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.A rope of length A is 1.5 times as long as a rope of length B. Write the ratio A : B in its simplest form.
- 2.A 1.5 kg bag of pasta costs £2.85 and a 2.4 kg bag of the same pasta costs £4.32. Work out which bag gives better value for money.
- 3.A concrete block exerts a force of 45 N on the ground through a base with an area of 0.05 m². Work out the pressure on the ground, in N/m².
- 4.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
- 5.A model car is built at a scale of 1 : 20 compared with the real car. Work out the factor by which the surface area to be painted on the real car is bigger than the surface area of the model.
- 6.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.
- 7.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 8.An acid solution has a concentration of 20%. The solution contains 40 g of acid, and the rest of the solution is water. Work out the mass of water in the solution.
- 9.An oven is preheating. Its temperature, T °C, is plotted against time, t minutes, on a straight-line graph. The line passes through the points (2, 60) and (6, 140). Work out the gradient of the line.
- 10.A shop's number of online orders is proportional to its number of in-store orders. Last month there were 45 in-store orders and 180 online orders. Write the ratio of online orders to in-store orders, in its simplest form.
- 11.5 decorators paint 3 rooms in 6 hours. All the decorators work at the same rate. Work out how long it takes 10 decorators to paint 5 rooms.
- 12.A sequence is defined by the iterative rule x_{n+1} = 0.8x_n + 50, with x_0 = 200. Work out x_2, the value of the sequence after two iterations.
- 13.A greenhouse's temperature, in °C, is recorded every hour during the day: 09:00, 15.0; 10:00, 18.4; 11:00, 20.1; 12:00, 19.8; 13:00, 17.2. Work out between which two consecutive readings the instantaneous rate of change of the temperature is most likely to have been zero.
- 14.A concert hall has 54 seats in total, of which 24 are reserved and the rest are free. Write the number of free seats as a fraction of the number of reserved seats. Give your answer in its simplest form.
- 15.A furniture designer makes a scale model of a wardrobe using a scale of 1 : 8. The real wardrobe is 2.4 m tall and 1.2 m wide. Work out the perimeter of the front face of the model wardrobe, in centimetres.
Answer key
- (c) 3 : 2 — Method: 'A is k times B' means that if B is taken as 1 part then A is k parts, and a ratio in its simplest form is written with whole numbers that have no common factor. Working: taking B as 1, A is 1.5, so A : B = 1.5 : 1; multiplying both parts by 2 clears the decimal and gives 3 : 2, and 3 and 2 share no factor. Answer: 3 : 2. The distractors: 2 : 3 comes from writing B before A, reversing the order the question asks for; 1.5 : 1 is the right relationship but is not in its simplest form, because a ratio in its simplest form uses whole numbers; 3 : 5 comes from comparing A with the combined length rather than with B, since A : (A + B) = 1.5 : 2.5 = 3 : 5.
- (b) The 2.4 kg bag, since it costs £1.80 per kg compared with £1.90 per kg for the 1.5 kg bag. — To compare value for money, work out the cost per kilogram for each bag. 1.5 kg bag: £2.85 ÷ 1.5 = £1.90 per kg. 2.4 kg bag: £4.32 ÷ 2.4 = £1.80 per kg. Since £1.80 is less than £1.90, the 2.4 kg bag gives better value. The option comparing £2.85 with £4.32 directly is wrong because it compares the total prices, not the price per kilogram — a bigger bag naturally costs more in total even if it is better value. The option that names the 1.5 kg bag with £1.80 per kg and the 2.4 kg bag with £1.90 per kg has the correct unit prices but has swapped which bag they belong to. The option giving £1.19 per kg and £2.88 per kg comes from dividing each price by the wrong bag's mass (£2.85 ÷ 2.4 and £4.32 ÷ 1.5).
- (b) 900 N/m² — Pressure = force ÷ area, so 45 ÷ 0.05 = 900 N/m². Working out 45 ÷ 5 = 9 misplaces the decimal point in the area, treating 0.05 m² as 5 m². Working out 45 × 0.05 = 2.25 multiplies force and area together instead of dividing. Working out 45 + 0.05 = 45.05 simply adds the two given numbers, which does not give a pressure at all. The pressure on the ground is 900 N/m².
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
- (d) 400 — The area scale factor is the length scale factor squared: 20² = 400, so the real car's surface area is 400 times the model's. 20 comes from using the length scale factor itself, without squaring it. 8000 comes from cubing the length scale factor (20³), instead of squaring it — cubing is the rule for volume, not area. 40 comes from doubling the length scale factor (2 × 20), instead of squaring it.
- (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.
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (a) 160 g — Method: use the ratio 20:100 to find the mass of the whole solution from the mass of acid, then take the acid away to leave the water. Working: 20:100 = 40:m, and 40 ÷ 20 = 2, so m = 2 × 100 = 200 g of solution; the water is 200 − 40 = 160 g. Answer: 160 g. The distractors: 200 g is the mass of the whole solution, which is the middle step and includes the acid the question asks you to leave out; 8 g comes from working out 20% of 40 g, which treats the 40 g as the whole solution rather than as the 20% inside it; 10 g comes from reading the 40 g as the 80% that is water, giving a solution of 50 g and a difference of 50 − 40.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (d) 4:1 — Write the ratio online : in-store using the numbers in the question: 180 : 45. Divide both parts by their highest common factor, 45, to give 4 : 1. (1:4 comes from writing the ratio the wrong way round, in-store : online. 3:1 comes from subtracting the in-store orders from the online orders, 180 − 45 = 135, and comparing that to the in-store orders, 135:45, instead of dividing. 4:5 comes from comparing the online orders to the total number of orders, 180:225.)
- (c) 5 hours — Method: measure the job in decorator-hours, which is in the same ratio as the number of rooms, then share the decorator-hours between the decorators available. Working: 5 decorators × 6 hours = 30 decorator-hours for 3 rooms, so one room takes 30 ÷ 3 = 10 decorator-hours; 5 rooms take 5 × 10 = 50 decorator-hours; shared between 10 decorators that is 50 ÷ 10 = 5 hours. Answer: 5 hours. The distractors: 3 hours comes from halving the 6 hours because the number of decorators doubles, while forgetting that there are also more rooms to paint; 10 hours comes from scaling the 6 hours up for the rooms only, 6 × 5 ÷ 3, and leaving the workforce at 5 decorators; 6 hours comes from assuming that doubling the decorators and increasing the rooms cancel each other out, which they do not, because the rooms rise by a factor of 5/3 and the workforce by a factor of 2.
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (a) 11:00 to 12:00 — Method: the instantaneous rate of change is zero at a turning point, where a rising trend becomes a falling trend; that lies within the first interval whose difference has changed sign from the interval before it. Working: the differences between consecutive readings are +3.4 °C (09:00 to 10:00), +1.7 °C (10:00 to 11:00), −0.3 °C (11:00 to 12:00) and −2.6 °C (12:00 to 13:00); the sign changes from positive to negative within 11:00 to 12:00, since the temperature is still rising up to 11:00 (20.1 °C, the highest recorded value) and has fallen by 12:00, so the instantaneous rate of change was zero somewhere within that interval. Choosing 09:00 to 10:00 picks out the interval with the largest positive difference, +3.4 °C, confusing the fastest rise with no change at all. Choosing 10:00 to 11:00 picks the last interval where the temperature was still rising, one interval too early, without checking that the very next interval turns negative. Choosing 12:00 to 13:00 picks out the interval with the largest-magnitude difference, −2.6 °C, the fastest fall, not where the change is zero. Zero instantaneous rate of change happens at a turning point, where the readings stop rising and start falling — find the FIRST interval whose difference has flipped sign from the one before it, not the biggest change or an interval where the old sign still held.
- (d) 5/4 — Work out the number of free seats: 54 − 24 = 30. Form the fraction 30/24; both numbers share a factor of 6, so 30 ÷ 6 = 5 and 24 ÷ 6 = 4, giving 5/4. 4/5 comes from writing the fraction the wrong way round, as reserved over free (24/30). 5/9 comes from comparing the free seats with the total number of seats (30/54), instead of with the reserved seats. 5/6 comes from miscalculating 54 − 24 as 20 instead of 30, then forming 20/24.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
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