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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- (c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total journey.
- (a) £14224 — Value after 2 years: £15000 × 1.04 × 1.04 = £16224. Money left after buying the trailer: £16224 − £2000 = £14224. £14200 comes from treating the two 4% increases as a single flat 8% increase applied once instead of compounding: £15000 × 1.08 = £16200, and £16200 − £2000 = £14200. £13600 comes from applying the 4% increase only once, for 1 year instead of 2: £15000 × 1.04 = £15600, and £15600 − £2000 = £13600. £18224 comes from adding the £2000 instead of subtracting it: £16224 + £2000 = £18224.
- (a) 1.375 — The multiplier is the new value divided by the original value: 176 000 ÷ 128 000 = 1.375. Choosing 0.375 comes from working out the increase as a fraction of the original (48 000 ÷ 128 000 = 0.375) and stopping there, without adding on the 1 whole that represents the original turnover. Choosing 0.727 comes from dividing the 2023 turnover by the 2024 turnover, the wrong way round (128 000 ÷ 176 000, rounded to 3 decimal places). Choosing 137.5 comes from writing the multiplier as a percentage — the 2024 turnover is 137.5% of the 2023 turnover — instead of converting it to the decimal multiplier 1.375.
- (d) 8/5 — Two masses can only be compared once they are in the same unit. Since 1 kg is 1000 g, the recipe needs 1200 g. The recipe's mass is being written as a fraction of Dan's mass, so 1200 goes on the top and 750 on the bottom, giving 1200/750. The highest common factor of the two is 150: 1200 ÷ 150 = 8 and 750 ÷ 150 = 5. The fraction is 8/5, which is greater than 1 because the recipe needs more flour than Dan has.
- (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.
- (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).
- (b) £250 — Since cost is proportional to the cube of the radius, C = kr³. Using r = 3, C = 54: 3³ = 27, so 54 = k × 27, giving k = 54 ÷ 27 = 2. The equation is C = 2r³. When r = 5: 5³ = 125, so C = 2 × 125 = 250. Treating the relationship as proportional to r² instead of r³ gives k = 54 ÷ 9 = 6 and then C = 6 × 25 = 150, which models area scaling, not volume scaling. Treating it as proportional to r itself gives k = 54 ÷ 3 = 18 and then C = 18 × 5 = 90. Finding k correctly from the cube but then multiplying it by the radius instead of by the cube of the radius gives 2 × 5 = 10, which applies the right constant to the wrong power of r. The cost of a container of radius 5 cm is £250.
- (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
- (a) 1/9 — The ratio copper : tin is 9:1, so write tin over copper: 1/9. (9/1 comes from writing the ratio the wrong way round, copper over tin. 1/10 comes from comparing the tin to the total mass of the alloy, 1 part out of 10. 9/10 comes from comparing the copper to the total mass of the alloy, 9 parts out of 10.)
- (d) Profit is decreasing by £10 per £1 rise in price. — The gradient of a tangent gives the instantaneous rate of change of profit with respect to price, found from the change in profit divided by the change in price between two points on the tangent. Here the tangent passes through (12, 540) and (18, 480), so the change in profit is 480 − 540 = −60 and the change in price is 18 − 12 = 6. The gradient is −60 ÷ 6 = −10. A negative gradient means profit is decreasing as price increases, so profit is decreasing at an instantaneous rate of £10 for every £1 rise in price. Subtracting the profits in the wrong order, 540 − 480 = 60, and dividing by the same change in price, 60 ÷ 6 = 10, gives a positive value and the wrong direction — profit is not increasing at £15. Stopping after finding only the change in profit, 480 − 540 = −60, without dividing by the change in price, is not a rate at all. Reading off the change in price, 6, and calling it the rate gives £6 per £1 rise in price, but 6 is only the width of the price interval — it is not a change in profit at all, and profit falls across that interval, so the direction is wrong too. The instantaneous rate of change of profit with respect to price at £15 is a decrease of £10 per £1 rise in price.
- (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.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (c) 14 litres — Method: find the amount of fuel used per km first, then use it to find the fuel needed for 175 km. Working: 24 ÷ 300 = 0.08 litres per km, and 0.08 × 175 = 14 litres. So 14 litres are needed. Distractor 24 litres comes from assuming the same amount of fuel is used no matter the distance, without scaling. Distractor 21 litres comes from misreading the original distance as 200 km instead of 300 km. Distractor 1.4 litres comes from a decimal-point slip, giving an answer ten times too small.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
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