Amusements in Mathematics | Page 9

to be valueless, and the salesman was requested by his neighbour to refund the amount he had received. To do this, he was compelled to borrow the ��25 from a friend, as the cyclist forgot to leave his address, and could not be found. Now, as the bicycle cost the salesman ��11, how much money did he lose altogether?
39.--THE COSTERMONGER'S PUZZLE.
"How much did yer pay for them oranges, Bill?"
"I ain't a-goin' to tell yer, Jim. But I beat the old cove down fourpence a hundred."
"What good did that do yer?"
"Well, it meant five more oranges on every ten shillin's-worth."
Now, what price did Bill actually pay for the oranges? There is only one rate that will fit in with his statements.

AGE AND KINSHIP PUZZLES.
"The days of our years are threescore years and ten."
--Psalm xc. 10.
For centuries it has been a favourite method of propounding arithmetical puzzles to pose them in the form of questions as to the age of an individual. They generally lend themselves to very easy solution by the use of algebra, though often the difficulty lies in stating them correctly. They may be made very complex and may demand considerable ingenuity, but no general laws can well be laid down for their solution. The solver must use his own sagacity. As for puzzles in relationship or kinship, it is quite curious how bewildering many people find these things. Even in ordinary conversation, some statement as to relationship, which is quite clear in the mind of the speaker, will immediately tie the brains of other people into knots. Such expressions as "He is my uncle's son-in-law's sister" convey absolutely nothing to some people without a detailed and laboured explanation. In such cases the best course is to sketch a brief genealogical table, when the eye comes immediately to the assistance of the brain. In these days, when we have a growing lack of respect for pedigrees, most people have got out of the habit of rapidly drawing such tables, which is to be regretted, as they would save a lot of time and brain racking on occasions.
40.--MAMMA'S AGE.
Tommy: "How old are you, mamma?"
Mamma: "Let me think, Tommy. Well, our three ages add up to exactly seventy years."
Tommy: "That's a lot, isn't it? And how old are you, papa?"
Papa: "Just six times as old as you, my son."
Tommy: "Shall I ever be half as old as you, papa?"
Papa: "Yes, Tommy; and when that happens our three ages will add up to exactly twice as much as to-day."
Tommy: "And supposing I was born before you, papa; and supposing mamma had forgot all about it, and hadn't been at home when I came; and supposing--"
Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll have a headache."
Now, if Tommy had been some years older he might have calculated the exact ages of his parents from the information they had given him. Can you find out the exact age of mamma?
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."
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?
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?
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