Complete Works of Lewis Carroll (99 page)

ANSWERS TO KNOT VIII.

§ 1.
The Pigs.

Problem.
—Place twenty-four pigs in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last.

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
= 5
x
;
i.e.
4
x
=
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
x
½,
x
½,
x
-1,
x
-1,
i.e.
4
x
-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 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 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.