Chapter 20, Problem 20P

(a)

To determine

Find the monthly payment amount.

Answer to Problem 20P

The monthly payment amount is $\$304.15$.

Explanation of Solution

Given data:

The normal interest rate $\left(i\right)$ is $8\%$,

The present value $\left(P\right)$ is $\$15000$,

The number of years $\left(n\right)$ is 60 months that is 5 years,

The number of interest compounding periods per year $\left(m\right)$ is given monthly therefore 12.

Formula used:

Formula to calculate the uniform series payment with the given present cost is,

$A=P\left[\frac{\left(\frac{i}{m}\right){\left(1+\frac{i}{m}\right)}^{nm}}{{\left(1+\frac{i}{m}\right)}^{nm}-1}\right]$ (1)

Here,

$P$ is the Present cost,

$i$ is the normal interest rate,

$n$ is the number of years,

$m$ is the number of interest compounding periods per year.

Calculation:

Substitute $\$15000$ for $P$, $8\%$ for $i$, $5$ for $n$, and $12$ for $m$ in equation (1) to find $A$.

$\begin{array}{c}A=\left(\$15000\right)\left[\frac{\left(\frac{\left(8\%\right)}{12}\right){\left(1+\frac{\left(8\%\right)}{12}\right)}^{\left(5\right)\left(12\right)}}{{\left(1+\frac{\left(8\%\right)}{12}\right)}^{\left(5\right)\left(12\right)}-1}\right]\\ =\left(\$15000\right)\left[\frac{\left(\frac{0.08}{12}\right){\left(1+\frac{0.08}{12}\right)}^{60}}{{\left(1+\frac{0.08}{12}\right)}^{60}-1}\right]\\ =\left(\$15000\right)\left[\frac{\left(\frac{0.08}{12}\right){\left(1+\frac{0.08}{12}\right)}^{60}}{{\left(1+\frac{0.08}{12}\right)}^{60}-1}\right]\\ =\$304.15\end{array}$

Therefore, the monthly payment amount is $\$304.15$.

Conclusion:

Thus, the monthly payment amount is $\$304.15$.

(b)

To determine

Find the effective interest rate.

Answer to Problem 20P

The effective interest rate is $10\%$.

Explanation of Solution

Given data:

The present value $\left(P\right)$ is $\$15000$,

The number of years $\left(n\right)$ is 60 months that is 5 years,

The number of interest compounding periods per year $\left(m\right)$ is given monthly therefore 12.

Formula used:

Formula to calculate the uniform series payment with the given present cost is,

$A=P\left[\frac{\left(\frac{i}{m}\right){\left(1+\frac{i}{m}\right)}^{nm}}{{\left(1+\frac{i}{m}\right)}^{nm}-1}\right]$ (2)

Here,

$P$ is the Present cost,

$i$ is the normal interest rate,

$n$ is the number of years,

$m$ is the number of interest compounding periods per year.

Formula to calculate the effective interest rate is,

$i_{\text{eff}}={\left(1+\frac{i}{m}\right)}^m-1$ (3)

Calculation:

It is given that the bank charges $4.5\%$ of the loan amount at the time the bank gives loan.  Therefore, the total loan amount given by the bank is,

$\text{Loan}\text{amount}=P-\left(4.5\%\right)P$ (4)

Substitute $\$15000$ for $P$ in equation (4) to find the $\text{Loan}\text{amount}$.

$\begin{array}{c}\text{Loan}\text{amount}=\left(\$15000\right)-\left(4.5\%\right)\left(\$15000\right)\\ =\left(\$15000\right)-\left(0.045\right)\left(\$15000\right)\\ =\left(\$15000\right)-\left(\$675\right)\\ =\$14325\end{array}$

Hence, the present value $\left(P\right)$ is $\$14325$.

Substitute $\$304.15$ for $A$, $\$14325$ for $P$, $5$ for $n$, and $12$ for $m$ in equation (2) to find $i$.

$\left(\$304.15\right)=\left(\$14325\right)\left[\frac{\left(\frac{i}{12}\right){\left(1+\frac{i}{12}\right)}^{\left(5\right)\left(12\right)}}{{\left(1+\frac{i}{12}\right)}^{\left(5\right)\left(12\right)}-1}\right]$

Reduce the equation as,

$\left[\frac{\left(\frac{i}{12}\right){\left(1+\frac{i}{12}\right)}^{60}}{{\left(1+\frac{i}{12}\right)}^{60}-1}\right]=0.0212$ (5)

By using trial and error method, calculate the value of $i$ in equation (5).

(i) Interest rate of $5\%$:

Substitute $5\%$ for $i$ in equation (5).

$\begin{array}{c}\left[\frac{\left(\frac{\left(5\%\right)}{12}\right){\left(1+\frac{\left(5\%\right)}{12}\right)}^{60}}{{\left(1+\frac{\left(5\%\right)}{12}\right)}^{60}-1}\right]=0.0212\\ \left[\frac{\left(\frac{0.05}{12}\right){\left(1+\frac{0.05}{12}\right)}^{60}}{{\left(1+\frac{0.05}{12}\right)}^{60}-1}\right]=0.0212\\ 0.0188=0.0212\end{array}$ (6)

From the equation (6), it is clear that $\text{L}\text{.H}\text{.S}\ne \text{R}\text{.H}\text{.S}$. Therefore, the value of $i$ should not be $5\%$.

(ii) Interest rate of $10\%$:

Substitute $10\%$ for $i$ in equation (5).

$\begin{array}{c}\left[\frac{\left(\frac{\left(10\%\right)}{12}\right){\left(1+\frac{\left(10\%\right)}{12}\right)}^{60}}{{\left(1+\frac{\left(10\%\right)}{12}\right)}^{60}-1}\right]=0.0212\\ \left[\frac{\left(\frac{0.1}{12}\right){\left(1+\frac{0.1}{12}\right)}^{60}}{{\left(1+\frac{0.1}{12}\right)}^{60}-1}\right]=0.0212\\ 0.0212=0.0212\end{array}$ (7)

From the equation (7), it is clear that $\text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S}$. Therefore, the value of $i$ is $10\%$.

Substitute $10\%$ for $i$, and $12$ for $m$ in equation (3) to find $i_{\text{eff}}$.

$\begin{array}{c}{i}_{\text{eff}}={\left(1+\frac{\left(10\%\right)}{12}\right)}^{12}-1\\ ={\left(1+\frac{0.1}{12}\right)}^{12}-1\\ =0.1\left(\text{or}\right)10\%\end{array}$

Therefore, the effective interest rate is $10\%$.

Conclusion:

Thus, the effective interest rate is $10\%$.