Question

Suppose a car that sells for $40,000 depreciates 20% per year. How many years would it take for the car to have a value less than $21,000?

a. Use the formula $A_n = a_1 (r)^n^-^1$ to determine the solution.

b. Use the formula $A_n=a_0(1+r)^n$ to determine the solution.

Answer 1

Answer:

principle, r is the decimal rate, and t is the time in years

 

     P = 12,500

     r = .2 ... the decimal equivalent of 20%

     t = 3  ... number of years

 

     Plugging these values into the equation, we obtain:

 

     A = 12,500(1 - .2)^3

     A = 12,500 (.8)^3

     A = 12,500 (.512)

     A = 6400

 

So after 3 years, the car will be worth $6400.