1. **Problem Statement:** A man buys a house for 275000. He pays 70000 down and takes a mortgage at 6.6% annual interest on the balance. We need to find the monthly payment and total interest for mortgage lengths of (a) 15 years, (b) 20 years, (c) 25 years, and (d) find when half the 20-year loan is paid off. 2. **Given:** - House price = 275000 - Down payment = 70000 - Loan principal $P = 275000 - 70000 = 205000$ - Annual interest rate $r = 6.6\% = 0.066$ - Monthly interest rate $i = \frac{0.066}{12} = 0.0055$ - Number of payments $n = \text{years} \times 12$ 3. **Formula for monthly payment $M$ on a fixed-rate mortgage:** $$ M = P \times \frac{i(1+i)^n}{(1+i)^n - 1} $$ This formula calculates the fixed monthly payment to fully amortize the loan over $n$ months. 4. **Calculate for each case:** (a) 15 years: $n = 15 \times 12 = 180$ $$ M = 205000 \times \frac{0.0055(1+0.0055)^{180}}{(1+0.0055)^{180} - 1} $$ Calculate $(1+0.0055)^{180} \approx 2.6533$ $$ M = 205000 \times \frac{0.0055 \times 2.6533}{2.6533 - 1} = 205000 \times \frac{0.014593}{1.6533} = 205000 \times 0.008828 = 1809.74 $$ Total paid = $1809.74 \times 180 = 325753.20$ Interest paid = $325753.20 - 205000 = 120753.20$ (b) 20 years: $n = 20 \times 12 = 240$ $$ (1+0.0055)^{240} \approx 3.6425 $$ $$ M = 205000 \times \frac{0.0055 \times 3.6425}{3.6425 - 1} = 205000 \times \frac{0.020034}{2.6425} = 205000 \times 0.007584 = 1555.68 $$ Total paid = $1555.68 \times 240 = 373363.20$ Interest paid = $373363.20 - 205000 = 168363.20$ (c) 25 years: $n = 25 \times 12 = 300$ $$ (1+0.0055)^{300} \approx 5.0001 $$ $$ M = 205000 \times \frac{0.0055 \times 5.0001}{5.0001 - 1} = 205000 \times \frac{0.0275005}{4.0001} = 205000 \times 0.006875 = 1409.38 $$ Total paid = $1409.38 \times 300 = 422814$ Interest paid = $422814 - 205000 = 217814$ 5. **(d) When is half the 20-year loan paid off?** Half the loan principal is $102500$. We find the time $t$ in months when the remaining balance equals $102500$. The remaining balance after $m$ payments is: $$ B_m = P \times \frac{(1+i)^n - (1+i)^m}{(1+i)^n - 1} $$ Set $B_m = 102500$, solve for $m$: $$ 102500 = 205000 \times \frac{(1.0055)^{240} - (1.0055)^m}{(1.0055)^{240} - 1} $$ Divide both sides by 205000: $$ 0.5 = \frac{3.6425 - (1.0055)^m}{3.6425 - 1} = \frac{3.6425 - (1.0055)^m}{2.6425} $$ Multiply both sides: $$ 0.5 \times 2.6425 = 3.6425 - (1.0055)^m $$ $$ 1.32125 = 3.6425 - (1.0055)^m $$ $$ (1.0055)^m = 3.6425 - 1.32125 = 2.32125 $$ Take natural log: $$ m \ln(1.0055) = \ln(2.32125) $$ $$ m = \frac{\ln(2.32125)}{\ln(1.0055)} = \frac{0.842}{0.00548} \approx 153.6 \text{ months} $$ Convert to years: $$ \frac{153.6}{12} = 12.8 \text{ years} $$ **Final answers:** - (a) Monthly payment = 1809.74, Interest = 120753.20 - (b) Monthly payment = 1555.68, Interest = 168363.20 - (c) Monthly payment = 1409.38, Interest = 217814 - (d) Half the 20-year loan is paid off after approximately 12.8 years.