by Dudeney, Henry Ernest, 1857-1930
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The party consisted of two little girls and a boy, their father and mother, and their father's father and mother.
Thos. Bloggs m
W. Snoggs m Kate Bloggs.
m Henry Eloggs.
Jane John Bloggs w Snoggs
Joseph Bloggs *n
A f. Mary
Snoggs tfj Bloggs
The letter m stands for "married." It will be seen that John Snoggs can say to Joseph Bloggs, "You are my father's brother-in-law, because my father married your sister Kate; you are my brother's father-in-law, because my brother Alfred married your daughter Mary; and you are my father-in-laWs brother, because my wife Jane was your brother Henry's daughter."
56.?WILSON'S POSER.? solution
If there are two men, each of whom marries the mother of the other, and there is a son of each marriage, then each of such sons will be at the same time uncle and nephew of the other. There are other ways in which the relationship may be brought about, but this is the simplest.
57.?WHAT WAS THE TIME?? solution
The time must have been 9.36 p.m. A quarter of the time since noon is 2 hr. 24 min., and a half of the time 1 noon next day is 7 hr. 12 min. These added together make 9 hr. 36 min.
58.?A TIME PUZZLE.? solution
Twenty-six minutes.
59.?A PUZZLING WATCH.? solution
If the 65 minutes be counted on the face of the same watch, then the problem would be impossible: for the hands must coincide every 65 5 /^ minutes as shown by its face, and it matters not whether it runs fast or slow; but if it is measured by true time, it gains 5 /^ of a minute in 65 minutes, or 6 %|43of a minute per hour.
60.?THE WAPSHAWS WHARF MYSTERY? solution
There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide 12 hours by 11 we get 1 hr. 5 min. 27 3 /^ sec, and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54 6 /^ sec. (twice the above time); next at 3 hr. 16 min. 21 9 /^ sec; next at 4 hr. 21 min. 49 1 /^ sec. This last is the only occasion on which the two hands are together with the second hand "just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his Across the World for a Wife, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. 49 1 /^ sec. out in his reckoning.
Pg154
61.?CHANGING PLACES.? solution
There are thirty-six pairs of times when the hands exactly change places between three p.m. and midnight. The number of pairs of times from any hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. In the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, the required answer.
The first pair of times is 3 hr. 21 57 A|43 m ' n - anc ' 4 hr. 16 112 A|43 m ' n -> anc ' the last pair is 10 hr. 59 83 A|43 min. and 11 hr. 54 138 /i43inin. I will not give all the remainder of the thirty-six pairs of times, but supply a formula by which any of the sixty-six pairs that occur from midday to midnight may be at once found:?
720b + 60a 720a + 60b min.
a hr min. and b hr. ?
143 143
For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10 (where nought stands for 12 o'clock midday); and b may represent any hour, later than a, up to 11.
By the aid of this formula there is no difficulty in discovering the answer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58 106 /-|43 min. and 11 hr. 44 128 A|43 min., the latter being the time when the minute hand is nearest of all to the point IX?in fact, it is only 15 /^43 of a minute distant.
Readers may find it instructive to make a table of all the sixty-six pairs of times when the hands of a clock change places. An easy way is as follows: Make a column for the first times and a second column for the second times of the pairs. By making a = 0 and b = 1 in the above expressions we find the first case, and enter hr. 5 5 A|43 min. at the head of the first column, and 1 hr. 0 6 %|43 min. at the head of the second column. Now, by successively adding 5 5 A|43 min. in the first, and 1 hr. 0 6 %|43 min. in the second column, we get all the
