Skip to main content
Advanced number and financial mathematics
Lesson 2 of 75
Easy practice

Reverse compound change

Objective. Recover original values from repeated growth or depreciation.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Reverse compound change finds an original value before a known percentage growth or decay was applied, by dividing the final amount by the overall multiplier rather than taking a percentage of the final amount. If a quantity decayed by r% to give amount A, the original value is A ÷ (1−r/100); for growth it is A ÷ (1+r/100). This works because the original value, multiplied by the multiplier, equals A, so dividing undoes the multiplication. For example, if a price after a 10% reduction is £63, the original price is 63 ÷ 0.9 = £70, not 63 × 1.1.

Worked example

After a 15% pay rise, Sam's salary is £27600. Find his salary before the rise.

£27600.00
Money illustration for £27600
  1. 1Growth multiplier = 1 + 0.15 = 1.15
  2. 2Original salary × 1.15 = 27600
  3. 3Original salary = 27600 ÷ 1.15
  4. 4Compute 27600 ÷ 1.15 = 24000

Answer: £24000

Concept mastery

Based on your past attempts at this lesson — the question types to practise again come first.

Complete a practice set to see your concept breakdown here.

Practice

Practice happens on its own screen, one question at a time. Answers stay hidden until you submit yours, and you can stop and pick up at the same question later.

Start practising

Common mistake. A common error is to add or subtract the percentage directly from the final value, calculating £63 + 10% = £69.30 instead of dividing by the decay multiplier 0.9 to get £70; percentage change must be undone by division, not addition.

Your progress

Tick each step as you finish it. Your exact reading position is saved automatically, so Continue drops you back on the same line.

0 of 5 steps complete

Your place saves automatically as you read.

My dashboard