SOLUTION: A tank contains 20 liters of a mixture of alcohol and water which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so t

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Question 49561: A tank contains 20 liters of a mixture of alcohol and water which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 25% alcohol by volume?
Answer by ptaylor(2198)   (Show Source): You can put this solution on YOUR website!
Let x=amt of mixture that should be removed and replaced by an equal volume of water.
Let's focus first on the alcohol since we are not adding any alcohol back in.
We know that the amount of pure alcohol in the 20 liter tank (20)(.40)minus the amt of pure alcohol drained out (x)(.40) plus the amt of pure alcohol that is added back (0) equals the amt of pure alcohol in the final mixture (20)(.25). Thus, our equation to solve is:
(20)(.40) -.40x=(20)(.25) simplifying, we get:
-.4x=-8+5=-3
4x=30
x=7.5 gal
Now, lets solve this problem, focusing on the water:
We know that the amount of water in the 20 liter tank (20)(.60) minus the amt of water drained out (x)(.60) plus the amt of water that is added back (x)(1) equals the amt of water in the final mixture (20)(.75). thus, our equation to solve is:
(20)(.60)-.60x+x=(20)(.75) simplifying, we get:
12+.4x=15 and
.4x=15-12=3
4x=30
x=7.5 gal
Hope this helps----ptaylor

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