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How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to obtain 40 liters of 8% alcohol solution?
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How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to obtain 40 liters of 8% alcohol solution?
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May 25, 2020
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How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to obtain 40 liters of 8%
alcohol solution?
Mathematics
middle-school
Bryanjonker
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Bryanjonker
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Answer:
x+y=40liters
10%x+5%y=8%×40
(2x+y)/20=8%×40
(2x+y)=64
x=24
y=16
Brad Pitcher
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Jun 1, 2020
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Brad Pitcher
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