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) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
- (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) 1:25 — Write the ratio time : volume using the numbers in the question: 8 : 200. Divide both parts by their highest common factor, 8, to give 1 : 25. (25:1 comes from writing the ratio the wrong way round, volume : time. 8:25 comes from dividing only the volume by 8 and leaving the time unchanged. 25:8 is that same mistake written the wrong way round.)
- (a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.
- (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.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (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) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (b) 80 — Method: the difference between 40% and 25% of the number is 15% of the number, and that difference is 12. Working: 15% of the number is 12, so 1% of the number is 12 ÷ 15 = 0.8, and the number is 0.8 × 100 = 80. Check: 40% of 80 is 32, 25% of 80 is 20, and 32 − 20 = 12. Answer: 80. The distractors: 30 comes from solving 40% of the number = 12; 48 comes from solving 25% of the number = 12; 15 is the percentage difference written as the answer.
- (a) 22p — First convert the hourly rate to pence: £13.20 = 1320p. Then convert from per hour to per minute by dividing by 60, since there are 60 minutes in an hour: 1320 ÷ 60 = 22p per minute. Working out 1320 × 60 = 79200p multiplies by 60 instead of dividing, going the wrong way between per hour and per minute. Working out 13.20 × 10 = 132p converts pounds to pence using the wrong power of ten, and dividing that by 60 carries the error through to give 2.2p. Working out 13.20 × 100 = 1320p converts the currency correctly but stops there, leaving the rate as pence per hour rather than completing the second conversion to pence per minute. The rate of pay is 22p per minute.
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
- (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (d) 12 — First find the y-values at the two ends of the chord. At x = 2, y = 2 × 2² + 5 = 2 × 4 + 5 = 8 + 5 = 13. At x = 4, y = 2 × 4² + 5 = 2 × 16 + 5 = 32 + 5 = 37. The gradient of the chord is the change in y divided by the change in x: 37 − 13 = 24, then 24 ÷ (4 − 2) = 24 ÷ 2 = 12. Stopping after finding the change in y and not dividing by the change in x gives 24, which is not a gradient at all. Inverting the fraction, change in x divided by change in y, gives 2 ÷ 24 ≈ 0.08. Averaging the two y-values instead of finding the change between them gives (13 + 37) ÷ 2 = 50 ÷ 2 = 25. A gradient is always change in y over change in x — never the other way round, and never a single y-value.
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
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