Question

Bessie took out a subsidized student loan of $5000 at a 2.4% APR, compounded monthly, to pay for her last semester of college. If she will begin paying off the loan in 10 months with monthly payments lasting for 20 years, what will be the total amount that she pays in interest on the loan?

Answer 1

is 10 months a defered period?

Answer 2

i think it is..., its for Personal Finance

Answer 3

compound the loan amount for 10 months to determine a value to calculate payments with

Answer 4

once we know the payments amount, we just add all the payments over 20 years and subtract the loan amount to determine the interest cost of the loan

Answer 5

how do i do that ?

Answer 6

oh wait....

Answer 7

would this be a correct part...?

Answer 8

i mean... am i going in the riht direction?

Answer 9

5000(1+.002) ... thats good, but we need to tack on an exponent right?

Answer 10

yup... should i multiply 20*12 to make the exponent?

Answer 11

lets worry about the 10 months first ... the balance to pay off is increasing during the first 10 months

5000(1.002)^10 would be the amount to payoff over 20 years

Answer 12

oh ok, should i input this into my calculator?

Answer 13

you can if you want .... but this only gets us to a balance that needs to be paid off; we would still need to calculate what the payment amount would be

Answer 14

ah ok

Answer 15

do you have a formula in your material for calculating payments?

Answer 16

amorzitation?

Answer 17

i was thinking annuity ....

Answer 18

oh

Answer 19

what is the formula for amortization? im not good with the terminology is all

Answer 20

lol ok

Answer 21

would this be it?

Answer 22

that does look like what ive seen floating about yes :)

i created my own formula based off of the reoccurence of the payments ...

\[B_n=B_ok^{12y}-P\frac{1-k^{12y}}{1-k}\]
when Bn is zero the loan is paid off so solving for P we get
\[0=B_ok^{12y}-P\frac{1-k^{12y}}{1-k}\]
\[B_ok^{12y}=P\frac{1-k^{12y}}{1-k}\]
\[B_ok^{12y}\frac{1-k}{1-k^{12y}}=P\]

Answer 23

this simplifies even more to the formula in the book, but i like this one better

Answer 24

urs? or the pic?

Answer 25

so:

P = (5000k^(10))*k^(12*20)(1-k)\(1-k^(12*20)), k=1+.024/12

http://www.wolframalpha.com/input/?i=%285000k%5E%2810%29%29*k%5E%2812*20%29%281-k%29%5C%281-k%5E%2812*20%29%29%2C+k%3D1%2B.024%2F12

and that is 12*20 payments to add up .... then subtract the 5000

Answer 26

i perfer my own formula, i can adapt it to sooo many other things rather than trying to memorize all the different formulas they like to throw at you