Loan Payment Ca39Db

1. **State the problem:** Calculate the monthly payment for a loan of $7100$ dollars at an annual interest rate of $6.5\%$ with a term of $10$ years. The student graduates $2$ years and $7$ months after the loan is acquired, and payments are deferred for $6$ months after graduation. 2. **Understand the timeline:** - Loan term: $10$ years = $120$ months. - Graduation time: $2$ years $7$ months = $31$ months. - Payment deferral after graduation: $6$ months. Total months before payments start = $31 + 6 = 37$ months. 3. **Calculate the loan balance at the time payments start:** Since payments are deferred, interest accrues during the $37$ months. The loan uses simple interest or compound interest? Usually, student loans accrue interest monthly compounded. Assuming monthly compounding: - Monthly interest rate $i = \frac{6.5}{100} \div 12 = 0.0054167$ (approx) - Number of months interest accrues before payments start $n = 37$ Future value of loan at payment start: $$ A = P(1 + i)^n = 7100 \times (1 + 0.0054167)^{37} $$ Calculate: $$ (1 + 0.0054167)^{37} \approx 1.214 $$ So, $$ A = 7100 \times 1.214 = 8619.4 $$ 4. **Calculate monthly payment over remaining term:** Payments start after $37$ months, so remaining term is $120 - 37 = 83$ months. Use the amortization formula for monthly payment $M$: $$ M = A \times \frac{i(1+i)^N}{(1+i)^N - 1} $$ where - $A = 8619.4$ - $i = 0.0054167$ - $N = 83$ Calculate: $$ (1+i)^N = (1.0054167)^{83} \approx 1.538 $$ Then numerator: $$ i(1+i)^N = 0.0054167 \times 1.538 = 0.00833 $$ Denominator: $$ (1+i)^N - 1 = 1.538 - 1 = 0.538 $$ So, $$ M = 8619.4 \times \frac{0.00833}{0.538} = 8619.4 \times 0.01549 = 133.5 $$ 5. **Final answer:** The monthly payment is approximately **$133.50$**. **Summary:** - Loan amount: $7100$ - Interest rate: $6.5\%$ annually compounded monthly - Payments deferred for $37$ months - Remaining term: $83$ months - Monthly payment: $\boxed{133.50}$