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 hiker's distance from the start of the trail, in kilometres, is plotted against time, in minutes, since she set off. At t = 45 minutes she stops to rest, so her distance from the start is neither increasing nor decreasing at that instant. Which statement about the tangent to the distance–time graph at t = 45 minutes is correct?
- 2.A loan of £3000 has interest added at 2% each month, and then a fixed repayment of £200 is made. This is modelled by the recurrence B_{n+1} = 1.02B_n − 200, where B_n is the balance after n months and B_0 = 3000. Work out the balance after 2 months.
- 3.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.
- 4.Two mathematically similar hexagonal tiles have areas 18 cm² and 50 cm². Work out the ratio of the side length of the smaller tile to the side length of the larger tile, in simplest form.
- 5.A tap fills a tank at a rate of 15 litres per minute. Given that 1 litre = 1000 cm³, work out the rate at which the tank fills in cm³ per second.
- 6.A car travels at a constant speed. It travels 150 miles in 3 hours. At the same speed, work out how far it travels in 5 hours.
- 7.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 8.A photo is enlarged so that its new width is 1.4 times its original width. Write the multiplier that would scale the new width back down to the original width, as a fraction in its simplest form.
- 9.Two mathematically similar logos are printed on a poster. They have areas 20 cm² and 45 cm². Nadia says that the length scale factor from the smaller logo to the larger logo is 45 ÷ 20 = 2.25. Give a reason why Nadia is incorrect, and work out the correct length scale factor.
- 10.A metal sample has a mass of 342.6 g and a volume of 18 cm³. Work out the density of the sample, in g/cm³, to 1 decimal place.
- 11.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.
- 12.A trust fund contains £8000. Each year, 5% interest is added, and then £600 is paid out to a charity. Work out the amount remaining in the fund after 2 years.
- 13.A charity collects donations from adults and children in the ratio 5:2. Altogether, £238 is collected. Work out how much more the adults donate than the children.
- 14.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 15.In a fruit bowl, the ratio of apples to oranges is 5 : 3. Work out what percentage of the fruit in the bowl is apples.
Answer key
- (c) The tangent is horizontal, so its gradient is 0. — Method: at any point where a distance–time graph is momentarily neither increasing nor decreasing, the tangent to the graph at that point is horizontal, and the gradient of a horizontal line is 0 — this is the instantaneous rate of change at that instant. Working: since the hiker's distance is neither increasing nor decreasing at t = 45 minutes, the tangent there is horizontal, so its gradient is 0. Claiming the tangent is vertical, with an undefined gradient, is the opposite of what the stem says: a vertical tangent would mean the distance was changing infinitely fast at that instant, not that it had stopped changing, and on a distance–time graph it cannot happen at all. Reading the gradient as 45, the time value given in the stem, mistakes a value used to LOCATE the point for the rate of change AT that point. Claiming the gradient cannot be found without also knowing the distance at t = 45 minutes overlooks that 'momentarily stationary' already tells you the rate of change directly, without needing to read any distance value at all. Whenever a stem tells you a quantity is momentarily not changing, that is telling you the instantaneous rate of change directly — it is 0, and no further data is needed to find it.
- (c) £2717.20 — The recurrence B_{n+1} = 1.02B_n − 200 must be applied once for each month, using the previous month's balance each time. Starting from B_0 = 3000: 3000 × 1.02 = 3060, so B_1 = 3060 − 200 = 2860. Then 2860 × 1.02 = 2917.2, so B_2 = 2917.2 − 200 = 2717.2. Stopping after one month leaves B_1 = £2860.00, not the balance after two months. Applying two months of interest together, 1.02² = 1.0404, and 3000 × 1.0404 = 3121.2, and then subtracting 400 in one go, 3121.2 − 400 = 2721.2, does not reproduce the recurrence, because the second month's interest should be earned on the balance after the first repayment, not on the original £3000. Subtracting £200 twice from B_1 without adding a second month of interest, 2860 − 200 = 2660, drops the interest for the second month altogether. The balance after 2 months is £2717.20.
- (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.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (d) 250 cm³/s — Method: first change litres per minute into cm³ per minute, then change per minute into per second. Working: 15 × 1000 = 15000 cm³ per minute, then 15000 ÷ 60 = 250 cm³ per second. So the tank fills at 250 cm³ per second. Distractor 15000 cm³/s comes from stopping after the first step and forgetting to change minutes into seconds. Distractor 900000 cm³/s comes from multiplying by 60 instead of dividing. Distractor 2500 cm³/s comes from dividing by 6 instead of 60.
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (b) 5/7 — The enlargement multiplier is 1.4, which as a fraction is 7/5. To reverse an enlargement, use the reciprocal of the multiplier: flip 7/5 to get 5/7. 7/5 comes from using the enlargement multiplier again, instead of reversing it. 3/5 comes from treating the reverse as 'give back the extra amount', working out 1 − (1.4 − 1) = 0.6, instead of using the reciprocal. 5/2 comes from ignoring the whole number in 1.4 and inverting only the decimal part, 0.4, as if it were the whole multiplier.
- (d) 1.5 — Method: the length scale factor is the square root of the area scale factor, not the area scale factor itself. Working: the area scale factor is 45 ÷ 20 = 2.25, and the square root of 2.25 is 1.5. Answer: 1.5. Nadia's answer, 2.25, is the AREA scale factor — she never took the square root to get back to the length scale factor. 4.5 comes from doubling the area scale factor instead of taking its square root. 0.67 comes from taking the square root in the wrong direction, finding the scale factor from the larger rug to the smaller rug instead of the other way round.
- (b) 19.0 g/cm³ — Density = mass ÷ volume. 342.6 ÷ 18 = 19.0333…, which rounds to 19.0 g/cm³ (1 d.p.). 6166.8 g/cm³ comes from multiplying the mass by the volume instead of dividing (342.6 × 18). 324.6 g/cm³ comes from subtracting the volume from the mass (342.6 − 18) instead of dividing. 0.1 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (18 ÷ 342.6 = 0.0525…, rounded to 1 d.p.).
- (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.
- (c) £7590 — Apply interest, then subtract the payment, once for each year. Year 1: 8000 × 1.05 = 8400, then 8400 − 600 = 7800. Year 2: 7800 × 1.05 = 8190, then 8190 − 600 = 7590, so £7590 remains after 2 years. Forgetting the payments altogether and only compounding the interest gives 8000 × 1.05 = 8400, then 8400 × 1.05 = 8820 — this ignores that £600 leaves the fund every year. Subtracting the £600 BEFORE adding interest each year, instead of after, gives (8000 − 600) × 1.05 = 7770, then (7770 − 600) × 1.05 = 7528.50, which changes the order the two operations happen in and so changes the amount that earns interest each year. Subtracting the two payments as one lump sum of £1200 at the very end, from the no-withdrawal total 8820 − 1200 = 7620, ignores that the first £600 withdrawal also stops earning interest during the second year. Always apply interest, then the withdrawal, in that order, once for every single year.
- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (b) 62.5% — Total parts = 5 + 3 = 8. Apples make up 5 parts, so the percentage is 5/8 × 100 = 62.5%. A student who finds the oranges' share instead gets 3/8 × 100 = 37.5%. A student who assumes an even split gets 50%. A student who inverts the fraction gets 8/5 × 100 = 160%.
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