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.The mass of a length of copper pipe is proportional to its length. A pipe of length 4.5 m has a mass of 12.6 kg. Work out the mass of a pipe of the same type of length 9.5 m.
- 2.A fruit punch is made from orange juice, pineapple juice and lemonade in the ratio 5:3:2. A jug holds 3.5 litres of punch in total. Work out the volume of pineapple juice needed.
- 3.A stall sells bottles of juice. Two bottles cost £3.00, five bottles cost £7.50 and eight bottles cost £12.00. The cost, C pounds, is in a fixed ratio to the number of bottles, n. Write down a formula for C in terms of n.
- 4.A scale drawing of a garden uses a scale of 1 : 20. Write down a formula for the real length, L metres, in terms of the length on the drawing, d centimetres, given that 1 metre = 100 centimetres.
- 5.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 6.The temperature of a chemical reaction, in °C, is modelled by T = 80 − 6t + 0.5t², where t is the time in minutes after the reaction starts. Which of these four statements about the reaction between t = 2 and t = 6 minutes is correct?
- 7.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
- 8.A force of 126 N acts on an area of 3.5 m². Work out the pressure on the area, in N/m².
- 9.The exchange rate is £1 = 1.28 US dollars. Convert £350 into US dollars.
- 10.A printer prints at a constant rate. The time taken to print a batch of forms is inversely proportional to the printer's speed, in pages per minute. Printing at 20 pages per minute takes 15 minutes. Work out how long the same batch takes to print at 25 pages per minute. Give your answer in minutes.
- 11.The depth of water in a tank, in cm, is recorded every 10 seconds: at t = 10, depth = 32; at t = 20, depth = 45; at t = 30, depth = 56. Use the most appropriate chord from these readings to estimate the instantaneous rate of change of depth at t = 20.
- 12.Before a charity campaign, donations were £6400 per month. Donations grew by 8% in the first year after the campaign, and then fell by 3% in the second year as interest faded. Work out the amount donated per month at the end of the second year, to the nearest pound.
- 13.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.
- 14.A map has a scale of 1 : 40 000. A cycle path measures 7.5 cm on the map. Work out the real length of the cycle path, in kilometres.
- 15.A cyclist rides 17.5 km on Saturday and 12.25 km on Sunday. Write the distance ridden on Saturday as a fraction of the distance ridden on Sunday, giving your answer in its simplest form.
Answer key
- (d) 26.6 — Find the constant multiplier — the mass of each metre of pipe: 12.6 ÷ 4.5 = 2.8, so the mass is always 2.8 times the length. For a length of 9.5 m, the mass is 9.5 × 2.8 = 26.6 kg. 17.6 comes from assuming an additive relationship instead of a multiplicative one — adding the increase in length (9.5 − 4.5 = 5) onto 12.6. 3.4 comes from using the multiplier the wrong way round (4.5 ÷ 12.6, rounded to 1 d.p.), then multiplying by 9.5. 12.6 comes from simply repeating the given mass, without applying the multiplier to the new length.
- (c) 1.05 litres — Method: find the value of one part of the ratio from the total volume, then find the share for pineapple juice. Working: the ratio 5:3:2 has 5 + 3 + 2 = 10 parts, so one part is 3.5 ÷ 10 = 0.35 litres, and the pineapple juice is 3 × 0.35 = 1.05 litres. So 1.05 litres of pineapple juice is needed. Distractor 1.75 litres is the volume of orange juice, not pineapple juice. Distractor 0.7 litres is the volume of lemonade, not pineapple juice. Distractor 0.35 litres is the value of one part, found correctly but never multiplied by 3.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (c) Average rate, t = 2 to 6, is −2°C/min — First find the temperature at each end of the interval. At t = 2, T = 80 − 6 × 2 + 0.5 × 2² = 80 − 12 + 2 = 70. At t = 6, T = 80 − 6 × 6 + 0.5 × 6² = 80 − 36 + 18 = 62. The average rate of change over the interval is the change in T divided by the change in t: 62 − 70 = −8, then −8 ÷ 4 = −2°C per minute, so the statement about the average rate is correct. The instantaneous rate at t = 6 is not −2: completing the square gives T = 0.5(t − 6)² + 62, so t = 6 is the turning point of the curve, where the tangent is horizontal and the rate is 0°C per minute — the reaction has stopped cooling by then. The instantaneous rate at t = 2 is not −2 either: a short chord centred on t = 2, from t = 1.9 (T = 70.405) to t = 2.1 (T = 69.605), gives −0.8 ÷ 0.2 = −4°C per minute, so the reaction is cooling twice as fast at the start of the interval as the average over it. Saying the temperature falls 2°C in total confuses the RATE, −2°C per minute, with a TOTAL drop, which is 70 − 62 = 8°C over the four minutes. Always check whether a figure is a rate, per minute, or a total change.
- (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.
- (a) 36 — Pressure = force ÷ area, so 126 ÷ 3.5 = 36 N/m². (0.03 comes from dividing the area by the force instead of the force by the area, the wrong way round. 129.5 comes from adding 126 and 3.5 instead of dividing. 441 comes from multiplying 126 by 3.5 instead of dividing.)
- (c) 448.00 US dollars — Method: multiply the amount in pounds by the exchange rate. Working: £350 × 1.28 = 448.00 US dollars. Wrong options: 273.44 US dollars comes from dividing by the rate instead of multiplying (350 ÷ 1.28); 351.28 US dollars comes from adding the rate to the amount instead of multiplying; 4,480.00 US dollars comes from a decimal-point slip, using 12.8 instead of 1.28.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
- (c) 1.2 cm/s — To estimate an instantaneous rate of change at a point from a table of readings, use the chord that spans the point symmetrically — equal steps either side — because the over-estimate on one side and the under-estimate on the other largely cancel. Here that is the chord from t = 10 to t = 30. The change in depth is 56 − 32 = 24 and the change in time is 30 − 10 = 20, so the estimate is 24 ÷ 20 = 1.2 cm/s. The one-sided chord from t = 20 to t = 30 gives (56 − 45) ÷ (30 − 20) = 11 ÷ 10 = 1.1 cm/s, which estimates the rate somewhere between t = 20 and t = 30 rather than at t = 20 itself. Dividing the 20-second change in depth by the 10-second gap between consecutive readings gives 24 ÷ 10 = 2.4, mixing the change from one interval with the time from another. Reporting the change in depth, 24, on its own is not a rate at all, because it has not been divided by a time. The best estimate of the instantaneous rate of change of depth at t = 20 is 1.2 cm/s.
- (d) £6705 — After the first year: £6400 × 1.08 = £6912. After the second year: £6912 × 0.97 = £6704.64, which rounds to £6705 (nearest pound). £6720 comes from treating the +8% and −3% changes as a single net +5% change applied to the original amount instead of applying each change in turn: £6400 × 1.05 = £6720. £6912 comes from applying only the first year's growth and stopping there, without applying the second year's fall. £7104 comes from adding the two percentages together as +11% and applying that to the original amount instead of applying each change to the correct starting amount in turn: £6400 × 1.11 = £7104.
- (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.
- (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.
- (a) 10/7 — Put Saturday's distance over Sunday's distance: 17.5/12.25. Multiply both numbers by 100 to clear the decimals: 1750/1225. Divide both by their highest common factor, 175: 1750÷175 = 10, 1225÷175 = 7, giving 10/7. (7/10 comes from writing the distances the wrong way round. 3/7 comes from finding the difference, 17.5 − 12.25 = 5.25 km, and writing it as a fraction of Sunday's distance, 5.25/12.25. 10/17 comes from comparing Saturday's distance to the total distance ridden, 17.5/29.75.)
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