you appeal to the policeman who was on duty that night. "Well, Mum, I
did see a chap getting out over your garden-wall: but I was a good bit
off, so I didn't chase him, like. I just cut down the short way to the
Chequers, and who should I meet but Bill Sykes, coming full split round
the corner. So I just ups and says 'My lad, you're wanted.' That's all I
says. And he says 'I'll go along quiet, Bobby,' he says, 'without the
darbies,' he says." There's your _Arithmetical_ policeman. Now try the
other method. "I seed somebody a running, but he was well gone or ever
_I_ got nigh the place. So I just took a look round in the garden. And I
noticed the foot-marks, where the chap had come right across your
flower-beds. They was good big foot-marks sure-ly. And I noticed as the
left foot went down at the heel, ever so much deeper than the other. And
I says to myself 'The chap's been a big hulking chap: and he goes lame
on his left foot.' And I rubs my hand on the wall where he got over, and
there was soot on it, and no mistake. So I says to myself 'Now where can
I light on a big man, in the chimbley-sweep line, what's lame of one
foot?' And I flashes up permiscuous: and I says 'It's Bill Sykes!' says
I." There is your _Algebraical_ policeman--a higher intellectual type,
to my thinking, than the other.
LITTLE JACK'S solution calls for a word of praise, as he has written out
what really is an algebraical proof _in words_, without representing any
of his facts as equations. If it is all his own, he will make a good
algebraist in the time to come. I beg to thank SIMPLE SUSAN for some
kind words of sympathy, to the same effect as those received from OLD
CAT.
HECLA and MARTREB are the only two who have used a method _certain_
either to produce the answer, or else to prove it impossible: so they
must share between them the highest honours.
CLASS LIST.
I.
HECLA.
MARTREB.
II.
Sec. 1 (2 _steps_).
ADELAIDE.
CLIFTON C....
E. K. C.
GUY.
L'INCONNU.
LITTLE JACK.
NIL DESPERANDUM.
SIMPLE SUSAN.
YELLOW-HAMMER.
WOOLLY ONE.
Sec. 2 (3 _steps_).
A. A.
A CHRISTMAS CAROL.
AFTERNOON TEA.
AN APPRECIATIVE SCHOOLMA'AM.
BABY.
BALBUS.
BOG-OAK.
THE RED QUEEN.
WALL-FLOWER.
Sec. 3 (4 _steps_).
HAWTHORN.
JORAM.
S. S. G.
Sec. 4 (5 _steps_).
A STEPNEY COACH.
Sec. 5 (6 _steps_).
BAY LAUREL.
BRADSHAW OF THE FUTURE.
Sec. 6 (9 _steps_).
OLD KING COLE.
Sec. 7 (14 _steps_).
THESEUS.
ANSWERS TO CORRESPONDENTS.
I have received several letters on the subjects of Knots II. and VI.,
which lead me to think some further explanation desirable.
In Knot II., I had intended the numbering of the houses to begin at one
corner of the Square, and this was assumed by most, if not all, of the
competitors. TROJANUS however says "assuming, in default of any
information, that the street enters the square in the middle of each
side, it may be supposed that the numbering begins at a street." But
surely the other is the more natural assumption?
In Knot VI., the first Problem was of course a mere _jeu de mots_, whose
presence I thought excusable in a series of Problems whose aim is to
entertain rather than to instruct: but it has not escaped the
contemptuous criticisms of two of my correspondents, who seem to think
that Apollo is in duty bound to keep his bow always on the stretch.
Neither of them has guessed it: and this is true human nature. Only the
other day--the 31st of September, to be quite exact--I met my old friend
Brown, and gave him a riddle I had just heard. With one great effort of
his colossal mind, Brown guessed it. "Right!" said I. "Ah," said he,
"it's very neat--very neat. And it isn't an answer that would occur to
everybody. Very neat indeed." A few yards further on, I fell in with
Smith and to him I propounded the same riddle. He frowned over it for a
minute, and then gave it up. Meekly I faltered out the answer. "A poor
thing, sir!" Smith growled, as he turned away. "A very poor thing! I
wonder you care to repeat such rubbish!" Yet Smith's mind is, if
possible, even more colossal than Brown's.
The second Problem of Knot VI. is an example in ordinary Double Rule of
Three, whose essential feature is that the result depends on the
variation of several elements, which are so related to it that, if all
but one be constant, it varies as that one: hence, if none be constant,
it varies as their product. Thus, for example, the cubical contents of a
rectangular tank vary as its length, if breadth and depth be constant,
and so on; hence, if none be constant, it varies as the product of the
length, breadth, and depth.
When the result is not thus connected with the varying elements, the
Problem ceases to be Double Rule of Three and often becomes one of great
complexity.
To illustrate this, let us take two candidates for a prize, _A_ and _B_,
who are to compete in French, German, and Italian:
(_a_) Let it be laid down that the result is to depend on their
_relative_ knowledge of each subject, so that, whether their marks, for
French, be "1, 2" or "100, 200," the result will be the same: and let it
also be laid down that, if they get equal marks on 2 papers, the final
marks are to have the same ratio as those of the 3rd paper. This is a
case of ordinary Double Rule of Three. We multiply _A_'s 3 marks
together, and do the same for _B_. Note that, if _A_ gets a single "0,"
his final mark is "0," even if he gets full marks for 2 papers while _B_
gets only one mark for each paper. This of course would be very unfair
on _A_, though a correct solution under the given conditions.
(_b_) The result is to depend, as before, on _relative_ knowledge; but
French is to have twice as much weight as German or Italian. This is an
unusual form of question. I should be inclined to say "the resulting
ratio is to be nearer to the French ratio than if we multiplied as in
(_a_), and so much nearer that it would be necessary to use the other
multipliers _twice_ to produce the same result as in (_a_):" _e.g._ if
the French Ratio were 9/10, and the others 4/9, 1/9 so that the ultimate
ratio, by method (_a_), would be 2/45, I should multiply instead by 2/3,
1/3, giving the result, 1/3 which is nearer to 9/10 than if he had used
method (_a_).
(_c_) The result is to depend on _actual_ amount of knowledge of the 3
subjects collectively. Here we have to ask two questions. (1) What is
to be the "unit" (_i.e._ "standard to measure by") in each subject? (2)
Are these units to be of equal, or unequal value? The usual "unit" is
the knowledge shown by answering the whole paper correctly; calling this
"100," all lower amounts are represented by numbers between "0" and
"100." Then, if these units are to be of equal value, we simply add
_A_'s 3 marks together, and do the same for _B_.
(_d_) The conditions are the same as (_c_), but French is to have double
weight. Here we simply double the French marks, and add as before.
(_e_) French is to have such weight, that, if other marks be equal, the
ultimate ratio is to be that of the French paper, so that a "0" in this
would swamp the candidate: but the other two subjects are only to affect
the result collectively, by the amount of knowledge shown, the two being
reckoned of equal value. Here I should add _A_'s German and Italian
marks together, and multiply by his French mark.
But I need not go on: the problem may evidently be set with many varying
conditions, each requiring its own method of solution. The Problem in
Knot VI. was meant to belong to variety (_a_), and to make this clear, I
inserted the following passage:
"Usually the competitors differ in one point only. Thus, last year, Fifi
and Gogo made the same number of scarves in the trial week, and they
were equally light; but Fifi's were twice as warm as Gogo's, and she was
pronounced twice as good."
What I have said will suffice, I hope, as an answer to BALBUS, who holds
that (_a_) and (_c_) are the only possible varieties of the problem, and
that to say "We cannot use addition, therefore we must be intended to
use multiplication," is "no more illogical than, from knowledge that one
was not born in the night, to infer that he was born in the daytime";
and also to FIFEE, who says "I think a little more consideration will
show you that our 'error of _adding_ the proportional numbers together
for each candidate instead of _multiplying_' is no error at all." Why,
even if addition _had_ been the right method to use, not one of the
writers (I speak from memory) showed any consciousness of the necessity
of fixing a "unit" for each subject. "No error at all!" They were
positively steeped in error!
One correspondent (I do not name him, as the communication is not quite
friendly in tone) writes thus:--"I wish to add, very respectfully, that
I think it would be in better taste if you were to abstain from the very
trenchant expressions which you are accustomed to indulge in when
criticising the answer. That such a tone must not be" ("be not"?)
"agreeable to the persons concerned who have made mistakes may possibly
have no great weight with you, but I hope you will feel that it would be
as well not to employ it, _unless you are quite certain of being correct
yourself_." The only instances the writer gives of the "trenchant
expressions" are "hapless" and "malefactors." I beg to assure him (and
any others who may need the assurance: I trust there are none) that all
such words have been used in jest, and with no idea that they could
possibly annoy any one, and that I sincerely regret any annoyance I may
have thus inadvertently given. May I hope that in future they will
recognise the distinction between severe language used in sober earnest,
and the "words of unmeant bitterness," which Coleridge has alluded to in
that lovely passage beginning "A little child, a limber elf"? If the
writer will refer to that passage, or to the preface to "Fire, Famine,
and Slaughter," he will find the distinction, for which I plead, far
better drawn out than I could hope to do in any words of mine.
The writer's insinuation that I care not how much annoyance I give to my
readers I think it best to pass over in silence; but to his concluding
remark I must entirely demur. I hold that to use language likely to
annoy any of my correspondents would not be in the least justified by
the plea that I was "quite certain of being correct." I trust that the
knot-untiers and I are not on such terms as those!
I beg to thank _G. B._ for the offer of a puzzle--which, however, is too
like the old one "Make four 9's into 100."
ANSWERS TO KNOT VIII.
Sec. 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"!
Sec. 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 12-1/2 minutes: when will he be overtaken by one?
_Answer._--In 6-1/4 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 2-1/2
minutes, it goes in that time as far as the traveller walks in 12-1/2;
_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 x 15/4. Hence he is overtaken in
18-3/4 minutes after starting, _i.e._ in 6-1/4 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 1-1/4 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 1-12th. The
_principle_ is correct, and might have been applied earlier.
CLASS LIST.
I.
BALBUS.
DELTA.
ANSWERS TO KNOT IX.
Sec. 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.
* * * * *
Sec. 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.
* * * * *
Sec. 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 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_ +
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 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 x 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 x 2 = 120-1/2. 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 x 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 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_ + 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 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 ETONAE attempts to prove what C. G. L. assumes 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."
CLASS LIST.
I.
VINDEX.
II.
ALUMNUS ETONAE.
OLD KING COLE.
III.
DINAH MITE.
JANET.
MAGPIE.
TAFFY.
ANSWERS TO KNOT X.
Sec. 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
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.
Sec. 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?
Sec. 3. THE SONS' AGES.