1. From point $A$ to point $B$, which are 24 km apart, a pedestrian and a cyclist set out simultaneously. The cyclist, who spent no less than two hours on the journey from $A$ to $B$, without stopping, turned back and started moving towards point $A$ at a speed twice the initial speed. After 24 minutes from his departure from point $B$, the cyclist met the pedestrian. Determine the greatest possible integer value of the pedestrian's speed (in km/h), and for this value of the pedestrian's speed, find the initial speed of the cyclist.
Solution: Let $x$ (km/h) be the speed of the cyclist, $y$ (km/h) be the initial speed of the truck, and $t$ (h) be the time spent by the truck traveling from $A$ to $B$. Then

$$
\left\{\begin{array}{rl}
x(t+0.4)+0.8 y=24, \\
y t & =24, \\
t & \geq 2,
\end{array} \Rightarrow x(24 / y+0.4)+0.8 y=24, \Rightarrow 2 y^{2}+(x-60) y+60 x=0\right.
$$

For the quadratic equation to have a solution, it is necessary that $D=(x-60)^{2}-480 x \geq 0$. Therefore, $x^{2}-600 x+3600 \geq 0, D / 4=(120 \sqrt{6})^{2}$, and

$x \in(-\infty ; 300-120 \sqrt{6}] \cup[300+120 \sqrt{6} ;+\infty)$. Since by the condition $x \in \mathbb{N}$, and $0.2 x<12$, i.e., $x<60$, then $\quad x \in[1 ; 300-120 \sqrt{6}] \cap \mathbb{N}$. Using the estimate $2.44<\sqrt{6}<2.45$, we get the estimate $293<120 \sqrt{6}<294$, and $6<300-120 \sqrt{6}<7$. The largest possible integer value of the speed $x=6$. We find the initial speed of the truck when $x=6$ from the equation $2 y^{2}-54 y+360=0$, or $y^{2}-27 y+180=0, \quad y_{1}=12, \quad y_{2}=15$. Since $t \geq 2, \quad t=\frac{24}{y} \geq 2$, and $y \leq 12$, then $y=12$.

Answer: 6 km/h, 12 km/h.