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.
Calculator
Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.A scale drawing of a park has a scale of 1 : 2500. On the drawing, the distance between the entrance and the lake is 4.4 cm. A jogger runs from the entrance to the lake and then back to the entrance. Work out the total distance the jogger runs, in kilometres.
- 2.A lorry is carrying a load of mass m tonnes. Given that 1 tonne = 1000 kg, write down an expression for the mass of the load in kilograms.
- 3.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 4.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 5.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
- 6.In a choir of 30 singers, 40% are altos and the rest are sopranos. Write the number of sopranos as a fraction of the number of altos. Give your answer in its simplest form.
- 7.A furniture maker builds two mathematically similar bookcases. The smaller bookcase is 40 cm tall and uses 2 m² of wood. The larger bookcase is 60 cm tall. Work out the area of wood needed for the larger bookcase, giving your answer in m².
- 8.The number of subscribers to a streaming app, in thousands, is plotted against time, in months since launch. A tangent to the graph at t = 6 months has gradient 4.8. A tangent at t = 18 months has gradient 1.1. A manager claims the app is growing faster at 18 months than it was at 6 months. Work out how the growth rate has changed, and decide whether the manager is correct.
- 9.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.
- 10.A factory machine produces bottles at a constant rate. In 45 minutes it produces 810 bottles. The factory needs 2,160 bottles for an order. Working at the same rate, work out how many minutes it will take to produce the order.
- 11.The density of a type of solid plastic is 0.9 g/cm³. Work out the mass of 0.5 m³ of the plastic, in kilograms.
- 12.A shop sells identical pens. The total cost, C pence, of n pens is given by C = 12n. Write the ratio n : C in its simplest form.
- 13.The height of a candle, in cm, is measured as it burns: t = 0 min, height = 20.0; t = 3 min, height = 18.7; t = 5 min, height = 17.5; t = 6 min, height = 16.6; t = 7 min, height = 15.4; t = 9 min, height = 13.4. Which pair of readings gives the best estimate of the instantaneous rate of change of the height at t = 6 minutes, and why?
- 14.The cost of manufacturing a spherical container is proportional to the cube of its radius. A container of radius 3 cm costs £54 to manufacture. Construct the equation connecting cost C and radius r, then work out the cost of a container of radius 5 cm.
- 15.A laptop costs £800 when new. Its value decreases by 25% of its value at the start of each year. Work out how much value the laptop loses in the second year.
Answer key
- (b) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
- (a) 1000m — Method: kilograms are a smaller unit than tonnes, so change tonnes into kilograms by multiplying by 1000. Working: m tonnes = m × 1000 kg = 1000m kg. So the expression is 1000m. Distractor m/1000 comes from dividing by 1000 instead of multiplying, which would make the number of kilograms smaller than the number of tonnes, the wrong way round. Distractor 1000 + m comes from adding the conversion factor instead of multiplying by it. Distractor m − 1000 comes from subtracting the conversion factor instead of multiplying by it.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
- (d) 3/2 — Work out the number of altos: 40% of 30 = 12. The rest are sopranos, so there are 30 − 12 = 18 sopranos. Form the fraction 18/12; both numbers share a factor of 6, so 18 ÷ 6 = 3 and 12 ÷ 6 = 2, giving 3/2. 2/3 comes from writing the fraction the wrong way round, as altos over sopranos (12/18). 3/5 comes from comparing the sopranos with the whole choir (18/30), instead of with the altos. 7/3 comes from miscalculating 30 − 12 as 28 instead of 18, then forming 28/12.
- (b) 4.5 m² — The height ratio is 40 : 60, which simplifies to 2 : 3, so the larger bookcase is 1.5 times as tall as the smaller one. Areas scale with the square of the length scale factor, so the wood needed scales by 1.5² = 2.25. 2 × 2.25 = 4.5, so the larger bookcase needs 4.5 m² of wood. Giving 3 m² uses the length scale factor, 1.5, without squaring it (2 × 1.5 = 3). Giving 6.75 m² cubes the scale factor, 1.5³ = 3.375, as if wood coverage were a volume (2 × 3.375 = 6.75). Giving 2.25 m² is the squared scale factor on its own, without multiplying by the smaller bookcase's wood area of 2 m².
- (b) No — rate fell by 3.7 thousand/month — Each tangent gradient is the instantaneous growth rate, in thousand subscribers per month. To compare them, subtract the later rate from the earlier one: 4.8 − 1.1 = 3.7. Since 1.1 is less than 4.8, the growth rate has fallen by 3.7 thousand subscribers per month, so the manager is wrong — the app is growing more slowly at 18 months, not faster. Subtracting the other way round and calling the result a rise, 'rate rose by 3.7 thousand/month', gets the direction backwards: the later gradient is the smaller of the two. Adding the two gradients, 4.8 + 1.1 = 5.9, and calling this a combined rate that shows speeding up, is the wrong operation for comparing two rates. Treating the difference 3.7 as a total number of subscribers lost, rather than a rate in thousands per month, confuses a rate with a count. Always subtract the two rates in a sensible order and keep the units in thousands per month.
- (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.
- (b) 120 minutes — Method: find the rate in bottles per minute, then divide the order size by the rate. Working: rate = 810 ÷ 45 = 18 bottles per minute. Time = 2,160 ÷ 18 = 120 minutes. Wrong options: 1,350 minutes comes from subtracting 810 from 2,160 instead of using the rate; 48 minutes comes from dividing the order size by the original time (2,160 ÷ 45) instead of the rate; 108 minutes comes from rounding the rate to 20 bottles per minute before dividing.
- (b) 450.00 kg — 1 m³ = 100 × 100 × 100 = 1,000,000 cm³, so 0.5 m³ = 500,000 cm³. Mass = density × volume = 0.9 × 500,000 = 450,000 g. Converting to kilograms by dividing by 1000 gives 450,000 ÷ 1000 = 450.00 kg. Skipping the m³-to-cm³ conversion and multiplying 0.9 × 0.5 = 0.45 treats the volume as if it were already 0.5 cm³, giving 0.45 kg. Finding the mass correctly in grams, 450,000 g, but not converting to kilograms leaves 450000.00 kg, out by a factor of 1000. Using the area conversion factor of 10,000, as if converting m² to cm², instead of the volume factor of 1,000,000 gives 0.5 × 10,000 = 5,000 'cm³', and a mass of 0.9 × 5,000 = 4,500 g, which is 4.50 kg.
- (b) 1:12 — C = 12n means that for every 1 pen there are 12 pence of cost, so n : C = 1 : 12, and the highest common factor of 1 and 12 is 1, so this is already in its simplest form. Writing C : n instead of n : C gives 12 : 1, the ratio the wrong way round. Reading C = 12n as '12 more than n' instead of '12 times n', so n = 1 gives C = 13, gives 1 : 13, from adding instead of multiplying. Choosing n = 12, so C = 12 × 12 = 144, gives the correct pair of values n : C = 12 : 144, but the ratio the right way round left unsimplified — 12 and 144 share a common factor of 12, which has not been cancelled.
- (b) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (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.
- (a) £150 — Value after year 1: £800 × 0.75 = £600. Value after year 2: £600 × 0.75 = £450. The loss during the second year alone is £600 − £450 = £150. £450 comes from giving the value remaining after 2 years, not the amount lost during the second year. £200 comes from working out the loss during the first year instead of the second: £800 − £600 = £200. £350 comes from working out the total loss over both years instead of just the second year's loss: £800 − £450 = £350.
Build your own mix at the worksheet builder.