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?

the few answers i read below differ from mine (and of course i think i am right!)

let volume of jug = j

let V stand for (pure) wine

let W stand for (pure) water

After first extraction/mixing, new liquid is made up of:

(10-j)V + jW

After 2nd extraction/mixing, new liquid is made up of:

(10-j) [ (10-j)V + jW] + jW

= (10-j)^2 V + (11-j)j W

= (j^2 - 20j + 100) V + (11j - j^2) W

And the mix is 50/50, so V and W coefficients must be equal:

j^2 - 20j +100 = 11j - j^2

=> 2j^2 - 31j +100 = 0

Now solving quadratic for j gives:

j= 10.9 (ignore as outside bounds) OR

j= 4.578

That is, Jug volume = 4.578 gallons

Posted by David on 2004-07-14 19:49:23