Question

WILL FAN + MEDAL Luella took out an unsubsidized student loan of $37,000 at a 4.8% APR, compounded monthly, to pay for her last six semesters of college. If she will begin paying off the loan in 33 months with monthly payments lasting for 20 years, what will be the amount of her monthly payment? A.$271.56 B.$240.11 C.$238.04 D.$273.92

Answer 1

i really need help here....

Answer 2

compound the 37k for 33 months, what do we get for balance to pay off?

Answer 3

oh and amistre? ur formula didn't really help me..... :P

Answer 4

First find out how much she owes (i.e. how much interest has accrued in addition to the initial loan amount) using the compound interest formula:
\[A=P\left(1+\frac{r}{n}\right)^{nt}\]
(if I'm not mistaken)

Then I believe the present value formula should be used to figure out payment size.

Answer 5

oo

Answer 6

i worked it out on the wolf and linked it ...

Answer 7

alright fine lol

Answer 8

u get an award for trying anyway xD

Answer 9

it even works for this one :)

Answer 10

\[P=37000k^{33}~k^{12*20}~\frac{1-k}{1-k^{12*20}},~k=1+.048/12\]
http://www.wolframalpha.com/input/?i=37000k%5E%2833%29k%5E%2812*20%29+%281-k%29%2F%281-k%5E%2812*20%29%29%2Ck%3D1%2B.048%2F12

Answer 11

how should i input that into my calculator?

Answer 12

depends on the calculator you have; i use a ti83 .... ive never used a financial calculator tho

Answer 13

i use a ti84 silver edition

Answer 14

you know how to store things into the alpha characters?

Answer 15

on tvm?

Answer 16

or the fracions?

Answer 17

i like using K as a clean up for the compounding mess:

1+.048/12, STO \(\to\), alpha, K ; the "(" key is "K"

37000, alpha, K, ^, (33), STO \(\to\), alpha, B

then im ready to run the setup

Answer 18

B K^(12*20) (1-k) \(\div\) (1-K^(12*20))

Answer 19

lol its a little too confusing here

Answer 20

then id just stick with the wolfs output :)

Answer 21

well how do you get the answers?! its vital!!!!

Answer 22

oh wait.... >_<

Answer 23

it was posted on htere :P

Answer 24

hallelu-friggin-jah finally... on to question 6

Answer 25

5*

Answer 26

id have to go thru some more confusing things .... like this:

lets start with a begining loan amount Bo\[B_o=B_o\]and take this thru each month with a compounding variable ... k
\[B_1=B_ok\]
\[B_2=B_o~k~k=B_ok^2\]
\[B_3=B_ok^3\]
do this for 33 months to get:\[B_{33}=B_ok^{33}\]then we can use that amount to start paying off
\[B_{34}=B_ok^{34}-P\]
\[B_{35}=B_ok^{35}-Pk-P\]
\[B_{36}=B_ok^{36}-Pk^2-Pk-P\]
this goes on and on and forms a geometric sum of the payments in terms of k
\[B_{n}=B_ok^{d+n}-P\frac{1-k^n}{1-k}\]
when Bn = 0 the loan is paid off
\[0=B_ok^{d+n}-P\frac{1-k^n}{1-k}\]
so solving for P is the payment amount
\[B_ok^{d+n}=P\frac{1-k^n}{1-k}\]
\[B_ok^{d+n}\frac{1-k}{1-k^n}=P\]

Answer 27

soo, with the specifics given

\[P=37000\left(1+\frac{.048}{12}\right)^{33+12(20)}~\frac{1-(1+\frac{.048}{12})}{1-(1+\frac{.048}{12})^{12(20)}}\]

Answer 28

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