12. (12 points) Person A and Person B travel from location A to location B. Person A starts 48 minutes before Person B, and as a result, Person B catches up with Person A when Person A has walked $\frac{2}{3}$ of the total distance. If Person B immediately returns at the same speed after reaching location B, then Person B meets Person A 6 minutes after leaving location B. How many more minutes will Person A need to walk to reach location B after Person B catches up with Person A again at the same spot?
【Solution】Solution: Let the speeds of A and B be $v_{\text {A }}$ and $v_{\text {B }}$, respectively. When A has walked $\frac{2}{3}$ of the total distance and is caught up by B, the time is $t$ hours, then,
$v_{\text {A }}\left(t+\frac{4}{5}\right)=v_{\text {B }} t=\frac{2}{3} S, \quad \therefore v_{\text {A }}=\frac{\frac{2}{3} S}{t+\frac{4}{5}}, v_{\text {B }}=\frac{\frac{2}{3} S}{t}$,
Also, $v_{\text {A }}\left(t+\frac{48}{60}+\frac{1}{2} t+\frac{6}{60}\right)+\frac{6}{60} v_{\text {B }}=S$
Substituting and simplifying, we get $t=\frac{2}{5}$ hours $=24$ minutes,
So A needs 108 minutes to travel the entire distance,
After meeting, B takes 24 minutes to return to the point where A was caught up, which means A has traveled another 24 minutes, totaling $72+24$ $=96$ minutes,

So A still needs to walk $108-96=12$ minutes.
The answer is 12 minutes.