Question 49561
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