SOLUTION: At one boat dealer, she found a boat she likes that sells for $15,000 and depreciates at a rate of 30% per year. What will be the value of the boat after 3 years?

Algebra ->  Finance -> SOLUTION: At one boat dealer, she found a boat she likes that sells for $15,000 and depreciates at a rate of 30% per year. What will be the value of the boat after 3 years?      Log On


   

Question 765749: At one boat dealer, she found a boat she likes that sells for $15,000 and depreciates at a rate of
30% per year. What will be the value of the boat after 3 years?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

you can do it this way:
In the problem you are given the rate of decrease of 30% or 0.30.
Each year, the boat is worth 100-30=70% of its value the previous year, so after 3years, its value will be
$15000%2A+0.30+=4500 will cost less than year before; so price firs year after is $15000-4500=10500
next year value 10500 depreciates again for 30%; so new value will be
10500%2A0.30=3150; so price second year after is $10500-3150=7350
and finally,
$7350%2A0.30=2205; so price in three years is
$7350-2205=5145

or use exponential growth/decay formula:
Since the decay equation takes on the form P=P%5B0%5D%281-k%29%5Et, we can take note that k=.30, P%5B0%5D=15000 and t=3.
Plugging this into the equation, we have
P=15000%281-.3%29%5E3
P=5145
So after 3 years, the car will be worth 5145