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) 4.7 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.
- (c) 1.25 cm — Method: find the real length using the first map's scale, then use the second map's scale to find its drawn length. Working: real length = 5 × 20 000 = 100 000 cm. On the second map: 100 000 ÷ 80 000 = 1.25 cm. Wrong options: 20 cm comes from inverting the ratio of the two scales (5 × 80 000 ÷ 20 000); 5 cm comes from wrongly assuming the length looks the same on both maps; 12.5 cm comes from dropping a zero from the second scale factor and dividing by 8 000 instead of 80 000 (100 000 ÷ 8 000).
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
- (d) 1.5 — Method: the length scale factor is the square root of the area scale factor, not the area scale factor itself. Working: the area scale factor is 45 ÷ 20 = 2.25, and the square root of 2.25 is 1.5. Answer: 1.5. Nadia's answer, 2.25, is the AREA scale factor — she never took the square root to get back to the length scale factor. 4.5 comes from doubling the area scale factor instead of taking its square root. 0.67 comes from taking the square root in the wrong direction, finding the scale factor from the larger rug to the smaller rug instead of the other way round.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (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.
- (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) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (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.
- (a) £11.70 — Rate of pay = total pay ÷ hours worked, so £105.30 ÷ 9 = £11.70 per hour. Working out £105.30 − 9 = £96.30 subtracts the number of hours from the total pay instead of dividing. Working out £105.30 × 9 = £947.70 multiplies total pay by hours worked instead of dividing. Misplacing the decimal point in the correct answer gives £117.00 instead of £11.70. Maya's rate of pay is £11.70 per hour.
- (c) Firm B — £3 per mile against Firm A's £2 per mile — Firm A's gradient is (14 − 4) ÷ 5 = 2, so it charges £2 per mile. Firm B's gradient is (21 − 6) ÷ 5 = 3, so it charges £3 per mile. £3 is more than £2, so Firm B charges more per mile. Swapping the two firms' gradients gives the answer with Firm A at £3 and Firm B at £2, which has the labels the wrong way round. Dividing the change in miles by the change in cost, instead of the other way round, gives 5 ÷ 10 = £0.50 for Firm A and 5 ÷ 15 = £0.33 for Firm B and so names Firm A — that is the gradient upside down. And a positive fixed charge does not mean two firms charge the same rate: the rate is found from the gradient, not from whether the intercept is positive.
- (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.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (c) 183 mg — Apply the decay, then add the new dose, once for each hour. Hour 1: 0.7 × 200 = 140, then 140 + 50 = 190. Hour 2: 0.7 × 190 = 133, then 133 + 50 = 183, so there is 183 mg after 2 hours. Forgetting the top-up dose and only applying the decay gives 0.7 × 200 = 140, then 0.7 × 140 = 98 — this ignores that a further 50 mg is given every hour. Adding the 50 mg BEFORE the decay is applied, instead of after, gives 0.7 × (200 + 50) = 175, then 0.7 × (175 + 50) = 157.5, which changes how much of the dose is eliminated in the same hour it is given. Multiplying by 0.3, the percentage ELIMINATED, instead of by 0.7, the percentage REMAINING, gives 0.3 × 200 + 50 = 110, then 0.3 × 110 + 50 = 83 — this mixes up the amount that leaves the bloodstream with the amount that stays in it. Always check whether a percentage describes what remains or what is removed before choosing the multiplier.
- (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.
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