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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- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (b) Takings rise about £14 per 1°C rise — The gradient here is positive, so as temperature rises, takings rise too: near 22°C, takings increase by about £14 for every 1°C rise in temperature. Reversing this to say takings rise for every 1°C FALL gets the direction of the independent variable backwards — a positive gradient means both quantities move the same way. Saying 'takings are £14 at 22°C' confuses the gradient, a rate of change, with the y-value on the graph, which is the takings itself. Saying takings 'rose £14 in total' from 0°C to 22°C treats the gradient at a single point as if it applied over the whole range from 0°C to 22°C, when it only describes the instant at 22°C. Always keep a rate, a total change and a single reading separate.
- (d) Faster at t = 8s — still accelerating — The gradient of a tangent on a distance-time graph is the instantaneous speed, in m/s. At t = 2 seconds the speed is 6 m/s; at t = 8 seconds it is 9.5 m/s, which is faster, so the sprinter is still accelerating between these two times. Saying the sprinter is slower at t = 8s reverses the comparison — 9.5 is greater than 6, not less. Writing 9.5 − 6 = 3.5 and calling this 'metres further covered' turns the difference of two speeds into a distance, which the units do not support: a difference of two speeds is itself a speed, not a distance. Taking 9.5 m/s, the larger of the two instantaneous speeds, as the average speed for the whole race confuses a speed at one instant with an average over the whole distance, which would need the total distance and total time, not two tangent gradients.
- (a) 11:00 to 12:00 — Method: the instantaneous rate of change is zero at a turning point, where a rising trend becomes a falling trend; that lies within the first interval whose difference has changed sign from the interval before it. Working: the differences between consecutive readings are +3.4 °C (09:00 to 10:00), +1.7 °C (10:00 to 11:00), −0.3 °C (11:00 to 12:00) and −2.6 °C (12:00 to 13:00); the sign changes from positive to negative within 11:00 to 12:00, since the temperature is still rising up to 11:00 (20.1 °C, the highest recorded value) and has fallen by 12:00, so the instantaneous rate of change was zero somewhere within that interval. Choosing 09:00 to 10:00 picks out the interval with the largest positive difference, +3.4 °C, confusing the fastest rise with no change at all. Choosing 10:00 to 11:00 picks the last interval where the temperature was still rising, one interval too early, without checking that the very next interval turns negative. Choosing 12:00 to 13:00 picks out the interval with the largest-magnitude difference, −2.6 °C, the fastest fall, not where the change is zero. Zero instantaneous rate of change happens at a turning point, where the readings stop rising and start falling — find the FIRST interval whose difference has flipped sign from the one before it, not the biggest change or an interval where the old sign still held.
- (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.
- (c) They are in direct proportion, because y ÷ x = 2.5 for both pairs. — Testing direct proportion means checking that y ÷ x is the same for every pair: 15 ÷ 6 = 2.5 and 25 ÷ 10 = 2.5, so the quantities are in direct proportion. Saying they are not in proportion because x + y differs uses addition, which is not the correct test for proportion. Saying they are not in proportion because y − x differs also uses the wrong test — subtraction, not division. Saying they are in proportion because x × y is 90 and 250 uses multiplication, which is the test for inverse proportion, and the two products are not even equal to each other, so this option also contradicts itself.
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
- (d) 100 — Rearrange L = 0.6L + 40 by collecting the L terms on one side: L − 0.6L = 40, which gives 0.4L = 40, then L = 40 ÷ 0.4 = 100. Subtracting the other way round, 0.6L − L = 40, gives −0.4L = 40, then L = 40 ÷ (−0.4) = −100 — a sign error that flips the answer negative even though a long-run value here must be positive. Ignoring the 0.6L term completely and solving L = 40 directly gives 40, which throws away the recurrence's own multiplier. Dividing 40 by 0.6 instead of by the correct coefficient 0.4 gives 40 ÷ 0.6 ≈ 66.7, a slip that comes from dividing by the coefficient of L on the RIGHT of the original equation rather than by what is left once the L terms are collected on one side. Always collect the L terms first, then divide by whatever coefficient of L remains.
- (c) £7617.60 — To decrease by 8% each year, multiply by 0.92 (100% − 8%) twice. £9000 × 0.92 × 0.92 = £7617.60. £7560.00 comes from treating the two 8% decreases as a single flat 16% decrease applied once instead of compounding: £9000 × 0.84 = £7560.00. £8280.00 comes from applying the 8% decrease only once, for 1 year instead of 2: £9000 × 0.92 = £8280.00. £10497.60 comes from multiplying by 1.08 twice, increasing the value instead of decreasing it: £9000 × 1.08 × 1.08 = £10497.60.
- (c) 4 m/s — The average rate of change of distance with respect to time over an interval is the change in distance divided by the change in time — the gradient of the chord joining the two endpoints, not the gradient of any tangent inside the interval. From t = 3 to t = 8 the change in time is 8 − 3 = 5 and the change in distance is 32 − 12 = 20, so the average speed is 20 ÷ 5 = 4 m/s. Reporting the change in distance on its own, as 20 m/s, is not a speed: those 20 metres were covered over the whole 5 seconds, not in one second, so the 20 still has to be divided by the 5. The tangent's gradient of 3 m/s is the instantaneous speed at the single moment t = 6, not the average over the whole 5-second interval, so it must not be used here. Adding the change in distance and the change in time instead of dividing gives 20 + 5 = 25, which is not a speed. The average speed of the cyclist over the interval is 4 m/s.
- (c) 1.05 litres — Method: find the value of one part of the ratio from the total volume, then find the share for pineapple juice. Working: the ratio 5:3:2 has 5 + 3 + 2 = 10 parts, so one part is 3.5 ÷ 10 = 0.35 litres, and the pineapple juice is 3 × 0.35 = 1.05 litres. So 1.05 litres of pineapple juice is needed. Distractor 1.75 litres is the volume of orange juice, not pineapple juice. Distractor 0.7 litres is the volume of lemonade, not pineapple juice. Distractor 0.35 litres is the value of one part, found correctly but never multiplied by 3.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (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.
- (a) 46.8 km/h — Method: 1 hour = 60 minutes, so a rate given in kilometres per minute is converted to kilometres per hour by multiplying by 60. Working: 0.78 × 60 = 46.8, so the instantaneous rate of change is 46.8 kilometres per hour. Keeping the given value unchanged and only relabelling the unit gives 0.78 km/h, which ignores that the time unit has changed. Dividing by 60 instead of multiplying — as you would when converting to a larger length unit — gives 0.78 ÷ 60 = 0.013 km/h, the wrong direction for a rate measured against a larger time unit. Adding 60 to the given rate instead of multiplying by it gives 0.78 + 60 = 60.78 km/h. Converting a rate always means multiplying or dividing by the conversion factor between the units, never adding it, and the direction depends on whether the new time unit is bigger or smaller than the old one.
- (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.
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