1. **Problem 2a:** Calculate the time to pay off the mortgage. Given: - House price = 300000 - Down payment = 25% of 300000 = 0.25 \times 300000 = 75000 - Loan amount (PV) = 300000 - 75000 = 225000 - Monthly payment (PMT) = 1500 - Interest rate = 4.15% compounded semi-annually 2. **Convert interest rate to monthly effective rate:** Annual nominal rate compounded semi-annually: 4.15% or 0.0415 Effective semi-annual rate = 0.0415 / 2 = 0.02075 Effective annual rate = \left(1 + 0.02075\right)^2 - 1 = 0.0419 (approx) Monthly interest rate (i) = \left(1 + 0.0419\right)^{\frac{1}{12}} - 1 \approx 0.00342 3. **Use the amortization formula to find number of months (N):** Formula for loan amortization: $$PV = PMT \times \frac{1 - (1 + i)^{-N}}{i}$$ Rearranged to solve for N: $$1 - (1 + i)^{-N} = \frac{PV \times i}{PMT}$$ $$ (1 + i)^{-N} = 1 - \frac{PV \times i}{PMT}$$ $$ -N \ln(1 + i) = \ln\left(1 - \frac{PV \times i}{PMT}\right)$$ $$ N = - \frac{\ln\left(1 - \frac{PV \times i}{PMT}\right)}{\ln(1 + i)}$$ Substitute values: $$ N = - \frac{\ln\left(1 - \frac{225000 \times 0.00342}{1500}\right)}{\ln(1 + 0.00342)}$$ Calculate numerator inside log: $$ \frac{225000 \times 0.00342}{1500} = \frac{769.5}{1500} = 0.513$$ So: $$ N = - \frac{\ln(1 - 0.513)}{\ln(1.00342)} = - \frac{\ln(0.487)}{0.00341}$$ Calculate logs: $$ \ln(0.487) \approx -0.719$$ $$ N = - \frac{-0.719}{0.00341} = 210.8 \text{ months}$$ 4. **Convert months to years and months:** $$ 210.8 \text{ months} = 17 \text{ years and } 6.8 \text{ months} \approx 17 \text{ years and } 7 \text{ months}$$ --- 5. **Problem 2b:** Calculate total interest paid. Total payments = PMT \times N = 1500 \times 210.8 = 316200 Interest paid = Total payments - Loan amount = 316200 - 225000 = 91200 --- 6. **Problem 3:** Calculate monthly deposit for Michael's retirement. Given: - Withdrawal amount per quarter (PMT) = 2500 - Withdrawal period = 20 years = 80 quarters - Interest rate = 5% compounded monthly - Saving period = 10 years = 120 months 7. **Calculate present value (PV) of withdrawals at retirement:** Quarterly interest rate: $$ i_q = \left(1 + \frac{0.05}{12}\right)^3 - 1 \approx (1 + 0.004167)^3 - 1 = 0.01255$$ PV of withdrawals: $$ PV = PMT \times \frac{1 - (1 + i_q)^{-80}}{i_q}$$ Calculate: $$ (1 + 0.01255)^{-80} = (1.01255)^{-80} \approx 0.364$$ So: $$ PV = 2500 \times \frac{1 - 0.364}{0.01255} = 2500 \times \frac{0.636}{0.01255} = 2500 \times 50.68 = 126700$$ 8. **Calculate monthly deposit (PMT) to accumulate PV in 10 years:** Monthly interest rate: $$ i_m = \frac{0.05}{12} = 0.004167$$ Number of months: $$ N = 120$$ Formula for future value of annuity: $$ FV = PMT \times \frac{(1 + i_m)^N - 1}{i_m}$$ We want FV = PV = 126700, solve for PMT: $$ PMT = \frac{PV \times i_m}{(1 + i_m)^N - 1}$$ Calculate: $$ (1 + 0.004167)^{120} = 1.647$$ So: $$ PMT = \frac{126700 \times 0.004167}{1.647 - 1} = \frac{528}{0.647} = 816.1$$ **Michael must deposit approximately 816 each month.**