What is the linear model for a tractor with a value of $151,000 assuming it will have a trade in value of $87,00 after 10 years?

Question

What I the depreciated value of the tractor after 6 years? When will the depreciated value fall below $40,000?

Answer 1

This is not a linear model but the formula for depreciation is

 

A=P(1-r)^t

A is the value after t years

P is the initial value

r is the rate of depreciation per year

 

Before we try to answer the first question, we need to determine the value for r using the information given to us.

Let's take the formula and put in the values we now at the ten year point.

 

87,000 = 151,000 (1 - r)¹⁰

 

87,000/151,000 = (1 - r)¹⁰

 

.576 = (1 - r)¹⁰

 

log .576 = log(1 - r)¹⁰ = 10*log(1 - r)

 

-.0239 = log(1 - r)

 

10^-.0239 = 1 - r

 

.9464 = 1 - r

 

r = .05 = 5%

 

A = 151,000(.9464)⁶ = $108,500 is the value after 6 years

 

40,000 = 151,000(.9464)^X

 

-.5769 = xlog.9464

 

x = 24.11

 

After 25 years the value will fall below $40,000