SOLUTION: how many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution

Question 467150: how many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
----------- percent ---------------- quantity
Alcohol I 50 ---------------- 40 L
Alcohol II 10 ---------------- x L
Total 40 ---------------- 40 + x L

50*40+10*x=40(40+x)
2000 + 10 x = 1600 + 40 x
10 x - 40 x = 1600 - 2000
-30 x = -400
/-30
x = 13.33 L 10 % Alcohol II

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