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 13 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, = 3_d._": 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 INERTIÆ 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 3_y_ + 7 × (3-_y_)/3-4_y_ + 10 × (3-_y_)/3 = 3. By making two mistakes in this line, she brings out _y_ = 3/2. Try it again, oh VIS INERTIÆ! Away with INERTIÆ: 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 7_d._ and 1_s._ 5_d._ He assumes, with Too Much Confidence, that biscuits were 1/2_d._ each, and that Clara paid for 8, though she only ate 7!