I showed the picture at the time to a few friends, and they expressed very different opinions on the matter. One said, "I don't believe he would marry a girl like Number 7." Another said, "I am sure a nice girl like Number 3 would not marry such a fellow!" Another said, "It must be Number 1, because she has got as far away as possible from the brute!" It was suggested, again, that it must be Number 11, because "he seems to be looking towards her;" but a cynic retorted, "For that very reason, if he is really looking at her, I should say that she is not his wife!" I now leave the question in the hands of my readers. Which is really Number 10's wife? The illustration is of necessity considerably reduced from the large scale on which it originally appeared in _The Weekly Dispatch_ (24th May 1903), but it is hoped that the details will be sufficiently clear to allow the reader to derive entertainment from its examination. In any case the solution given will enable him to follow the points with interest. SOLUTIONS. 1.--A POST-OFFICE PERPLEXITY. The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8 twopence-halfpenny stamps, which delivery exactly fulfils the conditions and represents a cost of five shillings. 2.--YOUTHFUL PRECOCITY. The price of the banana must have been one penny farthing. Thus, 960 bananas would cost L5, and 480 sixpences would buy 2,304 bananas. 3.--AT A CATTLE MARKET. Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether. 4.--THE BEANFEAST PUZZLE. The cobblers spent 35s., the tailors spent also 35s., the hatters spent 42s., and the glovers spent 21s. Thus, they spent altogether L6,13s., while it will be found that the five cobblers spent as much as four tailors, twelve tailors as much as nine hatters, and six hatters as much as eight glovers. 5.--A QUEER COINCIDENCE. Puzzles of this class are generally solved in the old books by the tedious process of "working backwards." But a simple general solution is as follows: If there are n players, the amount held by every player at the end will be m(2^n), the last winner must have held m(n + 1) at the start, the next m(2n + 1), the next m(4n + 1), the next m(8n + 1), and so on to the first player, who must have held m(2^{n - 1}n + 1). Thus, in this case, n = 7, and the amount held by every player at the end was 2^7 farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449 farthings. 6.--A CHARITABLE BEQUEST. There are seven different ways in which the money may be distributed: 5 women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women and 10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1 man. But the last case must not be counted, because the condition was that there should be "men," and a single man is not men. Therefore the answer is six years.