Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (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) 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.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (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.
- (c) The tangent is horizontal, so its gradient is 0. — Method: at any point where a distance–time graph is momentarily neither increasing nor decreasing, the tangent to the graph at that point is horizontal, and the gradient of a horizontal line is 0 — this is the instantaneous rate of change at that instant. Working: since the hiker's distance is neither increasing nor decreasing at t = 45 minutes, the tangent there is horizontal, so its gradient is 0. Claiming the tangent is vertical, with an undefined gradient, is the opposite of what the stem says: a vertical tangent would mean the distance was changing infinitely fast at that instant, not that it had stopped changing, and on a distance–time graph it cannot happen at all. Reading the gradient as 45, the time value given in the stem, mistakes a value used to LOCATE the point for the rate of change AT that point. Claiming the gradient cannot be found without also knowing the distance at t = 45 minutes overlooks that 'momentarily stationary' already tells you the rate of change directly, without needing to read any distance value at all. Whenever a stem tells you a quantity is momentarily not changing, that is telling you the instantaneous rate of change directly — it is 0, and no further data is needed to find it.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (d) 4:1 — Write the ratio online : in-store using the numbers in the question: 180 : 45. Divide both parts by their highest common factor, 45, to give 4 : 1. (1:4 comes from writing the ratio the wrong way round, in-store : online. 3:1 comes from subtracting the in-store orders from the online orders, 180 − 45 = 135, and comparing that to the in-store orders, 135:45, instead of dividing. 4:5 comes from comparing the online orders to the total number of orders, 180:225.)
- (d) 13/10 — First find the selling price: £150 + £45 = £195. Put the selling price over the cost price: 195/150. Divide both numbers by their highest common factor, 15: 195÷15 = 13, 150÷15 = 10, giving 13/10. (10/13 comes from writing the prices the wrong way round. 3/10 is just the profit written as a fraction of the cost price, 45/150, not the selling price. 13/23 comes from comparing the selling price to the combined total of the cost price and the selling price, 195/345.)
- (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.
- (d) A falling curve that never touches either axis — Method: inverse proportion means the product of the two quantities is constant, so P = k ÷ Q; as Q grows P shrinks, and P can never reach zero because k divided by a number is never zero. Working: taking k = 12 as an example, the pairs (1, 12), (2, 6), (3, 4), (6, 2) and (12, 1) drop steeply at first and then flatten out, so the graph is a curve that approaches both axes without meeting either of them. Answer: a falling curve that never touches either axis. The distractors: 'a straight line through the origin' is the graph of direct proportion, P = kQ, which is the opposite relationship; 'a straight line with a negative gradient' is the commonest error, reading 'P falls as Q rises' as a straight line, but on such a line P would drop by the same amount for every increase in Q and would cross the horizontal axis into negative values; 'a straight line crossing the vertical axis above zero' is a relationship of the form P = mQ + c, in which P and Q are not proportional at all.
- (a) 8 cm — Method: a scale of 1 : n means the real distance is n times the distance on the map, so to go from the real distance back to the map distance, put both lengths in the same unit and then divide by the scale. Working: 1 km = 100 000 cm, so 8 km = 8 × 100 000 = 800 000 cm; 800 000 ÷ 100 000 = 8. Answer: 8 cm. The distractors: 800000 cm comes from converting the 8 km into centimetres and stopping there, so the division by the scale — the inverse operation the question asks for — is never done; 80 cm comes from taking a metre to be 1000 cm, which turns 8 km into 8 × 1000 × 1000 = 8 000 000 cm and gives 8 000 000 ÷ 100 000 = 80; 0.08 cm comes from taking a kilometre to be 1000 cm, which turns 8 km into 8000 cm and gives 8000 ÷ 100 000 = 0.08.
- (c) 16% — Method: scale each concentration to its own mass to find the salt it contains, add the two masses of salt, then write the ratio of salt to mixture per 100 g. Working: 10:100 = x:300 gives 30 g of salt, and 25:100 = y:200 gives 50 g of salt; the mixture holds 30 + 50 = 80 g of salt in 300 + 200 = 500 g of solution; 80:500 = 16:100. Answer: 16%. The distractors: 17.5% is the mean of 10% and 25%, which ignores that there is more of the weaker solution than of the stronger one; 19% comes from swapping the two concentrations over, working out (300 × 25% + 200 × 10%) ÷ 500; 26.7% comes from dividing the 80 g of salt by the 300 g of the first solution rather than by the 500 g of mixture.
- (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.
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