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Compound interestAppreciation and depreciation

In compound interest the amount in interest is added to the original at the end of each year. So the next year the interest is worked out on a larger amount of money.

Part of Application of MathsFinancial skills

Appreciation and depreciation

Appreciation

Appreciation is the increase in value of an item. A common example of this is house prices. Each year, the value of a house increases, so we say that its value appreciates.

The rate of appreciation is often given as a percentage.

Example

A flat is bought for \(\pounds90,000\).

In each of the three years that follows, its value appreciates by \(8\%\).

How much is the flat worth after three years?

Year 1 appreciation: \(8\% of \pounds90,000 = \pounds7200\)

Flat Value = \(90,000 + 7,200 = \pounds97,200\)

Year 2 appreciation: \(8\%\,of\,\pounds97,200 = \pounds7776\) \(8\%\,of\,\pounds97,200 = \pounds7776\)

Flat Value = \(97,200 + 7,776 = \pounds104,976\)

Year 3 appreciation: \(8\%\,of\,\pounds104,976 = \pounds8398.08\)

Flat Value = \(104,976 + 8,398.08 = \pounds113,374.08\)

So after three years, the flat is worth \(\pounds113,374.08\).

Depreciation

Depreciation is the decrease in value of an item. A common example of this is car prices. Each year, the value of a car decreases, so we say that its value depreciates.

The rate of depreciation is often given as a percentage.

Example

A new car costs \(\pounds12,000\).

The car loses \(10\%\) of its value during the first year and \(15\%\) of its value during the second year.

How much is the car worth after two years?

Year 1 depreciation: \(10\% of \pounds12000 = \pounds1200\)

Car Value = \(12,000 - 1,200 = \pounds10,800\)

Year 2 depreciation: \(15\% of \pounds10800 = \pounds1620\)

Car Value = \(10,800 - 1,620 = \pounds9,180\)

So after 2 years, the car is worth \(\pounds9180\).

Percentage appreciation/depreciation

It is also possible to calculate how much an item has appreciated or depreciated simply by finding the difference in value as a percentage of the original value.

Example

A computer worth \(\pounds1500\) has depreciated in value to \(\pounds900\) in the past three years.

What is the percentage depreciation in the value of the computer?

Depreciation = \(1500 - 900 = \pounds600\)

Percentage depreciation = \(\frac{600}{1500}\times 100\)

\(= 40\%\)

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