A Tangled Tale

Simple Susan.
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.

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.

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 thus meets it at the end of 18 units, and so all the way round. A westerly train starting has 30 units between it and the first train it will meet: it does 3-5ths of this while the other does 2-5ths, and thus meets it at the end of 18 units, and so all the way round. Hence if the railway be divided, by 19 posts, into 20 parts, each containing 18 units, trains meet at every post, and, in (1), each traveller passes 19 posts in going round, and so meets 19 trains. But, in (2), the easterly traveller only begins to count after traversing 2-5ths of the journey, i.e., on reaching the 8th post, and so counts 12 posts: similarly the other counts 8. They meet at the end of 2-5ths of 3 hours, or 3-5ths of 2 hours, i.e., 72 minutes.

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 prize, they would have got no marks. [N.B.—I have not ventured to put E. A.'s name in full, as she only gave it provisionally, in case her answer should prove right.]

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 are never to be counted, even when met elsewhere. Q. Y. Z. tries a rather complex algebraical solution, and succeeds in finding the time of meeting correctly: all else is wrong. Sea-Gull seems to think that, in (1), the easterly train stood still for 3 hours; and says that, in (2), the travellers met at the end of 71 minutes 40 seconds. Thistledown nobly confesses to having tried no calculation, but merely having drawn a picture of the railway and counted the trains; in (1), she counts wrong; in (2) she makes them meet in 75 minutes. Tom-Quad omits (1): in (2) he makes Clara count the train she met on her arrival. The unsigned one is also unintelligible; it states that the travellers go "1-24th more than the total distance to be traversed"! The "Clara" theory, already referred to, is adopted by 5 of these, viz., Bo-Peep, Financier, Kate B., Tom-Quad, and the nameless writer.

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 existing between miles and hours? Do not be disheartened by your two friends' sarcastic remarks on your "roundabout ways." Their short method, of adding 12 and 8, has the slight disadvantage of bringing the answer wrong: even a "roundabout" method is better than that! M. A. H., in (2), makes the travellers count "one" after they met, not when they met. Cheshire Cat and Old Maid get "20" as answer for (1), by forgetting to strike out the train met on arrival. The others all get "18" in various ways. Bog-Oak, Guy, and R. W. divide the trains which the westerly traveller has to meet into 2 sets, viz., those already on the line, which they (rightly) make "11," and those which started during her 2 hours' journey (exclusive of train met on arrival), which they (wrongly) make "7"; and they make a similar mistake with the easterly train. Bridget (rightly) says that the westerly traveller met a train every 6 minutes for 2 hours, but (wrongly) makes the number "20"; it should be "21." G. E. B. adopts Bo-Peep's method, but (wrongly) strikes out (for the easterly traveller) the train which started at the commencement of the previous 2 hours. Mary thinks a train, met on arrival, must not be counted, even when met on a previous occasion.

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 solved the whole problem. Your names are Aix-les-Bains, Algernon Bray (thanks for a friendly remark, which comes with a heart-warmth that not even the Atlantic could chill), Arvon, Bradshaw of the Future, Fifee, H. L. R., J. L. O., Omega, S. S. G., and Waiting for the Train. Several of these have put Clara, provisionally, into the easterly train: but they seem to have understood that the data do not decide that point.

CLASS LIST.

I.

Aix-les-Bains.
Algernon Bray.
Bradshaw of the Future.
Fifee.
H. L. R.
Omega.
S. S. G.
Waiting for the train.

II.

Arvon.
J. L. O.

III.

F. Lee.
G. S. C.
X. A. B.

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 sort of clue to the process by which their answers were obtained. In guessing a conundrum, or in catching a flea, we do not expect the breathless victor to give us afterwards, in cold blood, a history of the mental or muscular efforts by which he achieved success; but a mathematical calculation is another thing. The names of this "mute inglorious" band are Common Sense, D. E. R., Douglas, E. L., Ellen, I. M. T., J. M. C., Joseph, Knot I, Lucy, Meek, M. F. C., Pyramus, Shah, Veritas.

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 highest marks to those who use the simplest machinery. I have put into Class I. those whose answers seemed specially short and neat, and into Class III. those that seemed specially long or clumsy. Of this last set, A. C. M., Furze-Bush, James, Partridge, R. W., and Waiting for the Train, have sent long wandering solutions, the substitutions having no definite method, but seeming to have been made to see what would come of it. Chilpome and Dublin Boy omit some of the working. Arvon Marlborough Boy only finds the weight of one sack.

CLASS LIST

I.

B. E. D.
C. H.
Constance Johnson.
Greystead.
Guy.
Hoopoe.
J. F. A.
M. A. H.
Number Five.
Pedro.
R. E. X.
Seven Old Men.
Vis Inertiæ.
Willy B.
Yahoo.

II.

American Subscriber.
An appreciative schoolma'am.
Ayr.
Bradshaw of the Future.
Cheam.
C. M. G.
Dinah Mite.
Duckwing.
E. C. M.
E. N. Lowry.
Era.
Euroclydon.
F. H. W.
Fifee.
G. E. B.
Harlequin.
Hawthorn.
Hough Green.
J. A. B.
Jack Tar.
J. B. B.
Kgovjni.
Land Lubber.
L. D.
Magpie.
Mary.
Mhruxi.
Minnie.
Money-Spinner.
Nairam.
Old Cat.
Polichinelle.
Simple Susan.
S. S. G.
Thisbe.
Verena.
Wamba.
Wolfe.
Wykehamicus.
Y. M. A. H.

III.

A. C. M.
Arvon Marlborough Boy.
Chilpome.
Dublin Boy.
Furze-Bush.
James.
Partridge.
R. W.
Waiting for the Train.

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 as close together as they will go, so as to get the minimum number of pictures——erasing optional marks where by so doing we can run them closer, but otherwise letting them stand. There are 10 necessary marks in the 1st row, and in the 3rd; but only 7 in the 2nd. Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the 2nd.

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 marked, with 30 marks; in one she gives 2 x's to 6 pictures; in another to 7; in the 3rd she gives 2 o's to 5; thus in every case ignoring the conditions. (I pause to remark that the condition "2 x's to 4 or 5 pictures" can only mean "either to 4 or else to 5": if, as one competitor holds, it might mean any number not less than 4, the words "or 5" would be superfluous.) I. E. A. (I am happy to say that none of these bloodless phantoms appear this time in the class-list. Is it IDEA with the "D" left out?) gives 2 x's to 6 pictures. She then takes me to task for using the word "ought" instead of "nought." No doubt, to one who thus rebels against the rules laid down for her guidance, the word must be distasteful. But does not I. E. A. remember the parallel case of "adder"? That creature was originally "a nadder": then the two words took to bandying the poor "n" backwards and forwards like a shuttlecock, the final state of the game being "an adder." May not "a nought" have similarly become "an ought"? Anyhow, "oughts and crosses" is a very old game. I don't think I ever heard it called "noughts and crosses."

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 trust I. E. A. will be merciful to the spelling), it is scarcely to be distinguished from "zero."

CLASS LIST.

I.

Guy.
Old Cat.
Sea-Breeze.

II.

Ayr.
Bradshaw of the Future.
F. Lee.
H. Vernon.

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 total result, we must multiply L's 3 marks together, and do the same for M and for Z. The final numbers are 1 × ¹⁄₅ × 3, ²⁄₅ × ²⁄₃ × 1, ²⁄₃ × 1 × ¼; i.e. ³⁄₅, ²⁄₃, ¹⁄₃; i.e. multiplying throughout by 15 (which will not alter the proportion), 9, 10, 5; showing the order of merit to be M, L, Z.

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: he, she, or it, is Addlepate. The other twenty-three may be divided into five groups.

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 shrouded in mystery—I scarcely feel competent to criticize it. Mrs. Gamp says "if Zuzu makes 4 while Lolo makes 3, Zuzu makes 6 while Lolo makes 5 (bad reasoning), while Mimi makes 2." From this she concludes "therefore Zuzu excels in speed by 1" (i.e. when compared with Lolo; but what about Mimi?). She then compares the 3 kinds of excellence, measured on this mystic scale. Janet takes the statement, that "Lolo makes 5 while Mimi makes 2," to prove that "Lolo makes 3 while Mimi makes 1 and Zuzu 4" (worse reasoning than Mrs. Gamp's), and thence concludes that "Zuzu excels in speed by ¹⁄₈"! Janet should have been Adeline, "mystery of mysteries!"

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 defective reasoning, the other for scanty working. Balbus gives two reasons for saying that addition of marks is not the right method, and then adds "it follows that the decision must be made by multiplying the marks together." This is hardly more logical than to say "This is not Spring: therefore it must be Autumn."

CLASS LIST.

I.

Dinah Mite.
E. B. D. L.
Joram.

II.

Balbus.
The Elder Traveller.

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 "Patience") before any solution would have been hit on by the most ingenious of them.

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 13 cases, the answer is right. Of the remaining 28, no less than 26 have sent in accidental solutions, and therefore fall short of the highest honours.

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. By making two mistakes in this line, she brings out y = ²⁄₂. Try it again, oh Vis Inertiæ! Away with Inertiæ: infuse a little more Vis: and you will bring out the correct (though uninteresting) result, 0 = 0! This will show you that it is hopeless to try to coax any one of these 3 unknowns to reveal its separate value. The other competitor, who is wrong throughout, is either J. M. C. or T. M. C.: but, whether he be a Juvenile Mis-Calculator or a True Mathematician Confused, he makes the answers 7d. and 1s. 5d. He assumes, with Too Much Confidence, that biscuits were ½d. each, and that Clara paid for 8, though she only ate 7!

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 circumstances? She begins by assuming that lemonade is 4d. a glass, and sandwiches 3d. each, (making with the 2 given equations, four conditions to be fulfilled by three miserable unknowns!). And, having (naturally) developed this into a contradiction, she then tries 5d. and 2d. with a similar result. (N.B. This process might have been carried on through the whole of the Tertiary Period, without gratifying one single Megatherium.) She then, by a "happy thought," tries half-penny biscuits, and so obtains a consistent result. This may be a good solution, viewing the problem as a conundrum: but it is not scientific. Janet identifies sandwiches with biscuits! "One sandwich + 3 biscuits" she makes equal to "4." Four what? Mayfair makes the astounding assertion that the equation, s + 3b = 3, "is evidently only satisfied by s = ²⁄₂, b = ½"! Old Cat believes that the assumption that a sandwich costs 1½d. is "the only way to avoid unmanageable fractions." But why avoid them? Is there not a certain glow of triumph in taming such a fraction? "Ladies and gentlemen, the fraction now before you is one that for years defied all efforts of a refining nature: it was, in a word, hopelessly vulgar. Treating it as a circulating decimal (the treadmill of fractions) only made matters worse. As a last resource, I reduced it to its lowest terms, and extracted its square root!" Joking apart, let me thank Old Cat for some very kind words of sympathy, in reference to a correspondent (whose name I am happy to say I have now forgotten) who had found fault with me as a discourteous critic. O. V. L. is beyond my comprehension. He takes the given equations as (1) and (2): thence, by the process [(2)-(1)] deduces (rightly) equation (3) viz. s + 3b = 3: and thence again, by the process [×33] (a hopeless mystery), deduces 3s + 4b = 4. I have nothing to say about it: I give it up. Sea-Breeze says "it is immaterial to the answer" (why?) "in what proportion 3d. is divided between the sandwich and the 3 biscuits": so she assumes s = l½d., b = ½d. Stanza is one of a very irregular metre. At first she (like Janet) identifies sandwiches with biscuits. She then tries two assumptions (s = 1, b = ²⁄₃, and s = ½ b = ²⁄₆), and (naturally) ends in contradictions. Then she returns to the first assumption, and finds the 3 unknowns separately: quod est absurdum. Stiletto identifies sandwiches and biscuits, as "articles." Is the word ever used by confectioners? I fancied "What is the next article, Ma'am?" was limited to linendrapers. Two Sisters first assume that biscuits are 4 a penny, and then that they are 2 a penny, adding that "the answer will of course be the same in both cases." It is a dreamy remark, making one feel something like Macbeth grasping at the spectral dagger. "Is this a statement that I see before me?" If you were to say "we both walked the same way this morning," and I were to say "one of you walked the same way, but the other didn't," which of the three would be the most hopelessly confused? Turtle Pyate (what is a Turtle Pyate, please?) and Old Crow, who send a joint answer, and Y. Y., adopt the same method. Y. Y. gets the equation s + 3b = 3: and then says "this sum must be apportioned in one of the three following ways." It may be, I grant you: but Y. Y. do you say "must"? I fear it is possible for Y. Y. to be two Y's. The other two conspirators are less positive: they say it "can" be so divided: but they add "either of the three prices being right"! This is bad grammar and bad arithmetic at once, oh mysterious birds!

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 data and the answers. In order to do them full justice I have arranged the 2nd class in sections, according to the number of steps. The two Kings are fearfully deliberate! I suppose walking quick, or taking short cuts, is inconsistent with kingly dignity: but really, in reading Theseus' solution, one almost fancied he was "marking time," and making no advance at all! The other King will, I hope, pardon me for having altered "Coal" into "Cole." King Coilus, or Coil, seems to have reigned soon after Arthur's time. Henry of Huntingdon identifies him with the King Coël who first built walls round Colchester, which was named after him. In the Chronicle of Robert of Gloucester we read:—