Q:

Vanessa can afford a $1405-per-month house loan payment. If she is being offered a 20-year house loan with an APR of 4.8%, compounded monthly, which of these expressions represents the value of the most money she can borrow?

Accepted Solution

A:
To solve this we are going to use the present value formula: [tex]PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ][/tex]
where 
[tex]PV[/tex] is the present value
[tex]P[/tex] is the periodic payment
[tex]r[/tex] is the interest rate in decimal form
[tex]n[/tex] is the number of times the interest is compounded per year 
[tex]k[/tex] is the number of payments per year
[tex]t[/tex] is the number of years

We know for our problem that she can afford a $1405-per-month house loan payment, so [tex]P=1405[/tex] and [tex]k=12[/tex]. We also know that the loan is a 20-year house loan, so [tex]t=20[/tex]. To convert the interest rate to decimal form, we are going to divide the rate by 100% [tex]r= \frac{4.8}{100}=0.048 [/tex]. Since the interest rate is compounded monthly, it is compounded 12 times per year; therefore, [tex]n=12[/tex]. Lets replace those values in our formula:

[tex]PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ][/tex]
[tex]PV=1405[ \frac{1-(1+ \frac{0.048}{12})^{-(12)(20)} }{ \frac{0.048}{12} } ][/tex]
[tex]PV=216501.09[/tex]

We can conclude that the expression that represents the value of the most money she can borrow is: [tex]PV=1405[ \frac{1-(1+ \frac{0.048}{12})^{-(12)(20)} }{ \frac{0.048}{12} } ][/tex]. Evaluating the expression we can prove that she can borrow $216,501.09