R9, R16Ratio, Proportion & Rates of ChangeFoundation & HigherGrades 4–7

Percentage Change & Growth

Q: Can I find the original amount by reversing each year separately?

Yes, but dividing once by multiplier^n is quicker and less error-prone.

This lesson takes multipliers further. You will measure a change as a percentage, work backwards from a new value to the original, and apply the same percentage change repeatedly — the maths behind savings accounts, car depreciation, population growth and inflation.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

Every question here comes back to one equation: $\text{original} \times \text{multiplier} = \text{new}$ . Finding the new value multiplies; finding the original (a reverse percentage) divides; finding the percentage compares new and original.

Repeated change multiplies repeatedly, so after $n$ years the value is $\text{original} \times \text{multiplier}^n$ . That is why compound interest grows faster than simple interest: each year's interest earns interest too.

Compound Interest and DepreciationCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Calculate a percentage increase or decrease
Find the original amount before a percentage change (reverse percentages)
Calculate compound interest and depreciation
Solve growth and decay problems, including finding the number of years

Percentage change

Percentage change $= \dfrac{\text{change}}{\text{original}} \times 100$ . The change is always compared with the $\textbf{original}$ value, not the new one.

Profit and loss are percentage changes too: percentage profit $= \dfrac{\text{profit}}{\text{cost price}} \times 100$ .

Example 1A house bought for £240,000 sells for £282,000. Find the percentage increase.

1Change

= 282{,}000 - 240{,}000 = 42{,}000

\frac{42{,}000}{240{,}000} \times 100 = 17.5

Answer

17.5\%

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

A phone falls in price from £450 to £369. What is the percentage decrease?

Your answer

Show worked solution

Solution

1Change $= 450 - 369 = 81$ .
2 $\frac{81}{450} \times 100 = 18\%$ .

Answer: $18\%$

Reverse percentages

When you know the value $\textbf{after}$ a percentage change, divide by the multiplier to find the original. Never find the percentage of the new value.

Example 1In a 20% off sale, a jacket costs £68. What was the original price?

1The sale price is 80% of the original: multiplier

0.8

2Original

= 68 \div 0.8 = 85

Answer

£85

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

After a 15% pay rise, Tom earns £29,900. What did he earn before the rise, in pounds?

Your answer

Show hint

His new pay is 115% of the old pay.

Show worked solution

Solution

1Multiplier $1.15$ .
2 $29{,}900 \div 1.15 = 26{,}000$ .

Answer: £26,000

Example 2A price including 20% VAT is £54. How much VAT is paid?

1Price before VAT:

54 \div 1.2 = 45

2VAT:

54 - 45 = 9

Answer

£9

Why you divide

Taking 20% off £68 and adding it back gives £81.60, not £85. The 20% was 20% of the original, which is bigger than the sale price — so 20% of the sale price is too small. Dividing by the multiplier avoids this trap completely.

Compound interest and depreciation

Compound growth: $\text{value} = \text{original} \times \left(1 + \frac{r}{100}\right)^n$ . Depreciation (compound decay): $\text{value} = \text{original} \times \left(1 - \frac{r}{100}\right)^n$ .

Simple interest adds the same amount each year; compound interest adds a percentage of the growing total, so the gap widens every year.

Simple interest grows by the same amount each year (a straight line); compound interest grows by the same percentage (a curve that steepens).

Example 1£3,000 is invested at 4% compound interest per year. Find its value after 5 years.

1Multiplier

1.04

3000 \times 1.04^5 = 3649.958\ldots

Answer

£3,649.96

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

A car worth £18,000 depreciates by 12% per year. Find its value after 3 years, to the nearest pound.

Your answer

Show worked solution

Solution

1Multiplier $0.88$ .
2 $18{,}000 \times 0.88^3 = 12{,}266.496$ , so £12,266.

Answer: £12,266

Example 2£5,000 is invested at 3% per year compound interest. How much interest is earned in 4 years?

1Value:

5000 \times 1.03^4 = 5627.54

(to the nearest penny).

2Interest:

5627.54 - 5000

Answer

£627.54

Tip — Use the power button — never add the same percentage of the original each year, which is simple interest.

Growth and decay problems

The same formula models populations, bacteria, radioactive decay and inflation. To find how many years it takes to pass a target, try successive powers (trial and improvement) or repeatedly multiply with the ANS key, and state the first whole number of years that works.

Example 1A town of 12,000 people grows by 6% per year. After how many years will it first exceed 15,000?

12{,}000 \times 1.06^3 = 14{,}292.2

— not yet.

12{,}000 \times 1.06^4 = 15{,}149.7

— exceeds 15,000.

Answer

4 years

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

A radioactive sample of 800 g loses 25% of its mass each hour. After how many whole hours is the mass first less than 200 g?

Your answer

Show worked solution

Solution

1Multiplier $0.75$ .
2 $800 \times 0.75^4 = 253.1$ g — not yet below 200.
3 $800 \times 0.75^5 = 189.8$ g — below 200. So 5 hours.

Answer: $5$ hours

Think like an examiner

In reverse percentage questions, working like "20% of £68 = £13.60" signals the wrong method and scores zero.
Write the multiplier and the power, e.g. $3000 \times 1.04^5$ — this is the method mark.
Read whether the question wants the final value or the interest earned.

Growth and decay

\%\text{ change} = \dfrac{\text{change}}{\text{original}} \times 100

Compare with the original.

\text{original} = \dfrac{\text{new}}{\text{multiplier}}

Reverse percentages.

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

Compound growth (+) and decay (−).

Watch out for these

Percentage change as

\frac{\text{change}}{\text{new}} \times 100

Divide by the original value.

Reverse: adding 20% back on to a sale price.

Divide by the multiplier

0.8

Compound interest as

3000 + 5 \times 120

That is simple interest. Use

3000 \times 1.04^5

Depreciation of 12% using the multiplier

1.12

0.12

Decay multiplier:

0.88

Stretch yourself

An investment of £P grows at 5% per year compound interest for 2 years and then 2% per year for 3 more years. It is then worth £9,000. Find $P$ to the nearest pound.

Hint — Apply both multipliers in sequence, then use a reverse calculation.

Reveal solution

Solution

P \times 1.05^2 \times 1.02^3 = 9000

1.05^2 \times 1.02^3 = 1.1025 \times 1.061208 = 1.169982\ldots

P = 9000 \div 1.169982\ldots = 7692.43\ldots

, i.e. £7,692.

\blacksquare

Questions students ask

What is the difference between depreciation and decay?

None mathematically. Depreciation is the word used for the value of objects such as cars; decay is used in science.

Can I find the original amount by reversing each year separately?

Yes, but dividing once by

\text{multiplier}^n

is quicker and less error-prone.

Should I round each year?

No. Keep the full calculator value until the end, then round the final answer — usually to the nearest penny for money.

Key takeaways

Percentage change = change ÷ original × 100.
Reverse percentage: divide the new value by the multiplier.
Compound change: original × multiplier $^n$ .
Growth uses $1 + \frac{r}{100}$ ; decay uses $1 - \frac{r}{100}$ .
For "how many years", test successive powers and give the first whole number that works.

Exam-style questions

Practise it the way it is examined

3 original questions · 10 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

In a sale, all prices are reduced by 35%. A TV costs £416 in the sale. Work out the price of the TV before the sale.[3]
Show mark scheme
- M1Recognises $416$ is $65\%$ of the original, or uses $0.65$
- M1 $416 \div 0.65$
- A1£640
Answer: £640

Question 2Grade 54 marks

Jo invests £4,500 at 2.8% compound interest per year. Raj invests £4,500 at 3% simple interest per year.

After 6 years, who has more money, and by how much?[4]
Show mark scheme
- M1Jo: $4500 \times 1.028^6$
- A1Jo £5,310.94 (5310.937…)
- M1Raj: $4500 + 6 \times 135 = £5,310$
- A1Jo has more, by 94p (accept £0.94)

Examiner note

Close answers like this are designed to test accuracy — do not round until the final comparison.

Question 3Grade 73 marks

The number of bacteria in a dish increases by 18% every hour. Initially there are 2,500 bacteria. After how many whole hours will there first be more than 10,000?[3]
Show mark scheme
- M1Uses $2500 \times 1.18^n$ for at least two values of $n$
- M1 $n = 8$ : 9,397… and $n = 9$ : 11,088…
- A19 hours
Answer: 9 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.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

A price rises from £40 to £50. Percentage increase?
- $25\%$
- $20\%$
- $10\%$
- $125\%$
Show answer and explanation
Answer: $25\%$
$\frac{10}{40} \times 100$ .
Multiplier for 3 years of 5% compound growth?
- $1.05^3$
- $1.15$
- $3 \times 1.05$
- $0.05^3$
Show answer and explanation
Answer: $1.05^3$
Apply 1.05 three times.
After a 10% increase, a price is £77. Original price?
- £70
- £69.30
- £84.70
- £67
Show answer and explanation
Answer: £70
$77 \div 1.1$ .
£1,000 at 10% compound interest for 2 years is worth…
- £1,210
- £1,200
- £1,100
- £1,020
Show answer and explanation
Answer: £1,210
$1000 \times 1.1^2$ .
A value falls from 80 to 68. Percentage decrease?
- $15\%$
- $12\%$
- $17.6\%$
- $85\%$
Show answer and explanation
Answer: $15\%$
$\frac{12}{80} \times 100$ .
Multiplier for a 7% annual depreciation?
- $0.93$
- $1.07$
- $0.07$
- $0.7$
Show answer and explanation
Answer: $0.93$
$1 - 0.07$ .
A price includes 20% VAT and is £90. Price without VAT?
- £75
- £72
- £70
- £108
Show answer and explanation
Answer: £75
$90 \div 1.2$ .
A value halves every year. What percentage remains after 3 years?
- $12.5\%$
- $50\%$
- $25\%$
- $0\%$
Show answer and explanation
Answer: $12.5\%$
$0.5^3 = 0.125$ .

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