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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- 1.3 kg of apples cost £6. Work out the cost of 5 kg of the same apples.
- 2.A scale drawing states '1 cm represents 25 m'. Which of these is the correct way to write this scale in the form 1 : n?
- 3.A map has a scale of 1 : 20 000. A different map of the same area has a scale of 1 : 80 000. A lake's shoreline is drawn 5 cm long on the first map. Work out the length of the same shoreline on the second map, in centimetres.
- 4.A candle burns down at a constant rate. Its height, h cm, is plotted against time, t minutes, on a straight-line graph. The gradient of the line is −0.3. Which statement correctly describes what the gradient means?
- 5.The density of a liquid is 0.92 g/cm³. Work out the density of the liquid in kg/m³.
- 6.A table shows three pairs of values: when x = 2, y = 6; when x = 5, y = 15; when x = 8, y = 24. In every pair the ratio y : x is the same. Write down a formula for y in terms of x.
- 7.A charity receives £360 in donations in January and £480 in donations in February. Write the amount received in January as a fraction of the amount received in February, giving your answer in its simplest form.
- 8.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 9.A cyclist travels d kilometres in t hours. Write down an expression, in terms of d and t, for the cyclist's average speed in km/h.
- 10.1 metre = 100 cm. Work out how many square centimetres there are in 1 square metre.
- 11.A plan is drawn to a scale of 1 : 500. Work out how many metres are represented by 1 cm on the plan.
- 12.6 identical printers can print a batch of exam papers in 40 minutes, all working at the same rate. Working at the same rate, work out how many minutes 4 of these printers would take to print the same batch.
- 13.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.
- 14.A car is bought for £12000. Its value falls by 15% in the first year and by 10% in each year after that. Work out the value of the car 3 years after it was bought.
- 15.A mobile phone tariff is shown on a straight-line graph with the monthly cost, C pounds, on the vertical axis and the amount of data used, g gigabytes, on the horizontal axis. The line passes through (0, 10) and (8, 26). Work out the gradient and say what it represents.
Answer key
- (b) £10 — Method: find the cost of 1 kg by dividing, then multiply by the mass wanted — the unitary method for direct proportion. Working: £6 ÷ 3 = £2 per kg, and £2 × 5 = £10. Answer: £10. The distractors: £11 comes from adding 5 to the £6 instead of scaling; £30 comes from multiplying £6 by 5 without first dividing by 3; £3.60 comes from turning the proportion upside down, dividing by 5 and multiplying by 3.
- (d) 1 : 2500 — Method: convert the real-world measurement to the same unit as the drawing (centimetres) before writing the ratio. Working: 25 m = 2500 cm, so the scale is 1 : 2500. Wrong options: 1 : 25 comes from not converting metres to centimetres at all; 1 : 250 comes from converting metres to centimetres using ×10 instead of ×100; 1 : 2.5 comes from converting in the wrong direction (treating 25 m as 2.5 cm).
- (c) 1.25 cm — Method: find the real length using the first map's scale, then use the second map's scale to find its drawn length. Working: real length = 5 × 20 000 = 100 000 cm. On the second map: 100 000 ÷ 80 000 = 1.25 cm. Wrong options: 20 cm comes from inverting the ratio of the two scales (5 × 80 000 ÷ 20 000); 5 cm comes from wrongly assuming the length looks the same on both maps; 12.5 cm comes from dropping a zero from the second scale factor and dividing by 8 000 instead of 80 000 (100 000 ÷ 8 000).
- (d) The candle's height decreases by 0.3 cm every minute. — A negative gradient means the quantity on the vertical axis decreases as the quantity on the horizontal axis increases. The size of the gradient, 0.3, gives the amount of decrease per minute.
- (c) 920 kg/m³ — Convert each unit in turn. Mass: 1 g = 0.001 kg. Volume: 1 m³ = 100 × 100 × 100 = 1 000 000 cm³. So a density of 0.92 g per cm³ is 0.92 × 1 000 000 = 920 000 g in every cubic metre, and 920 000 g = 920 000 × 0.001 = 920 kg. The two conversions leave a single factor of 1 000 000 × 0.001 = 1000, so in one step multiply g/cm³ by 1000: 0.92 × 1000 = 920 kg/m³. Multiplying by 100 instead of 1000 gives 92 kg/m³, using the factor for 1 m² rather than 1 m³ of volume. Multiplying by 10 instead of 1000 gives 9.2 kg/m³, moving the decimal point one place for a conversion that moves it three. Dividing by 1000 instead of multiplying gives 0.00092 kg/m³, going the wrong way between the units — a kilogram is heavier than a gram, but a cubic metre is a million times bigger than a cubic centimetre, so the number must get larger, not smaller. The liquid's density is 920 kg/m³.
- (c) y = 3x — Method: if the ratio y : x is the same in every pair then y is always the same multiple of x, and that multiple is found by dividing a y value by its own x value. Working: 6 ÷ 2 = 3, 15 ÷ 5 = 3 and 24 ÷ 8 = 3, so every y is 3 lots of its x, which gives y = 3x. Answer: y = 3x. The distractors: y = x + 4 comes from subtracting instead of dividing on the first pair, 6 − 2 = 4, and testing it no further; y = x + 16 comes from the same subtraction on the last pair, 24 − 8 = 16; y = x/3 comes from dividing the x value by the y value, 2 ÷ 6, which gives the ratio x : y and not y : x.
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (d) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (b) 10000 cm² — Method: an area conversion factor is the square of the length conversion factor, because both sides of the square are scaled. Working: a square metre is a square of side 100 cm, so its area is 100 × 100 = 10000 cm². Answer: 10000 cm². The distractors: 100 cm² comes from using the length factor without squaring it; 200 cm² comes from doubling the length factor instead of squaring it; 1000000 cm² comes from cubing the factor, which is the conversion for a volume, not an area.
- (b) 5 m — A scale of 1 : 500 means 1 cm on the plan represents 500 cm in real life. Converting to metres: 500 ÷ 100 = 5 m. 500 m comes from forgetting to convert the 500 cm into metres at all. 50 m comes from dividing by 10 instead of 100 when converting centimetres to metres: 500 ÷ 10 = 50. 0.5 m comes from dividing by 1000 instead of 100, as if converting to kilometres instead of metres: 500 ÷ 1000 = 0.5.
- (a) 60 minutes — Method: for inverse proportion, printers × time is constant. Working: 6 × 40 = 240 (the constant). With 4 printers: 240 ÷ 4 = 60 minutes. Wrong options: 26.7 minutes comes from treating the relationship as direct proportion, scaling the time down as printers decrease (40 × 4 ÷ 6); 24 minutes comes from multiplying the two printer counts together instead of using the constant; 40 minutes comes from not adjusting the time at all for the change in printers.
- (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.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (b) 2, the cost in pounds of each extra gigabyte — Method: the gradient is the change in cost divided by the change in data, so it is the cost of each extra gigabyte; the value where the line meets the vertical axis is the charge before any data is used, which is a different quantity. Working: from (0, 10) to (8, 26) the cost rises by 26 − 10 = 16 pounds while the data rises by 8 − 0 = 8 gigabytes, so the gradient is 16 ÷ 8 = 2, meaning each extra gigabyte costs £2. Answer: 2, the cost in pounds of each extra gigabyte. The distractors: '10, the cost in pounds of each extra gigabyte' reads the intercept as the gradient, but 10 is what the tariff costs when no data at all has been used; '3.25, the cost in pounds of each extra gigabyte' comes from 26 ÷ 8, treating the line as though it passed through the origin when it starts at 10; '2, the fixed monthly charge in pounds' has the gradient right but describes the intercept, and the fixed charge on this tariff is £10.
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