This problem is a complex financial problem that requires several skills, perhap

Question

This problem is a complex financial problem that requires several skills, perhaps some from previous sections.

Clark and Lana take a 30-year home mortgage of $128,000 at 7.8%, compounded monthly. They make their regular monthly payments for 5 years, then decide to pay $1500per month.

(a) Find their regular monthly payment. (Round your answer to the nearest cent.)
$   

(b) Find the unpaid balance when they begin paying the $1500. (Round your answer to the nearest cent.)
$  

(c) How many payments of $1500 will it take to pay off the loan? (Round your answer to two decimal places.)
monthly payments

(d) How much interest will they save by paying the loan using the number of payments from part (c)? (Round your answer to the nearest cent.)
$

Explanation / Answer

(a). The formula used to calculate the fixed monthly payment (P) required to fully amortize a loan of L dollars over a term of n months at a monthly interest rate of r is P = L [r(1 +r)n]/[(1 + r)n - 1].

Here, L=128000, r= 7.8/1200= 13/2000, and n=30*12=360.Therefore, P= 128000* (13/2000)* [(1+13/2000)360]/[(1+13/2000)360-1] = 832*10.30292458/9.30292458= $ 921.43 (On rounding off to the nearest cent).

(b). The formula used to calculate the remaining loan balance (B) of a fixed payment loan of $ L, after p months is B = L [(1+ r)n - (1+ r)p]/[(1 + r)n - 1]. Here, p= 5*12 = 60 so that B= 1280000[(1+13/2000)360-(1+13/2000)60] / (1+13/2000)360-1] = 128000*[10.30292458-1.475117969]/ 9.30292458= $ 121462.80 (On rounding off to the nearest cent). Thus, the unpaid balance when Clark and Lara begin paying $1500 per month is $121462.80.

(c). Let the amount of $ 121462.80 be amortized in n installments of $ 1500 each. Then 1500 =         (121462.80)(13/2000)[(1+13/2000)n]/ [(1+13/2000)n-1] or, [(2013/2000)n] = [(1500*2000)/ (13*121462.80)]*[(2013/2000)n-1] = 1.899916936 *[(2013/2000)n-1]

Now, let (2013/2000)n = x. Then, the above equation changes to x = 1.899916936 (x-1) or, 0.899916936 x = 1.899916936 so that x = 1.899916936 /0.899916936 =2.111213668. Thus, (2013/2000)n =2.111213668. Now, on taking log of both the sides, we get n log (2013/2000) = log 2.111213668 or, n (log 2013-log 2000) = log 2.111213668 so that n = log 2.111213668/(log 2013-log 2000) = 0.324532188/(3.303843775-3.301029996) = 0.324532188 /0.00281377934 =115.34 ( on rounding off to2 decimal places).

(d). The amount repaid by Clark and Lara since they started paying $ 1500 per month is 115.34* 1500 = $173010. Had Clark and Lara not started paying $ 1500 per month, they would have repaid 300*921.43 = $276429. Hence, the amount saved by Clark and Lara is =$276429-$173010 = $ 103419.

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