hms.tech said:
Hmm... I hadn't thought of that.
In that case, I would amend my method to :
10 - x/2 - x = x - x/2 +x
Left side : we removed x gallons of pure wine from the keg, and then afterwards we took another jug full of WINE and WATER (mixed in equal proportions since the question explicitly states "thoroughly mixed")
No, "thoroughly mixed" does NOT mean "equal proportions" unless your jug happened to be exactly 1/2 the size of the keg- since you are
assuming that, it should be no surprise that you get "5 gallons", half the size of the keg! For example, if the size of the jug were 1 gallon, you would have removed one gallon of wine so that the keg has 9 gallons of wine left. Adding one gallon of water, you now have 9 gallons of wine and 1 gallon of water so that, after "thoroughly mixing" each gallon contains 9/10 gallon of wine and 1/10 gallon of water.
If the jug were really 5 gallons, after the first removal of wine and addition of water, you would have 5 gallons of wine and 5 gallons of water- "thoroughly mixing" each gallon would be half wine and half water. Removing 5 more gallons would remove 5/2= 2.5 gallons of wine leaving 2.5 gallons of wine. Replacing that with water would give 2.5 gallons of wine, 7.5 gallons of water, NOT "equal quantities".
You are told, rather, that after the
second jug is removed, and replaced with water, you have "equal proportions" of wine and water.
Right side of the equation : we added x gallons of pure water to the keg, then were forced to remove x/2 water from the keg (unintentionally) and later added x gallons of pure water
The answer comes out to be 5 gallons .
Is it correct now ?
The keg initially contained 10 gal of wine. You remove x gallons and replace with water, you now have 10- x gallons of wine in the keg, x gallons of water. After "thoroughly mixing", each gallon in the keg contains (10- x)/10 gallons of wine, x/10 gallons of water.
The second jug, then, removes x(10- x)/10 gallons of wine, leaving (10-x)- x(10- x)/10= (100- 10x- 10x+ x^2)/10= (100- 20x+ x^2)/10 gallons of wine in the keg.
If the keg
then "contains equal quantities of water and wine", that is, 5 gallons of wine and 5 gallons of water, we must have (100- 20x+ x^2)/10= 5. So 100- 20x+ x^2= 50 and then x^2- 20x= -50. Solve that equation for x.
That quadratic equation has two roots, only one of which is valid for this problem.
