Intellect Sharpeners

New Zealand Herald, Volume LXXI, Issue 21882, 18 August 1934, Page 5 (Supplement)

Intellect Sharpeners

SHARING EXPENSES liV C. J. WHEP,KFORK A man hired a car to take him ft journey of 90 miles, and for this he was charged five guineas. \\ hen he had travelled 55 miles he picked up another man on the road and took him to a town 20 miles farther on. But live miles beyond where lie had picked up this man, A, he took on board a second man, B, and drove him all the way to the end of the journey, where he paid off the er.r himself. Each of .these men wished to pay for his own share of the co. t of the car. How much should they pay?

TRANSPORT PROBLEM The officer commanding a large body of troops found it necessary to move them across a rivor at once without waiting for repair i to be made to a bridge which had i>eeii blown up. Unfortunately the enemy had also destroyed nearly £. ; the available boats, and only five small ones could be found. These could hold only 5, 6, 7, 8 and 11 sold ers, in addition to the civilian employed in each boat to row. The operation of crossing in such small numbers of course took a long time, but the commanc it was at least grateful for the fact hat he had five boats instead of one. While commenting on this fact he noticed one result which interested him. If the largest boat had been the only one, it would have been capable of taking them all across in full boatloads, but any one of the others would have left one man behind at the last, "lor whom an extra trip would have been required. How many men of all ranks had to he ferried across?

DEAL IN LIVESTOCK A correspondent asks for explanation of a puzzle in which it is proposed to spend £2O in purchasing 20 animals, consisting of cows at £5, sheep at £1 and pigs at one shilling. "Are they ever as cheap as that?" one may ask. The answer is that the problem'given thus is impossible. The nearest approach to this, for which a solution can be found, is to buy 27 of them for £2O, and this is quite easy. A better problem, which is now offered, is to purchase 56 of these animals with £55, in which the price paid for the pigs is 2s Bd, while the cows and sheep cost the same as before. It is not more than pleasantly difficult.

ARMCHAIR PROBLEMS The price charged for admission to an amateur performance was 2s, but only one-third of the seats were filled. Next night the play was repeated and only Is was charged. Every seat was occupied, with the result that the money received was £lO more than it had been the night before. How many seats are there in the hall P In a race over a distance of 600 yards A was first and B was second, but the winner surpassed all other competitors by a very decided amount. In the first 40 yards run by A his average distance ahead of B was one yard. If these two runners maintained a -uniform pace, by how many yards did A win ?

THIRTEEN MONEY BOXES Thirteen children have each the same value of money in their money boxes, but no two of them have it in the same arrangement of coins. The numbers of the coins owned by any one child range from one to twelve, except that no child has three coins, but two have six and two also have seven. Halfpence are included, but no farthings. How much money does each - child possess ? LAST WEEK'S SOLUTIONS Two Tiled Floors. —Fourteen hours. Obtaining a Lilt.—Two and a-half miles. Word Change.—Head, held, hold, fold, food, foot. Counting the Ducks.—Both men missed the ducks which moved during the counting and the correct number was 38.

At a Flag Station.—The fare is 23 pence and is usually paid by giving a two-shilling piece and receiving one penny as change. The other two conditions are satisfied if the number of passengers was 73. Armchair Problems.—(l) The three totals are necessarily 25, 26, and 27. The first is given by 3, 4, 5, 6, 7, the second by 11, 12, 1, 2, and the third • by 8, 9, 10. . (2) The beauty of the problem is to make the breakage tako place id a way which seems the natural result of a fall. Solutions which require one number to be takfSn from the middle of a long sequence, as if punched out, ought not to be accepted if more plausible results can be found. The best one seems to be as follows: On the first piece 10, 11, 12, 1, 2, 3, on the second 4, 5, on the third 6, and on the fourth piece, which is missing, 7, 8, 9.