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
Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.An online shop's total number of orders received is recorded at the end of each day for five consecutive days: Day 1, 1,284 orders; Day 2, 1,509 orders; Day 3, 1,830 orders; Day 4, 2,296 orders; Day 5, 2,510 orders. Work out between which two consecutive days the average rate of increase in orders was greatest.
- 2.A photo is enlarged so that its new width is 1.4 times its original width. Write the multiplier that would scale the new width back down to the original width, as a fraction in its simplest form.
- 3.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
- 4.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 5.A car travels at a constant speed. It travels 150 miles in 3 hours. At the same speed, work out how far it travels in 5 hours.
- 6.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 7.A cyclist travels 18 km in 45 minutes. Work out the average speed, in km/h.
- 8.A patient is given a dose of 200 mg of a drug. Each hour, 30% of the drug remaining in the bloodstream is eliminated, and then a further dose of 50 mg is given. Using the recurrence D_{n+1} = 0.7D_n + 50, with D_0 = 200, work out the amount of drug in the bloodstream after 2 hours.
- 9.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 10.Maya is paid £105.30 for working 9 hours. Work out her rate of pay, in £ per hour.
- 11.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 12.A tank is filled at a rate of 3 litres every 2 minutes. Given that 1 litre = 1000 cm³, work out the flow rate in cm³ per minute.
- 13.Two broadband providers show their monthly cost on straight-line graphs, with the cost in pounds on the vertical axis and the data used in gigabytes on the horizontal axis. Provider X's line passes through (0, 15) and (100, 35). Provider Y's line passes through (0, 5) and (100, 45). Work out which provider is cheaper for a customer who uses 40 gigabytes in a month.
- 14.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 15.A spring's extension is directly proportional to the force applied to it. A force of 5 N produces an extension of 12 mm. Work out the extension produced by a force of 20 N.
Answer key
- (b) Day 3 to Day 4 — Method: the average rate of increase between two consecutive days is the difference in the number of orders divided by the number of days between them, which here is just the difference itself, since each gap is one day; comparing all four differences finds which is greatest. Working: the differences are 1,509 − 1,284 = 225 (Day 1 to Day 2), 1,830 − 1,509 = 321 (Day 2 to Day 3), 2,296 − 1,830 = 466 (Day 3 to Day 4), and 2,510 − 2,296 = 214 (Day 4 to Day 5); 466 is the greatest of the four, so the rate of increase was greatest from Day 3 to Day 4. Choosing Day 4 to Day 5 comes from picking the interval that ends on the highest total number of orders, 2,510, confusing the SIZE of the total with the RATE at which it grew. Choosing Day 1 to Day 2 comes from assuming the rate must be greatest at the very start, without working out any of the four differences. Choosing Day 2 to Day 3 comes from comparing only the first two differences, 225 and 321, and stopping there without checking Day 3 to Day 4 or Day 4 to Day 5. Finding the greatest rate of change from a table always means computing every difference between consecutive values and comparing them all — the day with the highest total, or the first pair you check, is not a shortcut.
- (b) 5/7 — The enlargement multiplier is 1.4, which as a fraction is 7/5. To reverse an enlargement, use the reciprocal of the multiplier: flip 7/5 to get 5/7. 7/5 comes from using the enlargement multiplier again, instead of reversing it. 3/5 comes from treating the reverse as 'give back the extra amount', working out 1 − (1.4 − 1) = 0.6, instead of using the reciprocal. 5/2 comes from ignoring the whole number in 1.4 and inverting only the decimal part, 0.4, as if it were the whole multiplier.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (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.
- (b) 24 km/h — First convert 45 minutes to hours: 45 ÷ 60 = 0.75 hours. Then divide the distance by the time: 18 ÷ 0.75 = 24 km/h. Reading 45 minutes as 0.45 hours (writing the minutes after the decimal point instead of dividing by 60) gives 18 ÷ 0.45 = 40 km/h. Working out 18 × 0.75 = 13.5 multiplies by the time instead of dividing. Working out 18 ÷ 45 = 0.4 divides by 45 without ever converting the minutes to hours. The cyclist's average speed is 24 km/h.
- (c) 183 mg — Apply the decay, then add the new dose, once for each hour. Hour 1: 0.7 × 200 = 140, then 140 + 50 = 190. Hour 2: 0.7 × 190 = 133, then 133 + 50 = 183, so there is 183 mg after 2 hours. Forgetting the top-up dose and only applying the decay gives 0.7 × 200 = 140, then 0.7 × 140 = 98 — this ignores that a further 50 mg is given every hour. Adding the 50 mg BEFORE the decay is applied, instead of after, gives 0.7 × (200 + 50) = 175, then 0.7 × (175 + 50) = 157.5, which changes how much of the dose is eliminated in the same hour it is given. Multiplying by 0.3, the percentage ELIMINATED, instead of by 0.7, the percentage REMAINING, gives 0.3 × 200 + 50 = 110, then 0.3 × 110 + 50 = 83 — this mixes up the amount that leaves the bloodstream with the amount that stays in it. Always check whether a percentage describes what remains or what is removed before choosing the multiplier.
- (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.
- (a) £11.70 — Rate of pay = total pay ÷ hours worked, so £105.30 ÷ 9 = £11.70 per hour. Working out £105.30 − 9 = £96.30 subtracts the number of hours from the total pay instead of dividing. Working out £105.30 × 9 = £947.70 multiplies total pay by hours worked instead of dividing. Misplacing the decimal point in the correct answer gives £117.00 instead of £11.70. Maya's rate of pay is £11.70 per hour.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (b) 1500 — The rate is 3 ÷ 2 = 1.5 litres per minute. Converting to cm³: 1.5 × 1000 = 1500 cm³ per minute. Getting 3000 comes from converting 3 litres to cm³ first (3000 cm³) and forgetting to divide by the 2 minutes. Getting 750 comes from dividing by the 2 minutes a second time after converting (1500 ÷ 2). Getting 2000 comes from converting the 2 minutes as if it were litres (2 × 1000) instead of using the correct rate of 1.5 litres per minute.
- (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (c) 48 mm — Method: extension = k × force, where k = extension ÷ force. Working: k = 12 ÷ 5 = 2.4 mm per N. At 20 N: extension = 2.4 × 20 = 48 mm. Wrong options: 32 mm comes from adding the extension and force numbers instead of scaling (12 + 20); 3 mm comes from treating the relationship as inverse proportion (12 × 5 ÷ 20); 36 mm comes from using an incorrect scale factor of 3 between the forces instead of the correct factor of 4 (20 ÷ 5).
Build your own mix at the worksheet builder.