24 questions across ratio, direct and inverse proportion, compound measures and area and volume scale factors.
⚗️ Ratio and proportion mastery — Higher
Ratio, proportion and rates of change is 20% of a Higher paper and it is the area that rewards method over recall. This sheet works through it in one sitting: simplifying and sharing in a ratio, including the questions that give you one share or the difference rather than the total; expressing one quantity as a fraction or percentage of another; direct and inverse proportion, both numerically and as an equation with a constant k; compound measures — speed, density and pressure — with unit conversions built in rather than avoided; and the similar-shapes work where lengths, areas and volumes scale by the factor, its square and its cube. That last group is where Higher candidates lose the most marks in this area, so it is deliberately over-represented here.
- 1.A shop sells ribbon by the metre. 2 m costs £3.00, 4 m costs £6.00, and 7 m costs £10.50. Does this data show that the cost is directly proportional to the length of ribbon bought? Choose the correct verdict and reason.
- 2.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.
- 3.An amount of money is shared in the ratio 1:2:3. The largest share is £90 more than the smallest share. Work out the total amount that was shared.
- 4.6 identical taps fill a paddling pool in 20 minutes. Each tap fills at the same steady rate. Work out how long 3 of these taps would take to fill the same pool.
- 5.A spring's extension is directly proportional to the force applied to it. A force of 5 N produces an extension of 12 mm. Work out the extension produced by a force of 20 N.
- 6.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 7.Paint costs £14.40 for every 20 m² of wall it covers. Assuming the same rate, work out the cost of the paint needed to cover 56 m² of wall.
- 8.A printer prints at a constant rate. The time taken to print a batch of forms is inversely proportional to the printer's speed, in pages per minute. Printing at 20 pages per minute takes 15 minutes. Work out how long the same batch takes to print at 25 pages per minute. Give your answer in minutes.
- 9.y is directly proportional to x. When x = 7, the value of y is 21. Work out the value of x when y = 12.
- 10.A 1.5 kg bag of pasta costs £2.85 and a 2.4 kg bag of the same pasta costs £4.32. Work out which bag gives better value for money.
- 11.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 12.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.
- 13.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 14.A gardener mixes 300 ml of plant feed concentrate with 1.2 litres of water to make a spray. Write the ratio of concentrate to water in its simplest form.
- 15.Two mathematically similar water bottles have a volume scale factor of 8 from the smaller bottle to the larger bottle. Work out the surface area scale factor from the smaller bottle to the larger bottle.
- 16.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
- 17.The ratio of Josh's savings to Mia's savings is 4:7. Mia has £39 more than Josh. Work out Josh's savings.
- 18.Simplify the ratio 45 : 30 : 75 to its simplest form.
- 19.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 20.y is directly proportional to x². When x = 3, y = 45. Construct the equation connecting x and y, then work out the value of y when x = 5.
- 21.A statue exerts a downward force of 68.4 N on its base, which has an area of 2.85 m². Work out the pressure that the statue exerts on its base, in N/m².
- 22.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 23.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 24.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
Answer key
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (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.
- (d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (c) 40 minutes — Method: this is inverse proportion — fewer taps means longer, not shorter — so the number of taps × the time taken stays constant. Working: 6 × 20 = 120, and with 3 taps the time is 120 ÷ 3 = 40 minutes. So 3 taps take 40 minutes. Distractor 10 minutes comes from treating it as direct proportion instead of inverse, working out 20 × 3 ÷ 6. Distractor 30 minutes comes from halving the number of taps and adding half the original time, 20 + 10, instead of doubling the time. Distractor 17 minutes comes from subtracting the number of taps removed, 3, directly from the original time, 20.
- (c) 48 mm — Method: extension = k × force, where k = extension ÷ force. Working: k = 12 ÷ 5 = 2.4 mm per N. At 20 N: extension = 2.4 × 20 = 48 mm. Wrong options: 32 mm comes from adding the extension and force numbers instead of scaling (12 + 20); 3 mm comes from treating the relationship as inverse proportion (12 × 5 ÷ 20); 36 mm comes from using an incorrect scale factor of 3 between the forces instead of the correct factor of 4 (20 ÷ 5).
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (a) £40.32 — Find the cost per square metre from the rate given: £14.40 ÷ 20 = £0.72 per m². Then multiply by the area to be covered: £0.72 × 56 = £40.32. Working out 14.40 × 20 ÷ 56 ≈ £5.14 uses the ratio the wrong way round, scaling down as if 56 m² needed less paint than 20 m². Stopping at £0.72 only gives the cost per square metre, not the cost for the whole wall. Working out 14.40 + (56 − 20) = £50.40 adds the extra square metres straight onto the cost in pounds, treating square metres and pounds as the same kind of quantity. Covering 56 m² costs £40.32.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
- (a) 4 — Method: find the constant of proportionality from the pair given, write the equation, then substitute the new value of y and solve. Working: k = 21 ÷ 7 = 3, so y = 3x; putting y = 12 gives 12 = 3x, and x = 12 ÷ 3 = 4. Answer: 4. The distractors: 36 comes from multiplying by the constant instead of dividing by it, 12 × 3, which is the proportion set up upside down; 84 comes from multiplying 12 by the 7 from the first pair, using a value of x as though it were the constant; 9 comes from working out 12 − 3, treating the equation as y = x + 3 rather than y = 3x.
- (b) The 2.4 kg bag, since it costs £1.80 per kg compared with £1.90 per kg for the 1.5 kg bag. — To compare value for money, work out the cost per kilogram for each bag. 1.5 kg bag: £2.85 ÷ 1.5 = £1.90 per kg. 2.4 kg bag: £4.32 ÷ 2.4 = £1.80 per kg. Since £1.80 is less than £1.90, the 2.4 kg bag gives better value. The option comparing £2.85 with £4.32 directly is wrong because it compares the total prices, not the price per kilogram — a bigger bag naturally costs more in total even if it is better value. The option that names the 1.5 kg bag with £1.80 per kg and the 2.4 kg bag with £1.90 per kg has the correct unit prices but has swapped which bag they belong to. The option giving £1.19 per kg and £2.88 per kg comes from dividing each price by the wrong bag's mass (£2.85 ÷ 2.4 and £4.32 ÷ 1.5).
- (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.
- (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.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (b) 4 — Method: find the length scale factor by taking the cube root of the volume scale factor, then square it to get the area scale factor. Working: 8 = 2³, so the length scale factor is 2, and the area scale factor is 2² = 4. Answer: 4. 8 comes from using the volume scale factor itself as if it were the area scale factor. 64 comes from squaring the volume scale factor, 8² = 64, instead of first taking its cube root. 2 comes from correctly finding the length scale factor but then forgetting to square it.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
- (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.
- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (a) 24 N/m² — Pressure = force ÷ area. 68.4 ÷ 2.85 = 24 N/m². 194.94 N/m² comes from multiplying the force by the area instead of dividing (68.4 × 2.85). 65.55 N/m² comes from subtracting the area from the force (68.4 − 2.85) instead of dividing. 0.04 N/m² comes from dividing the area by the force instead of the force by the area (2.85 ÷ 68.4).
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (c) 27 : 64 — For similar solids, the ratio of volumes is the ratio of lengths cubed: 3³ : 4³ = 27 : 64. 3 : 4 comes from using the height ratio itself as the volume ratio, without cubing it at all. 9 : 16 comes from squaring each part instead of cubing (3² : 4²) — squaring is the rule for area, not volume. 27 : 4 comes from cubing only the first part of the ratio (3³ = 27), and leaving the second part uncubed.
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
What is on this worksheet?
The sheet holds 24 questions drawn from the MathsUK bank — the content area covered: Ratio, proportion and rates of change (statements R4, R5, R9, R10, R11, R12, R13). It is pitched at GCSE Higher and takes about 45 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 24 questions before checking — about 45 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 24 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
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