38.—­THE BICYCLE THIEF.

People give all sorts of absurd answers to this question, and yet it is perfectly simple if one just considers that the salesman cannot possibly have lost more than the cyclist actually stole.  The latter rode away with a bicycle which cost the salesman eleven pounds, and the ten pounds “change;” he thus made off with twenty-one pounds, in exchange for a worthless bit of paper.  This is the exact amount of the salesman’s loss, and the other operations of changing the cheque and borrowing from a friend do not affect the question in the slightest.  The loss of prospective profit on the sale of the bicycle is, of course, not direct loss of money out of pocket.

39.—­THE COSTERMONGER’S PUZZLE.

Bill must have paid 8s. per hundred for his oranges—­that is, 125 for 10s.  At 8s. 4d. per hundred, he would only have received 120 oranges for 10s.  This exactly agrees with Bill’s statement.

40.—­MAMMA’S AGE.

The age of Mamma must have been 29 years 2 months; that of Papa, 35 years; and that of the child, Tommy, 5 years 10 months.  Added together, these make seventy years.  The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amount to 140 years, and Tommy will be just half the age of his father.

41.—­THEIR AGES.

The gentleman’s age must have been 54 years and that of his wife 45 years.

42.—­THE FAMILY AGES.

The ages were as follows:  Billie, 31/2 years; Gertrude, 13/4 year;
Henrietta, 51/4 years; Charlie, 101/2; years; and Janet, 21 years.

43.—­MRS. TIMPKINS’S AGE.

The age of the younger at marriage is always the same as the number of years that expire before the elder becomes twice her age, if he was three times as old at marriage.  In our case it was eighteen years afterwards; therefore Mrs. Timpkins was eighteen years of age on the wedding-day, and her husband fifty-four.

44.—­A CENSUS PUZZLE.

Miss Ada Jorkins must have been twenty-four and her little brother Johnnie three years of age, with thirteen brothers and sisters between.  There was a trap for the solver in the words “seven times older than little Johnnie.”  Of course, “seven times older” is equal to eight times as old.  It is surprising how many people hastily assume that it is the same as “seven times as old.”  Some of the best writers have committed this blunder.  Probably many of my readers thought that the ages 241/2 and 31/2 were correct.

45.—­MOTHER AND DAUGHTER.

In four and a half years, when the daughter will be sixteen years and a half and the mother forty-nine and a half years of age.

46.—­MARY AND MARMADUKE.

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.