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.
- 1.6 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.
- 2.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.
- 3.Two mathematically similar rectangles have lengths in the ratio 3 : 5. The area of the smaller rectangle is 27 cm². Work out the area of the larger rectangle.
- 4.It takes 8 painters 6 days to paint a fence. Working at the same rate, work out how many days 4 painters would take to paint the same fence.
- 5.Two quantities y and z are each in direct proportion to x, and are given by y = 3x and z = 7x. Work out the difference between the value of y and the value of z when x = 4.y = 3x
- 6.A necklace is made using gold beads and silver beads in the ratio 7:3. There are 40 more gold beads than silver beads. Work out the total number of beads in the necklace.
- 7.A coach travels the 54 miles from London to Brighton in 1 hour 30 minutes. Work out the average speed of the coach, in mph.
- 8.In a school choir the ratio of boys to girls is 3:4. When 6 more boys join the choir, the ratio of boys to girls becomes 1:1. Work out how many girls are in the choir.
- 9.A garden centre sells two mathematically similar sacks of grass seed. The amount of lawn a sack can treat is proportional to the volume of seed inside it. The smaller sack is 20 cm tall and treats a lawn of area 30 m². The larger sack is 40 cm tall. A gardener needs to treat a lawn with an area of 500 m² using only the larger sacks. Work out the minimum number of larger sacks needed.
- 10.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 11.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 12.Write 45 minutes : 2 hours as a ratio in its simplest form.
- 13.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.
- 14.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 15.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.
- 16.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.
- 17.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.
- 18.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.
- 19.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 20.Simplify the ratio 45 : 30 : 75 to its simplest form.
- 21.A metal sample has a mass of 342.6 g and a volume of 18 cm³. Work out the density of the sample, in g/cm³, to 1 decimal place.
- 22.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 23.Which of these ratios is equivalent to 6 : 10 : 14?
- 24.A cyclist rides 19.3 km in 47 minutes. Work out her average speed, in km/h, to 1 decimal place.
Answer key
- (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) 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.
- (b) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
- (a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
- (d) 36 mph — First convert 1 hour 30 minutes to hours: 30 minutes is half an hour, so the time is 1.5 hours. Then divide the distance by the time: 54 ÷ 1.5 = 36 mph. Reading 1 hour 30 minutes as 1.3 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 54 ÷ 1.3 ≈ 41.54 mph. Working out 54 ÷ 30 = 1.8 divides by the number of minutes only, ignoring the hour. Working out 54 × 1.5 = 81 multiplies by the time instead of dividing. The coach's average speed is 36 mph.
- (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
- (c) 3 — Method: find the height scale factor, cube it to find the volume (and coverage) scale factor, use it to find one large sack's coverage, then divide the total lawn area by this and round up to a whole number of sacks. Working: height scale factor = 40 ÷ 20 = 2, so coverage scale factor = 2³ = 8, and each large sack covers 30 × 8 = 240 m². 500 ÷ 240 = 2.08…, which rounds UP to 3 whole sacks. Answer: 3. 2 comes from correctly finding that each large sack covers 240 m², but then rounding 500 ÷ 240 down instead of up, which would leave part of the lawn untreated. 5 comes from squaring the height scale factor (2² = 4) instead of cubing it, giving a coverage of only 30 × 4 = 120 m² per sack. 17 comes from forgetting to scale the coverage at all and dividing 500 by the smaller sack's coverage of 30 m².
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (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.
- (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.
- (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) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (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.
- (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.
- (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.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (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.
- (b) 19.0 g/cm³ — Density = mass ÷ volume. 342.6 ÷ 18 = 19.0333…, which rounds to 19.0 g/cm³ (1 d.p.). 6166.8 g/cm³ comes from multiplying the mass by the volume instead of dividing (342.6 × 18). 324.6 g/cm³ comes from subtracting the volume from the mass (342.6 − 18) instead of dividing. 0.1 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (18 ÷ 342.6 = 0.0525…, rounded to 1 d.p.).
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (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.
- (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.
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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