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WORKED EXAMPLEV = P(1 − r)ⁿ
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worked exampleMaths6 min read

Reducing balance depreciation, step by step

A $30,000 car loses 15% of its value each year. Work out what it is worth after six years.

reducing-balance-depreciation.md
In short

Multiply the starting value by 0.85 for each year. After six years that is 30,000 × 0.85⁶, about $11,314.

The idea

Reducing balance means the loss each year is a percentage of what is left, not of the original price. So the loss shrinks as the value shrinks.

Losing 15% means keeping 85%. The multiplier each year is 0.85.

The formula

V = P × (1 − r)ⁿ

Here P = 30,000, r = 0.15 and n = 6.

Year by year

YearValue at startLoss (15%)Value at end
1$30,000.00$4,500.00$25,500.00
2$25,500.00$3,825.00$21,675.00
3$21,675.00$3,251.25$18,423.75
4$18,423.75$2,763.56$15,660.19
5$15,660.19$2,349.03$13,311.16
6$13,311.16$1,996.67$11,314.49

Check with the formula: 0.85⁶ ≈ 0.3771, and 30,000 × 0.3771 ≈ $11,314.

The common mistake

Try it: a $12,000 laptop loses 20% each year. What is it worth after 3 years? Answer: 12,000 × 0.8³ = $6,144.

AI explanations can contain mistakes. Check important answers with your teacher.

AI explanations can contain mistakes. Check important answers with your teacher.

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