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My solution is below. Richard offers a simpler solution here.
After removing the first jugful of wine, there are 10-J gallons of wine left in the 10 gallon keg. After removing the second gallon of the mixture, J*(10-J)/10 gallons of wine are removed. In the end, there are 5 gallons of wine, so 10 - J - J*(10 - J)/10 = 5
10 - J - J*(10 - J)/10 = 5
J^2/10 - 2*J + 5 = 0
J = 10 + 5*sqrt(2) or 10-5*sqrt(2)
Since 10+5*sqrt(2) is larger than the original keg, the answer is 10-5*sqrt(2) = 2.929 gallons.
SilverKnight expands the problem and considers the jug needed so 3, 4 and 5 draws yield a 50% mixture here. |
