Amusements in Mathematics, Henry Ernest Dudeney [books to read to be successful txt] 📗
- Author: Henry Ernest Dudeney
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Marmaduke's age must have been twenty-nine years and two-fifths, and Mary's nineteen years and three-fifths. When Marmaduke was aged nineteen and three-fifths, Mary was only nine and four-fifths; so Marmaduke was at that time twice her age.
47.—ROVER'S AGE.—solution
Rover's present age is ten years and Mildred's thirty years. Five years ago their respective ages were five and twenty-five. Remember that we said "four times older than the dog," which is the same as "five times as old." (See answer to No. 44.)
48.—CONCERNING TOMMY'S AGE.—solution
Tommy Smart's age must have been nine years and three-fifths. Ann's age was sixteen and four-fifths, the mother's thirty-eight and two-fifths, and the father's fifty and two-fifths.
49.—NEXT-DOOR NEIGHBOURS.—solution
Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. Simkin 40; Sophy 10; and Sammy 8.
50.—THE BAG OF NUTS.—solution
It will be found that when Herbert takes twelve, Robert and Christopher will take nine and fourteen respectively, and that they will have together taken thirty-five nuts. As 35 is contained in 770 twenty-two times, we have merely to multiply 12, 9, and 14 by 22 to discover that Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as the total of their ages is 17½ years or half the sum of 12, 9, and 14, their respective ages must be 6, 4½, and 7 years.
51.—HOW OLD WAS MARY?—solution
The age of Mary to that of Ann must be as 5 to 3. And as the sum of their ages was 44, Mary was 27½ and Ann 16½. One is exactly 11 years older than the other. I will now insert in brackets in the original statement the various ages specified: "Mary is (27½) twice as old as Ann was (13¾) when Mary was half as old (24¾) as Ann will be (49½) when Ann is three times as old (49½) as Mary was (16½) when Mary was (16½) three times as old as Ann (5½)." Now, check this backwards. When Mary was three times as old as Ann, Mary was 16½ and Ann 5½ (11 years younger). Then we get 49½ for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was 24¾. And at that time Ann must have been 13¾ (11 years younger). Therefore Mary is now twice as old—27½, and Ann 11 years younger—16½.
52.—QUEER RELATIONSHIPS.—solution
If a man marries a woman, who dies, and he then marries his deceased wife's sister and himself dies, it may be correctly said that he had (previously) married the sister of his widow.
The youth was not the nephew of Jane Brown, because he happened to be her son. Her surname was the same as that of her brother, because she had married a man of the same name as herself.
53.—HEARD ON THE TUBE RAILWAY.—solution
The gentleman was the second lady's uncle.
54.—A FAMILY PARTY.—solution
The party consisted of two little girls and a boy, their father and mother, and their father's father and mother.
55.—A MIXED PEDIGREE.—solution
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-law's 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 till 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 655/11 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 5/11 of a minute in 65 minutes, or 60/143 of a minute per hour.
60.—THE WAPSHAW'S 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. 273/11 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. 546/11 sec. (twice the above time); next at 3 hr. 16 min. 219/11 sec.; next at 4 hr. 21 min. 491/11 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. 491/11 sec. out in his reckoning.
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. 2157/143 min. and 4 hr. 16112/143 min., and the last pair is 10 hr. 5983/143 min. and 11 hr. 54138/143 min. 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:—
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. 58106/143 min. and 11 hr. 44128/143 min., the latter being the time when the minute hand is nearest of all to the point IX—in fact, it is only 15/143 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. 55/143 min. at the head of the first column, and 1 hr. 060/143 min. at the head of the second column. Now, by successively adding 55/143 min. in the first, and 1 hr. 060/143 min. in the second column, we get all the eleven pairs in which the first time is a certain number of minutes after nought, or mid-day. Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the ten pairs after 1 o'clock. Then there is another "jump," and you will be able to get by addition all the nine pairs after 2 o'clock. And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." In this way we get under the successive hours, 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees with the formula in the first paragraph of this article.
Some time ago the principal of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. At length some of his readers convinced him that the answer is, "At 55/143 min. past XII;" and this he finally gave as correct, together with the reason for it that at that time the time indicated is the same whichever hand you may assume as hour hand!
62.—THE CLUB CLOCK.—solution
The positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. 511143/1427 sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. 52496/1427 sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. 22106/1427 sec. after eleven; but the question applied to the hands, and the second hand would not be between the others at that time, but outside them.
63.—THE STOP-WATCH.—solution
The time indicated on the watch was 55/11 min. past 9, when the second hand would be at 273/11 sec. The next time the hands would be similar distances apart would be 546/11 min. past 2, when the second hand would be at 328/11 sec. But you need only hold the watch (or our previous illustration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.
64.—THE THREE CLOCKS.—solution
As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?
I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by four without a remainder is
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