28(1104). Two motorcyclists set off simultaneously from $A$ to $B$ and from $B$ to $A$. The first arrived in $B$ 2.5 hours after the meeting, while the second arrived in $A$ 1.6 hours after the meeting. How many hours was each motorcyclist on the road?
Solution. Let $AB = s$ (in km), the time until the motorcyclists meet $-t$ (in hours). Then the time of movement of the first motorcyclist (from $A$ to $B$) is $(t+2.5)$ hours, its speed is $\frac{s}{t+2.5}$ kilometers per hour, the time of movement of the second motorcyclist is $(t+1.6)$ hours, its speed is $\frac{s}{t+1.6}$ kilometers per hour.

Expressing the distance from the meeting place of the motorcyclists to $B$, we will have the equation

$$
\frac{s}{t+2.5} \cdot 2.5 = \frac{s}{t+1.6} \cdot t.
$$

Since $s \neq 0$, then $\frac{2.5}{t+2.5} = \frac{t}{t+1.6}$, from which $t^2 = 4, t = 2$.

Remark. You can express the distance from the meeting place of the motorcyclists to $A$ (not to $B$) or determine the entire distance $AB$. In the first case, we get the equation $\frac{s}{t+2.5} \cdot t = \frac{s}{t+1.6} \cdot 1.6$, in the second case $s = \frac{s}{t+2.5} \cdot t + \frac{s}{t+1.6} \cdot t$. Dividing both sides of the equations by $s \neq 0$ and solving the resulting equations, we find the value of $t (t=2)$. The time of movement of the first motorcyclist is 4.5 hours $(2+2.5=4.5)$, the time of movement of the second motorcyclist is 3.6 hours $(2+1.6=3.6)$.

Answer: 4.5 hours; 3.6 hours.

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