A used car was purchased in July 2000 for $11,900. If the car depreciates 13% of its value each year, what is the value of the car, to the nearest hundred dollars, in July 2003?

Question

3615 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T01:50:45

#$7200# to nearest $100 dollars

Original value depreciates by 13% annually. So for the 1st year, we have:

#$11900-$11900xx13/100=$10353#

#->11900xx1-11900xx0.13color(white)(88)#( 0.13 is 13% as a decimal )

Factoring:

#->11900(1-0.13)=>11900xx0.87#

For the 2nd year:

#10353xx0.87=907.11#

For the 3rd year:

#907.11xx0.87=7836.1857#

So this is following the pattern:

#11900xx0.87->10353xx0.87->907.11xx0.87#

This is the same as:

#11900xx0.87xx0.87xx0.87=>11900(0.87)^3#

This is for 3 years, so for #n# years:

#:.#

#11900(0.87)^n#

Since #0.87=(1-0.13)#, we have:

#11900(1-0.13)^n#

From this we can derive the formula:

#FV=PV(1-R)^n#

#FV= #Future Value

#PV=#Present Value

#r=# interest rate as a decimal

#n=# number of years

For the original problem:

#PV=11900#

#r=0.13#

#n=3 7/12=43/12#

#FV=11900(1-0.13)^(43/12)=7224.774281#

#$7200# to nearest $100 dollars