Question

Joe bought a car for $9,000 in 2013. His purchase has depreciated 2.5% every year since then. What was his car worth in 2019 to the nearest dollar?

a. $7650
b. $7732
C. $8350
d. $1602​

Answer 1

b. $7732

Explanation:

Value of a depreciating product:

The value of a depreciating product, after t years, is given by:

`V(t) = V(0)(1 - r)^t`

In which V(0) is the initial value and r is the decay rate, as a decimal.

Joe bought a car for $9,000 in 2013. His purchase has depreciated 2.5% every year since then.

This means that
`V(0) = 9000, r = 0.025`. So

`V(t) = V(0)(1 - r)^t`

`V(t) = 9000(1 - 0.025)^t`

`V(t) = 9000(0.975)^t`

What was his car worth in 2019 to the nearest dollar?

2019 is 2019 - 2013 = 6 years after 2013, so this is V(6).

`V(6) = 9000(0.975)^6 = 7732`

The correct answer is given by option b.