A servant has a method to steal wine. He removes 3 cups from a barrel of wine and replaces it with 3 cups of water. The next day he wants more wine, so he does the same thing: he removes 3 cups from the same barrel (now with diluted wine) and replaces it with 3 cups of water. The following day he repeats this one more time, so he has drawn 3 times from the same barrel and has poured back 9 cups of water. At this point, the barrel is 50% wine and 50% water. How many cups of wine were originally in the barrel?

Thanks for help.

Guest Jun 23, 2017

#1**+2 **

Let us start with a barrel of wine.

We have: Wine amount(in cups)=W, and its concentration is 100%. Then he removes 3 cups from diluted wine, so we have: W-3 and its concentration would be:[W-3] / W, or [1 - 3/W]. Then he removes 3 more cups of diluted wine, so we have:W-3-3[1-3/W]=W-6+9/W, and the concentration will be={W-6+9/W} / W =[1-3/W]^2. Then he removes 3 more cups of diluted wine, so we have:W-6+9/W-3[1-3/W]^2=[W-9+27/W-27/W^2] and the concentration would be:[1-3/W]^3. So, now we know the final concentration=

[1-3/W]^3 = 1/2, or 50%. Will solve for W.

[1-3/W^3] =1/2,

= [1-3/W]=[(1/2)^1/3]. Then we have:

W =[3(2)^1/3] / [(2)^1/3 - 1]

**W = ~14.5420 cups of wine in the barrel.**

Guest Jun 23, 2017

edited by
Guest
Jun 23, 2017

#2**+2 **

Let N be the number of cups of liquid in the barrel. This is pure wine to start with.

At any point in time, the concentration of wine is defined as the amount of pure wine in the barrel divided by N.

Let C1 be the concentration of wine in the barrel at the beginning of the first day before anything is removed that day.

Let C2 be the concentration of wine in the barrel at the beginning of the second day before anything is removed that day.

Let C3 be the concentration of wine in the barrel at the beginning of the third day before anything is removed that day.

We know that C1=1.

During the first day, three cups of liquid are removed from the barrel. Those three cups contain 3 *C1 cups of pure wine.

That means the concentration of wine in the barrel at the end of the first day (and the beginning of the second day) is

C2=(N - 3*C1) / N

During the second day three more cups of liquid (this time it is diluted) are removed from the barrel. Those three cups contain 3*C2 cups of pure wine. That means the concentration of wine in the barrel at the end of the second day (and the beginning of the third day) is

C3 = (N - 3*C1 - 3*C2) / N

During the third day, three more cups of diluted liquid are removed from the barrel. Those three cups contain 3*C3 cups of pure wine. That makes the concentration of wine in the barrel drop to

C4 = (N – 3*C1 – 3*C2 – 3*C3) / N

which we are told should be 0.5

Plug the equation for C1 into the one for C2, then plug that in to the equation for C3 then plug that into the equation for C4 .

Set C4 equal to 0.5 and solve the resulting cubic to get 14.542 cups of wine in the barrel to start with.

Guest Jun 23, 2017

#4**+1 **

Each at the end of it all the wine is diluted by 50%. And since each each iteration of the process will dillute the wine by the same factor each time we know that the dilution factor z is z = (0.5)^(1/3) which turns out to be z = 0.7937.

And the dilution of the first iteration is simply the wine left divided by the entire volume.(B = Barrel) z = wineleft / B

And wine left is just B - 3 cups. z = (B-3cups)/B. z = 1- 3cups/B. rearranging for B:

zB = B - 3 c

zB -B = -3c

B*(z-1) = -3c

B= -3c/(z-1)

And putting in the values you get

B = 14.512 cups

Man writing this out took me longer than actually calcualting it

Guest Jun 23, 2017

#5**+1 **

In a problem such as this one, you are only concerned with the final concentration. As framed by Guest #1, the first dilution was: [1 - 3/W].The second would be:[1 - 3/W]^2. And the third would be:

[1 - 3/W]^3. Since this last concentration =50% of the wine, it is a simple matter of solving:

[1 - 3/W]^3 = 1/2, and W=14.542 cups.

Guest Jun 24, 2017