## Ken's POTW

A Keg and a Jug
A man had a 10-gallon keg of wine and a jug. One day, he drew off a
jugful of wine and filled up the keg with water. Later on, when the
wine and water had got 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?
Source: Henry Ernest Dudeny's *536 Curious Problems & Puzzles*

Solutions were received from:
Kirk Bresniker,
Sorin Ionescu
Dorval School
Quebec, CANADA,
Bill Chapp,
George Song,
and
Geoff Brock.
Kirk Bresniker's solution:

We start off with a Keg of K gallons (K=10), and a jug of J gallons
(J=?).
After we remove the first jugful of wine, there are K-J gallons of wine
left in the K gallon keg. When we remove the second gallon of
the mixture, we are removing J*(K-J)/K gallons of wine. In the end, we
have K/2 gallons of wine, so
K - J - J * (K - J)/K = K/2
10 - J - J * (10 - J)/10 = 5
J^2/10 - 2*J + 5 = 0
Solving we obtain J = 10 +/- 5*sqrt(2)
Since 10+5*sqrt(2) is larger than our original keg, the answer is
** 10-5*sqrt(2) **[ ~= 2.92 gallons - KD ] .
The first removal leaves 10 - 10 + 5*sqrt(2) = 5*sqrt(2) gallons of
wine in the container, and the second removal leaves
10 - 10 + 5*sqrt(2) - (10 - 5*sqrt(2))(10 - 10 + 5*sqrt(2))/10
= 5*sqrt(2) - (10 - 5*sqrt(2))(sqrt(2)/2)
= 5*sqrt(2) - 5*sqrt(2) + 5*(sqrt(2)^2)/2
= 5

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