A recent college graduate buys a new car by borrowing $24,000 at 8.4%, compounde
ID: 2893618 • Letter: A
Question
A recent college graduate buys a new car by borrowing $24,000 at 8.4%, compounded monthly, for 4 years. She decides to pay an extra $25 per payment. (a) What is the monthly payment required by the loan? (Round your answer to the nearest cent.) How much does she decide to pay each month? (Round your answer to the nearest cent.) (b) How many payments (that include the extra $25) will she make? (Round your answer up to the next whole number.) payments (c) How much will she save by paying the extra $25? (Round your answer to the nearest cent.)Explanation / Answer
From the given question,
principal=$24000
rate of interest(i)=8.4%=0.084 compounded monthly
time (n)= 4 years
Amount=A
A=P(1+i/12)12n
=24000[1+(0.084/12)](12*4)
=33545
monthly payment=33545/12
=$699
Monthly payment required by the loan is $699.
She is paying =699+25=$724 every month.
b)Number of payments:
0.084/12=0.007
((24000(1.007))-724)(1.007)-724)(1.007)-724.....=0
if p=24000, r=1.007, x=724, then
((pr-x)r-x)r-x.....
(pr2-xr-x)r-x....
pr3-xr2-xr-x...
prn-x(rn-1+rn-2+....+1)
prn-x(rn-1)/(r-1)=0
24000(1.007)n-724(1.007n-1)/(1.007-1)=0
24000(1.007)n-103428.57(1.007n-1)=0
1.007n(24000-103428.57)=-103428.57
1.007n=1.032
n=37.84 years
=38 months
c) Total amount paid=724*38=$ 27512
amount saved=33545-27412
=$ 6033
Amount saved= $ 6033