Question

What is the monthly payment on a $40,000 student loan at 4% interest over 15 years?

Answer 1

si = ptr /100

Answer 2

p=40000
t=15
r=4

Answer 3

si = (40000*15*4)/100=1500

Answer 4

i tend to use a formula i developed since the text formulas never really made sense to me

Answer 5

a=p+i
a=amount
p=principal
si = simple interest
r= rate of interest
t = time

Answer 6

a= 40000+1500 =41500 dollars

Answer 7

Oh thank you so much, I had no idea how to do this problem. You helped a lot

Answer 8

rate of interest in %
time in years
a,p,i in dollars

Answer 9

Wait, so the answer is 1500?

Answer 10

41500

Answer 11

the balance due on a loan is the compounding loan amount, minus a geometric sum of payments made.

Bn = Bk^n - P(1-k^n)/(1-k)

since the ending balance is 0 over n periods or 15*12 periods we have

0 = Bk^n - P(1-k^n)/(1-k)

solving for P we get

P = Bk^n (1-k)/(1-k^n)

which can be simplified if needed to the standard text formula but i find this form easier

Answer 12

k is a clean up variable for the compounding interest of (1+i/12)

Answer 13

Wait I got confused again, how is 41,500 the answer if it's asking for the monthly payment

Answer 14

filling in, i get

P = 40000(1+.04/12)^(15*12) (1-(1+.04/12))/(1-(1+.04/12)^(15*12))

which the wolf or calculator easily determines

Answer 15

http://www.wolframalpha.com/input/?i=40000%281%2B.04%2F12%29%5E%2815*12%29+%281-%281%2B.04%2F12%29%29%2F%281-%281%2B.04%2F12%29%5E%2815*12%29%29

Answer 16

vee, im not sure what CGG was doing, so i cant comment on their approach, but can you undertsand what i did?

Answer 17

i think so

Answer 18

taking the formula for the balance of a loan, i solved for P

Answer 19

im getting 295.88 is the result of filling in the required parts

Answer 20

I got that too

Answer 21

then we agree :) yay!!

Answer 22

I need help with another problem

Answer 23

Oh that's how they did it! Thanks for posting this question. I'm curious on how to solve this problem. Now I know what to do next time!

As always, helpers in OS give very clear and nice explanations! Thanks again! (: