1MA1 · R16Ratio, Proportion & Rates of ChangeFoundation & HigherGrades 4–7

Growth & Decay

Put £4000 in a savings account paying 3% compound interest, and each year you earn interest on last year’s interest too. After 5 years you have not £4600 but £4637.10. $\textbf{Compound growth}$ and $\textbf{decay}$ apply the same percentage change repeatedly, and a multiplier raised to a power calculates it in a single step.

40 min Ratio & Proportion

The big picture

Repeated percentage change is repeated multiplication: $n$ years of 3% growth multiply by $1.03$ a total of $n$ times, which is $1.03^{n}$ . The same structure models depreciating cars, cooling drinks, spreading infections and radioactive decay.

Edexcel often contrasts simple interest (the same amount added each year) with compound interest, and asks how long it takes to reach a target — a question answered by trying successive powers.

What you'll learn

Calculate simple interest and compound interest
Use the formula $\text{amount} = \text{original} \times \text{multiplier}^{n}$
Solve depreciation and other decay problems
Find the number of periods needed to reach a target value

Simple and compound interest

$\textbf{Simple interest}$ is calculated on the original amount only, so the same interest is added each year. £5000 at 2.5% simple interest for 4 years earns $4 \times 125 = 500$ , giving £5500.

$\textbf{Compound interest}$ is calculated on the current amount, including previous interest. £5000 at 2.5% compound for 4 years gives $5000 \times 1.025^{4} =$ £5519.06.

Compound interest always overtakes simple interest over time, because the amount being multiplied keeps growing.

A = P\left(1 + \tfrac{r}{100}\right)^{n}

Compound growth: P original, r% per period, n periods.

Example 1£4000 is invested at 3% compound interest per year. Find the value after 5 years.

1Multiplier: 1.03.

4000 \times 1.03^{5} = 4637.096\dots

Answer£4637.10

Your turnSame method, new numbers — try it before moving on.

£2,500 is invested at 2.5% compound interest per year. Find its value after 4 years, to the nearest penny.

Your answer

Show worked solution

Solution

1Multiplier $1.025$ .
2 $2500 \times 1.025^4 = 2759.532\ldots$

Answer: £2,759.53

Depreciation and decay

Decay uses a multiplier below 1. A car losing 12% of its value each year is multiplied by $0.88$ annually.

After $n$ years its value is original $\times 0.88^{n}$ .

A = P\left(1 - \tfrac{r}{100}\right)^{n}

Compound decay.

Example 1A car worth £18 000 depreciates by 12% per year. Find its value after 3 years.

1Multiplier: 0.88.

18\,000 \times 0.88^{3} = 12\,266.496

Answer£12 266.50

Your turnSame method, new numbers — try it before moving on.

A machine worth £9,000 depreciates by 15% per year. Find its value after 3 years, to the nearest penny.

Your answer

Show worked solution

Solution

1Multiplier $0.85$ .
2 $9000 \times 0.85^3 = 5527.125$ , so £5,527.13.

Answer: £5,527.13

Why this works

After 3 years at 12% per year the car has lost about 32% of its value, not 36%. Each year’s loss is 12% of a smaller amount.

Reaching a target

To find how many periods until a value passes a target, try increasing values of $n$ in the formula until the target is crossed.

State the first whole number of periods that works, with the values either side as evidence.

Example 1A town’s population of 25 000 falls by 8% each year. After how many years will it first be less than half its current size?

1Need

0.92^{n} < 0.5

0.92^{8} = 0.513

— still above half.

0.92^{9} = 0.472

— below half.

AnswerAfter 9 years.

Your turnSame method, new numbers — try it before moving on.

A population of 40,000 grows by 5% per year. After how many whole years will it first exceed 50,000?

Your answer

Show worked solution

Solution

1 $40{,}000 \times 1.05^4 = 48{,}620$ — not yet.
2 $40{,}000 \times 1.05^5 = 51{,}051$ — exceeds 50,000.

Answer: $5$ years

Tip — Show the values for the last two powers you tried. " $n = 9$ " alone does not show that $n = 8$ was not enough.

Other growth models

Growth need not be yearly: a bacteria population doubling every hour is multiplied by $2^{t}$ after $t$ hours; interest paid monthly uses the monthly rate and number of months.

Working backwards, divide by the multiplier raised to the power: an investment worth £5304.50 after 2 years at 3% was $5304.50 \div 1.03^{2} =$ £5000.

Think like an examiner

Write the multiplier and the power used.
Round money to 2 decimal places only at the end.
For "how many years" questions, show values for consecutive powers.

Growth and decay

A = P \times m^{n}

Repeated change with multiplier m.

m = 1 + \tfrac{r}{100}

Growth multiplier.

m = 1 - \tfrac{r}{100}

Decay multiplier.

Watch out for these

Calculating compound interest as

n \times

one year’s interest.

That is simple interest; compound uses

P \times m^{n}

Using 0.12 as the multiplier for a 12% decrease.

The multiplier is 0.88.

Giving the interest earned when the question asks for the total amount.

Read carefully: total amount or interest only.

Rounding after each year and multiplying again.

Use

m^{n}

in one calculation to avoid rounding errors.

Stretch yourself

Asha invests £2000 at 6% compound interest per year. Ben invests £2600 at 4% simple interest per year. After how many whole years does Asha’s investment first exceed Ben’s?

Hint — Asha’s amount is $2000 \times 1.06^{n}$ ; Ben’s is $2600 + 104n$ . Compare them year by year.

Reveal solution

SolutionBen gains

4\%

of £2600, which is £104, each year. Year 10: Asha

2000 \times 1.06^{10} = 3581.70

, Ben

2600 + 1040 = 3640

— Ben ahead. Year 11: Asha

3796.60

, Ben

3744

— Asha ahead. So Asha’s investment first exceeds Ben’s after 11 years.

\blacksquare

Questions students ask

What is the difference between simple and compound interest?

Simple interest is always calculated on the original amount; compound interest is calculated on the growing total.

How do I find the original amount?

Divide the final amount by the multiplier raised to the number of periods.

Does compound decay ever reach zero?

No — multiplying by a number between 0 and 1 keeps making the value smaller but never zero.

Key takeaways

Simple interest adds the same amount each period.
Compound change: amount $=$ original $\times$ multiplier $^{n}$ .
Growth multipliers exceed 1; decay multipliers are between 0 and 1.
Find how long to reach a target by trying successive powers.
Work backwards by dividing by the multiplier to the power $n$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

£1,500 is invested for 3 years.

How much more interest is earned at 4% compound interest than at 4% simple interest?[3]
Show mark scheme
- M1Simple interest $1500 \times 0.04 \times 3 = 180$
- M1Compound $1500 \times 1.04^3 = 1687.30$
- A1£7.30
Answer: £7.30

Question 2Grade 63 marks

£2,000 is invested at a fixed rate of compound interest. After 3 years it is worth £2,382.03. Find the rate.[3]
Show mark scheme
- M1 $1.x^3 = \frac{2382.03}{2000} = 1.191015$
- M1Cube root: $1.06$
- A1 $6\%$
Answer: $6\%$

Question 3Grade 73 marks

A 5000 g radioactive sample loses 10% of its mass each hour. After how many whole hours is it first less than half its original mass?[3]
Show mark scheme
- M1 $5000 \times 0.9^n$ for at least two values of $n$
- M1 $n = 6$ : 2657 g; $n = 7$ : 2391 g
- A17 hours
Answer: 7 hours

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

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