Question

For accounting purposes, the value of assets (land, buildings, equipment) in a business depreciates at a set rate per year. The value, V(t), of $408,000 worth of assets after t years, which depreciate at 18% per year, is given by the formula V(t) = V0(b)t. What is the value of V0 and b, and when rounded to the nearest cent, what is the value of the assets after 8 years?

Answer 1

V(t) = $408,000 - ( $408,000 x 18%)
= $334,560 - ($334,560 x 18%)
= $274,339.20 - ($274,339.20 x 18%)
= $224,958.14 - ($224,958.14 x 18%)
= $184,465.68 - ($184,465.68 x 18%)
= $151,261.86 - ($151,261.86 x 18%)
= $124,034.73 - ($124,034.73 x 18%)
= $101,708.48 - ($101,708.48 x 18 %)
= $83,400.95

The new carrying value of the asset on the current year is deducted with the depreciation rate to get the carrying value of the next year

Answer 2

For this case we have an equation of the form:


`V (t) = V0 * (b) ^ t`
Where,
v0: initial value in assets
b: depreciation rate
t: time in years.
Substituting values we have:

`V (t) = 408000 * (0.82) ^ t`
For year 8 we have:

`V (t) = 408000 * (0.82) ^ 8 V (t) = 83400.94703`
Rounding off we have:
V (t) = 83401
Answer:
the value of V0 and b are:
V0 = $ 408,000
b = 0.82
the value of the assets after 8 years is:V (t) = 83401 $