Part 1: Complete the following showing your set-ups. All your comparisons will b
ID: 3116998 • Letter: P
Question
Part 1: Complete the following showing your set-ups. All your comparisons will be made to this first home. 1. $220,000 with a 20 % down and with an You are making all comparisons to this home, buying price interest rate of 6.5%. a) Find the mortgage amount a. 2 points b) Find the monthly payment if the term b is 30 years. (Show the set-up) You will be comparing problems 2 and 3 to this and using this for your amortization. 4 points c) Find the monthly payment if the te is 15 years. 4 points d) Find the total interest on the 30-yea d loan. 2 pointsExplanation / Answer
1.(a) The mortgage amount is $ 220000 *80 % = $176000.
(b). The formula to be used is P = L[r(1 + r)n]/[(1 + r)n - 1] where L is the loan amount, P is the periodic payment, r is the rate of interest per period and n is the no. of periods. Here, the loan amount is L = $ 220000 *80 % = $176000, r = 6.5/1200 = 13/2400( we have presumed monthly compounding, in absence of information to the contrary) and n = 30*12 = 360. Therefore, P =176000(13/2400) [(1+13/2400)360 ] /[(1+13/2400)360 -1] =(2860/3)(6.991797982)/( 5.991797982)= $1112.44 (on rounding off to the nearest cent). Thus, the monthly payment is $1112.44.
(c ). If the term of the loan is 15 years, then P = 176000(13/2400) [(1+13/2400)180] /[(1+13/2400)180 -1] =(2860/3)(2.6442000821)/ (1.6442000821) = $1533.15(on rounding off to the nearest cent). Thus, the monthly payment is $1533.15.
(d). The total interest on the 30 year loan is total repayment – loan amount = 360*$1112.44-$ 176000 = $ 224478.40
(e). The total repayment on the 15 year loan is 180*$1533.15 = $ 275967.The total interest paid is $ 275967- $176000 = $ 99967.
(f). The savings in interest $ 224478.40-$ 99967 = 124511.
2. (a). The formula to be used is P = L[r(1 + r)n]/[(1 + r)n - 1] where L is the loan amount, P is the periodic payment, r is the rate of interest per period and n is the no. of periods.
Here, the loan amount is L = $ 220000 *80 % = $176000, r = 5.5/1200 = 11/2400( we have presumed monthly compounding, in absence of information to the contrary) and n = 30*12 = 360. Therefore, P = 176000(11/2400) [(1+11/2400)360 ] /[(1+11/2400)360 -1] = (19360/24)(5.187387834/4.187387834) = $ 999.31(on rounding off to the nearest cent).Thus, the monthlypayment is $ 999.31.
(b).The total repayment is 360*$ 999.31= $ 359751.60 so that the amount of interest paid is $ 359751.60 -$ 176000 = $183751.60.
(c ). The saving in interest compared to 1.(d) is $ 224478.40 -$183751.60 = $ 40726.80