SOLUTION: I need help with this problem, I don't even know where to begin: A chemist has three different acid solutions. The first acid solution contains 30% acid, the second contains 40%,

Question 838757: I need help with this problem, I don't even know where to begin:
A chemist has three different acid solutions. The first acid solution contains 30% acid, the second contains 40%, and the third contains 60%. He wants to use all three solutions to obtain a mixture of 161 liters containing 50% acid, using twice as much of the 60% solution as the 40% solution. How many liters of each solution should be used?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
c=2b,
a+b+c=161,
.3a+.4b+.6c=.5*161
a=23., b=46., c=92.
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