1. **State the problem:** A student borrows 53900 at an annual interest rate of 10.8% compounded monthly, to be paid over 30 years. We need to find the monthly payment and total interest paid. 2. **Formula used:** The monthly payment for a loan with compound interest is given by the amortization formula: $$P = \frac{r \times PV}{1 - (1 + r)^{-n}}$$ where: - $P$ is the monthly payment, - $r$ is the monthly interest rate (annual rate divided by 12), - $PV$ is the present value or loan amount, - $n$ is the total number of payments (months). 3. **Calculate parameters:** - Annual interest rate = 10.8% = 0.108 - Monthly interest rate $r = \frac{0.108}{12} = 0.009$ - Number of payments $n = 30 \times 12 = 360$ - Loan amount $PV = 53900$ 4. **Calculate monthly payment:** $$P = \frac{0.009 \times 53900}{1 - (1 + 0.009)^{-360}}$$ 5. **Calculate denominator:** $$1 - (1 + 0.009)^{-360} = 1 - (1.009)^{-360}$$ Calculate $(1.009)^{-360}$: $$ (1.009)^{360} = e^{360 \times \ln(1.009)} \approx e^{360 \times 0.00896} = e^{3.2256} \approx 25.2$$ So, $$ (1.009)^{-360} = \frac{1}{25.2} \approx 0.03968$$ 6. **Substitute back:** $$1 - 0.03968 = 0.96032$$ 7. **Calculate numerator:** $$0.009 \times 53900 = 485.1$$ 8. **Calculate payment:** $$P = \frac{485.1}{0.96032} \approx 505.3$$ 9. **Monthly payment:** $505.30$ (rounded to nearest cent) 10. **Calculate total amount paid:** $$\text{Total paid} = P \times n = 505.3 \times 360 = 181908$$ 11. **Calculate total interest paid:** $$\text{Interest} = \text{Total paid} - PV = 181908 - 53900 = 128008$$ **Final answers:** - Monthly payment = $505.30$ - Total interest paid = $128008$