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.
Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.Order these three values from smallest to largest: 3/8, 0.43, 41%.
- 2.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 3.The price of a games console is reduced by 10%. In a later sale the reduced price is reduced by 10% again. Work out the overall percentage decrease.
- 4.A car is bought for £12000. Its value falls by 15% in the first year and by 10% in each year after that. Work out the value of the car 3 years after it was bought.
- 5.Two mathematically similar rectangles have lengths in the ratio 3 : 5. The area of the smaller rectangle is 27 cm². Work out the area of the larger rectangle.
- 6.A car is bought for £9000. Its value decreases by 8% each year. Work out its value after 2 years.
- 7.A process is modelled by the recurrence P_{n+1} = 0.6P_n + 40. As n increases, P_n approaches a long-run value L, which satisfies L = 0.6L + 40. Solve this equation to find L.
- 8.Write 400 g : 1.5 kg as a ratio in its simplest form.
- 9.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 10.Two straight-line graphs represent y = 2x and y = 5x. For the same value of x, where x ≠ 0, write down the ratio of the y-value on the first graph to the y-value on the second graph, in its simplest form.y = 2xy = 5x
- 11.Two mathematically similar logos are printed on a poster. They have areas 20 cm² and 45 cm². Nadia says that the length scale factor from the smaller logo to the larger logo is 45 ÷ 20 = 2.25. Give a reason why Nadia is incorrect, and work out the correct length scale factor.
- 12.A recipe for 6 muffins needs 150 g of flour. Aisha wants to make 15 muffins using the same recipe. Work out how much flour she needs.
- 13.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 14.A shop sells identical pens. The total cost, C pence, of n pens is given by C = 12n. Write the ratio n : C in its simplest form.
- 15.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
Answer key
- (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (d) 19% — Method: write each decrease as a multiplier, multiply the multipliers, then compare the result with 100%. Working: a 10% decrease is a multiplier of 0.9, so the two reductions together give 0.9 × 0.9 = 0.81; the final price is 81% of the original, so the price has fallen by 100% − 81% = 19%. Answer: an overall decrease of 19%. The distractors: 20% comes from adding the two reductions, 10% + 10%, which charges the second 10% against the original price instead of against the already reduced price; 21% comes from using the increase multiplier by mistake, since 1.1 × 1.1 = 1.21, and reading that 21% as a decrease; 81% is the percentage of the original price still being paid, not the percentage taken off.
- (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) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (c) £7617.60 — To decrease by 8% each year, multiply by 0.92 (100% − 8%) twice. £9000 × 0.92 × 0.92 = £7617.60. £7560.00 comes from treating the two 8% decreases as a single flat 16% decrease applied once instead of compounding: £9000 × 0.84 = £7560.00. £8280.00 comes from applying the 8% decrease only once, for 1 year instead of 2: £9000 × 0.92 = £8280.00. £10497.60 comes from multiplying by 1.08 twice, increasing the value instead of decreasing it: £9000 × 1.08 × 1.08 = £10497.60.
- (d) 100 — Rearrange L = 0.6L + 40 by collecting the L terms on one side: L − 0.6L = 40, which gives 0.4L = 40, then L = 40 ÷ 0.4 = 100. Subtracting the other way round, 0.6L − L = 40, gives −0.4L = 40, then L = 40 ÷ (−0.4) = −100 — a sign error that flips the answer negative even though a long-run value here must be positive. Ignoring the 0.6L term completely and solving L = 40 directly gives 40, which throws away the recurrence's own multiplier. Dividing 40 by 0.6 instead of by the correct coefficient 0.4 gives 40 ÷ 0.6 ≈ 66.7, a slip that comes from dividing by the coefficient of L on the RIGHT of the original equation rather than by what is left once the L terms are collected on one side. Always collect the L terms first, then divide by whatever coefficient of L remains.
- (a) 4 : 15 — Convert to the same unit first: 1.5 kg = 1500 g, since 1 kg = 1000 g. This gives the ratio 400 : 1500. Divide both parts by their highest common factor, 100, to get 4 : 15. Giving 40 : 150 divides by 10 only, which is not the highest common factor, so it is not fully simplified. Giving 15 : 4 swaps the order. Giving 4 : 1.5 has not converted 1.5 kg into grams, so the two parts are not measured in the same unit.
- (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.
- (d) 2:5 — The ratio of the y-values equals the ratio of the coefficients of x, since x cancels: 2x : 5x = 2 : 5.
- (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.
- (a) 375 g — Find the amount of flour needed per muffin: 150 ÷ 6 = 25 g. Multiply by the new number of muffins: 25 × 15 = 375 g. (60 g comes from using the scale factor the wrong way round, 6/15 × 150. 150 g comes from not scaling the recipe at all. 300 g comes from rounding the scale factor, 15 ÷ 6, down to 2 before multiplying.)
- (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) 1:12 — C = 12n means that for every 1 pen there are 12 pence of cost, so n : C = 1 : 12, and the highest common factor of 1 and 12 is 1, so this is already in its simplest form. Writing C : n instead of n : C gives 12 : 1, the ratio the wrong way round. Reading C = 12n as '12 more than n' instead of '12 times n', so n = 1 gives C = 13, gives 1 : 13, from adding instead of multiplying. Choosing n = 12, so C = 12 × 12 = 144, gives the correct pair of values n : C = 12 : 144, but the ratio the right way round left unsimplified — 12 and 144 share a common factor of 12, which has not been cancelled.
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
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