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 + z2 = 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-2x=5, i.e. x=53-2x. Now if we substitute these values of x, y, in the quantities whose values are required, the first becomes (5+2z)+(3-3x)+z, i.e. 8: and the second becomes 2(5+2x)+3(3-3z)+5z, i.e. 19. Hence the answers are (1) 8d., (2) 1s. 7d.
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 impossible, 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 12 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,=3d.”: 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 (3y+7)3-y/3-4y+10(3-y/3)=3. By making two mistakes in this line, she brings out y=3/2. Try it again, O Vis Inerniae! 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 7d. and 1s. 5d. He assumes with Too Much Confidence, that biscuits were 1/2d. 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 3d.” and proceeds, “therefore 1 sandwich = 1 1/2d., 3 biscuits = 1 1/2 d., 1 lemonade = 6d.” Dinah Mite begins like Anita: and thence proves (rightly) that a biscuit costs less than 1d.: whence she concludes (wrongly) that it must cost 1/2d. 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&rdqu............