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This lesson pulls together the whole Ratio, Proportion and Rates of Change strand of OCR GCSE Mathematics (J560) into a set of exam-style questions. Each question is labelled with a mark allocation and a tier — Foundation (grades 1–5) or Higher (grades 4–9, marked [H]) — and is followed by a full worked solution with notes on where the method marks fall. Work each question first, then read the solution. Throughout, remember the OCR command words: "Work out", "Calculate", "Show that", "Give a reason for your answer", "Write down" and "Find".
The questions are designed to build AO1 (use and apply standard techniques), AO2 (reason, interpret and communicate) and AO3 (solve problems) — the three assessment objectives every OCR paper rewards. The mark allocation is a strong hint about how much work is expected: a 2-mark question usually needs a single method step and an answer, while a 4- or 5-mark question expects several clearly shown stages. Showing your method matters enormously, because most marks in this strand are method marks — even if your final arithmetic slips, a clearly written multiplier, value of one part, or conversion factor will earn credit. As you work through, notice how often the same handful of techniques reappears: simplifying with the HCF, finding the value of one part, applying a multiplier, using a rate formula, and converting units. Mastering those few methods is what unlocks the whole topic.
A few habits make a real difference on ratio and proportion questions:
Write the ratio 18:24 in its simplest form.
The HCF of 18 and 24 is 6. Dividing both parts:
18:24=3:4.
Answer: 3:4.
Method marks: one mark for identifying a common factor and dividing; one for the fully simplified ratio.
Share £80 between Theo and Ruby in the ratio 3:5. Work out how much each receives.
Total parts =3+5=8. Value of one part =80÷8=10, i.e. £10.
Theo =3×10=30 (£30); Ruby =5×10=40 (£40).
Check: 30+40=80. Correct.
Answer: Theo £30, Ruby £40.
Method marks: one for the total parts, one for the value of one part, one for both correct shares.
5 identical chocolate bars cost £3.45. Work out the cost of 8 bars.
This is direct proportion, best handled with the unitary method. First find the cost of one bar, then scale up.
Cost of one bar =3.45÷5=0.69, i.e. £0.69.
Cost of 8 bars =8×0.69=5.52, i.e. £5.52.
Answer: £5.52.
Method marks: one for the unitary step (cost of one), one for multiplying by 8, one for the correct money answer. Make sure to scale up — 8 bars cost more than 5, so the answer should be larger than £3.45.
A jacket costs £64. In a sale it is reduced by 25%. Work out the sale price.
A 25% reduction means paying 75% of the price, so the multiplier is 0.75.
Sale price =64×0.75=48, i.e. £48.
Answer: £48.
Method marks: one for the multiplier 0.75 (or finding 25%), one for the method, one for the correct answer. On the non-calculator paper you could instead find 25% as 64÷4=16 and subtract: 64−16=48 — both routes are acceptable.
A car travels 180 km in 2 hours 30 minutes. Work out the average speed in km/h.
The time must be in hours to match "km/h". Since 30 minutes is half an hour, 2 hours 30 minutes =2.5 hours.
average speed=timedistance=2.5180=72 km/h.
Answer: 72 km/h.
Method marks: one for converting the time to 2.5 hours, one for the speed formula, one for the division, one for the correct answer with units. The most common error is writing the time as 2.3 hours; remember 30 minutes is 0.5 of an hour, not 0.3.
The ratio of staff to children at a nursery must be 1:4. There are 52 children. Work out the smallest number of staff needed.
For every 4 children, 1 member of staff is needed.
52÷4=13 staff.
Answer: 13 staff.
Method marks: one for recognising "children ÷4", one for the division, one for the correct answer.
After a 15% increase, a monthly rent is £575. Work out the rent before the increase.
This is a reverse percentage. A 15% increase has multiplier 1.15, so £575 represents the original rent ×1.15. To undo the increase, divide by the multiplier:
original=575÷1.15=500.
So the rent was £500.
Check: 500×1.15=575. Correct.
Answer: £500.
Method marks: one for the multiplier 1.15, one for dividing (not subtracting), one for the correct answer. The classic trap is to take 15% of £575 and subtract it, which gives the wrong answer because the 15% was applied to the original, not the final amount.
£4,000 is invested at 3% compound interest per year. Work out the value after 5 years, to the nearest penny.
The multiplier is 1.03, applied 5 times:
4,000×1.035=4,000×1.159274=4,637.10.
So the value is £4,637.10.
Answer: £4,637.10.
Method marks: one for the multiplier and power 1.035, one for the calculation, one for rounding correctly to the nearest penny.
y is inversely proportional to x. When x=4, y=15. Work out y when x=10.
For inverse proportion, y=xk, and the constant is the product of the known pair: k=xy=4×15=60.
So y=x60, and when x=10:
y=1060=6.
Answer: y=6.
Method marks: one for k=60, one for the equation y=x60, one for the correct value. Using xy to find k here would be the direct-proportion method and would give the wrong answer — for inverse proportion the constant is always the product.
A population of 8,000 decreases by 5% each year. Work out how many complete years until the population first falls below 6,000.
A 5% decrease leaves 95%, so the decay factor is 0.95. We need the smallest n with 8,000×0.95n<6,000. Dividing both sides by 8,000 removes the starting amount and leaves the simpler condition 0.95n<0.75.
Testing whole-number powers either side of the boundary: 0.955=0.7738 (still above 0.75), 0.956=0.7351 (now below 0.75).
So the population first drops below 6,000 after 6 years.
Answer: 6 years.
Method marks: one for the decay factor 0.95, one for setting up 0.95n<0.75, one for testing values either side of the boundary, one for the correct whole-number answer. Giving a decimal answer here would lose the final mark — the question asks for complete years.
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