A Tangled Tale - Lewis Carroll (ebook pdf reader for pc txt) 📗

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., Seagull, 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. Seagull 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.

Aix-les-Bains.

Algernon Bray.

Bradshaw of the Future.

Fifee.

H. L. R.

Omega.

S. S. G.

Waiting for the train.

Arvon.

J. L. O.

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