9. - +points 0/6 Submissions Used My Notes The Martinezes are planning to refina
ID: 3116951 • Letter: 9
Question
9. - +points 0/6 Submissions Used My Notes The Martinezes are planning to refinance their home (assuming that there are no additional finance charges). The outstanding balance on their original loan is $200,000. Their finance company has offered them two options: Option A: A fixed-rate mortgage at an interest rate of 6.5% per year compounded monthly, payable over a 30-year period in 350 equal month y installments. Option B: A fixed-rate mortgage at an interest rate of 6.25% per year compounded monthly, payable over a 12-year period in 144 equal monthly installments (a) Find the monthly payment required to amortize each of these loans over the life of the loan. (Round your answers to the nearest cent.) Option A: Option B: (b) How much interest would the Martinezes save if they chose the 12-year mortgage instead of the 30-year mortgage? Use the rounded monthly payment values from part (a). (Round your answer to the nearest cent.) 10. + 1 points 0/6 Submissions Used My Notes The Turners have purchased a house for $140,000. They made an initial down payment of $20,000 and secured a mortgage with interest charged at the rate of 10%/year compounded monthly on the unpaid balance. The loan is to be amortized over 30 yr. (Round your answers to the nearest cent.) (a) What monthly payment will the Turners be required to make? (b) How much total interest will they pay on the loan? (c) What will be their equity after 10 years? (d) What will be their equity after 22 years?Explanation / Answer
9. (a).The formula to be used is P = L[r(1 + r)n]/[(1 + r)n - 1], where L is the loan amount, r is rate of interest per period, n is the number of periods and P is the periodic payment.
Option A: Here, L = $ 200000, r=6.5/1200, n = 360, so that P = 200000*(6.5/1200)*[(1+6.5/1200)360]/ [(1+6.5/1200)360 -1] = 200000*(6.5/1200)*(6.991797982/5.991797982) = $ 1264.14.
Option B: Here, L = $ 200000, r=6.25/1200, n = 144, so that P = 200000*(6.25/1200)*[(1+6.25/1200)144]/ [(1+6.25/1200)144 -1] = 200000*(6.25/1200)*(2.112883558/1.112883558) = $ 1977.67.
(b). The amount repaid to the Bank in case of option A is $ 1264.14*360 = $ 455090.40 so that the interest paid to the Bank is $ 455090.40 -$ 200000= $ 255090.40. The amount repaid to the Bank in case of option B is $ 1977.67*144 = $ 284784.48 so that the interest paid to the Bank is $ 284784.48- $ 200000= $84784.48.
Thus, the amount of interest that the Martinezes would save, if they chose option B is $ 255090.40 -$84784.48 = $ 170305.92.
10.(a) Here, L = $ 140000-$ 20000 = $ 120000, n = 30*12 = 360 and r = 10/1200 = 1/120.The monthly payment required to be made by the Turners is P = 120000*(1/120)*[(1+1/120)360]/ [(1+1/120)360 -1] =1000*(19.83739935/18. 83739935) = $ 1053.09 ( on rounding off to the nearest cent).
(b). The total amount repaid to the Bank is 360*$ 1053.09 = $379112.40. Hence, the interest paid by the Turners is $379112.40 -$ 120000 = $ 259112.40.
(c ). The formula is used to calculate the remaining loan balance (B) of a fixed payment loan after p months is B = L[(1 + r)n - (1 + r)p]/[(1 + r)n - 1].
After 10 years, B = 120000* [(1+1/120)360-(1+1/120)120]/ [(1/120)360-1] = 120000*(19.83739935-2.70704149)/( 18.83739935) = 120000*(17.13035786/18.83739935) = $ 109125.62 so that the Turners’ equity after 10 years is $ 140000- $109125.62 = $ 30874.38.
After 22 years, B = 120000* [(1+1/120)360-(1+1/120)264]/ [(1/120)360-1] = 120000*(19.83739935-8.943114819)/( 18.83739935) = 120000*(10.89428453/18.83739935) = $ 69399.93 so that the Turners’ equity after 22 years is $ 140000- $69399.93 = $70600.07.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.