Question: Brad decides to purchase a $295,000 house. He wants to finance the entire balance. He has received…

Brad decides to purchase a $295,000 house. He wants to finance the entire balance. He has received an APR of 4.1% for a 30-year mortgage. What is Brad’s monthly payment? Round your answer to the nearest hundredth.

Formulas

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

where:

• \( P \) is the principal amount ($295,000)
• \( r \) is the monthly interest rate (4.1% / 12)
• \( n \) is the number of payments (30 years * 12 months/year)

Solution

Brad wants to calculate his monthly mortgage payment for a $295,000 house with an APR of 4.1% over a 30-year term. First, identify the principal amount \( P \), the annual interest rate, and the loan term. \[ P = \$295{,}000 & \] Convert the annual interest rate to a monthly interest rate \( r \): \[ r = \frac{4.1\%}{12} = \frac{0.041}{12} \approx 0.0034167 & \] Calculate the total number of monthly payments \( n \): \[ n = 30 \text{ years} \times 12 \text{ months/year} = 360 \text{ payments} & \] Use the mortgage payment formula: \[ M = P \times \frac{r(1 + r)^n}{(1 + r)^n - 1} & \] Plugging in the values: \[ M = 295{,}000 \times \frac{0.0034167 \times (1 + 0.0034167)^{360}}{(1 + 0.0034167)^{360} - 1} & \] Calculating the numerator and the denominator: \[ (1 + 0.0034167)^{360} \approx 3.4677 & \] \[ M = 295{,}000 \times \frac{0.0034167 \times 3.4677}{3.4677 - 1} \approx 295{,}000 \times \frac{0.011821}{2.4677} \approx 295{,}000 \times 0.004796 \approx 1{,}415.57 & \] Brad’s monthly payment is approximately \$1,415.57.