GCSE · Ratio, proportion and rates of change

Compound Interest, Growth and Decay

Put money in a savings account and the bank pays you interest. Leave it there and next year you earn interest on your interest too. That snowball effect is called compound growth, and the same maths describes populations, bacteria, car values and radioactive materials. One idea runs through all of it: multiply by the same number again and again.

GCSE subject content: set up, solve and interpret the answers in growth and decay problems, including compound interest; solve problems involving percentage change, including original value problems; (Higher) recognise, sketch and interpret graphs of exponential functions y = kˣ

Before you start: a quick recap

Percentages as decimals. 1% = 1/100 = 0.01, so 4% = 0.04 and 15% = 0.15. To find 4% of £250, work out 0.04 × 250 = £10.
Simple interest pays the same amount every year, worked out on the money you first put in. Revise it on the Simple Interest page.
Percentage increase and decrease with a multiplier: to increase by 20%, multiply by 1.2. Revise it on the Percentage Change page.
Words you'll need. The principal (P) is the amount you start with. The rate (r) is the percentage added each year (or each hour, day…). The amount (A) is what you have at the end.

1 · Simple vs compound: interest on interest

Two friends each save £1000 for 5 years at 10% a year.

Sam's account pays simple interest: 10% of the original £1000, which is £100 every year. Same amount, every year.
Cara's account pays compound interest: each year the bank works out 10% of whatever is in the account now and adds it on. Year 1: 10% of £1000 = £100, giving £1100. Year 2: 10% of £1100 = £110, giving £1210. Each year's interest is a bit bigger than the last, because it is worked out on a bigger amount.

The difference is the interest on interest: in year 2 Cara also earns 10% on the £100 she earned in year 1. Change the numbers and step through the years. Grey is the money put in, teal is simple interest, orange is the part of the compound interest that matches it, and plum is the extra that compounding produces.

principal P rate r years

The key difference. Simple interest grows by adding the same amount each year. Compound interest grows by multiplying by the same number each year. Over a few years the gap is small; over 10, 20 or 40 years it becomes huge. Almost all real savings accounts, loans and credit cards use compound interest.

2 · Multipliers: one number per year

Working out "10% of the amount, then add it on" every year is slow. Here's the shortcut. The amount you start the year with is 100% of itself. Adding 10% gives 100% + 10% = 110% of it. And 110% as a decimal is 110 ÷ 100 = 1.1. So

add 10% = × 1.1

The number you multiply by is called the multiplier. For a decrease you take away: losing 15% leaves 100% − 15% = 85%, so the multiplier is 0.85. A multiplier bigger than 1 means growth; smaller than 1 means a decrease.

Try it: write the multiplier

multiplier

Common mistake: 3% is 0.03, not 0.3. A 3% increase has multiplier 1.03. The multiplier 1.3 would be a 30% increase. Always check: "per cent" means "out of 100", so divide by 100.

3 · The compound interest formula

£2000 is saved at 4% compound interest. Each year the amount is multiplied by 1.04. Press the button to add years and watch the chain of multipliers grow.

After 3 years the money has been multiplied by 1.04 three times: 1.04 × 1.04 × 1.04, which is 1.04³ (see index notation). After n years it has been multiplied by 1.04ⁿ. In general, with principal P and a rate of r% per year:

A = P × (1 + r/100)ⁿ

Here (1 + r/100) is just the multiplier (1 + 4/100 = 1.04) and n is the number of years (or whatever time period the rate is per). The interest earned is the amount minus the principal: A − P. On a calculator, use the power key: 2000 × 1.04 ^ 3 (it may be labelled x^y or x^■).

Try it: use the formula

Exam traps with the formula

Amount or interest? "How much is in the account" means A. "How much interest is earned" means A − P. Read the last line of the question twice.
Power, not times. 3 years at 4% is × 1.04³, not × 1.04 × 3 and not × 1.12.
Round only at the end, to the nearest penny (2 decimal places for money). Rounding every year can make the last penny wrong.

4 · Depreciation and decay: multipliers below 1

Most things you buy lose value over time. A car might lose 15% of its value each year. That's called depreciation. It works exactly like compound interest, but the multiplier is less than 1: losing 15% leaves 85%, so multiply by 0.85 each year. Anything that shrinks by the same percentage each time is called exponential decay.

new car £ loses a year after years

Value

Total lost

Wrong: n × r%

Two things to notice.

3 years of 15% depreciation is not 45% off. Each year's 15% is taken from a smaller value, so the total loss is less: 1 − 0.85³ = 0.385875, a 38.6% loss (to 3 s.f.). The teal dashed line shows the wrong "take off 15% of the original every year" method.
Exponential decay never reaches zero. Taking 15% off always leaves 85% of something. The straight-line method would reach £0 after under 7 years.

5 · Population growth and other growth and decay

The same formula works for anything that changes by a fixed percentage each time period, not just money:

new amount = start × (multiplier)^{number of periods}

The time period doesn't have to be years. If bacteria increase by 30% per hour, then n is the number of hours. Pick a situation and slide the time. The orange square has an area equal to the amount now; the dashed square is the amount at the start.

situation time

answer

Rounding to suit the context. You can't have 26 530.2 people, so a population is rounded to a whole number (or to 3 s.f. if the question says so). Money is rounded to the nearest penny. A mass might be given to 3 s.f. Always keep the full calculator value until the last step.

6 · How many years? Trial and the bank-account table

Sometimes the question is turned round: "£5000 is invested at 3.5% compound interest. After how many years will it first be worth more than £6000?" You can't just divide, because the growth isn't the same each year. Instead, try values of n until the amount first passes the target. Type a number of years to test it:

£5000 at target

try n =

Or build the bank-account table one year at a time, like a bank statement:

So the answer is n = years

Show the two years either side. In an exam, write down the value for the last year that is below the target and the first year that is above it, e.g. "1.035⁵ × 5000 = £5938.43 (not enough), 1.035⁶ × 5000 = £6146.28 (more than £6000), so 6 years." That is your evidence.

7 · Reverse problems and repeated percentage change Higher

Working backwards. If you know the amount after compound growth and need the original, undo the multiplying by dividing by the same multiplier: original = A ÷ (multiplier)ⁿ. It's tempting to take 4% off for each year instead, but that doesn't work: 4% of the bigger, later amount is more than the 4% that was added.

final amount £ after of % a year for years

Repeated percentage change. When a price goes up 20% and then down 20%, it does not get back to where it started. The second change is a percentage of a different (bigger) amount. Multiply the multipliers to get one overall multiplier: 1.2 × 0.8 = 0.96, which is a 4% decrease overall.

1st 2nd 3rd

8 · Exponential growth vs linear growth Higher

Plot the amount against time and the two kinds of growth look completely different. Linear growth (adding the same amount each year, like simple interest) is a straight line. Exponential growth (multiplying by the same number each year, like compound interest) is a curve that gets steeper and steeper. Its equation has the form y = A × kˣ, where x is the time and k is the multiplier. Graphs like y = kˣ are called exponential functions: the variable x is in the power (the exponent).

Start both at 100. Change the linear "amount added per year" and the exponential percentage rate, then drag across the graph to compare them at any time.

linear: add per year exponential: per year show years

Linear at cursor

Exponential at cursor

Overtaking

Recognising y = kˣ (k positive).

It always passes through (0, 1), because k⁰ = 1. (With a start value A, y = A × kˣ passes through (0, A).)
If k > 1 it is growth: a curve that rises more and more steeply. Whatever straight line you compare it with, the exponential eventually overtakes it.
If 0 < k < 1 it is decay: it falls steeply at first, then levels off, getting closer and closer to the x-axis but never touching it. y is never 0 or negative.

Worked examples

£3000 is invested at 2.5% per year compound interest for 4 years. Find the amount and the interest earned.

Multiplier: 100% + 2.5% = 102.5% = 1.025.

Formula: A = P × (multiplier)ⁿ = 3000 × 1.025⁴.

Calculate: 1.025⁴ = 1.103812890…, so A = 3000 × 1.103812890… = 3311.438671…

Round to the nearest penny: A = £3311.44.

Interest earned: 3311.44 − 3000 = £311.44.

Sense check: simple interest would be 4 × 2.5% = 10% of £3000 = £300. Compound should be a little more ✓

Compare £800 for 3 years at 5% simple interest and at 5% compound interest.

Simple: 5% of £800 = £40 a year. 3 × £40 = £120. Total = 800 + 120 = £920.

Compound, year by year: 800 × 1.05 = 840; 840 × 1.05 = 882; 882 × 1.05 = 926.10.

Compound, in one go: 800 × 1.05³ = 800 × 1.157625 = £926.10. Same answer.

Difference: 926.10 − 920 = £6.10 more with compound interest: the interest earned on earlier interest.

A car costs £15 000. It depreciates by 12% per year. Find its value after 5 years, to the nearest pound.

Multiplier: 100% − 12% = 88% = 0.88. (Less than 1, because it's a decrease.)

Formula: value = 15 000 × 0.88⁵.

Calculate: 0.88⁵ = 0.527731916…, so value = 15 000 × 0.527731916… = 7915.978…

Answer: £7916 to the nearest pound.

Watch out: 5 × 12% = 60% off would give £6000. That's wrong, because each year's 12% is taken from a smaller value.

A town has 40 000 people. The population grows by 3% each year. Estimate the population after 10 years, to 3 s.f.

Multiplier: 1.03.

Formula: population = 40 000 × 1.03¹⁰.

Calculate: 1.03¹⁰ = 1.343916…, so population = 53 756.655…

Round to 3 s.f.: 53 800 people.

Interpret: this assumes the growth rate stays at 3% every year, which real populations may not do. That's why it's an estimate.

£2000 is invested at 6% compound interest. After how many whole years will it first be worth more than £2500?

Try values of n with 2000 × 1.06ⁿ:

n = 1: £2120.00 n = 2: £2247.20 n = 3: £2382.03 (still below £2500)

n = 4: 2000 × 1.06⁴ = £2524.95 (more than £2500)

Answer: 4 years. Write down both n = 3 and n = 4 as your evidence.

Trap: with simple interest it would take 500 ÷ 120 = 4.17 years, so 5 whole years. Compound is quicker, so don't use the simple-interest shortcut.

Higher After depreciating by 20% a year for 5 years, a machine is worth £4096. What was it worth when new?

Set up: original × 0.8⁵ = 4096.

Undo by dividing: original = 4096 ÷ 0.8⁵ = 4096 ÷ 0.32768.

Answer: £12 500.

Check: 12 500 × 0.8⁵ = 12 500 × 0.32768 = 4096 ✓

Wrong method: adding 20% back on each year (4096 × 1.2⁵ = £10 192.19) doesn't undo it. Divide by 0.8, don't multiply by 1.2.

Higher A jacket's price is increased by 10%, then reduced by 10% in a sale. What is the overall percentage change?

Multipliers: increase 10% → × 1.1; decrease 10% → × 0.9.

Overall multiplier: 1.1 × 0.9 = 0.99.

Interpret: 0.99 = 99%, so the price is now 1% lower: a 1% decrease overall.

Why not 0%? The 10% off is 10% of the higher price, so it takes off more than the 10% that was added. Check with £100: £100 → £110 → £99.

Quick check

What you've learned

Simple interest adds the same amount every year. Compound interest works out the interest on the current amount, so you also earn interest on interest.
A percentage change is one multiplier: increase by r% → × (1 + r/100); decrease by r% → × (1 − r/100). 3% is 0.03, so the multipliers are 1.03 and 0.97.
Compound growth over n periods: A = P × (1 + r/100)ⁿ. The interest is A − P. Round only at the end, to the nearest penny for money.
Depreciation and decay use a multiplier less than 1. The total loss is not n × r%, and the value never reaches zero.
The same method works for populations, bacteria and radioactive decay; the time period can be years, hours or days. Round to suit the context.
To find how many years, try values of n (or build the year-by-year table) and quote the year either side of the target.
Higher: to reverse compound change, divide by the multiplier to the power n. For repeated changes, multiply the multipliers together. Exponential graphs y = A × kˣ curve upwards for k > 1 and decay towards the x-axis for 0 < k < 1; linear growth is a straight line.