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.Which of these ratios is equivalent to 6 : 10 : 14?
- 2.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 3.In a science lesson Priya has 10 litres of a solution that is 30% salt. She adds water to make a solution that is 20% salt. Work out how many litres of water she adds.
- 4.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.
- 5.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 6.A car travels for 3 hours at an average speed of 80 km/h and then for 1 hour at an average speed of 40 km/h. Work out the average speed for the whole journey.
- 7.The strength of a radio signal, in units, is inversely proportional to the square of the distance from the transmitter, in km. At a distance of 2 km the signal strength is 20 units. Construct the equation connecting signal strength S and distance d, then work out the distance at which the signal strength is 5 units.
- 8.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.
- 9.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.
- 10.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 11.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.
- 12.A cyclist rides 19.3 km in 47 minutes. Work out her average speed, in km/h, to 1 decimal place.
- 13.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 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.y is inversely proportional to x². When x = 2, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 4.
- 16.A jug holds 3 litres of a drink that is 60% fruit juice. 1 litre of water is added to the jug. Work out the percentage of the new mixture that is fruit juice.
- 17.A rectangular garden has its length and width in the ratio 5:3. Given that the perimeter of the garden is 64 m, work out the width of the garden.
- 18.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 19.Maya is paid £105.30 for working 9 hours. Work out her rate of pay, in £ per hour.
- 20.A recipe for pastry uses flour and butter in the ratio 3:2. A baker has 180 g of butter and wants to make pastry using all of it. Work out the total mass of pastry the baker can make.
- 21.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.
- 22.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 23.A shop sells rope by the metre. 3 m costs £7.50 and 5 m costs £11.00. Does this data show that the cost is directly proportional to the length of rope bought? Choose the correct verdict and reason.
- 24.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.
Answer key
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (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.
- (b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.
- (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) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (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).
- (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.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (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.)
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (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.
- (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.
- (a) 2 — Since y is inversely proportional to x², y = k/x². Using x = 2, y = 8: 2² = 4, so 8 = k ÷ 4, giving k = 8 × 4 = 32. The equation is y = 32/x². When x = 4: 4² = 16, so y = 32 ÷ 16 = 2. Treating the relationship as inversely proportional to x itself, rather than to x², gives k = 8 × 2 = 16 and then y = 16 ÷ 4 = 4, a different relationship. Using x instead of x² in the new calculation gives y = 32 ÷ 4 = 8, skipping the square. Multiplying by x² instead of dividing by it gives y = 32 × 16 = 512, the wrong operation for an inverse relationship. When x = 4, y = 2.
- (c) 45% — Method: adding water changes the total volume but not the amount of fruit juice, so find the juice, find the new total volume, and write the first as a percentage of the second. Working: 3 × 0.6 = 1.8 litres of fruit juice; the new volume is 3 + 1 = 4 litres; 1.8 ÷ 4 = 0.45, which is 45%. Answer: 45%. The distractors: 60% is the strength before the water goes in, and assumes that adding water leaves the strength unchanged; 15% comes from dividing the 60% by the 4 litres of mixture instead of dividing the 1.8 litres of juice by the 4 litres; 75% is the fraction of the new mixture that came out of the original jug, 3 litres out of 4, which ignores that only 60% of that 3 litres was juice.
- (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
- (d) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (a) £11.70 — Rate of pay = total pay ÷ hours worked, so £105.30 ÷ 9 = £11.70 per hour. Working out £105.30 − 9 = £96.30 subtracts the number of hours from the total pay instead of dividing. Working out £105.30 × 9 = £947.70 multiplies total pay by hours worked instead of dividing. Misplacing the decimal point in the correct answer gives £117.00 instead of £11.70. Maya's rate of pay is £11.70 per hour.
- (c) 450 g — Method: use the amount of butter given to find the value of one part of the ratio, then find the mass of flour, and finally add flour and butter to get the total. Working: 180 g of butter is 2 parts, so one part is 180 ÷ 2 = 90 g. The flour is 3 parts, so 3 × 90 = 270 g, and the total mass is 270 + 180 = 450 g. So the baker can make 450 g of pastry. Distractor 270 g is only the mass of flour, forgetting to add the butter back on. Distractor 300 g comes from treating the 180 g as 3 parts instead of 2, swapping which ratio number matches the butter. Distractor 540 g comes from multiplying 180 by 3 directly instead of first finding the value of one part.
- (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) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (a) No — the cost per metre differs: £2.50/m vs £2.20/m — Method: divide cost by length for each pair and compare the unit rates. Working: £7.50 ÷ 3 = £2.50 per m; £11.00 ÷ 5 = £2.20 per m. The rates are different, so this is NOT direct proportion. Wrong options: 'Yes — both amounts increase' wrongly assumes any increasing relationship is proportional; 'No — because 5 m costs more in total' judges by total cost rather than the rate per metre, which is not valid reasoning on its own; 'Yes — the cost per metre is £2.50 in both cases' miscalculates the second rate (11.00 ÷ 5 is £2.20, not £2.50).
- (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).
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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