jmc

From $1$ to $9$, the times will be of the form $\boxed{a:ba}$. There are $9$ choices for $a$ and $6$ choices for $b$, so there are $9\cdot 6=54$ times in this period. From $10$ to $12$, the minutes are already determined, so there are only $3$ times in this case. In total, there are $54+3=57\to \boxed{57}$ palindromic times.