To calculate the remaining balance on a mortgage after a certain period, we can use the formula for the remaining balance on an amortizing loan. The formula for the remaining balance \( B \) after \( n \) payments is given by:
\[ B = P \times \frac{(1 + r)^N - (1 + r)^n}{(1 + r)^N - 1} \]
where:
- \( P \) is the principal of the loan (initial mortgage amount),
- \( r \) is the monthly interest rate,
- \( N \) is the total number of payments,
- \( n \) is the number of payments made.
Given the details of the mortgage:
- Principal (\( P \)) = $251,200.00
- Annual interest rate (APR) = 5.76% = 0.0576
- Monthly interest rate (\( r \)) = \( \frac{0.0576}{12} = 0.0048 \)
- Total payments (\( N \)) = 30 years × 12 months/year = 360 months
- Payments made (\( n \)) = 8 years × 12 months/year = 96 months
Now, we can substitute these values into the formula:
\[ B = 251200 \times \frac{(1 + 0.0048)^{360} - (1 + 0.0048)^{96}}{(1 + 0.0048)^{360} - 1} \]
First, we compute \( (1 + r)^{N} \) and \( (1 + r)^{n} \):
- \( (1 + 0.0048)^{360} \)
- \( (1 + 0.0048)^{96} \)
Using a calculator:
- \( (1 + 0.0048)^{360} \approx 5.061295 \)
- \( (1 + 0.0048)^{96} \approx 1.487708 \)
Substituting these values back into the balance formula gives:
\[ B = 251200 \times \frac{5.061295 - 1.487708}{5.061295 - 1} \]
Calculating the numerator:
\[ 5.061295 - 1.487708 \approx 3.573587 \]
And the denominator:
\[ 5.061295 - 1 \approx 4.061295 \]
Now substituting these results into the equation:
\[ B = 251200 \times \frac{3.573587}{4.061295} \approx 251200 \times 0.879128 \approx 220,925.04 \]
Thus, the family’s remaining balance on the mortgage after 8 years will be approximately $220,925.04.