SOLUTION: How many liters of a 60% alcohol solution must be mixed with 50 liters of a 90% solution to get a 70% solution?
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Question 335775: How many liters of a 60% alcohol solution must be mixed with 50 liters of a 90% solution to get a 70% solution?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many liters of a 60% alcohol solution must be mixed with 50 liters of a 90% solution to get a 70% solution?
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Equation:
alcohol+ alcohol = alcohol
0.60x + 0.90*50 = 0.70(x+50)
Multiply thru by 100 to get:
60x + 90*50 = 70x + 70*50
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-10x = -20*50
x = 100 liters (amt. of 60% solution needed in the mixture)
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Cheers,
Stan H.
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