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.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.
- 2.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.
- 3.A sculptor makes two mathematically similar statues. The smaller statue is 20 cm tall and 80 ml of varnish covers its surface. The larger statue is 50 cm tall. Work out how much varnish is needed to cover the surface of the larger statue.
- 4.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.
- 5.Write the ratio 3/4 : 1/2 as a ratio of whole numbers in its simplest form.
- 6.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.
- 7.A delivery driver travels 45 km in the first 30 minutes of a journey, and then a further 75 km in the next 1 hour. Work out her average speed for the whole journey, in km/h.
- 8.A jug of squash is made by mixing water and syrup in the ratio 6:1. Nia wants to make 8.4 litres of squash. Work out how much syrup she needs, in litres.
- 9.A recipe for 8 muffins needs 200 g of sugar. Sam wants to make 20 muffins for a bake sale, and he already has 350 g of sugar. Work out how many more grams of sugar he needs to buy.
- 10.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.
- 11.Write £3.60 : £2.40 as a ratio in its simplest form.
- 12.Order these three values from smallest to largest: 3/8, 0.43, 41%.
- 13.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.
- 14.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.
- 15.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.
- 16.Two mathematically similar cylinders have heights in the ratio 3 : 4. Write the ratio of their volumes in its simplest form.
- 17.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.
- 18.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.
- 19.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.
- 20.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.
- 21.Write 45 minutes : 2 hours as a ratio in its simplest form.
- 22.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.
- 23.Write 2 m : 150 cm : 50 cm as a ratio of whole numbers in its simplest form.
- 24.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.
Answer key
- (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.
- (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) 500 ml — Varnish covers a surface, so the amount needed scales with the area scale factor, which is the square of the length scale factor. The length scale factor is 50 ÷ 20 = 2.5, so the area scale factor is 2.5 × 2.5 = 6.25. The varnish needed for the larger statue is 80 × 6.25 = 500 ml. Using 2.5 on its own would scale a length, not a surface.
- (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) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (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) 80 km/h — Average speed = total distance ÷ total time. Total distance = 45 + 75 = 120 km. Total time = 30 minutes + 1 hour = 1.5 hours. 120 ÷ 1.5 = 80 km/h. 60 km/h comes from treating the 30 minutes as a whole hour, giving a total time of 2 hours instead of 1.5 (120 ÷ 2). 82.5 km/h comes from averaging the two separate speeds (45 ÷ 0.5 = 90 km/h and 75 ÷ 1 = 75 km/h, then (90 + 75) ÷ 2) instead of using total distance over total time. 75 km/h comes from using only the second part of the journey (75 km in 1 hour) and ignoring the first part.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (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) 3 : 2 — Convert both amounts to pence: £3.60 = 360p and £2.40 = 240p, giving the ratio 360 : 240. Divide both parts by their highest common factor, 120, to get 3 : 2. Giving 360 : 240 has not been simplified at all. Giving 2 : 3 swaps the order. Giving 36 : 24 has been divided by 10, which is a common factor but not the highest one, so it is not yet in simplest form.
- (a) 3/8, 41%, 0.43 — Method: convert every value to a decimal so they can be compared on the same scale, then order them. Working: 3/8 = 0.375, 41% = 0.41, and 0.43 stays as 0.43, so from smallest to largest the decimals are 0.375, 0.41, 0.43, giving the order 3/8, 41%, 0.43. Answer: 3/8, 41%, 0.43. The order 3/8, 0.43, 41% comes from comparing 0.43 and 41% as raw digits (43 versus 41) without converting 41% into the decimal 0.41 first, wrongly placing 0.43 before 41%. The order 0.43, 41%, 3/8 comes from placing the values in completely reversed order, from largest to smallest instead of smallest to largest. The order 41%, 0.43, 3/8 comes from ordering the values by their TYPE (percentage, then decimal, then fraction) rather than by their actual size.
- (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) 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.
- (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.
- (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.
- (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.
- (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).
- (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.
- (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.
- (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.
- (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.
- (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
- (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).
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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