smc

Consider the distance-time graph and use coordinate geometry, with time on the $x$-axis and distance on the $y$-axis. Thus Jack starts at $(0,0)$ and his initial motion, taking time in hours, is $y=15x$. This ends when $y=5$, giving the point $(\frac{1}{3},5)$. He now runs in the opposite direction at $20$ km/hr, so the equation is $y-5=-20(x-\frac{1}{3})$. Now Jill starts at $(\frac{1}{6},0)$ (as Jack has a head start of 10 minutes = $\frac{1}{6}$ hours), so her equation is $y=16(x-\frac{1}{6}).$ Solving simultaneously with Jack's equation gives $-20x+\frac{35}{3}=16x-\frac{8}{3}⟹\frac{43}{3}=36x⟹x=\frac{43}{108}$, so $y=16(\frac{25}{108})=\frac{400}{108}$, and thus the required distance is $5-\frac{400}{108}=\frac{140}{108}=\frac{70}{54}=\frac{35}{27}.$ $\boxed{\frac{35}{27}}$