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
1. OLD CAT OLD HEN POLAR STAR SIMPLE SUSAN
WHITE SUGAR
2. BRADSHAW OF THE FUTURE T.R.
3. ALGERNON BRAY DINAH MITE G.S.C. JANE E. J.D.W.
MAGPIE S.S.G. TOKIO
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?
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 venerable bird! And close after OLD HEN prowls, with hungry eyes, OLD CAT, who calmly assumes, to being with, that the son who comes of age is the eldest. Eat your bird, Puss, for you will get nothing from me!
There are yet two zeroes to dispose of. MINERVA assumes that, on every occasion, a son comes of age; and that it is only such a son who is `tipped with gold'. Is it wise thus to interpret, `Now, my boys, calculate your ages, and you shall have the money'? BRADSHAW OF THE FUTURE says `let' the ages at first be 9, 6, 3, then assumes that the second occasion was 6 years afterwards, and on these baseless assumptions brings out the right answers. Guide future travelers, an thou will; thou art no Bradshaw for this Age!
Of those who win honours, the merely `honourable' are two. DINAH MITE ascertains (rightly) the relationship between the three ages at first, but then assumes one of them to be `6', thus making the rest of her