Question 936149: How much water must be added to 400L of mixture that is 80% alcohol to reduce it to a 60% mixture? Found 3 solutions by josgarithmetic, srinivas.g, TimothyLamb:Answer by josgarithmetic(39618) (Show Source):
You can put this solution on YOUR website! initial quantity of mixture = 400 L
alcohol % = 80 %
quantity of alcohol in the mixture = 80 % of 400 L
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=
= L
Let X be the quantity of water added to the mixture
final quantity of the mixture = 400+ x
alcohol 5 in final mixture = 60 %
quantity of alcohol in final mixture = 60 % of (400+x)
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=
quantity of alcohol in the initial mixture and final mixture is same
multiply with 5 on both sides
move 1200 to the left
divide with 3 on both sides
Result: Quantity of water added = 133.33 L
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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