Complete Works of Lewis Carroll

by Lewis Carroll

  Answer.—Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6.

  This problem is noticed by only two correspondents. Balbus says "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble." Nolens Volens makes Her Radiancy change the direction of going round; and even then is obliged to add "the pigs must be carried in front of her"!

  § 2. The Grurmstipths.

  Problem.—Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12½ minutes: when will he be overtaken by one?

  Answer.—In 6¼ minutes.

  Solution.—Let "a" be the distance an omnibus goes in 15 minutes, and "x" the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2½ minutes, it goes in that time as far as the traveller walks in 12½; i.e. it goes 5 times as fast. Now the overtaking omnibus is "a" behind the traveller when he starts, and therefore goes "a + x" while he goes "x." Hence a + x = 5x; i.e. 4x = a, and x = a/4. This distance would be traversed by an omnibus in 15⁄4 minutes, and therefore by the traveller in 5 × 15⁄4. Hence he is overtaken in 18¾ minutes after starting, i.e. in 6¼ minutes after meeting the omnibus.

  Four answers have been received, of which two are wrong. Dinah Mite rightly states that the overtaking omnibus reached the point where they met the other omnibus 5 minutes after they left, but wrongly concludes that, going 5 times as fast, it would overtake them in another minute. The travellers are 5-minutes-walk ahead of the omnibus, and must walk 1-4th of this distance farther before the omnibus overtakes them, which will be 1-5th of the distance traversed by the omnibus in the same time: this will require 1¼ minutes more. Nolens Volens tries it by a process like "Achilles and the Tortoise." He rightly states that, when the overtaking omnibus leaves the gate, the travellers are 1-5th of "a" ahead, and that it will take the omnibus 3 minutes to traverse this distance; "during which time" the travellers, he tells us, go 1-15th of "a" (this should be 1-25th). The travellers being now 1-15th of "a" ahead, he concludes that the work remaining to be done is for the travellers to go 1-60th of "a," while the omnibus goes l-12th. The principle is correct, and might have been applied earlier.

  CLASS LIST.

  I.

  Balbus.

  Delta.

  ANSWERS TO KNOT IX.

  § 1. The Buckets.

  Problem.—Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one?

  Solution.—Lardner means, by "displaces," "occupies a space which might be filled with water without any change in the surroundings." If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner's statement.

  Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. Hecla says that "only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced." Hence, according to Hecla, a solid, whose weight was equal to that of an equal bulk of water, would not float till the whole of it was below "the original level" of the water: but, as a matter of fact, it would float as soon as it was all under water. Magpie says the fallacy is "the assumption that one body can displace another from a place where it isn't," and that Lardner's assertion is incorrect, except when the containing vessel "was originally full to the brim." But the question of floating depends on the present state of things, not on past history. Old King Cole takes the same view as Hecla. Tympanum and Vindex assume that "displaced" means "raised above its original level," and merely explain how it comes to pass that the water, so raised, is less in bulk than the immersed portion of bucket, and thus land themselves—or rather set themselves floating—in the same boat as Hecla.

  I regret that there is no Class-list to publish for this Problem.

  § 2. Balbus' Essay.

  Problem.—Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, &c., which series has no end. He concludes that the water will rise without limit. Is this true?

  Solution.—No. This series can never reach 4 inches, since, however many terms we take, we are always short of 4 inches by an amount equal to the last term taken.

  Three answers have been received—but only two seem to me worthy of honours.

  Tympanum says that the statement about the stick "is merely a blind, to which the old answer may well be applied, solvitur ambulando, or rather mergendo." I trust Tympanum will not test this in his own person, by taking the place of the man in Balbus' Essay! He would infallibly be drowned.

  Old King Cole rightly points out that the series, 2, 1, &c., is a decreasing Geometrical Progression: while Vindex rightly identifies the fallacy as that of "Achilles and the Tortoise."

  CLASS LIST.

  I.

  Old King Cole.

  Vindex.

  § 3. The Garden.

  Problem.—An oblong garden, half a yard longer than wide, consists entirely of a gravel-walk, spirally arranged, a yard wide and 3,630 yards long. Find the dimensions of the garden.

  Answer.—60, 60½.

  Solution.—The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square-yards and fractions of a square-yard, contained in that piece of walk: and the distance, traversed in passing through a square-yard at a corner, is evidently a yard. Hence the area of the garden is 3,630 square-yards: i.e., if x be the width, x (x + ½) = 3,630. Solving this Quadratic, we find x = 60. Hence the dimensions are 60, 60½.

  Twelve answers have been received—seven right and five wrong.

  C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.'s "working" consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow's is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60½"! Nabob's calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals the length plus the breadth. That is 60.25 × 2 = 120½. His first assertion is only true of a square garden. His second is irrelevant, since 60.25 is not the square-root of 3,630! Nay, Bob, this will not do! Tympanum says that, by extracting the square-root of 3,630, we get 60 yards with a remainder of 30/60, or half-a-yard, which we add so as to make the oblong 60 × 60½. This is very terrible: but worse remains behind. Tympanum proceeds thus:—"But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half-a-yard wide, the only space not occupied by walk." But Balbus expressly said that the walk "used up the whole of the area." Oh, Tympanum! My tympa is exhausted: my brain is num! I can say no more.

  Hecla indulges, again and again, in that most fatal of all habits in computation—the making two mistakes which cancel each other. She takes x as the width of the garden, in yards, and x + ½ as its length, and makes her first "coil" the sum of x½, x½, x-1, x-1, i.e. 4x-3: but the fourth term should be x-1½, so that her first coil is ½ a yard too long. Her second coil is the sum of x-2½, x-2½, x-3, x-3: here the first term should be x-2 and the last x-3½: these two mistakes cancel, and this coil is therefore right. And the same th
ing is true of every other coil but the last, which needs an extra half-yard to reach the end of the path: and this exactly balances the mistake in the first coil. Thus the sum total of the coils comes right though the working is all wrong.

  Of the seven who are right, Dinah Mite, Janet, Magpie, and Taffy make the same assumption as C. G. L. and Co. They then solve by a Quadratic. Magpie also tries it by Arithmetical Progression, but fails to notice that the first and last "coils" have special values.

  Alumnus Etonæ attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5½. He ought to have proved it generally: what is true of one number is not always true of others. Old King Cole solves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic.

  Vindex proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, "whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction."

  CLASS LIST.

  I.

  Vindex.

  II.

  Alumnus Etonæ.

  Old King Cole.

  III.

  Dinah Mite.

  Janet.

  Magpie.

  Taffy.

  ANSWERS TO KNOT X.

  § 1. The Chelsea Pensioners.

  Problem.—If 70 per cent. have lost an eye, 75 per cent. an ear, 80 per cent. an arm, 85 per cent. a leg: what percentage, at least, must have lost all four?

  Answer.—Ten.

  Solution.—(I adopt that of Polar Star, as being better than my own). Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10.

  Nineteen answers have been received. One is "5," but, as no working is given with it, it must, in accordance with the rule, remain "a deed without a name." Janet makes it "35 and 2⁄10ths." I am sorry she has misunderstood the question, and has supposed that those who had lost an ear were 75 per cent. of those who had lost an eye; and so on. Of course, on this supposition, the percentages must all be multiplied together. This she has done correctly, but I can give her no honours, as I do not think the question will fairly bear her interpretation, Three Score and Ten makes it "19 and 2⁄8ths." Her solution has given me—I will not say "many anxious days and sleepless nights," for I wish to be strictly truthful, but—some trouble in making any sense at all of it. She makes the number of "pensioners wounded once" to be 310 ("per cent.," I suppose!): dividing by 4, she gets 77 and a half as "average percentage:" again dividing by 4, she gets 19 and 2⁄8ths as "percentage wounded four times." Does she suppose wounds of different kinds to "absorb" each other, so to speak? Then, no doubt, the data are equivalent to 77 pensioners with one wound each, and a half-pensioner with a half-wound. And does she then suppose these concentrated wounds to be transferable, so that 2⁄4ths of these unfortunates can obtain perfect health by handing over their wounds to the remaining 1⁄4th? Granting these suppositions, her answer is right; or rather, if the question had been "A road is covered with one inch of gravel, along 77 and a half per cent. of it. How much of it could be covered 4 inches deep with the same material?" her answer would have been right. But alas, that wasn't the question! Delta makes some most amazing assumptions: "let every one who has not lost an eye have lost an ear," "let every one who has not lost both eyes and ears have lost an arm." Her ideas of a battle-field are grim indeed. Fancy a warrior who would continue fighting after losing both eyes, both ears, and both arms! This is a case which she (or "it?") evidently considers possible.

  Next come eight writers who have made the unwarrantable assumption that, because 70 per cent. have lost an eye, therefore 30 per cent. have not lost one, so that they have both eyes. This is illogical. If you give me a bag containing 100 sovereigns, and if in an hour I come to you (my face not beaming with gratitude nearly so much as when I received the bag) to say "I am sorry to tell you that 70 of these sovereigns are bad," do I thereby guarantee the other 30 to be good? Perhaps I have not tested them yet. The sides of this illogical octagon are as follows, in alphabetical order:—Algernon Bray, Dinah Mite, G. S. C., Jane E., J. D. W., Magpie (who makes the delightful remark "therefore 90 per cent. have two of something," recalling to one's memory that fortunate monarch, with whom Xerxes was so much pleased that "he gave him ten of everything!"), S. S. G., and Tokio.

  Bradshaw of the Future and T. R. do the question in a piecemeal fashion—on the principle that the 70 per cent. and the 75 per cent., though commenced at opposite ends of the 100, must overlap by at least 45 per cent.; and so on. This is quite correct working, but not, I think, quite the best way of doing it.

  The other five competitors will, I hope, feel themselves sufficiently glorified by being placed in the first class, without my composing a Triumphal Ode for each!

  CLASS LIST.

  I.

  Old Cat.

  Old Hen.

  Polar Star.

  Simple Susan.

  White Sugar.

  II.

  Bradshaw of the Future.

  T. R.

  III.

  Algernon Bray.

  Dinah Mite.

  G. S. C.

  Jane E.

  J. D. W.

  Magpie.

  S. S. G.

  Tokio.

  § 2. Change of Day.

  I must postpone, sine die, the geographical problem—partly because I have not yet received the statistics I am hoping for, and partly because I am myself so entirely puzzled by it; and when an examiner is himself dimly hovering between a second class and a third how is he to decide the position of others?

  § 3. The Sons' Ages.

  Problem.—"At first, two of the ages are together equal to the third. A few years afterwards, two of them are together double of the third. When the number of years since the first occasion is two-thirds of the sum of the ages on that occasion, one age is 21. What are the other two?

  Answer.—"15 and 18."

  Solution.—Let the ages at first be x, y, (x + y). Now, if a + b = 2c, then (a-n) + (b-n) = 2(c-n), whatever be the value of n. Hence the second relationship, if ever true, was always true. Hence it was true at first. But it cannot be true that x and y are together double of (x + y). Hence it must be true of (x + y), together with x or y; and it does not matter which we take. We assume, then, (x + y) + x = 2y; i.e. y = 2x. Hence the three ages were, at first, x, 2x, 3x; and the number of years, since that time is two-thirds of 6x, i.e. is 4x. Hence the present ages are 5x, 6x, 7x. The ages are clearly integers, since this is only "the year when one of my sons comes of age." Hence 7x = 21, x = 3, and the other ages are 15, 18.

  Eighteen answers have been received. One of the writers merely asserts that the first occasion was 12 years ago, that the ages were then 9, 6, and 3; and that on the second occasion they were 14, 11, and 8! As a Roman father, I ought to withhold the name of the rash writer; but respect for age makes me break the rule: it is Three Score and Ten. Jane E. also asserts that the ages at first were 9, 6, 3: then she calculates the present ages, leaving the second occasion unnoticed. Old Hen is nearly as bad; she "tried various numbers till I found one that fitted all the conditions"; but merely scratching up the earth, and pecking about, is not the way to solve a problem, oh venerable bird! And close after Old Hen prowls, with hungry eyes, Old Cat, who calmly assumes, to begin with, that the son who comes of age is the eldest. Eat your bird, Puss, for you will get nothing from me!

  There are yet two zeroes to dispose of. Minerva assumes that, on every occasion, a son comes of age; and that it is only such a son who is "tipped with gold." Is it wise thus to interpret "now, my boys, calculate your ages, and you shall have the money"? Bradshaw of the Future says "let" the ages at first be 9, 6, 3, then assumes that the second occasion was 6 years afterwards, and on these baseless assumptions br
ings out the right answers. Guide future travellers, an thou wilt: thou art no Bradshaw for this Age!

  Of those who win honours, the merely "honourable" are two. Dinah Mite ascertains (rightly) the relationship between the three ages at first, but then assumes one of them to be "6," thus making the rest of her solution tentative. M. F. C. does the algebra all right up to the conclusion that the present ages are 5z, 6z, and 7z; it then assumes, without giving any reason, that 7z = 21.

  Of the more honourable, Delta attempts a novelty—to discover which son comes of age by elimination: it assumes, successively, that it is the middle one, and that it is the youngest; and in each case it apparently brings out an absurdity. Still, as the proof contains the following bit of algebra, "63 = 7x + 4y; ∴ 21 = x + 4 sevenths of y," I trust it will admit that its proof is not quite conclusive. The rest of its work is good. Magpie betrays the deplorable tendency of her tribe—to appropriate any stray conclusion she comes across, without having any strict logical right to it. Assuming A, B, C, as the ages at first, and D as the number of the years that have elapsed since then, she finds (rightly) the 3 equations, 2A = B, C = B + A, D = 2B. She then says "supposing that A = 1, then B = 2, C = 3, and D = 4. Therefore for A, B, C, D, four numbers are wanted which shall be to each other as 1:2:3:4." It is in the "therefore" that I detect the unconscientiousness of this bird. The conclusion is true, but this is only because the equations are "homogeneous" (i.e. having one "unknown" in each term), a fact which I strongly suspect had not been grasped—I beg pardon, clawed—by her. Were I to lay this little pitfall, "A + 1 = B, B + 1 = C; supposing A = 1, then B = 2 and C = 3. Therefore for A, B, C, three numbers are wanted which shall be to one another as 1:2:3," would you not flutter down into it, oh Magpie, as amiably as a Dove? Simple Susan is anything but simple to me. After ascertaining that the 3 ages at first are as 3:2:1, she says "then, as two-thirds of their sum, added to one of them, = 21, the sum cannot exceed 30, and consequently the highest cannot exceed 15." I suppose her (mental) argument is something like this:—"two-thirds of sum, + one age, = 21; ∴ sum, + 3 halves of one age, = 31 and a half. But 3 halves of one age cannot be less than 1 and-a-half (here I perceive that Simple Susan would on no account present a guinea to a new-born baby!) hence the sum cannot exceed 30." This is ingenious, but her proof, after that, is (as she candidly admits) "clumsy and roundabout." She finds that there are 5 possible sets of ages, and eliminates four of them. Suppose that, instead of 5, there had been 5 million possible sets? Would Simple Susan have courageously ordered in the necessary gallon of ink and ream of paper?