Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

Let the total length of the circular track be $L$.
Let the speed of ant A be $v_A$ and the speed of ant B be $v_B$.

When they meet for the first time at 10:00 am:
Ant A has covered 60% of the track, which means distance covered by A is $0.6L$.
Ant B has covered the remaining 40% of the track, which means distance covered by B is $0.4L$.
Since they start at the same time and meet, the time taken by both to meet is the same. Let this time be $t$.
$t=\frac{0.6L}{v_A}=\frac{0.4L}{v_B}$
Thus, the ratio of their speeds is:
$\frac{v_A}{v_B}=\frac{0.6}{0.4}=\frac{3}{2}$

We are told A returns to P at 10:12 am. Since they met at 10:00 am, A takes 12 minutes to cover the remaining portion of the track to return to P.
Since A has already covered 60% of the track, the remaining portion to return to P (moving in the same direction) is 40% of the track, i.e., $0.4L$.
Time taken by A to cover this $0.4L$ is 12 minutes.
$\frac{0.4L}{v_A}=12\text{minutes}\Rightarrow \frac{L}{v_A}=30\text{minutes}$
So, A takes 30 minutes to complete the full loop.

Now, we find the time taken by B to complete the full loop ($T_B$):
$T_B=\frac{L}{v_B}$
Since $v_B=\frac{2}{3}$:
$T_B=\frac{L}{\frac{2}{3}}=\frac{3}{2}\left(\frac{L}{v_A}\right)=\frac{3}{2}\cdot 30=45\text{minutes}$

Since B started at the same time as A, let's determine when they started:
A takes 30 minutes to complete the track.
Since A completed 60% of the track at 10:00 am, the time taken to reach the meeting point is:
$0.6\cdot 30=18\text{minutes}$ before 10:00 am, which means they started at 9:42 am.
B takes 45 minutes to complete the track from the start time (9:42 am).
Therefore, B returns to P at:
9:42 am + 45 minutes = 10:27 am.