SOLUTION: How much water must be added to 400L of mixture that is 80% alcohol to reduce it to a 60% mixture?

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Question 936148: How much water must be added to 400L of mixture that is 80% alcohol to reduce it to a 60% mixture?
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
x = liters of 80% alcohol/water = liters of 20% water/alcohol
y = liters of pure water
z = liters of 60% alcohol/water = liters of 40% water/alcohol
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x = 400
z = x + y
z = 400 + y
0.20x + 1.00y = 0.40z
0.20*400 + 1.00y = 0.40*(400 + y)
0.20*400 + y = 0.40*400 + 0.40y
y - 0.40y = 0.40*400 - 0.20*400
0.60y = 0.40*400 - 0.20*400
y = (0.40*400 - 0.20*400)/0.60
y = 133.333333333
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answer:
y = liters of pure water = 133 1/3
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