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 Megatherium.) 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 [×33] (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!"