Solution

(related to Problem: The Gardener And The Cook)

Nobody succeeded in solving the puzzle, so I had to let the cat out of the bag — an operation that was dimly foreshadowed by the puss in the original illustration. But I first reminded the reader that this puzzle appeared on April $1,$ a day on which none of us ever resents being made an "April Fool;" though, as I practically "gave the thing away" by especially drawing attention to the fact that it was All Fools' Day, it was quite remarkable that my correspondents, without a single exception, fell into the trap.

One large body of correspondents held that what the cook loses in stride is exactly made up in greater speed; consequently both advance at the same rate, and the result must be a tie. But another considerable section saw that, though this might be so in a race $200$ ft. straight away, it could not really be, because they each go a stated distance at "every bound," and as $100$ is not an exact multiple of $3,$ the gardener at his thirty-fourth bound will go $2$ ft. beyond the mark. The gardener will, therefore, run to a point $102$ ft. straight away and return (204 ft. in all), and so lose by $4$ ft. This point certainly comes into the puzzle. But the most important fact of all is this, that it so happens that the gardener was a pupil from the Horticultural College for Lady Gardeners at, if I remember aright, Swanley; while the cook was a very accomplished French chef of the hemale persuasion! Therefore "she (the gardener) made three bounds to his (the cook's) two." It will now be found that while the gardener is running her $204$ ft. in $68$ bounds of $3$ ft., the somewhat infirm old cook can only make $45\frac{1}{3}$ of his $2$ ft. bounds, which equals $90$ ft. $8$ in. The result is that the lady gardener wins the race by $109$ ft. $4$ in. at a moment when the cook is in the air, one-third through his $46$th bound.

The moral of this puzzle is twofold: 1. Never take things for granted in attempting to solve puzzles; 1. always remember All Fools' Day when it comes round.

I was not writing of any gardener and cook, but of a particular couple, in "a race that I witnessed." The statement of the eye-witness must, therefore, be accepted: as the reader was not there, he cannot contradict it. Of course, the information supplied was insufficient, but the correct reply was: "Assuming the gardener to be the 'he,' the cook wins by $4$ ft.; but if the gardener is the 'she,' then the gardener wins by $109$ ft. $4$ in." This would have won the prize. Curiously enough, one solitary competitor got on to the right track but failed to follow it up. He said: "Is this a regular April $1$ catch, meaning that they only ran $6$ ft. each, and consequently the race was unfinished? If not, I think the following must be the solution, supposing the gardener to be the 'he' and the cook the 'she.'" Though his solution was wrong even in the case he supposed, yet he was the only person who suspected the question of sex.

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References

Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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