Question

A car cost $20,000 when it was purchased. The value of the car decreases by 8% each year. Find the rate of decay each month and select the correct answer below. −0.006924% −0.006667% −0.666667% −0.0081%

Answer 1

The rate of decay each month is -0.006924%

Explanation:

The cost of a car after 1 year at a 8% decreased

Decreased price = Sale price - (Sale price x porcentage of decreased/100%)

Decreased price = $20,000 - ($20,000 x 8%/100%)

Decreased price = $20,000 - ($20,000 x 0.08)

Decreased price = $20,000 - $1,600 = $18,400

So, the value of the car decreased by 8% in a year is $18,400.

The general equation for exponential decay is:

`y=C(1-r)^(t)`

Where

y = final amount

C = Sale price

r = rate of decay

t = time

We know that the sale price was $20,000. After a year the decreased price is $18,400.

From the general equation for exponential decay, our sale price is $20,000 decreased over a time of 12 months resulting the final amount of $18,400.

substituting the values

`18,400=20,000(1-r)^(12)`

Solving the equation for r

`(18,400)/(20,000)=(20,000)/(20,000)(1-r)^(12)`

`(18,400)/(20,000)=(1-r)^(12)`

`(18,400)/(20,000)^{(1)/(12) } =[(1-r)^(12)]^{(1)/(12)}`

`(18,400)/(20,000)^{(1)/(12) } =1-r`

`r+(18,400)/(20,000)^{(1)/(12) } =1-r+r`

`r+(18,400)/(20,000)^{(1)/(12) } =1`

`r+(18,400)/(20,000)^{(1)/(12) }-(18,400)/(20,000)^{(1)/(12) } =1-(18,400)/(20,000)^{(1)/(12) }`

`r=1-(18,400)/(20,000)^{(1)/(12)}`

`r=1-\sqrt[12]{(18,400)/(20,000)}`

`r=1-\sqrt[12]{0.92}\\r=1-0.993076\\r= -0.006924`