427.—­PHEASANT-SHOOTING.

The arithmetic of this puzzle is very easy indeed.  There were clearly 24 pheasants at the start.  Of these 16 were shot dead, 1 was wounded in the wing, and 7 got away.  The reader may have concluded that the answer is, therefore, that “seven remained.”  But as they flew away it is clearly absurd to say that they “remained.”  Had they done so they would certainly have been killed.  Must we then conclude that the 17 that were shot remained, because the others flew away?  No; because the question was not “how many remained?” but “how many still remained?” Now the poor bird that was wounded in the wing, though unable to fly, was very active in its painful struggles to run away.  The answer is, therefore, that the 16 birds that were shot dead “still remained,” or “remained still.”

428.—­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 specially 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+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 46th bound.