【Question 27】  
At 8:00 AM, Jia departs from location $A$ to location $B$, and at 9:00 AM, Yi departs from $A$ to catch up with Jia. By 11:00, Yi catches up with Jia, then immediately turns around and doubles his speed to return to $A$. Upon reaching $A$, Yi immediately turns around again and doubles his speed once more to catch up with Jia, and just catches up with Jia at $B$. At this point, it is $\qquad$ hour $\qquad$ minute.
【Analysis and Solution】
Travel problem.
The distance covered by A in 3 hours is equal to the distance covered by B in 2 hours;
Therefore, the speed ratio of A to B is $2: 3$;
Let the speed of A be $2x$ km/h, and the speed of B be $3x$ km/h;
From $8:00 \sim 9:00$, A travels from $A$ to $C, AC=2x$ km;
From 9:00 to 11:00, A travels from $C$ to $D, CD=4x$ km, and B travels from $A$ to $D, AD=6x$ km;
When B returns from $D$ to $A$, it takes $6x \div 6x=1$ hour, and A travels from $D$ to $E, DE=2x$ km;
Finally, B catches up with A, with a speed difference of $12x-2x=10x$ km/h, $AE=8x$ km, B travels from $A$ to $B$ in $8x \div 10x=0.8$ hours $=48$ minutes; at this point, it is already 12:48.