Question

brady took out a loan for $2800 at an 18% APR, compounded monthly, to buy a computer. if he will make monthly payments of $201.50 to pay off the loan, how many total payments will have to make?

16
17
14
15

Answer 1

To determine how many monthly payments Brady will need to make to pay off his loan, we can use the formula for monthly payments on an amortized loan, which is given by:

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

Where:
- \( P \) is the monthly payment,
- \( PV \) is the present value (loan amount),
- \( r \) is the monthly interest rate,
- \( n \) is the number of payments.

We need to find \( n \) in this scenario:

1. **Convert the APR to a monthly interest rate**:
\[
\text{APR} = 18\% \Rightarrow r = \frac{18\%}{12} = 0.015
\]

2. **Plug in the values**:
\[
201.50 = \frac{0.015 \times 2800}{1 - (1 + 0.015)^{-n}}
\]

3. **Rearranging for \( n \)**:
\[
201.50 = \frac{42}{1 - (1.015)^{-n}}
\]
\[
201.50 \cdot [1 - (1.015)^{-n}] = 42
\]
\[
201.50 - 201.50 \cdot (1.015)^{-n} = 42
\]
\[
201.50 \cdot (1.015)^{-n} = 201.50 - 42
\]
\[
201.50 \cdot (1.015)^{-n} = 159.50
\]
\[
(1.015)^{-n} = \frac{159.50}{201.50}
\]
\[
(1.015)^{-n} \approx 0.7901
\]

4. **Taking the natural logarithm**:
\[
-n \cdot \ln(1.015) = \ln(0.7901)
\]
\[
n = \frac{\ln(0.7901)}{-\ln(1.015)}
\]

5. **Calculating the logarithms**:
- \(\ln(0.7901) \approx -0.2325\)
- \(\ln(1.015) \approx 0.0149\)

Therefore:
\[
n = \frac{-0.2325}{-0.0149} \approx 15.6
\]

Since \( n \) must be a whole number, we round up to the next whole number, which means Brady will need to make 16 monthly payments.

Thus, the answer is:
**16 payments**.