Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.The number of biscuits, b, that each guest at a party receives is inversely proportional to the number of guests, g. Which equation could show this relationship?
- 2.A scale drawing of a garden uses a scale of 1 : 20. Write down a formula for the real length, L metres, in terms of the length on the drawing, d centimetres, given that 1 metre = 100 centimetres.
- 3.The length of ribbon A is 5/12 of the length of ribbon B. Write the length of ribbon B as a fraction of the length of ribbon A.
- 4.The number of pupils who walk to school is 3 times the number who cycle. Write the ratio of the number who cycle to the number who walk, in its simplest form.
- 5.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.
- 6.A block of wood has a mass of 60 g and a volume of 100 cm³. Work out the density of the block, in g/cm³.
- 7.A cyclist rides 17.5 km on Saturday and 12.25 km on Sunday. Write the distance ridden on Saturday as a fraction of the distance ridden on Sunday, giving your answer in its simplest form.
- 8.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 9.A scale model of a bus is built so that 5 cm on the model represents 60 cm on the real bus. Write the ratio of the length of the model to the length of the real bus, in its simplest form.
- 10.Paint costs £14.40 for every 20 m² of wall it covers. Assuming the same rate, work out the cost of the paint needed to cover 56 m² of wall.
- 11.Write 1/5 as a percentage.
- 12.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.
- 13.An oven is preheating. Its temperature, T °C, is plotted against time, t minutes, on a straight-line graph. The line passes through the points (2, 60) and (6, 140). Work out the gradient of the line.
- 14.3 builders put up a fence in 12 days. All the builders work at the same rate, so the number of days is inversely proportional to the number of builders. Work out how long 9 builders take to put up the same fence.
- 15.The ratio of two lengths is 0.6 : 1.5. Write this ratio in its simplest whole-number form.
Answer key
- (d) b = k ÷ g — Inverse proportion means b × g stays constant (equal to k), so b = k ÷ g. 'b = kg' is the equation for direct proportion, not inverse. 'b = k + g' treats the relationship as additive, which is not proportion at all. 'b = g ÷ k' is direct proportion in disguise: rearranging it gives g = kb, so b still grows as g grows. For inverse proportion the g must be underneath the constant, not above it.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (a) 12/5 — If A is 5/12 of B, then B is the reciprocal of that fraction times A: flip 5/12 to get 12/5, so B is 12/5 of A. 5/12 comes from keeping the same fraction without flipping it, treating the relationship as if it works the same way in both directions. 7/12 comes from computing 1 − 5/12 = 7/12, which is not how a fraction reverses. 12/7 comes from subtracting 5 from 12 to get 7, and writing 12 over that, instead of swapping the numerator and denominator of 5/12.
- (b) 1:3 — Let the number who cycle be 1 part. Since 3 times as many pupils walk, the number who walk is 3 parts. The ratio of cycle to walk is 1:3. (3:1 comes from writing the ratio the wrong way round, walk to cycle. 1:4 comes from treating “3 times as many” as “3 more”, so walk = cycle + 3, giving parts 1 and 4. 3:4 comes from comparing the number who walk with the total number of pupils, 3 parts out of 4 parts altogether, instead of comparing cycle with walk.)
- (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) 0.6 g/cm³ — Density = mass ÷ volume. 60 ÷ 100 = 0.6 g/cm³. 1.67 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (100 ÷ 60). 6000 g/cm³ comes from multiplying the mass by the volume instead of dividing (60 × 100). 40 g/cm³ comes from subtracting the mass from the volume (100 − 60) instead of dividing.
- (a) 10/7 — Put Saturday's distance over Sunday's distance: 17.5/12.25. Multiply both numbers by 100 to clear the decimals: 1750/1225. Divide both by their highest common factor, 175: 1750÷175 = 10, 1225÷175 = 7, giving 10/7. (7/10 comes from writing the distances the wrong way round. 3/7 comes from finding the difference, 17.5 − 12.25 = 5.25 km, and writing it as a fraction of Sunday's distance, 5.25/12.25. 10/17 comes from comparing Saturday's distance to the total distance ridden, 17.5/29.75.)
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (d) 1:12 — Write the ratio model : real bus using the numbers in the question: 5 : 60. Divide both parts by their highest common factor, 5: 5÷5 = 1, 60÷5 = 12, giving 1 : 12. (12:1 comes from writing the ratio the wrong way round, real bus : model. 1:55 comes from dividing the first part correctly, 5÷5 = 1, but subtracting on the second part, 60 − 5 = 55, instead of dividing it by 5 as well. 5:12 comes from dividing only the real length by 5 and leaving the model length unchanged — both parts of a ratio must be divided by the same number.)
- (a) £40.32 — Find the cost per square metre from the rate given: £14.40 ÷ 20 = £0.72 per m². Then multiply by the area to be covered: £0.72 × 56 = £40.32. Working out 14.40 × 20 ÷ 56 ≈ £5.14 uses the ratio the wrong way round, scaling down as if 56 m² needed less paint than 20 m². Stopping at £0.72 only gives the cost per square metre, not the cost for the whole wall. Working out 14.40 + (56 − 20) = £50.40 adds the extra square metres straight onto the cost in pounds, treating square metres and pounds as the same kind of quantity. Covering 56 m² costs £40.32.
- (a) 20% — Method: a percentage is a number of parts per hundred, so divide the numerator by the denominator and multiply by 100. Working: 1 ÷ 5 = 0.2, and 0.2 × 100 = 20; the same result comes from scaling 1/5 to 20/100. Answer: 20%. The distractors: 5% comes from reading the denominator 5 straight off as the percentage; 25% comes from confusing 1/5 with 1/4, the fraction met most often in percentage work; 50% comes from a place value slip in the division, writing 1 ÷ 5 as 0.5 instead of 0.2.
- (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.
- (d) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (a) 4 days — Method: the fence is a fixed amount of work, so builders × days is constant; find that product and divide it by the new number of builders. Working: 3 × 12 = 36 builder-days of work, so with 9 builders the time is 36 ÷ 9 = 4 days. Answer: 4 days. The distractors: 6 days comes from halving the 12 days because there are more builders, rather than dividing by the factor of 3 by which the workforce has grown; 36 days is the constant product of builders and days, given as a number of days instead of being shared between the builders; 9 days comes from taking 3 days off the 12, treating three extra builders as three fewer days, which is additive rather than proportional.
- (d) 2:5 — Multiply both parts by 10 to clear the decimals: 6 : 15. Divide both parts by their common factor 3: 2 : 5.
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