Answers 239- FIND THE DISTANCE
The distance between the two places must have been 18 miles. The meeting
points were 10 miles from A and 12 miles from B. Simply
multiply 10 (the first distance) by 3 and deduct the second distance, 12.
Could anything be simpler? Try other distances for the meeting points (taking
care that the first meeting distance is more than two-thirds of the second) and
you will find the little rule will always work.- THE MAN AND THE DOG
The dog's speed was 16 miles per hour. The following facts will give the
reader clues to the general solution. The distance remaining to be walked side
by side with the dog was 81 feet, the fourth power of 3 (for the dog returned
four times), and the distance to the end of the road was 625 feet, the fourth
power of 5. Then the difference between the speeds (in miles per hour)
of man and dog (that is, 12) and the sum of the speeds (20) must be in the
same ratio, 3 to 5, as is the case.- BAXTER'S DOG
It is obvious that Baxter will overtake Anderson in one hour, for each will
be 4 miles from the hotel in the same direction. Then, as the dog has been
running uniformly at 10 miles an hour during that hour, he must have run 10
miles! When a friend put this problem before a French professor of mathe-
matics, he exclaimed: "Man Dieu, queUe serle!" quite overlooking the simple
manner of solution.- THE RUNNER'S REFRESHMENT
As the radius is t, the diameter is 2t. The diameter multiplied by 'TT (the
Greek letter pi) gives us the circumference, 2t'TT miles. As he goes round
n times, 2t'TTn equals the number of miles run, and, as he drinks s quarts per
mile, he consumes 2t'TTns quarts. We can put the factors in any order: there-
fore the answer is 2'TTnts (two pints) or one quart.