Growth and Decay – GCSE Maths

Growth and decay problems all come from the same idea: the same percentage change applied over and over again. Each time, you multiply by the same multiplier.

The growth and decay formula

$A = P \times M^{\,n}$
$P$ = starting amount · $M$ = multiplier · $n$ = number of time periods · $A$ = final amount

Situation	Multiplier $M$
Growth of $3\%$ per year	$1.03$
Growth of $12\%$ per year	$1.12$
Decay of $8\%$ per year	$0.92$
Decay of $25\%$ per year	$0.75$
Doubling each period	$2$
Halving each period	$0.5$

Never multiply the percentage by the number of years. $5\%$ growth for $3$ years is not $15\%$ growth — it is $1.05^3 = 1.157625$, that is $15.76\%$.

Compound interest means you earn interest on the interest already earned. Each year the interest is calculated on the new balance.

Compound interest

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

Write down $P$ (the starting amount), the rate and $n$ (the number of years).
Work out the multiplier: $1 + \dfrac{r}{100}$.
Raise it to the power $n$ using the $x^y$ button on your calculator.
Multiply by $P$ and round sensibly (money to the nearest penny).
If the question wants the interest, remember to subtract $P$ at the end.

Worked Example 1 — Compound interest

£$4000$ is invested at $3.5\%$ compound interest per year for $6$ years. Find the value of the investment and the total interest earned.

①Multiplier $= 1 + 0.035 = 1.035$

②$A = 4000 \times 1.035^{6}$

③$1.035^6 = 1.229255\ldots$

④$A = 4000 \times 1.229255 = £4917.02$ (nearest penny)

⑤Interest $= 4917.02 - 4000 = £917.02$

Worked Example 2 — Simple vs compound

Compare £$5000$ invested for $4$ years at $6\%$ simple interest and at $6\%$ compound interest.

①Simple: $5000 \times 0.06 = £300$ per year, so $300 \times 4 = £1200$ interest. Total $= £6200$.

②Compound: $5000 \times 1.06^4 = 5000 \times 1.26247696 = £6312.38$.

③Interest $= £1312.38$.

Compound interest earns £$112.38$ more over the four years — and the gap widens rapidly as the years go on.

Depreciation is compound decrease: cars, laptops and machinery all lose a percentage of their value each year.

Decay

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

Worked Example 3 — Car depreciation

A car costing £$22\,000$ loses $18\%$ of its value each year. Find its value after $5$ years, to the nearest pound.

①Multiplier $= 1 - 0.18 = 0.82$

②$A = 22\,000 \times 0.82^{5}$

③$0.82^5 = 0.3707398\ldots$

④$A = 22\,000 \times 0.3707398 = £8156$ (nearest pound)

The car has lost about $63\%$ of its value in five years.

Worked Example 4 — Half-life

A radioactive substance has a half-life of $3$ days: its mass halves every $3$ days. A sample starts at $640$ g. Find the mass after $15$ days.

①Number of half-lives $= 15 \div 3 = 5$

②$A = 640 \times 0.5^{5}$

③$= 640 \times 0.03125 = 20$ g

Check by halving: $640 \to 320 \to 160 \to 80 \to 40 \to 20$ ✓

When $n$ is the unknown, the GCSE method is trial and improvement: try values of $n$ until you cross the target.

Worked Example 5 — Finding the number of years

£$3000$ is invested at $4\%$ compound interest. After how many complete years will the investment first exceed £$4000$?

①$A = 3000 \times 1.04^n$; we need $A > 4000$.

$n$	$5$	$6$	$7$	$8$
$A$ (£)	$3649.96$	$3795.96$	$3947.80$	$4105.71$

②After $7$ years the value is still below £$4000$; after $8$ years it is above.

Answer: $8$ years

Calculator tip: type $3000 \times 1.04$, press $=$, then keep pressing $=$ (or use the ANS key) to step through the years one at a time.

Worked Example 6 — Decay over time

A population of $12\,000$ fish is falling by $15\%$ per year. After how many years will the population first drop below $5000$?

①$A = 12\,000 \times 0.85^n$

②$n = 4$: $12\,000 \times 0.85^4 = 6264$

③$n = 5$: $12\,000 \times 0.85^5 = 5324$

④$n = 6$: $12\,000 \times 0.85^6 = 4526$

Answer: $6$ years

Worked Example 7 — Finding the starting value

After $3$ years of $5\%$ compound growth, an investment is worth £$9261$. What was invested at the start?

①$9261 = P \times 1.05^3$

②$1.05^3 = 1.157625$

③$P = 9261 \div 1.157625 = £8000$

Worked Example 8 — Finding the rate

A painting bought for £$2000$ is worth £$3456$ after $4$ years. Find the annual percentage growth rate.

①$3456 = 2000 \times M^4$

②$M^4 = \dfrac{3456}{2000} = 1.728$

③$M = \sqrt[4]{1.728} = 1.14657\ldots$ (use the $\sqrt[x]{\;}$ button, or raise to the power $0.25$)

④$M - 1 = 0.1466$, so the rate is about $14.7\%$ per year.

An iterative process is any rule you apply repeatedly, feeding each answer back in as the next input. Growth and decay are just one example.

Iteration notation

$x_{n+1} = f(x_n)$ — "the next term is found by applying the rule to the one before"

Worked Example 9 — An iterative formula

Use $x_{n+1} = \sqrt{5x_n + 3}$ with $x_1 = 4$ to find $x_2$, $x_3$ and $x_4$, each to 4 decimal places.

①$x_2 = \sqrt{5 \times 4 + 3} = \sqrt{23} = 4.7958$

②$x_3 = \sqrt{5 \times 4.7958 + 3} = \sqrt{26.9790} = 5.1941$

③$x_4 = \sqrt{5 \times 5.1941 + 3} = \sqrt{28.9705} = 5.3824$

The values are settling towards about $5.54$, which is the solution of $x^2 = 5x + 3$.

Calculator method: type $4$ and press $=$. Then type $\sqrt{5 \times \text{ANS} + 3}$ and press $=$ repeatedly. Each press gives the next term.

Worked Example 10 — Iteration with a real context

A savings account pays $2\%$ interest per year, and the saver adds £$500$ at the end of each year. She starts with £$2000$. Write an iterative rule and find the balance after $3$ years.

①Rule: $B_{n+1} = 1.02 \times B_n + 500$

②Year 1: $1.02 \times 2000 + 500 = 2040 + 500 = £2540$

③Year 2: $1.02 \times 2540 + 500 = 2590.80 + 500 = £3090.80$

④Year 3: $1.02 \times 3090.80 + 500 = 3152.62 + 500 = £3652.62$

Balance after $3$ years $= £3652.62$

Note that the simple formula $A = P \times M^n$ does not work here, because extra money is added each year. An iterative approach is essential.

The formula

$A = P \times M^{\,n}$ for repeated percentage change.

Growth

$M = 1 + \dfrac{r}{100}$, e.g. $+7\% \Rightarrow 1.07$.

Decay

$M = 1 - \dfrac{r}{100}$, e.g. $-7\% \Rightarrow 0.93$.

Interest only

Subtract $P$ from the final amount.

Simple vs compound

Simple is a straight line; compound is a curve and always ends up larger.

Finding $n$

Trial and improvement — press $=$ repeatedly and count.

Finding $P$

Divide the final amount by $M^{\,n}$.

Finding $M$

$M = \sqrt[n]{\dfrac{A}{P}}$, then subtract $1$ for the rate.

Iteration

$x_{n+1} = f(x_n)$; use the ANS key to repeat the rule.

Question 1

£$6000$ is invested at $4\%$ compound interest per year. Find the value after $3$ years, to the nearest penny.

▶ Show solution

$A = 6000 \times 1.04^3$

$1.04^3 = 1.124864$

$A = 6000 \times 1.124864 = £6749.18$

Question 2

A motorbike bought for £$7500$ depreciates by $22\%$ each year. Find its value after $4$ years, to the nearest pound.

▶ Show solution

Multiplier $= 0.78$

$A = 7500 \times 0.78^4 = 7500 \times 0.37015056$

$= £2776$ (nearest pound)

Question 3

£$2500$ is invested for $5$ years at $3\%$ per year. Find the total interest earned with (a) simple interest, (b) compound interest.

▶ Show solution

(a) $2500 \times 0.03 = £75$ per year; $75 \times 5 = £375$.

(b) $A = 2500 \times 1.03^5 = 2500 \times 1.159274 = £2898.19$.

Interest $= 2898.19 - 2500 = £398.19$.

Compound earns £$23.19$ more.

Question 4

A bacterial colony of $500$ cells grows by $30\%$ every hour. How many cells are there after $6$ hours, to the nearest whole number?

▶ Show solution

$A = 500 \times 1.3^6$

$1.3^6 = 4.826809$

$A = 500 \times 4.826809 = 2413.4$

About $2413$ cells

Question 5

A radioactive isotope has a half-life of $8$ hours. A $960$ g sample is left for $2$ days. How much remains?

▶ Show solution

$2$ days $= 48$ hours; number of half-lives $= 48 \div 8 = 6$.

$A = 960 \times 0.5^6 = 960 \times \dfrac{1}{64} = 15$ g

Question 6

£$1500$ is invested at $5\%$ compound interest. After how many complete years does the investment first exceed £$2000$?

▶ Show solution

$A = 1500 \times 1.05^n$

$n = 5$: $1914.42$ — still below.

$n = 6$: $2010.14$ — above.

Answer: $6$ years

Question 7

After $2$ years of $8\%$ annual compound growth, a fund is worth £$11\,664$. What was the original investment?

▶ Show solution

$11\,664 = P \times 1.08^2 = P \times 1.1664$

$P = 11\,664 \div 1.1664 = £10\,000$

Question 8

A house bought for £$180\,000$ is worth £$239\,580$ after $3$ years. Find the annual percentage increase.

▶ Show solution

$M^3 = \dfrac{239\,580}{180\,000} = 1.331$

$M = \sqrt[3]{1.331} = 1.1$

$M - 1 = 0.1$, so the house increased by $10\%$ per year.

Check: $180\,000 \times 1.1^3 = 180\,000 \times 1.331 = £239\,580$ ✓

Question 9

Bank A offers $3.5\%$ compound interest per year. Bank B offers $4\%$ for the first year and then $3\%$ for each following year. £$8000$ is invested for $3$ years. Which bank gives more, and by how much?

▶ Show solution

Bank A: $8000 \times 1.035^3 = 8000 \times 1.108718 = £8869.74$

Bank B: $8000 \times 1.04 \times 1.03 \times 1.03 = 8000 \times 1.103336 = £8826.69$

Bank A gives more, by $8869.74 - 8826.69 = £43.05$.

Question 10

A lake contains $80\,000$ litres of water. Each week $12\%$ evaporates, but $6000$ litres of rain are added at the end of the week.

(a) Write an iterative formula for the volume $V_n$ after $n$ weeks. (b) Find the volume after $3$ weeks, to the nearest litre. (c) Explain why the volume will eventually settle at $50\,000$ litres.

▶ Show solution

(a) $V_{n+1} = 0.88 V_n + 6000$, with $V_0 = 80\,000$.

(b) Week 1: $0.88 \times 80\,000 + 6000 = 70\,400 + 6000 = 76\,400$

Week 2: $0.88 \times 76\,400 + 6000 = 67\,232 + 6000 = 73\,232$

Week 3: $0.88 \times 73\,232 + 6000 = 64\,444.16 + 6000 = \mathbf{70\,444}$ litres

(c) The volume settles when it stops changing, i.e. when $V = 0.88V + 6000$.

$0.12V = 6000 \Rightarrow V = 50\,000$ litres.

At that level the $12\%$ lost ($6000$ litres) exactly balances the $6000$ litres of rain added.