HECLA indulges, again and again, in that most fatal of all habits in computation--the making _two_ mistakes which cancel each other. She takes _x_ as the width of the garden, in yards, and _x_ + 1/2 as its length, and makes her first "coil" the sum of _x_-1/2, _x_-1/2, _x_-1, _x_-1, _i.e._ 4_x_-3: but the fourth term should be _x_-1-1/2, so that her first coil is 1/2 a yard too long. Her second coil is the sum of _x_-2-1/2, _x_-2-1/2, _x_-3, _x_-3: here the first term should be _x_-2 and the last _x_-3-1/2: these two mistakes cancel, and this coil is therefore right. And the same thing is true of every other coil but the last, which needs an extra half-yard to reach the _end_ of the path: and this exactly balances the mistake in the first coil. Thus the sum total of the coils comes right though the working is all wrong. Of the seven who are right, DINAH MITE, JANET, MAGPIE, and TAFFY make the same assumption as C. G. L. and Co. They then solve by a Quadratic. MAGPIE also tries it by Arithmetical Progression, but fails to notice that the first and last "coils" have special values. ALUMNUS ETONÆ attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5-1/2. He ought to have proved it generally: what is true of one number is not always true of others. OLD KING COLE solves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic. VINDEX proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, "whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction." CLASS LIST. I. VINDEX. II. ALUMNUS ETONÆ. OLD KING COLE. III. DINAH MITE. JANET. MAGPIE. TAFFY. ANSWERS TO KNOT X. § 1. THE CHELSEA PENSIONERS. _Problem._--If 70 per cent. have lost an eye, 75 per cent. an ear, 80 per cent. an arm, 85 per cent. a leg: what percentage, _at least_, must have lost all four? _Answer._--Ten. * * * * * _Solution._--(I adopt that of POLAR STAR, as being better than my own). Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10. * * * * *