Chapter 20, Problem 22P

To determine

Find the interest rates charged by the banks.

Answer to Problem 22P

The interest rate of the first bank and second bank is $8\%$ and $6\%$.

Explanation of Solution

Given data:

Total amount borrowed $\left(T\right)$ from bank is $\$12000$,

In First bank, uniform series payment at the end of each year $\left(A_1\right)$ is $\$2595.78$, and the number of years $\left(n_1\right)$ is $6$,

In Second bank, uniform series payment at the end of each month $\left(A_2\right)$ is $\$198.87$, and the number of periods $\left(n_2\right)$ is $6$ years that is $72$ months.

Formula used:

Consider the following expression obtained from the payment details of first bank,

$T=A_1\left(\frac{P}{A},i,n_1\right)$ (1)

Here,

$T$ is the total borrowed amount,

$A_1$ is the uniform series payment in the first bank,

$P$ is the present cost,

$A$ is the uniform series payment,

$i$ is the normal interest rate,

$n_1$ is the number of years.

The equation (1) is rewritten with respect to the given values is,

$\$12000=\left(\$2595.78\right)\left(\frac{P}{A},i,6\right)$ (2)

Consider the following expression obtained from the payment details of second bank,

$T=A_2\left(\frac{P}{A},i,n_2\right)$ (3)

Here,

$A_1$ is the uniform series payment in the second bank,

$n_2$ is the number of periods in months.

The equation (3) is rewritten with respect to the given values is,

$\$12000=\left(\$198.87\right)\left(\frac{P}{A},i,72\right)$ (4)

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

$\left(\frac{P}{A},i,n\right)=\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]$ (5)

Formula to calculate the present cost for the given uniform series payment when the interest compounds $m$ times per year is,

$\left(\frac{P}{A},i,n\right)=\left[\frac{{\left(1+\frac{i}{m}\right)}^{nm}-1}{\left(\frac{i}{m}\right){\left(1+\frac{i}{m}\right)}^{nm}}\right]$ (6)

Calculation:

Case 1: Interest rate of first bank

Substitute the equation (5) in equation (2) and substitute $6$ for $n$ to find $i$.

$\begin{array}{l}\$12000=\left(\$2595.78\right)\left[\frac{{\left(1+i\right)}^6-1}{i{\left(1+i\right)}^6}\right]\\ \left[\frac{{\left(1+i\right)}^6-1}{i{\left(1+i\right)}^6}\right]=\frac{\$12000}{\$2595.78}\\ \left[\frac{{\left(1+i\right)}^6-1}{i{\left(1+i\right)}^6}\right]=4.623\end{array}$ (7)

By using trial and error method, calculate the value of $i$ in equation (7).

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

Substitute $5\%$ for $i$ in equation (7).

$\begin{array}{r}\left[\frac{{\left(1+\left(5\%\right)\right)}^6-1}{\left(5\%\right){\left(1+\left(5\%\right)\right)}^6}\right]=4.623\\ \left[\frac{{\left(1+0.05\right)}^6-1}{\left(0.05\right){\left(1+0.05\right)}^6}\right]=4.623\end{array}$

Reduce the equation as,

$5.076=4.623$ (8)

From the equation (8), 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 $8\%$:

Substitute $8\%$ for $i$ in equation (7).

$\begin{array}{r}\left[\frac{{\left(1+\left(8\%\right)\right)}^6-1}{\left(8\%\right){\left(1+\left(8\%\right)\right)}^6}\right]=4.623\\ \left[\frac{{\left(1+0.08\right)}^6-1}{\left(0.08\right){\left(1+0.08\right)}^6}\right]=4.623\end{array}$

Reduce the equation as,

$4.623=4.623$ (9)

From the equation (9), it is clear that $\text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S}$. Therefore, the value of $i$ is $8\%$.

Case 2: Interest rate of second bank:

Substitute the equation (6) in equation (4), and substitute $6$ for $n$, and $12$ for $m$ to find $i$.

$\$12000=\left(\$198.87\right)\left[\frac{{\left(1+\frac{i}{12}\right)}^{\left(6\right)\left(12\right)}-1}{\left(\frac{i}{12}\right){\left(1+\frac{i}{12}\right)}^{\left(6\right)\left(12\right)}}\right]$

Rewrite the equation as follows,

$\begin{array}{l}\left[\frac{{\left(1+\frac{i}{12}\right)}^{72}-1}{\left(\frac{i}{12}\right){\left(1+\frac{i}{12}\right)}^{72}}\right]=\frac{\$12000}{\$198.87}\\ \left[\frac{{\left(1+\frac{i}{12}\right)}^{72}-1}{\left(\frac{i}{12}\right){\left(1+\frac{i}{12}\right)}^{72}}\right]=60.34\end{array}$ (10)

By using trial and error method, calculate the value of $i$ in equation (10).

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

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

$\begin{array}{l}\left[\frac{{\left(1+\frac{\left(5\%\right)}{12}\right)}^{72}-1}{\left(\frac{\left(5\%\right)}{12}\right){\left(1+\frac{\left(5\%\right)}{12}\right)}^{72}}\right]=60.34\\ \left[\frac{{\left(1+\frac{0.05}{12}\right)}^{72}-1}{\left(\frac{0.05}{12}\right){\left(1+\frac{0.05}{12}\right)}^{72}}\right]=60.34\end{array}$

Reduce the equation as,

$62.1=60.34$ (11)

From the equation (11), 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 $6\%$:

Substitute $6\%$ for $i$ in equation (10).

$\begin{array}{l}\left[\frac{{\left(1+\frac{\left(6\%\right)}{12}\right)}^{72}-1}{\left(\frac{\left(6\%\right)}{12}\right){\left(1+\frac{\left(6\%\right)}{12}\right)}^{72}}\right]=60.34\\ \left[\frac{{\left(1+\frac{0.06}{12}\right)}^{72}-1}{\left(\frac{0.06}{12}\right){\left(1+\frac{0.06}{12}\right)}^{72}}\right]=60.34\end{array}$

Reduce the equation as,

$60.34=60.34$ (12)

From the equation (12), it is clear that $\text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S}$. Therefore, the value of $i$ is $6\%$.

Therefore, from the analysis the interest rate of the first bank and second bank is $8\%$ and $6\%$.

The second bank is highly preferable because it provides less interest rate than the first bank.

Conclusion:

Thus, the interest rate of the first bank and second bank is $8\%$ and $6\%$.