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Topic: wine and water (Read 856 times) |
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skogen
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wine and water
« on: Apr 27th, 2007, 6:36am » |
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Hi this is my first time posting in this forum, and that is because i want to know the answer of the riddle with the water and the wine
"Suppose a 1-liter bottle of wine is hanging from the ceiling, and from it hangs a 1-liter bottle of water. On the bottom of both bottles there is a small hole, through which a constant amount of fluid pours out (neglecting pressure differences). So the wine bottle is becoming empty and the water bottle remains full, though the concentration of wine is increasing.
Assuming that wine and water mix instantly and completely, find out the concentration of wine in the water bottle after the top one is empty. Give two different methods of solution."
I have searched for the solution but i cant find it, and I cant solve it myself, anyway here able to help?
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towr
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Re: wine and water
« Reply #1 on: Apr 27th, 2007, 8:05am » |
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hmm, well. Consider there are X drops in a liter.
You can then consider the change in concentration at each step. The recursion for the concentration C will be
C(n+1)=[1+(X-1)*C(n) ]/X starting at C(0)=0.
You can turn that into a closed formula, and look at C(X) (the concentration for the last drop of wine)
Now possibly, you want an answer for the continuous case (i.e. not drops but a stream), in which case take the limit of X  
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Wikipedia, Google, Mathworld, Integer sequence DB
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Obob
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Re: wine and water
« Reply #2 on: Apr 27th, 2007, 9:56am » |
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I think this method works.
Say water drips out of each container at a rate of r liters/second. Denote by C(t) the concentration of wine in the lower bottle at time t. So C(0)=0. Now the rate at which the concentration is changing is given by C'(t) = r(1-C(t)). The first term on the right comes from the wine pouring in on the top, which increases the concentration, while the second term accounts for the decrease in concentration. One can solve this differential equation, and the solution is
C(t) = 1-exp(-rt).
Now the wine bottle is empty at the time t_0 that rt_0 = 1 liter, and at this time
C(t_0) = 1 - exp(-1).
So the concentration should be 1-e^{-1}, which is approximately .63. The only step which I'm a little shaky on is justifying that the differential equation is the right one, but seeing as this answer is physically plausible I think it probably is.
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skogen
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Re: wine and water
« Reply #3 on: Apr 27th, 2007, 9:57am » |
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I must say I did not quite understand that:S What is the n? the number of drops?
I have a teacher in math, he said I would get the answer by typing 1/e1 on a calculator, will that be right?
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| « Last Edit: Apr 27th, 2007, 9:58am by skogen » |
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Obob
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Re: wine and water
« Reply #4 on: Apr 27th, 2007, 10:06am » |
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I think 1/e would be the concentration of water in the water bottle after the wine has all dripped in. If you think about it, until there is as much wine in the water bottle as there is water, there will be more water that is lost than wine from the bottom bottle. So the concentration of wine must be at least 50%, seeing as more water is lost than wine.
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skogen
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Re: wine and water
« Reply #5 on: Apr 27th, 2007, 10:21am » |
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I do not know how to get the 1/e1, I do not know what to do at all really, at least I think it is very hard to think now
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Obob
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Re: wine and water
« Reply #6 on: Apr 27th, 2007, 10:52am » |
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Chances are if you haven't heard of the number e yet, then this problem is too advanced for you at this point. Both towr and my solutions require some knowledge of calculus.
The explicit solution for towr's recurrence is
c(n) = 1 - (1-1/x)^n.
(Mathematica's RSolve command gives this, I didn't know this nice command before!) Taking n=x and letting x-> infinity gives the same answer 1-1/e as I got by solving a differential equation.
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| « Last Edit: Apr 27th, 2007, 2:59pm by Obob » |
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JiNbOtAk
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Re: wine and water
« Reply #7 on: Apr 30th, 2007, 4:44am » |
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e
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