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

Answer:

b. $7732

Step-by-step 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.