How can I solve this mathematically?

[kingwinner]

A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water. Is there more wine in the water or is there more water in the wine? :)

 

How can I solve this question mathematically? :shrug:

Do I first have to assume that the glass of water taken and the glass of the mixture taken have equal volumes?

[Jay-qu]

yes they would have to be to make it solvable, try assigning values and just keep them constant - say 1L for the jug and 250mL for the glass, it shouldnt matter so long as you are consistant ;) then it is just a matter of fractions, Im sure you can get it from here

:love:

[Four Dogs]

Well I don't know if Mathematics are necessary to prove this. There is certainly more water in the wine. If two glasses were used and each filled with an equal amount of water or wine, then poured back into the opposite container, then there would be an equal amount of water in the wine and vise versa. In your example, pure wine was diluted with pure water. The wine is now a mixture, but the remaining water is pure. You now draw a glass of wine water mixture and add it to the pure water. The wine in the mixture is less concentrated then the water originally added to the wine, therefore the new water wine mixture has a lesser concentration of wine than the wine water mixture has of water.

But just for fun, lets plug in some values. Lets pretend that 1 Jug = 5 Glasses. At the start, our water to wine ratio in Jug 1 (the water) is 0:5

We draw 1 glass of water and add it to Jug 2 (the wine) the wine to water ratio in Jug 2 is now 1:5 or 1/6th and Jug 1 is 0:4. Now we draw 1 glass of 1:5 and pour it back into the remaining pure water. the amount of wine introduced into the water is 1:5 of 1:4 =1/6 of 1/5th or 1/30th =1:29

The amount of water in the wine stays the same at 1:5.

In conclusion, the concentration of water in the wine is 5 times greater than the concentration of wine in the water.

 

Of course, the actual concentrations would vary depending apon the volume of the glass compared to that of the jug.

[Jay-qu]

Its all well and good to solve the problem, but this is the homework forum.. Its much more useful to the questioner if you just help them along instead of doing the whole thing for them :eek_big: sometimes I will but it is normally after I can see that they have tried doing it already and are stuck - not cant be bothered ;)

[CraigD]

…How can I solve this question mathematically? :eek_big:

Do I first have to assume that the glass of water taken and the glass of the mixture taken have equal volumes?

You solve this problem mathematically by writing expressions 4 quantities (you can make up any symbolic names for them you like):

• the amount (volume) of water in the water jug
  
• the amount of wine in the water jug
  
• the amount of water in the wine jug
  
• the amount of wine in the wine jug
  

Just before and after each of the mixing actions. To do this, you’ll need to make up a new symbol for the volume of the glass, or, if you want to see what happens if you use a different glass for each action, a symbol for each of them.

 

You can compare the resulting final expression for the wine or water in the wine or water jug to find the question’s answer with simple algebra.

 

In this problem, though, unless you’re specifically asked to, or need to describe the amounts exactly, you’re probably better off answering the question using less mathematical reasoning. Just ask yourself if the amount of water returned to the water jug is greater or less than the amount removed from it – you should be able to answer the question intuitively.

 

This sort of problem is in a famous class of Science homework problems – it’s intended to confuse the sort of thoughtful student who tries to get an exact answer, but knows that adding a volume of alcohol A to a volume of water B doesn’t produce a mixture of volume A+B, but slightly less – the alcohol and water molecules fit together more compactly than alcohol or water alone. To solve the problem without requiring all sorts of additional data, you must make the simplifying assumption that, for ordinary water and wine, you can ignore this complicating factor.

[kingwinner]

It seems more complicated with mathemical solution......

 

And why do we have to consider the concentrations? We are interested in the amounts only, right?

 

How can it solve the problem intuitively?

Let jug A be water jug

Let jug B be wine jug

 

I know that the glass of mixture taken from jug B has part wine, part water. So only part of the water went back to the jug A. But at the same time, the amount of water in jug B decreased, right? So I don't know whether there is more wine in the water or more water in the wine.

This really got my mind confused...I can't think...it is so difficult to sort out this...help

[Buffy]

You should think about the ratios between the water and the wine in each jug as you do the transfers with the glasses. It is an algebraic solution (not too complicated a one at that!), and you need to do formulas of the concentrations at each stage. The concentrations are ratios, since you don't know the exact size of the jugs and glass, but you do know that the jugs and glass are the same size and you can use them as your variables for the equations you build...

 

Hint, Hint,

Buffy

[Qfwfq]

Do I first have to assume that the glass of water taken and the glass of the mixture taken have equal volumes?

Obviously, "a glass of" is meant as a unit of measure.

 

I know that the glass of mixture taken from jug B has part wine, part water. So only part of the water went back to the jug A. But at the same time, the amount of water in jug B decreased, right?

Now that shows that you're thinking. :eek:

 

The second point is true but, to see if it's relevant, try reasoning on the final total volume instead. ;)

[kingwinner]

Some one else in another forum states that:

"The glass you scoop from B contains x ml of wine, the rest is water. This means you must have left x ml of water behind in Jug B...

So now, you've got x ml of water in the wine, and x ml (the same!) of wine in the water."

 

And this implies that "there are more water in the wine" is a WRONG statement?

 

I am completely mixed up and confused...:D

[C1ay]

I know that the glass of mixture taken from jug B has part wine, part water. So only part of the water went back to the jug A. But at the same time, the amount of water in jug B decreased, right? So I don't know whether there is more wine in the water or more water in the wine.

This really got my mind confused...I can't think...it is so difficult to sort out this...help

Then step back, assign some arbitary values to the volumes and work through it to see what happens. Use nice, round even numbers. Let the jugs be 100 units and the glass be 10 units. Now move through the mixtures and compare the percentages. Try letting the glass be half the size of a jug and look at the percentage or ratio.

[Qfwfq]

Most of all, compare the final volumes, of jug A and B. Initially, they were equal. Afterwards, they are... (?)

[CraigD]

I am completely mixed up and confused...:lol:

I feel your pain, kingwinner.

 

Let’s see if we can derive some useful general formulae for “taking C units from a mixture of A units of one thing and B or another”.

 

First, what’s the concentration of StuffA in a mixture of A units of StuffA and B units of StuffB? Concentration (we’ll assume we mean by volume, and nothing complicated happens to volumes in this case) is defined as (quantity of part)/(quantity of whole), so:

ConcentrationStuffA = A/(A+;)

Note that concentration must be between 0 and 1.

Next, realize that when on takes a volume of C units from a mixture, one gets equal parts of the stuff in it, as determined by their concentration, so:

NewA = C*A/(A+B)

NewB = C*B/(A+B)

 

With these formulae, we can now answer a question like “if we take 100 cm^3 from a mixture of 2000 cm^3 of wine and 8000 cm^3 of water, how much of each do we get, and what’s left?”

NewA = 100*2000/(2000+8000) = 20

NewB = 100*8000/(2000+8000) = 80

So, we get 20 cm^3 of wine, 80 cm^3 of water. What’s left is

A = OldA – NewA = 2000-20 = 1980

B = OldB – NewB = 8000-80 = 7920

 

We now have the tools to answer any question about swapping mixtures between jugs. As several folk have suggested, just make up some numbers for the volumes held by the jugs (remember the problem says they hold equal amounts) and held by the glass, and go through the process step-by-step, finding how much water and wine is in each jug and the glass at each step.

 

I think you’ll be delighted (and maybe surprised) at the answer you get. I was.