Depreciation
Objective. Model repeated percentage decrease over time.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Depreciation models repeated percentage decrease over time, such as a car losing value each year, using a multiplier less than 1 raised to a power: final value = initial value × (1 − r/100)ⁿ. For example, a car worth £12,000 depreciating at 15% per year for 2 years is worth 12000 × 0.85² = 12000 × 0.7225 = £8,670. Each year's decrease is calculated on the value remaining after the previous year's decrease, not on the original value, which is why the multiplier approach correctly compounds the percentage decreases together. This contrasts with a straight-line decrease, which would subtract the same fixed amount every year.
Worked example
A laptop is bought for £900 and depreciates at a rate of 20% per year. Find its value after 3 years.
- 1Find the multiplier for a 20% decrease: 1 − 20/100 = 0.8.
- 2Raise the multiplier to the power of the number of years: 0.8³.
- 30.8³ = 0.512.
- 4Multiply by the original value: 900 × 0.512 = 460.8.
Answer: £460.80
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 practisingCommon mistake. Pupils sometimes calculate the percentage decrease from the original value every year instead of from the previous year's reduced value, effectively modelling simple rather than compound decrease. Always raise the single-year multiplier to the power of the number of years.
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.