Appreciation/depreciation
The value of a house is usually said to increase with time. Therefore its value is said to appreciate.
Example 1
A flat bought for \(\pounds74\,000\) in 2005 appreciated in value each year by \(1.5\%\).
Calculate the value of the flat after four years.
\({(1.015)^4} \times 74000\)
\(= \pounds78540.90\)
The value of the flat after four years is \(\pounds78\,540.90\).
The value of a car however is usually said to decrease in value with time. Therefore its value is said to depreciate.
Example 2
Jack and Trisha bought a new car for \(\pounds8500\) in 2008. In the first year, its value depreciated by \(20\%\), in the second year by \(15\%\) and in the third by \(10\%\).
Calculate the value of the car at the end of each year.
Value at the end of Year 1:
Percentage has gone down by \(20\%\), therefore \(100\% - 20\% = 80\%\)
\(80\% \,of\,8500\)
\(= 0.8 \times 8500\)
\(= \pounds6800\)
Value at the end of Year 2:
Percentage has gone down by \(15\%\), therefore \(100\% - 15\% = 85\%\)
\(85\% \,of\,6800\)
\(= 0.85 \times 6800\)
\(= \pounds5780\)
Value at the end of Year 3:
Percentage has gone down by \(10\%\), therefore \(100\% - 10\% = 90\%\)
\(90\% \,of\,5780\)
\(= 0.9 \times 5780\)
\(= \pounds5202\)
Now try this:
Question
Jean bought an antique brooch for \(\pounds200\). Its value appreciates by \(10\%\) each year.
Calculate the value of the brooch after 10 years.
\({(1.1)^{10}} \times 200\)
\(= \pounds518.75\)