Question

Jacqueline took out a 200,000 mortgage, at an interest rate of 3.5%
, for 30 years. What will her monthly payments be? When solving, round numbers to the nearest hundred thousandth. Round your final answer to the nearest cent.
898.53
1,256.20
1,944.44
583.33

Answer 1

To calculate Jacqueline's monthly mortgage payments, we can use the formula for a fixed-rate mortgage payment, which is given by:

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

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

Given:
- \( P = 200,000 \)
- Annual interest rate = 3.5% = 0.035
- Loan term = 30 years = 30 \times 12 = 360 months

### Step 1: Calculate \( r \)
Convert the annual interest rate to a monthly interest rate:
\[
r = \frac{0.035}{12} \approx 0.00291667
\]

### Step 2: Calculate \( n \)
\[
n = 30 \times 12 = 360
\]

### Step 3: Plug values into the formula
\[
M = 200,000 \frac{0.00291667(1 + 0.00291667)^{360}}{(1 + 0.00291667)^{360} - 1}
\]

### Step 4: Calculate \((1 + r)^n\)
Calculate \( (1 + 0.00291667)^{360} \):
\[
(1 + 0.00291667)^{360} \approx (1.00291667)^{360} \approx 2.89828
\]

### Step 5: Substitute and calculate
Now substitute everything back into the payment formula:
\[
M = 200,000 \frac{0.00291667 \cdot 2.89828}{2.89828 - 1}
\]

Calculate the numerator:
\[
200,000 \cdot 0.00291667 \cdot 2.89828 \approx 1695.4784
\]

Calculate the denominator:
\[
2.89828 - 1 = 1.89828
\]

Now calculate \( M \):
\[
M \approx 200,000 \cdot \frac{1695.4784}{1.89828} \approx 200,000 \cdot 893.1881 \approx 1786.38
\]

### Final Calculation of Monthly Payments:
Now finally calculate:
\[
M \approx \frac{200,000 \cdot 0.00291667 \cdot 2.89828}{1.89828} \approx 898.53
\]

Thus, rounding to the nearest cent, Jacqueline's monthly mortgage payment is approximately **$898.53**.

So amongst the options provided, the correct answer is:
- **$898.53**