SOLUTION: A solution containing 30% alcohol is to be mixed with a solution containing 50%
alcohol to make 200 L of a solution containing 42% alcohol. How much of the 50%
solution should be
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Question 231929: A solution containing 30% alcohol is to be mixed with a solution containing 50%
alcohol to make 200 L of a solution containing 42% alcohol. How much of the 50%
solution should be used?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A solution containing 30% alcohol is to be mixed with a solution containing 50%
alcohol to make 200 L of a solution containing 42% alcohol. How much of the 50%
solution should be used?
----------------------
Equation:
alcohol + alcohol = alcohol
0.30x + 0.50(200-x) = 0.42*200
Multiply thru by 100 to get:
30x + 50*200 - 50x = 42*200
-20x = -8*200
x = 80 liters (amount of 30% solution needed in the mixture)
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Cheers,
Stan H.
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