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.£2000 is invested in a savings account that pays 5% compound interest each year. Work out the value of the investment at the end of 2 years.
- 2.Triangle ABC is mathematically similar to triangle PQR, with AB corresponding to PQ and BC corresponding to QR. AB = 6 cm, BC = 8 cm and PQ = 12 cm. Work out the length of QR.
- 3.A straight-line graph shows the distance travelled, d kilometres, against the time taken, t hours, for a cyclist. The line passes through the origin and has a gradient of 18. Write down what the gradient of this line represents.
- 4.Write 3 hours 45 minutes as a decimal number of hours.
- 5.A runner's speed is 8 m/s. Given that there are 3600 seconds in an hour and 1000 metres in a kilometre, work out the runner's speed in km/h.
- 6.A charity collects donations from adults and children in the ratio 5:2. Altogether, £238 is collected. Work out how much more the adults donate than the children.
- 7.A straight-line graph shows the cost, C pounds, of hiring a bike for h hours. The line passes through the points (1, 12) and (4, 27). Which of these statements about the line is true?
- 8.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 9.Square P has sides of length 3 cm. Square Q has sides of length 12 cm. Write the ratio of the area of square P to the area of square Q in its simplest form.
- 10.Two quantities x and y are in direct proportion. When x = 8, the value of y is 20. Work out the value of y when x = 14.
- 11.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 12.A tap fills a tank at a rate of 15 litres per minute. Given that 1 litre = 1000 cm³, work out the rate at which the tank fills in cm³ per second.
- 13.A printer's ink cartridge level, I millilitres, is plotted against the number of pages printed, p, and the points lie on a straight line. The line passes through (0, 24) and (300, 9). The printer has now printed 300 pages. Work out how many more pages it can print before the cartridge is empty.
- 14.Maya buys a vintage car for £15000. Its value is expected to increase by 4% each year for 2 years. She then plans to sell the car and use £2000 of the money from the sale to buy a trailer. Work out how much money she will have left after selling the car and buying the trailer.
- 15.Water flows into a tank at a constant rate. After 8 minutes the tank holds 200 litres. Write the ratio of time in minutes to volume in litres, in its simplest form.
Answer key
- (b) £2205.00 — With compound interest each year's interest is worked out on the value at the start of that year, so a 5% rise is a multiplier of 1.05 applied once per year. After the first year: 2000 × 1.05 = 2100. After the second year: 2100 × 1.05 = 2205. The question asks for the value of the investment, not for the interest earned, so the answer is £2205.00.
- (c) 16 cm — Corresponding sides of similar triangles are all in the same ratio. Use the pair whose lengths are both known: the scale factor from triangle ABC to triangle PQR is 12 ÷ 6 = 2. Since QR corresponds to BC, multiply BC by that scale factor: 8 × 2 = 16, so QR = 16 cm.
- (d) The cyclist's speed, in kilometres per hour — Method: on any straight-line graph the gradient is the change in the quantity on the vertical axis for each 1 unit of the quantity on the horizontal axis, so its meaning is read off the two axis labels. Working: the vertical axis is distance in kilometres and the horizontal axis is time in hours, so the gradient counts kilometres for each hour, and distance for each hour is speed. Answer: the cyclist's speed, in kilometres per hour. The distractors: 'the total distance the cyclist travels, in kilometres' reads the gradient as a value taken off the vertical axis, but a gradient is a rate and no end point of the journey has been given; 'the time the cyclist takes, in hours' names the horizontal axis, which is the quantity the gradient divides by rather than the gradient itself; 'the distance the cyclist travels in 18 hours' treats the 18 as a value of t, when 18 is the steepness of the line and not a point on it.
- (d) 3.75 — 45 minutes is 45 ÷ 60 = 0.75 of an hour, so 3 hours 45 minutes = 3.75 hours. Getting 3.45 comes from writing the minutes directly after the decimal point instead of converting them to a fraction of an hour. Getting 3.67 comes from misreading 45 minutes as 40 minutes and converting 40 ÷ 60 instead. Getting 4.15 comes from rounding 45 minutes up to the next whole hour and adding the remainder as if it were more minutes past that hour.
- (a) 28.8 km/h — Method: first change metres per second into metres per hour, then change metres into kilometres. Working: 8 × 3600 = 28800 metres per hour, then 28800 ÷ 1000 = 28.8 km/h. So the runner's speed is 28.8 km/h. Distractor 28800 km/h comes from stopping after the first step and forgetting to change metres into kilometres. Distractor 2.22 km/h comes from dividing by 3600 instead of multiplying, then multiplying by 1000. Distractor 2.88 km/h comes from using 360 instead of 3600 seconds in an hour, missing a zero.
- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (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.
- (c) 1 : 16 — Work out each area before comparing. Square P has area 3 × 3 = 9 cm². Square Q has area 12 × 12 = 144 cm². The ratio of the areas is 9 : 144, and both parts divide by 9: 9 ÷ 9 = 1 and 144 ÷ 9 = 16, giving 1 : 16. Notice that the sides are in the ratio 1 : 4, and the areas are in the ratio of the squares of those parts, which is what always happens when a length is scaled. Stopping at 1 : 4 would compare the sides and never the areas, and cubing the parts to reach 1 : 64 is the rule for volumes rather than for areas. The order matters too: the question names square P first, so its area must be the first part of the ratio.
- (d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (d) 250 cm³/s — Method: first change litres per minute into cm³ per minute, then change per minute into per second. Working: 15 × 1000 = 15000 cm³ per minute, then 15000 ÷ 60 = 250 cm³ per second. So the tank fills at 250 cm³ per second. Distractor 15000 cm³/s comes from stopping after the first step and forgetting to change minutes into seconds. Distractor 900000 cm³/s comes from multiplying by 60 instead of dividing. Distractor 2500 cm³/s comes from dividing by 6 instead of 60.
- (c) 180 pages — The gradient is (9 − 24) ÷ (300 − 0) = −15 ÷ 300 = −0.05, so the cartridge uses 0.05 ml of ink per page. At p = 300 there are 9 ml left. The extra pages before the cartridge is empty is 9 ÷ 0.05 = 180 pages. Giving 9 as the answer confuses the millilitres of ink remaining with the number of pages remaining — they are different quantities with different units. Multiplying instead of dividing, 9 × 0.05 = 0.45, does not undo the rate correctly. Working out the total number of pages a full cartridge lasts, 24 ÷ 0.05 = 480 pages, answers how many pages the cartridge prints in total from full, not how many more pages it can print from the 300-page point.
- (a) £14224 — Value after 2 years: £15000 × 1.04 × 1.04 = £16224. Money left after buying the trailer: £16224 − £2000 = £14224. £14200 comes from treating the two 4% increases as a single flat 8% increase applied once instead of compounding: £15000 × 1.08 = £16200, and £16200 − £2000 = £14200. £13600 comes from applying the 4% increase only once, for 1 year instead of 2: £15000 × 1.04 = £15600, and £15600 − £2000 = £13600. £18224 comes from adding the £2000 instead of subtracting it: £16224 + £2000 = £18224.
- (b) 1:25 — Write the ratio time : volume using the numbers in the question: 8 : 200. Divide both parts by their highest common factor, 8, to give 1 : 25. (25:1 comes from writing the ratio the wrong way round, volume : time. 8:25 comes from dividing only the volume by 8 and leaving the time unchanged. 25:8 is that same mistake written the wrong way round.)
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