Question

A car was valued at $38,000 in the year 2007. By 2013, the value had depreciated to $11,000. If the car's value continues to drop by the same percentage, what will it be worth by 2017?

Answer 1

From the depreciation formula:

`A=P(1-(r)/(100))^n`

we know the final value A=$11,000, the initial values P=$38000 and the number of years n=6. Then, we need to find the rate r. By substituting these values into the formula, we have

`11000=38000(1-(r)/(100))^6`

Then, by dividing both sides by 38000, we get

`\begin{gathered} (11000)/(38000)=(1-(r)/(100))^6 \\ or\text{ equivalently,} \\ (1-(r)/(100))^6=0.28947 \end{gathered}`

Now, by applying 6th root to both sides, we have

`(1-(r)/(100))=0.813332`

by subtracting 1 to both sides, we have

`-(r)/(100)=-0.186667`

or equivalently,

`(r)/(100)=0.186667`

Then, the rate is given by

`\begin{gathered} r=100*0.186667 \\ r=18.6667\text{ \%} \end{gathered}`

Once we know the rate, we can find the car's value by 2017. Then, we have

`A=38000(1-(18.667)/(100))^(10)`

where n=10 years (from 2007 to 2017). By computing the term into the paranthesis, we have

`\begin{gathered} A=38000(1-0.18667)^(10) \\ A=38000(0.813332)^(10) \end{gathered}`

which gives

`\begin{gathered} A=38000(0.12667) \\ A=4813.549 \end{gathered}`

Therefore, the answer is $4813.549