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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- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (a) 15.3 litres — Squash : water = 2 : 9, so water is 9 ÷ 2 = 4.5 times the amount of squash. Multiply: 3.4 × 4.5 = 15.3 litres. Using the multiplier upside down — treating squash as 9 ÷ 2 times water, when it is water that is 9 ÷ 2 times squash — and calculating 3.4 × (2 ÷ 9) gives about 0.8 litres (to 1 d.p.); that would be the squash needed for 3.4 litres of water, not the water needed for 3.4 litres of squash. Adding the difference between the ratio parts, 9 − 2 = 7, to the squash amount, 3.4 + 7 = 10.4, mistakes a ratio for a fixed extra amount. Using the total number of parts, 2 + 9 = 11, so the multiplier 11 ÷ 2 = 5.5, gives 3.4 × 5.5 = 18.7 litres — that finds the total mix from the squash amount, not the water alone.
- (c) 500 g — Method: a concentration of 10% is the ratio 10:100, and the salt and the solution in the beaker must be in that same ratio, so write 10:100 = 50:m and scale. Working: 50 ÷ 10 = 5, so the salt is 5 times the 10 of the ratio; the solution must be 5 times the 100 of the ratio, giving 5 × 100 = 500 g. Answer: 500 g. The distractors: 5 g comes from working out 10% of 50 g, which treats the 50 g as the whole solution when it is the salt inside it; 450 g comes from scaling correctly and then taking the 50 g of salt away, which gives the mass of water rather than the mass of the whole solution; 5000 g comes from dividing by 0.01 instead of 0.1, that is from writing 10% as 0.01.
- (b) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (a) 2.17 litres per minute — There are 60 minutes in an hour, so to convert litres per hour to litres per minute you divide by 60: 130 ÷ 60 = 2.1666..., which rounds to 2.17 litres per minute. Multiplying by 60 instead of dividing gives 130 × 60 = 7800.00 litres per minute, using the conversion factor the wrong way round. Leaving the rate unchanged, 130.00, ignores that 'per hour' and 'per minute' are different units. Dividing by 50 instead of 60, misremembering the number of minutes in an hour, gives 130 ÷ 50 = 2.60 litres per minute.
- (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.
- (a) 40 m² — A scale of 1 : 10 is a length scale factor of 10 from model to real. Areas scale with the square of the length scale factor: 10² = 100. 0.4 × 100 = 40, so the real staircase covers 40 m². Giving 4 m² uses the length scale factor, 10, without squaring it (0.4 × 10 = 4). Giving 0.04 m² divides by the scale factor instead of multiplying by its square (0.4 ÷ 10 = 0.04). Giving 400 m² cubes the scale factor, 10³ = 1000, as if area scaled like a volume (0.4 × 1000 = 400).
- (b) 5 years — The recurrence P_{n+1} = 1.1P_n − 30 must be applied once per year, checking after each application whether the population has passed 460. Starting from P_0 = 400: 400 × 1.1 − 30 = 410, so P_1 = 410. Then 410 × 1.1 − 30 = 421, so P_2 = 421. Then 421 × 1.1 − 30 = 433.1, so P_3 = 433.1. Then 433.1 × 1.1 − 30 = 446.41, so P_4 = 446.41, which is still below 460. Then 446.41 × 1.1 − 30 = 461.051, so P_5 = 461.051, the first value above 460. The population first exceeds 460 after 5 complete years. Stopping at P_4 = 446.41 and reporting 4 years reports the last year the population was still below 460, not the first year it was above. Counting the starting value P_0 = 400 as a year of growth makes P_5 the sixth number in the list and gives 6 years, but P_0 is the population before any year has passed, so P_5 is reached after 5 years, not 6. Reading "exceed 460" as "exceed the starting population of 400" instead gives P_1 = 410, already above 400, and 1 year — but the threshold named in the question is 460, not the starting value, so always check every value against the number actually stated in the question.
- (a) £150 — Value after year 1: £800 × 0.75 = £600. Value after year 2: £600 × 0.75 = £450. The loss during the second year alone is £600 − £450 = £150. £450 comes from giving the value remaining after 2 years, not the amount lost during the second year. £200 comes from working out the loss during the first year instead of the second: £800 − £600 = £200. £350 comes from working out the total loss over both years instead of just the second year's loss: £800 − £450 = £350.
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
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