Part 66

34.--THE GROCER AND DRAPER.

The grocer was delayed half a minute and the draper eight minutes and a half (seventeen times as long as the grocer), making together nine minutes. Now, the grocer took twenty-four minutes to weigh out the sugar, and, with the half-minute delay, spent 24 min. 30 sec. over the task; but the draper had only to make forty-seven cuts to divide the roll of cloth, containing forty-eight yards, into yard pieces! This took him 15 min. 40 sec., and when we add the eight minutes and a half delay we get 24 min. 10 sec., from which it is clear that the draper won the race by twenty seconds. The majority of solvers make forty-eight cuts to divide the roll into forty-eight pieces!

35.--JUDKINS'S CATTLE.

As there were five droves with an equal number of animals in each drove, the number must be divisible by 5; and as every one of the eight dealers bought the same number of animals, the number must be divisible by 8. Therefore the number must be a multiple of 40. The highest possible multiple of 40 that will work will be found to be 120, and this number could be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3 oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement that the animals consisted of "oxen, pigs, and sheep," because a single ox is not oxen. Therefore the second grouping is the correct answer.

36.--BUYING APPLES.

As there were the same number of boys as girls, it is clear that the number of children must be even, and, apart from a careful and exact reading of the question, there would be three different answers. There might be two, six, or fourteen children. In the first of these cases there 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 halfpenny apple and 2 third-penny apples. The value of these 3 apples is one penny and one-sixth, which multiplied by six makes sevenpence. Consequently, the correct answer is that there were six children--three girls and three boys.

37.--BUYING CHESTNUTS.

In 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.

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.