# From point A to point B, the distance between which is 4 km, two pedestrians set off. The second pedestrian left point A

**From point A to point B, the distance between which is 4 km, two pedestrians set off. The second pedestrian left point A 10 minutes later than the first, but arrived at point B 2 minutes earlier. Find the speed of the second pedestrian if it is known that it is 1 km / h faster than the speed of the first pedestrian.**

Let’s denote by x the speed (in km / h) of the second pedestrian.

Then the first pedestrian was moving at a speed of (x – 1) km / h.

Let’s express 10 minutes and 2 minutes in hours. We have: 10 min = (10/60) h = 1/6 h; 2 minutes = (2/60) h = 1/30 h.

From point A to point B, the first pedestrian walked (4 km / (x – 1) km / h) = (4 / (x – 1)) h, and the second pedestrian walked (4 / x) h.

According to the condition of the task 4 / x + 10/60 + 1/30 = 4 / (x – 1).

We get the quadratic equation x ^ 2 – x – 20 = 0, which has two roots: x1 = 5 and x2 = –4 (subsidiary root).

Answer: 5 km / h.