# From point A to point B, the distance between which is 50 km, a motorist and a cyclist left at the same time.

**From point A to point B, the distance between which is 50 km, a motorist and a cyclist left at the same time. It is known that an auto-bike rider travels 25 km more per hour than a cyclist. Determine the speed of the cyclist if it is known that it arrived at point B 2 hours 5 minutes later than the motorist.**

Let the speed of the cyclist be x km / h, then the speed of the motorist is (x + 25) km / h. A cyclist covered a distance of 50 kilometers in 50 / x hours, and a motorist in 50 / (x + 25) hours. By the condition of the problem, it is known that the travel time of a cyclist is longer than the travel time of a motorist by (50 / x – 50 / (x + 25)) hours or by 2 hours 5 minutes = 2 5/60 hours = 2 1/12 hours = 25/12 h. Let’s compose the equation and solve it.

50 / x – 50 / (x + 25) = 25/12;

(50 * 12 (x + 25) – (50 * 12x) / (12x (x + 25)) = (25 * x (x + 25)) / (12x (x + 25));

O.D.Z. x ≠ 0, x ≠ -25;

50 * 12 (x + 25) – (50 * 12x) = 25x (x + 25);

600x + 15000 – 600x = 25x ^ 2 + 625x;

25x ^ 2 + 625x – 15000 = 0;

x ^ 2 + 25x – 600 = 0;

D = b ^ 2 – 4ac;

D = 25 ^ 2 – 4 * 1 * (-600) = 625 + 2400 = 3025; √D = 55;

x = (-b ± √D) / (2a);

x1 = (-25 + 55) / 2 = 30/2 = 15 (km / h) – cyclist’s speed;

x2 = (-25 – 55) / 2 = -80/2 = -40 – the speed cannot be negative.

Answer. 15 km / h