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98 The seventeen whose solutions are practically valueless are, and. tries it by a sort of "rule of false," assuming experimentally that. 1, 2, weigh 6 lbs. each, and having thus produced $17 1⁄2$, instead of 16, as the weight of 1, 3, and 5, she removes "the superfluous pound and a half," but does not explain how she knows from which to take it. says that (when in that peculiar state) "it seemed perfectly clear" to her that, "3 out of the 5 sacks being weighed twice over, $3⁄5$ of 45 = 27, must be the total weight of the 5 sacks." As to which I can only say, with the Captain, "it beats me entirely!" , on the plea that "one must have a starting-point," assumes (what I fear is a mere guess) that. 1 weighed 5½ lbs. The rest all do it, wholly or partly, by guess-work.

The problem is of course (as any Algebraist sees at once) a case of "simultaneous simple equations." It is, however, easily soluble by Arithmetic only; and, when this is the case, I hold that it is bad workmanship to use the more complex method. I have not, this time, given more credit to arithmetical solutions; but in future problems I shall (other things being equal) give the