A Tangled Tale, Lewis Carroll [free ebook novel .txt] 📗
- Author: Lewis Carroll
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Class List.
A Marlborough Boy.
Putney Walker.
Blithe.
E. W.
L. B.
O. V. L.
Rose.
Sea Breeze.
Simple Susan.
Money-Spinner.
Blithe has made so ingenious an addition to the problem, and Simple Susan and Co. have solved it in such tuneful verse, that I record both their answers in full. I have altered a word or two in Blithe’s—which I trust she will excuse; it did not seem quite clear as it stood.
“Yet stay,” said the youth, as a gleam of inspiration lighted up the relaxing muscles of his quiescent features. “Stay. Methinks it matters little when we reached that summit, the crown of our toil. For in the space of time wherein we clambered up one mile and bounded down the same on our return, we could have trudged the twain on the level. We have plodded, then, four-and-twenty miles in these six mortal hours; for never a moment did we stop for catching of fleeting breath or for gazing on the scene around!”
“Very good,” said the old man. “Twelve miles out and twelve miles in. And we reached the top some time between six and seven of the clock. Now mark me! For every five minutes that had fled since six of the clock when we stood on yonder peak, so many miles had we toiled upwards on the dreary mountainside!”
The youth moaned and rushed into the hostel.
Blithe.The elder and the younger knight,
They sallied forth at three;
How far they went on level ground
It matters not to me;
What time they reached the foot of hill,
When they began to mount,
Are problems which I hold to be
Of very small account.
The moment that each waved his hat
Upon the topmost peak—
To trivial query such as this
No answer will I seek.
Yet can I tell the distance well
They must have travelled o’er:
On hill and plain, ’twixt three and nine,
The miles were twenty-four.
Four miles an hour their steady pace
Along the level track,
Three when they climbed—but six when they
Came swiftly striding back
Adown the hill; and little skill
It needs, methinks, to show,
Up hill and down together told,
Four miles an hour they go.
For whether long or short the time
Upon the hill they spent,
Two thirds were passed in going up,
One third in the descent.
Two thirds at three, one third at six,
If rightly reckoned o’er,
Will make one whole at four—the tale
Is tangled now no more.
Money Spinner. Answers to Knot II §1: The Dinner Party
Problem.—“The Governor of Kgovjni wants to give a very small dinner party, and invites his father’s brother-in-law, his brother’s father-in-law, his father-in-law’s brother, and his brother-in-law’s father. Find the number of guests.”
Answer.—“One.”
In this genealogy, males are denoted by capitals, and females by small letters. The Governor is E and his guest is C.Ten answers have been received. Of these, one is wrong, Galanthus Nivalis Major, who insists on inviting two guests, one being the Governor’s wife’s brother’s father. If she had taken his sister’s husband’s father instead, she would have found it possible to reduce the guests to one.
Of the nine who send right answers, Sea-Breeze is the very faintest breath that ever bore the name! She simply states that the Governor’s uncle might fulfill all the conditions “by intermarriages”! “Wind of the western sea,” you have had a very narrow escape! Be thankful to appear in the Class-list at all! Bog-Oak and Bradshaw of the Future use genealogies which require 16 people instead of 14, by inviting the Governor’s father’s sister’s husband instead of his father’s wife’s brother. I cannot think this so good a solution as one that requires only 14. Caius and Valentine deserve special mention as the only two who have supplied genealogies.
Class List.
Bee.
Caius.
M. M.
Matthew Matticks.
Old Cat.
Valentine.
Bog-Oak.
Bradshaw of the Future.
Sea-Breeze.
§2: The LodgingsProblem.—“A Square has 20 doors on each side, which contains 21 equal parts. They are numbered all round, beginning at one corner. From which of the four, Nos. 9, 25, 52, 73, is the sum of the distances, to the other three, least?”
Answer.—“From No. 9.”
Let A be No. 9, B No. 25, C No. 52, and D No. 73.
Then AB=(122+52)=169=13;
AC=21;
AD=(92+82)=145=12+; (N.B. i.e. “between 12 and 13.”)
BC=(162+122)=400=20;
BD=(32+212)=450=21+;
CD=(92+132)=250=15+;
Hence sum of distances from A is between 46 and 47; from B, between 54 and 55; from C, between 56 and 57; from D, between 48 and 51. (Why not “between 48 and 49”? Make this out for yourselves.) Hence the sum is least for A.
Twenty-five solutions have been received. Of these, 15 must be marked “0,” 5 are partly right, and 5
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