The distance must have been sixty miles. If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock--an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock--an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock--the time appointed. 72.--THE HYDROPLANE QUESTION. The machine must have gone at the rate of seven-twenty-fourths of a mile per minute and the wind travelled five-twenty-fourths of a mile per minute. Thus, going, the wind would help, and the machine would do twelve-twenty-fourths, or half a mile a minute, and returning only two-twenty-fourths, or one-twelfth of a mile per minute, the wind being against it. The machine without any wind could therefore do the ten miles in thirty-four and two-sevenths minutes, since it could do seven miles in twenty-four minutes. 73.--DONKEY RIDING. The complete mile was run in nine minutes. From the facts stated we cannot determine the time taken over the first and second quarter-miles separately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each. 74.--THE BASKET OF POTATOES. Multiply together the number of potatoes, the number less one, and twice the number less one, then divide by 3. Thus 50, 49, and 99 multiplied together make 242,550, which, divided by 3, gives us 80,850 yards as the correct answer. The boy would thus have to travel 45 miles and fifteen-sixteenths--a nice little recreation after a day's work. 75.--THE PASSENGER'S FARE. Mr. Tompkins should have paid fifteen shillings as his correct share of the motor-car fare. He only shared half the distance travelled for L3, and therefore should pay half of thirty shillings, or fifteen shillings. 76.--THE BARREL OF BEER. Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these 20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels). 77.--DIGITS AND SQUARES. The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given. 78.--ODD AND EVEN DIGITS. As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both equal 84+1/3. Without any use of fractions it is obviously impossible. 79.--THE LOCKERS PUZZLE.