§ 1. The Chelsea Pensioners

Problem. — If 70 per cent have lost an eye, 75 per cent an ear, 80 per cent an arm, 85 per cent a leg: what percentage, at least, must have lost all four?

Answer. — Ten.

Solution. — (I adopt that of Polar Star, as being better than my own.) Adding the wounds together, we get 70+75+80+85=310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10.

Nineteen answers have been received. One is “5” but, as no working is given with it, it must, in accordance with the rule, remain “a deed without a name”. Janet makes it “35 7/10”. I am sorry she has misunderstood the question, and has supposed that those who had lost an ear were 75 per cent of those who had lost an eye; and so on. Of course, on this supposition, the percentages must all be multiplied together. This she has done correctly, but I can give her no honours, as I do not think the question will fairly bear her interpretation. Three Score and Ten makes it “19 3/8” Her solution has given me — I will not say “many anxious days and sleepless nights”, for I wish to be strictly truthful, but — some trouble in making any sense at all of it. She makes the number of “pensioners wounded once” to be 310 (“per cent,” I supposd): dividing by 4, she gets 77½ as “average percentage”: again dividing by 4, she gets 19 3/8 as “percentage wounded four times”. Does she suppose wounds of different kinds to “absorb” each other, so to speak! Then, no doubt, the data are equivalent to 77 pensioners with one wound each and a half-pensioner with a half-wound. And does she then suppose these concentrated wounds to be transferable, so that 3/4 of these unfortunates can obtain perfect health by handing over their wounds to the remaining 1/4? Granting these suppositions, her answer is right; or rather if the question had been, “A road is covered with one inch of gravel, along 77½ per cent of it. How much of it could be covered 4 inches deep with the same material?” her answer would have been right. But alas, that wasn’t the question! Delta makes some most amazing assumptions: “let every one who has not lost an eye have lost an ear,” “let every one who has not lost both eyes and ears have lost an arm.” Her ideas of a battlefield are grim indeed. Fancy a warrior who would continue fighting after losing both eyes, both ears, and both arms! This is a case which she (or “it “?) evidently considers possible.

Next come eight writers who have made the unwarrantable assumption that, because 70 per cent have lost an eye, therefore 30 per cent have not lost one, so that they have both eyes. This is illogical. If you give me a bag containing 100 sovereigns, and if in an hour I come to you (my face not beaming with gratitude nearly so much as when I received the bag) to say, “I am sorry to tell you that 70 of these sovereigns are bad,” do I thereby guarantee the other 30 to be good? Perhaps I have not tested them yet. The sides of this illogical octagon are as follows, in alphabetical order: Algernon Bray, Dinah Mite, G. S. C., Jane E., J. D. W., Magpie (who makes the delightful remark, “Therefore 90 per cent have two of something,” recalling to one’s memory that fortunate monarch with whom Xerxes was so much pleased that “he gave him ten of everything”!), S. S G., and Tokio.

Bradshaw of the Future and T. R. do the question in a piecemeal fashion — on the principle that the 70 per cent and the 75 per cent, though commenced at opposite ends of the 100, must overlap by at least 45 per cent; and so on. This is quite correct working, but not, I think, quite the best way of doing it.

The other five competitors will, I hope, feel themselves sufficiently glorified by being placed in the first class, without my composing a Triumphal Ode for each!

Class List.

I.

Old Cat. Polar Straw. Old Hen. Simple Susan. White Sugar.

II.

Bradshaw of the Future. T.R.

III.

Algernon Bray. J. D. W. Dinah Mite. Magpie. G. S. C. Jane E. S. S. G. Tokio.

§ 2. Change of Day

I must postpone, sine die, the geographical problem — partly because I have not yet received the statistics I am hoping for, and partly because I am myself so entirely puzzled by it; and when an examiner is himself dimly hovering between a second class and a third, how is he to decide the position of others?

§ 3. The Son’s Ages

Problem. — At first, two of the ages are together equal to the third. A few years afterwards, two of them are together double of the third. When the number of years since the first occasion is two-thirds of the sum of the ages On that occasion, one age is 21. What are the other two?

Answer. — 15 and 18.

Solution. — Let the ages at first be x, y, (x + y) Now, if a+b=2c, then (a-n) + (b-n)=2(c-n), whatever be the value of n. Hence the second relationship, if ever true, was always true. Hence it was true at first. But it cannot be true that x and y are together double of (x +y). Hence it must be true of (x +y), together with x or y; and it does not matter which we take. We assume, then, (x +y) +x = 2y,. i.e. y = 2x. Hence the three ages were, at first, x, 2x, 3x, and the number of years since that time is two-thirds of 6x, i.e. is 4x. Hence the present ages are 5x, 6x, 7x. The ages are clearly integers, since this is only “the year when one of my sons comes of age”. Hence 7x=21, x=3, and the other ages are 15, 18.

Eighteen answers have been received. One of the writers merely asserts that the first occasion was 12 years ago, that the ages were then 9, 6, and 3; and that on the second occasion they were 14, 11, and 8! As a Roman father, I ought to withhold the name of the rash writer; but respect for age makes me break the rule: it is Three Score and Ten. Jane E. also asserts that the ages at first were 9, 6, 3: then she calculates the present ages, leaving the second occasion unnoticed. Old Hen is nearly as bad; she “tried various numbers till I found one that fitted all the conditions”; but merely scratching up the earth, and pecking about, is not the way to solve a problem, O vene............