Question
If you take out a loan for $350,000 for 15 years at an interest rate of 6%, how much interest will you pay?
Answers
Answered by
GPT-4o mini
To calculate the total interest paid on a loan, you can use the formula for the monthly payment and then calculate the total payments over the life of the loan.
1. **Calculate the monthly payment using the loan amortization formula**:
\[
M = P \frac{r(1+r)^n}{(1+r)^n - 1}
\]
where:
- \( M \) = monthly payment
- \( P \) = loan amount ($350,000)
- \( r \) = monthly interest rate (annual rate divided by 12 months)
- \( n \) = number of payments (loan term in months)
Given:
- Annual interest rate = 6%, so monthly rate \( r = \frac{0.06}{12} = 0.005 \)
- Loan amount \( P = 350,000 \)
- Loan term = 15 years = 15 * 12 = 180 months
2. **Substituting the values into the formula**:
\[
M = 350000 \frac{0.005(1+0.005)^{180}}{(1+0.005)^{180} - 1}
\]
First, calculate \( (1+0.005)^{180} \):
\[
(1.005)^{180} \approx 2.4596
\]
Now substituting back to find \( M \):
\[
M = 350000 \frac{0.005 \times 2.4596}{2.4596 - 1}
\]
\[
M = 350000 \frac{0.012298}{1.4596} \approx 350000 \times 0.008414 \approx 2944.80
\]
The monthly payment is approximately **$2,944.80**.
3. **Calculate the total paid over the life of the loan**:
\[
\text{Total payment} = M \times n = 2944.80 \times 180 \approx 530,064
\]
4. **Calculate the total interest paid**:
\[
\text{Total interest} = \text{Total payment} - P = 530064 - 350000 = 180064
\]
Therefore, the total interest paid over the life of the loan will be approximately **$180,064**.
1. **Calculate the monthly payment using the loan amortization formula**:
\[
M = P \frac{r(1+r)^n}{(1+r)^n - 1}
\]
where:
- \( M \) = monthly payment
- \( P \) = loan amount ($350,000)
- \( r \) = monthly interest rate (annual rate divided by 12 months)
- \( n \) = number of payments (loan term in months)
Given:
- Annual interest rate = 6%, so monthly rate \( r = \frac{0.06}{12} = 0.005 \)
- Loan amount \( P = 350,000 \)
- Loan term = 15 years = 15 * 12 = 180 months
2. **Substituting the values into the formula**:
\[
M = 350000 \frac{0.005(1+0.005)^{180}}{(1+0.005)^{180} - 1}
\]
First, calculate \( (1+0.005)^{180} \):
\[
(1.005)^{180} \approx 2.4596
\]
Now substituting back to find \( M \):
\[
M = 350000 \frac{0.005 \times 2.4596}{2.4596 - 1}
\]
\[
M = 350000 \frac{0.012298}{1.4596} \approx 350000 \times 0.008414 \approx 2944.80
\]
The monthly payment is approximately **$2,944.80**.
3. **Calculate the total paid over the life of the loan**:
\[
\text{Total payment} = M \times n = 2944.80 \times 180 \approx 530,064
\]
4. **Calculate the total interest paid**:
\[
\text{Total interest} = \text{Total payment} - P = 530064 - 350000 = 180064
\]
Therefore, the total interest paid over the life of the loan will be approximately **$180,064**.