Reverse percentage problems

Recover the original value after a percentage change — classic exam reverse-percentage items.

Key formulas

Original = New \div (1 \pm r/100)

\text{If new is after} 12\% rise: \text{divide by} 1.12

If a price rises 10% to $220, the original is not 220 − 10% of 220. The 10% was taken of the unknown original, so divide 220 by 1.10 to get 200.

After an increase of r%, divide by (1 + r/100). After a decrease of r%, divide by (1 − r/100). State the multiplier before dividing.

If two changes happened, divide by each multiplier in reverse order (or by their product). Keep exact fractions until the end when possible.

Write “original × multiplier = new” then rearrange. Never take the percentage of the new value unless the question says so.

Problem. After a 15% increase a fee is $460. Find the original fee.

Solution. Original × 1.15 = 460 ⇒ original = 460 ÷ 1.15 = 400.

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After 15% rise a fee is $460. Original?

After a 15% increase a fee is $460. Find the original fee.
A phone sells for $3600 after a 20% discount. Find the marked price.
A value falls 8% to 184. Find the original value.
After a 10% rise and then a 5% rise, a salary is $23100. Find the salary before both rises.

No — percentage change finds the rate; reverse percentage finds the missing original.

Yes — dividing by a factor less than 1 increases the number.