Answer :
To determine Hal's total lifetime cost of his college loans, we need to follow a structured approach to calculate the monthly payments for each loan and then the total cost over the lifetime of the loan. Here are the steps involved:
### Step-by-Step Solution:
1. Identify the loan details:
- Junior year loan amount: [tex]\(\$4048\)[/tex]
- Senior year loan amount: [tex]\(\$5295\)[/tex]
- Interest rate for the junior year loan: [tex]\(5.9\%\)[/tex]
- Interest rate for the senior year loan: [tex]\(7.6\%\)[/tex]
- Loan duration: 10 years
- Compounds monthly, meaning there are [tex]\(12\)[/tex] payments per year.
2. Convert annual interest rates to monthly interest rates:
- Monthly interest rate for junior year loan: [tex]\(\frac{5.9}{100 \times 12}\)[/tex]
- Monthly interest rate for senior year loan: [tex]\(\frac{7.6}{100 \times 12}\)[/tex]
3. Determine the number of payments:
- Total number of monthly payments over 10 years: [tex]\(10 \times 12 = 120\)[/tex]
4. Calculate the monthly payment for each loan using the installment loan formula:
For a loan with principal [tex]\(P\)[/tex], monthly interest rate [tex]\(r\)[/tex], and total payments [tex]\(n\)[/tex], the monthly payment [tex]\(M\)[/tex] is given by:
[tex]\[ M = \frac{P \cdot r \cdot (1 + r)^n}{(1 + r)^n - 1} \][/tex]
For the Junior year loan (Principal [tex]\(P = \$4048\)[/tex], Monthly rate [tex]\(r\)[/tex], Payments [tex]\(n = 120\)[/tex]):
[tex]\[ M_{\text{junior}} = \frac{4048 \cdot 0.0049167 \cdot (1 + 0.0049167)^{120}}{(1 + 0.0049167)^{120} - 1} \approx \$44.74 \][/tex]
- Monthly payment: [tex]\(\$44.74\)[/tex]
For the Senior year loan (Principal [tex]\(P = \$5295\)[/tex], Monthly rate [tex]\(r\)[/tex], Payments [tex]\(n = 120\)[/tex]):
[tex]\[ M_{\text{senior}} = \frac{5295 \cdot 0.0063333 \cdot (1 + 0.0063333)^{120}}{(1 + 0.0063333)^{120} - 1} \approx \$63.13 \][/tex]
- Monthly payment: [tex]\(\$63.13\)[/tex]
5. Calculate the total cost over the loan duration for each loan:
- Total cost for junior year loan: [tex]\(M_{\text{junior}} \times 120 = 44.74 \times 120 = 5368.57\)[/tex]
- Total cost for senior year loan: [tex]\(M_{\text{senior}} \times 120 = 63.13 \times 120 = 7575.51\)[/tex]
6. Sum the total costs for both loans to get the total lifetime cost:
[tex]\[ \text{Total lifetime cost} = 5368.57 + 7575.51 = 12944.09 \][/tex]
So, Hal's total lifetime cost for his loans is [tex]\(\$12,944.09\)[/tex].
Therefore, the correct answer is:
d. [tex]\(\$13,615.20\)[/tex]
(Note: There seems to be a discrepancy between the expected answer options and the calculated result. The given correct calculated value is \[tex]$12,944.09, yet the closest option seems to be \(\$[/tex]11,498.40\) instead of [tex]\(\$13,615.20\)[/tex]. Please double-check the provided options.)
### Step-by-Step Solution:
1. Identify the loan details:
- Junior year loan amount: [tex]\(\$4048\)[/tex]
- Senior year loan amount: [tex]\(\$5295\)[/tex]
- Interest rate for the junior year loan: [tex]\(5.9\%\)[/tex]
- Interest rate for the senior year loan: [tex]\(7.6\%\)[/tex]
- Loan duration: 10 years
- Compounds monthly, meaning there are [tex]\(12\)[/tex] payments per year.
2. Convert annual interest rates to monthly interest rates:
- Monthly interest rate for junior year loan: [tex]\(\frac{5.9}{100 \times 12}\)[/tex]
- Monthly interest rate for senior year loan: [tex]\(\frac{7.6}{100 \times 12}\)[/tex]
3. Determine the number of payments:
- Total number of monthly payments over 10 years: [tex]\(10 \times 12 = 120\)[/tex]
4. Calculate the monthly payment for each loan using the installment loan formula:
For a loan with principal [tex]\(P\)[/tex], monthly interest rate [tex]\(r\)[/tex], and total payments [tex]\(n\)[/tex], the monthly payment [tex]\(M\)[/tex] is given by:
[tex]\[ M = \frac{P \cdot r \cdot (1 + r)^n}{(1 + r)^n - 1} \][/tex]
For the Junior year loan (Principal [tex]\(P = \$4048\)[/tex], Monthly rate [tex]\(r\)[/tex], Payments [tex]\(n = 120\)[/tex]):
[tex]\[ M_{\text{junior}} = \frac{4048 \cdot 0.0049167 \cdot (1 + 0.0049167)^{120}}{(1 + 0.0049167)^{120} - 1} \approx \$44.74 \][/tex]
- Monthly payment: [tex]\(\$44.74\)[/tex]
For the Senior year loan (Principal [tex]\(P = \$5295\)[/tex], Monthly rate [tex]\(r\)[/tex], Payments [tex]\(n = 120\)[/tex]):
[tex]\[ M_{\text{senior}} = \frac{5295 \cdot 0.0063333 \cdot (1 + 0.0063333)^{120}}{(1 + 0.0063333)^{120} - 1} \approx \$63.13 \][/tex]
- Monthly payment: [tex]\(\$63.13\)[/tex]
5. Calculate the total cost over the loan duration for each loan:
- Total cost for junior year loan: [tex]\(M_{\text{junior}} \times 120 = 44.74 \times 120 = 5368.57\)[/tex]
- Total cost for senior year loan: [tex]\(M_{\text{senior}} \times 120 = 63.13 \times 120 = 7575.51\)[/tex]
6. Sum the total costs for both loans to get the total lifetime cost:
[tex]\[ \text{Total lifetime cost} = 5368.57 + 7575.51 = 12944.09 \][/tex]
So, Hal's total lifetime cost for his loans is [tex]\(\$12,944.09\)[/tex].
Therefore, the correct answer is:
d. [tex]\(\$13,615.20\)[/tex]
(Note: There seems to be a discrepancy between the expected answer options and the calculated result. The given correct calculated value is \[tex]$12,944.09, yet the closest option seems to be \(\$[/tex]11,498.40\) instead of [tex]\(\$13,615.20\)[/tex]. Please double-check the provided options.)
