INTELLECT SHARPENERS

Otago Daily Times, Issue 22609, 28 June 1935, Page 5

INTELLECT SHARPENERS

Written for the Otago Daily Times. By C. J. Whebefobk. [Correspondence should be addressed to Box 1177, Wellington.] ARMCHAIR PROBLEM. Three men are anxious to take up a small matter of business, for which £l2O will be required. At present B has 20 per cent, more money than A. C cannot find any capital at the present moment, because his bank account is already overdrawn. But he will be able to cash the coupons of some debentures in a few days, and these will bring in £4O, so that he can extinguish his overdraft and will also have sufficient money to make up the sums held by A and B to the amount required. How much money have A and B, and what is the amount of C's overdraft? A FEW YARDS OF RIBBON. A girl bought a certain number ot yards of ribbon at twopence per yard, and again the same number of yards at one penny halfpenny per yard. Her sister spent the same amount of money, but she spent half of it on the first sample of ribbon and the other half on the second. The result was that she obtained two yards more than the first girl. How many yards did each girl buy? i i A CYCLIST'S HOLIDAY. At a tourist resort there are five roads leading from the hotel to one of the beauty spots of the neighbourhood, and each one in its turn is one mile longer than the one before it. A cyclist who was staying there visited this place on seven different occasions, and each time he chose a different pair of roads. Each day's journey was one mile longer than that which he rode the previous dayThere was one road on which he travelled only once, because he- found it so rough that he decided not to take it ..again. For convenience let these roads in the order of their length be named A, B, O, D, and E, of which A is the shortest and E the longest. His week's riding made a total of 70 miles. What are the lengths of these roads, and winch is the one on which he cycled only once? A PROBLEM IN AGES. Mr Richard was standing in front of me at the Post Office a few days ago and paying in a sum of money, which was all in bundles of notes, containing £IOO in each. He explained to me afterwards that he had opened an account tor his three children together, and had just deposited what he considered to be a suitable amount. " But you will never guess how I calculated that sum," he added. 1 accepted the challenge. " You took the mean of the three ages and multip bed. by 100," I said. "Wrong," he replied brieflv. "Then you -took the sum ot all three ages, increased it by one, and multiplied by 32." " Wrong again, he replied. I thought again, then I made a third attempt. "You have taken the sum of the eldest and youngest and multiplied by 50." "That is a foolish answer, he said. "I see I shall have to tell you. I added together the age of the youngest, the square of the age of the next, and the cube of the age of the eldest." What are the ages of Mr Richard's children? A SHOPPING PROBLEM. Four men, Tom, Dick, Harry, and Algernon, went shopping with their wyes. Tom spent twice as much money as Dick, three times as much as Harry, but only half as much as Algernon. Ihey spent these sums in different departments ot the same establishment. Tom and Mary went together to one department, Dick and Jane to another, Harry and Isabel to a third, and Algernon and Kate to a fourth. In this way no man was in company with his own wife, but each man. spent just twice as much money as. the woman who went to the same counter with him. ' No wife spent the same amount as her husband. Kate is JJK'U s sister, and is two years younger than he is. Isabel is Jane's younger sister, Tom is the youngest of the four men, and his wife, is the youngest of the four women present. The problem is to arrange these eight persons as four married pairs. ' T SUPPLIED TO ORJ)ER. ' V Several correspondents have asked for non-mathematical problems, and the following may be regarded as written to their orders. In a: certain small township five of the shops are occupied _by: tenants, whose names are Ritchie, Edwards, Adams, Clark and Harris. Their businesses are respectively tnose ot storekeeper, draper, fruiterer, confectioner and stationer. The storekeeper and the fruiterer are the only ones who sell tobacco, and the stationer is the only one who sells magazines and books. Kecently a car passed through this town, and stayed for an hour or more. In addition to the driver, it contained a man. a woman, a boy and a girl. Each of these five persons visited four of the shops described, and made one purchase in each, but no two of them chose the same four. The initials of the persons from whom they obtained the goods made words of four letters, except those to whom the girl went. The driver and the boy bought cigarettes at different shops. The man bought a book, and the woman bought a magazine. The boy did not go to the draper. The girl bought some cakes at the confectioner's, and shared them with the boy, who gave her in return some of the fruit he had bought. In fact, the boy and the girl walked ' round in one another's company most of the time, and visited three shops together.—To which four shops did the driver go, and where did he buy his cigarettes? SOLUTIONS OF LAST WEEK'S PROBLEMS. Substitution.—l67B .plus 3742 equals 5420. The book is "In Scotland Again. Armchair Problems.—(l) 33 in yard, 56 and 7 in paddocks; (2) 38 miles per hour. Mountaineering.—4s36 feet. _ One guinea for the photos and three guineas for the painting. Casualties. —There were 9999 men at first, and the prisoners made 97 companies of 127 men each. Nine Schoolgirls.—First day: 1,2, 3, and 4. 5,6, and 7,8, 9. Second day: 1,4, ,7, and 2,5, 8. and 3,6, 9. Third day: 1,6, 8, and 2,4, 9, and 3,5, 7. Fourth day: 1,5, 9, and 2,6, 7, and 3. 4, 8. Lessons in Driving.—Tom must be an only son, because the father's remarks about his sister's behaviour toward him would be incomplete without referring to brothers, if any existed. There must be at least three daughters, otherwise he would speak of his "elder daughter" instead of his " eldest" one. But 3is evidently the correct number, because it is obvious that there cannot be 4 or 6, and 5 leads to a nonsensical result in the velocity at which the car was driven. Hence, there were 5 persons in the car, so that the distance driven must be a multiple of 10. Twenty is not enough, find 30 is not a factor of 80, but 40 miles fits perfectly. Therefore, each person drove for eight miles.