Respuesta :
The answer to the question above is letter B. The new car depreciates at the rate of 15% per year. The expected value of the car after 5 years is $11,093. to explain the calculation of the answer :
year amount % interest total
1 25000 0.15 3750 21250
2 21250 0.15 3187.5 18062.5
3 18062.5 0.15 2709.375 15353.13
4 15353.13 0.15 2302.969 13050.16
5 13050.16 0.15 1957.523 11092.63
year amount % interest total
1 25000 0.15 3750 21250
2 21250 0.15 3187.5 18062.5
3 18062.5 0.15 2709.375 15353.13
4 15353.13 0.15 2302.969 13050.16
5 13050.16 0.15 1957.523 11092.63
You might have lacking given information so you can re-check the given.
You are lacking the value of L (life of the car in years).
There are several types of depreciation formulas. The straight-line method, sinking fund method, declining balance and double-declining balance. For the other methods you can search it up.
I used the sinking fund method for this case:
d = \frac{(25000 - 0)*0.15}{(1+0.15)^{L} -1
where d is the annual cost of depreciation
Co is the original cost,
[tex] C_{L} [/tex] is the value at the end of the life of object or salvage value
L is useful life of the property
and i is the interest rate.
The salvage value can be assumed as zero in this case, therefore,
[tex]d = \frac{(25000 - 0)*0.15}{(1+0.15)^{L} -1}[/tex]
In solving for the expected depreciation one can use this formula:
[tex] D_n = d*\frac{(1 + i)^{n}-1 }{i}[/tex]
where n is the number of years of the expected depreciation
D_n is the depreciation up to the number of years (n)
You are lacking the value of L (life of the car in years).
There are several types of depreciation formulas. The straight-line method, sinking fund method, declining balance and double-declining balance. For the other methods you can search it up.
I used the sinking fund method for this case:
d = \frac{(25000 - 0)*0.15}{(1+0.15)^{L} -1
where d is the annual cost of depreciation
Co is the original cost,
[tex] C_{L} [/tex] is the value at the end of the life of object or salvage value
L is useful life of the property
and i is the interest rate.
The salvage value can be assumed as zero in this case, therefore,
[tex]d = \frac{(25000 - 0)*0.15}{(1+0.15)^{L} -1}[/tex]
In solving for the expected depreciation one can use this formula:
[tex] D_n = d*\frac{(1 + i)^{n}-1 }{i}[/tex]
where n is the number of years of the expected depreciation
D_n is the depreciation up to the number of years (n)