A normal percentage question gives you the original amount and asks for the amount after a change. A reverse percentage question does the opposite: it gives you the amount after a change (e.g. a sale price) and asks for the original (100%) amount before the change. Since the forward calculation is a multiplication, undoing it requires division.

For a change of $r\%$:

• Increase: multiplier $p = 1 + \dfrac{r}{100}$ — e.g. a $5\%$ pay rise gives $p=1.05$.
• Decrease: multiplier $p = 1 - \dfrac{r}{100}$ — e.g. a $15\%$ discount gives $p=0.85$ (the price left is $85\%$ of the original).

The governing relationship is always:

so to reverse it:

Example: A DVD player now costs $£242$ after a $15\%$ reduction. The multiplier is $p=0.85$, so $\text{Original} = 242 \div 0.85 = £284.71$.

The logic is identical — only the direction of $p$ changes. If Jennie's new salary is $£31{,}500$ after a $5\%$ rise: $p=1.05$, so her old salary is $31{,}500 \div 1.05 = £30{,}000$. If a trader's final sale price of $£465$ includes a $15\%$ profit margin on the cost price: $p = 1.15$, so cost price $= 465 \div 1.15 = £404.35$ (to 2 d.p.).

The method is unchanged — division still applies. If $150\%$ of a number is $690$: $p = 1.50$, so the number is $690 \div 1.50 = 460$.

Worked Example

A shop sells trainers at $£68$ after reducing the original price by $20\%$. Find the original price.

1. A $20\%$ reduction means the sale price is $100\%-20\%=80\%$ of the original: $p=0.80$.
2. $\text{Original}\times0.80 = 68$.
3. $\text{Original} = 68\div0.80 = £85$.
4. Check: $85$ is bigger than $68$, which makes sense for a price before a reduction. ✓

Single reverse percentages are the foundation for reverse compound interest/depreciation, where a change is applied repeatedly over $n$ periods: $\text{Final} = \text{Original}\times p^n$. Reversing it uses exactly the same division logic, just with $p$ raised to a power: $\text{Original} = \text{Final}\div p^n$.

Example: An investment is worth $£432$ after 3 years of $20\%$ annual compound interest. $p=1.20$, $n=3$, so $\text{Original}=432\div(1.20)^3$.