Hey, this is Presh Talwalkar.
A servant has a method to steal wine.
He removes three 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 three cups from the same barrel that now contains diluted wine and replaces it with 3 cups of water.
The following day he repeats this one more time.
So he has drawn a total of three times from the same barrel and has poured back a total of nine cups of water.
At this point the barrel is 50% wine and fifty percent water.
The question is how many cups of wine were originally in the barrel?
This problem is a classic mathematical puzzle from the 16th century.
Can you figure it out?
Give this problem a try and when you're ready keep watching the video for the solution.
We'll get started with the barrel of wine.
We're going to keep track of the number of cups of wine in the barrel and the concentration of wine in the barrel.
At the beginning we'll denote the unknown amount of wine by the variable x cups.
The barrel at first contains only wine, so it's concentration will be one or it's a hundred percent wine.
The servant then removes three cups from the barrel and replaces it with 3 cups of water.
The barrel will have a diluted amount of wine.
We know that three cups of wine are removed so the total amount of wine remaining will be x minus 3.
Since the servant has placed back in three cups of water
the total amount of liquid in the barrel remains the same at x.
So the concentration of wine will be the amount of wine x minus 3 divided by the total volume of liquid x.
If we divide this through by x we have 1 minus 3 over x. It will be convenient to write it in this form.
The next day the servant does the same thing
so the wine and the barrel gets diluted even more.
How much wine will be in the barrel?
When we already have x minus 3 cups that are in there and then he removes 3 cups of the diluted wine.
So the amount of wine remaining will be x minus 3 minus 3
times the diluted wine that he is removing. The concentration of the diluted wine
allows us to figure out the exact amount of wine that he's removing as opposed to water.
So we can now simplify this fraction to figure out the amount of wine that's remaining in the barrel.
This will be x minus 6 plus 9 over x.
Once again the total volume of liquid remains at x because the servant is replacing the wine that he removes with water.
So the concentration of wine will be this amount of wine
divided by x.
We can then divide through
and then we can actually see that we can factor this term.
We end up with the concentration of 1 minus 3 over x, the quantity squared.
So the following day the servant does this one more time, so the wine gets even more diluted?
So from the amount of wine in the previous day,
we subtract out three times the concentration of wine from the previous day.
So we can simplify this fraction to figure out the amount of wine that's remaining in the barrel.
This will be 9 minus x plus 27 over x minus 27 over x squared.
This amount divided by the total volume x is how much of the wine concentration is remaining in the barrel.
I'll go ahead and simplify this and it actually ends up being one minus 3 over x cubed.
So that's the final concentration of wine now.
We've gone through all this algebra, and I just want to point out there was a shortcut
we could have used if we had realized earlier.
After we had gone through the first step of figuring out the concentration of wine,
we could have noticed that the concentration going from one hundred percent
to the first step is found by multiplying by 1 minus 3 divided by x.
Now in order to get to the next concentration. We actually could have just multiplied by the same percentage
to get to the next concentration.
We could have done the same thing once more when we're looking at the concentration and we're iterating the same process the
concentrations get multiplied by each other.
In any case we figured out the concentration of wine at the very end.
And we are given this is equal to fifty percent.
So we now have an algebra problem to solve.
We have the quantity 1 minus 3 over x cubed equal to one-half.
We can take the cube root of each side, and then we can isolate and solve for the term x.
We end up with x being equal to three times the cube root of 2 all divided by the cube root of 2 minus 1.
And this gives us our answer that there were approximately
14.54 cups of wine in the barrel to start with.
Did you figure this out?
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