Question 466638: How many liters (L) of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution? Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! How many liters (L) of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution?
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x=liters of 20% alcohol solution to be mixed with 60 liters of 90% alcohol solution.
x+60 =liters of 50% alcohol solution
20%x+90%(60)=50%(x+60)
.2x+54=.5x+30
.3x=24
x=80
ans:
80 liters of a 20% alcohol solution must be mixed with 60 L of a 90% alcohol solution to get a 50% alcohol solution.
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