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?
(In reply to
re(2): the CORRECT solution by David)
At first i thought yours was correct, because i had the same thing. However the answer is clearly wrong so i looked for mistake.
It's in the 2nd removal from the keg, basically you have multiplied by (10 - j), which was the amount left after j had been removed. If you think logically this is like mutiplying by the amount left after 1 jug has been removed, so the amount you are left with in the keg is not 10 gallons.
Using your method :
After Monday, there is (10-j)V + jW
On Friday, j/10[(10-j)V + jW] is removed, and replaced with water. This means you are left with (1-j/10)[(10-j)V + jW] + jW/10[(10-j) + j], which simplifies to V(10-j) - V(j - [j^2]/10) + 2jW - W(j^2)/10. As it is 50/50 mixture V coefficients equal W coefficients. Therefore:
10 - j - j + (j^2)/10 = 2j - (j^2)/10
(j^2)/5 - 4j + 10 = 0
j^2 - 20j + 50 = 0
Solve this quadratic in j to give j = 17.071... (invalid as >10) and j = 2.9289... (correct solution)
Other way sure is easier though i wish i had tried that first =)
re(3): the CORRECT solution
