Here, \(r\) is the percentage change. If it’s an increase, \(r>0\); if it’s a decrease, \(r<0\).
Why it’s true
A percentage increase means: Final = Original × (1 + r/100).
Rearrange to find Original = Final ÷ (1 + r/100).
Similarly, for a percentage decrease: Final = Original × (1 - r/100).
Rearranging works the same way.
Recipe (how to use it)
Identify the final amount and the percentage change \(r\).
Add/subtract the percentage in multiplier form: \(1 + r/100\).
Divide the final by this multiplier.
Simplify to find the original.
Spotting it
Look for wording like “after a 20% increase, the price is …” or “after a 15% reduction, the value is …”. These are reverse percentage problems.
Common pairings
Price increases (shops, VAT, discounts).
Population growth/decline questions.
Financial exam problems with “before” and “after” values.
Mini examples
Final price £120 after 20% increase → Original = 120 ÷ 1.2 = £100.
Final £85 after 15% decrease → Original = 85 ÷ 0.85 = £100.
Pitfalls
Adding/subtracting percentages instead of using multipliers.
Forgetting that a decrease means dividing by less than 1.
Mixing up original and final values.
Exam strategy
Always write the multiplier first: (1 + r/100).
Check whether it’s an increase or decrease.
Double-check by recalculating the percentage change forwards.
Summary
Reverse percentages let us work backwards to the original value. Divide the final value by the multiplier (1 ± r/100) depending on whether it was an increase or decrease.
Worked examples
Show / hide (10) — toggle with E
After a 20% increase, the price is £120. What was the original price?
\( Multiplier = 1.2 \)
\( Original = 120/1.2=100 \)
Answer:
100
After a 10% decrease, the value is £180. What was the original value?
\( Multiplier = 0.9 \)
\( Original = 180/0.9=200 \)
Answer:
200
After a 25% increase, the population is 500. What was it before?
\( Multiplier=1.25 \)
\( Original=500/1.25=400 \)
Answer:
400
After a 30% decrease, the number of students is 70. What was it before?
\( Multiplier=0.7 \)
\( Original=70/0.7=100 \)
Answer:
100
After a 12% increase, a shirt costs £56. Find the original price.
\( Multiplier=1.12 \)
\( Original=56/1.12=50 \)
Answer:
50
After a 40% increase, the value is 350. What was it before?
\( Multiplier=1.4 \)
\( Original=350/1.4=250 \)
Answer:
250
After a 15% decrease, the price is £85. What was it originally?
\( Multiplier=0.85 \)
\( Original=85/0.85=100 \)
Answer:
100
After a 60% increase, the cost is £800. What was it before?
\( Multiplier=1.6 \)
\( Original=800/1.6=500 \)
Answer:
500
After a 22% increase, the population is 4884. What was it before?
\( Multiplier=1.22 \)
\( Original=4884/1.22=4003.28≈4003 \)
Answer:
4003
After a 5% decrease, the salary is £38,000. What was the original?