Question

Farmer buys a tractor for 159,000 and assumes the trade in value of 87,000 after 10 years . The farmer uses a constant rate of depreciation to determine the annual value of the tractor . Find the linear model

Answer 1

Hello there. To solve this question, we have to find the linear model of the annual value of the tractor.

In this case, we want to find a line equation as follows:

$y=mt+b$

Where m is the slope of the line, b is the y-intercept and t is the time in year.

We know the farmer bought a tractor for $159.000 and assumes the trade in value of $87.000 after 10 years.

Think of these values as points in a graph: (0, 159000) and (10, 87000)

Since the graph of this function is on the ty plane, we can use the slope formula to find the value of m:

$m=\frac{87000-159000}{10-0}=-\frac{72000}{10}=-7200$

Now, plugging in the value of m and any of the points in the equation of the line, we calculate the value of b:

$\begin{gathered} 159000=-7200\cdot0+b \\ b=159000 \end{gathered}$

Therefore the equation of the line is:

$y=-7200t+159000$

B) What is the depreciated value of the tractor after 6 years?

Plugging in t = 6, we get:

$y=-7200\cdot6+159000=115800$

C) When will the depreciated value fall below $80.000?

In this case, we solve for the inequality:

$\begin{gathered} y<80000 \\ -7200t+159000<80000 \end{gathered}$

Subtract 159000 on both sides of the inequality

$-7200t<-79000$

Divide both sides of the inequality by a factor of -7200. Remember this flips the inequality sign.

$t>10.97\text{ years}$

Since we need to round to the nearest integer:

The depreciated value will fall below $80.000 on the 11th year.

The graph will look like the following:

Out of scale. The correct option is indeed C.