“A knot!” said Alice. “Oh, do let me help to undo it!”

**ANSWERS TO KNOT I.**

*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 farewel-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

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* *INERTIAE*,
and YAK. Of these, F. K., FIFEE, T. B. C, and
*VIS* *INERTIAE* 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.**

Se. 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

+------+-+----+

| | |

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.

Se. 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, No,
25, 52, 73, is the sum of the distances, to the other
three, least?”

*Answer.* “From N.”

Let A be N, B N, C N, and D N.

Then AB = [ sqrt](12^ + 5^)
= [ sqrt]169 = 13;

AC = 21;

AD = [ sqrt](9^ + 8^) =
[ sqrt]145 = 12 +

(N.B. *i.e.* “between
12 and 13.”)

BC = [ sqrt](16^ + 12^)
= [ sqrt]400 = 20;

*BD* = [ sqrt](3^ + 21^)
= [ sqrt]450 = 21+;

*CD* = [ sqrt](9^ + 13^)
= [ 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 N is the house, “as the diagram shows.”
I cannot see *how* it does so. OLD CAT assumes
that the house *must* be N or N.
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 No, 21, 60, 70: this will make your premiss
true, and your conclusion false. Similarly, for
the second, try No, 30, 51, 71.)

Of the five partly-right solutions,
RAGS AND TATTERS and MAD HATTER (who send one answer
between them) make N 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 No, 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 N, 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 No, 2, weigh 12 lbs.; No, 3, 13-1/2 lbs.; No, 4, 11-1/2 lbs.; No,
5, 8 lbs.; No, 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 N *thrice* and each other *twice*.
Deducting twice the sum of the 1st and 4th weighings,
we get 21 lbs. for *thrice* N, *i.e.*,
7 lbs. for N. Hence, the 2nd and 3rd weighings
give 6-1/2 lbs., 4-1/2 lbs. for No, 4; and hence
again, the 1st and 4th weighings give 5-1/2 lbs., 3-1/2
lbs., for No, 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!

The seventeen whose solutions are
practically valueless are ARDMORE, A READY RECKONER,
ARTHUR, BOG-LARK, BOG-OAK, BRIDGET, FIRST ATTEMPT,
J. L. C., M. E. T., ROSE, ROWENA, SEA-BREEZE, SYLVIA,
THISTLEDOWN, THREE-FIFTHS ASLEEP, *VENDREDI*, and
WINIFRED. BOG-LARK tries it by a sort of “rule
of false,” assuming experimentally that No, 2, weigh 6 lbs. each, and having thus produced
17-1/2, instead of 16, as the weight of 1, 3, and
5, she removes “the superfluous pound and a half,”
but does not explain how she knows from which to take
it. THREE-FIFTHS ASLEEP says that (when in that
peculiar state) “it seemed perfectly clear”
to her that, “3 out of the 5 sacks being weighed
twice over, 3/5 of 45 = 27, must be the total weight
of the 5 sacks.” As to which I can only
say, with the Captain, “it beats me entirely!”
WINIFRED, on the plea that “one must have a
starting-point,” assumes (what I fear is a mere
guess) that N weighed 5-1/2 lbs. The rest
all do it, wholly or partly, by guess-work.

The problem is of course (as any Algebraist
sees at once) a case of “simultaneous simple
equations.” It is, however, easily soluble
by Arithmetic only; and, when this is the case, I
hold that it is bad workmanship to use the more complex
method. I have not, this time, given more credit
to arithmetical solutions; but in future problems I
shall (other things being equal) give the highest
marks to those who use the simplest machinery.
I have put into Class I. those whose answers seemed
specially short and neat, and into Class III. those
that seemed specially long or clumsy. Of this
last set, A. C. M., FURZE-BUSH, JAMES, PARTRIDGE,
R. W., and WAITING FOR THE TRAIN, have sent long wandering
solutions, the substitutions having no definite method,
but seeming to have been made to see what would come
of it. CHILPOME and DUBLIN BOY omit some of the
working. ARVON MARLBOROUGH BOY only finds the
weight of *one* sack.

**CLASS LIST**

**I.**

B. E. D.

C. H.

CONSTANCE JOHNSON.

GREYSTEAD.

GUY.

HOOPOE.

J. F. A.

M. A. H.

NUMBER FIVE.

PEDRO.

R. E. X.

SEVEN OLD MEN.

*VIS* *INERTIAE*.

WILLY B.

YAHOO.

**II.**

AMERICAN SUBSCRIBER.

AN APPRECIATIVE SCHOOLMA’AM.

AYR.

BRADSHAW OF THE FUTURE.

CHEAM.

C. M. G.

DINAH MITE.

DUCKWING.

E. C. M.

E. N. Lowry.

ERA.

EUROCLYDON.

F. H. W.

FIFEE.

G. E. B.

HARLEQUIN.

HAWTHORN.

HOUGH GREEN.

J. A. B.

JACK TAR.

J. B. B.

KGOVJNI.

LAND LUBBER.

L. D.

MAGPIE.

MARY.

MHRUXI.

MINNIE.

MONEY-SPINNER.

NAIRAM.

OLD CAT.

*POLICHINELLE*.

SIMPLE SUSAN.

S. S. G.

THISBE.

VERENA.

WAMBA.

WOLFE.

WYKEHAMICUS.

Y. M. A. H.

**III.**

A. C. M.

ARVON MARLBOROUGH BOY.

CHILPOME.

DUBLIN BOY.

FURZE-BUSH.

JAMES.

PARTRIDGE.

R. W.

WAITING FOR THE TRAIN.

**ANSWERS TO KNOT V.**

*Problem.* To mark
pictures, giving 3 x’s to 2 or 3, 2 to 4 or 5,
and 1 to 9 or 10; also giving 3 o’s to 1 or
2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the
smallest possible number of pictures, and to give them
the largest possible number of marks.

*Answer.* 10 pictures; 29 marks; arranged
thus:

x x x x x x x x x o

x x x x x o o o o

x x o o o o o o o o

*Solution.* By giving
all the x’s possible, putting into brackets the
optional ones, we get 10 pictures marked thus:

x x x x x x x x x (x)

x x x x (x)

x x (x)

By then assigning o’s in the same way, beginning at the other end, we get 9 pictures marked thus:

(o) o

(o) o o o

(o) o o o o o o o o

All we have now to do is to run these two wedges as close together as they will go, so as to get the minimum number of pictures erasing optional marks where by so doing we can run them closer, but otherwise letting them stand. There are 10 necessary marks in the 1st row, and in the 3rd; but only 7 in the 2nd. Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the 2nd.

Twenty-two answers have been received. Of these 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong.

Of the eleven answers with which some
working is supplied, 3 are wrong. C. H. begins
with the rash assertion that under the given conditions
“the sum is impossible. For,” he or
she adds (these initialed correspondents are dismally
vague beings to deal with: perhaps “it”
would be a better pronoun), “10 is the least
possible number of pictures” (granted):
“therefore we must either give 2 x’s to
6, or 2 o’s to 5.” Why “must,”
oh alphabetical phantom? It is nowhere ordained
that every picture “must” have 3 marks!
FIFEE sends a folio page of solution, which deserved
a better fate: she offers 3 answers, in each of
which 10 pictures are marked, with 30 marks; in one
she gives 2 x’s to 6 pictures; in another to
7; in the 3rd she gives 2 o’s to 5; thus in
every case ignoring the conditions. (I pause to remark
that the condition “2 x’s to 4 or 5 pictures”
can only mean “*either* to 4 *or else*
to 5”: if, as one competitor holds, it might
mean *any* number not less than 4, the words
“*or* 5” would be superfluous.) I.
E. A. (I am happy to say that none of these bloodless
phantoms appear this time in the class-list.
Is it IDEA with the “D” left out?) gives
2 x’s to 6 pictures. She then takes me
to task for using the word “ought” instead
of “nought.” No doubt, to one who
thus rebels against the rules laid down for her guidance,
the word must be distasteful. But does not I.
E. A. remember the parallel case of “adder”?
That creature was originally “a nadder”:
then the two words took to bandying the poor “n”
backwards and forwards like a shuttlecock, the final
state of the game being “an adder.”
May not “a nought” have similarly become
“an ought”? Anyhow, “oughts
and crosses” is a very old game. I don’t
think I ever heard it called “noughts and crosses.”

In the following Class-list, I hope the solitary occupant of III. will sheathe her claws when she hears how narrow an escape she has had of not being named at all. Her account of the process by which she got the answer is so meagre that, like the nursery tale of “Jack-a-Minory” (I trust I. E. A. will be merciful to the spelling), it is scarcely to be distinguished from “zero.”

**CLASS LIST.**

**I.**

GUY.

OLD CAT.

SEA-BREEZE.

**II.**

AYR.

BRADSHAW OF THE FUTURE.

F. LEE.

H. VERNON.

**III.**

CAT.

**ANSWERS TO KNOT VI.**

*Problem 1.* *A*
and *B* began the year with only 1,000_l._ a-piece.
They borrowed nought; they stole nought. On the
next New-Year’s Day they had 60,000_l._ between
them. How did they do it?

*Solution.* They went
that day to the Bank of England. *A* stood in
front of it, while *B* went round and stood behind
it.

Two answers have been received, both
worthy of much honour. ADDLEPATE makes them borrow
“0” and steal “0,” and uses
both cyphers by putting them at the right-hand end
of the 1,000_l._, thus producing 100,000_l._, which
is well over the mark. But (or to express it in
Latin) AT *SPES* *INFRACTA* has solved it even
more ingeniously: with the first cypher she turns
the “1” of the 1,000_l._ into a “9,”
and adds the result to the original sum, thus getting
10,000_l._: and in this, by means of the other
“0,” she turns the “1” into
a “6,” thus hitting the exact 60,000_l._

**CLASS LIST**

**I.**

*AT* *SPES* *INFRACTA*.

**II.**

ADDLEPATE.

*Problem 2.* *L*
makes 5 scarves, while *M* makes 2: *Z*
makes 4 while *L* makes 3. Five scarves
of *Z*’s weigh one of *L*’s;
5 of *M*’s weigh 3 of *Z*’s.
One of *M*’s is as warm as 4 of *Z*’s:
and one of *L*’s as warm as 3 of *M*’s.
Which is best, giving equal weight in the result to
rapidity of work, lightness, and warmth?

*Answer.* The order is *M*, *L*,
*Z*.

*Solution.* As to
rapidity (other things being constant) *L*’s
merit is to *M*’s in the ratio of 5 to
2: *Z*’s to *L*’s in the
ratio of 4 to 3. In order to get one set of 3
numbers fulfilling these conditions, it is perhaps
simplest to take the one that occurs *twice* as
unity, and reduce the others to fractions: this
gives, for *L*, *M*, and *Z*, the marks
1, 2/5, 4/3. In estimating for *lightness*,
we observe that the greater the weight, the less the
merit, so that *Z*’s merit is to *L*’s
as 5 to 1. Thus the marks for *lightness*
are 1/5, 5/3, 1. And similarly, the marks for
warmth are 3, 1, 1/4. To get the total result,
we must *multiply* *L*’s 3 marks together,
and do the same for *M* and for *Z*.
The final numbers are 1 x 1/5 x 3, 2/5 x 5/3 x 1, 4/3
x 1 x 1/4; *i.e.* 3/5, 2/3, 1/3; *i.e.*
multiplying throughout by 15 (which will not alter
the proportion), 9, 10, 5; showing the order of merit
to be *M*, *L*, *Z*.

Twenty-nine answers have been received,
of which five are right, and twenty-four wrong.
These hapless ones have all (with three exceptions)
fallen into the error of *adding* the proportional
numbers together, for each candidate, instead of *multiplying*.
*Why* the latter is right, rather than the former,
is fully proved in text-books, so I will not occupy
space by stating it here: but it can be *illustrated*
very easily by the case of length, breadth, and depth.
Suppose *A* and *B* are rival diggers of
rectangular tanks: the amount of work done is
evidently measured by the number of *cubical feet*
dug out. Let *A* dig a tank 10 feet long,
10 wide, 2 deep: let *B* dig one 6 feet long,
5 wide, 10 deep. The cubical contents are 200,
300; *i.e.* *B* is best digger in the ratio
of 3 to 2. Now try marking for length, width,
and depth, separately; giving a maximum mark of 10
to the best in each contest, and then *adding*
the results!

Of the twenty-four malefactors, one gives no working, and so has no real claim to be named; but I break the rule for once, in deference to its success in Problem 1: he, she, or it, is ADDLEPATE. The other twenty-three may be divided into five groups.

First and worst are, I take it, those
who put the rightful winner *last*; arranging
them as “Lolo, Zuzu, Mimi.” The names
of these desperate wrong-doers are AYR, BRADSHAW OF
THE FUTURE, FURZE-BUSH and POLLUX (who send a joint
answer), GREYSTEAD, GUY, OLD HEN, and SIMPLE SUSAN.
The latter was *once* best of all; the Old Hen
has taken advantage of her simplicity, and beguiled
her with the chaff which was the bane of her own chickenhood.

Secondly, I point the finger of scorn
at those who have put the worst candidate at the top;
arranging them as “Zuzu, Mimi, Lolo.”
They are *GRAECIA*, M. M., OLD CAT, and R. E. X.
“’Tis Greece, but .”

The third set have avoided both these
enormities, and have even succeeded in putting the
worst last, their answer being “Lolo, Mimi,
Zuzu.” Their names are AYR (who also appears
among the “quite too too"), CLIFTON C., F. B.,
FIFEE, GRIG, JANET, and MRS. SAIREY GAMP. F. B.
has not fallen into the common error; she *multiplies*
together the proportionate numbers she gets, but in
getting them she goes wrong, by reckoning warmth as
a *de*-merit. Possibly she is “Freshly
Burnt,” or comes “From Bombay.”
JANET and MRS. SAIREY GAMP have also avoided this
error: the method they have adopted is shrouded
in mystery I scarcely feel competent to
criticize it. MRS. GAMP says “if Zuzu makes
4 while Lolo makes 3, Zuzu makes 6 while Lolo makes
5 (bad reasoning), while Mimi makes 2.”
From this she concludes “therefore Zuzu excels
in speed by 1” (*i.e.* when compared with
Lolo; but what about Mimi?). She then compares
the 3 kinds of excellence, measured on this mystic
scale. JANET takes the statement, that “Lolo
makes 5 while Mimi makes 2,” to prove that “Lolo
makes 3 while Mimi makes 1 and Zuzu 4” (worse
reasoning than MRS. GAMP’S), and thence concludes
that “Zuzu excels in speed by 1/8”!
JANET should have been ADELINE, “mystery of mysteries!”

The fourth set actually put Mimi at
the top, arranging them as “Mimi, Zuzu, Lolo.”
They are MARQUIS AND CO., MARTREB, S. B. B. (first
initial scarcely legible: *may* be meant
for “J"), and STANZA.

The fifth set consist of AN ANCIENT
FISH and CAMEL. These ill-assorted comrades,
by dint of foot and fin, have scrambled into the right
answer, but, as their method is wrong, of course it
counts for nothing. Also AN ANCIENT FISH has
very ancient and fishlike ideas as to *how* numbers
represent merit: she says “Lolo gains 2-1/2
on Mimi.” Two and a half *what*?
Fish, fish, art thou in thy duty?

Of the five winners I put *BALBUS*
and THE ELDER TRAVELLER slightly below the other three *BALBUS*
for defective reasoning, the other for scanty working.
*BALBUS* gives two reasons for saying that *addition*
of marks is *not* the right method, and then
adds “it follows that the decision must be made
by *multiplying* the marks together.”
This is hardly more logical than to say “This
is not Spring: *therefore* it must be Autumn.”

**CLASS LIST.**

**I.**

DINAH MITE.

E. B. D. L.

JORAM.

**II.**

*BALBUS*.

THE ELDER TRAVELLER.

With regard to Knot V., I beg to express
to *VIS* *INERTIAE* and to any others who, like
her, understood the condition to be that *every*
marked picture must have *three* marks, my sincere
regret that the unfortunate phrase “*fill*
the columns with oughts and crosses” should have
caused them to waste so much time and trouble.
I can only repeat that a *literal* interpretation
of “fill” would seem to *me* to require
that *every* picture in the gallery should be
marked. *VIS* *INERTIAE* would have been
in the First Class if she had sent in the solution
she now offers.

**ANSWERS TO KNOT VII.**

*Problem.* Given that
one glass of lemonade, 3 sandwiches, and 7 biscuits,
cost 1_s._ 2_d._; and that one glass of lemonade, 4
sandwiches, and 10 biscuits, cost 1_s._ 5_d._:
find the cost of (1) a glass of lemonade, a sandwich,
and a biscuit; and (2) 2 glasses of lemonade, 3 sandwiches,
and 5 biscuits.

*Answer.* (1) 8_d._; (2) 1_s._ 7_d._

*Solution.* This is
best treated algebraically. Let *x* = the
cost (in pence) of a glass of lemonade, *y* of
a sandwich, and *z* of a biscuit. Then we
have *x* + 3_y_ + 7_z_ = 14, and *x* + 4_y_
+ 10_z_ = 17. And we require the values of *x*
+ *y* + *z*, and of 2_x_ + 3_y_ + 5_z_.
Now, from *two* equations only, we cannot find,
*separately*, the values of *three* unknowns:
certain *combinations* of them may, however, be
found. Also we know that we can, by the help
of the given equations, eliminate 2 of the 3 unknowns
from the quantity whose value is required, which will
then contain one only. If, then, the required
value is ascertainable at all, it can only be by the
3rd unknown vanishing of itself: otherwise the
problem is impossible.

Let us then eliminate lemonade and
sandwiches, and reduce everything to biscuits a
state of things even more depressing than “if
all the world were apple-pie” by
subtracting the 1st equation from the 2nd, which eliminates
lemonade, and gives *y* + 3_z_ = 3, or *y*
= 3-3_z_; and then substituting this value of *y*
in the 1st, which gives *x*-2_z_ = 5, *i.e.*
*x* = 5 + 2_z_. Now if we substitute these
values of *x*, *y*, in the quantities whose
values are required, the first becomes (5 + 2_z_) +
(3-3_z_) + *z*, *i.e.* 8: and the second
becomes 2(5 + 2_z_) + 3(3-3_z_) + 5_z_, *i.e.*
19. Hence the answers are (1) 8_d._, (2) 1_s._
7_d._

The above is a *universal* method:
that is, it is absolutely certain either to produce
the answer, or to prove that no answer is possible.
The question may also be solved by combining the quantities
whose values are given, so as to form those whose
values are required. This is merely a matter
of ingenuity and good luck: and as it *may*
fail, even when the thing is possible, and is of no
use in proving it *im*possible, I cannot rank
this method as equal in value with the other.
Even when it succeeds, it may prove a very tedious
process. Suppose the 26 competitors, who have
sent in what I may call *accidental* solutions,
had had a question to deal with where every number
contained 8 or 10 digits! I suspect it would
have been a case of “silvered is the raven hair”
(see “Patience”) before any solution would
have been hit on by the most ingenious of them.

Forty-five answers have come in, of
which 44 give, I am happy to say, some sort of *working*,
and therefore deserve to be mentioned by name, and
to have their virtues, or vices as the case may be,
discussed. Thirteen have made assumptions to
which they have no right, and so cannot figure in
the Class-list, even though, in 10 of the 13 cases,
the answer is right. Of the remaining 28, no
less than 26 have sent in *accidental* solutions,
and therefore fall short of the highest honours.

I will now discuss individual cases, taking the worst first, as my custom is.

FROGGY gives no working at
least this is all he gives: after stating the
given equations, he says “therefore the difference,
1 sandwich + 3 biscuits, = 3_d._”: then
follow the amounts of the unknown bills, with no further
hint as to how he got them. FROGGY has had a *very*
narrow escape of not being named at all!

Of those who are wrong, *VIS* *INERTIAE*
has sent in a piece of incorrect working. Peruse
the horrid details, and shudder! She takes *x*
(call it “*y*”) as the cost of a
sandwich, and concludes (rightly enough) that a biscuit
will cost (3-*y*)/3. She then subtracts the
second equation from the first, and deduces 3_y_ +
7 x (3-*y*)/3-4_y_ + 10 x (3-*y*)/3 = 3.
By making two mistakes in this line, she brings out
*y* = 3/2. Try it again, *oh* *VIS*
*INERTIAE*! Away with *INERTIAE*: infuse
a little more VIS: and you will bring out the
correct (though uninteresting) result, 0 = 0!
This will show you that it is hopeless to try to coax
any one of these 3 unknowns to reveal its *separate*
value. The other competitor, who is wrong throughout,
is either J. M. C. or T. M. C.: but, whether he
be a Juvenile Mis-Calculator or a True Mathematician
Confused, he makes the answers 7_d._ and 1_s._ 5_d._
He assumes, with Too Much Confidence, that biscuits
were 1/2_d._ each, and that Clara paid for 8, though
she only ate 7!

We will now consider the 13 whose
working is wrong, though the answer is right:
and, not to measure their demerits too exactly, I will
take them in alphabetical order. ANITA finds
(rightly) that “1 sandwich and 3 biscuits cost
3_d._,” and proceeds “therefore 1 sandwich
= 1-1/2_d._, 3 biscuits = 1-1/2_d._, 1 lemonade =
6_d._” DINAH MITE begins like ANITA: and
thence proves (rightly) that a biscuit costs less than
a 1_d._: whence she concludes (wrongly) that
it *must* cost 1/2_d._ F. C. W. is so beautifully
resigned to the certainty of a verdict of “guilty,”
that I have hardly the heart to utter the word, without
adding a “recommended to mercy owing to extenuating
circumstances.” But really, you know, where
*are* the extenuating circumstances? She
begins by assuming that lemonade is 4_d._ a glass,
and sandwiches 3_d._ each, (making with the 2 given
equations, *four* conditions to be fulfilled by
*three* miserable unknowns!). And, having
(naturally) developed this into a contradiction, she
then tries 5_d._ and 2_d._ with a similar result. (N.B.
*This* process might have been carried on through
the whole of the Tertiary Period, without gratifying
one *single* *Mégathérium*.) She then, by a
“happy thought,” tries half-penny biscuits,
and so obtains a consistent result. This may
be a good solution, viewing the problem as a conundrum:
but it is *not* scientific. JANET identifies
sandwiches with biscuits! “One sandwich
+ 3 biscuits” she makes equal to “4.”
Four *what*? MAYFAIR makes the astounding
assertion that the equation, *s* + 3_b_ = 3, “is
evidently only satisfied by *s* = 3/2, *b*
= 1/2”! OLD CAT believes that the assumption
that a sandwich costs 1-1/2_d._ is “the only
way to avoid unmanageable fractions.” But
*why* avoid them? Is there not a certain
glow of triumph in taming such a fraction? “Ladies
and gentlemen, the fraction now before you is one
that for years defied all efforts of a refining nature:
it was, in a word, hopelessly vulgar. Treating
it as a circulating decimal (the treadmill of fractions)
only made matters worse. As a last resource,
I reduced it to its lowest terms, and extracted its
square root!” Joking apart, let me thank OLD
CAT for some very kind words of sympathy, in reference
to a correspondent (whose name I am happy to say I
have now forgotten) who had found fault with me as
a discourteous critic. O. V. L. is beyond my
comprehension. He takes the given equations as
(1) and (2): thence, by the process [(2)-(1)]
deduces (rightly) equation (3) *viz*. *s*
+ 3_b_ = 3: and thence again, by the process
[x3] (a hopeless mystery), deduces 3_s_ + 4_b_ = 4.
I have nothing to say about it: I give it up.
SEA-BREEZE says “it is immaterial to the answer”
(why?) “in what proportion 3_d._ is divided between
the sandwich and the 3 biscuits”: so she
assumes *s* = l-1/2_d._, *b* = 1/2_d._ STANZA
is one of a very irregular metre. At first she
(like JANET) identifies sandwiches with biscuits.
She then tries two assumptions (*s* = 1, *b*
= 2/3, and *s* = 1/2 *b* = 5/6), and (naturally)
ends in contradictions. Then she returns to the
first assumption, and finds the 3 unknowns separately:
*quod est absurdum*. STILETTO identifies
sandwiches and biscuits, as “articles.”
Is the word ever used by confectioners? I fancied
“What is the next article, Ma’am?”
was limited to linendrapers. TWO SISTERS first
assume that biscuits are 4 a penny, and then that
they are 2 a penny, adding that “the answer
will of course be the same in both cases.”
It is a dreamy remark, making one feel something like
Macbeth grasping at the spectral dagger. “Is
this a statement that I see before me?” If you
were to say “we both walked the same way this
morning,” and *I* were to say “*one*
of you walked the same way, but the other didn’t,”
which of the three would be the most hopelessly confused?
TURTLE PYATE (what *is* a Turtle Pyate, please?)
and OLD CROW, who send a joint answer, and Y. Y., adopt
the same method. Y. Y. gets the equation *s*
+ 3_b_ = 3: and then says “this sum must
be apportioned in one of the three following ways.”
It *may* be, I grant you: but Y. Y. do you
say “must”? I fear it is *possible*
for Y. Y. to be *two* Y’s. The other
two conspirators are less positive: they say
it “can” be so divided: but they add
“either of the three prices being right”!
This is bad grammar and bad arithmetic at once, oh
mysterious birds!

Of those who win honours, THE SHETLAND
SNARK must have the 3rd class all to himself.
He has only answered half the question, *viz*. the
amount of Clara’s luncheon: the two little
old ladies he pitilessly leaves in the midst of their
“difficulty.” I beg to assure him
(with thanks for his friendly remarks) that entrance-fees
and subscriptions are things unknown in that most
economical of clubs, “The Knot-*Untiers*.”

The authors of the 26 “accidental”
solutions differ only in the number of steps they
have taken between the *data* and the answers.
In order to do them full justice I have arranged the
2nd class in sections, according to the number of
steps. The two Kings are fearfully deliberate!
I suppose walking quick, or taking short cuts, is
inconsistent with kingly dignity: but really,
in reading THESEUS’ solution, one almost fancied
he was “marking time,” and making no advance
at all! The other King will, I hope, pardon me
for having altered “Coal” into “Cole.”
King Coilus, or Coil, seems to have reigned soon after
Arthur’s time. Henry of Huntingdon identifies
him with the King *Coel* who first built walls
round Colchester, which was named after him.
In the Chronicle of Robert of Gloucester we read:

“Aftur Kyng Aruirag, of wam
we habbeth y told,

Marius ys sone was kyng, quoynte
*mon* & bold.

And ys sone was aftur hym, *Coil*
was ys name,

Bothe it were quoynte men, & of
noble fame.”

*BALBUS* lays it down as a general
principle that “in order to ascertain the cost
of any one luncheon, it must come to the same amount
upon two different assumptions.” (*Query.*
Should not “it” be “we”?
Otherwise the *luncheon* is represented as wishing
to ascertain its own cost!) He then makes two assumptions one,
that sandwiches cost nothing; the other, that biscuits
cost nothing, (either arrangement would lead to the
shop being inconveniently crowded!) and
brings out the unknown luncheons as 8_d._ and 19_d._,
on each assumption. He then concludes that this
agreement of results “shows that the answers
are correct.” Now I propose to disprove
his general law by simply giving *one* instance
of its failing. One instance is quite enough.
In logical language, in order to disprove a “universal
affirmative,” it is enough to prove its contradictory,
which is a “particular negative.” (I must
pause for a digression on Logic, and especially on
Ladies’ Logic. The universal affirmative
“everybody says he’s a duck” is crushed
instantly by proving the particular negative “Peter
says he’s a goose,” which is equivalent
to “Peter does *not* say he’s a duck.”
And the universal negative “nobody calls on
her” is well met by the particular affirmative
“*I* called yesterday.” In short,
either of two contradictories disproves the other:
and the moral is that, since a particular proposition
is much more easily proved than a universal one, it
is the wisest course, in arguing with a Lady, to limit
one’s *own* assertions to “particulars,”
and leave *her* to prove the “universal”
contradictory, if she can. You will thus generally
secure a *logical* victory: a *practical*
victory is not to be hoped for, since she can always
fall back upon the crushing remark “*that*
has nothing to do with it!” a move
for which Man has not yet discovered any satisfactory
answer. Now let us return to *BALBUS*.) Here
is my “particular negative,” on which to
test his rule. Suppose the two recorded luncheons
to have been “2 buns, one queen-cake, 2 sausage-rolls,
and a bottle of Zoedone: total, one-and-ninepence,”
and “one bun, 2 queen-cakes, a sausage-roll,
and a bottle of Zoedone: total, one-and-fourpence.”
And suppose Clara’s unknown luncheon to have
been “3 buns, one queen-cake, one sausage-roll,
and 2 bottles of Zoedone:” while the two
little sisters had been indulging in “8 buns,
4 queen-cakes, 2 sausage-rolls, and 6 bottles of Zoedone.”
(Poor souls, how thirsty they must have been!) If
*BALBUS* will kindly try this by his principle of
“two assumptions,” first assuming that
a bun is 1_d._ and a queen-cake 2_d._, and then that
a bun is 3_d._ and a queen-cake 3_d._, he will bring
out the other two luncheons, on each assumption, as
“one-and-nine-pence” and “four-and-ten-pence”
respectively, which harmony of results, he will say,
“shows that the answers are correct.”
And yet, as a matter of fact, the buns were 2_d._
each, the queen-cakes 3_d._, the sausage-rolls 6_d._,
and the Zoedone 2_d._ a bottle: so that Clara’s
third luncheon had cost one-and-sevenpence, and her
thirsty friends had spent four-and-fourpence!

Another remark of *BALBUS* I will
quote and discuss: for I think that it also may
yield a moral for some of my readers. He says
“it is the same thing in substance whether in
solving this problem we use words and call it Arithmetic,
or use letters and signs and call it Algebra.”
Now this does not appear to me a correct description
of the two methods: the Arithmetical method is
that of “synthesis” only; it goes from
one known fact to another, till it reaches its goal:
whereas the Algebraical method is that of “analysis”:
it begins with the goal, symbolically represented,
and so goes backwards, dragging its veiled victim with
it, till it has reached the full daylight of known
facts, in which it can tear off the veil and say “I
know you!”

Take an illustration. Your house
has been broken into and robbed, and 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.**

Se (2 *steps*).

ADELAIDE.

CLIFTON C....

E. K. C.

GUY.

*L’INCONNU*.

LITTLE JACK.

NIL DESPERANDUM.

SIMPLE SUSAN.

YELLOW-HAMMER.

WOOLLY ONE.

Se (3 *steps*).

A. A.

A CHRISTMAS CAROL.

AFTERNOON TEA.

AN APPRECIATIVE SCHOOLMA’AM.

BABY.

*BALBUS*.

BOG-OAK.

THE RED QUEEN.

WALL-FLOWER.

Se (4 *steps*).

HAWTHORN.

JORAM.

S. S. G.

Se (5 *steps*).

A STEPNEY COACH.

Se (6 *steps*).

BAY LAUREL.

BRADSHAW OF THE FUTURE.

Se (9 *steps*).

OLD KING COLE.

Se (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.**

Se. 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”!

Se. 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 l-12th. The *principle* is correct,
and might have been applied earlier.

**CLASS LIST.**

**I.**

*BALBUS*.

DELTA.

**ANSWERS TO KNOT IX.**

Se. 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.

Se. *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*.

Se. 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.**

Se. 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.

Se. 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?

Se. 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* = 2_c_, 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* = 2_y_; *i.e.*
*y* = 2_x_. Hence the three ages were, at
first, *x*, 2_x_, 3_x_; and the number of years,
since that time is two-thirds of 6_x_, *i.e.*
is 4_x_. Hence the present ages are 5_x_, 6_x_,
7_x_. The ages are clearly *integers*, since
this is only “the year when one of my sons comes
of age.” Hence 7_x_ = 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 brings 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 5_z_, 6_z_, and 7_z_; it then assumes, without
giving any reason, that 7_z_ = 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 = 7_x_ + 4_y_; [ therefore]
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,
2_A_ = *B*, *C* = *B* + *A*, *D*
= 2_B_. 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; [ therefore] 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?

The solution sent in by C. R. is, like that of SIMPLE SUSAN, partly tentative, and so does not rise higher than being Clumsily Right.

Among those who have earned the highest
honours, ALGERNON BRAY solves the problem quite correctly,
but adds that there is nothing to exclude the supposition
that all the ages were *fractional*. This
would make the number of answers infinite. Let
me meekly protest that I *never* intended my
readers to devote the rest of their lives to writing
out answers! E. M. RIX points out that, if fractional
ages be admissible, any one of the three sons might
be the one “come of age”; but she rightly
rejects this supposition on the ground that it would
make the problem indeterminate. WHITE SUGAR is
the only one who has detected an oversight of mine:
I had forgotten the possibility (which of course ought
to be allowed for) that the son, who came of age that
*year*, need not have done so by that *day*,
so that he *might* be only 20. This gives
a second solution, *viz*., 20, 24, 28. Well
said, pure Crystal! Verily, thy “fair discourse
hath been as sugar”!

**CLASS LIST.**

**I.**

ALGERNON BRAY.

AN OLD FOGEY.

E. M. RIX.

G. S. C.

S. S. G.

TOKIO.

T. R.

WHITE SUGAR.

**II.**

C. R.

DELTA.

MAGPIE.

SIMPLE SUSAN.

**III.**

DINAH MITE.

M. F. C.

I have received more than one remonstrance
on my assertion, in the Chelsea Pensioners’
problem, that it was illogical to assume, from the
*datum* “70 p. c. have lost an eye,”
that 30 p. c. have *not*. ALGERNON BRAY
states, as a parallel case, “suppose Tommy’s
father gives him 4 apples, and he eats one of them,
how many has he left?” and says “I think
we are justified in answering, 3.” I think
so too. There is no “must” here,
and the *data* are evidently meant to fix the
answer *exactly*: but, if the question were
set me “how many *must* he have left?”,
I should understand the *data* to be that his
father gave him 4 *at least*, but *may*
have given him more.

I take this opportunity of thanking
those who have sent, along with their answers to the
Tenth Knot, regrets that there are no more Knots to
come, or petitions that I should recall my resolution
to bring them to an end. I am most grateful for
their kind words; but I think it wisest to end what,
at best, was but a lame attempt. “The stretched
metre of an antique song” is beyond my compass;
and my puppets were neither distinctly *in* my
life (like those I now address), nor yet (like Alice
and the Mock Turtle) distinctly *out* of it.
Yet let me at least fancy, as I lay down the pen,
that I carry with me into my silent life, dear reader,
a farewell smile from your unseen face, and a kindly
farewell pressure from your unfelt hand! And
so, good night! Parting is such sweet sorrow,
that I shall say “good night!” till it
be morrow.