70. The Runner's Puzzle. Two men are running around a circle in opposite directions. Brown, the better runner, gave Tomkins a start of $\frac{1}{8}$ of the distance, but overestimated his own strength: after running $\frac{1}{6}$ of the distance, he met Tomkins and realized that his own chances of success were very slim.

How much faster must Brown now run to catch up with his opponent? This puzzle will be very simple if you thoroughly understand its conditions.

* In 1 English mile $(1609.315$ m) there are 5280 feet.
70. When Brown had left behind only \(\frac{1}{6}\), or \(\frac{4}{24}\), of the entire distance, Tomkins had already traveled \(\frac{5}{6}\) minus \(\frac{1}{8}\), or \(\frac{17}{24}\), of the entire distance. Therefore, Tomkins' speed is \(\frac{17}{4}\) times greater than Brown's speed. Brown had \(\frac{5}{6}\) of the distance left to run, while Tomkins had only \(\frac{1}{6}\) of the distance left. Therefore, for Brown to arrive at least simultaneously, he must increase his speed to 5 times that of Tomkins, which is 5 times greater than \(\frac{17}{4}\), or run \(\frac{85}{4}\) times faster than his initial speed. However, the question asked not "how many times," but "by how much," and "at \(\frac{85}{4}\) times faster" is the same as being faster by \(\frac{81}{4}\) of Brown's initial speed. The correct answer, therefore, is: by \(20 \frac{1}{4}\) times the initial speed of Brown, although it seems that such a recommendation is practically unfeasible.