# 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(57375)   (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) ================================================================== Cheers, Stan H.