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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Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (b) £52,000 — Method: convert 130% to a decimal multiplier and multiply it by last year's profit. Working: 130% = 1.3, so this year's profit is £40,000 × 1.3 = £52,000. Answer: £52,000. £12,000 comes from using only the extra 30% (130% − 100%) and forgetting to include the original 100%, £40,000 × 0.3 = £12,000. £40,130 comes from simply adding 130 onto £40,000, treating the percentage as an amount of money rather than a multiplier. £5,200 comes from misreading 130% as 13%, giving £40,000 × 0.13 = £5,200.
- (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.
- (c) £32 — First find the gradient: (26 − 14) ÷ (50 − 20) = 12 ÷ 30 = £0.40 per minute. Using the point (20, 14), the charge for 65 minutes is 14 + 0.40 × (65 − 20) = 14 + 18 = £32. Choosing £26 comes from treating the charge as directly proportional to the time, multiplying the gradient by 65 minutes and ignoring the fixed part of the charge (0.40 × 65 = 26). Choosing £40 comes from treating £14 as if it were the charge at 0 minutes, then adding the gradient multiplied by the full 65 minutes (14 + 0.40 × 65 = 40), instead of multiplying by the extra time past 20 minutes. Choosing £33.80 comes from assuming the charge is directly proportional to the minutes already known, scaling up from the point (50, 26) in the ratio 65:50 (65 ÷ 50 × 26 = 33.80).
- (a) 22p — First convert the hourly rate to pence: £13.20 = 1320p. Then convert from per hour to per minute by dividing by 60, since there are 60 minutes in an hour: 1320 ÷ 60 = 22p per minute. Working out 1320 × 60 = 79200p multiplies by 60 instead of dividing, going the wrong way between per hour and per minute. Working out 13.20 × 10 = 132p converts pounds to pence using the wrong power of ten, and dividing that by 60 carries the error through to give 2.2p. Working out 13.20 × 100 = 1320p converts the currency correctly but stops there, leaving the rate as pence per hour rather than completing the second conversion to pence per minute. The rate of pay is 22p per minute.
- (d) 3:8 — Convert 2 hours to minutes: 2 hours = 120 minutes. The ratio is 45 : 120. The highest common factor of 45 and 120 is 15. Divide both parts by 15: 45 ÷ 15 = 3 and 120 ÷ 15 = 8, giving 3 : 8. Leaving the hours unconverted gives 45 : 2 — the units on each side are different, so this does not compare like with like. Dividing by 5 instead of 15 gives 9 : 24, which still shares a common factor of 3, so it is not fully simplified. Swapping the order gives 8 : 3, hours to minutes instead of minutes to hours.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
- (d) 3/8 — The ratio 3 : 5 has 3 + 5 = 8 parts in total. The first part as a fraction of the whole is 3 out of 8, or 3/8. Writing 3/5 gives the first part compared to the second part, not to the whole. Writing 5/8 gives the second part as a fraction of the whole, not the first. Writing 8/3 has the fraction upside down — the whole must be on the bottom.
- (c) 5 : 9 — Write the ratio mass : cost = 3 : 5.4. Multiply both parts by 10 to clear the decimal: 30 : 54. Both numbers share a factor of 6, so 30 ÷ 6 = 5 and 54 ÷ 6 = 9, giving 5 : 9. 3 : 5 comes from ignoring the decimal point and treating £5.40 as £5. 9 : 5 comes from writing the ratio the wrong way round, cost to mass instead of mass to cost. 1 : 18 comes from multiplying only the cost by 10 instead of both parts, giving 3 : 54, and then cancelling that correctly to 1 : 18 — the cancelling is fine, but the ratio being cancelled is not the right one.
- (d) 4:3 — Boys = 28 − 16 = 12. Girls to boys is 16 : 12. The highest common factor of 16 and 12 is 4. Divide both parts by 4: 16 ÷ 4 = 4 and 12 ÷ 4 = 3, so the ratio is 4 : 3. Dividing by 2 instead of 4 gives 8 : 6, which still shares a common factor of 2, so it is not fully simplified. Swapping the order gives 3 : 4, boys to girls instead of girls to boys. Writing girls to the whole class, 16 : 28, simplifies to 4 : 7 — that compares girls to everyone, not girls to boys.
- (c) 2:3 — Divide both parts by the common factor 3a: 6a ÷ 3a = 2 and 9a ÷ 3a = 3, giving the ratio 2 : 3.
- (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.
- (c) 1 : 25 — The radii are in the ratio 4 : 20, which simplifies to 1 : 5. Areas scale with the square of the length ratio, so the area ratio is 1² : 5² = 1 : 25. Giving 1 : 5 uses the radius ratio without squaring it. Giving 1 : 10 doubles the radius ratio instead of squaring it. Giving 25 : 1 has the areas the right way round for larger to smaller, not smaller to larger.
- (a) 18 — Method: write both numbers with the same multiplier, turn the second ratio into an equation by cross-multiplying, solve for the multiplier and then build A from it. Working: let A = 3k and B = 5k, so 3k : (5k + 6) = 1 : 2; cross-multiplying gives 2 × 3k = 5k + 6, so 6k = 5k + 6 and k = 6; A = 3 × 6 = 18. Answer: 18, and the check works, because B = 30, B + 6 = 36 and 18:36 = 1:2. The distractors: 9 comes from reading the 6 as the difference between the two numbers — 5 − 3 = 2 parts, so one part is 3 and A is 3 × 3 — but the 6 is added to B, it is not the gap between A and B; 6 comes from solving A : (A + 6) = 1 : 2, adding the 6 to A instead of to B; 30 is the value of B, found from the correct multiplier but given in place of A.
- (d) A falling curve that never touches either axis — Method: inverse proportion means the product of the two quantities is constant, so P = k ÷ Q; as Q grows P shrinks, and P can never reach zero because k divided by a number is never zero. Working: taking k = 12 as an example, the pairs (1, 12), (2, 6), (3, 4), (6, 2) and (12, 1) drop steeply at first and then flatten out, so the graph is a curve that approaches both axes without meeting either of them. Answer: a falling curve that never touches either axis. The distractors: 'a straight line through the origin' is the graph of direct proportion, P = kQ, which is the opposite relationship; 'a straight line with a negative gradient' is the commonest error, reading 'P falls as Q rises' as a straight line, but on such a line P would drop by the same amount for every increase in Q and would cross the horizontal axis into negative values; 'a straight line crossing the vertical axis above zero' is a relationship of the form P = mQ + c, in which P and Q are not proportional at all.
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