Answer to Question #159144 in Algebra for Jan

Question

The value of a car after it is purchased depreciates according to the

A formula V(n) 5 25 000(0.85)n, where V(n) is the car’s value in the

nth year after it was purchased.

a) What is the purchase price of the car?

b) What is the annual rate of depreciation?

c) What is the car’s value at the end of 3 years?

d) How much value does the car lose in its first year?

e) How much value does it lose in its fifth year?

f) After how many years will the value of the car be half of the

original purchase price?

Answer 1

The value of the car in the year n is given by: $V(n)=25000(0.85)^n$ . We have to determine

a) The purchase price of the car: This is done by replacing n by 0 which gives V = 25000.

b) The annual rate of depreciation:

annual change in value is 25000(0.85)⁰ - 25000(0.85)¹ = 25000 (1-0.85)

= 25000(0.15)

the annual depreciation rate is = $\frac{annual (change)}{total (amount)}*100$ = $\frac{25000(0.15)}{25000}*100=15$ %

c) The car's value at the end of 3 years: The car's value at the end of 3 years is

$V(4) = 25000*(0.85)^3 = 15,353.125$

d) The value lost by the car in its first year: The value lost in the first year is V(0) - V(1) = 25000*(1.0 - 0.85) = 3750

e) The value lost in the fifth year: This is equal to

$V(4) - V(5) =25000[(0.85)^4 - (0.85)^5]= 25000[0.52200625-0.443705312]=1,957.52345$

the value lost in fifth year is 1957.52345

f) The number of years for the value to become equal to half of the original purchase price: If the number of years required for this is N, V(N) = 1/2*V(0)

=> $25000( 0.85)^N = (1/2)*25000(0.85)^0$

=> $0.85^N=1/2$

=>  taking log (base 10) both side

=> $log(0.85)^N=log(1/2)$

=> N* log(0.85)=log(0.5)

=> $N=\frac{log(0.5)}{log(0.85)}=4.265$ years.