Answer to Question #292525 in Algebra for jose

Question

A new car is purchased for 23700 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be, to the nearest cent, after 10 years?

Answer 1

For The given problem, The initial value of the car is 23700 dollars.

After one year, the price of the car depreciates by 13.5% per year. So , the new price of car after one year becomes:

$$P_1 =23700-{23700\cdot(13.5/100)} = {23700\cdot(1-13.5/100)} dollars$$

Again, taking P₁ as base price, the price of the car now depreciates by 13.5% per year. So , the new price (in dollars) of car after second year becomes:

$P_2 =P_1-{P_1\cdot(13.5/100)} = {P_1\cdot(1-13.5/100)} = 23700\cdot (1-13.5/100)\cdot (1-13.5/100) = 23700\cdot (1-13.5/100)^2 dollars$

Following the same, the new price of car after third year becomes:

$P_3 =P_2-{P_2\cdot(13.5/100)} = {P_2\cdot(1-13.5/100)} = 23700\cdot (1-13.5/100)^2\cdot (1-13.5/100) = 23700\cdot (1-13.5/100)^3 dollars$

And, so on as we continue to move to tenth year, the new price of car at the end of tenth year becomes:

$P_{10}=23700\cdot (1-13.5/100)^{10} = 23700\cdot (1-0.135)^{10}=23700\cdot 0.865^{10} = (23700\cdot 0.2345098) dollars = (23700\cdot 0.2345098\cdot100) cents = 555788 cents$

So the value of the car to the nearest cent after 10 years becomes 555788 cents.