A young couple buying their first home borrow $80,000 for 30 years at 7.3%, comp

Question

A young couple buying their first home borrow $80,000 for 30 years at 7.3%, compounded monthly, and make payments of $548.46. After 5 years, they are able to make a one-time payment of $2000 along with their 60th payment.

(a) Find the unpaid balance immediately after they pay the extra $2000 and their 60th payment. (Round your answer to the nearest cent.)
$  

(b) How many regular payments of $548.46 will amortize the unpaid balance from part (a)? (Round your answer to the nearest whole number.)
payments

(c) How much will the couple save over the life of the loan by paying the extra $2000? (Use your answer from part (b). 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=80000, r=7.3/1200=73/12000, and n=30*12=360.Therefore, P=80000* (73/12000) [(1+73/12000)360]/[(1+73/12000)360-1] = (20*73/3)*8.876129911 /7.876129911 = $ 548.46(On rounding off to the nearest cent).

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= 80000[(1+73/12000)360-(1+73/12000)60] / (1+73/12000)360-1] = 80000*[8.876129911-1.438922074]/ 7.876129911 = $ 75541.75 (On rounding off to the nearest cent). After a further payment of $ 2000, the balance left is $ 73541.75.

(b). Let the amount of $ 73541.75 be amortized in n installments of $ 548.46 each. Then 548.46 =            ( 75541.75)(73/12000)[(1+73/12000)n]/ [(1+73/12000)n-1] or, [(12073/12000)n] = [(548.46*12000)/ (73*73541.75)] * [(1207312000)n-1] = 1.225940479*[(12073/12000)n-1]

Now, let (12073/12000)n = x. Then, the above equation changes to x = 1.225940479 (x-1) or, 0.225940479 x = 1.225940479 so that x = 1.225940479/0.225940479 = 5.42594441. Thus, (12073/12000)n = 5.42594441. Now, on taking log of both the sides, we get n log (12073/12000) = log 5.42594441or, n (log 12073-log 12000) = log 5.42594441so that n = log 5.42594441/(log 12073-log 12000) = 0.73447534/(4.081815201-4.079181246) = 0.73447534/0.00263395495 =278.85, say 279 ( on rounding off to the nearest whole number).

(c). The amount repaid by the couple after the additional payment of $ 2000 is 279*548.46 = $153020.34. Had the couple not paid the additional amount of $ 2000, they would have repaid 300*548.46 =$164538. Hence, the amount saved by the couple is =$164538- $153020.34-$ 2000 = $ 9517.66

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