SOLUTION: A container has 20 L 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 th

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Question 482134: A container has 20 L 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?
Found 2 solutions by mananth, mathstutor494:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
percent percentage ---------------- quantity
Alcohol 40 ---------------- x liters
water 0 ---------------- 20 - x liters
Mixture 25.00% ---------------- 20

40x+0(20-x) = 25 * 20

40 x = 500

/ 40
x=12.5 liters 40.00% Alcohol
7.5liters water
7.5 liters of mixture has to be replaced with water

Answer by mathstutor494(120) About Me  (Show Source):
You can put this solution on YOUR website!
Current composition of mixture......
Alcohol : 20*40/100 = 8 L
So water 20-8 : 12 L
New composition of mixture required....
Alcohol : 20*25/100 = 5 L
So Water : 15 L
So mixture to be removed and replaced by an equal volume of water = 3L