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148 answers. Guide future travellers, an thou wilt: thou art no Bradshaw for this Age!

Of those who win honours, the merely "honourable" are two. ascertains (rightly) the relationship between the three ages at first, but then assumes one of them to be "6," thus making the rest of her solution tentative. M. F. C. does the algebra all right up to the conclusion that the present ages are 5z, 6z, and 7z; it then assumes, without giving any reason, that 7z = 21.

Of the more honourable, attempts a novelty—to discover which son comes of age by elimination: it assumes, successively, that it is the middle one, and that it is the youngest; and in each case it apparently brings out an absurdity. Still, as the proof contains the following bit of algebra, "63 = 7x + 4y; ∴ 21 = x + 4 sevenths of y," I trust it will admit that its proof is not quite conclusive. The rest of its work is good. betrays the deplorable tendency of her tribe—to appropriate any stray conclusion she comes across, without having any strict logical right to it. Assuming A, B, C, as the ages at first, and D as the number of the years that have elapsed since then, she finds (rightly) the 3 equations, 2A = B, C = B + A, D = 2B. She then says "supposing that A = 1, then B = 2, C = 3, and D = 4. Therefore for A, B C, D, four numbers are wanted which shall be to