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.
Calculator
Ratio, proportion and rates of change worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
- 2.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.
- 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.Map A has a scale of 1 : 25000 and Map B has a scale of 1 : 50000, both showing the same area. A real distance of 10 km is measured on each map. On which map does this distance appear as the longer length, and how long is it on that map, in centimetres?
- 5.A metal cylinder has a mass of 356.5 g and a volume of 47 cm³. Work out the density of the cylinder, in g/cm³, to 1 decimal place.
- 6.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 7.A plan is drawn to a scale of 1 : 500. Work out how many metres are represented by 1 cm on the plan.
- 8.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 9.The time taken for a train journey is inversely proportional to the average speed of the train. At an average speed of 60 km/h the journey takes 2 hours. Work out the time taken at an average speed of 40 km/h.
- 10.A toy manufacturer makes a model aircraft that is mathematically similar to the real aircraft, at a scale of 1 : 48. The wingspan of the model is 15 cm. Work out the wingspan of the real aircraft, giving your answer in metres.
- 11.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.
- 12.A scale model of a shipping container is built at a scale of 1 : 30, using material with the same density as the real container. The model has a mass of 400 g. Work out the mass of the real container, giving your answer in kilograms.
- 13.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.
- 14.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?
- 15.A shade of paint is made by mixing blue paint and white paint. To make 5 litres of the shade, 2 litres of blue paint is used and the rest is white paint. Write the ratio of blue paint to white paint in its simplest form.
Answer key
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (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.
- (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).
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
- (c) 7.6 — Density = mass ÷ volume, so 356.5 ÷ 47 = 7.585..., which rounds to 7.6 g/cm³ (1 d.p.). (0.1 comes from dividing the volume by the mass instead of the mass by the volume, the wrong way round. 7.5 comes from rounding 7.585 down instead of up to 1 decimal place. 403.5 comes from adding the mass and the volume instead of dividing.)
- (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (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.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
- (b) 7.2 m — Multiply the model wingspan by the scale factor: 15 × 48 = 720. This is in centimetres, and 720 cm = 7.2 m, since 1 m = 100 cm. Giving 0.31 m divides by the scale factor instead of multiplying (15 ÷ 48 ≈ 0.31), scaling the model down rather than the real aircraft up. Giving 72 m converts centimetres to metres by dividing by 10 instead of 100. Giving 0.72 m converts by dividing by 1000 instead of 100.
- (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) 10,800 kg — Method: since the model and the real container are similar and made of the same material, mass scales with volume, so the mass scale factor is the length scale factor cubed. Working: 30³ = 27,000, so the real container's mass is 400 × 27,000 = 10,800,000 g, which is 10,800,000 ÷ 1,000 = 10,800 kg. Answer: 10,800 kg. 12 kg comes from using the length scale factor directly, 400 × 30 = 12,000 g, without cubing it. 360 kg comes from squaring the length scale factor instead of cubing it, 400 × 30² = 360,000 g. 10,800,000 kg comes from correctly cubing the scale factor but then forgetting to convert the mass from grams into kilograms.
- (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.
- (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 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
Build your own mix at the worksheet builder.