Complete Works of Lewis Carroll (100 page)

III.

Dinah Mite.

Janet.

Magpie.

Taffy.

ANSWERS TO KNOT X.

§ 1.
The Chelsea Pensioners.

Problem.
—If 70 per cent.
have lost an eye, 75 per cent.
an ear, 80 per cent.
an arm, 85 per cent.
a leg: what percentage,
at least
, must have lost all four?

Answer.
—Ten.

Solution.
—(I adopt that of Polar Star, as being better than my own).
Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men.
Therefore the least percentage is 10.

Nineteen answers have been received.
One is "5," but, as no working is given with it, it must, in accordance with the rule, remain "a deed without a name."
Janet makes it "35 and
2
⁄
10
ths."
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
2
⁄
8
ths."
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
2
⁄
8
ths 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
2
⁄
4
ths of these unfortunates can obtain perfect health by handing over their wounds to the remaining
1
⁄
4
th?
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.

§ 2.
Change of Day.

I must postpone,
sine die
, the geographical problem—partly because I have not yet received the statistics I am hoping for, and partly because I am myself so entirely puzzled by it; and when an examiner is himself dimly hovering between a second class and a third how is he to decide the position of others?

§ 3.
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
;
∴
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;
∴
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?