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

This guide follows the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), sections 5.02 (direct and inverse proportion) and 5.03 (discrete growth and decay). Most of it applies to both tiers. Anything the specification lists only for Higher tier is labelled Higher tier only.

Use it with the revision notes and the practice questions. The course hub is OCR GCSE Mathematics, the printable list is the OCR GCSE Mathematics checklist, and the free diagnostics help you find weak spots.

What this unit covers

Reference	What you must be able to do	Tier
5.02a	Solve direct proportion problems, including equality of ratios and currency conversion	Both tiers
5.02a	Use y = kx and recognise that y is proportional to x	Both tiers
5.02a	Formulate and solve equations where a quantity is proportional to a power or root of another	Higher tier only
5.02b	Solve inverse proportion problems, such as speed and time	Both tiers
5.02b	Use y = k/x and recognise that y is inversely proportional to x	Both tiers
5.02b	Formulate and solve equations for inverse proportion to a power or root	Higher tier only
5.03a	Calculate simple interest	Both tiers
5.03a	Solve step-by-step multiplier problems: compound interest, depreciation	Both tiers
5.03a	Express exponential growth or decay as a formula, then solve and interpret problems	Higher tier only

Direct proportion (5.02a)

Two quantities are in direct proportion when they rise and fall together in the same ratio. Double one and the other doubles. Halve one and the other halves. The ratio y : x never changes, so y/x is a constant.

Equality of ratios

If y is proportional to x, then y1/x1 = y2/x2. Fill in the three values you know and solve for the fourth.

Worked example 1. 9 metres of cable cost £14.40. Find the cost of 25 metres.

Cost per metre = 14.40 ÷ 9 = £1.60
Cost of 25 m   = 1.60 × 25 = £40.00

Check with ratios: 14.40/9 = 40/25, and both sides equal 1.6.

Currency conversion

An exchange rate is a direct proportion. Multiply to go into the foreign currency and divide to come back.

Worked example 2. The rate is £1 = €1.14. (a) Convert €300 to pounds. (b) Convert £450 to euros.

(a) 300 ÷ 1.14 = £263.16 (to the nearest penny)
(b) 450 × 1.14 = €513.00

Sense check: a pound buys more than one euro, so the pound amount must be the smaller number.

The equation y = kx

Writing y ∝ x means y is directly proportional to x. This is the same as y = kx, where k is a constant (the constant of proportionality). Graphically, y = kx is a straight line through the origin.

Method.

1. Write y = kx.
2. Substitute a known pair of values and find k.
3. Write the full equation with k as a number.
4. Substitute the new value and solve.

Worked example 3. y is directly proportional to x. When x = 14, y = 35. Find y when x = 22, and x when y = 80.

35 = k × 14        so k = 2.5
y = 2.5x
x = 22:  y = 2.5 × 22 = 55
y = 80:  80 = 2.5x, so x = 80 ÷ 2.5 = 32

Direct proportion to a power or root (Higher tier only)

Here y ∝ xⁿ, so y = kxⁿ. Common cases are y = kx² (square), y = kx³ (cube) and y = k√x (square root). The method is identical, but you must apply the power or root to x before using the value.

Worked example 4 (Higher tier only). The cost C (£) of a circular rug is proportional to the square of its diameter d (metres). A rug with d = 1.5 costs £90. Find the cost of a rug with d = 2.4.

C = kd²
90 = k × 1.5² = 2.25k   so k = 40
C = 40d²
d = 2.4:  C = 40 × 5.76 = £230.40

Notice the effect of the power. Doubling d multiplies C by 2² = 4, not 2.

Worked example 5 (Higher tier only). The period T (seconds) of a pendulum is proportional to √L, where L is its length in metres. T = 1.4 when L = 0.49. Find T when L = 1.69, and L when T = 3.

T = k√L
1.4 = k × 0.7           so k = 2
T = 2√L
L = 1.69:  T = 2 × 1.3 = 2.6 seconds
T = 3:     3 = 2√L, so √L = 1.5 and L = 1.5² = 2.25 metres

When you are finding x from y, undo the power or root in the opposite order: square both sides to remove a root, take a root to remove a square.

Inverse proportion (5.02b)

Two quantities are in inverse proportion when one rises as the other falls, so that their product stays the same. Double one and the other halves. The classic setting is speed and time over a fixed distance: speed × time = distance.

Worked example 6. A cyclist rides at 12 km/h and takes 2.5 hours. How long would the same trip take at 20 km/h?

Distance = 12 × 2.5 = 30 km
Time at 20 km/h = 30 ÷ 20 = 1.5 hours (1 hour 30 minutes)

Faster speed, shorter time. If an inverse answer moves the wrong way, you have probably used direct proportion.

The equation y = k/x

Writing y ∝ 1/x means y is inversely proportional to x. This is the same as y = k/x, so xy = k. The graph is a curve that never touches either axis.

Worked example 7. y is inversely proportional to x. When x = 8, y = 5. Find y when x = 10, and x when y = 1.6.

y = k/x
5 = k/8      so k = 40
y = 40/x
x = 10:  y = 40/10 = 4
y = 1.6: 1.6 = 40/x, so x = 40 ÷ 1.6 = 25

Inverse proportion to a power or root (Higher tier only)

Typical forms are y = k/x², y = k/x³ and y = k/√x.

Worked example 8 (Higher tier only). The loudness I of a sound varies inversely with d², where d is the distance from the source. I = 80 when d = 3. Find I when d = 12, and d when I = 45.

I = k/d²
80 = k/9          so k = 720
I = 720/d²
d = 12:  I = 720/144 = 5
I = 45:  45 = 720/d², so d² = 16 and d = 4

Worked example 9 (Higher tier only). y is inversely proportional to √x. y = 3 when x = 16. Find y when x = 36, and x when y = 8.

y = k/√x
3 = k/4        so k = 12
y = 12/√x
x = 36:  y = 12/6 = 2
y = 8:   8 = 12/√x, so √x = 1.5 and x = 2.25

Growth and decay (5.03a)

Simple interest

Simple interest is paid on the original sum only, so the same amount is added every year.

Simple interest = P × r × n / 100, where P is the amount invested, r is the yearly percentage rate and n is the number of years.

Worked example 10. Find the simple interest on £2,400 at 3.5% per year for 4 years, and the total amount at the end.

Interest = 2400 × 3.5 × 4 / 100 = £336
Total    = 2400 + 336 = £2,736

You can also work backwards. If £1,500 earns £180 simple interest in 5 years, then 180 = 1500 × r × 5 / 100, so r = 180 × 100 ÷ 7500 = 2.4% per year.

Multipliers

A percentage change becomes one multiplication.

Change	Multiplier
Increase by 7%	1.07
Increase by 2.5%	1.025
Decrease by 18%	0.82
Decrease by 12%	0.88

For repeated changes, multiply the multipliers. You do not add percentages.

Compound interest

Compound interest is added to the balance each year, so the next year’s interest is calculated on a larger amount. For a fixed rate over n years, multiply by the same multiplier n times.

Worked example 11. £5,000 earns 2.8% compound interest per year. Find the value after 3 years and the interest earned.

Multiplier = 1.028
After 3 years: 5000 × 1.028³ = £5,431.87
Interest = 5431.87 − 5000 = £431.87

Round only at the end.

Depreciation with different yearly changes

When the rate changes each year, use a different multiplier for each year.

Worked example 12. A van is worth £22,000 new. It loses 25% of its value in year 1, 15% of the remaining value in year 2 and 10% in year 3. Find its value after three years and the overall percentage decrease.

Multipliers: 0.75, 0.85, 0.90
Value = 22000 × 0.75 × 0.85 × 0.90 = £12,622.50
Overall multiplier = 0.75 × 0.85 × 0.90 = 0.57375
Overall decrease = 1 − 0.57375 = 0.42625 = 42.6% (3 s.f.)

The overall decrease is not 25 + 15 + 10 = 50%. Each percentage applies to a smaller amount than the one before.

Growth and decay as a formula (Higher tier only)

Higher tier only. When the same multiplier is applied n times, write the result as a formula:

A = A₀ × rⁿ

A₀ is the starting amount, r is the multiplier per period and n is the number of periods. If r > 1 you have growth. If 0 < r < 1 you have decay. The specification’s own example is A = 100 × 1.1ⁿ for £100 at 10% a year, so the starting amount is 100, the multiplier is 1.1 and the growth rate is 10%.

Worked example 13 (Higher tier only). A culture starts with 800 bacteria and grows by 15% every hour. (a) Write a formula for the number N after t hours. (b) Find N after 6 hours. (c) Find the first whole hour at which N exceeds 2,000.

(a) N = 800 × 1.15ᵗ
(b) N = 800 × 1.15⁶ = 1850.45..., so about 1,850 bacteria
(c) Try values of t:
    t = 6:  1,850   (not enough)
    t = 7:  2,128   (more than 2,000)
    So t = 7 hours.

Part (c) is solved by trial. Put the answer in context: “after 7 hours” is the first whole hour at which the count is above 2,000.

Worked example 14 (Higher tier only). A machine’s value V (£) after n years is V = 18,000 × 0.88ⁿ. Interpret the numbers, then find when it first falls below £10,000.

18,000 is the starting value.
0.88 = 1 − 0.12, so the value falls by 12% each year.
n = 4: 18000 × 0.88⁴ = 10,794.52   (still above 10,000)
n = 5: 18000 × 0.88⁵ = 9,499.17    (below 10,000)
It first falls below £10,000 during the fifth year, so after 5 whole years.

Reading a formula the other way is just as common. For A = 250 × 1.04ⁿ, the starting amount is 250 and the growth is 4% per period.

Common errors

• Using direct proportion when the situation is inverse (more workers means fewer days, not more).
• Dividing by the exchange rate when you should multiply, or the reverse. Check which currency should give the larger number.
• Finding k and then forgetting to use it in the final line.
• In a power or root problem, squaring k as well, or forgetting to square the new x.
• Treating a 3-year compound problem as 3 × the yearly percentage. Use the multiplier raised to the power 3.
• Adding percentages in successive changes. Multiply multipliers instead.
• Rounding an interim value early, which moves the final answer.
• Writing a decay multiplier as 0.12 instead of 0.88 when the value falls by 12%.

Next steps

Recap with the revision notes, then try the practice questions (both linked at the top). For earlier percentage work, see fractions, decimals and percentages.

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.