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.A cordial drink is made by mixing cordial concentrate and water in the ratio 1 : 6. Freya says that a 700 ml jug of this drink contains 100 ml of concentrate. Is she correct? Give a reason for your answer.
- 2.Which of these ratios is equivalent to 6 : 10 : 14?
- 3.The ratio of Josh's savings to Mia's savings is 4:7. Mia has £39 more than Josh. Work out Josh's savings.
- 4.A straight-line graph passes through the origin and the point (4, 10). The line represents y = kx. Write down the ratio x : y in its simplest form.
- 5.A charity receives £360 in donations in January and £480 in donations in February. Write the amount received in January as a fraction of the amount received in February, giving your answer in its simplest form.
- 6.A candle burns down at a constant rate. Its height, h cm, is plotted against time, t minutes, on a straight-line graph. The gradient of the line is −0.3. Which statement correctly describes what the gradient means?
- 7.A river 120 km long is drawn as a line 6 cm long on a map. Work out the scale of the map.
- 8.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 9.Water flows into a tank at a constant rate. After 8 minutes the tank holds 200 litres. Write the ratio of time in minutes to volume in litres, in its simplest form.
- 10.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 11.The number of tickets a group can afford is inversely proportional to the price per ticket. At £4 per ticket, the group can afford 12 tickets. Work out how many tickets the group can afford at £6 per ticket.
- 12.A tangent to a curve at the point where x = 3 passes through the points (1, 4) and (5, 20). Work out the gradient of the tangent at x = 3, and hence the instantaneous rate of change of y with respect to x at that point.
- 13.A process is modelled by the recurrence P_{n+1} = 0.6P_n + 40. As n increases, P_n approaches a long-run value L, which satisfies L = 0.6L + 40. Solve this equation to find L.
- 14.A garden centre has 90 rose bushes in stock. It receives a delivery that increases its stock of rose bushes by 20%. In the same week it sells 150 lavender bushes. Work out the new number of rose bushes, then write it as a fraction of the number of lavender bushes sold, giving your answer in its simplest form.
- 15.On a straight-line graph the volume of water, V litres, in a tank is plotted on the vertical axis and the time, t minutes, on the horizontal axis. The line passes through (2, 50) and (6, 130). Work out the gradient of the line and give its units.
Answer key
- (c) Yes — 700 ÷ 7 = 100 ml for the 1 part of concentrate — Method: add the ratio parts to find the total number of parts, divide the total volume by this, then use the ratio to find concentrate's share. Working: 1 + 6 = 7 parts. 700 ÷ 7 = 100 ml per part. Concentrate = 1 part = 100 ml, so Freya is correct. Wrong options: 'divide 700 by 6' uses only one of the ratio numbers instead of the total of 7 parts, giving about 117 ml; '600 ml is concentrate' swaps which ratio number belongs to the concentrate and which belongs to the water; 'half of 700 ml should be concentrate' ignores the ratio altogether and assumes an equal split.
- (c) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (c) 2 : 5 — The point (4, 10) gives x = 4, y = 10, so x : y = 4 : 10. Dividing both parts by their highest common factor, 2, gives 2 : 5 in simplest form. Inverting the whole ratio gives 5 : 2, which is y : x instead of x : y. Dividing only the x-part by 2 and leaving the y-part as 10 gives 2 : 10, but scaling one part on its own changes the ratio: 2 : 10 is the same as 1 : 5, not 4 : 10. Dividing only the y-part by 2 and leaving the x-part as 4 gives 4 : 5, the same one-sided mistake made on the other part of the ratio.
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (d) The candle's height decreases by 0.3 cm every minute. — A negative gradient means the quantity on the vertical axis decreases as the quantity on the horizontal axis increases. The size of the gradient, 0.3, gives the amount of decrease per minute.
- (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) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (b) 1:25 — Write the ratio time : volume using the numbers in the question: 8 : 200. Divide both parts by their highest common factor, 8, to give 1 : 25. (25:1 comes from writing the ratio the wrong way round, volume : time. 8:25 comes from dividing only the volume by 8 and leaving the time unchanged. 25:8 is that same mistake written the wrong way round.)
- (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.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (d) 100 — Rearrange L = 0.6L + 40 by collecting the L terms on one side: L − 0.6L = 40, which gives 0.4L = 40, then L = 40 ÷ 0.4 = 100. Subtracting the other way round, 0.6L − L = 40, gives −0.4L = 40, then L = 40 ÷ (−0.4) = −100 — a sign error that flips the answer negative even though a long-run value here must be positive. Ignoring the 0.6L term completely and solving L = 40 directly gives 40, which throws away the recurrence's own multiplier. Dividing 40 by 0.6 instead of by the correct coefficient 0.4 gives 40 ÷ 0.6 ≈ 66.7, a slip that comes from dividing by the coefficient of L on the RIGHT of the original equation rather than by what is left once the L terms are collected on one side. Always collect the L terms first, then divide by whatever coefficient of L remains.
- (a) 18/25 — First find the new number of rose bushes: 90 × 1.2 = 108 (a 20% increase multiplies by 1.2). Then write 108 over 150 and divide top and bottom by 6 to get 18/25. Choosing 3/5 comes from using the original 90 rose bushes without applying the 20% increase (90/150 = 3/5). Choosing 25/18 comes from writing the number of lavender bushes over the new number of rose bushes, the wrong way round. Choosing 3/25 comes from multiplying 90 by 0.2 instead of 1.2, finding only the increase (18) rather than the new total, then writing 18/150 = 3/25.
- (a) 20 litres per minute — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, and its units are the vertical unit for each one of the horizontal unit. Working: from (2, 50) to (6, 130) the volume changes by 130 − 50 = 80 litres and the time changes by 6 − 2 = 4 minutes, so the gradient is 80 ÷ 4 = 20, measured in litres for each minute. Answer: 20 litres per minute. The distractors: 25 litres per minute comes from using one point on its own, 50 ÷ 2, which assumes the line starts at the origin when the tank already held 50 litres at 2 minutes; 0.05 litres per minute comes from dividing the change in time by the change in volume, 4 ÷ 80, which gives the time for each litre but is then labelled as litres for each minute; 20 minutes for each litre has the right value with the units the wrong way round, and a tank that needed 20 minutes to gain a single litre would be filling far more slowly than this one.
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