% Percentages and Percentage Change

GCSE Maths · Ratio, Proportion & Rates of Change (R9)

Ages 15–16 · Foundation & Higher

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1 What a Percentage Actually Is

Per cent means "per hundred". A percentage is simply a fraction whose denominator is $100$.

Definition
$p\% = \dfrac{p}{100}$
PercentageFractionDecimal
$10\%$$\dfrac{1}{10}$$0.1$
$25\%$$\dfrac{1}{4}$$0.25$
$50\%$$\dfrac{1}{2}$$0.5$
$75\%$$\dfrac{3}{4}$$0.75$
$100\%$$1$$1$
$150\%$$\dfrac{3}{2}$$1.5$
$0.5\%$$\dfrac{1}{200}$$0.005$
Percentages above $100\%$ are perfectly normal. $250\%$ of a quantity just means $2.5$ times it. This happens all the time with growth figures.
2 Finding a Percentage of an Amount
Calculator method
$p\%$ of $A = \dfrac{p}{100} \times A$
Worked Example 1 — Calculator method

Find $23\%$ of £$640$.

$23\% = 0.23$
$0.23 \times 640 = 147.2$

Answer: £$147.20$

Without a calculator, build the answer from easy percentages:

PercentageNon-calculator method
$50\%$halve
$25\%$halve, then halve again
$10\%$divide by $10$
$5\%$half of $10\%$
$1\%$divide by $100$
Worked Example 2 — Building up without a calculator

Find $37\%$ of $840$ g.

$10\% = 84$, so $30\% = 84 \times 3 = 252$
$5\% = 42$
$1\% = 8.4$, so $2\% = 16.8$
$37\% = 252 + 42 + 16.8 = 310.8$

Answer: $310.8$ g

3 One Quantity as a Percentage of Another
The rule
$A$ as a percentage of $B = \dfrac{A}{B} \times 100$
Convert to the same units first, just as with fractions and ratios.
Worked Example 3 — Test score

Nadia scored $57$ out of $75$. What percentage is this?

$\dfrac{57}{75} = 0.76$
$0.76 \times 100 = 76$

Answer: $76\%$

Worked Example 4 — Different units

Express $360$ g as a percentage of $1.5$ kg.

$1.5$ kg $= 1500$ g
$\dfrac{360}{1500} \times 100 = 24$

Answer: $24\%$

4 Percentage Increase and Decrease — Use Multipliers

The single most important technique in this whole subtopic is the percentage multiplier. Learn it and almost every percentage question becomes one calculation.

Multipliers
Increase by $p\%$:  $\times \left(1 + \dfrac{p}{100}\right)$
Decrease by $p\%$:  $\times \left(1 - \dfrac{p}{100}\right)$
ChangeMultiplierChangeMultiplier
$+5\%$$1.05$$-5\%$$0.95$
$+12\%$$1.12$$-12\%$$0.88$
$+20\%$$1.2$$-20\%$$0.8$
$+7.5\%$$1.075$$-7.5\%$$0.925$
$+100\%$$2$$-100\%$$0$
£240 × 1.15 (up 15%) × 0.85 (down 15%) £276 £204
Worked Example 5 — Increase

A season ticket costing £$680$ rises by $8\%$. Find the new price.

Multiplier $= 1 + 0.08 = 1.08$
$680 \times 1.08 = 734.4$

Answer: £$734.40$

Worked Example 6 — Decrease

A laptop costing £$540$ is reduced by $35\%$. Find the sale price.

Multiplier $= 1 - 0.35 = 0.65$
$540 \times 0.65 = 351$

Answer: £$351$

5 Calculating a Percentage Change
Percentage change
$\text{percentage change} = \dfrac{\text{change}}{\text{original}} \times 100$
Always divide by the ORIGINAL value — the value you started with — never by the new one.
Worked Example 7 — Percentage profit

A trader buys a painting for £$450$ and sells it for £$603$. Find the percentage profit.

Profit $= 603 - 450 = £153$
$\dfrac{153}{450} = 0.34$
$0.34 \times 100 = 34$

Answer: $34\%$ profit

Worked Example 8 — Percentage loss

A car bought for £$18\,500$ is sold three years later for £$11\,470$. Find the percentage loss.

Loss $= 18\,500 - 11\,470 = £7030$
$\dfrac{7030}{18\,500} = 0.38$

Answer: a $38\%$ loss

6 Reverse Percentages — Finding the Original

These are the questions students lose most marks on. The giveaway wording is "before the increase", "original price", "pre-VAT" or "what was it worth last year".

The rule
$\text{original} = \dfrac{\text{new value}}{\text{multiplier}}$
Never take the percentage off the new value. Taking $20\%$ off a price that has had $20\%$ added does not get you back to the start.
Worked Example 9 — Reverse increase

A television costs £$564$ including $20\%$ VAT. Find the price before VAT.

The £$564$ represents $120\%$ of the pre-VAT price.
Multiplier $= 1.2$
Original $= 564 \div 1.2 = £470$

Check: $470 \times 1.2 = 564$ ✓

The wrong method would give $564 \times 0.8 = £451.20$ — over £$18$ out.

Worked Example 10 — Reverse decrease

In a sale, all prices are cut by $30\%$. A coat now costs £$87.50$. What was its original price?

Multiplier $= 1 - 0.30 = 0.7$
Original $= 87.50 \div 0.7 = £125$

Check: $125 \times 0.7 = 87.50$ ✓

How to tell the difference: if the question gives you a starting value and asks for the end value, multiply. If it gives you the end value and asks for the start, divide.
7 Simple Interest

With simple interest the interest is calculated on the original amount every year, so you earn the same amount each year.

Simple interest
$I = \dfrac{P \times r \times n}{100}$
$P$ = principal, $r$ = rate per year (%), $n$ = number of years
Worked Example 11 — Simple interest

£$3200$ is invested at $4.5\%$ simple interest per year for $6$ years. Find the interest earned and the final total.

Interest in one year $= 3200 \times 0.045 = £144$
Over $6$ years: $144 \times 6 = £864$
Total $= 3200 + 864 = £4064$
Simple vs compound. Simple interest adds the same amount each year. Compound interest pays interest on the interest, so it grows faster — see the Growth and Decay page.
8 Comparing Quantities Using Percentages

Percentages let you compare fairly when the totals are different.

Worked Example 12 — Fair comparison

In Maths, Tom scored $34$ out of $40$. In Physics he scored $51$ out of $60$. In which subject did he do better?

Maths: $\dfrac{34}{40} \times 100 = 85\%$
Physics: $\dfrac{51}{60} \times 100 = 85\%$

Equally well — both are $85\%$, even though the raw marks differ.

Percentage points vs percentages. If support rises from $40\%$ to $50\%$, that is a rise of $10$ percentage points, but a percentage increase of $\dfrac{10}{40} \times 100 = 25\%$. Examiners love this distinction.
9 Quick Reference

Definition

$p\% = \dfrac{p}{100}$ — a fraction out of one hundred.

Of an amount

$p\%$ of $A = \dfrac{p}{100} \times A$.

As a percentage

$\dfrac{A}{B} \times 100$, with matching units.

Increase / decrease

$\times (1 + \tfrac{p}{100})$ or $\times(1 - \tfrac{p}{100})$.

Percentage change

$\dfrac{\text{change}}{\text{original}} \times 100$. Divide by the original.

Reverse

Original $=$ new $\div$ multiplier.

Simple interest

$I = \dfrac{Prn}{100}$ — the same amount every year.

Combining

Multiply multipliers; $1.1 \times 0.9 = 0.99$, a $1\%$ fall.

10 Practice Questions
Question 1

Find (a) $18\%$ of £$250$   (b) $6.5\%$ of $840$ kg.

▶ Show solution

(a) $0.18 \times 250 = £45$

(b) $0.065 \times 840 = 54.6$ kg

Question 2

Express $27$ minutes as a percentage of $2$ hours.

▶ Show solution

$2$ hours $= 120$ minutes.

$\dfrac{27}{120} \times 100 = 22.5$

Answer: $22.5\%$

Question 3

A phone contract costs £$28$ per month. The price rises by $7.5\%$. Find the new monthly cost.

▶ Show solution

Multiplier $= 1.075$

$28 \times 1.075 = 30.1$

Answer: £$30.10$ per month

Question 4

A bicycle priced at £$425$ is reduced by $16\%$ in a sale. Find the sale price.

▶ Show solution

Multiplier $= 1 - 0.16 = 0.84$

$425 \times 0.84 = £357$

Question 5

The population of a village grew from $1250$ to $1425$. Find the percentage increase.

▶ Show solution

Change $= 1425 - 1250 = 175$

$\dfrac{175}{1250} \times 100 = 14$

Answer: a $14\%$ increase

Question 6

A meal costs £$46.20$ including a $10\%$ service charge. Find the cost of the meal before the service charge was added.

▶ Show solution

This is a reverse percentage. Multiplier $= 1.1$.

$46.20 \div 1.1 = £42$

Check: $42 \times 1.1 = 46.20$ ✓

Question 7

After a $12\%$ pay rise, Aisha earns £$29\,120$ per year. What did she earn before the rise?

▶ Show solution

Multiplier $= 1.12$.

$29\,120 \div 1.12 = £26\,000$

Question 8

£$4500$ is invested for $5$ years at $3.2\%$ simple interest per year. Find the total value at the end.

▶ Show solution

Interest per year $= 4500 \times 0.032 = £144$

Over $5$ years $= 144 \times 5 = £720$

Total $= 4500 + 720 = £5220$

Question 9

A shop increases a price by $25\%$, then later reduces the new price by $25\%$. Show that the final price is not the same as the original, and find the overall percentage change.

▶ Show solution

Overall multiplier $= 1.25 \times 0.75 = 0.9375$

Since $0.9375 \neq 1$, the price has changed.

$0.9375 = 93.75\%$, so the overall change is a decrease of $100 - 93.75 = \mathbf{6.25\%}$.

Why? The $25\%$ rise is calculated on the small original, but the $25\%$ fall is calculated on the larger increased price, so more is taken off than was put on.

Question 10

A jeweller buys a ring for £$820$. She adds $65\%$ to get the shop price, then in a sale reduces the shop price by $20\%$.

(a) Find the sale price.   (b) Find her percentage profit on the original £$820$.   (c) What single multiplier takes £$820$ to the sale price?

▶ Show solution

(a) Shop price $= 820 \times 1.65 = £1353$

Sale price $= 1353 \times 0.8 = £1082.40$

(b) Profit $= 1082.40 - 820 = £262.40$

$\dfrac{262.40}{820} \times 100 = 32$, so a $\mathbf{32\%}$ profit.

(c) $1.65 \times 0.8 = \mathbf{1.32}$ — which confirms the $32\%$ increase ✓

Percentages & Percentage Change (R9) · GCSE Maths Revision · Created with MathJax