The following special catch puzzle appeared in the issue of The Weekly Dispatch for All Fools' Day, 1900. It caused considerable amusement; for out of a very large body of competitors, many quite expert, not a single person solved it, though it ran for nearly a month.

" A race between a man and a woman that I happened to witness one All Fools' Day has fixed itself indelibly on my memory. It happened at a country-house, where the gardener and the cook decided to run a race to a point 100 feet straight away and return. I found that the gardener ran 3 feet at every bound and the cook only 2 feet, but then she made three bounds to his two. Now, what was the result of the race?"

A fortnight after publication the editor added the following note: "It has been suggested that perhaps there is a catch in the 'return,' but there is not. The race is to a point 100 feet away and home again—that is, a distance of 200 feet. One correspondent asks whether they take exactly the same time in turning, to which I reply that they do. Another seems to suspect that it is really a conundrum, and that the answer is that 'the result of the race was a (matrimonial) tie.' But I had no such intention. The puzzle is an arithmetical one, as it purports to be."

As the first half of the race is to a point 100 feet away, the gardener, who runs 3 feet per bound will have to travel 102 feet, before turning, while the cook will only have to run 100 feet.  As the problem states that even some experts did not come up with the solution, the problem would most likely lie in that the gardener was not the man, but the woman, and vice versa.  Making this the assumption, the cook only made two bounds (a total of 4 feet) for every three of the gardener’s (a total of 9 feet).  When the gardener had travelled the 204 feet to complete the race, the cook would have only traveled 4/9 the distance, or 90 2/3 feet -- i.e, not yet to the turning point.

Posted by Dej Mar on 2006-03-01 18:12:19