  "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); 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 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 CLASS LIST. I. A Marlborough Boy. II. Blithe. 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. 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. II. Bog-Oak. 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 = v(122 + 52) = v169 = 13; 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 v169 + v442 + v130 = 741. (I suppose you mean v741, which would be a little nearer the truth. But roots cannot be added in this manner. Do you think v9 + v16 is 25, or even v25?) 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 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 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 II. Matthew Matticks. III. Cheam. 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. ANSWERS TO KNOT III.Problem.—(1) "Two travellers, starting at the same time, went opposite ways round a circular railway. Trains start each way every 15 minutes, the easterly ones going round in 3 hours, the westerly in 2. How many trains did each meet on the way, not counting trains met at the terminus itself?" (2) "They went round, as before, each traveller counting as 'one' the train containing the other traveller. How many did each meet?" Answers.—(1) 19. (2) The easterly traveller met 12; the other 8. The trains one way took 180 minutes, the other way 120. Let us take the L. C. M., 360, and divide the railway into 360 units. Then one set of trains went at the rate of 2 units a minute and at intervals of 30 units; the other at the rate of 3 units a minute and at intervals of 45 units. An easterly train starting has 45 units between it and the first train it will meet: it does 2-5ths of this while the other does 3-5ths, and Forty-five answers have been received. Of these 12 are beyond the reach of discussion, as they give no working. I can but enumerate their names. Ardmore, E. A., F. A. D., L. D., Matthew Matticks, M. E. T., Poo-Poo, and The Red Queen are all wrong. Beta and Rowena have got (1) right and (2) wrong. Cheeky Bob and Nairam give the right answers, but it may perhaps make the one less cheeky, and induce the other to take a less inverted view of things, to be informed that, if this had been a competition for a Of the 33 answers for which the working is given, 10 are wrong; 11 half-wrong and half-right; 3 right, except that they cherish the delusion that it was Clara who travelled in the easterly train—a point which the data do not enable us to settle; and 9 wholly right. The 10 wrong answers are from Bo-Peep, Financier, I. W. T., Kate B., M. A. H., Q. Y. Z., Sea-Gull, Thistledown, Tom-Quad, and an unsigned one. Bo-Peep rightly says that the easterly traveller met all trains which started during the 3 hours of her trip, as well as all which started during the previous 2 hours, i.e., all which started at the commencements of 20 periods of 15 minutes each; and she is right in striking out the one she met at the moment of starting; but wrong in striking out the last train, for she did not meet this at the terminus, but 15 minutes before she got there. She makes the same mistake in (2). Financier thinks that any train, met for the second time, is not to be counted. I. W. T. finds, by a process which is not stated, that the travellers met at the end of 71 minutes and 26½ seconds. Kate B. thinks the trains which are met on starting and on arriving The 11 half-right answers are from Bog-Oak, Bridget, Castor, Cheshire Cat, G. E. B., Guy, Mary, M. A. H., Old Maid, R. W., and Vendredi. All these adopt the "Clara" theory. Castor omits (1). Vendredi gets (1) right, but in (2) makes the same mistake as Bo-Peep. I notice in your solution a marvellous proportion-sum:—"300 miles: 2 hours :: one mile: 24 seconds." May I venture to advise your acquiring, as soon as possible, an utter disbelief in the possibility of a ratio The 3, who are wholly right but for the unfortunate "Clara" theory, are F. Lee, G. S. C., and X. A. B. And now "descend, ye classic Ten!" who have CLASS LIST. I.
II. Arvon. III. F. Lee. ANSWERS TO KNOT IV.Problem.—"There are 5 sacks, of which Nos. 1, 2, weigh 12 lbs.; Nos. 2, 3, 13½ lbs.; Nos. 3, 4, 11½ lbs.; Nos. 4, 5, 8 lbs.; Nos. 1, 3, 5, 16 lbs. Required the weight of each sack." Answer.—"5½, 6½, 7, 4½, 3½." The sum of all the weighings, 61 lbs., includes sack No. 3 thrice and each other twice. Deducting twice the sum of the 1st and 4th weighings, we get 21 lbs. for thrice No. 3, i.e., 7 lbs. for No. 3. Hence, the 2nd and 3rd weighings give 6½ lbs., 4½ lbs. for Nos. 2, 4; and hence again, the 1st and 4th weighings give 5½ lbs., 3½ lbs., for Nos. 1, 5. Ninety-seven answers have been received. Of these, 15 are beyond the reach of discussion, as they give no working. I can but enumerate their names, and I take this opportunity of saying that this is the last time I shall put on record the names of competitors who give no Of the eighty-two answers with which the working, or some approach to it, is supplied, one is wrong: seventeen have given solutions which are (from one cause or another) practically valueless: the remaining sixty-four I shall try to arrange in a Class-list, according to the varying degrees of shortness and neatness to which they seem to have attained. The solitary wrong answer is from Nell. To be thus "alone in the crowd" is a distinction—a painful one, no doubt, but still a distinction. I am sorry for you, my dear young lady, and I seem to hear your tearful exclamation, when you read these lines, "Ah! This is the knell of all my hopes!" Why, oh why, did you assume that the 4th and 5th bags weighed 4 lbs. each? And why did you not test your answers? However, please try again: and please don't change your nom-de-plume: let us have Nell in the First Class next time! The seventeen whose solutions are practically valueless are Ardmore, A ready Reckoner, Arthur, Bog-Lark, Bog-Oak, Bridget, First Attempt, J. L. C., M. E. T., Rose, Rowena, Sea-Breeze, Sylvia, Thistledown, Three-Fifths Asleep, Vendredi, and Winifred. Bog-Lark tries it by a sort of "rule of false," assuming experimentally that Nos. 1, 2, weigh 6 lbs. each, and having thus produced 17½, instead of 16, as the weight of 1, 3, and 5, she removes "the superfluous pound and a half," but does not explain how she knows from which to take it. Three-fifths Asleep says that (when in that peculiar state) "it seemed perfectly clear" to her that, "3 out of the 5 sacks being weighed twice over, 2/5 of 45 = 27, must be the total weight of the 5 sacks." As to which I can only say, with the Captain, "it beats me entirely!" Winifred, on the plea that "one must have a starting-point," assumes (what I fear is a mere guess) that No. 1 weighed 5½ lbs. The rest all do it, wholly or partly, by guess-work. The problem is of course (as any Algebraist sees at once) a case of "simultaneous simple equations." It is, however, easily soluble by Arithmetic only; and, when this is the case, I hold that it is bad workmanship to use the more complex method. I have not, this time, given more credit to arithmetical solutions; but in future problems I shall (other things being equal) give the CLASS LIST I. B. E. D. II. American Subscriber. III. A. C. M. ANSWERS TO KNOT V.Problem.—To mark pictures, giving 3 x's to 2 or 3, 2 to 4 or 5, and 1 to 9 or 10; also giving 3 o's to 1 or 2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the smallest possible number of pictures, and to give them the largest possible number of marks. Answer.—10 pictures; 29 marks; arranged thus:—
Solution.—By giving all the x's possible, putting into brackets the optional ones, we get 10 pictures marked thus:—
By then assigning o's in the same way, beginning at the other end, we get 9 pictures marked thus:—
All we have now to do is to run these two wedges Twenty-two answers have been received. Of these 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong. Of the eleven answers with which some working is supplied, 3 are wrong. C. H. begins with the rash assertion that under the given conditions "the sum is impossible. For," he or she adds (these initialed correspondents are dismally vague beings to deal with: perhaps "it" would be a better pronoun), "10 is the least possible number of pictures" (granted): "therefore we must either give 2 x's to 6, or 2 o's to 5." Why "must," oh alphabetical phantom? It is nowhere ordained that every picture "must" have 3 marks! Fifee sends a folio page of solution, which deserved a better fate: she offers 3 answers, in each of which 10 pictures are In the following Class-list, I hope the solitary occupant of III. will sheathe her claws when she hears how narrow an escape she has had of not being named at all. Her account of the process by which she got the answer is so meagre that, like the nursery tale of "Jack-a-Minory" (I CLASS LIST. I. Guy. II. Ayr. III. Cat. ANSWERS TO KNOT VI.Problem 1.—A and B began the year with only 1,000l. a-piece. They borrowed nought; they stole nought. On the next New-Year's Day they had 60,000l. between them. How did they do it? Solution.—They went that day to the Bank of England. A stood in front of it, while B went round and stood behind it. Two answers have been received, both worthy of much honour. Addlepate makes them borrow "0" and steal "0," and uses both cyphers by putting them at the right-hand end of the 1,000l., thus producing 100,000l., which is well over the mark. But (or to express it in Latin) At Spes infracta has solved it even more ingeniously: with the first cypher she turns the "1" of the 1,000l. into a "9," and adds the result to the original sum, thus getting 10,000l.: and in this, by means of the other "0," she turns the "1" into a "6," thus hitting the exact 60,000l. CLASS LIST I. At Spes Infracta. II. Addlepate. Problem 2.—L makes 5 scarves, while M makes 2: Z makes 4 while L makes 3. Five scarves of Z's weigh one of L's; 5 of M's weigh 3 of Z's. One of M's is as warm as 4 of Z's: and one of L's as warm as 3 of M's. Which is best, giving equal weight in the result to rapidity of work, lightness, and warmth? Answer.—The order is M, L, Z. Solution.—As to rapidity (other things being constant) L's merit is to M's in the ratio of 5 to 2: Z's to L's in the ratio of 4 to 3. In order to get one set of 3 numbers fulfilling these conditions, it is perhaps simplest to take the one that occurs twice as unity, and reduce the others to fractions: this gives, for L, M, and Z, the marks 1, 2/5, 2/3. In estimating for lightness, we observe that the greater the weight, the less the merit, so that Z's merit is to L's as 5 to 1. Thus the marks for lightness are 1/5, 2/3, 1. And similarly, the marks for warmth are 3, 1, ¼. To get the Twenty-nine answers have been received, of which five are right, and twenty-four wrong. These hapless ones have all (with three exceptions) fallen into the error of adding the proportional numbers together, for each candidate, instead of multiplying. Why the latter is right, rather than the former, is fully proved in text-books, so I will not occupy space by stating it here: but it can be illustrated very easily by the case of length, breadth, and depth. Suppose A and B are rival diggers of rectangular tanks: the amount of work done is evidently measured by the number of cubical feet dug out. Let A dig a tank 10 feet long, 10 wide, 2 deep: let B dig one 6 feet long, 5 wide, 10 deep. The cubical contents are 200, 300; i.e. B is best digger in the ratio of 3 to 2. Now try marking for length, width, and depth, separately; giving a maximum mark of 10 to the best in each contest, and then adding the results! Of the twenty-four malefactors, one gives no working, and so has no real claim to be named; but I break the rule for once, in deference to its success in Problem 1: First and worst are, I take it, those who put the rightful winner last; arranging them as "Lolo, Zuzu, Mimi." The names of these desperate wrong-doers are Ayr, Bradshaw of the Future, Furze-bush and Pollux (who send a joint answer), Greystead, Guy, Old Hen, and Simple Susan. The latter was once best of all; the Old Hen has taken advantage of her simplicity, and beguiled her with the chaff which was the bane of her own chickenhood. Secondly, I point the finger of scorn at those who have put the worst candidate at the top; arranging them as "Zuzu, Mimi, Lolo." They are Graecia, M. M., Old Cat, and R. E. X. "'Tis Greece, but——." The third set have avoided both these enormities, and have even succeeded in putting the worst last, their answer being "Lolo, Mimi, Zuzu." Their names are Ayr (who also appears among the "quite too too"), Clifton C., F. B., Fifee, Grig, Janet, and Mrs. Sairey Gamp. F. B. has not fallen into the common error; she multiplies together the proportionate numbers she gets, but in getting them she goes wrong, by reckoning warmth as a de-merit. Possibly she is "Freshly Burnt," or comes "From Bombay." Janet and Mrs. Sairey Gamp have also avoided this error: the method they have adopted is The fourth set actually put Mimi at the top, arranging them as "Mimi, Zuzu, Lolo." They are Marquis and Co., Martreb, S. B. B. (first initial scarcely legible: may be meant for "J"), and Stanza. The fifth set consist of An ancient Fish and Camel. These ill-assorted comrades, by dint of foot and fin, have scrambled into the right answer, but, as their method is wrong, of course it counts for nothing. Also An ancient Fish has very ancient and fishlike ideas as to how numbers represent merit: she says "Lolo gains 2½ on Mimi." Two and a half what? Fish, fish, art thou in thy duty? Of the five winners I put Balbus and The elder Traveller slightly below the other three—Balbus for CLASS LIST. I. Dinah Mite. II. Balbus. With regard to Knot V., I beg to express to Vis InertiÆ and to any others who, like her, understood the condition to be that every marked picture must have three marks, my sincere regret that the unfortunate phrase "fill the columns with oughts and crosses" should have caused them to waste so much time and trouble. I can only repeat that a literal interpretation of "fill" would seem to me to require that every picture in the gallery should be marked. Vis InertiÆ would have been in the First Class if she had sent in the solution she now offers. ANSWERS TO KNOT VII.Problem.—Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1s. 2d.; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1s. 5d.: find the cost of (1) a glass of lemonade, a sandwich, and a biscuit; and (2) 2 glasses of lemonade, 3 sandwiches, and 5 biscuits. Answer.—(1) 8d.; (2) 1s. 7d. Solution.—This is best treated algebraically. Let x = the cost (in pence) of a glass of lemonade, y of a sandwich, and z of a biscuit. Then we have x + 3y + 7z = 14, and x + 4y + 10z = 17. And we require the values of x + y + z, and of 2x + 3y + 5z. Now, from two equations only, we cannot find, separately, the values of three unknowns: certain combinations of them may, however, be found. Also we know that we can, by the help of the given equations, eliminate 2 of the 3 unknowns from the quantity whose value is required, which will then contain one only. If, then, the required value is ascertainable at all, it can only be by the 3rd unknown vanishing of itself: otherwise the problem is impossible. Let us then eliminate lemonade and sandwiches, and reduce everything to biscuits—a state of things even more depressing than "if all the world were apple-pie"—by subtracting the 1st equation from the 2nd, which eliminates lemonade, and gives y + 3z = 3, or y = 3-3z; and then substituting this value of y in the 1st, which gives x-2z = 5, i.e. x = 5 + 2z. Now if we substitute these values of x, y, in the quantities whose values are required, the first becomes (5 + 2z) + (3-3z) + z, i.e. 8: and the second becomes 2(5 + 2z) + 3(3-3z) + 5z, i.e. 19. Hence the answers are (1) 8d., (2) 1s. 7d. The above is a universal method: that is, it is absolutely certain either to produce the answer, or to prove that no answer is possible. The question may also be solved by combining the quantities whose values are given, so as to form those whose values are required. This is merely a matter of ingenuity and good luck: and as it may fail, even when the thing is possible, and is of no use in proving it impossible, I cannot rank this method as equal in value with the other. Even when it succeeds, it may prove a very tedious process. Suppose the 26 competitors, who have sent in what I may call accidental solutions, had had a question to deal with where every number contained 8 or 10 digits! I suspect it would have been a case of "silvered is the raven hair" (see Forty-five answers have come in, of which 44 give, I am happy to say, some sort of working, and therefore deserve to be mentioned by name, and to have their virtues, or vices as the case may be, discussed. Thirteen have made assumptions to which they have no right, and so cannot figure in the Class-list, even though, in 10 of the I will now discuss individual cases, taking the worst first, as my custom is. Froggy gives no working—at least this is all he gives: after stating the given equations, he says "therefore the difference, 1 sandwich + 3 biscuits, = 3d.": then follow the amounts of the unknown bills, with no further hint as to how he got them. Froggy has had a very narrow escape of not being named at all! Of those who are wrong, Vis InertiÆ has sent in a piece of incorrect working. Peruse the horrid details, and shudder! She takes x (call it "y") as the cost of a sandwich, and concludes (rightly enough) that a biscuit will cost (3-y)/3. She then subtracts the second equation from the first, and deduces 3y + 7 × (3-y)/3-4y + 10 × (3-y)/3 = 3. We will now consider the 13 whose working is wrong, though the answer is right: and, not to measure their demerits too exactly, I will take them in alphabetical order. Anita finds (rightly) that "1 sandwich and 3 biscuits cost 3d.," and proceeds "therefore 1 sandwich = 1½d., 3 biscuits = 1½d., 1 lemonade = 6d." Dinah Mite begins like Anita: and thence proves (rightly) that a biscuit costs less than a 1d.: whence she concludes (wrongly) that it must cost ½d. F. C. W. is so beautifully resigned to the certainty of a verdict of "guilty," that I have hardly the heart to utter the word, without adding a "recommended to mercy owing to extenuating circumstances." But really, you know, where are the extenuating Of those who win honours, The Shetland Snark must have the 3rd class all to himself. He has only answered half the question, viz. the amount of Clara's luncheon: the two little old ladies he pitilessly leaves in the midst of their "difficulty." I beg to assure him (with thanks for his friendly remarks) that entrance-fees and subscriptions are things unknown in that most economical of clubs, "The Knot-Untiers." The authors of the 26 "accidental" solutions differ only in the number of steps they have taken between the Balbus lays it down as a general principle that "in order to ascertain the cost of any one luncheon, it must come to the same amount upon two different assumptions." (Query. Should not "it" be "we"? Otherwise the luncheon is represented as wishing to ascertain its own cost!) He then makes two assumptions—one, that sandwiches cost nothing; the other, that biscuits cost nothing, (either arrangement would lead to the shop being inconveniently crowded!)—and brings out the unknown Another remark of Balbus I will quote and discuss: for I think that it also may yield a moral for some of my readers. He says "it is the same thing in substance whether in solving this problem we use words and call it Arithmetic, or use letters and signs and call it Algebra." Take an illustration. Your house has been broken into and robbed, and you appeal to the policeman who was on duty that night. "Well, Mum, I did see a chap getting out over your garden-wall: but I was a good bit off, so I didn't chase him, like. I just cut down the short way to the Chequers, and who should I meet but Bill Sykes, coming full split round the corner. So I just ups and says 'My lad, you're wanted.' That's all I says. And he says 'I'll go along quiet, Bobby,' he says, 'without the darbies,' he says." There's your Arithmetical policeman. Now try the other method. "I seed somebody a running, but he was well gone or ever I got nigh the place. So I just took a look round in the garden. And I noticed the foot-marks, where the chap had come right across your flower-beds. They was good big foot-marks sure-ly. And I noticed as the left foot went down at the heel, ever so much deeper than the other. And I says to myself Little Jack's solution calls for a word of praise, as he has written out what really is an algebraical proof in words, without representing any of his facts as equations. If it is all his own, he will make a good algebraist in the time to come. I beg to thank Simple Susan for some kind words of sympathy, to the same effect as those received from Old Cat. Hecla and Martreb are the only two who have used a method certain either to produce the answer, or else to prove it impossible: so they must share between them the highest honours. CLASS LIST. I. Hecla. II. § 1 (2 steps). Adelaide. § 2 (3 steps). A. A. § 3 (4 steps). Hawthorn. § 4 (5 steps). A Stepney Coach. § 5 (6 steps). Bay Laurel. § 6 (9 steps). Old King Cole. § 7 (14 steps). Theseus. ANSWERS TO CORRESPONDENTS.I have received several letters on the subjects of Knots II. and VI., which lead me to think some further explanation desirable. In Knot II., I had intended the numbering of the houses to begin at one corner of the Square, and this was assumed by most, if not all, of the competitors. Trojanus however says "assuming, in default of any information, that the street enters the square in the middle of each side, it may be supposed that the numbering begins at a street." But surely the other is the more natural assumption? In Knot VI., the first Problem was of course a mere jeu de mots, whose presence I thought excusable in a series of Problems whose aim is to entertain rather than to instruct: but it has not escaped the contemptuous criticisms of two of my correspondents, who seem to think that Apollo is in duty bound to keep his bow always on the stretch. Neither of them has guessed it: and this is true human nature. Only the other day—the 31st of September, to be quite exact—I met my old friend Brown, and gave him a riddle I had just heard. With one great effort of his colossal mind, Brown guessed it. "Right!" said I. "Ah," said The second Problem of Knot VI. is an example in ordinary Double Rule of Three, whose essential feature is that the result depends on the variation of several elements, which are so related to it that, if all but one be constant, it varies as that one: hence, if none be constant, it varies as their product. Thus, for example, the cubical contents of a rectangular tank vary as its length, if breadth and depth be constant, and so on; hence, if none be constant, it varies as the product of the length, breadth, and depth. When the result is not thus connected with the varying elements, the Problem ceases to be Double Rule of Three and often becomes one of great complexity. To illustrate this, let us take two candidates for a prize, A and B, who are to compete in French, German, and Italian: (a) Let it be laid down that the result is to depend (b) The result is to depend, as before, on relative knowledge; but French is to have twice as much weight as German or Italian. This is an unusual form of question. I should be inclined to say "the resulting ratio is to be nearer to the French ratio than if we multiplied as in (a), and so much nearer that it would be necessary to use the other multipliers twice to produce the same result as in (a):" e.g. if the French Ratio were 2/10, and the others 2/9, 1/9 so that the ultimate ratio, by method (a), would be 2/45, I should multiply instead by 2/3, 1/3, giving the result, 1/3 which is nearer to 2/10 than if he had used method (a). (c) The result is to depend on actual amount of knowledge of the 3 subjects collectively. Here we have (d) The conditions are the same as (c), but French is to have double weight. Here we simply double the French marks, and add as before. (e) French is to have such weight, that, if other marks be equal, the ultimate ratio is to be that of the French paper, so that a "0" in this would swamp the candidate: but the other two subjects are only to affect the result collectively, by the amount of knowledge shown, the two being reckoned of equal value. Here I should add A's German and Italian marks together, and multiply by his French mark. But I need not go on: the problem may evidently be set with many varying conditions, each requiring its own method of solution. The Problem in Knot VI. was meant to belong to variety (a), and to make this clear, I inserted the following passage: "Usually the competitors differ in one point only. Thus, last year, Fifi and Gogo made the same number of What I have said will suffice, I hope, as an answer to Balbus, who holds that (a) and (c) are the only possible varieties of the problem, and that to say "We cannot use addition, therefore we must be intended to use multiplication," is "no more illogical than, from knowledge that one was not born in the night, to infer that he was born in the daytime"; and also to Fifee, who says "I think a little more consideration will show you that our 'error of adding the proportional numbers together for each candidate instead of multiplying' is no error at all." Why, even if addition had been the right method to use, not one of the writers (I speak from memory) showed any consciousness of the necessity of fixing a "unit" for each subject. "No error at all!" They were positively steeped in error! One correspondent (I do not name him, as the communication is not quite friendly in tone) writes thus:—"I wish to add, very respectfully, that I think it would be in better taste if you were to abstain from the very trenchant expressions which you are accustomed to indulge in when criticising the answer. That such a tone must not be" ("be not"?) "agreeable to The writer's insinuation that I care not how much annoyance I give to my readers I think it best to pass over in silence; but to his concluding remark I must entirely demur. I hold that to use language likely to annoy any of my correspondents would not be in the least justified by the plea that I was "quite certain of I beg to thank G. B. for the offer of a puzzle—which, however, is too like the old one "Make four 9's into 100." ANSWERS TO KNOT VIII.§ 1. The Pigs. Problem.—Place twenty-four pigs in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last. Answer.—Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6. This problem is noticed by only two correspondents. Balbus says "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble." Nolens Volens makes Her Radiancy change the direction of going round; and even then is obliged to add "the pigs must be carried in front of her"! § 2. The Grurmstipths. Problem.—Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on Answer.—In 6¼ minutes. Solution.—Let "a" be the distance an omnibus goes in 15 minutes, and "x" the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2½ minutes, it goes in that time as far as the traveller walks in 12½; i.e. it goes 5 times as fast. Now the overtaking omnibus is "a" behind the traveller when he starts, and therefore goes "a + x" while he goes "x." Hence a + x = 5x; i.e. 4x = a, and x = a/4. This distance would be traversed by an omnibus in 15/4 minutes, and therefore by the traveller in 5 × 15/4. Hence he is overtaken in 18¾ minutes after starting, i.e. in 6¼ minutes after meeting the omnibus. Four answers have been received, of which two are wrong. Dinah Mite rightly states that the overtaking omnibus reached the point where they met the other omnibus 5 minutes after they left, but wrongly concludes that, going 5 times as fast, it would overtake them in another minute. The travellers are 5-minutes-walk ahead CLASS LIST. I. Balbus. ANSWERS TO KNOT IX.§ 1. The Buckets. Problem.—Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one? Solution.—Lardner means, by "displaces," "occupies a space which might be filled with water without any change in the surroundings." If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner's statement. Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. Hecla says that "only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced." Hence, according to I regret that there is no Class-list to publish for this Problem. § 2. Balbus' Essay. Problem.—Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, &c., which series has no end. He concludes that the water will rise without limit. Is this true? Solution.—No. This series can never reach 4 inches, Three answers have been received—but only two seem to me worthy of honours. Tympanum says that the statement about the stick "is merely a blind, to which the old answer may well be applied, solvitur ambulando, or rather mergendo." I trust Tympanum will not test this in his own person, by taking the place of the man in Balbus' Essay! He would infallibly be drowned. Old King Cole rightly points out that the series, 2, 1, &c., is a decreasing Geometrical Progression: while Vindex rightly identifies the fallacy as that of "Achilles and the Tortoise." CLASS LIST. I. Old King Cole. § 3. The Garden. Problem.—An oblong garden, half a yard longer than wide, consists entirely of a gravel-walk, spirally arranged, a yard wide and 3,630 yards long. Find the dimensions of the garden. Answer.—60, 60½. Solution.—The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square-yards and fractions of a square-yard, contained in that piece of walk: and the distance, traversed in passing through a square-yard at a corner, is evidently a yard. Hence the area of the garden is 3,630 square-yards: i.e., if x be the width, x (x + ½) = 3,630. Solving this Quadratic, we find x = 60. Hence the dimensions are 60, 60½. Twelve answers have been received—seven right and five wrong. C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.'s "working" consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow's is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60½"! Nabob's calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals the Hecla indulges, again and again, in that most fatal of all habits in computation—the making two mistakes which cancel each other. She takes x as the width of the garden, in yards, and x + ½ as its length, and makes her first "coil" the sum of x½, x½, x-1, x-1, i.e. 4x-3: but the fourth term should be x-1½, so that her first coil is ½ a yard too long. Her second coil is the sum of x-2½, x-2½, x-3, x-3: here the first term should be x-2 and the last x-3½: these two Of the seven who are right, Dinah Mite, Janet, Magpie, and Taffy make the same assumption as C. G. L. and Co. They then solve by a Quadratic. Magpie also tries it by Arithmetical Progression, but fails to notice that the first and last "coils" have special values. Alumnus EtonÆ attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5½. He ought to have proved it generally: what is true of one number is not always true of others. Old King Cole solves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic. Vindex proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, "whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction." CLASS LIST. I. Vindex. II. Alumnus EtonÆ. III. Dinah Mite. 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 2/10ths." 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 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. II. Bradshaw of the Future. III. Algernon Bray. § 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 = 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, 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 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 5z, 6z, and 7z; it then assumes, without giving any reason, that 7z = 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 = 7x + 4y; ? 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, 2A = B, C = B + A, D = 2B. 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 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. II. C. R. III. Dinah Mite. 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 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. |
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