It is a brain-teaser problem. You can replace any liquid name (e.g. coffer, alcohol) with wine or water in the title. The original problem is explained as follows: http://en.wikipedia.org/wiki/Wine/water_mixing_problem
"In the wine/water mixing problem, one starts with two jars, one holding wine and the other an equal volume of water. A spoon of water is taken from the water jar and added to the wine. A spoon of the wine/water mixture is then returned to the water jar, so that the volumes in the jars are again equal. The question is then posed—which is more, the wine volume in the water jar or the water volumn in the wine jar?"
Here is my solution.
We assume:
Originally, the volume of water in jar of water (jar W): Vw
Originally, The volume of wine in jar of wine (jar A): Va
The volume of spoon: Vs
We first a get a spoon of water from jar W and put it into jar A. Then we get a spoon of liquid from jar A and put it back into jar W.
p is percentage of wine in the second spoon transferred from jar A to jar W. (It is interesting that we do not need to mix the water and wine thoroughly.)
1) Situation 1: Vw == Va
(1) Case 1: Vs <= Vw Water in jar A: S1 = Vs – (1- p) Vs = p * Vs Wine in jar W: S2 = p * Vs In this case, the water in jar A has the same amount as the wine in jar W. (S1 = S2). (2) Case 2: Vw < Vs <= Vw + Va Water in jar A: S1 = Vw – (1-p)Vs = Vw – Vs + p * Vs Wine in jar W: S2 = p*Vs In this case, the water in jar A is less than the wine in jar W (S1 < S2) (3) Case 3: Vw + Va < Vs Water in jar A: S1 = Vw – (1-p)(Vw + Va) = 0 , where p= (Va/(Vw+Va)) Wine in Jar W: S2 = (Vw + Va) * p = Va, where p = (Va/Vw+Va) In this case, the water in jar A is less than the wine in jar W (S1 < S2) We can combine case 2 and case 3 together. S1 is same as S2 ; when Vs≤Vw
S1 is less then S2 ; when else
where S1 represents the water in jar A (original wine jar); S2 represents the wine in jar W (original water jar)
2) Situation 2: Vw != Va , without loss of generality, we assume Vw < Va
(1) Case 1: Vs <= Vw Water in jar A: S1 = Vs – (1- p) Vs = p * Vs Wine in jar W: S2 = p * Vs In this case, the water in jar A has the same amount as the wine in jar W (S1 = S2). (2) Case 2: Vw < Vs <= Va + Vw Water in Jar A: S1 = Vw – (1-p) Vs = Vw – Vs + p * Vs Wine in Jar W: S2 = p * Vs In this case, the water in jar A is less than the wine in jar W (S1 < S2) (3) Case 3: Vw + Va < Vs Water in jar A: S1 = Vw – (1-p)(Vw + Va) = 0 ; where p= (Va/(Vw+Va)) Wine in jar W: S2 = (Vw + Va) * p = Va; where p = (Va/Vw+Va) In this case, the water in jar A is less than the wine in jar W (S1 < S2) We can combine case 2 and case 3 together. We assume Vw < Va. S1 is the same as S2 ; when Vs≤Vw
S1 is less than S2 ; when else
where S1 represents the water in jar A (original wine jar); S2 represents the wine in jar W (original water jar)
"In the wine/water mixing problem, one starts with two jars, one holding wine and the other an equal volume of water. A spoon of water is taken from the water jar and added to the wine. A spoon of the wine/water mixture is then returned to the water jar, so that the volumes in the jars are again equal. The question is then posed—which is more, the wine volume in the water jar or the water volumn in the wine jar?"
Here is my solution.
We assume:
Originally, the volume of water in jar of water (jar W): Vw
Originally, The volume of wine in jar of wine (jar A): Va
The volume of spoon: Vs
We first a get a spoon of water from jar W and put it into jar A. Then we get a spoon of liquid from jar A and put it back into jar W.
p is percentage of wine in the second spoon transferred from jar A to jar W. (It is interesting that we do not need to mix the water and wine thoroughly.)
1) Situation 1: Vw == Va
(1) Case 1: Vs <= Vw Water in jar A: S1 = Vs – (1- p) Vs = p * Vs Wine in jar W: S2 = p * Vs In this case, the water in jar A has the same amount as the wine in jar W. (S1 = S2). (2) Case 2: Vw < Vs <= Vw + Va Water in jar A: S1 = Vw – (1-p)Vs = Vw – Vs + p * Vs Wine in jar W: S2 = p*Vs In this case, the water in jar A is less than the wine in jar W (S1 < S2) (3) Case 3: Vw + Va < Vs Water in jar A: S1 = Vw – (1-p)(Vw + Va) = 0 , where p= (Va/(Vw+Va)) Wine in Jar W: S2 = (Vw + Va) * p = Va, where p = (Va/Vw+Va) In this case, the water in jar A is less than the wine in jar W (S1 < S2) We can combine case 2 and case 3 together. S1 is same as S2 ; when Vs≤Vw
S1 is less then S2 ; when else
where S1 represents the water in jar A (original wine jar); S2 represents the wine in jar W (original water jar)
2) Situation 2: Vw != Va , without loss of generality, we assume Vw < Va
(1) Case 1: Vs <= Vw Water in jar A: S1 = Vs – (1- p) Vs = p * Vs Wine in jar W: S2 = p * Vs In this case, the water in jar A has the same amount as the wine in jar W (S1 = S2). (2) Case 2: Vw < Vs <= Va + Vw Water in Jar A: S1 = Vw – (1-p) Vs = Vw – Vs + p * Vs Wine in Jar W: S2 = p * Vs In this case, the water in jar A is less than the wine in jar W (S1 < S2) (3) Case 3: Vw + Va < Vs Water in jar A: S1 = Vw – (1-p)(Vw + Va) = 0 ; where p= (Va/(Vw+Va)) Wine in jar W: S2 = (Vw + Va) * p = Va; where p = (Va/Vw+Va) In this case, the water in jar A is less than the wine in jar W (S1 < S2) We can combine case 2 and case 3 together. We assume Vw < Va. S1 is the same as S2 ; when Vs≤Vw
S1 is less than S2 ; when else
where S1 represents the water in jar A (original wine jar); S2 represents the wine in jar W (original water jar)
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