97.—­THE SPOT ON THE TABLE.

The ordinary schoolboy would correctly treat this as a quadratic equation.  Here is the actual arithmetic.  Double the product of the two distances from the walls.  This gives us 144, which is the square of 12.  The sum of the two distances is 17.  If we add these two numbers, 12 and 17, together, and also subtract one from the other, we get the two answers that 29 or 5 was the radius, or half-diameter, of the table.  Consequently, the full diameter was 58 in. or 10 in.  But a table of the latter dimensions would be absurd, and not at all in accordance with the illustration.  Therefore the table must have been 58 in. in diameter.  In this case the spot was on the edge nearest to the corner of the room—­to which the boy was pointing.  If the other answer were admissible, the spot would be on the edge farthest from the corner of the room.

98.—­ACADEMIC COURTESIES.

There must have been ten boys and twenty girls.  The number of bows girl to girl was therefore 380, of boy to boy 90, of girl with boy 400, and of boys and girls to teacher 30, making together 900, as stated.  It will be remembered that it was not said that the teacher himself returned the bows of any child.

99.—­THE THIRTY-THREE PEARLS.

The value of the large central pearl must have been L3,000.  The pearl at one end (from which they increased in value by L100) was L1,400; the pearl at the other end, L600.

100.—­THE LABOURER’S PUZZLE.

The man said, “I am going twice as deep,” not “as deep again.”  That is to say, he was still going twice as deep as he had gone already, so that when finished the hole would be three times its present depth.  Then the answer is that at present the hole is 3 ft. 6 in. deep and the man 2 ft. 4 in. above ground.  When completed the hole will be 10 ft. 6 in. deep, and therefore the man will then be 4 ft. 8 in. below the surface, or twice the distance that he is now above ground.

101.—­THE TRUSSES OF HAY.

Add together the ten weights and divide by 4, and we get 289 lbs. as the weight of the five trusses together.  If we call the five trusses in the order of weight A, B, C, D, and E, the lightest being A and the heaviest E, then the lightest, no lbs., must be the weight of A and B; and the next lightest, 112 lbs., must be the weight of A and C. Then the two heaviest, D and E, must weigh 121 lbs., and C and E must weigh 120 lbs.  We thus know that A, B, D, and E weigh together 231 lbs., which, deducted from 289 lbs. (the weight of the five trusses), gives us the weight of C as 58 lbs.  Now, by mere subtraction, we find the weight of each of the five trusses—­54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62 lbs. respectively.

102.—­MR. GUBBINS IN A FOG.

The candles must have burnt for three hours and three-quarters.  One candle had one-sixteenth of its total length left and the other four-sixteenths.

103.—­PAINTING THE LAMP-POSTS.