# Homework Help: Basic Algebra based question

1. Feb 2, 2013

### hms.tech

1. The problem statement, all variables and given/known data

A man had a ten gallon keg of wine and a jug. One day, he drew off a jug full of wine and filled up the keg with water. Later on, when the wine and the 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 water and wine. What was the capacity of the Jug ?

2. Relevant equations

Algebra

3. The attempt at a solution

Here is what I think should be the correct solution :
Let x be the capacity of the jug
10-2x = 2x
∴x= 2.5 Gallons

Am I correct ?

2. Feb 2, 2013

### HallsofIvy

It's impossible to know if what you are doing makes sense (whether or not "2.5 gallons" is correct) because you have not said what you are doing. What do "10- 2x" and "2x" represent? If you intend "10- 2x" to represent the the amount of wine in the keg you are wrong because the second jug drawn off is NOT all wine.

3. Feb 2, 2013

### hms.tech

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")

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 ?

Last edited: Feb 2, 2013
4. Feb 2, 2013

### HallsofIvy

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.

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.

Last edited by a moderator: Feb 2, 2013
5. Feb 2, 2013

### hms.tech

Got it . So the underlying concept was to prioritize "concentration" over quantity

6. Feb 2, 2013

### HallsofIvy

I'm not sure I like the word "prioritize" here but, yes, you need to work with the concentration.

By the way, what answer did you get?

7. Feb 2, 2013

10-5(2)^0.5

≈2.9 gallons