A car valued at $ 600,000.00 depreciates by 10% each year. What will be the value of the car at the end of two years?
- $ 120,000.00
- $ 480,000.00
-
$ 486,000.00
- $ 540,000.00
The correct answer is: C
Explanation
Using P\(_n\) = P\(_o\)(1 - \(\frac{r}{100})^n\)
= 600,000(1 - \(\frac{10}{100})^2\)
= 600,000(0.90)\(^2\) = $486,000.00