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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (a) 8/5 — Scaling down from 8 people to 5 people uses the multiplier 5/8. To reverse this and scale back up from 5 people to 8 people, use the reciprocal of that multiplier: flip 5/8 to get 8/5. 5/8 comes from using the forward (scaling down) multiplier again, instead of reversing it. 3/5 comes from writing the difference in people (8 − 5 = 3) over 5, instead of using the reciprocal of 5/8. 25/64 comes from multiplying the forward multiplier by itself (5/8 × 5/8), instead of finding its reciprocal.
- (b) The ratio C : m is not constant because the formula includes a fixed charge of £3 as well as the charge per mile. — For the ratio C : m to stay constant, C must be directly proportional to m, i.e. C = km with no constant term. Because of the +3 fixed charge, C is not directly proportional to m: for example m = 1 gives C = 5.5 (ratio 5.5 : 1), while m = 10 gives C = 28 (ratio 2.8 : 1) — the ratio has changed.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
- (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) £14224 — Value after 2 years: £15000 × 1.04 × 1.04 = £16224. Money left after buying the trailer: £16224 − £2000 = £14224. £14200 comes from treating the two 4% increases as a single flat 8% increase applied once instead of compounding: £15000 × 1.08 = £16200, and £16200 − £2000 = £14200. £13600 comes from applying the 4% increase only once, for 1 year instead of 2: £15000 × 1.04 = £15600, and £15600 − £2000 = £13600. £18224 comes from adding the £2000 instead of subtracting it: £16224 + £2000 = £18224.
- (d) 2:5 — The ratio of the y-values equals the ratio of the coefficients of x, since x cancels: 2x : 5x = 2 : 5.
- (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.
- (a) 1 : 2 000 000 — Method: a scale is a ratio between two lengths written in the same unit, reduced so that the map distance is 1. Working: 120 km = 120 × 1000 × 100 = 12 000 000 cm, so the ratio is 6 : 12 000 000, and dividing both parts by 6 gives 1 : 2 000 000. Answer: 1 : 2 000 000. The distractors: 1 : 12 000 000 comes from writing the real length in centimetres without dividing by the 6 cm on the map; 1 : 200 000 comes from taking 120 km as 1 200 000 cm, one conversion step short, before dividing by 6; 1 : 20 000 comes from converting 120 km to 120 000 m and treating those metres as centimetres.
- (b) 75 cm² — Areas of similar shapes are in the ratio of the squares of their lengths. Squaring both parts of 3 : 5 gives an area ratio of 9 : 25, so the larger area is 25/9 of the smaller one. Working with the smaller area: 27 ÷ 9 = 3, and 3 × 25 = 75. The area of the larger rectangle is 75 cm².
- (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.
- (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.
- (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.
- (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.
Build your own mix at the worksheet builder.