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 straight-line graph shows the cost, C pounds, of buying n litres of paint. The line passes through the origin and through the point (5, 40). Work out the gradient of the line.
- 2.A model railway is built at a scale where the ratio of model length to real length is 1 : 76. Given that a real carriage is 20.6 metres long, work out the length of the model carriage in centimetres, to 1 decimal place.
- 3.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.
- 4.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?
- 5.A garden centre has 90 rose bushes in stock. It receives a delivery that increases its stock of rose bushes by 20%. In the same week it sells 150 lavender bushes. Work out the new number of rose bushes, then write it as a fraction of the number of lavender bushes sold, giving your answer in its simplest form.
- 6.Two numbers are in the ratio 5 : 8. The larger number is 40. Work out the smaller number.
- 7.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.
- 8.A cartographer is choosing a scale for a map to show a small park in as much detail as possible. Which of these scales would show the MOST detail — 1 : 500, 1 : 5000, or 1 : 50 000? Give the correct scale and reason.
- 9.Four graphs are described below. Each shows one quantity plotted against another. Write down the description of the graph that shows two quantities in direct proportion.
- 10.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 11.A scale model of a bridge is built at a scale of 1 : 120. The real bridge is 84 m long. Work out the length of the model, in centimetres.
- 12.Write 45 minutes : 2 hours as a ratio in its simplest form.
- 13.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 14.A force of 250 N acts on an area of 0.5 m². Using pressure = force ÷ area, work out the pressure, in pascals.
- 15.A 750 g box of cereal costs £2.70. A 500 g box of the same cereal costs £1.95. Work out which box is better value, and its cost per 100 g.
Answer key
- (b) 8 — Method: the gradient of a straight line is the change in the vertical value divided by the change in the horizontal value between two points on the line. Working: from the origin (0, 0) to (5, 40) the vertical change is 40 − 0 = 40 and the horizontal change is 5 − 0 = 5, so the gradient is 40 ÷ 5 = 8, which here means a cost of £8 for each litre. Answer: 8. The distractors: 0.125 comes from dividing the horizontal change by the vertical change, 5 ÷ 40, which gives litres per pound instead of the gradient; 40 comes from reading off the vertical value of the point and calling it the gradient, ignoring the 5 litres it took to reach that cost; 35 comes from subtracting the two coordinates, 40 − 5, instead of dividing them.
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (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) 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.
- (a) 18/25 — First find the new number of rose bushes: 90 × 1.2 = 108 (a 20% increase multiplies by 1.2). Then write 108 over 150 and divide top and bottom by 6 to get 18/25. Choosing 3/5 comes from using the original 90 rose bushes without applying the 20% increase (90/150 = 3/5). Choosing 25/18 comes from writing the number of lavender bushes over the new number of rose bushes, the wrong way round. Choosing 3/25 comes from multiplying 90 by 0.2 instead of 1.2, finding only the increase (18) rather than the new total, then writing 18/150 = 3/25.
- (c) 25 — Method: divide the larger number by its ratio part to find the value of one part, then multiply by the smaller number's ratio part. Working: 40 ÷ 8 = 5 (value of one part). Smaller number = 5 × 5 = 25. Wrong options: 64 comes from dividing by the smaller ratio part instead of the larger (40 ÷ 5 × 8); 45 comes from adding the value of one part onto 40 instead of scaling down (40 + 5); 35 comes from subtracting the value of one part from 40 (40 − 5) instead of multiplying it by the smaller ratio part.
- (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.
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (c) A straight line through the origin, rising from left to right — Two quantities are in direct proportion when one is a constant multiple of the other, so their graph is a straight line through the origin: when one quantity is 0 the other is 0 as well, and doubling one doubles the other. A straight line crossing the vertical axis at 5 has a fixed amount added on, so when the horizontal quantity is 0 the vertical quantity is 5, not 0 — a straight line on its own is not enough for direct proportion. A curve that falls steeply and then levels off without touching either axis shows inverse proportion: one quantity grows as the other shrinks, and their product stays the same. A horizontal line at a height of 3 shows a quantity that does not change at all as the other one grows, so it is not proportional to it.
- (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) 70 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 84 m = 8400 cm. 8400 ÷ 120 = 70 cm. Wrong options: 0.7 cm comes from dividing 84 by 120 without converting metres to centimetres; 1,008,000 cm comes from multiplying instead of dividing (8400 × 120); 7 cm comes from converting 84 m to 840 cm (using ×10 instead of ×100) before dividing.
- (d) 3:8 — Convert 2 hours to minutes: 2 hours = 120 minutes. The ratio is 45 : 120. The highest common factor of 45 and 120 is 15. Divide both parts by 15: 45 ÷ 15 = 3 and 120 ÷ 15 = 8, giving 3 : 8. Leaving the hours unconverted gives 45 : 2 — the units on each side are different, so this does not compare like with like. Dividing by 5 instead of 15 gives 9 : 24, which still shares a common factor of 3, so it is not fully simplified. Swapping the order gives 8 : 3, hours to minutes instead of minutes to hours.
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
- (c) 500 — Pressure = force ÷ area = 250 ÷ 0.5 = 500 pascals. Getting 125 comes from multiplying the force by the area instead of dividing (250 × 0.5 = 125). Getting 249.5 comes from subtracting the area from the force instead of dividing. Getting 50 comes from misreading the area as 5 m² instead of 0.5 m² and dividing 250 by 5.
- (d) The 750 g box, at 36p per 100 g — Work out the cost per 100 g of each box. 750 g box: 270p ÷ 7.5 = 36p per 100 g. 500 g box: 195p ÷ 5 = 39p per 100 g. The lower cost per 100 g is the better value, so the 750 g box at 36p per 100 g is the answer. Choosing the 500 g box at 39p per 100 g gets the maths right but picks the higher unit price, not realising a smaller cost per 100 g is the better deal. Choosing the 500 g box because £1.95 is lower than £2.70 compares the total prices without allowing for the different pack sizes at all. Working out 270 ÷ 5 = 54p divides the 750 g box's price by the wrong number of hundred-grams (the 500 g box's), giving a rate that belongs to neither box. The 750 g box, at 36p per 100 g, is the better value.
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