Question #292525

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?


Expert's answer

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:



P1=2370023700(13.5/100)=23700(113.5/100)dollarsP_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:


P2=P1P1(13.5/100)=P1(113.5/100)=23700(113.5/100)(113.5/100)=23700(113.5/100)2dollarsP_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:


P3=P2P2(13.5/100)=P2(113.5/100)=23700(113.5/100)2(113.5/100)=23700(113.5/100)3dollarsP_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:


P10=23700(113.5/100)10=23700(10.135)10=237000.86510=(237000.2345098)dollars=(237000.2345098100)cents=555788centsP_{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.


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