Use \( \frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} \) to determine the regular payment amount, rounded to the nearest dollar. Consider the following pair of mortgage loan options for a \( \$ 185,000 \) mortgage. Which mortgage loan has the larger total cost (closing costs + the amount paid for points + total cost of interest)? By how much? Mortgage A: 20-year fixed at \( 12.25 \% \) with closing costs of \( \$ 1600 \) and 1 point. Mortgage B: 20-year fixed at \( 10.5 \% \) with closing costs of \( \$ 1600 \) and 4 points. Choose the correct answer below, and fill in the answer box to complete your choice. (Do not round until the final answer. Then round to the nearest dollar as needed.) A. Mortgage B has a larger total cost than mortgage A by \( \$ \square \). B. Mortgage A has a larger total cost than mortgage B by \( \$ \square \).

To determine the regular payment amount for each mortgage, we will utilize the formula provided. First, let's calculate the monthly payment for both Mortgage A and Mortgage B. ### Mortgage A: - Principal \( P = 185,000 \) - Annual Interest Rate \( r = 12.25\% = 0.1225 \) - Number of Payments per Year \( n = 12 \) - Total Payments \( t = 20 \) Calculating \( \frac{r}{n} = \frac{0.1225}{12} \approx 0.01020833 \). Now, we apply the payment formula: \[ \text{Payment}_A = \frac{P\left(\frac{r}{n}\right)}{1-\left(1+\frac{r}{n}\right)^{-nt}} = \frac{185,000 \times 0.01020833}{1-\left(1+0.01020833\right)^{-240}} \] Calculating the denominator: \[ 1 - \left(1.01020833\right)^{-240} \approx 1 - 0.11506086 \approx 0.88493914 \] Now substituting back into the payment equation: \[ \text{Payment}_A \approx \frac{185,000 \times 0.01020833}{0.88493914} \approx \frac{1,891.54105}{0.88493914} \approx 2,136.77 \] Rounding gives a monthly payment of approximately **$2,137**. Now calculate the total cost of Mortgage A: \[ \text{Total Payments} = \text{Payment} \times 12 \times t = 2,137 \times 12 \times 20 = 512,880 \] Add closing costs and points (\( 1\% \) of \( 185,000 = 1,850 \)): \[ \text{Total Cost}_A = 512,880 + 1,600 + 1,850 = 516,330 \] ### Mortgage B: - Principal \( P = 185,000 \) - Annual Interest Rate \( r = 10.5\% = 0.105 \) Calculating \( \frac{r}{n} = \frac{0.105}{12} \approx 0.00875 \). Using the same payment formula: \[ \text{Payment}_B = \frac{185,000 \times 0.00875}{1-\left(1+0.00875\right)^{-240}} \] Calculating the denominator: \[ 1 - \left(1.00875\right)^{-240} \approx 1 - 0.21779468 \approx 0.78220532 \] Now substituting back into the payment equation: \[ \text{Payment}_B \approx \frac{185,000 \times 0.00875}{0.78220532} \approx \frac{1,618.75}{0.78220532} \approx 2,066.79 \] Rounding gives a monthly payment of approximately **$2,067**. Calculating the total cost of Mortgage B: \[ \text{Total Payments} = \text{Payment} \times 12 \times t = 2,067 \times 12 \times 20 = 496,080 \] Add closing costs and points (\( 4\% \) of \( 185,000 = 7,400 \)): \[ \text{Total Cost}_B = 496,080 + 1,600 + 7,400 = 505,080 \] ### Summary of Total Costs: - Total Cost A: **$516,330** - Total Cost B: **$505,080** Now, let's find out which mortgage has a larger total cost: \[ 516,330 - 505,080 = 11,250 \] So, the answer is: B. Mortgage A has a larger total cost than mortgage B by \(\$ 11,250\).