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.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 2.A river 120 km long is drawn as a line 6 cm long on a map. Work out the scale of the map.
- 3.Write the ratio 2/3 : 1/2 as a ratio of whole numbers, in its simplest form.
- 4.Ffion is paid £11.20 per hour. She works 6 hours on Monday and 4.5 hours on Tuesday. Work out her total pay for the two days.
- 5.The height of a candle, in cm, is measured as it burns: t = 0 min, height = 20.0; t = 3 min, height = 18.7; t = 5 min, height = 17.5; t = 6 min, height = 16.6; t = 7 min, height = 15.4; t = 9 min, height = 13.4. Which pair of readings gives the best estimate of the instantaneous rate of change of the height at t = 6 minutes, and why?
- 6.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.
- 7.A rectangular garden has its length and width in the ratio 5:3. Given that the perimeter of the garden is 64 m, work out the width of the garden.
- 8.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.
- 9.1 metre = 100 cm. Work out how many square centimetres there are in 1 square metre.
- 10.A tangent to a curve at the point where x = 3 passes through the points (1, 2) and (5, 14). A candidate says: "The gradient of the tangent at x = 3 is 3, because 14 − 2 = 12 and 12 ÷ 4 = 3." Which statement is correct?
- 11.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 12.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.
- 13.The height of a ball, in metres, above the ground is plotted against time, in seconds, after it is thrown. The tangent to the graph at t = 1.5 seconds has gradient 2. Which of these four statements about the ball at t = 1.5 seconds is correct?
- 14.Two quantities y and z are each in direct proportion to x, and are given by y = 3x and z = 7x. Work out the difference between the value of y and the value of z when x = 4.y = 3x
- 15.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
Answer key
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (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.
- (a) 4:3 — Multiply both parts by the lowest common denominator, 6: (2/3) × 6 = 4 and (1/2) × 6 = 3, giving the ratio 4 : 3, which is already in simplest form.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (b) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (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.
- (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
- (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.
- (b) 10000 cm² — Method: an area conversion factor is the square of the length conversion factor, because both sides of the square are scaled. Working: a square metre is a square of side 100 cm, so its area is 100 × 100 = 10000 cm². Answer: 10000 cm². The distractors: 100 cm² comes from using the length factor without squaring it; 200 cm² comes from doubling the length factor instead of squaring it; 1000000 cm² comes from cubing the factor, which is the conversion for a volume, not an area.
- (a) The gradient is 3; the candidate's method is right. — The gradient of a tangent, like any straight line, is the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 2) and (5, 14), so the change in y is 14 − 2 = 12 and the change in x is 5 − 1 = 4. The gradient is 12 ÷ 4 = 3, so the candidate's calculation is correct. Subtracting in the wrong order, (2 − 14) ÷ (5 − 1), gives −12 ÷ 4 = −3, the wrong sign. Adding the two changes instead of dividing them, 12 + 4 = 16, does not find a gradient at all. Dividing the change in x by the change in y instead of the other way round, 4 ÷ 12 = 1/3, inverts the calculation completely. Before accepting or rejecting a claimed gradient, always redo the calculation yourself in the same order — change in y over change in x — rather than trusting the arithmetic as given.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (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.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (b) 16 — Method: substitute x into each equation separately, then subtract the smaller value from the larger. Working: y = 3 × 4 = 12 and z = 7 × 4 = 28, so the difference is 28 − 12 = 16. Answer: 16. The distractors: 4 comes from subtracting the constants, 7 − 3, which is the difference between the two gradients rather than the difference between the values at x = 4; 40 comes from adding the two values, 12 + 28, instead of subtracting them; 28 is the value of z on its own, given instead of being compared with the value of y.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
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