Question 30971: A company wants to invest $180,000 in three accounts, one bearing 4.5%, another 6%, and another 8% interest. Twice as much will be invested at 4.5% as at 8%, and the total interest earned the first year needs to be $10,400. How much should be invested at each rate?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Let amount invested at 8% be "x"; interest is 0.08x
Then amount invested at 4% is"2x"; interest is 004(2x)=0.08x
And amount invested at 6% is 0.06(180,000-3x); int. is 10,800-0.18x
EQUATION:
interest + interest + interest = $10,400
0.08x+0.08x+10,800-0.18x=10,400
Multiply thru by 100 to get:
8x+8x+1080000-18x=1040000
-2x-40000
x=$20,000 (amount invested at 8%)
2x=$40,000 (amount invested at 4%)
180,000-3x=$120,000 (amount invested at 6%)
Cheers,
Stan H.
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