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.
Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.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.
- 2.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
- 3.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 4.The value of a delivery van, in £, is plotted against its age, in years, since it was bought. At age 2 years, the gradient of the tangent to the graph is −950. What does this tell you about the van at age 2 years?
- 5.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.
- 6.A statue exerts a downward force of 68.4 N on its base, which has an area of 2.85 m². Work out the pressure that the statue exerts on its base, in N/m².
- 7.A patient is given a dose of 200 mg of a drug. Each hour, 30% of the drug remaining in the bloodstream is eliminated, and then a further dose of 50 mg is given. Using the recurrence D_{n+1} = 0.7D_n + 50, with D_0 = 200, work out the amount of drug in the bloodstream after 2 hours.
- 8.A crate exerts a force of 84 N on the floor through a base with an area of 2.4 m². Work out the pressure on the floor, in N/m².
- 9.A water tank is a cuboid measuring 30 cm × 20 cm × 10 cm. The tank is full of water. Work out how many litres of water it holds.
- 10.Jamal invests £600 in a savings account paying 3% simple interest per year. Work out the total amount in the account after 4 years.
- 11.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.
- 12.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.
- 13.A machine is bought for £8500. Its value depreciates by 6% each year. Work out the value of the machine after 3 years, to the nearest pound.
- 14.y is directly proportional to x². When x = 3, y = 45. Construct the equation connecting x and y, then work out the value of y when x = 5.
- 15.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
Answer key
- (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.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (c) Falling at £950 per year — The gradient of a tangent on a value-age graph is a rate, in pounds per year, so −950 means the van's value is falling at £950 per year at that instant. Writing this as 950% per year mistakes a rate measured in pounds per year for a percentage — the units of a gradient come from the units on the two axes, £ and years, not from a percentage. Saying the value 'falls by £950 over the next year' treats the instantaneous rate at age 2 as if it stayed constant for a whole year, which finds an average future change, not the instantaneous rate at age 2 itself. Reading the sign the wrong way round gives 'rising at £950 per year', which would mean the van is gaining value. Always match the units of a gradient to the units on the two axes of the graph.
- (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.
- (a) 24 N/m² — Pressure = force ÷ area. 68.4 ÷ 2.85 = 24 N/m². 194.94 N/m² comes from multiplying the force by the area instead of dividing (68.4 × 2.85). 65.55 N/m² comes from subtracting the area from the force (68.4 − 2.85) instead of dividing. 0.04 N/m² comes from dividing the area by the force instead of the force by the area (2.85 ÷ 68.4).
- (c) 183 mg — Apply the decay, then add the new dose, once for each hour. Hour 1: 0.7 × 200 = 140, then 140 + 50 = 190. Hour 2: 0.7 × 190 = 133, then 133 + 50 = 183, so there is 183 mg after 2 hours. Forgetting the top-up dose and only applying the decay gives 0.7 × 200 = 140, then 0.7 × 140 = 98 — this ignores that a further 50 mg is given every hour. Adding the 50 mg BEFORE the decay is applied, instead of after, gives 0.7 × (200 + 50) = 175, then 0.7 × (175 + 50) = 157.5, which changes how much of the dose is eliminated in the same hour it is given. Multiplying by 0.3, the percentage ELIMINATED, instead of by 0.7, the percentage REMAINING, gives 0.3 × 200 + 50 = 110, then 0.3 × 110 + 50 = 83 — this mixes up the amount that leaves the bloodstream with the amount that stays in it. Always check whether a percentage describes what remains or what is removed before choosing the multiplier.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (b) 6 litres — Method: find the volume of the cuboid in cm³, then change cm³ into litres using 1 litre = 1000 cm³. Working: 30 × 20 × 10 = 6000 cm³, and 6000 ÷ 1000 = 6. Answer: 6 litres. The distractors: 60 litres comes from using 1 litre = 100 cm³; 600 litres comes from using 1 litre = 10 cm³; 0.6 litres comes from using 1 litre = 10 000 cm³.
- (c) £672 — Simple interest per year = 3% of £600 = £18. Over 4 years the interest is 18 × 4 = £72. Total in the account = £600 + £72 = £672. A student who gives just the interest, without adding it to the principal, writes £72. A student who adds only one year's interest instead of four gets £600 + £18 = £618. A student who wrongly compounds the interest each year gets 600 × 1.03⁴ = £675.31.
- (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.
- (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.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
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