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.

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!

Of those who win honours, The Shetland Snark must have the Third 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 Second 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 Coël 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.

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!" Take an illustration: Your house has been broken into and robbed, and you appeal to the policeman who was on duty that night. "Well, mum, I did see a chap getting out over your garden wall: but I was a good bit off, so I didn't chase him, like. I just cut down the short way to the Chequers", and who should I meet but Bill Sykes, coming full split round the corner. So I just ups and says, 'My lad, you're wanted.' That's all I says. And he says, 'I'll go along quiet, Bobby,' he says,'without the darbies,' he says." There's your Arithmetical policeman. Now try the other method: "I seed somebody a-running, but he was well gone or ever I got nigh the place. So I just took a look round in the garden. And I noticed the footmarks, where the chap had come right across your flower-beds. They was good big footmarks sure-ly. And I noticed as the left foot went down at the heel, ever so much deeper than the other. And I says to myself, 'The chap's been a big hulking chap: and he goes lame on his left foot'. And I rubs my hand on the wall where he got over, and there was soot on it, and no mistake. So I says to myself, 'Now where can I light on a big man, in the chimbley-sweep line, what's lame of one foot?" And I flashes up permiscuous: and I says, 'It's Bill Sykes!' says I" There is your Algebraical policeman--a higher intellectual type, to my thinking, than the other.

Little Jack's solution calls for a word of praise, as he has written out what really is an algebraical proof in words, without representing any of his facts as equations. If it is all his own, he will make a good algebraist in the time to come. I beg to thank Simple Susan for some kind words of sympathy, to the same effect as those received from Old Cat.

Hecla and Martreb are the only two who have used a method certain either to produce the answer, or else to prove it impossible: so they must share between them the highest honours.

Class List.

I.

 Hecla. Martreb.

II.

1 (2 steps)

 Adelaide. Little Jack. Clifton C.... Nil Desperandum. E. K. C. Simple Susan. Guy. Yellow-Hammer. L'Inconnu. Woolly One.

2 (3 steps)

3 (4 steps)

 A. A. Baby. A Christmas Carol. Balbus. Afternoon Tea. Bog-Oak. An Appreciative School-Ma'am. The Red Queen. Wall-Flower.
 Hawthorn. Joram. S. S. G.

4 (5 steps)
A Stepney Coach.

5 (6 steps)

 Bay Laurel. Bradshaw of the Future.

6 (9 steps)
Old King Cole.

7 (14 steps)
Theseus.