Let's break down the problem and solve it step by step: --- ### **Step 1: Find the Purchase Price and Down Payment** - **Purchase price:** \$1,250,000 - **Down payment:** 20% of \$1,250,000 \[ \text{Down payment} = 0.20 \times 1,250,000 = \$250,000 \] - **Loan amount:** \[ \text{Loan amount} = 1,250,000 - 250,000 = \$1,000,000 \] --- ### **Step 2: Find Monthly Payment (\(M\))** A conventional fixed-rate mortgage monthly payment is calculated as: \[ M = P \cdot \frac{r(1+r)^n}{(1+r)^n-1} \] Where: - \(P\) = loan amount = \$1,000,000 - \(r\) = monthly interest rate = \(6.0\% \div 12 = 0.5\% = 0.005\) - \(n\) = total payments = \(15 \times 12 = 180\) Plug in the numbers: \[ M = 1,000,000 \cdot \frac{0.005(1+0.005)^{180}}{(1+0.005)^{180} - 1} \] Calculate \((1+0.005)^{180}\): Let's do this step-by-step: - \(\ln(1.005) \approx 0.004987\) - \(180 \times 0.004987 \approx 0.8977\) - \(e^{0.8977} \approx 2.453\) So, \((1.005)^{180} \approx 2.453\) Now, plug these into the formula: \[ M = 1,000,000 \cdot \frac{0.005 \times 2.453}{2.453 - 1} \] \[ = 1,000,000 \cdot \frac{0.012265}{1.453} \] \[ = 1,000,000 \cdot 0.008443 \] \[ = \$8,443 \text{ (rounded to nearest dollar)} \] --- ### **Step 3: Find Remaining Balances at Year 5 and Year 10** #### **A. Calculate balance after \(k\) payments:** The formula for remaining loan balance after \(k\) payments: \[ B_k = P \cdot \frac{(1+r)^n - (1+r)^k}{(1+r)^n - 1} \] - After 5 years: \(k_1 = 5 \times 12 = 60\) - After 10 years: \(k_2 = 10 \times 12 = 120\) \[ (1.005)^{60} \approx e^{60 \times 0.004987} = e^{0.2992} \approx 1.3488 \] \[ (1.005)^{120} \approx e^{120 \times 0.004987} = e^{0.5984} \approx 1.8194 \] Now plug in: #### **After 5 years (\(k = 60\)):** \[ B_{60} = 1,000,000 \cdot \frac{2.453 - 1.3488}{2.453 - 1} \] \[ = 1,000,000 \cdot \frac{1.1042}{1.453} \] \[ = 1,000,000 \cdot 0.760\ (\text{rounded to three decimals}) \] \[ = \$760,000 \] #### **After 10 years (\(k = 120\)):** \[ B_{120} = 1,000,000 \cdot \frac{2.453 - 1.8194}{2.453 - 1} \] \[ = 1,000,000 \cdot \frac{0.6336}{1.453} \] \[ = 1,000,000 \cdot 0.436\ (\text{rounded to three decimals}) \] \[ = \$436,000 \] --- ### **Step 4: Calculate Total Payments Made between Years 5 and 10** - Number of months between year 5 and year 10: \(120 - 60 = 60\) months - Total payment over these 60 months: \[ \text{Total payments} = 60 \times 8,443 = \$506,580 \] --- ### **Step 5: Calculate Principal Paid between Years 5 and 10** - Principal at end of year 5: \$760,000 - Principal at end of year 10: \$436,000 - Principal paid = \$760,000 - \$436,000 = \$324,000 --- ### **Step 6: Calculate Total Interest Paid between Years 5 and 10** \[ \text{Total interest paid} = \text{Total payments} - \text{Principal paid} \] \[ = 506,580 - 324,000 = \$182,580 \] --- ## **Final Answer** \[ \boxed{\$182,580} \] **Total interest paid between year 5 and year 10 is approximately \$182,580.** --- **If you want more exactness, you can use a financial calculator or spreadsheet for precise amortization, but this estimate is within a few hundred dollars.**