Percentage Change
Ever wondered how shops calculate sale discounts, or how banks work out the interest on your savings? All of these involve percentage change — measuring how much a quantity has increased or decreased as a proportion of the original amount. It’s one of the most widely used maths skills in everyday life.
The Core Concept
Calculating Percentage Change
To express any change as a percentage of the original value:
If the new value is greater than the original → percentage increase.
If the new value is less than the original → percentage decrease.
Multipliers: The Faster Way
A multiplier lets you find the new value in a single step — and is essential for reverse percentages.
- Increase by $p\%$: multiplier $= 1 + \frac{p}{100}$
e.g. 15% increase → $1 + 0.15 = \mathbf{1.15}$ - Decrease by $p\%$: multiplier $= 1 – \frac{p}{100}$
e.g. 20% decrease → $1 – 0.20 = \mathbf{0.80}$
Then simply: New Value = Original × Multiplier.
Visualising Multipliers
Two More Key Techniques
Reverse Percentages (Finding the Original)
Sometimes you’re given the new value after a change and need to find the original value. Rearrange the multiplier formula:
Example: Trainers cost £63 after a 30% reduction.
Multiplier $= 1 – 0.30 = 0.70$.
Original $= £63 \div 0.70 = \mathbf{£90}$.
⚠️ Never apply the percentage to the new (sale) value — it is already 70% of the original.
Simple Interest
Simple interest is calculated only on the original principal — interest doesn’t compound.
Where P = principal, R = annual rate (%), T = time (years).
Total Amount = $P + I$.
Note: $T$ must be in years. Convert months to years before substituting (e.g. 6 months = 0.5 years).
Worked Examples
Example 1: Percentage Increase
A football shirt costs £80. Its value increases by 15%. What is its new value?
- Multiplier for +15%: $1 + 0.15 = 1.15$.
- New value $= £80 \times 1.15 = \mathbf{£92}$.
Answer: £92
Example 2: Percentage Decrease
A laptop costs £450 and has a 20% discount. What is the sale price?
- Multiplier for −20%: $1 – 0.20 = 0.80$.
- Sale price $= £450 \times 0.80 = \mathbf{£360}$.
Answer: £360
Example 3: Reverse Percentage
Trainers are on sale for £63 — a 30% reduction. What was the original price?
- Multiplier for −30%: $0.70$.
- Original $= £63 \div 0.70 = \mathbf{£90}$.
Answer: £90
Example 4: Simple Interest
Jacob invests £2,000 at 4% simple interest for 3 years. Find the interest and total amount.
- $I = \dfrac{2000 \times 4 \times 3}{100} = \dfrac{24000}{100} = \mathbf{£240}$.
- Total $= £2{,}000 + £240 = \mathbf{£2{,}240}$.
Interest: £240 | Total: £2,240
Tutor Insights
🤔 Common Misunderstandings
- Confusing increase and decrease multipliers. Increase means more than 100% (multiplier > 1); decrease means less than 100% (multiplier < 1). Never use 0.9 for a 10% increase or 1.1 for a 10% decrease.
- Applying the percentage to the wrong value in reverse percentage problems. The new/sale value is already the reduced amount — always divide by the multiplier.
- Time period for simple interest. The rate (R) is annual, so time (T) must be in years.
📝 Common Exam Mistakes
- Lack of working out. Examiners want to see your method, not just the final answer.
- Rounding mid-calculation. Keep full precision until the very last step — for money, round to 2 d.p. at the end.
- Not reading the question. Is it asking for the new value, the original value, just the interest, or the total amount?
- Calculator input errors. Convert percentages to decimals yourself rather than relying on the % key.
Practice Questions
- A bike costs £240. Its price increases by 10%. What is the new price?
- A smart TV is reduced by 25%. The original price was £600. What is the sale price?
- After a 5% pay rise, Sarah’s new weekly wage is £315. What was her original weekly wage?
- A car depreciates by 18% in its first year. Its value after one year is £8,200. What was its original value?
- Liam invests £5,000 for 4 years at 3% simple interest per year.
(a) How much interest will he earn?
(b) What will the total amount be? - A school’s student numbers rose from 800 to 920. Calculate the percentage increase.
- A shopkeeper buys a coat for £70 and sells it for £105. What is the percentage profit?
- A phone contract costs £35/month. Prices increase by 2%. What is the new monthly cost?
- An antique vase sold for £156, which was a 30% loss for the seller. How much did the seller originally pay?
Show Answers
- $£240 \times 1.10 = \mathbf{£264}$.
- $£600 \times 0.75 = \mathbf{£450}$.
- $£315 \div 1.05 = \mathbf{£300}$.
- $£8{,}200 \div 0.82 = \mathbf{£10{,}000}$.
- (a) $I = \frac{5000 \times 3 \times 4}{100} = \mathbf{£600}$. (b) $£5{,}000 + £600 = \mathbf{£5{,}600}$.
- Change $= 120$. $\frac{120}{800} \times 100 = \mathbf{15\%}$.
- Profit $= £35$. $\frac{35}{70} \times 100 = \mathbf{50\%}$.
- $£35 \times 1.02 = \mathbf{£35.70}$.
- $£156 \div 0.70 = \mathbf{£222.86}$ (2 d.p.).
FAQs
Q: What’s the difference between finding a percentage of an amount and percentage change?
A: Finding a percentage of an amount gives you a part of a whole (e.g. 25% of £100 = £25). Percentage change measures how much a value has changed relative to its original, expressed as a percentage (e.g. £100 rising to £120 is a 20% increase).
Q: When should I use multipliers?
A: Always use multipliers for percentage increase or decrease questions where you need to find the new value or the original value. They make calculations faster and less error-prone than multi-step methods.
Q: Are reverse percentages really that hard?
A: Not once you’ve got the multiplier concept. The instinct is to apply the percentage to the given (sale) value — but that value is already the reduced amount. Always divide by the multiplier, and check your answer by working forwards again.
Q: Is simple interest always calculated yearly?
A: The rate R is always an annual rate. The time T can be any period, but it must be converted to years. So 6 months = 0.5 years, and 18 months = 1.5 years.
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