INTELLECT SHARPENERS.

Otago Daily Times, Issue 21022, 10 May 1930, Page 21

INTELLECT SHARPENERS.

By T. L. Briton. SOMEWHAT MIXED. A correspondent has forwarded some particulars of a family “ relationship ” tangle which he describes as “ somewhat mixed,” stating that it is a case of a New Zealand family known to him. A, widower who may be called “A” married again when his own son 44 B ” was 14 years of age, and by his second wife " C ” he had two daughters “ D ” apd “ E.” Now “C ” had a younger sister “ F ” who later on married 44 B,” they having a son “ G ” and the latter, though some years younger than “D ” desired to marry that lady, but the law, however, prevented the union owing to their relationship, notwithstanding that the two Were entitled to marry had the law viewed the position in the light of another relationship which they bore to each other. The whole of these curious details make an interesting puzzle, and the questions for the reader to answer are, first, in how many relationships does “A” stand to 44 B ” and what are they 7 Second, how does “ F ” stand in relation to the two girls “D” and “E”? and third, why is it not legally possible for ‘ D ” and “ G” to marry? The situation is certainly “ somewhat mixed ” as the correspondent states, but no doubt the reader will succeed jn straightening out the tangle. VARYING TIME-PIECES. If three clocks showing the same time are set going simultaneously, and after a given time it was found that, one had gained, that another had lost as much as the first gained, whilst the third clock had kept exact time, it is obvious that at some future time, ff kept going at .these relative speeds, the hands of the three Clocks would again indicate the same time. Here is an arithmetically-simple problem on. the point which, however, contains a “ pitfall ” easily avoided if the question be given a little thought. Thirty years ago (to be exact on New Years Day, 1900), at noon, three clocks showed the same time. At noon next day one had gained half a minute, another had lost that much, while the third had kept correct time. If the three clocks were kept goin" uniformly in this way, when did they next show the same.time together? WEIGHING ALLUVIAL GOLD. Here Is a weighing problem that may give the reader a little thinking to find an easy method of solving it, the calculation itself not being difficult. ' ' There are five small parcels of fine gold, the weights being stamped on each of the packets A, B, C, D, and E, and these weights are required to be found by the reader when told how much they weigh In their several pairs. The figures given are in pennyweights, A and B together weighing 30, A and C 32, A and D 33, B and C 34, B and D 35, A and E 30, C and D 37., B and E 38, C and E 40, and D and E 41.' There is a very simple way to proceed in cases like this, and it is useful i'°iT now me thod, which wilJ be published next Saturday, In the meantime can the reader find the five separate weights! IN FOUR WAYS. There are four different ways in which the nine figures, I to 9, can be arranged m three rows of three figures each, so that the number represented by the figures in the. top row of any of the f ° ur arrangements is exactly one-half of that in the middle row and one-third of that in the bottom row, the middle row being therefore exactly one-third of the sum of the three rows. The example given below is one arrangement,’which when added up gives the highest total of any of the four sets. Can the reader arrange the nine figures in combination in the manner stated so that the total of the three rows when added up v will be the lowest one possible? . 327 654 981 1962 SILVER AND COPPER COINS. As readers are aware, a silver or a copper coin of the -realm becomes prac- - tically valueless when broken, though a mutilated or fractured gold coin retains its value, less the portion lost or destroyed. But for- problem purposes it is assumed that the silver and copper coins, if broken, retain their value similarly to gold coins, the broken ones not being rendered totally valueless. Let us take a shilling, a threepenny piece, and a penny, and break off a small fractional part, the same fraction in each instance. What would be the respective values of the major parts remaining if the difference between their total face values before mutilation and their values after being broken represents fourpence, assuming as stated above that' the coins retain their intrinsic value less the worth of the fragments broken off? This little problem should not require the solver to seek the aid of writing material in his effort to find the correct answer. LAST WEEK’S , SOLUTIONS. POOLING THEIR RESOURCES. The deposit on the bicycle was Bs, the change being therefore 6d. A WIFE’S FINANCE. The wife’s bills totalled £0 Bs, the money in the cash box when cleared being 12 £1 notes and 16 single shillings. Twelve 10s notes and four 2s pieces were returned. A NOVEL SCHEME. Twenty ladies, 10 gentlemen, and the matron. , AN INCUBATOR AN?; SOME EGGS. The eggs were 2s per dozen, the incubator costing £4. ON THE DRAUGHT BOARD. I If the eight vertical rows are lettered , A to H, and the horizontal rows manI bered 1 to 3, the following positions will fulfil the conditions of the problem;IC, 2E, SB, 4H, SA, 6G, 7D, and BF. ANSWERS'TO CORRESPONDENTS. Left School.” It could be treated as a quadratic equation, but it is quite siniple by arithmetic. The difficulty will be explained if sent along. Rex.” Seven divided by decimal seven is the same as expressing it, 70 divided by seven, viz., 10, as was explained at the time. J. Allan, J. A. R., L. W., E. M. M.— Thanks. Will be looked into.