jmc

The equation of the first line is $y=10x+b$ where $b$ is the $y$-intercept of the two lines. Since $(s,0)$ lies on the line, we can plug this into the line's equation to get $0=10s+b\Rightarrow s=-\frac{b}{10}$. Similarly, the second line has equation $y=6x+b$. Plugging $(t,0)$ into this equation gives $0=6t+b\Rightarrow t=-\frac{b}{6}$. Thus $\frac{s}{t}=-\frac{b}{10}\cdot -\frac{6}{b}=\boxed{\frac{3}{5}}$.