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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- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (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) They are in direct proportion, because y ÷ x = 2.5 for both pairs. — Testing direct proportion means checking that y ÷ x is the same for every pair: 15 ÷ 6 = 2.5 and 25 ÷ 10 = 2.5, so the quantities are in direct proportion. Saying they are not in proportion because x + y differs uses addition, which is not the correct test for proportion. Saying they are not in proportion because y − x differs also uses the wrong test — subtraction, not division. Saying they are in proportion because x × y is 90 and 250 uses multiplication, which is the test for inverse proportion, and the two products are not even equal to each other, so this option also contradicts itself.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total journey.
- (d) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (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) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (c) £7590 — Apply interest, then subtract the payment, once for each year. Year 1: 8000 × 1.05 = 8400, then 8400 − 600 = 7800. Year 2: 7800 × 1.05 = 8190, then 8190 − 600 = 7590, so £7590 remains after 2 years. Forgetting the payments altogether and only compounding the interest gives 8000 × 1.05 = 8400, then 8400 × 1.05 = 8820 — this ignores that £600 leaves the fund every year. Subtracting the £600 BEFORE adding interest each year, instead of after, gives (8000 − 600) × 1.05 = 7770, then (7770 − 600) × 1.05 = 7528.50, which changes the order the two operations happen in and so changes the amount that earns interest each year. Subtracting the two payments as one lump sum of £1200 at the very end, from the no-withdrawal total 8820 − 1200 = 7620, ignores that the first £600 withdrawal also stops earning interest during the second year. Always apply interest, then the withdrawal, in that order, once for every single year.
- (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.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (a) 4 weeks — Apply the recurrence week by week. C_1 = 0.75 × 500 + 40 = 375 + 40 = 415. C_2 = 0.75 × 415 + 40 = 311.25 + 40 = 351.25. C_3 = 0.75 × 351.25 + 40 = 263.4375 + 40 = 303.4375. C_4 = 0.75 × 303.4375 + 40 = 227.578125 + 40 = 267.578125. C_3 = 303.4375 is still above 300, but C_4 = 267.58 has dropped below it, so the lake first becomes safe after 4 weeks. Taking 25% of the ORIGINAL 500 every week instead of 25% of the current amount, a flat 125 each time, gives 500 − 125 + 40 = 415, then 415 − 125 + 40 = 330, then 330 − 125 + 40 = 245, which crosses 300 a week too early and gives the wrong answer of 3 weeks. Continuing one extra step to C_5 = 0.75 × 267.578125 + 40 = 200.68 + 40 = 240.68 and calling it 5 weeks overshoots, since the concentration had already dropped below 300 at C_4. Forgetting the 40 units of run-off each week and only applying the decay gives C_1 = 0.75 × 500 = 375, then C_2 = 0.75 × 375 = 281.25 — this is already below 300 after only 2 weeks, because without the run-off the concentration falls much faster.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
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