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?

Almost all the difficulty comes from expressing the process correctly in algebraic terms. The key insight  is that (as always) only the wine matters.

After the first draw, we have (10-j) wine, where j represents the amount in the jug.

On the second draw, we are taking some water out with the wine, so only a proportion of the remaining wine is depleted. That proportion is (10-j)*(j/10).

All told, (10-j)-(10-j)*j/10=5, so (10-j)^2=50.

There are actually two solutions: j=5(2-2^(1/2)), and j=5(2+2^(1/2)). The latter solution requires the use of minuswine and correspondingly increased quantities of water. Unlikely? Possibly; but perhaps no harder than finding a jug of irrational volume.

Edited on March 8, 2016, 12:09 am

Posted by broll on 2016-03-08 00:07:23