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.)
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.)
$

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) .

(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= 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.

( c). Now, on using the 1^st formula again, we have P = 59439.87*(37/6000)[(1+37/6000)³⁰⁰]/           [(1+37/6000)³⁰⁰-1] =59439.87*(37/6000)* 6.32379294/5.32379294 = $ 435.40(On rounding off to the nearest cent) . The amount saved by the couple in 300 installments is 300( 450.05-435.40)-2000 = $ (4395-2000) = $ 2395.