by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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Calling the three villages by their initial letters, it is clear that the three roads form a triangle, A, B, C, with a perpendicular, measuring twelve miles, dropped from C to the base A, B. This divides our triangle into two right-angled triangles with a twelve-mile side in common. It is then found that the distance from A to C is 15 miles, from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These figures are easily proved, for the square of 12 added to the square of 9 equals the square of 15, and the square of 12 added to the square of 16 equals the square of 20.
70.?DRAWING HER PENSION.? solution
The distance must be 6% miles.
71.?SIR EDWYN DE TUDOR.? solution
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.? solution
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.? solution
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.?solution
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.?solution
Mr. Tompkins should have paid fifteen shillings as his correct share of the motor-car fare. He only shared half the distance travelled for £3, and therefore should pay half of thirty shillings, or fifteen shillings.
76.?THE BARREL OF BEER.? solution
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.? solution
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.?solution Pg 156
As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5V3 and 84 + 2 /g, both equal 84V3. Without any use of fractions it is obviously impossible.
79.?THE LOCKERS PUZZLE.? solution
The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657 + 324. The middle sum may be either 720 =134+586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and anyone of three arrangements in the case of the middle locker. Here is one solution:?
107 134 235 249 586 746
356 720 981
