*Problem*. — (1) Two travelers, 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 traveler counting as “one” the train containing the other traveler. How many did each meet?

*Answers*. — (1) 19. (2) The easterly traveler 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 2/5 of this while the other does 3/5, 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 3/5 of this while the other does 2/5, 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 traveler passes 19 posts in going round, and so meets 19 trains. But, in (2), the easterly traveler only begins to count after traversing 2/5 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 2/5 of 3 hours, or 3/5 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 traveled 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, Thistle-Down, Tom-Quad, and an unsigned one. Bo-Peep rightly says that the easterly traveler 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 travelers met at the end of 71 minutes and 26½ seconds. Kate B. thinks the trains which are met on starting and arriving are never to be counted, even when met elsewhere. Q. Y. Z. tries a rather complex algebraic 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 travelers meet 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 travelers go “1/24 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 travelers 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 traveler 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 the (wrongly) make “7”; and they make a similar mistake with the easterly train. Bridget (rightly) says that the westerly traveler 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 traveler) 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.

Aix-les-Bains.

H. L. R.

Algernon Bray.

Omega.

Bradshaw of the Future.

S. S. G.

Fifee.

Waiting for the Train.

Arvon.

J. L. O.

F. Lee.

G. S. C.

X. A. B.

https://ebooks.adelaide.edu.au/c/carroll/lewis/tangled/answers3.html

Last updated Tuesday, August 25, 2015 at 14:06