraction 150/500; both numbers share a factor of 50, so 150 ÷ 50 = 3 and 500 ÷ 50 = 10, giving 3/10. 7/10 comes from writing the fraction of the tank that is full (350/500), instead of the empty space. 1/2 comes from miscalculating 500 − 350 as 250 instead of 150. 3/7 comes from comparing the empty space with the water held (150/350), instead of with the tank's total capacity.
- (c) 1 : 27 — The edge lengths are in the ratio 2 : 6, which simplifies to 1 : 3. Volumes scale with the cube of the length ratio, so the volume ratio is 1³ : 3³ = 1 : 27. Giving 1 : 3 uses the length ratio without cubing it. Giving 1 : 9 squares the length ratio, which is the rule for areas, instead of cubing it, which is the rule for volumes. Giving 27 : 1 has the ratio the right way round for larger to smaller, not smaller to larger as the question asks.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (d) 75 pages — Method: find the number of pages printed in one minute, then scale up to 10 minutes. Working: 45 ÷ 6 = 7.5 pages per minute, so 7.5 × 10 = 75 pages. Answer: 75 pages. 27 pages comes from using the ratio upside down, 45 × 6 ÷ 10, instead of finding the rate per minute first. 55 pages comes from simply adding the extra minutes, 10, onto the original number of pages, 45. 70 pages comes from rounding the rate down to 7 pages per minute before multiplying by 10, instead of using the exact rate of 7.5.
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