Question

Question41. Ingrid will start college next year. She wasapproved for 10-year unsubsidized Federal Loanfor the amount of $15,000 at 4.29%a) How much interest will Ingrid accrue for 4.5 years nonpayment period?Hence the interest Ingrid will accrued for 4.5 years non-payment period is $3,602.91b) What will the new principal be when she begins making loan payments?$18,602.91c) How much interest will she pay over the life of the loan? $9,201.713. Ingrid made her last monthly interest only payment on September 1 Her next payment is due on October 1. What will be the amount of interest only payment?4. Suppose Ingrid has decided to apply for a privateloan rather then a federal loan. She has beenapproved for a 10 year loan with APR of 7.8%a) What is her monthly payments?b) What is the total amount she will pay back?c) What is total interest amount?

Answer 1

.Explanation

Question 4

Part A

This part wants us to get Ingrid's monthly payments

To do so, let us first give the formula to be used

The next step will be to define all parameters to be used

In our case

`\begin{gathered} r=(APR)/(12)=\frac{7.8\text{ \%}}{12}=(0.078)/(12)=6.5*10^(-3) \\ n=since\text{ there are 12 months in a year, In 10 years, there will be 120 months, so that} \\ n=120 \\ P=\text{ \$15,000} \end{gathered}`

So, upon substituting the formula

`EMI=15000*(6.5*10^(-3)(1+6.5*10^(-3))^(120))/((1+6.5*10^(-3))^(120)-1)`

Simplifying, we will have

`EMI=\text{ \$15,000}*0.012027`
`EMI=\text{ \$180.41}`

Therefore, the monthly payment will be $180.41

Part B

To determine the total amount she will pay back, we will multiply her monthly payments by

the number of times she will pay back.

Since she has to pay $180.41 for 120 months, Her total payback amount will be

`\text{ \$180.41}*120=\text{ \$21,649.21}`

Hence, she will pay back $21,649.21

Part C

To determine the interest amount, we will simply subtract the amount of loan collected from the total payback amount

This will be

`\text{ \$21,649.21- \$15,000= \$6,649.21}`

Thus, the interest amount is $6,649.21