"NUTS!" A PROBLEM IN MEAN VALUES

Evening Post, Volume CXX, Issue 12, 13 July 1935, Page 28

"NUTS!"

A PROBLEM IN MEAN VALUES

INTELLECT SHARPENERS

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(By C. J. Wbercfore.)

Readeri wit* a little Ingenuity will find In this col'imii an abundant store ef entertainment and amusemon;, and tho solving of the problems should provide excellent menial exhilaration. While some of tne 'nuts" may appear harder that others, it will be found that none ■vill require n sledge-hammer to eraeV them. Address correspondence t» P.O. Bon 11/7, Wtl> lington.

The ages of a family of five children, reckoned in years without fractions, have a mean value of 14$. The mean value of a group of three of them is 14, of another such group it is 16, and two groups of three, each having mean values of 15, can be made. It is also possible to make a group of four, which has a mean value of 14, but no other groups of this number can be formed without giving either halves or quarters of years in their mean values. There are no twins in this family. What are these five ages? IN A TEA SHOP. [ Dorothy and Ethel went, into partnership over a tea-shop, and had no ■ reason to regret their venture. They served sixpenny and ninepenny orders, and for some time they could not agree whether the lower priced business should be continued. As Dorothy pointed out, the net profit on the smaller lot was only three farthings, whereas the larger one brought in three times as much. Ethel maintained that they were making more money out of the sixpenny customers, so Dorothy took one day's cash and worked out the profits for that day. She found that this came to £2, and that Ethel's statement was correct, but only by the narrowest possible margin. How many customers, had visited the tearooms that day? CAN SHEEP OWNERS ANSWER THIS? "I have just asked the owner of those sheep how many there are in each of the three lots into which he has drafted them," said the Incurable problemist. "There are a hundred altogether, and he has put the ewes in one yard, the hoggets into another, and there are three ewes for every hogget. The third yard holds the wethers. He is going to take out one of the hoggets, because it does not belong to him, and then the number of wethers will be a multiple of the hoggets that are left." The Ordinary Man, to whom he spoke, protested that this told him nothing, because it could happen in quite a large number of ways. The Problemist replied triumphantly: "I was waiting for you to say that. The number of wethers is the smallest that can be under the conditions I have given, so now you have a choice little problem to solve. How many slieep are there in each of the three yards?" TWO WATCHES. Two boys met at school after the holidays, and each could show the other a watch he had just purchased. Tom had received a tip from an uncle, and had spent a certain fraction of the' money on his watch. Reginald's watch was a much more expensive article, but the tip he had received was twice as much as that given to his friend, j The fraction of this windfall which { he had spent was equal to the fraction previously mentioned with one-! tenth added to it. Even if Tom had spent twice as much as 'he did, the price of his watch would still be two shillings less than that paid by Reginald. How much money had each of these boys received as presents? A COMBINATION LOCK. An eccentric man had only the sum of £800 to leave and only four distant relations in whom he took very little interest. It amused him to state in his 'will that this money should be placed in a box with a combination lock, and each person in turn should be allowed a chance to win the whole sum by finding the word which opened it. There were six cylinders, each of which revolved independently of the others, and each of these had eight | letters on it. To open the lock a six-letter, word had to- be formed by selecting one letter from each cylinder and- bringing it in line opposite to an index. The letters on the six cylinders were as follows:— 1. Trembled 4. Brittany 2. Failsyou 5. Torments 3. Wetbacks 6. Returned The. signature-to'the will was dated April 21, 1935. The four beneficiaries were two brothers, Charles and WilJiam, and two sisters, Theresa and Beatrice. Theresa is married to a dairy farmer, and should be very glad to have the "unusual experience of becoming the owner of a little money, but neither she, nor any of the others guessed correctly. Evidently they were all influenced by their professions when setting up their' six-letter words, but "it is not clear whether- they thought of their own names. Thus Charles did not use a C, and William did hot use a W, but Theresa used two T's, and Beatrice used two B's. What are the professions of these four persons, and which words did they select? A second problem from the same data is found in a' codicil to the will, which is also dated April 21. It states that if none of the four beneficiaries are successful, the box shall "be opened by the executors, and the money divided into a sufficient number of equal shares. Each person shall receive three. shares for every cylinder set correctly, but shall also receive one share for every cylinder set incorrectly.. Thus no person could receive more than sixteen shares or less than six. Beatrice received.£l2o. How much did each of the others receive? SOLUTIONS. Three Men.—The distance is 20 miles, and. A.walked two miles between the two meeting places. • Armchair Problems.—(l) The first month must be one of 31 days, and the other of 30. He drove 390 miles, and walked 120. (2) Harry lost one shilling. ■ ■ ■ ■ ; Rhyme.—The missing words arc: Love, live, hive, have, hate. The endings of the lines should be: To let her • know, meet,, to work we go, street, v spare, have more to . do, anywhere, rare, have to share, swear, few. Anterior to Other Story.—Cinderella collected £1 os.sd, and the sisters 15 shillings, arid 14 shillings, 7 pence. The-five pence is the minimum possible. If anyone is interested in the problem of the maximum difference, the answer is £1 0s lOd, which means, of course, that one box is empty. . Crossing: River.—lt is clear that Wo. 1 lost the sum of 7£ pence and one penny, and No. 16 gained the difference of these two quantities. The intermediate numbers are readily interpolated, and by tabulating the last seven, it is made evident that the person speaking is No. 13, and she gained 3J pence. Gamble. —The sum of all numbers up to 28 equals half sum of 28 and 784, which is 406. As only 400 persons were supplied on. 26 days, it is clear that two numbers which make 6 are absent, and these must be 2 and 4. Therefore the supply of odd numbers is two more than the supply of even numbers, so that the girl .won two pence, / . ■*