A man had a 10-gallon keg full of wine, and an empty jug. On Monday he drew off a jugful of wine and filled up the keg with water. On Friday, after the wine and water had been thorougly mixed, he drew off another jugful and again filled up the keg with water. The keg then contained equal quantities of wine and water.

What was the capacity of the jug?

(In reply to Different Solution - Where Am I going wrong? by David Bate)

"This jug contains 50% wine and 50% water" is not in the problem statement and is not supported by the facts -- this is what leads to your different (and wrong!) answer.

Monday's final result is that the keg will contain wine and water in the respective proportions w and 1-w, say. Friday's removal will remove the proportions (1-w)^2 of water and (1-w)*w of wine and leave the proportions w*(1-w) of water and w^2 of wine. These proportions add up to 1 as they must:

(1-w)^2+2*(1-w)*w+w^2=1-2*w+w^2+2*w-2*w^2+w^2=1.

Hence Friday's final result is that the keg will contain the proportion w^2 of wine and the proportion 1-w^2 of water. According to the problem statement, then, w^2=1/2 (and also 1-w^2=1/2) so that w=sqrt(1/2). Thus 1-w=1-sqrt(1/2) which x10 gallons gives us approximately 2.9 gallons as the size of the jug.

Posted by Richard on 2004-07-19 15:32:48