Tangled Tale — Answers to Knot X
§ 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 and 7/10ths."
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 and 3/8ths."
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 suppose!
): dividing by 4, she gets 77 and a half as "average percentage:" again dividing by 4, she gets 19 and 3/8ths 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 3/4ths of these unfortunates can obtain perfect health by handing over their wounds to the remaining 1/4th?
Granting these suppositions, her answer is right; or rather, _if_ the question had been "A road is covered with one inch of gravel, along 77 and a half 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 battle-field 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.
OLD HEN.
POLAR STAR.
SIMPLE SUSAN.
WHITE SUGAR.
II. BRADSHAW OF THE FUTURE.
T.
R.
III. ALGERNON BRAY.
DINAH MITE.
G.
S.
C. JANE E.
J.
D. W.
MAGPIE.
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 SONS' 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_ = 2_c_, 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_ = 2_y_; _i.
e. _ _y_ = 2_x_.
Hence the three ages were, at first, _x_, 2_x_, 3_x_; and the number of years, since that time is two-thirds of 6_x_, _i.
e. _ is 4_x_.
Hence the present ages are 5_x_, 6_x_, 7_x_.
The ages are clearly _integers_, since this is only "the year when one of my sons comes of age."
Hence 7_x_ = 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, oh venerable bird!
And close after OLD HEN prowls, with hungry eyes, OLD CAT, who calmly assumes, to begin 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_ travellers, an thou wilt: 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 solution tentative.
M. F.
C.
does the algebra all right up to the conclusion that the present ages are 5_z_, 6_z_, and 7_z_; it then assumes, without giving any reason, that 7_z_ = 21. Of the more honourable, DELTA attempts a novelty--to discover _which_ son comes of age by elimination: it assumes, successively, that it is the middle one, and that it is the youngest; and in each case it _apparently_ brings out an absurdity.
Still, as the proof contains the following bit of algebra, "63 = 7_x_ + 4_y_; [** therefore] 21 = _x_ + 4 sevenths of _y_," I trust it will admit that its proof is not _quite_ conclusive.
The rest of its work is good.
MAGPIE betrays the deplorable tendency of her tribe--to appropriate any stray conclusion she comes across, without having any _strict_ logical right to it.
Assuming _A_, _B_, _C_, as the ages at first, and _D_ as the number of the years that have elapsed since then, she finds (rightly) the 3 equations, 2_A_ = _B_, _C_ = _B_ + _A_, _D_ = 2_B_.
She then says "supposing that _A_ = 1, then _B_ = 2, _C_ = 3, and _D_ = 4. Therefore for _A_, _B_, _C_, _D_, four numbers are wanted which shall be to each other as 1:2:3:4." It is in the "therefore" that I detect the unconscientiousness of this bird.
The conclusion _is_ true, but this is only because the equations are "homogeneous" (_i.
e. _ having one "unknown" in each term), a fact which I strongly suspect had not been grasped--I beg pardon, clawed--by her.
Were I to lay this little pitfall, "_A_ + 1 = _B_, _B_ + 1 = _C_;
supposing _A_ = 1, then _B_ = 2 and _C_ = 3. _Therefore_ for _A_, _B_, _C_, three numbers are wanted which shall be to one another as 1:2:3," would you not flutter down into it, oh MAGPIE, as amiably as a Dove?
SIMPLE SUSAN is anything but simple to _me_.
After ascertaining that the 3 ages at first are as 3:2:1, she says "then, as two-thirds of their sum, added to one of them, = 21, the sum cannot exceed 30, and consequently the highest cannot exceed 15." I suppose her (mental) argument is something like this:--"two-thirds of sum, + one age, = 21;
[** therefore] sum, + 3 halves of one age, = 31 and a half.
But 3 halves of one age cannot be less than 1 and-a-half (here I perceive that SIMPLE SUSAN would on no account present a guinea to a new-born baby!
) hence the sum cannot exceed 30." This is ingenious, but her proof, after that, is (as she candidly admits) "clumsy and roundabout."
She finds that there are 5 possible sets of ages, and eliminates four of them.
Suppose that, instead of 5, there had been 5 million possible sets?
Would SIMPLE SUSAN have courageously ordered in the necessary gallon of ink and ream of paper?
The solution sent in by C. R.
is, like that of SIMPLE SUSAN, partly tentative, and so does not rise higher than being Clumsily Right.
Among those who have earned the highest honours, ALGERNON BRAY solves the problem quite correctly, but adds that there is nothing to exclude the supposition that all the ages were _fractional_.
This would make the number of answers infinite.
Let me meekly protest that I _never_ intended my readers to devote the rest of their lives to writing out answers!
E.
M.
RIX points out that, if fractional ages be admissible, any one of the three sons might be the one "come of age"; but she rightly rejects this supposition on the ground that it would make the problem indeterminate.
WHITE SUGAR is the only one who has detected an oversight of mine: I had forgotten the possibility (which of course ought to be allowed for) that the son, who came of age that _year_, need not have done so by that _day_, so that he _might_ be only 20. This gives a second solution, viz.
, 20, 24, 28. Well said, pure Crystal!
Verily, thy "fair discourse hath been as sugar"!
CLASS LIST.
I. ALGERNON BRAY.
AN OLD FOGEY.
E.
M. RIX.
G.
S.
C. S.
S.
G.
TOKIO.
T.
R.
WHITE SUGAR.
II. C. R.
DELTA.
MAGPIE.
SIMPLE SUSAN.
III. DINAH MITE.
M. F.
C.
* * * * * I have received more than one remonstrance on my assertion, in the Chelsea Pensioners' problem, that it was illogical to assume, from the _datum_ "70 p. c.
have lost an eye," that 30 p. c.
have _not_.
ALGERNON
BRAY states, as a parallel case, "suppose Tommy's father gives him 4 apples, and he eats one of them, how many has he left?"
and says "I think we are justified in answering, 3." I think so too.
There is no "must" here, and the _data_ are evidently meant to fix the answer _exactly_: but, if the question were set me "how many _must_ he have left?"
, I should understand the _data_ to be that his father gave him 4 _at least_, but _may_ have given him more.
I take this opportunity of thanking those who have sent, along with their answers to the Tenth Knot, regrets that there are no more Knots to come, or petitions that I should recall my resolution to bring them to an end.
I am most grateful for their kind words; but I think it wisest to end what, at best, was but a lame attempt.
"The stretched metre of an antique song" is beyond my compass; and my puppets were neither distinctly _in_ my life (like those I now address), nor yet (like Alice and the Mock Turtle) distinctly _out_ of it.
Yet let me at least fancy, as I lay down the pen, that I carry with me into my silent life, dear reader, a farewell smile from your unseen face, and a kindly farewell pressure from your unfelt hand!
And so, good night!
Parting is such sweet sorrow, that I shall say "good night!"
till it be morrow.
THE END
LONDON: RICHARD CLAY AND SONS, PRINTERS.
[TURN OVER.
WORKS BY LEWIS CARROLL.
ALICE'S ADVENTURES IN WONDERLAND.
With Forty-two Illustrations by TENNIEL.
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AND REASON?
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Mr. Frost's pictures are new.
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_ Fifty Thousand.
A TANGLED TALE.
Reprinted from _The Monthly Packet_.
With Six Illustrations by Arthur B. Frost.
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_ * * * * * N.
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In selling the above-mentioned books to the Trade, Messrs. Macmillan and Co. will abate 2_d.
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* * * * * MR.
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* * * * * Transcriber's note The following changes have been made to the text: Page 88: "he corners of the" changed to "the corners of the".
Page 95: "Aix-le-Bains" changed to "Aix-les-Bains".
Page 108: "3/5, 2, 1/3" changed to "3/5, 2/3, 1/3".
Page 114: "10 of the 12 cases" changed to "10 of the 13 cases".
Page 121: "four-and fourpence" changed to "four-and-fourpence".
Last page: "Fifth Thousand" changed to "Fifty Thousand".
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