1. From point A to point B, which are 12 km apart, a pedestrian and a bus set out simultaneously. Arriving at point B in less than one hour, the bus, without stopping, turned around and started moving back towards point A at a speed twice its initial speed. After 12 minutes from its departure from point B, the bus 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, determine the initial speed of the bus (in km/h). In the answer, write the sum of the found values of the pedestrian's and the bus's speeds. (5 points)
Solution. Let $x$ be the pedestrian's speed (in km/h), $y$ be the car's speed (in km/h) on the way from $A$ to $B$, $2y$ be the car's speed on the way from $B$ to $A$, and $t$ be the time (in hours) the car spends traveling from $A$ to $B$.

$\left\{\begin{aligned} y t & =12, \\ x(t+0.2) & =12-0.4 y, \\ t & <1,\end{aligned}\right.$

$D=x^{2}-600 x+3600 \geq 0 ; \quad x^{2}-600 x+3600=0, D / 4=86400, x_{1 / 2}=300 \pm 120 \sqrt{6}$.

Since

$$
\frac{x}{5}<x\left(t+\frac{1}{5}\right)<12 \Rightarrow x<60, \quad \text { then } \quad x^{2}-600 x+3600 \geq 0 \quad \text { when }
$$

$x \leq 300-120 \sqrt{6} \approx 6.06$. Therefore, the greatest possible integer value of the pedestrian's speed $x=6$. Let's find $y$ when $x=6: 2 y^{2}-54 y+3600=0, y_{1}=12, y_{2}=15$. Since $t<1$, then $y<12 \Rightarrow y=15$.

## Answer: 21