Answer:
The mortgage chosen is option A;
15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470
Step-by-step explanation:
The details of the home purchase are;
The price of the home = $275,000
The mode of purchase of the home = Mortgage
The percentage of the loan amount payed as down payment = 20%
The amount used as down payment for the loan = $55,000
The principal of the mortgage borrowed, P = The price of the house - The down payment
∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000
The principal of the mortgage, P = $220,000
The formula for the total amount paid which is the cost of the loan is given as follows;
![Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n - \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }](https://tex.z-dn.net/?f=Outstanding%20%5C%20Loan%20%5C%20Balance%20%3D%20%5Cdfrac%7BP%20%5Ccdot%20%5Cleft%5B%5Cleft%281%2B%5Cdfrac%7Br%7D%7B12%7D%20%5Cright%29%5En%20-%20%20%5Cleft%281%2B%5Cdfrac%7Br%7D%7B12%7D%20%5Cright%29%5Em%20%5Cright%5D%20%7D%7B1%20-%20%5Cleft%281%2B%5Cdfrac%7Br%7D%7B12%7D%20%5Cright%29%5En%20%7D)
The formula for monthly payment on a mortgage, 'M', is given as follows;
A. When the mortgage term, t = 15-years,
The interest rate, r = 3%
The number of months over which the loan is payed, n = 12·t
∴ n = 12 months/year × 15 years = 180 months
n = 180 months
The monthly payment, 'M', is given as follows;
M =
The total amount paid over the loan term = Cost of the mortgage
Therefore, we have;
220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)
![M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1} \approx 1,519.28](https://tex.z-dn.net/?f=M%20%3D%20%5Cdfrac%7B220%2C000%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B0.03%7D%7B12%7D%20%5Cright%29%20%5Ccdot%20%5Cleft%281%2B%5Cdfrac%7B0.03%7D%7B12%7D%20%5Cright%29%5E%7B180%7D%20%7D%7B%5Cleft%281%2B%5Cdfrac%7B0.03%7D%7B12%7D%20%5Cright%29%5E%7B180%7D%20-%201%7D%20%20%5Capprox%201%2C519.28)
The minimum monthly payment for the loan, M ≈ $1,519.28
The total amount paid over loan term, A = n × M
∴ A ≈ 180 × $1,519.28 = $273,470
The total amount paid over loan term, A ≈ $270,470
B. When t = 20 year and r = 6%, we have;
n = 12 × 20 = 240
![\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1} \approx 1,576.15](https://tex.z-dn.net/?f=%5Ctherefore%20M%20%3D%20%5Cdfrac%7B220%2C000%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B0.06%7D%7B12%7D%20%5Cright%29%20%5Ccdot%20%5Cleft%281%2B%5Cdfrac%7B0.06%7D%7B12%7D%20%5Cright%29%5E%7B240%7D%20%7D%7B%5Cleft%281%2B%5Cdfrac%7B0.06%7D%7B12%7D%20%5Cright%29%5E%7B240%7D%20-%201%7D%20%20%5Capprox%201%2C576.15)
The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276
The monthly payment, M = $1,576.15
C. When t = 30 year and r = 5%, we have;
n = 12 × 30 = 360
![\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1} \approx 1,181.01](https://tex.z-dn.net/?f=%5Ctherefore%20M%20%3D%20%5Cdfrac%7B220%2C000%20%5Ccdot%20%5Cleft%28%5Cdfrac%7B0.05%7D%7B12%7D%20%5Cright%29%20%5Ccdot%20%5Cleft%281%2B%5Cdfrac%7B0.05%7D%7B12%7D%20%5Cright%29%5E%7B360%7D%20%7D%7B%5Cleft%281%2B%5Cdfrac%7B0.05%7D%7B12%7D%20%5Cright%29%5E%7B360%7D%20-%201%7D%20%20%5Capprox%201%2C181.01)
The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163
The monthly payment, M ≈ $1,181.01
The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability
Therefore;
The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;
The total amount paid over the loan term = $270,470