Question

A young couple buying their first home borrow $65,000 for 30 years at 7.4%, compounded monthly, and make payments of $450.05. 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 $450.05 will amortize the unpaid balance from part (a)? (Round your answer to the nearest whole number.) -this answer is in payments not $

(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.) the answer is not  2395.

Answer 1

(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=65000, r=7.4/1200=37/6000,and n=30*12=360.Therefore, P=65000* (37/6000) [(1+37/6000)³⁶⁰]/[(1+37/6000)³⁶⁰-1] = (65*37/6)*9.144778057/8.144778057 = $450.05 (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= 65000[(1+37/6000)³⁶⁰-(1+37/6000)⁶⁰] / (1+37/6000)³⁶⁰-1] = 65000*[9.144778057-1.446099684]/ 8.144778057= $ 61439.87(On rounding off to the nearest cent). After a further payment of $ 2000, the balance left is $ 59439.87.

( b). Let the amount of $ 59439.87 be amortized in n installments of $ 450.05 each. Then 450.05 = (59439.87)(37/6000)[(1+37/6000)ⁿ]/ [(1+37/6000)ⁿ-1] or, [(6037/6000)ⁿ] = [(450.05*6000)/ (37*59439.87)] * [(6037/6000)ⁿ-1] = 1.227813605*[(6037/6000)ⁿ-1]

Now, let (6037/6000)ⁿ = x. Then x = 1.227813605(x-1) or, 0.227813605 x = 1.227813605 so that x = 1.227813605/0.227813605 = 5.38955347. Thus, (6037/6000)ⁿ =5.38955347. Now, on taking log of both the sides, we get n log (6037/6000) = log 5.38955347 or, n (log 6037-log 6000) = log 5.38955347 so that n = log 5.38955347 /(log 6037-log 6000) = 0.731552784/(3.780821176-3.77815125) =0.731552784/0.002669926 = 274 (approximately).

( c). The amount repaid by the couple after the additional payment of $ 2000 is 274*450.05 = $123313.70. Had the couple not paid the additional amount of $ 2000, they would have repaid 300*450.05 =$135015. Hence, the amount saved by the couple is =$135015- $123313.70-$ 2000 = $ 9701.30