Find the monthly payment and total interest for a mortgage of \( \$ 96,500 \) at \( 6^{\frac{3}{4}} \% \) for 25 years, Select one: a. \( \$ 99,000.00 \) b. \( \$ 103,544.50 \) c. \( \$ 125,876.54 \) d. \( \$ 241,393.60 \)

We use the fixed‐payment mortgage formula \[ M=\frac{rP}{1-\left(1+r\right)^{-n}} \] where - \( M \) is the monthly payment, - \( P \) is the principal (\( \$96,\!500 \)), - \( r \) is the monthly interest rate, - \( n \) is the total number of payments. **Step 1. Calculate the monthly interest rate.** The annual interest rate is given as \(6^{\frac{3}{4}}\%\). Converting this to a decimal: \[ 6^{\frac{3}{4}}\% = 6.75\% = 0.0675. \] Thus, the monthly rate is \[ r=\frac{0.0675}{12}\approx 0.005625. \] **Step 2. Calculate the total number of payments.** For a 25-year term with monthly payments: \[ n=25\times 12=300. \] **Step 3. Substitute into the formula.** First, compute the numerator: \[ rP=0.005625\times 96,\!500\approx 542.8125. \] Next, compute the denominator: \[ 1-\left(1+r\right)^{-n}=1-\left(1.005625\right)^{-300}. \] We calculate \(\left(1.005625\right)^{300}\) using logarithms: \[ \ln\left(1.005625\right)\approx 0.005609, \quad 300\times 0.005609\approx 1.6827, \] \[ \left(1.005625\right)^{300}\approx e^{1.6827}\approx 5.376. \] Then, \[ \left(1.005625\right)^{-300}\approx \frac{1}{5.376}\approx 0.186. \] Thus, the denominator becomes \[ 1-0.186\approx 0.814. \] Now, the monthly payment is \[ M\approx\frac{542.8125}{0.814}\approx 667. \] So, the monthly payment is approximately \( \$667\). **Step 4. Calculate the total interest paid.** Total amount paid over 25 years is \[ \text{Total Payment}=667\times 300\approx \$200,\!100. \] Subtracting the principal gives the total interest: \[ \text{Total Interest}\approx 200,\!100-96,\!500\approx \$103,\!600. \] This value is very close to option (b) which is \( \$103,\!544.50 \). Thus, the answer is (b): \( \$103,\!544.50 \).