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.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.
- 2.A crate exerts a force of 84 N on the floor through a base with an area of 2.4 m². Work out the pressure on the floor, in N/m².
- 3.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 4.In one batch of shortbread, a bakery uses 240 g of butter and 160 g of sugar. Another batch, made in the same ratio, uses 90 g of sugar. Work out how much butter is needed for the second batch.
- 5.Noah runs 2 km in 10 minutes. Work out how long he takes to run 5 km at the same speed.
- 6.300 g of a salt solution of concentration 10% is poured into 200 g of a salt solution of concentration 25%. Work out the concentration of salt in the mixture.
- 7.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.
- 8.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 9.There are 400 students at a school. 25% of them have a brother, 40% have a sister and 15% have both a brother and a sister. Work out how many of the students have neither a brother nor a sister.
- 10.Write the ratio 2 : 1/4 as a ratio of whole numbers in its simplest form.
- 11.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
- 12.A map has a scale of 1 : 50 000. Two villages are 4 cm apart on the map. Work out the real distance between the villages, in kilometres.
- 13.Write the ratio 3/4 : 1/2 as a ratio of whole numbers in its simplest form.
- 14.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 15.A sprinter's distance from the start line, in metres, is plotted against time, in seconds. A tangent to the graph at t = 2 seconds has gradient 6. A tangent at t = 8 seconds has gradient 9.5. Which statement correctly compares the sprinter's speed at these two times?
Answer key
- (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.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (c) 135 g — Find the ratio of butter to sugar in the first batch: 240:160, which simplifies to 3:2. For the second batch, sugar = 90 g, so butter = 90 × 3/2 = 135 g. (60 g comes from using the ratio the wrong way round, 90 × 2/3. 170 g comes from subtracting the drop in sugar, 160 − 90 = 70 g, from the original butter amount, 240 − 70, instead of scaling. 240 g comes from not scaling the butter amount at all.)
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (c) 16% — Method: scale each concentration to its own mass to find the salt it contains, add the two masses of salt, then write the ratio of salt to mixture per 100 g. Working: 10:100 = x:300 gives 30 g of salt, and 25:100 = y:200 gives 50 g of salt; the mixture holds 30 + 50 = 80 g of salt in 300 + 200 = 500 g of solution; 80:500 = 16:100. Answer: 16%. The distractors: 17.5% is the mean of 10% and 25%, which ignores that there is more of the weaker solution than of the stronger one; 19% comes from swapping the two concentrations over, working out (300 × 25% + 200 × 10%) ÷ 500; 26.7% comes from dividing the 80 g of salt by the 300 g of the first solution rather than by the 500 g of mixture.
- (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.
- (b) 12 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
- (a) 8:1 — Multiply both parts of the ratio by 4 to clear the fraction: 2 × 4 = 8 and 1/4 × 4 = 1, giving 8 : 1. Getting 1 : 8 has the two parts the wrong way round. Getting 2 : 4 comes from writing down the denominator of the fraction (4) as the second part instead of multiplying through by it. Getting 8 : 4 comes from multiplying only the first part of the ratio by 4 and leaving the second part as the fraction's denominator.
- (d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
- (a) 2 km — Method: multiply the map distance by the scale to get the real distance in centimetres, then convert centimetres to kilometres using 100 cm = 1 m and 1000 m = 1 km. Working: 4 × 50 000 = 200 000 cm; 200 000 ÷ 100 = 2000 m; 2000 ÷ 1000 = 2. Answer: 2 km. The distractors: 200 km comes from dividing the 200 000 cm by 1000 in a single step, as if a kilometre were 1000 cm rather than the 100 000 cm it is; 20 km comes from converting to metres correctly, 200 000 ÷ 100 = 2000 m, and then dividing those metres by 100 instead of by 1000; 0.2 km comes from dividing by 1000 to reach metres, as if a metre were 1000 cm, and then dividing by 1000 again, so 200 000 is divided by 1 000 000 altogether.
- (b) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (d) Faster at t = 8s — still accelerating — The gradient of a tangent on a distance-time graph is the instantaneous speed, in m/s. At t = 2 seconds the speed is 6 m/s; at t = 8 seconds it is 9.5 m/s, which is faster, so the sprinter is still accelerating between these two times. Saying the sprinter is slower at t = 8s reverses the comparison — 9.5 is greater than 6, not less. Writing 9.5 − 6 = 3.5 and calling this 'metres further covered' turns the difference of two speeds into a distance, which the units do not support: a difference of two speeds is itself a speed, not a distance. Taking 9.5 m/s, the larger of the two instantaneous speeds, as the average speed for the whole race confuses a speed at one instant with an average over the whole distance, which would need the total distance and total time, not two tangent gradients.
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