Chapter 3.6, Problem 56E

Solution

a.

To find: Whenthe mortgage is completely paid.

The mortgage will be paid in $172$ monthly payments or it will take $14\text{ years and }4\text{ months}$ .

Given information:

The mortgage value is given $pv=\$86,000$ for $30$ years.

Monthly paymentis $\$884.61$ .

  $R=1050$

Interest $i=\frac{0.12}{12}=0.01$

Calculation:

The borrowed amount is,

  $pv=\$86,000$

  $86,000=1050\cdot \frac{1-{(1+0.01)}^{-n}}{0.01}$

Multiply the above equation on both sides to solve the exponential part,

  $860=1050\cdot [1-{1.01}^{-n}]$

Divide on both sides by $1050$ ,

  $\frac{86}{105}=1-{1.01}^{-n}$

Subtract by $1$ on both sides,

  $\begin{array}{c}-\frac{19}{105}=-{1.01}^{-n}\\ \frac{19}{105}={1.01}^{-n}\end{array}$

Taking log on both sides to get,

  $\log \left[\frac{19}{105}\right]=\log {1.01}^{-n}$

Solving the above equation to find the value of $n$

  $\begin{array}{c}\log 19-\log 105\text{ =}-n\log 1.01\\ n=\frac{\log 105-\log 19}{\log 1.01}\\ =171.8\text{ payments }\end{array}$

Therefore, the mortgage will be paid in $172$ monthly payments or it will take $\text{14 years and 4 months}$ .

b.

To find: How much the amount gets saved compared with the actual plan.

The amount get saved compared to the original plan is $\$138,069.60$ .

Given information:

The mortgage value is given $pv=\$86,000$ for $30$ years.

Monthly payment is $\$884.61$ .

  $R=1050$

Interest,

  $\begin{array}{c}i=\frac{0.12}{12}\\ =0.01\end{array}$

Calculation:

The total amount paid comparing with the actual plan is,

  $30\text{ years}\times 12\text{ months}\times \$884.61=\$318,459.60$

From part (a) it is known that the payment is increased by $\$1050$ is paid by,

  $n=171.8\text{ payments}$

So the total amount for the increased payment is,

  $171.8\text{ payments}\times \$1050=\$180,390.00$

Thus, the total amount saved is,

  $\$318,459.60-\$180,390.00=\$138,069,60$

Therefore, the amount get saved compared to the original plan is $\$138,069.60$ .