OCR GCSE Mathematics: Direct and inverse proportion, and growth and decay (J560) – Revision Notes

Quick recall for OCR GCSE (9-1) Mathematics J560 (Version 2.1, August 2026), sections 5.02 and 5.03. For full explanations and longer worked examples, read the study guide. When you are ready to practise, use the practice questions.

Related pages: OCR GCSE Mathematics hub, printable checklist and the free diagnostics. Content marked Higher tier only is listed in the third column of the specification.

Key results

Idea	Rule	Spec
Direct proportion	y = kx, so y/x is constant and y1/x1 = y2/x2	5.02a
Inverse proportion	y = k/x, so xy is constant and x1y1 = x2y2	5.02b
Direct, power or root (Higher tier only)	y = kx², y = kx³, y = k√x	5.02a
Inverse, power or root (Higher tier only)	y = k/x², y = k/√x	5.02b
Simple interest	P × r × n / 100	5.03a
Multiplier	Increase r%: 1 + r/100. Decrease r%: 1 − r/100	5.03a
Compound change	Start × multiplierⁿ	5.03a
Growth or decay formula (Higher tier only)	A = A₀ × rⁿ	5.03a

Direct proportion (5.02a)

Spot it: both quantities rise together, and zero of one means zero of the other. Doubling x doubles y.

Method in steps

1. Write y = kx.
2. Put in the known pair, then find k.
3. Write the equation with k as a number.
4. Substitute the new value.

Reminder. y is proportional to x. When x = 5, y = 12. Find y when x = 8.

12 = 5k, so k = 2.4
y = 2.4 × 8 = 19.2

Currency. Foreign currency from pounds: multiply by the rate. Pounds from foreign currency: divide by the rate. Check that the answer is sensible for the currency with the larger numbers.

Unitary method. Find the value of one unit first, then multiply. It is the same idea as finding k.

Higher tier only: powers and roots

Use y = kxⁿ with the same four steps. Apply the power or root to x before using the value.

Statement	Equation	If x doubles
y ∝ x²	y = kx²	y × 4
y ∝ x³	y = kx³	y × 8
y ∝ √x	y = k√x	y × √2

Reminder (Higher tier only). y is proportional to x². When x = 4, y = 48. Find x when y = 75.

48 = 16k, so k = 3
75 = 3x², so x² = 25 and x = 5

(Take the positive root when x is a length or a time.)

Inverse proportion (5.02b)

Spot it: one quantity rises as the other falls, with a fixed total job, distance or amount to share. Doubling x halves y.

Method in steps

1. Write y = k/x, or use xy = k.
2. Put in the known pair and find k by multiplying.
3. Write y = k/x with k as a number.
4. Substitute the new value.

Speed and time. Distance is fixed, so speed × time is constant. Doubling the speed halves the time.

Workers. 8 painters take 6 days. The job is 8 × 6 = 48 painter-days. With 12 painters it takes 48 ÷ 12 = 4 days. This assumes everyone works at the same rate.

Reminder. y is inversely proportional to x. When x = 5, y = 12. Find y when x = 8.

k = 5 × 12 = 60
y = 60/8 = 7.5

Compare this with the direct reminder: the same pair of values gives 19.2 (direct) and 7.5 (inverse). When x goes up, y goes up in one case and down in the other.

Higher tier only: powers and roots

Statement	Equation	If x doubles
y ∝ 1/x²	y = k/x²	y ÷ 4
y ∝ 1/√x	y = k/√x	y ÷ √2

Reminder (Higher tier only). y is inversely proportional to √x. When x = 9, y = 4. Find x when y = 3.

4 = k/3, so k = 12
3 = 12/√x, so √x = 4 and x = 16

Direct or inverse: decide before you calculate

Question to ask	Direct	Inverse
What happens to y when x increases?	y increases	y decreases
What stays constant?	y ÷ x	x × y
Straight line through the origin on a graph?	Yes (for y = kx)	No, it is a curve
Typical context	Cost and quantity, currency	Speed and time, workers and days

Growth and decay (5.03a)

Simple interest

Interest = P × r × n / 100

The same amount is added every year. Total = P + interest. To find the rate or time, rearrange the same formula.

Reminder. £1,800 at 4% simple for 3 years: 1800 × 4 × 3 / 100 = £216.

Compound interest and depreciation

Interest is added to the balance, so the base grows each year. Depreciation is a repeated percentage fall, so the base shrinks each year.

Method in steps

1. Convert each percentage to a multiplier.
2. Multiply the start value by one multiplier for each period.
3. Round at the end only.
4. Give the answer with units, and as money to the nearest penny.

Reminder. £6,000 at 3% compound for 2 years: 6000 × 1.03² = £6,365.40.

Different rates each year. Multiply different multipliers. For falls of 10% then 20%, the multiplier is 0.9 × 0.8 = 0.72. That is a 28% fall overall, not 30%.

Reversing a change. Divide by the multiplier. If a price after a 20% rise is £360, the original is 360 ÷ 1.2 = £300.

Higher tier only: exponential formulas

A = A₀ × rⁿ

Part	Meaning
A₀	Starting amount
r	Multiplier per period. Above 1 is growth, between 0 and 1 is decay
n	Number of periods

Reading a formula. For V = 900 × 0.9ⁿ the start is 900 and the value falls by 10% each period. After 3 periods V = 900 × 0.9³ = 656.1.

Solving “how long until” questions. Use trial. Calculate A for n = 1, 2, 3 and so on until the target is passed. State the whole number of periods that answers the question, and say what that means in context.

Must-know distinctions

• Simple and compound. Simple adds the same amount each year. Compound multiplies, so the amounts get bigger each year.
• Percentage and multiplier. A 12% fall is a multiplier of 0.88, not 0.12.
• Proportional and equal. y ∝ x does not mean y = x. It means y = kx for a constant k.
• Power on x only. In y = kx², the square applies to x and not to k.
• Interest rate and growth rate. Both can be read from r. For 1.035 the rate is 3.5%.

Quick self-test

Do these without notes. Questions 8 to 10 are Higher tier only.

1. 7 identical pens cost £9.45. Find the cost of 12 pens.
2. y is proportional to x. When x = 4.5, y = 18. Find y when x = 7.
3. y is inversely proportional to x. When x = 12, y = 5. Find x when y = 8.
4. 8 painters take 6 days. How long do 12 painters take?
5. Find the simple interest on £1,800 at 4% per year for 3 years.
6. Write the multiplier for (a) a 7% decrease and (b) a 2.5% increase.
7. Find the value of £6,000 after 2 years at 3% compound interest per year.
8. (Higher tier only) y is proportional to x². When x = 2, y = 20. Find y when x = 6.
9. (Higher tier only) y is inversely proportional to √x. When x = 9, y = 4. Find x when y = 3.
10. (Higher tier only) V = 900 × 0.9ⁿ. State the starting value and the percentage fall per period, then find V when n = 3.

Answers

1. 9.45 ÷ 7 = £1.35 each, so 12 pens cost £16.20.
2. k = 18 ÷ 4.5 = 4, so y = 4 × 7 = 28.
3. k = 12 × 5 = 60, so x = 60 ÷ 8 = 7.5.
4. Total = 8 × 6 = 48 painter-days, so 48 ÷ 12 = 4 days.
5. 1800 × 4 × 3 / 100 = £216.
6. (a) 0.93 (b) 1.025.
7. 6000 × 1.03² = 6000 × 1.0609 = £6,365.40.
8. k = 20 ÷ 4 = 5, so y = 5 × 36 = 180.
9. k = 4 × 3 = 12, so 3 = 12/√x, √x = 4 and x = 16.
10. Start 900, fall 10% per period, and V = 900 × 0.729 = 656.1.

Where marks are usually lost

• Choosing direct proportion for an inverse context, or the reverse.
• Not showing the step where k is found, so a wrong final number has nothing to support it.
• Writing y = kx when the question says proportional to x², or applying the power to k.
• Multiplying by the exchange rate when the rate should divide, then not noticing the answer is the wrong size.
• Adding percentages across several years instead of multiplying multipliers.
• Using 0.12 for a 12% decrease, or 1.5 for a 5% increase.
• Stopping a trial-and-improvement answer at the first value above the target without checking the one before it.
• Leaving money to many decimal places, or giving 12622.5 without the pound sign and second decimal place.
• Not saying what the answer means when the question asks you to interpret it.

Official syllabus

OCR, GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), first assessment 2017, sections 5.02a, 5.02b and 5.03a.