_Solution._--Lardner means, by "displaces," "occupies a s.p.a.ce 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 a.s.sumption that one body can displace another from a place where it isn't," and that Lardner's a.s.sertion 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 a.s.sume that "displaced" means "raised above its original level," and merely explain how it comes to pa.s.s 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 Cla.s.s-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."

CLa.s.s 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, 60-1/2.

_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 pa.s.sing 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_ + 1/2) = 3,630. Solving this Quadratic, we find _x_ = 60. Hence the dimensions are 60, 60-1/2.

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

C. G. L., NABOB, OLD CROW, and TYMPANUM a.s.sume 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 60-1/2"! 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 = 120-1/2. His first a.s.sertion 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 60-1/2. 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 s.p.a.ce 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 s.p.a.ce 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_ + 1/2 as its length, and makes her first "coil" the sum of _x_-1/2, _x_-1/2, _x_-1, _x_-1, _i.e._ 4_x_-3: but the fourth term should be _x_-1-1/2, so that her first coil is 1/2 a yard too long. Her second coil is the sum of _x_-2-1/2, _x_-2-1/2, _x_-3, _x_-3: here the first term should be _x_-2 and the last _x_-3-1/2: these two mistakes cancel, and this coil is therefore right. And the same thing 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 a.s.sumption 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 ETONae attempts to prove what C. G. L. a.s.sumes by a particular instance, taking a garden 6 by 5-1/2. 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."

CLa.s.s LIST.

I.

VINDEX.

II.

ALUMNUS ETONae.

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 7/10ths." 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 3/8ths." 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 3/8ths 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 3/4ths 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 a.s.sumptions: "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_.