A Tangled Tale Part 10

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 compet.i.tion 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 pa.s.ses 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 compet.i.tion 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 26-1/2 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 n.o.bly 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, CHEs.h.i.+RE 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. CHEs.h.i.+RE 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 cla.s.sic 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.

CLa.s.s 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, 13-1/2 lbs.; Nos. 3, 4, 11-1/2 lbs.; Nos. 4, 5, 8 lbs.; Nos. 1, 3, 5, 16 lbs. Required the weight of each sack."

_Answer._--"5-1/2, 6-1/2, 7, 4-1/2, 3-1/2."

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 6-1/2 lbs., 4-1/2 lbs. for Nos. 2, 4; and hence again, the 1st and 4th weighings give 5-1/2 lbs., 3-1/2 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 compet.i.tors 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 Cla.s.s-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 a.s.sume 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 Cla.s.s next time!