SOLUTION: How many liters of a 30% alcohol solution must be mixed with 80 liters of a 90% solution to get a 80% solution

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Question 558138: How many liters of a 30% alcohol solution must be mixed with 80 liters of a 90% solution to get a 80% solution
Answer by mananth(16946) About Me  (Show Source):
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percent ---------------- quantity
Alcohol type I 90 ---------------- 80 liters
Alcohol type II 30 ---------------- x liters
Mixture 80 ---------------- 80 + x liters

90*80+30x=80(80 +x)
7200+30x=6400+80x
30x-80x=6400-7200
-50 x = -800
/ -50
x= 16 liters of Alcohol type II