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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- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (b) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
- (a) 20 people/year — The gradient of a tangent to a graph at a point equals the instantaneous rate of change of the quantity there. A straight line's gradient is the change in the vertical value divided by the change in the horizontal value between two points on it. Here the tangent passes through (2, 180) and (6, 260), so the change in population is 260 − 180 = 80 and the change in time is 6 − 2 = 4. The gradient is 80 ÷ 4 = 20. Reporting the change in population, 80, on its own is not a rate, because that growth happened over 4 years and has not been divided by them. Adding the two changes instead of dividing gives 80 + 4 = 84, which is not a rate. Subtracting the coordinates in the wrong order, (180 − 260) ÷ (6 − 2), gives −80 ÷ 4 = −20, the wrong sign. The instantaneous rate of change of the population at t = 4 is 20 people per year.
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (c) 4/5 — Find 20% of £45: 10% is £4.50, so 20% is £9. The sale price is £45 − £9 = £36. Form the fraction 36/45; both numbers share a factor of 9, so 36 ÷ 9 = 4 and 45 ÷ 9 = 5, giving 4/5. 1/5 comes from writing the discount itself as a fraction of the normal price (9/45), instead of the sale price. 6/5 comes from adding the 20% instead of subtracting it, giving a sale price of £54, then 54/45 = 6/5. 5/9 comes from treating 'reduced by 20%' as 'reduced by £20', giving a sale price of £25, then 25/45 = 5/9.
- (b) 1:25 — Write the ratio time : volume using the numbers in the question: 8 : 200. Divide both parts by their highest common factor, 8, to give 1 : 25. (25:1 comes from writing the ratio the wrong way round, volume : time. 8:25 comes from dividing only the volume by 8 and leaving the time unchanged. 25:8 is that same mistake written the wrong way round.)
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (c) 36 cm — A perimeter is built from lengths, so it is multiplied by the length scale factor and not by the square of it. The scale factor from the smaller rectangle to the larger one is 10 ÷ 5 = 2. The larger perimeter is therefore 18 × 2 = 36 cm.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (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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