Learn4024Paper 2 · CalculatorNumberPercentages, interest, growth and decay

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Percentages, Interest, Growth and Decay

Turn every percentage change into a multiplier: a $15\%$ increase is $\times1.15$ , a $12\%$ decrease is $\times0.88$ . Repeated changes multiply, so compound interest and depreciation use powers of the multiplier. To undo a change (reverse percentage), divide by the multiplier.

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HEdexcel 4MA1 Papers 1F, 2FAQA 8300 Papers 1–3≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find a percentage of an amount and express one quantity as a percentage of another.
Use multipliers for percentage increase and decrease.
Solve reverse percentage problems.
Calculate simple and compound interest, growth and depreciation.

Pause and think

A price rises by $10\%$ and then falls by $10\%$ . Is it back to where it started?

Section 2 of 14

Core idea

Start hereThe sale trap

A shirt is 20% off and now costs 48. Many people add 20% back and say it was 57.60. It was actually 60. Percentages always refer to a starting amount, and the multiplier method keeps track of which one.

The big idea

Increase $r\%$ : × $(1+\frac r{100})$ . Decrease: × $(1-\frac r{100})$ . Reverse: ÷ multiplier. Compound: × multiplier $^n$ .

1. Multipliers

\text{new}=\text{original}\times\left(1\pm\frac r{100}\right)

$r$: percentage change.

Worked exampleChanges

Increase 80 by 15%; decrease 250 by 12%; % change from 40 to 52.

Goal: Values.
Strategy: Multipliers; change ÷ original.

$80\times1.15=92$
$100\%+15\%$ .
$250\times0.88=220$
$100\%-12\%$ .
$\frac{12}{40}\times100=30\%$
Divide by the original.

Answer

92

220

30\%

✓ Check

15\%

80

12

;

80+12=92

✓.

2. Reverse percentages

Worked exampleThe shirt

After 20% off, 48. Original?

Goal: Original price.
Strategy: original × 0.8 = 48.

$x\times0.8=48$
Multiplier.
$x=48\div0.8=60$
Undo.

Answer

60

✓ Check

20\%

60

12

;

60-12=48

✓ (

48\times1.2=57.60

is wrong).

3. Simple and compound interest

Simple vs compound interest: 1000 dollars at 10% a year

Compound interest earns interest on its interest, so it grows faster and faster.

Compound growth

$P\to P(1+r)$
After year 1.
$P(1+r)\to P(1+r)^2$
Interest on the new total.
$A=P(1+r)^n$
After $n$ years.

Worked example1000 at 10% for 3 years

Simple vs compound; and a 20 000 car depreciating 15% a year.

Goal: Values.
Strategy: Simple adds the same; compound multiplies.

$1000+3(100)=1300$
Simple.
$1000\times1.1^3=1331$
Compound.
$20000\times0.85^3=12282.50$
Depreciation.

Answer

1300

;

1331

;

12\,282.50

✓ CheckYear by year:

1100,1210,1331

✓.

Pause and think

Prices rise 10% then fall 10%. Net change?

Summary

Multiplier method.
Reverse: divide.
Compound: $P(1+r)^n$ .

Multiplier: Number to multiply by for a percentage change.
Depreciation: Loss of value.

HistoryBernoulli and compound interest. In 1683 Jacob Bernoulli asked what happens if interest is compounded more and more often. With

100\%

interest compounded infinitely often,

\$1

grows to

\$2.718\ldots

: the first appearance of the number

e

Fun factThe Rule of 72. To estimate how long money takes to double, divide

72

by the interest rate: at

6\%

, about

12

years. It works because

1.06^{12}\approx2.01

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