CAT 2023 Slot 3 Solved Paper

It is given that none of the streets has more than one team traveling along it in any direction at any point in time (point 1), which implies at $9.00\mathrm{hrs}$, all 4 teams have chosen different roots from the starting point.
It is also known that Teams 2 and 3 are the only ones in stations $\mathit{E}$ and $\mathit{D}$ respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations $\mathrm{A}$ and $\mathrm{E}$.
It is only possible when Team 2 traveled (A-E) via F, and Team 3 reached station D via station C.
It is also known that Teams 1 and 3 are the only ones in Station $\mathit{E}$ at 10:30 hrs, and Team 4 never passes through Stations B, D, or F. Hence, Team 1 must have chosen the (A-B) root at the starting point, and Team 4 has chosen the $\left(\mathit{A}-\mathit{E}\right)$ root at 9.00 hrs.
Hence, Team 1 will reach $\mathit{B}$ at 9.30 , and come to $\mathit{A}$ at 10.00 hrs. After that, they will go to $\mathit{E}$ at 10.30 hrs.
Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations $\mathit{A},\mathit{E}$, and $\mathit{C}$.
Hence, the roots of team 4 to reach station $\mathit{E}$ at 11.30 will be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Since team 1 is already traveling to $\mathit{E}$ from $\mathit{A}$ at $10.00\mathrm{hrs}$, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach $\mathit{E}$ at 11.30 is (A-E-A-C-A-E), and at $12.00\mathrm{hrs}$, team 4 will come back to station $\mathit{A}$.
Hence, the complete route diagram for team 4 is (A-E-A-C-A-E-A)

We can see that team 1 is at station $\mathit{E}$ at $10.30\mathrm{hrs}$, and they will reach station $\mathit{B}$ at $11.30\mathrm{hrs}$, which is only possible when they travel to $\mathit{B}$ via $\mathit{A}$.
Hence, the complete route diagram for team 1 is (A-B-A-E-A-B-A). It is also known that Teams 1 and 3 are the only ones in station E at 10:30 hrs.
The only possible root for Team 2 at 10.00 hrs is from $\mathit{E}$ to $\mathit{F}$ since they can't choose $\mathit{E}$ to $\mathit{D}$ because Team 3 is already on this route. Since team 3 has to reach A at 12.00 . The only possible combination for team 3 is E-D-C-ANow the roots for team 2 going back to $\mathrm{A}$ is from $\mathrm{F}$ at 10.30 hrs (F-A-F-A) or (F-E-F-A).
Hence, the final table is given below:

From the table, we can see that 2 teams (teams 3 and 4) have passed through station C on the given day.
The correct option is D