A Tangled Tale, Lewis Carroll [free ebook novel .txt] 📗

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 wrongdoers 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 ⅛”! 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 2½ 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.

Dinah Mite.

E. B. D. L.

Joram.

Balbus.

The Elder Traveller.

With regard to Knot V, I beg to express to Vis Inertiæ 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 Inertiæ 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 1s. 2d.; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1s. 5d.: 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) 8d.; (2) 1s. 7d.

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+3y+7z=14, and x+4y+10z=17. And we require the values of x+y+z, and of 2x+3y+5z. 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+3z=3, or y=3−3z; and then substituting this value of y in the 1st, which gives x−2z=5, i.e. x=5+2z. Now if we substitute these values of x, y, in the quantities whose values are required, the first becomes (5+2z)+(3−3z)+z, i.e. 8: and the second becomes 2(5+2z)+3(3−3z)+5z, i.e. 19. Hence the answers are (1) 8d., (2) 1s. 7d.

The