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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (c) 1.05 litres — Method: find the value of one part of the ratio from the total volume, then find the share for pineapple juice. Working: the ratio 5:3:2 has 5 + 3 + 2 = 10 parts, so one part is 3.5 ÷ 10 = 0.35 litres, and the pineapple juice is 3 × 0.35 = 1.05 litres. So 1.05 litres of pineapple juice is needed. Distractor 1.75 litres is the volume of orange juice, not pineapple juice. Distractor 0.7 litres is the volume of lemonade, not pineapple juice. Distractor 0.35 litres is the value of one part, found correctly but never multiplied by 3.
- (a) 60 minutes — Method: for inverse proportion, printers × time is constant. Working: 6 × 40 = 240 (the constant). With 4 printers: 240 ÷ 4 = 60 minutes. Wrong options: 26.7 minutes comes from treating the relationship as direct proportion, scaling the time down as printers decrease (40 × 4 ÷ 6); 24 minutes comes from multiplying the two printer counts together instead of using the constant; 40 minutes comes from not adjusting the time at all for the change in printers.
- (b) 2, the cost in pounds of each extra gigabyte — Method: the gradient is the change in cost divided by the change in data, so it is the cost of each extra gigabyte; the value where the line meets the vertical axis is the charge before any data is used, which is a different quantity. Working: from (0, 10) to (8, 26) the cost rises by 26 − 10 = 16 pounds while the data rises by 8 − 0 = 8 gigabytes, so the gradient is 16 ÷ 8 = 2, meaning each extra gigabyte costs £2. Answer: 2, the cost in pounds of each extra gigabyte. The distractors: '10, the cost in pounds of each extra gigabyte' reads the intercept as the gradient, but 10 is what the tariff costs when no data at all has been used; '3.25, the cost in pounds of each extra gigabyte' comes from 26 ÷ 8, treating the line as though it passed through the origin when it starts at 10; '2, the fixed monthly charge in pounds' has the gradient right but describes the intercept, and the fixed charge on this tariff is £10.
- (d) 400 — The area scale factor is the length scale factor squared: 20² = 400, so the real car's surface area is 400 times the model's. 20 comes from using the length scale factor itself, without squaring it. 8000 comes from cubing the length scale factor (20³), instead of squaring it — cubing is the rule for volume, not area. 40 comes from doubling the length scale factor (2 × 20), instead of squaring it.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (d) The 750 g box, at 36p per 100 g — Work out the cost per 100 g of each box. 750 g box: 270p ÷ 7.5 = 36p per 100 g. 500 g box: 195p ÷ 5 = 39p per 100 g. The lower cost per 100 g is the better value, so the 750 g box at 36p per 100 g is the answer. Choosing the 500 g box at 39p per 100 g gets the maths right but picks the higher unit price, not realising a smaller cost per 100 g is the better deal. Choosing the 500 g box because £1.95 is lower than £2.70 compares the total prices without allowing for the different pack sizes at all. Working out 270 ÷ 5 = 54p divides the 750 g box's price by the wrong number of hundred-grams (the 500 g box's), giving a rate that belongs to neither box. The 750 g box, at 36p per 100 g, is the better value.
- (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) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (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) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (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.
- (c) Yes, correct, because 24 ÷ 8 × 3 = 9. — Method: divide the count you know by its ratio part to find the value of one part, then multiply by the part you want. Working: one part = 24 ÷ 8 = 3 counters, so green = 3 × 3 = 9 counters, and the student is correct. Dividing 24 by 3 instead of 8 gives 24 ÷ 3 × 8 = 64, the ratio parts the wrong way round. Stopping after 24 ÷ 8 = 3 gives 3, forgetting to multiply by the green ratio part. Adding 24 + 3 = 27 confuses adding a ratio part with scaling by it.
- (a) 2/3 — Put the laptop bag's mass over the school bag's mass: 2.4/3.6. Multiply both numbers by 10 to clear the decimals: 24/36. Divide both by their highest common factor, 12: 24÷12 = 2, 36÷12 = 3, giving 2/3. (3/2 comes from writing the masses the wrong way round. 1/3 comes from finding the difference in the masses, 3.6 − 2.4 = 1.2 kg, and writing it as a fraction of the school bag's mass, 1.2/3.6. 2/5 comes from comparing the laptop bag's mass to the total mass of both bags, 2.4/6.)
- (c) £7590 — Apply interest, then subtract the payment, once for each year. Year 1: 8000 × 1.05 = 8400, then 8400 − 600 = 7800. Year 2: 7800 × 1.05 = 8190, then 8190 − 600 = 7590, so £7590 remains after 2 years. Forgetting the payments altogether and only compounding the interest gives 8000 × 1.05 = 8400, then 8400 × 1.05 = 8820 — this ignores that £600 leaves the fund every year. Subtracting the £600 BEFORE adding interest each year, instead of after, gives (8000 − 600) × 1.05 = 7770, then (7770 − 600) × 1.05 = 7528.50, which changes the order the two operations happen in and so changes the amount that earns interest each year. Subtracting the two payments as one lump sum of £1200 at the very end, from the no-withdrawal total 8820 − 1200 = 7620, ignores that the first £600 withdrawal also stops earning interest during the second year. Always apply interest, then the withdrawal, in that order, once for every single year.
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