Chapter 4
_Problem._--"Two travellers spend from 3 o'clock till 9 in walking along a level road, up a hill, and home again: their pace on the level being 4 miles an hour, up hill 3, and down hill 6. Find distance walked: also (within half an hour) time of reaching top of hill."
_Answer._--"24 miles: half-past 6."
* * * * *
_Solution._--A level mile takes 1/4 of an hour, up hill 1/3, down hill 1/6. Hence to go and return over the same mile, whether on the level or on the hill-side, takes 1/2 an hour. Hence in 6 hours they went 12 miles out and 12 back. If the 12 miles out had been nearly all level, they would have taken a little over 3 hours; if nearly all up hill, a little under 4. Hence 3-1/2 hours must be within 1/2 an hour of the time taken in reaching the peak; thus, as they started at 3, they got there within 1/2 an hour of 1/2 past 6.
* * * * *
Twenty-seven answers have come in. Of these, 9 are right, 16 partially right, and 2 wrong. The 16 give the _distance_ correctly, but they have failed to grasp the fact that the top of the hill might have been reached at _any_ moment between 6 o'clock and 7.
The two wrong answers are from GERTY VERNON and A NIHILIST. The former makes the distance "23 miles," while her revolutionary companion puts it at "27." GERTY VERNON says "they had to go 4 miles along the plain, and got to the foot of the hill at 4 o'clock." They _might_ have done so, I grant; but you have no ground for saying they _did_ so. "It was 7-1/2 miles to the top of the hill, and they reached that at 1/4 before 7 o'clock." Here you go wrong in your arithmetic, and I must, however reluctantly, bid you farewell. 7-1/2 miles, at 3 miles an hour, would _not_ require 2-3/4 hours. A NIHILIST says "Let _x_ denote the whole number of miles; _y_ the number of hours to hill-top; [** therefore] 3_y_ = number of miles to hill-top, and _x_-3_y_ = number of miles on the other side." You bewilder me. The other side of _what_? "Of the hill," you say. But then, how did they get home again? However, to accommodate your views we will build a new hostelry at the foot of the hill on the opposite side, and also assume (what I grant you is _possible_, though it is not _necessarily_ true) that there was no level road at all. Even then you go wrong.
You say
"_y_ = 6 - (_x_ - 3_y_)/6, ..... (i);
_x_/4-1/2 = 6 ..... (ii)."
I grant you (i), but I deny (ii): it rests on the assumption that to go _part_ of the time at 3 miles an hour, and the rest at 6 miles an hour, comes to the same result as going the _whole_ time at 4-1/2 miles an hour. But this would only be true if the "_part_" were an exact _half_, i.e., if they went up hill for 3 hours, and down hill for the other 3: which they certainly did _not_ do.
The sixteen, who are partially right, are AGNES BAILEY, F. K., FIFEE, G. E. B., H. P., KIT, M. E. T., MYSIE, A MOTHER'S SON, NAIRAM, A REDRUTHIAN, A SOCIALIST, SPEAR MAIDEN, T. B. C., VIS INERTIÆ, and YAK. Of these, F. K., FIFEE, T. B. C., and VIS INERTIÆ do not attempt the second part at all. F. K. and H. P. give no working. The rest make particular assumptions, such as that there was no level road--that there were 6 miles of level road--and so on, all leading to _particular_ times being fixed for reaching the hill-top. The most curious assumption is that of AGNES BAILEY, who says "Let _x_ = number of hours occupied in ascent; then _x_/2 = hours occupied in descent; and 4_x_/3 = hours occupied on the level." I suppose you were thinking of the relative _rates_, up hill and on the level; which we might express by saying that, if they went _x_ miles up hill in a certain time, they would go 4_x_/3 miles on the level _in the same time_. You have, in fact, assumed that they took _the same time_ on the level that they took in ascending the hill. FIFEE assumes that, when the aged knight said they had gone "four miles in the hour" on the level, he meant that four miles was the _distance_ gone, not merely the rate. This would have been--if FIFEE will excuse the slang expression--a "sell," ill-suited to the dignity of the hero.
And now "descend, ye classic Nine!" who have solved the whole problem, and let me sing your praises. Your names are BLITHE, E. W., L. B., A MARLBOROUGH BOY, O. V. L., PUTNEY WALKER, ROSE, SEA BREEZE, SIMPLE SUSAN, and MONEY SPINNER. (These last two I count as one, as they send a joint answer.) ROSE and SIMPLE SUSAN and CO. do not actually state that the hill-top was reached some time between 6 and 7, but, as they have clearly grasped the fact that a mile, ascended and descended, took the same time as two level miles, I mark them as "right." A MARLBOROUGH BOY and PUTNEY WALKER deserve honourable mention for their algebraical solutions being the only two who have perceived that the question leads to _an indeterminate equation_. E. W. brings a charge of untruthfulness against the aged knight--a serious charge, for he was the very pink of chivalry! She says "According to the data given, the time at the summit affords no clue to the total distance. It does not enable us to state precisely to an inch how much level and how much hill there was on the road." "Fair damsel," the aged knight replies, "--if, as I surmise, thy initials denote Early Womanhood--bethink thee that the word 'enable' is thine, not mine. I did but ask the time of reaching the hill-top as my _condition_ for further parley. If _now_ thou wilt not grant that I am a truth-loving man, then will I affirm that those same initials denote Envenomed Wickedness!"
CLASS LIST.
I.
A MARLBOROUGH BOY. PUTNEY WALKER.
II.
BLITHE. E. W. L. B. O. V. L. ROSE. SEA BREEZE. {SIMPLE SUSAN. {MONEY-SPINNER.
BLITHE has made so ingenious an addition to the problem, and SIMPLE SUSAN and CO. have solved it in such tuneful verse, that I record both their answers in full. I have altered a word or two in BLITHE'S--which I trust she will excuse; it did not seem quite clear as it stood.
* * * * *
"Yet stay," said the youth, as a gleam of inspiration lighted up the relaxing muscles of his quiescent features. "Stay. Methinks it matters little _when_ we reached that summit, the crown of our toil. For in the space of time wherein we clambered up one mile and bounded down the same on our return, we could have trudged the _twain_ on the level. We have plodded, then, four-and-twenty miles in these six mortal hours; for never a moment did we stop for catching of fleeting breath or for gazing on the scene around!"
"Very good," said the old man. "Twelve miles out and twelve miles in. And we reached the top some time between six and seven of the clock. Now mark me! For every five minutes that had fled since six of the clock when we stood on yonder peak, so many miles had we toiled upwards on the dreary mountainside!"
The youth moaned and rushed into the hostel.
BLITHE.
The elder and the younger knight, They sallied forth at three; How far they went on level ground It matters not to me; What time they reached the foot of hill, When they began to mount, Are problems which I hold to be Of very small account.
The moment that each waved his hat Upon the topmost peak-- To trivial query such as this No answer will I seek. Yet can I tell the distance well They must have travelled o'er: On hill and plain, 'twixt three and nine, The miles were twenty-four.
Four miles an hour their steady pace Along the level track, Three when they climbed--but six when they Came swiftly striding back Adown the hill; and little skill It needs, methinks, to show, Up hill and down together told, Four miles an hour they go.
For whether long or short the time Upon the hill they spent, Two thirds were passed in going up, One third in the descent. Two thirds at three, one third at six, If rightly reckoned o'er, Will make one whole at four--the tale Is tangled now no more.
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.
A = a | +------+-+----+ | | | b = B D = d C = c | | | | +---++--+ +-+-+ | | | | | | e = E | g = G | F ========= f
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 = [** sqrt](12^{2} + 5^{2}) = [** sqrt]169 = 13; AC = 21; AD = [** sqrt](9^{2} + 8^{2}) = [** sqrt]145 = 12+ (N.B. _i.e._ "between 12 and 13.") BC = [** sqrt](16^{2} + 12^{2}) = [** sqrt]400 = 20; BD = [** sqrt](3^{2} + 21^{2}) = [** sqrt]450 = 21+; CD = [** sqrt](9^{2} + 13^{2}) = [** sqrt]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 [** sqrt]169 + [** sqrt]442 + [** sqrt]130 = 741. (I suppose you mean [** sqrt]741, which would be a little nearer the truth. But roots cannot be added in this manner. Do you think [** sqrt]9 + [** sqrt]16 is 25, or even [** sqrt]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 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 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, 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 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!