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1. 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 thoroughly 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: unknown
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=10Âħsqrt(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.
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