Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation 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 Foundation
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- 1.A crate exerts a force of 84 N on the floor through a base with an area of 2.4 m². Work out the pressure on the floor, in N/m².
- 2.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.
- 3.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 4.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 5.Tap A fills a swimming pool in 6 hours. Tap B pours water twice as fast as tap A. The time taken to fill the pool is inversely proportional to the rate of flow. Work out how long tap B takes to fill the pool.
- 6.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.
- 7.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.
- 8.A fruit squash is made by mixing squash and water. In a jug, 2/9 of the total volume is squash. A caterer makes up 4.5 litres of the squash mixture. Work out how many litres of squash are in the mixture.
- 9.A mortar mix is made from sand and cement in the ratio 5 : 3. Write the ratio of sand to the total mix in its simplest form.
- 10.A car travels 150 km in 3 hours at a constant speed. Work out the average speed of the car, in km/h.
- 11.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
- 12.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 13.Write £3.60 : £2.40 as a ratio in its simplest form.
- 14.5 tins of soup cost £8 in total. Work out the multiplier that changes the number of tins into the total cost in pounds.
- 15.Last week Priya worked 5 shifts of 7 hours. This week she worked 4 shifts of 8 hours. Write the number of hours she worked last week as a fraction of the number of hours she worked this week.
Answer key
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (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.
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (a) 3 hours — Method: for a fixed pool the rate of flow multiplied by the time taken is constant, so multiplying the rate by a factor divides the time by that same factor. Working: tap B's rate is 2 times tap A's rate, so tap B's time is 6 ÷ 2 = 3 hours. Answer: 3 hours. The distractors: 12 hours comes from multiplying the time by 2 as well, which treats the time as directly proportional to the rate and has the faster tap taking longer; 4 hours comes from reading ‘twice as fast’ additively, as two hours quicker, and working out 6 − 2 instead of scaling the time by a factor of 2; 1.5 hours comes from applying the factor of 2 twice, halving 6 to 3 and then halving again.
- (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) 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.
- (d) 1.00 litres — Squash is 2/9 of the mixture, so the squash volume is 4.5 × 2/9 = 1.00 litres. Using the water's fraction, 7/9, instead of squash's gives 4.5 × 7/9 = 3.50 litres — the volume of water, not squash. Dividing 4.5 by 9 but forgetting to multiply by the numerator 2 gives 4.5 ÷ 9 = 0.50 litres, which is only 1/9 of the mixture. Halving the total volume instead of applying the fraction 2/9 gives 4.5 ÷ 2 = 2.25 litres, which assumes the mixture is half squash.
- (b) 5:8 — The mix has 5 parts sand and 3 parts cement, so 5 + 3 = 8 parts in total. Sand to total is 5 : 8, and since the highest common factor of 5 and 8 is 1, this is already in its simplest form. Giving 5 : 3 answers sand to cement, not sand to the total mix. Giving 3 : 8 is cement to total, the wrong part of the mix. Giving 8 : 5 has the total and the sand swapped round.
- (d) 50 — Speed = distance ÷ time, so 150 ÷ 3 = 50 km/h. (450 comes from multiplying 150 by 3 instead of dividing. 153 comes from adding 150 and 3. 147 comes from subtracting 3 from 150 — none of these use the correct compound-unit calculation, distance ÷ time.)
- (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.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (d) 3 : 2 — Convert both amounts to pence: £3.60 = 360p and £2.40 = 240p, giving the ratio 360 : 240. Divide both parts by their highest common factor, 120, to get 3 : 2. Giving 360 : 240 has not been simplified at all. Giving 2 : 3 swaps the order. Giving 36 : 24 has been divided by 10, which is a common factor but not the highest one, so it is not yet in simplest form.
- (c) 1.6 — The multiplier from tins to pounds is found by dividing the cost by the number of tins: £8 ÷ 5 = 1.6. 0.625 comes from dividing the wrong way round, tins ÷ cost (5 ÷ 8), which is the multiplier from pounds back to tins. 3 comes from subtracting the number of tins from the cost (8 − 5), instead of dividing. 40 comes from multiplying the cost by the number of tins (8 × 5), instead of dividing.
- (a) 35/32 — Work out each weekly total first. Last week: 5 × 7 = 35 hours. This week: 4 × 8 = 32 hours. Last week's total is being written as a fraction of this week's total, so last week goes on the top and this week goes on the bottom, giving 35/32. The two totals share no common factor, so the fraction cannot be cancelled. It is greater than 1, which says that Priya worked more hours last week than this week.
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