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 map has a scale of 2 cm : 5 km. A footpath measures 8 cm on the map. Work out the real length of the footpath, in kilometres.
- 2.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.
- 3.A block of wood has a mass of 60 g and a volume of 100 cm³. Work out the density of the block, in g/cm³.
- 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 recipe uses flour, sugar and butter in the ratio 8 : 3 : 5. Write the ratio of flour to the rest of the mixture (sugar and butter combined) in its simplest form.
- 6.Jamal invests £600 in a savings account paying 3% simple interest per year. Work out the total amount in the account after 4 years.
- 7.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 8.A garden plan is enlarged by a scale factor of 3. A flower bed on the original plan is 5 cm long. Work out the length of the flower bed on the enlarged plan, in centimetres.
- 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 furniture designer makes a scale model of a wardrobe using a scale of 1 : 8. The real wardrobe is 2.4 m tall and 1.2 m wide. Work out the perimeter of the front face of the model wardrobe, in centimetres.
- 11.A shop sells ribbon by the metre. 2 m costs £3.00, 4 m costs £6.00, and 7 m costs £10.50. Does this data show that the cost is directly proportional to the length of ribbon bought? Choose the correct verdict and reason.
- 12.Four graphs are described below. Each shows one quantity plotted against another. Write down the description of the graph that shows two quantities in direct proportion.
- 13.Priya is paid £58.50 for working 7.5 hours on a Saturday. Work out her rate of pay, in £ per hour.
- 14.8 identical pens cost £6.40 in total. Work out the cost of 5 of the same pens.
- 15.£2000 is invested in a savings account that pays 5% compound interest each year. Work out the value of the investment at the end of 2 years.
Answer key
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (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) 0.6 g/cm³ — Density = mass ÷ volume. 60 ÷ 100 = 0.6 g/cm³. 1.67 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (100 ÷ 60). 6000 g/cm³ comes from multiplying the mass by the volume instead of dividing (60 × 100). 40 g/cm³ comes from subtracting the mass from the volume (100 − 60) instead of dividing.
- (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) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (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.
- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (d) 15 cm — An enlargement by scale factor 3 means every length is multiplied by 3: 5 × 3 = 15 cm. Dividing by the scale factor instead, 5 ÷ 3 = 1.7 (to 1 d.p.), gives 1.7 cm — dividing shrinks the plan rather than enlarging it. Adding the scale factor instead of multiplying, 5 + 3 = 8, gives 8 cm. Squaring the scale factor, as you would for an area, 5 × 3² = 45, gives 45 cm — length scales by the factor itself, not its square.
- (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.)
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (c) A straight line through the origin, rising from left to right — Two quantities are in direct proportion when one is a constant multiple of the other, so their graph is a straight line through the origin: when one quantity is 0 the other is 0 as well, and doubling one doubles the other. A straight line crossing the vertical axis at 5 has a fixed amount added on, so when the horizontal quantity is 0 the vertical quantity is 5, not 0 — a straight line on its own is not enough for direct proportion. A curve that falls steeply and then levels off without touching either axis shows inverse proportion: one quantity grows as the other shrinks, and their product stays the same. A horizontal line at a height of 3 shows a quantity that does not change at all as the other one grows, so it is not proportional to it.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (c) £4.00 — Find the cost of one pen: £6.40 ÷ 8 = £0.80. Then multiply by 5 pens: £0.80 × 5 = £4.00. Dividing £6.40 by 5 and multiplying by 8 gives £10.24 — that uses the ratio the wrong way round, scaling as if 5 pens were more expensive than 8. Stopping at £0.80 only gives the price of one pen. Multiplying the price of one pen by the difference in the number of pens, (8 − 5) × £0.80, gives £2.40 — the cost of the pens NOT bought, not the cost of the 5 pens bought. 5 pens cost £4.00.
- (b) £2205.00 — With compound interest each year's interest is worked out on the value at the start of that year, so a 5% rise is a multiplier of 1.05 applied once per year. After the first year: 2000 × 1.05 = 2100. After the second year: 2100 × 1.05 = 2205. The question asks for the value of the investment, not for the interest earned, so the answer is £2205.00.
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