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[leyenda_asturias]

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!

[LIGER20498]

the wine that barrel originally hold would be 9 liters

[RobBofMD]

-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 literthat means the barrel originally contains 18 liters of wineand i guess im drunk now after all this wine

[Hatlady]

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

[Legend0113]

Amount of wine in the initial barrel = W:f(0) = WAfter 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)/WAnd finally the next 3 litres have a ratio of f(2)/W wine, so we have:f(3) = f(2) - 3 * f(2)/WIn 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) = Wf(1) = W-3f(2) = (W-3)² / Wf(3) = (W-3)^3 / W²And we are told the final ratio is equivalent to 1/2f(3)/W = 1/2Now just solve for W by taking the cube root of both sides:1/2 = (W-3 / W)^3W-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) = 3Factor out W:W(1 - cube root(1/2)) = 3Divide to get W alone:W = 3 / (1 - cube root(1/2) )W ≈ 14.5419663 litresAs a double check:At the beginning we have 14.54197 litres of wineAfter 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.

[LisaC]

Letw = liters of wine in barrel originallyHe removes 3 liters the first time. Now we have:w - 3 liters of wine3 liters of waterConcentration of wine is now(w - 3)/wHe 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/w²Which is not surprising. After the third time the concentration will be:(w - 3)³/w³But(w - 3)³/w³ = 1/2(w - 3)/w = (1/2)^(1/3)w - 3 = {(1/2)^(1/3)}ww{1 - (1/2)^(1/3)} = 3w = 3/{1 - (1/2)^(1/3)} ≈ 14.541966 liters of wine originally

[Becky]

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

[SpellCheck5343]

My result confirms with PuzzlingYou start with 14.55 Litres.Remove 3 Liters, the wine strength is 79% after adding back the water:(14.55-3) / 14.55 = 0.79Remove another 3 Liters, you in essence remove 2.4 Liters of Pure Alcohol plus 0.6 Liters of water.3 x 0.79 = 2.4Add back 3 L of water, and now your strength is 63%.(14.55 - 3-2.4) / 14.55 = 0.63Finally you remove another 3 Liters, in essence removing 1.9 L of alcohol and 1.1 L of water.3 x 0.63 = 1.9Your strength now should be 50%(14.55 - 3 - 2.4 - 1.9) / 14.55 = 0.5I did this in Excell by guessing and checking instead of working out the math.

[James]

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).

[xoxoxoxoxoxoxoxoxo]

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)/xDividing 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).