Chapter 12.3, Problem 71E

Depreciation A construction company purchases a bulldozer for $ 160,000. Each year the value of the bulldozer depreciates by 20% of its value in the preceding year. Let V_n be the value of the bulldozer in the nth year. (Let n = 1 be the year the bulldozer is purchased.)

1. (a) Find a formula for V_n.
2. (b) In what year will the value of the bulldozer be less than $100,000?

To determine

To find: The formula for the value of bulldozer in the nth year.

Answer to Problem 71E

The formula for the value of bulldozer in the nth year is $V_n=160000{\left(\frac{4}{5}\right)}^{n-1}$.

Explanation of Solution

Given:

The purchase value of bulldozer is $160,000. Each year the value of bulldozer decreases by 20% of its preceding year.

Calculation:

The purchase value of bulldozer is $160,000. The value decreased by 20% each year.

The value of bulldozer in the second year is calculated as Purchase value minus the depreciated value.

$\begin{array}{c}\text{second}\text{year}\text{value}=160000-\left(\frac{20}{100}\right)160000\\ =160000\left(1-\frac{20}{100}\right)\end{array}$

The value of bulldozer in the third year is,

$\text{third}\text{year}\text{value}=160000{\left(1-\frac{20}{100}\right)}^2$

In general the value of bulldozer denoted by $V_n$ in the nth year is given by,

$\begin{array}{c}{V}_n=160000{\left(1-\frac{20}{100}\right)}^{n-1}\\ =160000{\left(\frac{80}{100}\right)}^{n-1}\\ =160000{\left(\frac{4}{5}\right)}^{n-1}\end{array}$

Thus, the formula for the value of bulldozer in the nth year is $V_n=160000{\left(\frac{4}{5}\right)}^{n-1}$.

To determine

To find: The year at which the value of bulldozer is less than $100,000.

Answer to Problem 71E

Solution:

The value of bulldozer is less than $100,000 in 4^th year.

Explanation of Solution

Given:

The purchase value of bulldozer is $160,000. Each year the value of bulldozer decreases by 20% of its preceding year.

Calculation:

Use the formula from part (a) of the solution to find the value of bulldozer in the 3rd year.

The formula for the value of bulldozer in the nth year is,

$V_n=160000{\left(\frac{4}{5}\right)}^{n-1}$

Substitute 3 for n in the above formula.

$\begin{array}{c}{V}_n=160000{\left(\frac{4}{5}\right)}^{3-1}\\ =160000{\left(\frac{4}{5}\right)}^2\\ =6400\times 16\\ =102400\end{array}$

The value of bulldozer is $102400 that is more than $100000.

Again use the formula from part (a) of the solution to find the value of bulldozer in the 4^th year.

The formula for the value of bulldozer in the nth year is,

$V_n=160000{\left(\frac{4}{5}\right)}^{n-1}$

Substitute 4 for n in the above formula.

$\begin{array}{c}{V}_n=160000{\left(\frac{4}{5}\right)}^{4-1}\\ =160000{\left(\frac{4}{5}\right)}^3\\ =1280\times 64\\ =81920\end{array}$

The value of bulldozer is $81920 that is less than $100000.

Thus, the value of bulldozer is less than $100000 in 4^th year.