A Posteriori

Attempts to grapple with and elucidate empirical knowledge

Mixes made equal November 9, 2009

Filed under: Uncategorized — Rāhul @ 16:38
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Today, I shall explore a simple puzzle which illustrates the advantages of extreme conditions in getting pointers towards the answer. Consider this scenario found at the Times’ Tierneylab blog a few weeks ago. Essentially, the question is-

If we have two vessels, X and Y, X containing a volume “a” of liquid A and Y containing  a volume “b” of liquid B which mixes with liquid A uniformly, and we pour a volume x of liquid A from X to Y first followed by pouring a volume x of the mixture of liquids in Y back to X, will there now be more of liquid A in Y than of liquid B in X? Assume that b is greater than a.

Before solving this algebraically, let’s think about the problem. Since the volume b is more than a, there is more of liquid B than liquid A. When x of b is poured into X, X will now have a+x amount of a mixture of A and B whereas Y will have b-x amount of just liquid B. When x amount of the mixture in X is poured back to Y, X will go back to having just a amount, but of the mixture of A and B, and Y will go back to having b amount, again of the mixture of A and B. There doesn’t seem to be any immediately compelling reason why the amount of B in X and A in Y should be the same or for the percentages to be the same either. So, let’s consider an extreme case for some help with the arguments.

Let us consider the case where x=0. Then, we can see that no liquid is exchanged between X and Y and so at the end the amount of B in X and A in Y are the same, which is zero.  Now consider the case where x=a. After the first exchange, we’ll have 2a amount of liquid in X, which is made of a amount of A and a amount of b, i.e. 50% of each. When x=a amount is poured back from X, this would involve a/2 amount each of A and B. Adding this to the b-a amount of B that was left in Y, we’ll have b-(a/2) amount of B and a/2 amount of A there. At the same time, X will have a/2 amount each of A and B. So, the amount of B in X and A in Y are the same- a/2.

In both the special cases we considered, the amount of B in X and amount of A in Y are the same. It is interesting to see that the concentration of B in X and of A in Y are not the same because the volumes a and b are not the same. Given these two special cases, which give the same result for very different values of x, it seems likely that the result might hold for all x. These examples do not prove so, but it does motivate us as to the answer to look for. So, while we solve it algebraically, not only will we have a higher confidence in the result if we prove it, it is likely that we are more likely to get to it since we know what we are looking for!

Getting to the algebra now, in the general case, after the first exchange of liquids, X will now have a-x amount of A and Y will have a total amount of b+x, of which there is b of B and x of A. When x amount of the liquid from Y is now poured back to X, since the liquids in Y mixed well, the amount poured back will involve [b/(b+x)]*x of B and [x/(b+x)]*x of A. In the proportion of their volumes, this now adds up to x amount poured back from Y to X.  So, the amount of A left in Y is x-[(x*x)/(b+x)] which is bx/(b+x). The amount of B now in X would just be the amount of B poured from Y to X, which again is bx/(b+x). Voila, we have the same number!

This means that the concentration of A in Y is much less than the concentration of B in X. So, if I gulp in a little sea water and spit out the same amount, there will be the same amount of sea water left in my mouth as the amount of my spit in the sea, but my mouth will taste much saltier than the sea will of my spit!


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