Chapter 5
APPENDIX.
"A knot!" said Alice. "Oh, do let me help to undo it!"
ANSWERS TO KNOT I.
Problem.—"Two travellers spend from 3 o’clock till 9 in walking along a level road, up a hill, and home again: their pace on the level being 4 miles an hour, up hill 3, and down hill 6. Find distance walked: also (within half an hour) time of reaching top of hill."
Answer.—"24 miles: half-past 6."
Solution.—A level mile takes ¼ of an hour, up hill 1⁄3, down hill 1⁄6. Hence to go and return over the same mile, whether on the level or on the hill-side, takes ½ an hour. Hence in 6 hours they went 12 miles out and 12 back. If the 12 miles out had been nearly all level, they would have taken a little over 3 hours; if nearly all up hill, a little under 4. Hence 3½ hours must be within ½ an hour of the time taken in reaching the peak; thus, as they started at 3, they got there within ½ an hour of ½ past 6.
Twenty-seven answers have come in. Of these, 9 are right, 16 partially right, and 2 wrong. The 16 give the distance correctly, but they have failed to grasp the fact that the top of the hill might have been reached at any moment between 6 o’clock and 7.
The two wrong answers are from Gerty Vernon and A Nihilist. The former makes the distance "23 miles," while her revolutionary companion puts it at "27." Gerty Vernon says "they had to go 4 miles along the plain, and got to the foot of the hill at 4 o’clock." They might have done so, I grant; but you have no ground for saying they did so. "It was 7½ miles to the top of the hill, and they reached that at ¼ before 7 o’clock." Here you go wrong in your arithmetic, and I must, however reluctantly, bid you farewell. 7½ miles, at 3 miles an hour, would not require 2¾ hours. A Nihilist says "Let x denote the whole number of miles; y the number of hours to hill-top; ∴ 3y = number of miles to hill-top, and x-3y = number of miles on the other side." You bewilder me. The other side of what? "Of the hill," you say. But then, how did they get home again? However, to accommodate your views we will build a new hostelry at the foot of the hill on the opposite side, and also assume (what I grant you is possible, though it is not necessarily true) that there was no level road at all. Even then you go wrong.
You say
"y = 6 - (x - 3y)⁄6, ..... (i);
x⁄4½ = 6 ..... (ii)."
I grant you (i), but I deny (ii): it rests on the assumption that to go part of the time at 3 miles an hour, and the rest at 6 miles an hour, comes to the same result as going the whole time at 4½ miles an hour. But this would only be true if the "part" were an exact half, i.e., if they went up hill for 3 hours, and down hill for the other 3: which they certainly did not do.
The sixteen, who are partially right, are Agnes Bailey, F. K., Fifee, G. E. B., H. P., Kit, M. E. T., Mysie, A Mother’s Son, Nairam, A Redruthian, A Socialist, Spear Maiden, T. B. C, Vis Inertiæ, and Yak. Of these, F. K., Fifee, T. B. C, and Vis Inertiæ do not attempt the second part at all. F. K. and H. P. give no working. The rest make particular assumptions, such as that there was no level road—that there were 6 miles of level road—and so on, all leading to particular times being fixed for reaching the hill-top. The most curious assumption is that of Agnes Bailey, who says "Let x = number of hours occupied in ascent; then x⁄2 = hours occupied in descent; and 4x⁄3 = hours occupied on the level." I suppose you were thinking of the relative rates, up hill and on the level; which we might express by saying that, if they went x miles up hill in a certain time, they would go 4x⁄3 miles on the level in the same time. You have, in fact, assumed that they took the same time on the level that they took in ascending the hill. Fifee assumes that, when the aged knight said they had gone "four miles in the hour" on the level, he meant that four miles was the distance gone, not merely the rate. This would have been—if Fifee will excuse the slang expression—a "sell," ill-suited to the dignity of the hero.
And now "descend, ye classic Nine!" who have solved the whole problem, and let me sing your praises. Your names are Blithe, E. W., L. B., A Marlborough Boy, O. V. L., Putney Walker, Rose, Sea Breeze, Simple Susan, and Money Spinner. (These last two I count as one, as they send a joint answer.) Rose and Simple Susan and Co. do not actually state that the hill-top was reached some time between 6 and 7, but, as they have clearly grasped the fact that a mile, ascended and descended, took the same time as two level miles, I mark them as "right." A Marlborough Boy and Putney Walker deserve honourable mention for their algebraical solutions being the only two who have perceived that the question leads to an indeterminate equation. E. W. brings a charge of untruthfulness against the aged knight—a serious charge, for he was the very pink of chivalry! She says "According to the data given, the time at the summit affords no clue to the total distance. It does not enable us to state precisely to an inch how much level and how much hill there was on the road." "Fair damsel," the aged knight replies, "—if, as I surmise, thy initials denote Early Womanhood—bethink thee that the word ’enable’ is thine, not mine. I did but ask the time of reaching the hill-top as my condition for further parley. If now thou wilt not grant that I am a truth-loving man, then will I affirm that those same initials denote Envenomed Wickedness!"
CLASS LIST.
I.
A Marlborough Boy.
Putney Walker.
II.
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.
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.
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.
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.
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.
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.
I.
Bee.
Caius.
M. M.
Matthew Matticks.
Old Cat.
Valentine.
II.
Bog-Oak.
Bradshaw of the Future.
III.
Sea-Breeze.
§ 2. The Lodgings.
Problem.—"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 right. Of the 15, I may dismiss Alphabetical Phantom, Bog-Oak, Dinah Mite, Fifee, Galanthus Nivalis Major (I fear the cold spring has blighted our Snowdrop), Guy, H.M.S. Pinafore, Janet, and Valentine with the simple remark that they insist on the unfortunate lodgers keeping to the pavement. (I used the words "crossed to Number Seventy-three" for the special purpose of showing that short cuts were possible.) Sea-Breeze does the same, and adds that "the result would be the same" even if they crossed the Square, but gives no proof of this. M. M. draws a diagram, and says that No. 9 is the house, "as the diagram shows." I cannot see how it does so. Old Cat assumes that the house must be No. 9 or No. 73. She does not explain how she estimates the distances. Bee’s Arithmetic is faulty: she makes √169 + √442 + √130 = 741. (I suppose you mean √741, which would be a little nearer the truth. But roots cannot be added in this manner. Do you think √9 + √16 is 25, or even √25?) But Ayr’s state is more perilous still: she draws illogical conclusions with a frightful calmness. After pointing out (rightly) that AC is less than BD she says, "therefore the nearest house to the other three must be A or C." And again, after pointing out (rightly) that B and D are both within the half-square containing A, she says "therefore" AB + AD must be less than BC + CD. (There is no logical force in either "therefore." For the first, try Nos. 1, 21, 60, 70: this will make your premiss true, and your conclusion false. Similarly, for the second, try Nos. 1, 30, 51, 71.)
Of the five partly-right solutions, Rags and Tatters and Mad Hatter (who send one answer between them) make No. 25 6 units from the corner instead of 5. Cheam, E. R. D. L., and Meggy Potts leave openings at the corners of the Square, which are not in the data: moreover Cheam gives values for the distances without any hint that they are only approximations. Crophi and Mophi make the bold and unfounded assumption that there were really 21 houses on each side, instead of 20 as stated by Balbus. "We may assume," they add, "that the doors of Nos. 21, 42, 63, 84, are invisible from the centre of the Square"! What is there, I wonder, that Crophi and Mophi would not assume?
Of the five who are wholly right, I think Bradshaw Of the Future, Caius, Clifton C., and Martreb deserve special praise for their full analytical solutions. Matthew Matticks picks out No. 9, and proves it to be the right house in two ways, very neatly and ingeniously, but why he picks it out does not appear. It is an excellent synthetical proof, but lacks the analysis which the other four supply.
CLASS LIST.
I.
Bradshaw of the Future
Caius.
Clifton C.
Martreb.
II.
Matthew Matticks.
III.
Cheam.
Crophi and Mophi.
E. R. D. L.
Meggy Potts.
{Rags and Tatters.
{Mad Hatter.
A remonstrance has reached me from Scrutator on the subject of Knot I., which he declares was "no problem at all." "Two questions," he says, "are put. To solve one there is no data: the other answers itself." As to the first point, Scrutator is mistaken; there are (not "is") data sufficient to answer the question. As to the other, it is interesting to know that the question "answers itself," and I am sure it does the question great credit: still I fear I cannot enter it on the list of winners, as this competition is only open to human beings.