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
MathsUKwww.geekhero.co.uk
- (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.
- (c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
- (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.
- (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).
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (d) 26.6 — Find the constant multiplier — the mass of each metre of pipe: 12.6 ÷ 4.5 = 2.8, so the mass is always 2.8 times the length. For a length of 9.5 m, the mass is 9.5 × 2.8 = 26.6 kg. 17.6 comes from assuming an additive relationship instead of a multiplicative one — adding the increase in length (9.5 − 4.5 = 5) onto 12.6. 3.4 comes from using the multiplier the wrong way round (4.5 ÷ 12.6, rounded to 1 d.p.), then multiplying by 9.5. 12.6 comes from simply repeating the given mass, without applying the multiplier to the new length.
- (a) 2.17 litres per minute — There are 60 minutes in an hour, so to convert litres per hour to litres per minute you divide by 60: 130 ÷ 60 = 2.1666..., which rounds to 2.17 litres per minute. Multiplying by 60 instead of dividing gives 130 × 60 = 7800.00 litres per minute, using the conversion factor the wrong way round. Leaving the rate unchanged, 130.00, ignores that 'per hour' and 'per minute' are different units. Dividing by 50 instead of 60, misremembering the number of minutes in an hour, gives 130 ÷ 50 = 2.60 litres per minute.
- (d) 250 cm³/s — Method: first change litres per minute into cm³ per minute, then change per minute into per second. Working: 15 × 1000 = 15000 cm³ per minute, then 15000 ÷ 60 = 250 cm³ per second. So the tank fills at 250 cm³ per second. Distractor 15000 cm³/s comes from stopping after the first step and forgetting to change minutes into seconds. Distractor 900000 cm³/s comes from multiplying by 60 instead of dividing. Distractor 2500 cm³/s comes from dividing by 6 instead of 60.
- (a) 35/32 — Work out each weekly total first. Last week: 5 × 7 = 35 hours. This week: 4 × 8 = 32 hours. Last week's total is being written as a fraction of this week's total, so last week goes on the top and this week goes on the bottom, giving 35/32. The two totals share no common factor, so the fraction cannot be cancelled. It is greater than 1, which says that Priya worked more hours last week than this week.
- (c) 88 km/h — Multiply the speed in mph by the conversion factor: 55 × 1.6 = 88 km/h. Dividing by 1.6 instead of multiplying gives 55 ÷ 1.6 ≈ 34.38 km/h, going the wrong way between the units. Adding the conversion factor instead of multiplying gives 55 + 1.6 = 56.6 km/h, treating the factor as an amount rather than a multiplier. Multiplying by 0.6 instead of 1.6 gives 55 × 0.6 = 33 km/h, using only part of the conversion factor. 55 mph is equal to 88 km/h.
- (a) 20 litres per minute — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, and its units are the vertical unit for each one of the horizontal unit. Working: from (2, 50) to (6, 130) the volume changes by 130 − 50 = 80 litres and the time changes by 6 − 2 = 4 minutes, so the gradient is 80 ÷ 4 = 20, measured in litres for each minute. Answer: 20 litres per minute. The distractors: 25 litres per minute comes from using one point on its own, 50 ÷ 2, which assumes the line starts at the origin when the tank already held 50 litres at 2 minutes; 0.05 litres per minute comes from dividing the change in time by the change in volume, 4 ÷ 80, which gives the time for each litre but is then labelled as litres for each minute; 20 minutes for each litre has the right value with the units the wrong way round, and a tank that needed 20 minutes to gain a single litre would be filling far more slowly than this one.
- (a) 46.8 km/h — Method: 1 hour = 60 minutes, so a rate given in kilometres per minute is converted to kilometres per hour by multiplying by 60. Working: 0.78 × 60 = 46.8, so the instantaneous rate of change is 46.8 kilometres per hour. Keeping the given value unchanged and only relabelling the unit gives 0.78 km/h, which ignores that the time unit has changed. Dividing by 60 instead of multiplying — as you would when converting to a larger length unit — gives 0.78 ÷ 60 = 0.013 km/h, the wrong direction for a rate measured against a larger time unit. Adding 60 to the given rate instead of multiplying by it gives 0.78 + 60 = 60.78 km/h. Converting a rate always means multiplying or dividing by the conversion factor between the units, never adding it, and the direction depends on whether the new time unit is bigger or smaller than the old one.
- (d) 13/10 — First find the selling price: £150 + £45 = £195. Put the selling price over the cost price: 195/150. Divide both numbers by their highest common factor, 15: 195÷15 = 13, 150÷15 = 10, giving 13/10. (10/13 comes from writing the prices the wrong way round. 3/10 is just the profit written as a fraction of the cost price, 45/150, not the selling price. 13/23 comes from comparing the selling price to the combined total of the cost price and the selling price, 195/345.)
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
Build your own mix at the worksheet builder.