Let x be the size of the jug. After the man drew off his first jugful of wine, the keg contained 10-x gallons of wine.
When he filled up the keg with water, the proportion of wine was reduced to (10-x)/10.
The man's second jugful contained x(10-x)/10 gallons of wine, so the keg's wine content was reduced to
10-x-x(10-x)/10 gallons of wine.
Since the keg now contains equal quantities of wine and water,
10-x-x(10-x)/10=5. Now just solve for x.
10-x-x(10-x)/10=5
100-10x-(10x-x2)=50
x2-20x+50=0
x=10±sqrt(50)
Of these two values , 10+sqrt(50) is greater than 10, so it can't be the
capacity of the jug in this story. So x=10-sqrt(50), which is about 2.928932188 gallons.
Let's check this answer.
After the first jug of wine is taken, the keg contains
10-x = 10-10+sqrt(50) = sqrt(50) gallons of wine. The proportion of wine in the keg is
sqrt(50)/10. After the next jugful is taken, the keg contains (10-10+sqrt(50))(sqrt(50))/10=5 gallons of wine.
Do you see what's going on here? Every time the man draws off a jugfull,
and then fills up the keg with water, he dilutes the wine by a factor of
sqrt(50)/10 = 1/sqrt(2). So after two trips the wine in the keg is at
1/sqrt(2)2=1/2 strength. After three trips, the wine in the keg
would be 1/sqrt(2)3 of its original strength, and after four trips the wine
would be at
1/sqrt(2)4=1/4 strength.
