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152 are ten different ways in which the apples could be bought. But we were told there was an equal number of "boys and girls," and one boy and one girl are not boys and girls, so this case has to be excluded. In the case of fourteen children, the only possible distribution is that each child receives one halfpenny apple. But we were told that each child was to receive an equal distribution of "apples," and one apple is not apples, so this case has also to be excluded. We are therefore driven back on our third case, which exactly fits in with all the conditions. Three boys and three girls each receive 1 half-penny apple and 2 third-penny apples. The value of these 3 apples is one penny and one-sixth, which multiplied by six makes seven-pence. Consequently, the correct answer is that there were six children—three girls and three boys.

37.—BUYING CHESTNUTS.

solving this little puzzle we are concerned with the exact interpretation of the words used by the buyer and seller. I will give the question again, this time adding a few words to make the matter more clear. The added words are printed in italics.

"A man went into a shop to buy chestnuts. He said he wanted a pennyworth, and was given five chestnuts. 'It is not enough; I ought to have a sixth of a chestnut more,' he remarked. 'But if I give you one chestnut more,' the shopman replied, "you will have five-sixths too many.' Now, strange to say, they were both right. How many chestnuts should the buyer receive for half a crown?"

The answer is that the price was 155 chestnuts for half a crown. Divide this number by 30, and we find that the buyer was entitled to $5 1⁄6$ chestnuts in exchange for his penny. He was, therefore, right when he said, after receiving five only, that he still wanted a sixth. And the salesman was also correct in saying that if he gave one chestnut more (that is, six chestnuts in all) he would be giving five-sixths of a chestnut in excess.

38.—THE BICYCLE THIEF.

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.

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.

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.

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

42.—THE FAMILY AGES.

ages were as follows: Billie, 3½ years; Gertrude, 1¾ year; Henrietta, 5¼ years; Charlie, 10½ years; and Janet, 21 years.

43.—MRS. TIMPKINS'S AGE.

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.

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 24½ and 3½ were correct.

45.—MOTHER AND DAUGHTER.

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.

'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.

's present age is ten years and Mildred's thirty years. Five years ago their respective