Question 1080926: how much must you deposit in an account that pays 8% interest, compounded monthly, to have a balance of $1000 after three years
Found 3 solutions by Fombitz, jorel1380, Alan3354:
Answer by Fombitz(32388)   (Show Source):
Answer by jorel1380(3719)  (Show Source):
You can put this solution on YOUR website! In order to have a balance of $1000, after three years of 8% interest, compounded monthly, we have:
1000=N(1 + .08/12)^36 where N is the initial deposit. So
1000=N(1.00666667)^36
N=1000/(1.00666667)^36
ln N=ln 1000/(1.00666667)^36=ln 1000-36 ln (1.00666667)
ln N=6.6685517411100682158933244766027
N=e^6.6685517411100682158933244766027=$787.25463 as the initial deposit. ☺☺☺☺
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! how much must you deposit in an account that pays 8% interest, compounded monthly, to have a balance of $1000 after three years
----------
FV = Future value
PV = Present value
-----
3 years = 36 months, the compounding period.
8% per year = 8/12 % per month
--------
FV = PV*(1 + 0.08/12)^(3*12)
1000 = PV*1.00666667^36
PV = 1000/(1.00666667^36)
PV =~ $787.25
================
Sounds like a bad deal.
OK, not a bad deal, just a waste of time.
$213 would last about 3 minutes at the free clinic.
|
|
|
