Question

Gary bought a car for $40,000. If V = 40,000(.85)t represents the value of the car after t years, how long will it take the car to be worth less than one-fourth of its purchase price? A) 4 years B) 6 years C) 8 years D) 9 years

Answer 1

D) 9 years.

Explanation:

We have been given that Gary bought a car for $40,000 and equation
`V=40,000(0.85)^t` represents the value of the car after t years.

First of all we will find the one-fourth of 40,000.

`\text{One-forth of car's purchase price}=(\$40,000)/(4)`

`\text{One-forth of car's purchase price}=\$10,000`

To find the time it will take the car to be worth less than one-fourth of its purchase price, we will substitute V=10,000 in our given equation.

`10,000=40,000(0.85)^t`

Let us divide both sides of our equation by 40,000.

`(10,000)/(40,000)=(40,000(0.85)^t)/(40,000)`

`0.25=0.85^t`

Let us take natural log of both sides of our equation.

`ln(0.25)=ln(0.85^t)`

Using natural log property
`ln(a^b)=b*ln(a)` we will get,

`ln(0.25)=t*ln(0.85)`

`(ln(0.25))/(ln(0.85))=(t*ln(0.85))/(ln(0.85))`

`(-1.3862943611198906)/(-0.1625189294977749)=t`

`8.530048563597=t`

Upon rounding our answer to the nearest year we will get,

`t\approx 9`

Therefore, it will take 9 years the car to be worth less than one-fourth of its purchase price and option D is the correct choice.