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 train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 2.Two mathematically similar cylinders have volumes 64 cm³ and 216 cm³. Work out the ratio of the height of the smaller cylinder to the height of the larger cylinder, in simplest form.
- 3.A map has a scale of 2 cm : 5 km. A footpath measures 8 cm on the map. Work out the real length of the footpath, in kilometres.
- 4.Leah puts £4000 into a savings account paying 3% compound interest each year. At the end of 2 years she takes out all of the money and spends £1500 of it on a laptop. Work out how much of the money she has left.
- 5.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.
- 6.A map has a scale of 1 : 40 000. A cycle path measures 7.5 cm on the map. Work out the real length of the cycle path, in kilometres.
- 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.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 9.A jug of squash is made by mixing water and syrup in the ratio 6:1. Nia wants to make 8.4 litres of squash. Work out how much syrup she needs, in litres.
- 10.A metal has a density of 7.8 g/cm³. Work out the density of the metal in kg/m³.
- 11.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.
- 12.A water butt is being filled from a hosepipe at a constant rate while a small leak drains water out at a constant rate, giving a constant net rate of change. The volume of water in the butt, V litres, is shown on a straight-line graph against time, t minutes. The line passes through the points (5, 20) and (15, 60). The butt is empty at t = 0. Work out how many minutes it takes to reach a volume of 100 litres.
- 13.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?
- 14.A furniture maker builds two mathematically similar bookcases. The smaller bookcase is 40 cm tall and uses 2 m² of wood. The larger bookcase is 60 cm tall. Work out the area of wood needed for the larger bookcase, giving your answer in m².
- 15.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
Answer key
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (a) 2 : 3 — Simplify the volume ratio first: 64 : 216 divides by 8 to give 8 : 27. Volumes scale with the cube of the height ratio, so take the cube root of each part: the cube root of 8 is 2, and the cube root of 27 is 3, giving a height ratio of 2 : 3. Giving 3 : 2 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 8 : 27 is the simplified volume ratio, without cube-rooting it. Giving 64 : 216 is the volume ratio before it has even been simplified.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (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.
- (b) 3 km — Method: multiply by the scale factor to get the real length in centimetres, then convert to kilometres. Working: 7.5 × 40 000 = 300 000 cm. 300 000 ÷ 100 000 = 3 km. Wrong options: 30 km comes from dividing by 10 000 instead of 100 000 when converting to kilometres; 3000 km comes from dividing by 100 instead of 100 000; 0.3 km comes from dividing by 1 000 000, an extra factor of 10 too many.
- (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) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (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.
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
- (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).
- (b) 4.5 m² — The height ratio is 40 : 60, which simplifies to 2 : 3, so the larger bookcase is 1.5 times as tall as the smaller one. Areas scale with the square of the length scale factor, so the wood needed scales by 1.5² = 2.25. 2 × 2.25 = 4.5, so the larger bookcase needs 4.5 m² of wood. Giving 3 m² uses the length scale factor, 1.5, without squaring it (2 × 1.5 = 3). Giving 6.75 m² cubes the scale factor, 1.5³ = 3.375, as if wood coverage were a volume (2 × 3.375 = 6.75). Giving 2.25 m² is the squared scale factor on its own, without multiplying by the smaller bookcase's wood area of 2 m².
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
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