Since, minute needle takes $60$ minutes to cover $360^{\circ}$, So, in $1$ minute, it will cover $6^{\circ}.$ Similarly, hour needle takes $12 \times 60 = 720$ minutes to cover $360^{\circ}$, So, in $1$ minute, it will cover $(\frac{1}{2})^{\circ}.$
When both needles are at $12$ and starts traveling in their direction. Suppose, after $t$ minutes, minute needle covers $x^{\circ}$ with the speed of $6^{\circ}/min$ and in the same time, hour needle covers $(360-x)^{\circ}$ with the speed of $(\frac{1}{2})^{\circ}/min$.
When they meet, then $t = \frac{x^{\circ}}{6^{\circ}/min} = \frac{(360-x)^{\circ}}{(\frac{1}{2})^{\circ}/min}$
$\Rightarrow x = (\frac{12*360}{13})^{\circ}$
Since, $t = \frac{x^{\circ}}{6^{\circ}/min}$
So, $t = \frac{(\frac{12*360}{13})^{\circ}}{6^{\circ}}\; min$
$t = \frac{12}{13} \times 60\; min$
$t = \frac{12}{13} Hour$
