11.1. A pedestrian left point A for point B, which are 40 km apart, at 4:00 AM, and a cyclist set off from A at 7:20 AM, catching up with the pedestrian exactly halfway between A and B, after which both continued their journey. A second cyclist, traveling at the same speed as the first, left B for A at 8:30 AM and met the pedestrian one hour after the pedestrian's meeting with the first cyclist. Find the speeds of the pedestrian and the cyclists.
Answer: 5 km/h and 30 km/h.

Solution: Let the speeds of the pedestrian and the cyclists be $x$ and $y$ km/h, respectively. By 7:20 AM, the pedestrian has walked $\frac{10}{3} \cdot x$ km. After that, the cyclist started chasing him and caught up after $\frac{10 x}{3(y-x)}$ hours, during which the pedestrian continued walking at a speed of $x$ km/h and was caught 20 km from A, hence $\frac{10 x}{3}+\frac{10 x}{3(y-x)} \cdot x=20$. Simplifying, we get $x y=6(y-x)$. On the other hand, the meeting with the second cyclist occurred $\frac{10}{3}+\frac{10 x}{3(y-x)}+1$ hours after 4:00 AM, which should equal $\frac{9}{2}+\frac{40-\frac{9}{2} x}{x+y}$—the sum of 4.5 hours and the time it took for the pedestrian and the second cyclist, moving towards each other, to cover the remaining distance between them by 8:30 AM. From this, simplifying, we get $\frac{9 y+80}{2(x+y)}=\frac{13 y-3 x}{3(y-x)}$, which leads to $27 y^2 - 27 xy + 240(y-x) = 26 y^2 + 20 xy - 6 x^2$. Replacing $40 x y = 240(y-x)$, we get $y^2 - 7 x y + 6 x^2 = 0$, from which, considering $y > x$, we get $y = 6 x$. Substituting this into the equation $x y = 6(y-x)$, we have $x = 5, y = 30$.

Criterion: Answer with verification: 1 point.

Correct formulas: 3 points.