1. A train leaves $K$ for $L$ at 09:30 while another train leaves $L$ for $K$ at 10:00. The first train arrives in $\mathrm{L} 40$ minutes after the trains pass each other. The second train arrives in K 1 hour and 40 minutes after the trains pass.
Each train travels at a constant speed.
At what time did the trains pass each other?
SolUtion
Suppose that the first train covers a distance $u$ in one minute and the second train a distance $v$ in one minute, and that the trains pass $T$ minutes after 09:30. Call the crossing point $\mathrm{X}$.

The first train takes $T$ minutes to go from $\mathrm{K}$ to $\mathrm{X}$, and the second train takes 100 minutes to travel from $\mathrm{X}$ to $\mathrm{K}$. Hence, equating distances, $T u=100 v$.
The first train takes 40 minutes to travel from $\mathrm{X}$ to $\mathrm{L}$, and the second train takes $T-30$ minutes to travel from $L$ to $X$, since it started 30 minutes later than the first train. Hence, again equating distances, $40 u=(T-30) v$.
Each of these two equations allows us to find $\frac{u}{v}$. Equating the resulting expressions, we obtain a quadratic equation for $T$.
We have $\frac{u}{v}=\frac{T-30}{40}=\frac{100}{T}$.
Rearranging, we get $T^{2}-30 T-4000=0$, and factorising, we obtain $(T-80)(T+50)=0$ so that $T=80$ (ignoring the negative answer, which is not practicable here).

Hence the trains pass at $10: 50$.