A Tangled Tale by Lewis Carroll (best novels for beginners TXT) 📕

must either give 2 ×’s to 6, or 2 ○’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 ×’s to 6 pictures; in another to 7; in the 3rd she gives 2 ○’s to 5; thus in every case ignoring the conditions. (I pause to remark that the condition “2 ×’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 ×’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.

Guy.

Old Cat.

Sea-Breeze.

Ayr.

Bradshaw of the Future.

F. Lee.

H. Vernon.

Cat.

Answers to Knot VI

Problem 1.⁠—A and B began the year with only £1,000 apiece. They borrowed nought; they stole nought. On the next New-Year’s Day they had £60,000 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 ciphers by putting them at the right-hand end of the £1,000, thus producing £100,000, 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 cipher she turns the “1” of the £1,000 into a “9,” and adds the result to the original sum, thus getting £10,000: and in this, by means of the other “0,” she turns the “1” into a “6,” thus hitting the exact £60,000.

Class List.

At Spes Infracta.

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, ⅖, ⅔. 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 ⅕, ⅔, 1. And similarly, the marks for warmth are 3, 1, ¼. 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 1×15×3, 25×23×1, 23×1×14; i.e. ⅗, ⅔, ⅓; 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 textbooks, 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