How to Find the Original Value Using Inverse Percentages
Inverse percentages help you find the original value after a percentage increase or decrease. You work backwards using a percentage multiplier, which makes this topic essential for GCSE and IGCSE questions on discounts, markups, depreciation and growth. This lesson shows two reliable methods, the multiplier method and the back to 100 percent method, with clear worked examples and practice.
What Are Inverse Percentages?
A percentage change multiplies the original value by a multiplier \(m\). If you are given the final value and need to recover the original, you reverse the process by dividing:
\[ x = \frac{\text{Final}}{m} \]
This works because multiplying and dividing by the same number are inverse operations. Dividing by \(m\) exactly undoes the percentage change.
Core Ideas
\(m = 1 - \tfrac{R}{100}\) (decrease)
Finding the Original Value After an Increase
💡 Example A1: Multiplier Method
A pair of speakers sells for £230 after a 15% markup. Find the original cost.
\[ m = 1.15 \]
\[ x \times 1.15 = 230 \]
\[ x = \frac{230}{1.15} \]
\[ x = 200 \]
What's happening?
A 15% increase means the final price is 115% of the original, so the multiplier is \(1.15\).
Dividing by \(1.15\) reverses the increase exactly.
Original cost: £200
💡 Example A2: Back to 100%
A pair of speakers sells for £230 after a 15% markup. Find the original cost.
\[ 115\% = 230 \]
\[ 1\% = \frac{230}{115} = 2 \]
\[ 100\% = 2 \times 100 = 200 \]
What's happening?
£230 is 115% of the original, not 100%. Dividing by 115 finds what 1% is worth. Multiplying by 100 then gives the full original.
Original cost: £200
Finding the Original Value After a Decrease
💡 Example B1: Multiplier Method
A bicycle sells for £252 after a 16% reduction. Find the original price.
\[ m = 0.84 \]
\[ y \times 0.84 = 252 \]
\[ y = \frac{252}{0.84} \]
\[ y = 300 \]
What's happening?
A 16% decrease means the final price is 84% of the original, so the multiplier is \(0.84\).
Dividing by \(0.84\) reverses the decrease exactly.
Original price: £300
💡 Example B2: Back to 100%
A bicycle sells for £252 after a 16% reduction. Find the original price.
\[ 84\% = 252 \]
\[ 1\% = \frac{252}{84} = 3 \]
\[ 100\% = 3 \times 100 = 300 \]
What's happening?
£252 is only 84% of the original: 16% is missing. Find 1%, then scale back up to 100%.
Original price: £300
Multiple Successive Changes
💡 Example C: Reversing Two Changes
A bicycle was increased in price by 10%, then reduced by 20%. The final price is £704. Find the original price.
\[ m_1 = 1.10 \qquad m_2 = 0.80 \]
\[ \text{Undo the 20\% decrease first:} \]
\[ \frac{704}{0.80} = 880 \]
\[ \text{Then undo the 10\% increase:} \]
\[ \frac{880}{1.10} = 800 \]
What's happening?
Two changes were applied in order: first a 10% increase, then a 20% decrease.
To reverse them, work backwards: undo the last change first (divide by \(0.80\)), then undo the first change (divide by \(1.10\)).
Think of it like putting on a coat then a hat: to undress, the hat comes off first.
Original price: £800
🔑 Key Ideas
- Build the correct multiplier first: \(m = 1 + \tfrac{R}{100}\) for an increase, \(m = 1 - \tfrac{R}{100}\) for a decrease.
- Dividing by \(m\) undoes the change because \(x \times m \div m = x\).
- "Back to 100%" and the multiplier method always give the same answer.
- For multiple changes, divide by each multiplier in reverse order.
⚠️ Common Pitfalls
- Dividing by the rate \(R\) instead of the multiplier \(m\) (e.g. dividing by 15 instead of 1.15).
- Using the increase formula when the change was a decrease (and vice versa).
- Undoing multiple changes in the wrong order.
- Rounding too early: keep full precision until the final answer.