# Two cyclists leave at the same time from points A and B towards each other and meet in 2 hours 40 minutes.

**Two cyclists leave at the same time from points A and B towards each other and meet in 2 hours 40 minutes. If they both left point A and went to point B, and the second would leave 3 hours later than the first, then the second cyclist would catch up with the first, having covered 3/4 of the distance from A to B. How long will it take for the first cyclist to travel from A to B?**

Let the speed of the first cyclist be x km / h and the speed of the second cyclist y km / h. It is known from the condition of the problem that when two cyclists leave points A and B at the same time towards each other, they meet after 2 hours 40 minutes (2 hours 40 minutes = 8/3 hours), which means that the speed of convergence of cyclists will be (x + y) km / h, and the distance between points is 8 ∙ (x + y) / 3. Knowing that if they both left point A and went to point B, and the second would leave 3 hours later than the first, then the second cyclist would catch up with the first, having covered 3/4 of the distance from A to B, that is, the speed of their convergence was would be (y – x) km / h, and the distance traveled to the meeting is 3 ∙ (y – x) km, and the entire distance from A to B could be found using the quotient 3 ∙ (y – x): (3/4 ) = 4 ∙ (y – x) (km). From here we get the equation:

8 ∙ (x + y) / 3 = 4 ∙ (y – x);

y = 5 ∙ x.

To determine how long it will take for the first cyclist to travel from A to B, the distance traveled is 8 ∙ (x + y) / 3 = 8 ∙ (x + 5 ∙ x) / 3 = 16 ∙ x (km) divided by his speed x km / h:

(16 ∙ x): x = 16 (hours).

Answer: The first cyclist from A to B will take 16 hours.