SOLUTION: A container has 6 liters of a 40% alcohol solution in it. How much pure alcohol should be added to raise it to a 60% solution?
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Question 338751: A container has 6 liters of a 40% alcohol solution in it. How much pure alcohol should be added to raise it to a 60% solution?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let the 100% alcohol required be x liters
40 % 6 liters
100% x liters
60%% (x+6 ) liters
..
100x+40*6 = 60*(x+6)
100x+240=60x+360
100x-60x=360-240
40x= 120
divide by 40
x= 3 liters
alcohol required
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