Page:Amusements in mathematics.djvu/165

Rh 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.

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.

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.

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 $17 1⁄2$ years or half the sum of 12, 9, and 14, their respective ages must be 6, $4 1⁄2$, and 7 years.

51.—HOW OLD WAS MARY?

age of Mary to that of Ann must be as 5 to 3. And as the sum of their ages was 44, Mary was $27 1⁄2$ and Ann $16 1⁄2$. 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 ($27 1⁄2$) twice as old as Ann was (13|3|4) when Mary was half as old ($24 3⁄4$) as Ann will be ($49 1⁄2$) when Ann is three times as old ($49 1⁄2$) as Mary was ($16 1⁄2$) when Mary was ($16 1⁄2$) three times as old as Ann ($5 1⁄2$)." Now, check this backwards. When Mary was three times as old as Ann, Mary was $16 1⁄2$ and Ann $5 1⁄2$ (11 years younger). Then we get $49 1⁄2$ for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was $24 3⁄4$. And at that time Ann must have been $13 3⁄4$ (11 years younger). Therefore Mary is now twice as old—$27 1⁄2$, and Ann 11 years younger—$16 1⁄2$.

52.—QUEER RELATIONSHIPS.

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.

gentleman was the second lady's uncle.

54.—A FAMILY PARTY.

party consisted of two little girls and a boy, their father and mother, and their father's father and mother.

55.—A MIXED PEDIGREE.



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.

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?

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.

minutes.

59.—A PUZZLING WATCH.

the 65 minutes be counted on the face of the same watch, then the problem would be impossible: for the hands must coincide every $65 5⁄11$ 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.

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