not give at least four heads or four tails. Therefore the chances are only 12 to 20 in your favour, or (which is the same thing) 3 to 5. Put another way, you have only 3 chances out of 8.
The amount that should be paid for a draw from the bag that contains three sovereigns and one shilling is 153. 3d. Many persons will say that, as one's chances of drawing a sovereign were 3 out of 4, one should pay three-fourths of a pound, or 15s., overlooking the fact that one must draw at least a shilling—there being no blanks.
31.—DOMESTIC ECONOMY.
Without the hint that I gave, my readers would probably have been unanimous in deciding that Mr. Perkins's income must have been £1,710. But this is quite wrong. Mrs. Perkins says, "We have spent a third of his yearly income in rent," etc., etc.—that is, in two years they have spent an amount in rent, etc., equal to one-third of his yearly income. Note that she does not say that they have spent each year this sum, whatever it is, but that during the two years that amount has been spent. The only possible answer, according to the exact reading of her words, is, therefore, that his income was £180 per annum. Thus the amount spent in two years, during which his income has amounted to £360, will be £60 in rent, etc., £90 in domestic expenses, £20 in other ways, leaving the balance of £190 in the bank as stated.
32.—THE EXCURSION TICKET PUZZLE.
Nineteen shillings and ninepence may be paid in 458,908,622 different ways.
I do not propose to give my method of solution. Any such explanation would occupy an amount of space out of proportion to its interest or value. If I could give within reasonable limits a general solution for all money payments, I would strain a point to find room; but such a solution would be extremely complex and cumbersome, and I do not consider it worth the labour of working out.
Just to give an idea of what such a solution would involve, I will merely say that I find that, dealing only with those sums of money that are multiples of threepence, if we only use bronze coins any sum can be paid in ways where always represents the number of pence. If threepenny-pieces are admitted, there are ways. If sixpences are also used there are ways, when the sum is a multiple of sixpence, and the constant, 216, changes to 324 when the money is not such a multiple. And so the formulas increase in complexity in an accelerating ratio as we go on to the other coins.
I will, however, add an interesting little table of the possible ways of changing our current coins which I believe has never been given in a book before. Change may be given for a
| Farthing in | 0 way. |
| Halfpenny in | 1 way. |
| Penny in | 3 ways. |
| Threepenny-piece in | 16 ways. |
| Sixpence in | 66 ways. |
| Shilling in | 402 ways. |
| Florin in | 3,818 ways. |
| Half-crown in | 8,709 ways. |
| Double florin in | 60,239 ways. |
| Crown in | 166,651 ways. |
| Half-sovereign in | 6,261,622 ways. |
| Sovereign in | 500,291,833 ways. |
It is a little surprising to find that a sovereign may be changed in over five hundred million different ways. But I have no doubt as to the correctness of my figures.
33.—A PUZZLE IN REVERSALS.
(1) £13 (2) £23, 19s. 11d. The words "the number of pounds exceeds that of the pence" exclude such sums of money as £2, 16s. 2d. and all sums under £1.
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
