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.
Calculator
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (b) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
- (d) 500 ml — Varnish covers a surface, so the amount needed scales with the area scale factor, which is the square of the length scale factor. The length scale factor is 50 ÷ 20 = 2.5, so the area scale factor is 2.5 × 2.5 = 6.25. The varnish needed for the larger statue is 80 × 6.25 = 500 ml. Using 2.5 on its own would scale a length, not a surface.
- (c) 10,800 kg — Method: since the model and the real container are similar and made of the same material, mass scales with volume, so the mass scale factor is the length scale factor cubed. Working: 30³ = 27,000, so the real container's mass is 400 × 27,000 = 10,800,000 g, which is 10,800,000 ÷ 1,000 = 10,800 kg. Answer: 10,800 kg. 12 kg comes from using the length scale factor directly, 400 × 30 = 12,000 g, without cubing it. 360 kg comes from squaring the length scale factor instead of cubing it, 400 × 30² = 360,000 g. 10,800,000 kg comes from correctly cubing the scale factor but then forgetting to convert the mass from grams into kilograms.
- (c) 40 — Working out successive terms shows where the sequence is heading, but the terms themselves keep changing — the limit is the value where the sequence stops changing, so x_{n+1} = x_n = L there. Substituting into the rule: L = 0.5L + 20. Subtracting 0.5L from both sides: L − 0.5L = 20, so 0.5L = 20, and L = 20 ÷ 0.5 = 40. The individual terms are x_1 = 0.5 × 0 + 20 = 20, x_2 = 0.5 × 20 + 20 = 30 and x_3 = 0.5 × 30 + 20 = 35, getting closer to this value but not equal to it — 35 is only the third term, not the limit. Multiplying by 0.5 instead of dividing at the final step, 20 × 0.5 = 10, undoes the rearrangement rather than completing it, and gives a value smaller than terms the sequence has already passed. Writing the fixed-point equation with the wrong sign, L = 0.5L − 20, gives 0.5L = −20 and L = −40, which cannot be right since every term in the sequence is positive and increasing. The value the sequence is approaching is 40.
- (d) Gradient = acceleration; area = distance travelled. — Method: on a speed–time graph, the gradient of the graph at an instant is the rate of change of speed with time, which is acceleration; the area between the graph and the time-axis over an interval is the total distance covered in that interval, because it accumulates speed × time. Working: gradient = acceleration and area = distance travelled is the correct pairing. Swapping the two quantities completely, gradient = distance travelled and area = acceleration, is the reverse of what each actually measures. Keeping gradient = acceleration correct but then also claiming area = acceleration too is wrong because the area is a different physical quantity, distance, not a second way of finding the same rate. Claiming the gradient itself gives the speed confuses the RATE OF CHANGE of the plotted quantity with the plotted quantity itself — the gradient is how fast the speed is changing, not the speed. On any rate graph, the gradient of the graph is always the RATE at that instant, and the area under the graph is always the TOTAL AMOUNT accumulated — keep straight which of the two questions each one answers.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (a) 1/9 — The ratio copper : tin is 9:1, so write tin over copper: 1/9. (9/1 comes from writing the ratio the wrong way round, copper over tin. 1/10 comes from comparing the tin to the total mass of the alloy, 1 part out of 10. 9/10 comes from comparing the copper to the total mass of the alloy, 9 parts out of 10.)
- (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) No — rate fell by 3.7 thousand/month — Each tangent gradient is the instantaneous growth rate, in thousand subscribers per month. To compare them, subtract the later rate from the earlier one: 4.8 − 1.1 = 3.7. Since 1.1 is less than 4.8, the growth rate has fallen by 3.7 thousand subscribers per month, so the manager is wrong — the app is growing more slowly at 18 months, not faster. Subtracting the other way round and calling the result a rise, 'rate rose by 3.7 thousand/month', gets the direction backwards: the later gradient is the smaller of the two. Adding the two gradients, 4.8 + 1.1 = 5.9, and calling this a combined rate that shows speeding up, is the wrong operation for comparing two rates. Treating the difference 3.7 as a total number of subscribers lost, rather than a rate in thousands per month, confuses a rate with a count. Always subtract the two rates in a sensible order and keep the units in thousands per month.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (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.
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
Build your own mix at the worksheet builder.