Problem 293

André runs for 10 seconds at a speed of $y \mathrm{~m} / \mathrm{s}$. Therefore, André runs $10y \mathrm{~m}$. Carl runs for 20 seconds before André starts to run and then 10 seconds while André is running. Thus, Carl runs for 30 seconds. Since Carl runs at a speed of $x \mathrm{~m} / \mathrm{s}$, then Carl runs $30x \mathrm{~m}$. Since André and Carl run the same distance, then $30x \mathrm{~m} = 10y \mathrm{~m}$, which means that $\frac{y}{x} = 3$. Thus, $y: x = 3: 1$.