SOLUTION: How many liters of a 20% alcohol solution must be mixed with 60 L of a 70% solution to get a 50% solution?
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Question 380450: How many liters of a 20% alcohol solution must be mixed with 60 L of a 70% solution to get a 50% solution?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many liters of a 20% alcohol solution must be mixed with 60 L of a 70% solution to get a 50% solution?
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Equation:
alc + alc = alc
0.20x + 0.70*60 = 0.50(x+60)
Multiply thru by 100 to get:
20x + 70*60 = 50x + 50*60
30x = 20*60
x = 40 L (amt of 20% solution needed in the mixture)
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Using 2 equations with 2 variables:
Quantity Eq: x + 60 = y
Alcohol Equation: 0.20x + 0.70*60 = 0.50y
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Substitute for "y" in the Alcohol Equation then
solve as I did using one variable.
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If this is still confusing let me know.
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CHeers,
Stan H.
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