SOLUTION: how many liters of 14% alcohol solution must be mixed with 20 liters of a 50% solution to get a 30% solution

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Question 463610: how many liters of 14% alcohol solution must be mixed with 20 liters of a 50% solution to get a 30% solution
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
------ percent ---------------- quantity
Alcohol I 50 ---------------- 20 L
Alcohol II 14 ---------------- x L
Total 30 ---------------- 20 + x L

50*20+14*x=30(20+x)
1000+14x=600+30x
14x-30x=600-1000
-16x=-400
/-16
x=25L 14 % Alcohol II