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.Maya is paid £105.30 for working 9 hours. Work out her rate of pay, in £ per hour.
- 2.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
- 3.Two quantities x and y are in direct proportion. When x = 8, the value of y is 20. Work out the value of y when x = 14.
- 4.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.
- 5.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.
- 6.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.
- 7.A car uses fuel at a constant rate. It uses 24 litres of fuel to travel 300 km. Assuming the same rate, work out how many litres of fuel are needed for a journey of 175 km.
- 8.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.
- 9.Two mathematically similar triangular flags have areas in the ratio 4 : 25. The height of the smaller flag is 6 cm. Work out the height of the larger flag.
- 10.Write the ratio 8 : 15 in the form 1 : n.
- 11.Work out the value of x when 4/x = 2/5.
- 12.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 13.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 14.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.
- 15.y is directly proportional to x. When x = 7, the value of y is 21. Work out the value of x when y = 12.
- 16.Priya is paid £58.50 for working 7.5 hours on a Saturday. Work out her rate of pay, in £ per hour.
- 17.y is inversely proportional to x. When x = 5, y = 8. Work out the value of y when x = 10.
- 18.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.
- 19.Simplify the ratio 45 : 30 : 75 to its simplest form.
- 20.Two mathematically similar circles have radii 4 cm and 20 cm. Write the ratio of the area of the smaller circle to the area of the larger circle in its simplest form.
- 21.The cost of manufacturing a spherical container is proportional to the cube of its radius. A container of radius 3 cm costs £54 to manufacture. Construct the equation connecting cost C and radius r, then work out the cost of a container of radius 5 cm.
- 22.Triangle ABC is mathematically similar to triangle PQR, with AB corresponding to PQ and BC corresponding to QR. AB = 6 cm, BC = 8 cm and PQ = 12 cm. Work out the length of QR.
- 23.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.
- 24.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.
Answer key
- (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.
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
- (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) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (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) 14 litres — Method: find the amount of fuel used per km first, then use it to find the fuel needed for 175 km. Working: 24 ÷ 300 = 0.08 litres per km, and 0.08 × 175 = 14 litres. So 14 litres are needed. Distractor 24 litres comes from assuming the same amount of fuel is used no matter the distance, without scaling. Distractor 21 litres comes from misreading the original distance as 200 km instead of 300 km. Distractor 1.4 litres comes from a decimal-point slip, giving an answer ten times too small.
- (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.
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (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.
- (a) 10 — Method: two equal fractions can be rearranged by cross-multiplying, multiplying each numerator by the other denominator. Working: 4 × 5 = 2 × x, so 2x = 20 and x = 20 ÷ 2 = 10. Answer: 10. The distractors: 20 comes from cross-multiplying to 4 × 5 = 20 and stopping there, without dividing by the 2; 8 comes from multiplying the two numerators, 4 × 2; 2.5 comes from working only with the right-hand fraction, 5 ÷ 2, and ignoring the 4.
- (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) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (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.
- (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.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (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.
- (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) 1 : 25 — The radii are in the ratio 4 : 20, which simplifies to 1 : 5. Areas scale with the square of the length ratio, so the area ratio is 1² : 5² = 1 : 25. Giving 1 : 5 uses the radius ratio without squaring it. Giving 1 : 10 doubles the radius ratio instead of squaring it. Giving 25 : 1 has the areas the right way round for larger to smaller, not smaller to larger.
- (b) £250 — Since cost is proportional to the cube of the radius, C = kr³. Using r = 3, C = 54: 3³ = 27, so 54 = k × 27, giving k = 54 ÷ 27 = 2. The equation is C = 2r³. When r = 5: 5³ = 125, so C = 2 × 125 = 250. Treating the relationship as proportional to r² instead of r³ gives k = 54 ÷ 9 = 6 and then C = 6 × 25 = 150, which models area scaling, not volume scaling. Treating it as proportional to r itself gives k = 54 ÷ 3 = 18 and then C = 18 × 5 = 90. Finding k correctly from the cube but then multiplying it by the radius instead of by the cube of the radius gives 2 × 5 = 10, which applies the right constant to the wrong power of r. The cost of a container of radius 5 cm is £250.
- (c) 16 cm — Corresponding sides of similar triangles are all in the same ratio. Use the pair whose lengths are both known: the scale factor from triangle ABC to triangle PQR is 12 ÷ 6 = 2. Since QR corresponds to BC, multiply BC by that scale factor: 8 × 2 = 16, so QR = 16 cm.
- (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.
- (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.
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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