INTELLECT SHARPENERS.

New Zealand Herald, Volume LXIV, Issue 19792, 12 November 1927, Page 5 (Supplement)

INTELLECT SHARPENERS.

JIX T. L. BHITON.

LAST WEEK'S SOLUTIONS.

Readers are requested nofc to send in their solutions unless these are specially asked for, but to keep them for comparison with those published on the Saturday following the publication of tha problems.

SHEEP FENS. This problem is based on the centuryold question of the shepherd who had constructed a . pen of thirty-two hurdles of uniform size, which held a certain number of sheep and could hold no more. Being desirous of penning at least double tho number, under exactly the same conditions as to sheep-room, he was told, in answer to a query as to the fewest number of extra hurdles of the same size that would bo required, that, as his pen was fifteen hurdles by one, an extra hurdle at each end would enable double tho number of sheep to be penned under the same conditions as to space for each animal. But that reply is hardly correct or complete, because he could accomplish what he required without any extra hurdles. Here is a little problem on tho point. How should a pen be constructed with thirty-two of the samo sized hurdles, so that exactly six hundred and forty sheep could occupy it, allowing each animal one and a-half square feet of space, and what length should each ol the hurdles be ?

COMPETITORS IN PARTNERSHIP.

Six athletes, three of either sex, took part in a series of athletic events, and for scoring purposes each of the men selected a female partner. It was an individual competition, the three girls, A, B and C contesting against one anothoc in three separato events, and similarly, with regard to the three men, X, Y and Z } who competed against each other, the winners being decided by the greatest number of points gained by the selected couples. Out of a total of thirty-six ooints won by the fair maids, A gained three more than B, and C three more than A. Among the men, X scored xactly the same number of points as his fair partner, and Y only one-half the number of his associate's total, while Z gained exactly twice as many as his fair companion managed to obtain. If the total points gained by the six athletes were eightyone, can the reader discover how the competitors were partnered, and their respective scores? A FOINT ON THE CIRCUMFERENCE. The circumference of a circle wholly within a rectangle, fifteen feet by twelve feet, touches its base and a perpendicular, these two sides of the rectangle thus forming tangents of the circle. A point on the circumference is distant eight feet in a direct line from the one side mentioned, and nine feet from the other side of the rectangular figure. With these meagre particulars, can the reader determine the length of the diameter of the circle ? It will, of course, be noted that the actual length of the sides of the rectangle are not necessary to be known in order that a solution may be reached. But without limiting the area in which the circle is enclosed, it would be possible to have two circles of different sizes, which would i conform to the distances that the point on the circumference is from the sides of the rectangle. But with the lengths of the figure as given, ■ there can be only one solution of the problem, as the circle falls wholly within the rectangle. This is a very simple and practical every day problem. :

DARBY AND JOAN.

Darby and Joan were discussing the number of years they' had been . merrier!, and incidentally ( their ages,: when 'Darby 1 reminded his wife that, in ieven ysats? • time their combined ages would be 100. and that would be the year in which they hoped to celebrate their .silver * wedding. They were at school together, ' and in 1892, when neither had reached their " teens," Darby was exactly twice, the age of Joan. A son was born two years after their marriage, and if in that year his mother w;;s four-fifths the age of her husband, in what year did tho conversation take place.

DOMINOES PROBLEMS.

Here is a problem concerning the ancient game of dominoes, which may require a little patience and ingenuity to solve. Take the following five dominoes and play them in the order given—double-one; one-two; two-blank; doubts-blank, and blank-three. The total number of spots on the two-end dominoes, exactly equal those on the remaining three pieces. There are only a few other ways that five dominoes can be arranged to total five in this manner, and the reader will find it interesting to discover them, for of course the dominoes must bs played in accordance with the accepted rules of the game. Then try with the same number of dominoes to total six in a similar manner, and it will be found a very refreshing mental exercise.

A New Zealand Aeroplane.

The distance must have been 480 miles, a speed of 90 miles an hour, uniformly, enabling the journey to be accomplished in 5 hours 20 minutes, which is 20 minutes slower and 40 minutes faster than it would have taken had the uniform speed been 96 miles and 80 miles an hour respectively, as stated in the problem.

Newcomes and Esmonds. The youngest in the two families respectively was 14 and 13 years old, and the eldest of the eight children was a Newcome. The several ages of the respective familes were 54, 42. 19, 16, 15, 14, and 57, 39, 18, 17, 16, 13.

Mixed Sheep. D started with 300 wethers, sold none, but bought 100 hoggets and 50 ewes. E started with -400 hoggets, sold 200 and bought 250 ewes. A started with 500 ewes, sold 350, and bought 100 hoggets and 200 lambs;. L started with 600 lambs, sold 200 and bought 50 ewes.

Slow Cricket. The whole team only made 11 runs, being in all day, the " not out " batsman therefore must have scored 31 runs, it being stated in the problem that 10 men only scored 46 rnns between them, and that the top scorer made 24 runs, more than the average for the 11.

Weighing Potatoes, The separate weights of the five bags were 108, 112, 116, 118 and 1241b.

ANSWERS TO CORRESPONDENTS.

C.T.F.—The problem assumed that each coin had an equal clianco of being ex° pelled. so that any disparity in their weights does not affect it. W.T.O.—A " knot " in the sense you mention is equal to 1.151 ordinary m:e, though it is often used to denote the speed of a vessel per . nautical mile. Thanks for your comment.

equidistant.

The distance was . one-eighth