A Tangled Tale, Lewis Carroll [free ebook novel .txt] 📗
- Author: Lewis Carroll
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Next come eight writers who have made the unwarrantable assumption that, because 70 percent have lost an eye, therefore 30 percent 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 percent 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 Tokyo.
Bradshaw of the Future and T. R. do the question in a piecemeal fashion—on the principle that the 70 percent and the 75 percent, though commenced at opposite ends of the 100, must overlap by at least 45 percent; 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.
Old Cat.
Old Hen.
Polar Star.
Simple Susan.
White Sugar.
Bradshaw of the Future.
T. R.
Algernon Bray.
Dinah Mite.
G. S. C.
Jane E.
J. D. W.
Magpie.
S. S. G.
Tokyo.
§2: Change of DayI 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 Sons’ AgesProblem.—“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
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