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 2-1/2 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."
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.
_Problem._--Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1_s._ 2_d._; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1_s._ 5_d._: 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_ + 3_y_ + 7_z_ = 14, and _x_ + 4_y_ + 10_z_ = 17. And we require the values of _x_ + _y_ + _z_, and of 2_x_ + 3_y_ + 5_z_. 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.