Given :
At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already en route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | |||
2 | A | F | E | ||||
3 | A | C | D | ||||
4 | A | E | A | C | A | E | A |
Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | ||||
3 | A | C | D | E | |||
4 | A | E | A | C | A | E | A |
At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | |||
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows :
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | A/E | F | A |
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
From the above table , we can see that among the options given, Station E is the station visited the largest number of times.
So, the correct option is (C) : Station E.
Given :
At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already en route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | |||
2 | A | F | E | ||||
3 | A | C | D | ||||
4 | A | E | A | C | A | E | A |
Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | ||||
3 | A | C | D | E | |||
4 | A | E | A | C | A | E | A |
At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | |||
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows :
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | A/E | F | A |
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
From the data available in the above table , we can see that the teams have crossed the B Station 2 times in the given time period.
So, the correct answer is 2.
Given :
At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already en route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | |||
2 | A | F | E | ||||
3 | A | C | D | ||||
4 | A | E | A | C | A | E | A |
Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | ||||
3 | A | C | D | E | |||
4 | A | E | A | C | A | E | A |
At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | |||
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows :
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | A/E | F | A |
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
From the above table, we can infer that at 10:15am Team 3 is travelling from Station D to Station E.
So, the correct option is (C) : Team 3.
Given :
At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already en route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | |||
2 | A | F | E | ||||
3 | A | C | D | ||||
4 | A | E | A | C | A | E | A |
Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | ||||
3 | A | C | D | E | |||
4 | A | E | A | C | A | E | A |
At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | |||
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows :
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | A/E | F | A |
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
From the above table, we can easily see that only the Team 4 is passing through the Station E twice a day i.e 2 times in a day.
So, the correct answer is 2.
Given :
At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already en route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | |||
2 | A | F | E | ||||
3 | A | C | D | ||||
4 | A | E | A | C | A | E | A |
Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | ||||
3 | A | C | D | E | |||
4 | A | E | A | C | A | E | A |
At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | |||
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows :
Teams | 9.00 | 9.30 | 10.00 | 10.30 | 11.00 | 11.30 | 12.00 |
1 | A | B | A | E | A | B | A |
2 | A | F | E | F | A/E | F | A |
3 | A | C | D | E | D | C | A |
4 | A | E | A | C | A | E | A |
From the above table, we can infer that Team 3 and Team 4 are passing through the Station C on the mentioned day.
So, the correct option is (A) : 2.