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) Falling at £950 per year — The gradient of a tangent on a value-age graph is a rate, in pounds per year, so −950 means the van's value is falling at £950 per year at that instant. Writing this as 950% per year mistakes a rate measured in pounds per year for a percentage — the units of a gradient come from the units on the two axes, £ and years, not from a percentage. Saying the value 'falls by £950 over the next year' treats the instantaneous rate at age 2 as if it stayed constant for a whole year, which finds an average future change, not the instantaneous rate at age 2 itself. Reading the sign the wrong way round gives 'rising at £950 per year', which would mean the van is gaining value. Always match the units of a gradient to the units on the two axes of the graph.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (d) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (d) 25% — Method: percentage decrease = decrease ÷ original amount × 100. Working: the reduction is £60 − £45 = £15, and 15 ÷ 60 = 0.25, so 0.25 × 100 = 25%. Answer: 25%. The distractors: 15% comes from quoting the £15 reduction as though pounds and per cent were the same thing; 33% comes from dividing the £15 by the new price £45 instead of by the original £60, which gives 33% to the nearest per cent; 75% is the new price written as a percentage of the old one, which is what is still paid rather than what has been taken off.
- (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) 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.
- (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.
- (c) 40 cm — First find the real pitch length using Plan A's scale: 8 × 1000 = 8000 cm. Now find its length on Plan B using Plan B's scale: 8000 ÷ 200 = 40 cm. Assuming both plans use the same scale and leaving the length as 8 cm ignores that Plan B has a different scale. Using the ratio of the two scales upside down, 200 ÷ 1000 = 0.2, then 8 × 0.2 = 1.6, gives 1.6 cm — Plan B is drawn to a larger scale than Plan A, so lengths on it should be bigger, not smaller. Dividing by 2000 instead of 200 gives 4 cm, ten times too small.
- (b) 1500 m — Method: multiply the map length by the scale to get the real length in centimetres, then divide by 100 to change centimetres into metres. Working: 3 × 50 000 = 150 000 cm, and 150 000 ÷ 100 = 1500. Answer: 1500 m. The distractors: 15000 m comes from dividing the 150 000 cm by 10 instead of 100; 150 m comes from dividing by 1000, the conversion for kilometres; 50000 m comes from writing the scale itself as the answer and ignoring the 3 cm measured on the map.
- (a) y is always 4 times x. — Check the multiplier for each pair: 12 ÷ 3 = 4, 20 ÷ 5 = 4, 32 ÷ 8 = 4 — the same multiplier every time, so y is always 4 times x. 'x is always 4 times y' comes from writing the multiplier the wrong way round. 'y is always x plus 9' only works for the first pair (3 + 9 = 12); it gives 14 for x = 5 and 17 for x = 8, not 20 and 32. 'y is always double x, plus 6' also only works for the first pair (3 × 2 + 6 = 12); it gives 16 for x = 5, not 20.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
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