The speed of one cyclist is 3% higher than the speed of the other. After the start, one of them drove ten

The speed of one cyclist is 3% higher than the speed of the other. After the start, one of them drove ten laps along the inner track of the cycle track, then five laps along the outer track. The second cyclist rode ten laps on the outer lane of the cycle track, then five laps on the inner lane. How many times is the outside track longer than the inside track if the cyclists finished at the same time?

1. Let V be the speed of the second cyclist.

3% = 3/100 = 0.03 part of the speed.

Then (1.03 * V) is the speed of the first cyclist.

2. Let X be the length of the inner track, Y the length of the outer track.

Then one cyclist rode 5 * X + 10 * Y.

The second drove 10 * X + 5 * Y.

The travel time of the first was (5 * X + 10 * Y) / (1.03 * V).

The time of the second is (10 * X + 5 * Y) / V.

3. The cyclists finished at the same time.

(5 * X + 10 * Y) / (1.03 * V) = (10 * X + 5 * Y) / V.

5 * X + 10 * Y = 1.03 * (10 * X + 5 * Y).

10 * Y – 5.15 * Y = 10.3 * X – 5 * X.

4.85 * Y = 5.3 * X.

Y = 5.3 / 4.85 * X.

Y = 106/97 * X.

Answer: The outer path is 106/97 times longer than the inner one.