Chapter D, Problem 76P

To determine

The minimum salvage value as a percentage of first cost.

Answer to Problem 76P

Therefore the correction option is $\left(b\right)36\%$.

Explanation of Solution

Given data:

Cost of the tank is $\$150,000$.

Salvage value is $\$7,500$.

Time period is $10\text{years}$.

Annual repair cost for every 2 year is $\$30,000$.

Cost of the new tank is $\$325,000$.

Interest rate is $12\%$.

Number of years is $20$.

Concept used:

Write the expression to calculate the salvage value of the technology.

$SV=\left[\begin{array}{l}-\left(\text{cost}\right)-\left({\text{1}}^{\text{st}}\text{cost}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)-\left({\text{2}}^{\text{nd}}\text{cost}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\\ -\left({\text{3}}^{\text{rd}}\text{cost}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)-\left({\text{4}}^{\text{th}}\text{cost}\right)\left(\frac{1}{{\left(1+i\right)}^n}\right)\\ +\text{salvage}\text{value}\left(\frac{1}{{\left(1+i\right)}^n}\right)+\cos t_{\text{technology}}\end{array}\right]$ ...... (I)

Here, salvage value is $SV$, interest rate is $i$, and time period is $n$.

Calculation:

Calculate the salvage value for the technology,

Substitute, $\$150,000$ for cost, $\$30,000$ for ${\text{1}}^{\text{st}}$ cost, $\$30,000$ for ${\text{2}}^{\text{nd}}$ cost, $\$30,000$ for ${\text{3}}^{\text{rd}}$ cost, $\$7,500$ for ${\text{4}}^{\text{th}}$ cost, $12\%$ for $i$ and in increments of $2$ for $n$ in Equation (I).

$\begin{array}{c}SV=\left[\begin{array}{l}-\left(\text{\$150,000}\right)-\left(\text{\$30,000}\right)\left(\frac{1}{{\left(1+0.12\right)}^2}\right)-\left(\text{\$30,000}\right)\left(\frac{1}{{\left(1+0.12\right)}^4}\right)\\ -\left(\text{\$30,000}\right)\left(\frac{1}{{\left(1+0.12\right)}^6}\right)-\left(\text{\$30,000}\right)\left(\frac{1}{{\left(1+0.12\right)}^8}\right)\\ -\left(\text{\$7,500}\right)\left(\frac{1}{{\left(1+0.12\right)}^{10}}\right)+\$325,000\end{array}\right]\\ =\left[\begin{array}{l}-\left(\text{\$150,000}\right)-\left(\text{\$30,000}\right)\left(0.7972\right)-\left(\text{\$30,000}\right)\left(0.6355\right)\\ -\left(\text{\$30,000}\right)\left(0.5066\right)-\left(\$30,000\right)\left(0.4038\right)\\ -\left(\$7,500\right)\left(0.3222\right)+\$325,000\end{array}\right]\\ =\left[\begin{array}{l}-\text{\$150,000}-\$23,916-\$19,065\\ -\$15,198-\$12,114\\ -\$2,416.5+\$325,000\end{array}\right]\\ =\$102,290.5\end{array}$

Write the expression to calculate percentage to the first cost.

$\text{PC}=\frac{\text{Salvage}{\text{value}}_{\text{technology}}}{{\text{Cost}}_{\text{technology}}}\times 100$ ....... (II)

Substitute, $\$102,290.5$ for salvage value and $\$325,000$ for cost in Equation (II)

$\begin{array}{c}\text{PC}=\frac{\$102,290.5}{\$325,000}\times 100\\ =32\%\end{array}$.

Conclusion:

The nearest value to the answer is $36\%$.

Therefore, the correction option is $\left(b\right)36\%$.