Question 86251: An investment club wants to invest $50,000 in two simple interest accounts. One account earns 8.4% annual simple interest and the other account earns 11.9% annual simple interest. How much should be invested in each account so that both accounts earn the same annual interest?
Found 2 solutions by scianci, ankor@dixie-net.com:
Answer by scianci(186)   (Show Source):
You can put this solution on YOUR website! Interest = (Principal)(rate)(time), or I = Prt [r as a decimal, not a percent] In this case, we want to split the principal, $50,000, into two accounts, one with a rate of 8.4% and the other with a rate of 11.9%, in such a way that the two interest amounts will be equal. [Which account should more be deposited in to, the 8.4% or the 11.9%?] So, I = I, or Prt = Prt. r is given for each case [but change them to their decimal equivalents] and t = 1 year. Let the P for one account be x, then the P for the other account is 50,000 - x. Plug in the parameters into the equation Prt = Prt and solve for x.
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! An investment club wants to invest $50,000 in two simple interest accounts. One account earns 8.4% annual simple interest and the other account earns 11.9% annual simple interest. How much should be invested in each account so that both accounts earn the same annual interest?
:
Let x = amt invested at 8.4$
Then
(50000 - x) = amt invested at 11.9%
:
Interest at 8.4% = interest 11.9%
.084x = .119(50000-x)
.084x = 5950 - .119x
.084x + .119x = 5950
.203x = 5950
x = 5950/.203
x = $29,310.34 invested at 8.4%
:
50000 - 29310.34 = $20,689.66 invested at 11.9%
:
:
Check our solutions to see if interests are equal:
.084 * 29310.34 = 2462.07
.119 * 20689.66 = 2462.07
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