Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation 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 Foundation
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- 1.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.
- 2.A stall sells bottles of juice. Two bottles cost £3.00, five bottles cost £7.50 and eight bottles cost £12.00. The cost, C pounds, is in a fixed ratio to the number of bottles, n. Write down a formula for C in terms of n.
- 3.A savings account pays 3.5% compound interest each year. Bilal invests £1200. Work out how much interest, in total, he earns after 2 years, to the nearest penny.
- 4.A tank is filled at a rate of 3 litres every 2 minutes. Given that 1 litre = 1000 cm³, work out the flow rate in cm³ per minute.
- 5.A dog walker's charge, C pounds, for walking a dog for m minutes is shown on a straight-line graph. The line passes through the points (20, 14) and (50, 26). Work out the charge for a 65-minute walk.
- 6.Triangle ABC is mathematically similar to triangle PQR, with AB corresponding to PQ and BC corresponding to QR. AB = 6 cm, BC = 8 cm and PQ = 12 cm. Work out the length of QR.
- 7.Two mathematically similar tins have heights 8 cm and 12 cm. Jayden says that the volume of the larger tin is 1.5 times the volume of the smaller tin. Give a reason why Jayden is incorrect, and work out the correct volume scale factor.
- 8.y is inversely proportional to x. When x = 5, y = 8. Work out the value of y when x = 10.
- 9.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.
- 10.Priya invests £750 in a savings account that pays simple interest. After 3 years, the account contains £840. Work out the annual rate of simple interest.
- 11.A metal cylinder has a mass of 356.5 g and a volume of 47 cm³. Work out the density of the cylinder, in g/cm³, to 1 decimal place.
- 12.A cartographer is choosing a scale for a map to show a small park in as much detail as possible. Which of these scales would show the MOST detail — 1 : 500, 1 : 5000, or 1 : 50 000? Give the correct scale and reason.
- 13.Map A has a scale of 1 : 25000 and Map B has a scale of 1 : 50000, both showing the same area. A real distance of 10 km is measured on each map. On which map does this distance appear as the longer length, and how long is it on that map, in centimetres?
- 14.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 15.Last week Priya worked 5 shifts of 7 hours. This week she worked 4 shifts of 8 hours. Write the number of hours she worked last week as a fraction of the number of hours she worked this week.
Answer key
- (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.).
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (a) £85.47 — Value after 2 years: £1200 × 1.035 × 1.035 = £1285.47 (nearest penny). Interest earned = £1285.47 − £1200 = £85.47. £1285.47 is the total value of the account, not the interest earned on top of the original £1200. £84.00 comes from using simple interest instead of compound interest: £1200 × 0.035 × 2 = £84.00. £42.00 comes from working out only the first year's interest and stopping there: £1200 × 0.035 = £42.00.
- (b) 1500 — The rate is 3 ÷ 2 = 1.5 litres per minute. Converting to cm³: 1.5 × 1000 = 1500 cm³ per minute. Getting 3000 comes from converting 3 litres to cm³ first (3000 cm³) and forgetting to divide by the 2 minutes. Getting 750 comes from dividing by the 2 minutes a second time after converting (1500 ÷ 2). Getting 2000 comes from converting the 2 minutes as if it were litres (2 × 1000) instead of using the correct rate of 1.5 litres per minute.
- (c) £32 — First find the gradient: (26 − 14) ÷ (50 − 20) = 12 ÷ 30 = £0.40 per minute. Using the point (20, 14), the charge for 65 minutes is 14 + 0.40 × (65 − 20) = 14 + 18 = £32. Choosing £26 comes from treating the charge as directly proportional to the time, multiplying the gradient by 65 minutes and ignoring the fixed part of the charge (0.40 × 65 = 26). Choosing £40 comes from treating £14 as if it were the charge at 0 minutes, then adding the gradient multiplied by the full 65 minutes (14 + 0.40 × 65 = 40), instead of multiplying by the extra time past 20 minutes. Choosing £33.80 comes from assuming the charge is directly proportional to the minutes already known, scaling up from the point (50, 26) in the ratio 65:50 (65 ÷ 50 × 26 = 33.80).
- (c) 16 cm — Corresponding sides of similar triangles are all in the same ratio. Use the pair whose lengths are both known: the scale factor from triangle ABC to triangle PQR is 12 ÷ 6 = 2. Since QR corresponds to BC, multiply BC by that scale factor: 8 × 2 = 16, so QR = 16 cm.
- (a) 3.375 — Method: the volume scale factor between similar shapes is the length scale factor cubed, not the length scale factor itself. Working: the length scale factor is 12 ÷ 8 = 1.5, and 1.5³ = 3.375. Answer: 3.375. Jayden's answer, 1.5, is only the LENGTH scale factor — he never cubed it to get the volume scale factor. 2.25 comes from squaring the length scale factor instead of cubing it, which would give the area scale factor. 4.5 comes from multiplying the length scale factor by 3 (the number of dimensions) instead of cubing it.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (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% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (c) 7.6 — Density = mass ÷ volume, so 356.5 ÷ 47 = 7.585..., which rounds to 7.6 g/cm³ (1 d.p.). (0.1 comes from dividing the volume by the mass instead of the mass by the volume, the wrong way round. 7.5 comes from rounding 7.585 down instead of up to 1 decimal place. 403.5 comes from adding the mass and the volume instead of dividing.)
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (a) 35/32 — Work out each weekly total first. Last week: 5 × 7 = 35 hours. This week: 4 × 8 = 32 hours. Last week's total is being written as a fraction of this week's total, so last week goes on the top and this week goes on the bottom, giving 35/32. The two totals share no common factor, so the fraction cannot be cancelled. It is greater than 1, which says that Priya worked more hours last week than this week.
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