8.3. From points $A$ and $B$, two cyclists set off towards each other at the same time. They traveled at constant speeds. After meeting, the first cyclist took 40 minutes to reach point $B$, and the second cyclist took one and a half hours to reach point $A$. Find the time from the start of the journey until they met and the ratio of the cyclists' speeds.
Answer. Time until the meeting - 1 hour. The speed of the first cyclist is 1.5 times greater than the speed of the second. Solution. Let $v_{1}, v_{2}$ be the speeds of the cyclists, $t$ - the time until the meeting. Then the first cyclist traveled the distance $\mathrm{v}_{1} t$ before the meeting, and the second - the distance $\mathrm{v}_{2} t$. From the conditions of the problem, we then have $\frac{\mathrm{v}_{2} t}{\mathrm{v}_{1}}=40$ (min.) and $\frac{\mathrm{v}_{1} t}{\mathrm{v}_{2}}=90$ (min.). From these equations, by eliminating $t$, we get that $\left(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\right)^{2}=\frac{9}{4}$, i.e., $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{3}{2}$. Then the value of $t=60$ (min).