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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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.Triangle ABC is mathematically similar to triangle PQR, with AB corresponding to PQ and BC corresponding to QR. AB = 6 cm, BC = 8 cm and PQ = 12 cm. Work out the length of QR.
- 2.The ratio of the number of pages Mia has read to the number of pages Leo has read is 7 : 4. Mia has read 84 pages. Work out how many pages they have read altogether.
- 3.Two mathematically similar hexagonal tiles have areas 18 cm² and 50 cm². Work out the ratio of the side length of the smaller tile to the side length of the larger tile, in simplest form.
- 4.A coach journey of 240 km takes 3 hours. For this fixed distance the average speed needed is inversely proportional to the time taken. Work out the average speed needed to complete the same journey in 2 hours.
- 5.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
- 6.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 7.A shop's takings were £4500 in May. In June the takings fell by £900. Write the June takings as a fraction of the May takings, in its simplest form.
- 8.A train journey takes 2 hours 15 minutes. A bus journey to the same place takes 3 hours 45 minutes. Write the time for the train journey as a fraction of the time for the bus journey, giving your answer in its simplest form.
- 9.The strength of a radio signal, in units, is inversely proportional to the square of the distance from the transmitter, in km. At a distance of 2 km the signal strength is 20 units. Construct the equation connecting signal strength S and distance d, then work out the distance at which the signal strength is 5 units.
- 10.A tangent to a curve at the point where x = 3 passes through the points (1, 2) and (5, 14). A candidate says: "The gradient of the tangent at x = 3 is 3, because 14 − 2 = 12 and 12 ÷ 4 = 3." Which statement is correct?
- 11.£5000 is invested in an account paying 4% compound interest each year. Work out the total interest earned after 3 years.
- 12.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 13.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 14.Two mathematically similar triangular flags have areas in the ratio 4 : 25. The height of the smaller flag is 6 cm. Work out the height of the larger flag.
- 15.The value of a delivery van, in £, is plotted against its age, in years, since it was bought. At age 2 years, the gradient of the tangent to the graph is −950. What does this tell you about the van at age 2 years?
Answer key
- (c) 16 cm — Corresponding sides of similar triangles are all in the same ratio. Use the pair whose lengths are both known: the scale factor from triangle ABC to triangle PQR is 12 ÷ 6 = 2. Since QR corresponds to BC, multiply BC by that scale factor: 8 × 2 = 16, so QR = 16 cm.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (d) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (a) 4/5 — Method: find the June takings first, then write them over the May takings and cancel. Working: the takings fell by £900, so June is £4500 − £900 = £3600; the fraction is 3600/4500, and dividing the numerator and the denominator by 900 gives 4/5. Answer: 4/5 of the May takings. The distractors: 1/5 comes from writing the fall over the May takings, 900/4500, which answers how far the takings dropped rather than what June's takings are compared with May's; 5/4 comes from writing May over June, 4500/3600, reversing the order the question asks for; 4/9 comes from writing June over the two months added together, 3600/8100, a part-to-whole fraction when the comparison asked for is with May alone.
- (d) 3/5 — Convert both times to minutes: 2 hours 15 minutes = 135 minutes; 3 hours 45 minutes = 225 minutes. Put the train time over the bus time: 135/225. Divide both numbers by their highest common factor, 45: 135÷45 = 3, 225÷45 = 5, giving 3/5. (5/3 comes from writing the times the wrong way round. 2/5 comes from finding the difference, 225 − 135 = 90 minutes, and writing it as a fraction of the bus time, 90/225. 3/8 comes from comparing the train time to the total time for both journeys, 135/360.)
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (a) The gradient is 3; the candidate's method is right. — The gradient of a tangent, like any straight line, is the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 2) and (5, 14), so the change in y is 14 − 2 = 12 and the change in x is 5 − 1 = 4. The gradient is 12 ÷ 4 = 3, so the candidate's calculation is correct. Subtracting in the wrong order, (2 − 14) ÷ (5 − 1), gives −12 ÷ 4 = −3, the wrong sign. Adding the two changes instead of dividing them, 12 + 4 = 16, does not find a gradient at all. Dividing the change in x by the change in y instead of the other way round, 4 ÷ 12 = 1/3, inverts the calculation completely. Before accepting or rejecting a claimed gradient, always redo the calculation yourself in the same order — change in y over change in x — rather than trusting the arithmetic as given.
- (c) £624.32 — A 4% rise is a multiplier of 1.04, applied once each year. After year 1: 5000 × 1.04 = 5200. After year 2: 5200 × 1.04 = 5408. After year 3: 5408 × 1.04 = 5624.32. The question asks for the interest, not the value of the account, so take away the amount invested at the start: 5624.32 − 5000 = 624.32. The total interest earned is £624.32.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (b) 12 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (c) Falling at £950 per year — The gradient of a tangent on a value-age graph is a rate, in pounds per year, so −950 means the van's value is falling at £950 per year at that instant. Writing this as 950% per year mistakes a rate measured in pounds per year for a percentage — the units of a gradient come from the units on the two axes, £ and years, not from a percentage. Saying the value 'falls by £950 over the next year' treats the instantaneous rate at age 2 as if it stayed constant for a whole year, which finds an average future change, not the instantaneous rate at age 2 itself. Reading the sign the wrong way round gives 'rising at £950 per year', which would mean the van is gaining value. Always match the units of a gradient to the units on the two axes of the graph.
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