Another quite interesting math puzzle...?

Question

A dishonest barman removes three litres of wine from a barrel, replacing them with water and mixing the contents. He repeats the theft twice more, removing in total nine litres and replacing them with water. As a result the wine remaining in the barrel is half its former strength. How much wine did the barrel originally hold? (The answer will run into decimals, I think)

Can anyone help? 10 points goes to the correct answer with the clearest explanation.

Thank you!

Answer 1

Amount of wine in the initial barrel = W:
f(0) = W

After the first replacement, the amount of wine is reduced by 3 litres. The ratio of these litres is 100% wine or f(0)/W, so this becomes:
f(1) = f(0) - 3 * f(0)/W

The next removal, the 3 litres have a ratio of f(1)/W wine, so we have:
f(2) = f(1) - 3 * f(1)/W

And finally the next 3 litres have a ratio of f(2)/W wine, so we have:
f(3) = f(2) - 3 * f(2)/W

In general, we can rewrite f(n):
f(n) = f(n-1) -3/W * f(n-1)
f(n) = [ 1-3/W ] * f(n-1)
f(n) = [ (W-3)/W ] * f(n-1)

So each time the concentration changes by a ratio of (W-3)/W. We could jump ahead to realize that we'll have this ratio cubed, but let's see the steps:

f(0) = W
f(1) = W-3
f(2) = (W-3)² / W
f(3) = (W-3)^3 / W²

And we are told the final ratio is equivalent to 1/2
f(3)/W = 1/2

Now just solve for W by taking the cube root of both sides:
1/2 = (W-3 / W)^3
W-3 / W = cube root(1/2)

Then multiply by W:
W - 3 = W cube root(1/2)

Put W terms on the left:
W - W cube root(1/2) = 3

Factor out W:
W(1 - cube root(1/2)) = 3

Divide to get W alone:
W = 3 / (1 - cube root(1/2) )

W ≈ 14.5419663 litres

As a double check:
At the beginning we have 14.54197 litres of wine
After removing 3 liters (100% concentration of wine) we have:
11.54197 litres (this is now a 79.37% mixture).
After removing 2.3811 litres (79.37% concentration) we have:
9.160865 litres (this is now a 63% mixture)
After removing 1.8899 litres (63% concentration) we have:
7.270983 litres. And this is a 50% mixture of the original.

So the final answer is:
The barrel originally held approximately 14.5419663 litres.

Answer 2

You have to use a recursive method. At the end, the barrel contains less than 9 liters of water, since some water was removed along with the wine the second two times.

The first removal is 3 liters, leaving you with x-3 liters of wine.

The second removal is 3 liters of fluid with a concentration of (x-3)/x. In other words, you have:

(x-3) - 3*(x-3)/x

Dividing that by x gives you your new concentration.

The third removal is 3 liters of [(x-3) - 3* (x-3)/x]/x and will leave you with x/2 liters of wine in the barrel or:

(x-3) - 3*(x-3)/x -3*[(x-3) - 3* (x-3)/x]/x = x/2.

The rest is some simple algebra to simplify the equation into a cubic formula with three roots. Except not so simple that I didn't make an error somewhere along the line and wind up with 15.338 liters instead of the correct amount of (re-edit:14.542 liters - no wonder I had a typo in my Excel formula). Microsoft Excel allows you to brute strength these type of problems, sometimes with less chance of error (provided you enter the formula correctly and don't always start with 15 liters - heh).

Answer 3

Let

w = liters of wine in barrel originally

He removes 3 liters the first time. Now we have:

w - 3 liters of wine
3 liters of water

Concentration of wine is now

(w - 3)/w

He removes 3 liters the second time. But it was 3 liters of a mixture of wine and water. Now we have:

(w - 3) - 3*(w - 3)/w = (w - 3)(1 - 3/w)

And the concentration is now

{(w - 3)(1 - 3/w)}/w = {(w - 3)(w - 3)}/w²
= (w - 3)²/w²

Which is not surprising. After the third time the concentration will be:

(w - 3)³/w³

But

(w - 3)³/w³ = 1/2
(w - 3)/w = (1/2)^(1/3)
w - 3 = {(1/2)^(1/3)}w
w{1 - (1/2)^(1/3)} = 3
w = 3/{1 - (1/2)^(1/3)} ≈ 14.541966 liters of wine originally

Answer 4

Let X be the capacity of the barrel - i.e. how much wine the barrel originally holds. Let W(n) be the amount of wine in the barrel after the nth theft. This means that W(0) must be equal to X. During the (n+1)th theft, the amount of wine removed is equal to (9 liters)*(wine concentration before that theft) = 9*W(n)/X. Therefore, the amount of wine left after the (n+1)th theft is W(n+1) = W(n) - 9W(n)/X = W(n)*(1 - 9/x). Simplifying the recursive series, W(n) = (1 - 9/x)^n. We're told that W(3) = 0.5*W(0), so we have 0.5 = (1 - 9/x)^3 ==> 1 - 9/x = 0.7937 ==> 9/x = 0.2063 ==> x = 43.6258 liters

Answer 5

Heh, three diifferent answers...

Before I solved this, I tried some numbers, and 15 liters came close. Based on these trials, the exact formula should be 3/(1 - 1/2^1/3), which is what Puzzling has. So, I agree with him.

* * * * *

OK, checking back, I see a few posters between my post and Puzzling now have that exact formula, which I know wasn't there before. Well, Puzzling definitely worked it out first (along with Amber, who worked it out by interpolating).

Answer 6

-the barrel now containg 9 liters of water and unknown liters of wine
-the wine now is half the strength
-that means if the barrel at the begining contains pure wine (100%)and the wine now is half the original strength (50%)
that means equal volume of water and wine.
so the barre volume would be 9 liters(water)+9liters(wine)=18 liter
that means the barrel originally contains 18 liters of wine

and i guess im drunk now after all this wine

Answer 7

My result confirms with Puzzling

You start with 14.55 Litres.

Remove 3 Liters, the wine strength is 79% after adding back the water:
(14.55-3) / 14.55 = 0.79

Remove another 3 Liters, you in essence remove 2.4 Liters of Pure Alcohol plus 0.6 Liters of water.
3 x 0.79 = 2.4

Add back 3 L of water, and now your strength is 63%.
(14.55 - 3-2.4) / 14.55 = 0.63

Finally you remove another 3 Liters, in essence removing 1.9 L of alcohol and 1.1 L of water.
3 x 0.63 = 1.9

Your strength now should be 50%
(14.55 - 3 - 2.4 - 1.9) / 14.55 = 0.5

I did this in Excell by guessing and checking instead of working out the math.

Answer 8

One has (x-3)^3/x^3=1/2. So x=3/(1- cubic root of (1/2) ) so
x= 14. 542

Answer 9

the wine that barrel originally hold would be 9 liters