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."

First, assuming that the gardener's and cook's bounds are completed in the same amount of time, and that we are speaking in whole bounds, and that when "but then" is used, we are referring to turn point (for the gardener) of race. 

So when the Gardener reached his 34th bound, or 102 feet at the turn, the cook was at 68 feet, at which point the cook increased (her) pace to 3 bounds per his 2 bounds, which is actually the same pace as him.  (3 bounds at 2ft vs 2 bounds at 3ft)  as she is already behind by 34 feet (give or take depending on whole or partial bounds at the turn) at this point, she will lose the race by that amount. 

The Gardener Wins!  By a (0.00643939)Mile!

 

Posted by Adam on 2006-03-10 05:05:54