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.Write 250 g : 2 kg as a ratio in its simplest form.
- 2.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.
- 3.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.
- 4.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.
- 5.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.
- 6.A toy manufacturer makes a model aircraft that is mathematically similar to the real aircraft, at a scale of 1 : 48. The wingspan of the model is 15 cm. Work out the wingspan of the real aircraft, giving your answer in metres.
- 7.The manufacturing cost of a specialist lens, in pounds, is proportional to the square root of its diameter, in millimetres. A lens of diameter 9 mm costs £12 to manufacture. Construct the equation connecting cost C and diameter d, then work out the cost of manufacturing a lens of diameter 16 mm.
- 8.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.
- 9.Write 400 g : 1.5 kg as a ratio in its simplest form.
- 10.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
- 11.Write the ratio 5 : 8 in the form 1 : n.
- 12.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.
- 13.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 14.Priya invests £750 in a savings account that pays simple interest. After 3 years, the account contains £840. Work out the annual rate of simple interest.
- 15.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.
- 16.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.
- 17.A 1.5 kg bag of pasta costs £2.85 and a 2.4 kg bag of the same pasta costs £4.32. Work out which bag gives better value for money.
- 18.y is directly proportional to √x. When x = 25, y = 20. Construct the equation connecting x and y, then work out the value of x when y = 32.
- 19.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 20.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
- 21.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
- 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.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 24.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.
Answer key
- (a) 1:8 — Convert 2 kg to grams: 2 kg = 2000 g. The ratio is 250 : 2000. The highest common factor of 250 and 2000 is 250. Divide both parts by 250: 250 ÷ 250 = 1 and 2000 ÷ 250 = 8, so the ratio is 1 : 8. Leaving the kilograms unconverted gives 250 : 2, which simplifies to 125 : 1 — the units on each side are different, so this does not compare like with like. Dividing by 50 instead of 250 gives 5 : 40, which still shares a common factor of 5, so it is not fully simplified. Swapping the order gives 8 : 1, grams to kilograms the wrong way round.
- (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.)
- (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) 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.)
- (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).
- (b) 7.2 m — Multiply the model wingspan by the scale factor: 15 × 48 = 720. This is in centimetres, and 720 cm = 7.2 m, since 1 m = 100 cm. Giving 0.31 m divides by the scale factor instead of multiplying (15 ÷ 48 ≈ 0.31), scaling the model down rather than the real aircraft up. Giving 72 m converts centimetres to metres by dividing by 10 instead of 100. Giving 0.72 m converts by dividing by 1000 instead of 100.
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (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.
- (a) 4 : 15 — Convert to the same unit first: 1.5 kg = 1500 g, since 1 kg = 1000 g. This gives the ratio 400 : 1500. Divide both parts by their highest common factor, 100, to get 4 : 15. Giving 40 : 150 divides by 10 only, which is not the highest common factor, so it is not fully simplified. Giving 15 : 4 swaps the order. Giving 4 : 1.5 has not converted 1.5 kg into grams, so the two parts are not measured in the same unit.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (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) 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.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (b) 4% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (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.
- (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.
- (b) The 2.4 kg bag, since it costs £1.80 per kg compared with £1.90 per kg for the 1.5 kg bag. — To compare value for money, work out the cost per kilogram for each bag. 1.5 kg bag: £2.85 ÷ 1.5 = £1.90 per kg. 2.4 kg bag: £4.32 ÷ 2.4 = £1.80 per kg. Since £1.80 is less than £1.90, the 2.4 kg bag gives better value. The option comparing £2.85 with £4.32 directly is wrong because it compares the total prices, not the price per kilogram — a bigger bag naturally costs more in total even if it is better value. The option that names the 1.5 kg bag with £1.80 per kg and the 2.4 kg bag with £1.90 per kg has the correct unit prices but has swapped which bag they belong to. The option giving £1.19 per kg and £2.88 per kg comes from dividing each price by the wrong bag's mass (£2.85 ÷ 2.4 and £4.32 ÷ 1.5).
- (d) 64 — Since y is directly proportional to √x, y = k√x. Using x = 25, y = 20: √25 = 5, so 20 = k × 5, giving k = 20 ÷ 5 = 4. The equation is y = 4√x. When y = 32: √x = 32 ÷ 4 = 8, and x = 8² = 64. Stopping at √x = 8 without squaring leaves the square root of x, not x itself. Treating the relationship as if y were proportional to x itself gives k = 20 ÷ 25 = 0.8 and then x = 32 ÷ 0.8 = 40, which is a different relationship entirely. Multiplying instead of dividing when isolating √x gives √x = 32 × 4 = 128, far too large to be a square root here. When y = 32, x = 64.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (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.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
- (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 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (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.
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.
Similar worksheets worth a look
- 🧮 Paper 1 non-calculator warm-up — Higher · 20 questions · ~25 min
- 🎯 Grade 4 pass booster · 25 questions · ~40 min
- 📈 Quadratics: factorise, complete the square, formula · 24 questions · ~45 min
- ⭕ Circle theorems practice · 18 questions · ~40 min