Hal has just graduated from four years of college. For the last two years, he took out a Stafford loan to pay for his tuition. Each loan had a duration of ten years and interest compounded monthly. Hal will pay each of them back by making monthly payments, starting as he graduates. Hal's loans are detailed in the table below.

| Year | Loan Amount (\[tex]$) | Interest Rate (\%) | Subsidized? |
|-------|-------------------|-------------------|-------------|
| Junior| 4,048 | 5.9 | N |
| Senior| 5,295 | 7.6 | Y |

Once all of his loans are paid off, what will Hal's total lifetime cost be? Round all dollar values to the nearest cent.

A. \$[/tex]9,023.28
B. \[tex]$8,467.20
C. \$[/tex]11,498.40
D. \$13,615.20



Answer :

To determine Hal's total lifetime cost for his loans, we need to calculate the total amount he will pay for each loan over the duration of the repayment period. Each loan has its own principal amount, interest rate, and a duration of 10 years with monthly compounding interest. We'll use the formula for the monthly payment of an amortizing loan to calculate the payments and then sum up the total cost.

### Junior Year Loan
- Loan Amount (P): \[tex]$4048 - Interest Rate (r): 5.9% annually, or 5.9 / 100 / 12 = 0.0049167 monthly - Duration (n): 10 years, or 10 * 12 = 120 months Using the amortizing loan formula: \[ \text{Monthly Payment} = P \cdot \frac{r \cdot (1 + r)^n}{(1 + r)^n - 1} \] Plugging in the values for the junior year loan: \[ P_{\text{junior}} = 4048 \] \[ r_{\text{junior}} = 0.0049167 \] \[ n = 120 \] Calculating the monthly payment: \[ \text{Monthly Payment}_{\text{junior}} = 4048 \cdot \frac{0.0049167 \cdot (1 + 0.0049167)^{120}}{(1 + 0.0049167)^{120} - 1} \approx 44.7380872 \] Total payment over 120 months: \[ \text{Total Payment}_{\text{junior}} = 44.7380872 \cdot 120 \approx 5368.57 \] ### Senior Year Loan - Loan Amount (P): \$[/tex]5295
- Interest Rate (r): 7.6% annually, or 7.6 / 100 / 12 = 0.0063333 monthly
- Duration (n): 10 years, or 10 * 12 = 120 months

Using the amortizing loan formula:
[tex]\[ \text{Monthly Payment} = P \cdot \frac{r \cdot (1 + r)^n}{(1 + r)^n - 1} \][/tex]

Plugging in the values for the senior year loan:
[tex]\[ P_{\text{senior}} = 5295 \][/tex]
[tex]\[ r_{\text{senior}} = 0.0063333 \][/tex]
[tex]\[ n = 120 \][/tex]

Calculating the monthly payment:
[tex]\[ \text{Monthly Payment}_{\text{senior}} = 5295 \cdot \frac{0.0063333 \cdot (1 + 0.0063333)^{120}}{(1 + 0.0063333)^{120} - 1} \approx 63.1292878 \][/tex]

Total payment over 120 months:
[tex]\[ \text{Total Payment}_{\text{senior}} = 63.1292878 \cdot 120 \approx 7575.51 \][/tex]

### Total Lifetime Cost
Sum of the total junior and senior payments:
[tex]\[ \text{Total Lifetime Cost} = 5368.57 + 7575.51 = 12944.08 \][/tex]

Rounding to the nearest cent:
[tex]\[ \text{Total Lifetime Cost} = 12944.09 \][/tex]

Thus, once all of his loans are paid off, Hal's total lifetime cost will be approximately \[tex]$12944.09. The correct answer is \(d. \$[/tex]13,615.20\). However, based on our calculation of [tex]\( \$12944.09 \)[/tex] and the closest option provided, the proper answer should have been:
[tex]\[ \boxed{\$12944.09} \][/tex]

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