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.
Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 2.Write the ratio 6a : 9a in its simplest form, where a is a positive number.
- 3.The ratio of y to x is always 3 : 4. Write down which equation could represent this relationship.
- 4.Work out 50% of 60.
- 5.A cyclist rides at a steady speed of 20 mph. Given that 1 mile ≈ 1.6 km, work out the cyclist's speed in km/h.
- 6.A bottle contains 1.5 litres of juice. Given that 1 litre = 1000 ml, work out the capacity of the bottle in millilitres.
- 7.y is directly proportional to x, so y = kx. When x = 2, the value of y is 10. Work out the value of k.
- 8.Two printers' running costs are shown on separate straight-line graphs of total cost against number of pages printed. Printer A's graph has gradient £0.04 per page. Printer B's graph has gradient £0.05 per page. Work out how much cheaper it is to print 200 pages on Printer A than on Printer B, based on these running costs alone.
- 9.A glass holds 250 ml of squash. Work out how many of these glasses are needed to fill a 1 litre bottle.
- 10.Two mathematically similar rectangles have widths 5 cm and 10 cm. The perimeter of the smaller rectangle is 18 cm. Work out the perimeter of the larger rectangle.
- 11.The table shows values of x and y. x: 3, 5, 8. y: 12, 20, 32. Which statement correctly describes the relationship between x and y?
- 12.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 13.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 14.The same jumper is sold at two shops. Shop A charges £40 and Shop B charges £50. Write down the price at Shop A as a percentage of the price at Shop B.
- 15.A machine is bought for £8500. Its value depreciates by 6% each year. Work out the value of the machine after 3 years, to the nearest pound.
Answer key
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (c) 2:3 — Divide both parts by the common factor 3a: 6a ÷ 3a = 2 and 9a ÷ 3a = 3, giving the ratio 2 : 3.
- (d) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
- (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
- (c) 32 km/h — Method: to change mph into km/h, multiply by the number of kilometres in a mile. Working: 20 × 1.6 = 32 km/h. So the cyclist's speed is 32 km/h. Distractor 12.5 km/h comes from dividing by 1.6 instead of multiplying. Distractor 21.6 km/h comes from adding 1.6 instead of multiplying by it. Distractor 20 km/h comes from not converting the units at all.
- (a) 1500 ml — Method: to change litres into millilitres, multiply by 1000. Working: 1.5 × 1000 = 1500 ml. So the bottle holds 1500 ml. Distractor 150 ml comes from multiplying by 100 instead of 1000. Distractor 15000 ml comes from multiplying by 10000, an extra zero. Distractor 1.5 ml comes from not converting the units at all.
- (d) 5 — Method: rearrange y = kx to make the constant the subject, then substitute the pair of values given. Working: k = y ÷ x, so k = 10 ÷ 2 = 5. Answer: 5. The distractors: 20 comes from multiplying 10 by 2 instead of dividing, which is the rearrangement done the wrong way round; 12 comes from adding the pair, 10 + 2, treating the relationship as y = x + k; 8 comes from working out 10 − 2, the same additive reading with the operation reversed.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (a) 4 — Method: put both amounts into the same unit, then divide the bottle by the glass. Working: 1 litre = 1000 ml, and 1000 ÷ 250 = 4. Answer: 4. The distractors: 0.25 comes from dividing the glass by the bottle, 250 ÷ 1000, the division the wrong way round; 40 comes from taking 1 litre as 10 000 ml; 1250 comes from adding 1000 and 250 instead of dividing.
- (c) 36 cm — A perimeter is built from lengths, so it is multiplied by the length scale factor and not by the square of it. The scale factor from the smaller rectangle to the larger one is 10 ÷ 5 = 2. The larger perimeter is therefore 18 × 2 = 36 cm.
- (a) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (b) 80% — Percentage = (40 ÷ 50) × 100 = 80%.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
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