Wine/water mixing problem

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In the wine/water mixing problem, one starts with two containers, one holding wine and the other an equal volume of water. A cup of wine is taken from the wine container and added to the water. A cup of the wine/water mixture is then returned to the wine container, so that the volumes in the containers are again equal. The question is then posed—which of the two mixtures is purer?[1]

The problem can be solved without resorting to computation. It is not necessary to state the volumes of wine and water, as long as they are equal. The volume of the cup is irrelevant, as is the degree of stirring of the wine/water mixture. Also, any number of transfers can be made, as long as the volumes of liquid in each cup are the same at the end.[2]

Contents

[edit] Solution

The mixtures should be visualized as separated into their water and wine components. Conservation of substance then implies that the volume of wine in the container holding mostly water has to be equal to the volume of water in the container holding mostly wine.[2] To help in grasping this, the wine and water may be represented by 80 red and 80 white marbles, respectively. If 20, say, red marbles are mixed in with the white marbles, and 20 marbles of any color are returned to the red container, then there will again be 80 marbles in each container. If there are now x white marbles in the red container, then there must be x red marbles in the white container. The mixtures will therefore be of equal purity. An example is shown below.

Time Step Red Marble Container Action White Marble Container
0 80 (all red) 80 (all white)
1 20 (all red) →
2 60 (all red) 100 (80 white, 20 red)
3 ← 20 (16 white, 4 red)
4 80 (64 red, 16 white) 80 (64 white, 16 red)

[edit] General Case

Suppose container 1 has r1 red marbles and container 2 has r2 red marbles. Both containers have the same amount of marbles. Let f be the fraction of marbles transferred during step 1. (In the example above, f = 20 / 80.) After 4n time steps, the number of red marbles in each container is

r_1^{(n)} = \frac{r_1+r_2}{2} + \frac{r_1-r_2}{2}\left(\frac{1-f}{1+f}\right)^n,
r_2^{(n)} = \frac{r_1+r_2}{2} - \frac{r_1-r_2}{2}\left(\frac{1-f}{1+f}\right)^n;

assuming the marbles are divisible.

[edit] History

This puzzle was mentioned by W. W. Rouse Ball in the third, 1896, edition of his book Mathematical Recreations And Problems Of Past And Present Times, and is said to have been a favorite problem of Lewis Carroll.[3][4]

[edit] References

  1. ^ Gamow, George & Stern, Marvin (1958), Puzzle math, New York: Viking Press, pp. 103-104 
  2. ^ a b Chapter 10, Hexaflexagons And Other Mathematical Diversions: the first Scientific American book of puzzles & games, Martin Gardner, Chicago and London: The University of Chicago Press, 1988. ISBN 0-226-28254-6.
  3. ^ p. 662, Mathematical Recreations And Problems Of Past And Present Times, David Singmaster, pp. 653–663 in Landmark Writings in Western Mathematics 1640-1940, edited by Ivor Grattan-Guinness, Roger Cooke, et al., Elsevier: 2005, ISBN 0444508716.
  4. ^ Mathematical recreations and problems of past and present times, W. W. Rouse Ball, London and New York: Macmillan, 1896, 3rd ed.