jmc

The circumference of the larger circle, $C_1$, is $2\cdot 5\pi =10\pi$. The circumference of the smaller circle, $C_2$, is $2\cdot 2\pi =4\pi$. The bug on $C_1$ crawls the circumference in $\frac{10\pi }{3\pi }=\frac{10}{3}$ minutes, while the bug on $C_2$ crawls the circumference in $\frac{4\pi }{2.5\pi }=\frac{8}{5}$ minutes. The two bugs will meet at point P in some $t$ minutes, when $t÷\frac{10}{3}=\frac{3t}{10}$ and $t÷\frac{8}{5}=\frac{5t}{8}$ are both integers. We have $\text{GCD}(3,10)=\text{GCD}(5,8)=1$, so we have to find the LCM of $10=2\cdot 5$ and $8=2^3$. The LCM is $2^3\cdot 5=40$, so the bugs will next meet in $t=\boxed{40}$ minutes.