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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Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (c) 24p — Method: change the price to pence, then find what one tenth of 1 kg costs, since 100 g is one tenth of 1 kg. Working: £2.40 = 240p per kg, and 240 ÷ 10 = 24p per 100 g. So the cost is 24p per 100 g. Distractor 2.4p comes from dividing by 100 instead of 10. Distractor 2400p comes from multiplying by 10 instead of dividing. Distractor 240p comes from using the price per kg without scaling it down to 100 g.
- (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.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (c) 3 : 4 — Method: a fraction compares a part with the whole, while this ratio compares one part with the other part, so find the fraction that is black before writing the ratio. Working: if 3/7 are blue then the black pens make up 7/7 − 3/7 = 4/7 of the box, so out of every 7 pens 3 are blue and 4 are black, and blue : black = 3 : 4. Answer: 3 : 4. The distractors: 3 : 7 comes from reading the numerator and the denominator of 3/7 straight off as the two parts, which compares the blue pens with the whole box rather than with the black pens; 4 : 3 comes from writing the black pens before the blue pens, reversing the order asked for; 4 : 7 is the same numerator-and-denominator reading applied to the black fraction 4/7, again comparing a part with the whole box.
- (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.
- (d) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (a) 18/25 — First find the new number of rose bushes: 90 × 1.2 = 108 (a 20% increase multiplies by 1.2). Then write 108 over 150 and divide top and bottom by 6 to get 18/25. Choosing 3/5 comes from using the original 90 rose bushes without applying the 20% increase (90/150 = 3/5). Choosing 25/18 comes from writing the number of lavender bushes over the new number of rose bushes, the wrong way round. Choosing 3/25 comes from multiplying 90 by 0.2 instead of 1.2, finding only the increase (18) rather than the new total, then writing 18/150 = 3/25.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (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) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (c) Falling at 3.2°C per minute — The gradient of a tangent gives the instantaneous rate of change, in °C per minute here, not a temperature and not a total change. The negative sign means the temperature is falling, not rising, so the tea is cooling at a rate of 3.2°C per minute at the instant t = 4. Reading the sign the wrong way round gives 'rising at 3.2°C per minute', which would mean the tea is heating up. Treating −3.2 as a total drop since the tea was poured confuses a rate with an accumulated change, which would need the temperatures at two different times, not the gradient at one instant. Treating −3.2 as the temperature reading itself confuses the gradient, a rate of change, with the y-value on the graph. Always check whether a number is a rate, a total, or a single reading before you use it.
- (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.
- (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.
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