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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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.Write the ratio 8 : 15 in the form 1 : n.
- 2.A company had 8000 employees. The number of employees decreased by 5% in the first year, and then increased by 5% in the second year. Work out the number of employees at the end of the second year, to the nearest whole number.
- 3.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
- 4.A coach journey is 372 miles in total. After a stop, the coach has travelled 217 miles. Write the distance still to travel as a fraction of the total journey. Give your answer in its simplest form.
- 5.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.
- 6.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 7.The temperature of a chemical reaction, in °C, is modelled by T = 80 − 6t + 0.5t², where t is the time in minutes after the reaction starts. Which of these four statements about the reaction between t = 2 and t = 6 minutes is correct?
- 8.A charity fun run raises money through entry fees and donations. Entry fees raise £1,260, which is 60% of the total amount raised. Work out how much money was raised through donations.
- 9.A jug of squash is made by mixing water and syrup in the ratio 6:1. Nia wants to make 8.4 litres of squash. Work out how much syrup she needs, in litres.
- 10.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 11.The value of a rare coin increases by 12% each year. The coin is currently worth £270. Work out the value of the coin after 2 years, giving your answer to the nearest penny.
- 12.Two scale plans of the same football pitch are drawn: Plan A has a scale of 1 : 1000 and Plan B has a scale of 1 : 200. A pitch length is 8 cm on Plan A. Work out the length of the same pitch on Plan B, in centimetres.
- 13.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.
- 14.The amount of fuel left in a car's tank, F litres, is plotted against the distance travelled, d miles, and the points lie on a straight line. The line passes through (0, 45) and (150, 15). Work out the gradient of the line and say what it tells you.
- 15.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
Answer key
- (b) 1 : 1.875 — To write a ratio in the form 1 : n, divide both parts by the first part, 8: 8 ÷ 8 = 1 and 15 ÷ 8 = 1.875, giving 1 : 1.875. Giving 1 : 0.53 divides the wrong way round, computing 8 ÷ 15 instead of 15 ÷ 8. Giving 1.875 : 1 has the two parts of the answer swapped, which is the form n : 1, not 1 : n. Giving 8 : 1.875 divides only the second part by 8, so the first part is still 8, not 1.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total journey.
- (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) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (c) Average rate, t = 2 to 6, is −2°C/min — First find the temperature at each end of the interval. At t = 2, T = 80 − 6 × 2 + 0.5 × 2² = 80 − 12 + 2 = 70. At t = 6, T = 80 − 6 × 6 + 0.5 × 6² = 80 − 36 + 18 = 62. The average rate of change over the interval is the change in T divided by the change in t: 62 − 70 = −8, then −8 ÷ 4 = −2°C per minute, so the statement about the average rate is correct. The instantaneous rate at t = 6 is not −2: completing the square gives T = 0.5(t − 6)² + 62, so t = 6 is the turning point of the curve, where the tangent is horizontal and the rate is 0°C per minute — the reaction has stopped cooling by then. The instantaneous rate at t = 2 is not −2 either: a short chord centred on t = 2, from t = 1.9 (T = 70.405) to t = 2.1 (T = 69.605), gives −0.8 ÷ 0.2 = −4°C per minute, so the reaction is cooling twice as fast at the start of the interval as the average over it. Saying the temperature falls 2°C in total confuses the RATE, −2°C per minute, with a TOTAL drop, which is 70 − 62 = 8°C over the four minutes. Always check whether a figure is a rate, per minute, or a total change.
- (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) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (c) 40 cm — First find the real pitch length using Plan A's scale: 8 × 1000 = 8000 cm. Now find its length on Plan B using Plan B's scale: 8000 ÷ 200 = 40 cm. Assuming both plans use the same scale and leaving the length as 8 cm ignores that Plan B has a different scale. Using the ratio of the two scales upside down, 200 ÷ 1000 = 0.2, then 8 × 0.2 = 1.6, gives 1.6 cm — Plan B is drawn to a larger scale than Plan A, so lengths on it should be bigger, not smaller. Dividing by 2000 instead of 200 gives 4 cm, ten times too small.
- (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.
- (c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
- (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.
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