Chapter 6

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.