Source: Henry Ernest Dudeny's 536 Curious Problems & Puzzles
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