Chapter 11.2, Problem 59PPE

To determine

Find the monthly payment on a loan of $60000 at 6 % annual interest paid over 30 years.

Answer to Problem 59PPE

Monthly payment is $359.73

Explanation of Solution

Given:

The formula below gives the monthly payment m on a loan as a function of the amount borrowed A, the annual rate of interest r(expressed as a decimal) and the number of months n of the loan. Use this formula and a calculator.

  $m=\frac{\text{A}(\frac{r}{12}){(1+\frac{r}{12})}^n}{{(1+\frac{r}{12})}^n-1}$

Suppose a family wants to buy the house advertised at the right. They have $60,000 for a down payment. Their mortgage will have an annual interest rate of 6%. The loan is to be repaid over a 30 − yr period. How much will it cost the family to repay this mortgage over the 30 yr?

Concept Used:

  $30\text{yrs}=30·12=360\text{months}$

Substitute the value: A = $60000; r = 6% = 0.06 and n = 360.

Calculate for ‘m ‘

Calculation:

The formula: $m=\frac{\text{A}(\frac{r}{12}){(1+\frac{r}{12})}^n}{{(1+\frac{r}{12})}^n-1}$

  $\begin{array}{l}\text{A}\text{=}\text{\$60000;}\text{r=}\text{6\%}\text{=0}\text{.06}\text{,}\text{n}\text{=}\text{360}\text{month;}\text{Find }\text{m}\\ m=\frac{\text{A}(\frac{r}{12}){(1+\frac{r}{12})}^n}{{(1+\frac{r}{12})}^n-1}=\\ m=\frac{\text{60000}(\frac{0.06}{12}){(1+\frac{0.06}{12})}^{360}}{{(1+\frac{0.06}{12})}^{360}-1}\\ \\ m=\frac{(300)·\text{(6}\text{.0225752122632162)}}{\text{5}\text{.0225752122632162}}\\ \\ m=\frac{\text{1806}\text{.77256367896486}}{\text{5}\text{.0225752122632162}}\approx 359.73\end{array}$

m = $359.73

Thus, monthly payment is $359.73