SOLUTION: A vessel contains 49 gallons of alcohol. A certain amount of it is drawn and the replaced by the same amount of water. Then the same amount of the mixture is drawn, after which the

Question 353721: A vessel contains 49 gallons of alcohol. A certain amount of it is drawn and the replaced by the same amount of water. Then the same amount of the mixture is drawn, after which the vessel contains only 36 gallons of pure alcohol. How many gallons was drawn each time?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount that was drawn each time
After this amount is drawn and replaced by water the percent of pure alcohol remaining in the mixture is (49-x)/49; after we draw x amount from the mixture again, the amount of pure alcohol is ((49-x)/49)*(49-x) and we are told that this equals 36 gal. Soooo our equation to solve is:
((49-x)/49)*(49-x)=36 multiply each side by 49
(49-x)^2=1764 take the square root of both sides (we will only deal in positive #'s)
49-x=42 or
x=7 gal ----amount that was drawn each time
CK
((49-7)/49)*(49-7)=36
36=36
Hope this helps---ptaylor
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