a) Hand in in class Fri Oct. 4: The Turners have purchased a house for \( \$ 150,000 \). They made an initial down payment of \( 20 \% \) and secured a mortgage with interest charged at a rate of \( 3.5 \% / \) year on the unpaid balance. Assume that the loan is amortized over 30 years. (Round all answers to the nearest cent.) a) How much is the down payment? b) how much is borrowed from the bank? c) What monthly payment will the Turners be required to make? d) What will be their total price of the house? e) How much of the total amount is interest?

To solve the problem, we will go through each part step by step. ### a) How much is the down payment? The down payment is calculated as a percentage of the purchase price of the house. \[ \text{Down Payment} = \text{Purchase Price} \times \text{Down Payment Percentage} \] Given: - Purchase Price = \( \$150,000 \) - Down Payment Percentage = \( 20\% = 0.20 \) \[ \text{Down Payment} = 150,000 \times 0.20 = 30,000 \] **Down Payment: \( \$30,000 \)** ### b) How much is borrowed from the bank? The amount borrowed is the purchase price minus the down payment. \[ \text{Amount Borrowed} = \text{Purchase Price} - \text{Down Payment} \] \[ \text{Amount Borrowed} = 150,000 - 30,000 = 120,000 \] **Amount Borrowed: \( \$120,000 \)** ### c) What monthly payment will the Turners be required to make? To calculate the monthly payment for a fixed-rate mortgage, we can use the formula: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] Where: - \( M \) = monthly payment - \( P \) = principal (amount borrowed) - \( r \) = monthly interest rate (annual rate / 12) - \( n \) = number of payments (loan term in months) Given: - \( P = 120,000 \) - Annual interest rate = \( 3.5\% = 0.035 \) - Monthly interest rate \( r = \frac{0.035}{12} \approx 0.00291667 \) - Loan term = 30 years = \( 30 \times 12 = 360 \) months Now, substituting the values into the formula: \[ M = 120,000 \frac{0.00291667(1 + 0.00291667)^{360}}{(1 + 0.00291667)^{360} - 1} \] Calculating \( (1 + r)^{360} \): \[ (1 + 0.00291667)^{360} \approx 2.89828 \] Now substituting back into the formula: \[ M = 120,000 \frac{0.00291667 \times 2.89828}{2.89828 - 1} \] Calculating the numerator and denominator: \[ M = 120,000 \frac{0.008466}{1.89828} \approx 120,000 \times 0.00446 \approx 535.20 \] **Monthly Payment: \( \$535.20 \)** ### d) What will be their total price of the house? The total price of the house is the sum of the total payments made over the life of the loan plus the down payment. \[ \text{Total Payments} = M \times n \] \[ \text{Total Payments} = 535.20 \times 360 \approx 192,672 \] Now, adding the down payment: \[ \text{Total Price of the House} = \text{Total Payments} + \text{Down Payment} \] \[ \text{Total Price of the House} = 192,672 + 30,000 = 222,672 \] **Total Price of the House: \( \$222,672 \)** ### e) How much of the total amount is interest? To find the total interest paid, we subtract the amount borrowed from the total payments made. \[ \text{Total Interest} = \text{Total Payments} - \text{Amount Borrowed} \] \[ \text{Total Interest} = 192,672 - 120,000 = 72,672 \] **Total Interest: \( \$72,672 \)** ### Summary of Answers: a) Down Payment: \( \$30,000 \) b) Amount Borrowed: \( \$120,000 \) c) Monthly Payment: \( \$535.20 \) d) Total Price of the House: \( \$222,672 \) e) Total Interest: \( \$72,672 \)