Growth and decay | AQA GCSE Maths Notes

Visual model

Repeated percentage change compounds

same multiplier each period

\times r

Gold-standard guide

20 mins

What you will learn

Model repeated percentage change, including compound interest.

Use a clear step-by-step method for growth and decay.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier ratio, proportion and rates of change skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods

Step 1

Write the multiplier

For 4% growth: multiplier = 1 + $\frac{4}{1}00$ = 1.04

Step 2

Apply the compound formula

Value = 3500 × 1.04⁵

Step 3

Calculate 1.04^5

1.04⁵ ≈ 1.2166529

Watch out

Students apply simple interest (adding the same amount each year) instead of compound interest

Compound growth

$final = start \times multiplier^n.$

Decay

a decrease uses a multiplier less than 1.

Worked example

£3,500 is invested at 4% compound interest per year. Find its value after 5 years.

Write the multiplier: For 4% growth: multiplier = 1 + $\frac{4}{1}00$ = 1.04.

Apply the compound formula: Value = 3500 × 1.04⁵.

Calculate 1.04⁵: 1.04⁵ ≈ 1.2166529. Value = 3500 × 1.2166529 ≈ £4258.29.

Final answer

£4258.29 (to the nearest penny)

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

£3,500 is invested at 4% compound interest per year. Find its value after 5 years.

3 marks4 minsgrowth-and-decay-worked

Show solution

Worked solution

1.Write the multiplier: For 4% growth: multiplier = 1 + $\frac{4}{1}00$ = 1.04.
2.Apply the compound formula: Value = 3500 × 1.04⁵.
3.Calculate 1.04⁵: 1.04⁵ ≈ 1.2166529. Value = 3500 × 1.2166529 ≈ £4258.29.

Final answer

£4258.29 (to the nearest penny)

Mark points

M1: write the multiplier
M1: apply the compound formula
M1: calculate 1.04⁵
A1: £4258.29 (to the nearest penny)

Watch out

Students apply simple interest (adding the same amount each year) instead of compound interest.Compound: each year's interest is added to the running total.After 5 years, the amounts are very different — always use the power formula for compound.

Diagnosticrecall

A car worth £12,000 depreciates at 15% per year. Find its value after 3 years.

1 mark2 minsgrowth-and-decay-q1

Show solution

Worked solution

1.Spot the skill: Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.
2.Use the write the multiplier stage first, then apply the compound formula.
3.Keep the final answer visible: £7345.50.

Final answer

£7345.50

Mark points

M1: use the correct compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.this models repeated percentage change — interest building on interest.
A1: £7345.50

Watch out

Easyprocedure

Population of 50,000 grows at 2.5% per year. Find it after 4 years.

2 marks3 minsgrowth-and-decay-q2

Show solution

Worked solution

1.Spot the skill: Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.
2.Use the apply the compound formula stage first, then calculate 1.04⁵.
3.Keep the final answer visible: 55,190 (to nearest whole number).

Final answer

55,190 (to nearest whole number)

Mark points

M1: use the correct compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.this models repeated percentage change — interest building on interest.
A1: 55,190 (to nearest whole number)

Watch out

Mediumreasoning

An investment of £2000 grows to £2315.25 in 3 years. Find the annual interest rate.

3 marks4 minsgrowth-and-decay-q3

Show solution

Worked solution

1.Spot the skill: Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.
2.Use the calculate 1.04⁵ stage first, then write the multiplier.
3.Keep the final answer visible: 5%.

Final answer

Mark points

M1: use the correct compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.this models repeated percentage change — interest building on interest.
A1: 5%

Watch out

Hardproblem solving

A radioactive sample of 800 g loses 10% of its mass each year. Find the mass after 6 years.

3 marks5 minsgrowth-and-decay-q4

Show solution

Worked solution

1.Spot the skill: Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.
2.Use the write the multiplier stage first, then apply the compound formula.
3.Keep the final answer visible: ≈ 425.7 g.

Final answer

≈ 425.7 g

Mark points

M1: use the correct compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.this models repeated percentage change — interest building on interest.
A1: ≈ 425.7 g

Watch out

Exam-stylemulti-step

How many years does it take £1000 to exceed £1500 at 6% compound interest?

4 marks6 minsgrowth-and-decay-q5

Show solution

Worked solution

1.Spot the skill: Compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.
2.Use the apply the compound formula stage first, then calculate 1.04⁵.
3.Keep the final answer visible: 7 years.

Final answer

7 years

Mark points

M1: use the correct compound growth formula: final = initial × (multiplier)ⁿ, where multiplier = 1 + rate/100 for growth or 1 − rate/100 for decay, and n = number of periods.this models repeated percentage change — interest building on interest.
A1: 7 years

Watch out

Grade 9 stretchproblem solving

A car loses 18% of its value each year. It is worth £13,448 after 2 years. Find its original value.

4 marks7 minsgrowth-g9

Show solution

Worked solution

1.Use the multiplier 0.82.
2.Divide by 0.82² to reverse two years of depreciation.

Final answer

£20,000

Mark points

M1: use 13448 / 0.82²
A1: £20,000

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Growth and decay - 2 marksA car worth £12,000 depreciates at 15% per year. Find its value after 3 years.Mark answer

Answer

£7345.50

2Ratio and sharing - 2 marksTwo people share profit in ratio 5:3. Total profit £640. Find each share.Mark answer

Answer

£400 and £240

3Fractions and ratios - 2 marksA map scale is 1:25,000. Express as a fraction.Mark answer

Answer

$\frac{1}{2}5$ ,000

4Direct and inverse proportion - 3 marksy ∝ 1/x². When x = 2, y = 9. Find y when x = 6.Mark answer

Answer

Mastery check

I can explain the method for growth and decay.
I can show clear working without skipping key steps.
I can avoid this mistake: Students apply simple interest (adding the same amount each year) instead of compound interest.Compound: each year's interest is added to the running total.After 5 years, the amounts are very different — always use the power formula for compound.

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