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.Noah runs 2 km in 10 minutes. Work out how long he takes to run 5 km at the same speed.
- 2.Write the ratio 5 : 8 in the form 1 : n.
- 3.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 4.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.
- 5.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.
- 6.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 7.Two mathematically similar cubes have edge lengths 2 cm and 6 cm. Write the ratio of the volume of the smaller cube to the volume of the larger cube in its simplest form.
- 8.A journey takes 4 hours at an average speed of 60 km/h. The time taken is inversely proportional to the average speed. Work out how long the same journey takes at an average speed of 80 km/h.
- 9.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.
- 10.A recipe uses flour, sugar and butter in the ratio 8 : 3 : 5. Write the ratio of flour to the rest of the mixture (sugar and butter combined) in its simplest form.
- 11.Two cars leave the same point at the same time and travel in opposite directions. One travels at 80 km/h and the other at 60 km/h. Work out how long it takes until the cars are 350 km apart.
- 12.A cyclist travels 18 km in 45 minutes. Work out the average speed, in km/h.
- 13.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.
- 14.Ffion is paid £11.20 per hour. She works 6 hours on Monday and 4.5 hours on Tuesday. Work out her total pay for the two days.
- 15.Write the ratio 8 : 15 in the form 1 : n.
- 16.Two mathematically similar garden ponds have surface areas of 12 m² and 27 m². The fencing needed to go around the smaller pond costs £96. Assuming the cost of fencing is proportional to the perimeter of the pond, work out the cost of fencing the larger pond.
- 17.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 18.Two investors put money into a business in the ratio 3:5. The first investor puts in £1,200. Work out the total amount invested by both investors.
- 19.A coach journey of 240 km takes 3 hours. For this fixed distance the average speed needed is inversely proportional to the time taken. Work out the average speed needed to complete the same journey in 2 hours.
- 20.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.
- 21.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.
- 22.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.
- 23.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 24.Two numbers a and b are in the ratio a : b = 3 : 4. Given that a = 15, work out the value of b.
Answer key
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (b) 1:1.6 — To write 5 : 8 in the form 1 : n, divide both parts by 5, the first number, so that it becomes 1: 5 ÷ 5 = 1 and 8 ÷ 5 = 1.6, giving 1 : 1.6. Dividing both parts by 8 instead gives 0.6 : 1 (5 ÷ 8 = 0.625, rounded to 0.6) — the first part is no longer 1, so this is not in the required form. Dividing 5 by 8 but writing the result after the 1 gives 1 : 0.6, which divides in the wrong direction: n must come from 8 ÷ 5, not 5 ÷ 8. A slip in the division 8 ÷ 5, rounding it to 1.5 instead of the correct 1.6, gives 1 : 1.5.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (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.
- (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) 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 : 27 — The edge lengths are in the ratio 2 : 6, which simplifies to 1 : 3. Volumes scale with the cube of the length ratio, so the volume ratio is 1³ : 3³ = 1 : 27. Giving 1 : 3 uses the length ratio without cubing it. Giving 1 : 9 squares the length ratio, which is the rule for areas, instead of cubing it, which is the rule for volumes. Giving 27 : 1 has the ratio the right way round for larger to smaller, not smaller to larger as the question asks.
- (c) 3 hours — Method: in inverse proportion the product of the two quantities is constant, and here that product is the distance. Working: 60 × 4 = 240 km, so at 80 km/h the time is 240 ÷ 80 = 3. Answer: 3 hours. The distractors: 5 hours 20 minutes comes from treating the relationship as direct, working out 4 × 80 ÷ 60; 2 hours 40 minutes comes from cutting the time by the fraction the speed rose by — the speed went up by one third, so the time was cut by one third — which is not how inverse proportion works; 4 hours comes from dividing the 240 km by the original speed of 60 km/h again instead of by the new speed.
- (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.
- (d) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (b) 2.5 hours — Method: when two objects move in opposite directions the gap between them grows at the sum of their speeds, so divide the required gap by that combined rate. Working: 80 + 60 = 140 km of gap each hour, and 350 ÷ 140 = 2.5. Answer: 2.5 hours. The distractors: 17.5 hours comes from subtracting the speeds, 80 − 60 = 20, which is the rate for two cars travelling in the same direction; 5 hours comes from using the mean of the two speeds, 70 km/h, instead of their sum; 4.375 hours comes from dividing 350 by 80 and ignoring the second car altogether.
- (b) 24 km/h — First convert 45 minutes to hours: 45 ÷ 60 = 0.75 hours. Then divide the distance by the time: 18 ÷ 0.75 = 24 km/h. Reading 45 minutes as 0.45 hours (writing the minutes after the decimal point instead of dividing by 60) gives 18 ÷ 0.45 = 40 km/h. Working out 18 × 0.75 = 13.5 multiplies by the time instead of dividing. Working out 18 ÷ 45 = 0.4 divides by 45 without ever converting the minutes to hours. The cyclist's average speed is 24 km/h.
- (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.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (b) 1 : 1.875 — To write a ratio in the form 1 : n, divide both parts by the first part, 8: 8 ÷ 8 = 1 and 15 ÷ 8 = 1.875, giving 1 : 1.875. Giving 1 : 0.53 divides the wrong way round, computing 8 ÷ 15 instead of 15 ÷ 8. Giving 1.875 : 1 has the two parts of the answer swapped, which is the form n : 1, not 1 : n. Giving 8 : 1.875 divides only the second part by 8, so the first part is still 8, not 1.
- (d) £144 — Method: find the length (perimeter) scale factor by taking the square root of the area ratio, then apply it to the cost. Working: 12 : 27 simplifies to 4 : 9, and the square root of each part gives the length ratio 2 : 3, so the scale factor from the smaller to the larger pond is 3 ÷ 2 = 1.5. Cost = £96 × 1.5 = £144. Answer: £144. £216 comes from using the area ratio itself as the cost ratio, £96 × (27 ÷ 12) = £216, without taking the square root. £64 comes from using the length ratio the wrong way round, £96 × (2 ÷ 3) = £64. £111 comes from simply adding the difference in area, 27 − 12 = 15, onto the original cost, £96 + £15 = £111, instead of scaling proportionally.
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
- (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) 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) 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.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 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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