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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- (d) £4 — The gradient of the tangent gives the instantaneous rate of change of cost with respect to the number of passengers, in pounds per passenger. The tangent passes through (16, 184) and (24, 216), so the change in cost is 216 − 184 = 32 and the change in passengers is 24 − 16 = 8. The gradient is 32 ÷ 8 = 4. Stopping after finding the change in cost, without dividing by the change in passengers, leaves 32, not a rate. Adding the two changes instead of dividing gives 32 + 8 = 40, which is not a rate either. Reading off only the change in passengers, 8, is not a rate at all — a rate needs the change in cost as well. The instantaneous rate is £4 per extra passenger.
- (d) 10 — First find the height at each end of the chord. At t = 1, h = 30 × 1 − 5 × 1² = 30 − 5 = 25. At t = 3, h = 30 × 3 − 5 × 3² = 90 − 45 = 45. The gradient of the chord estimates the instantaneous rate at the midpoint t = 2: 45 − 25 = 20, then 20 ÷ (3 − 1) = 20 ÷ 2 = 10 m/s. Finding the change in height but forgetting to divide by the change in time gives 20, which is a distance, not a rate. Averaging the two heights instead of finding the difference gives (25 + 45) ÷ 2 = 70 ÷ 2 = 35. Subtracting in the wrong order, 25 − 45 = −20, then −20 ÷ 2 = −10, gives the correct size with the sign flipped — the rocket is rising, not falling, at t = 2 seconds, so a negative rate cannot be right here.
- (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) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (c) 40 — Working out successive terms shows where the sequence is heading, but the terms themselves keep changing — the limit is the value where the sequence stops changing, so x_{n+1} = x_n = L there. Substituting into the rule: L = 0.5L + 20. Subtracting 0.5L from both sides: L − 0.5L = 20, so 0.5L = 20, and L = 20 ÷ 0.5 = 40. The individual terms are x_1 = 0.5 × 0 + 20 = 20, x_2 = 0.5 × 20 + 20 = 30 and x_3 = 0.5 × 30 + 20 = 35, getting closer to this value but not equal to it — 35 is only the third term, not the limit. Multiplying by 0.5 instead of dividing at the final step, 20 × 0.5 = 10, undoes the rearrangement rather than completing it, and gives a value smaller than terms the sequence has already passed. Writing the fixed-point equation with the wrong sign, L = 0.5L − 20, gives 0.5L = −20 and L = −40, which cannot be right since every term in the sequence is positive and increasing. The value the sequence is approaching is 40.
- (a) 80 km/h — Average speed = total distance ÷ total time. Total distance = 45 + 75 = 120 km. Total time = 30 minutes + 1 hour = 1.5 hours. 120 ÷ 1.5 = 80 km/h. 60 km/h comes from treating the 30 minutes as a whole hour, giving a total time of 2 hours instead of 1.5 (120 ÷ 2). 82.5 km/h comes from averaging the two separate speeds (45 ÷ 0.5 = 90 km/h and 75 ÷ 1 = 75 km/h, then (90 + 75) ÷ 2) instead of using total distance over total time. 75 km/h comes from using only the second part of the journey (75 km in 1 hour) and ignoring the first part.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (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.
- (b) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
- (a) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (a) 2 : 3 — Simplify the volume ratio first: 64 : 216 divides by 8 to give 8 : 27. Volumes scale with the cube of the height ratio, so take the cube root of each part: the cube root of 8 is 2, and the cube root of 27 is 3, giving a height ratio of 2 : 3. Giving 3 : 2 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 8 : 27 is the simplified volume ratio, without cube-rooting it. Giving 64 : 216 is the volume ratio before it has even been simplified.
- (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.
- (d) 36 mph — First convert 1 hour 30 minutes to hours: 30 minutes is half an hour, so the time is 1.5 hours. Then divide the distance by the time: 54 ÷ 1.5 = 36 mph. Reading 1 hour 30 minutes as 1.3 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 54 ÷ 1.3 ≈ 41.54 mph. Working out 54 ÷ 30 = 1.8 divides by the number of minutes only, ignoring the hour. Working out 54 × 1.5 = 81 multiplies by the time instead of dividing. The coach's average speed is 36 mph.
- (b) 3 km — Method: multiply by the scale factor to get the real length in centimetres, then convert to kilometres. Working: 7.5 × 40 000 = 300 000 cm. 300 000 ÷ 100 000 = 3 km. Wrong options: 30 km comes from dividing by 10 000 instead of 100 000 when converting to kilometres; 3000 km comes from dividing by 100 instead of 100 000; 0.3 km comes from dividing by 1 000 000, an extra factor of 10 too many.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
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