Amusements in Mathematics, Henry Ernest Dudeney

  • Question 41 - THEIR AGES

    "My husband's age," remarked a lady the other day, "is represented by the figures of my own age reversed. He is my senior, and the difference between our ages is one-eleventh of their sum."

  • Question 42 - THE FAMILY AGES

    When the Smileys recently received a visit from the favourite uncle, the fond parents had all the five children brought into his presence. First came Billie and little Gertrude, and the uncle was informed that the boy was exactly twice as old as the girl. Then Henrietta arrived, and it was pointed out that the combined ages of herself and Gertrude equalled twice the age of Billie. Then Charlie came running in, and somebody remarked that now the combined ages of the two boys were exactly twice the combined ages of the two girls. The uncle was expressing his astonishment at these coincidences when Janet came in. "Ah! uncle," she exclaimed, "you have actually arrived on my twenty-first birthday!" To this Mr. Smiley added the final staggerer: "Yes, and now the combined ages of the three girls are exactly equal to twice the combined ages of the two boys." Can you give the age of each child?

  • Question 43 - MRS. TIMPKINS'S AGE

    Edwin: "Do you know, when the Timpkinses married eighteen years ago Timpkins was three times as old as his wife, and to-day he is just twice as old as she?"

    Angelina: "Then how old was Mrs. Timpkins on the wedding day?"

    Can you answer Angelina's question?

  • Question 44 - A CENSUS PUZZLE

    Mr. and Mrs. Jorkins have fifteen children, all born at intervals of one year and a half. Miss Ada Jorkins, the eldest, had an objection to state her age to the census man, but she admitted that she was just seven times older than little Johnnie, the youngest of all. What was Ada's age? Do not too hastily assume that you have solved this little poser. You may find that you have made a bad blunder!

  • Question 45 - MOTHER AND DAUGHTER

    "Mother, I wish you would give me a bicycle," said a girl of twelve the other day.

    "I do not think you are old enough yet, my dear," was the reply. "When I am only three times as old as you are you shall have one."

    Now, the mother's age is forty-five years. When may the young lady expect to receive her present?

  • Question 46 - MARY AND MARMADUKE

    Marmaduke: "Do you know, dear, that in seven years' time our combined ages will be sixty-three years?"

    Mary: "Is that really so? And yet it is a fact that when you were my present age you were twice as old as I was then. I worked it out last night."

    Now, what are the ages of Mary and Marmaduke?

  • Question 47 - ROVER'S AGE

    "Now, then, Tommy, how old is Rover?" Mildred's young man asked her brother.

    "Well, five years ago," was the youngster's reply, "sister was four times older than the dog, but now she is only three times as old."

    Can you tell Rover's age?

  • Question 48 - CONCERNING TOMMY'S AGE

    Tommy Smart was recently sent to a new school. On the first day of his arrival the teacher asked him his age, and this was his curious reply: "Well, you see, it is like this. At the time I was born—I forget the year—my only sister, Ann, happened to be just one-quarter the age of mother, and she is now one-third the age of father." "That's all very well," said the teacher, "but what I want is not the age of your sister Ann, but your own age." "I was just coming to that," Tommy answered; "I am just a quarter of mother's present age, and in four years' time I shall be a quarter the age of father. Isn't that funny?"

    This was all the information that the teacher could get out of Tommy Smart. Could you have told, from these facts, what was his precise age? It is certainly a little puzzling.

  • Question 49 - NEXT-DOOR NEIGHBOURS

    There were two families living next door to one another at Tooting Bec—the Jupps and the Simkins. The united ages of the four Jupps amounted to one hundred years, and the united ages of the Simkins also amounted to the same. It was found in the case of each family that the sum obtained by adding the squares of each of the children's ages to the square of the mother's age equalled the square of the father's age. In the case of the Jupps, however, Julia was one year older than her brother Joe, whereas Sophy Simkin was two years older than her brother Sammy. What was the age of each of the eight individuals?

  • Question 50 - THE BAG OF NUTS

    Three boys were given a bag of nuts as a Christmas present, and it was agreed that they should be divided in proportion to their ages, which together amounted to `17 1/2` years. Now the bag contained `770` nuts, and as often as Herbert took four Robert took three, and as often as Herbert took six Christopher took seven. The puzzle is to find out how many nuts each had, and what were the boys' respective ages.