Question

Derek borrows $38,179.00 to buy a car. He will make monthly payments for 6 years. The car loan has an interest rate of 5.17%. What will the payments be?

Answer 1

To calculate the monthly payments for a loan, you can use the formula for an amortizing loan:

\[
M = P \times \frac{r(1 + r)^n}{(1 + r)^n - 1}
\]

Where:
- \( M \) is the monthly payment
- \( P \) is the loan amount (principal)
- \( r \) is the monthly interest rate (annual interest rate divided by 12)
- \( n \) is the number of payments (loan term in months)

In this case:
- \( P = 38,179.00 \)
- Annual interest rate = 5.17\%, so the monthly interest rate \( r = \frac{5.17}{100 \times 12} \approx 0.00430833 \)
- The loan term is 6 years, so \( n = 6 \times 12 = 72 \)

Now, let's substitute these values into the formula:

1. Calculate the monthly interest rate:
\[
r = \frac{5.17}{100} \div 12 \approx 0.00430833
\]

2. Calculate \( (1 + r)^n \):
\[
(1 + r)^{72} = (1 + 0.00430833)^{72}
\]

Calculating this gives approximately:
\[
(1 + 0.00430833)^{72} \approx 1.348835395
\]

3. Substitute the values into the mortgage payment formula:

\[
M = 38,179 \times \frac{0.00430833 \times 1.348835395}{1.348835395 - 1}
\]

Calculating the numerator:

\[
0.00430833 \times 1.348835395 \approx 0.0058170496
\]

And calculating the denominator:

\[
1.348835395 - 1 \approx 0.348835395
\]

Now substitute these values into the formula for \( M \):

\[
M \approx 38,179 \times \frac{0.0058170496}{0.348835395}
\]

Calculating this gives:

\[
M \approx 38,179 \times 0.01666889076 \approx 636.71
\]

Thus, the monthly payment for Derek’s car loan will be approximately **$636.71**.