CAT Routes and Networks Questions & Solutions

A sample of real CAT Routes and Networks past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 13 Routes and Networks questions in all — sign up free to practise them timed.

1. Passage

   Comprehension:
   A, B, C, D, E and F are the six police stations in an area, which are connected by streets as shown below. Four teams - Team 1, Team 2, Team 3 and Team 4 - patrol these streets continuously between 09:00 hrs. and 12:00 hrs. each day.
   

   

   The teams need 30 minutes to cross a street connecting one police station to another. All four teams start from Station A at 09:00 hrs. and must return to Station A by 12:00 hrs. They can
   also pass via Station A at any point on their journeys.
   The following facts are known.
   1. None of the streets has more than one team traveling along it in any direction at any point in time.
   2. Teams 2 and 3 are the only ones in stations E and D respectively at 10:00 hrs.
   3. Teams 1 and 3 are the only ones in station E at 10:30 hrs.
   4. Teams 1 and 4 are the only ones in stations B and E respectively at 11:30 hrs.
   5. Team 1 and Team 4 are the only teams that patrol the street connecting stations A and E.
   6. Team 4 never passes through Stations B, D or F.

   Q1.CAT 2023

   Which one among the following stations is visited the largest number of times?

   • Station C

   • Station E

   • Station D

   • Station F

   Show solution

   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 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 E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and 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 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 (A-E) root at 9.00 hrs.

   Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs. 

   Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.

   Hence, the roots of team 4 to reach station 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 E from A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station 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 E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via 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 E to F since they can't choose E to 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-A

   

   Now the roots for team 2 going back to A is from 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 among the options station E is visited the largest number of times.
   

2. Q2.CAT 2023

   How many times do the teams pass through Station B in a day?

   Answer: 2

   Show solution

   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 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 E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and 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 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 (A-E) root at 9.00 hrs.

   Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs.

   Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.

   Hence, the roots of team 4 to reach station 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 E from A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station 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 E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via 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 E to F since they can't choose E to 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-A

   

   Now the roots for team 2 going back to A is from 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 the teams have passed through B 2 times in this given period.

3. Q3.CAT 2023

   Which team patrols the street connecting Stations D and E at 10:15 hrs?

   • Team 4

   • Team 1

   • Team 2

   • Team 3

   Show solution

   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 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 E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and 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 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 (A-E) root at 9.00 hrs.

   Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs.

   Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.

   Hence, the roots of team 4 to reach station 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 E from A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station 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 E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via 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 E to F since they can't choose E to 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-A

   

   Now the roots for team 2 going back to A is from 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 a 10.15 hrs, team 3 is travelling from station D to station E.

   The correct option is D

4. Q4.CAT 2023

   How many times does Team 4 pass through Station E in a day?

   Answer: 2

   Show solution

   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 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 E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and 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 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 (A-E) root at 9.00 hrs.

   Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs.

   Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.

   Hence, the roots of team 4 to reach station 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 E from A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station 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 E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via 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 E to F since they can't choose E to 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-A

   

   Now the roots for team 2 going back to A is from 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 team 4 passed station E 2 times in a day

5. Q5.CAT 2023

   How many teams pass through Station C in a day?

   • 4

   • 3

   • 1

   • 2

   Show solution

   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 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 E and D respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations A and 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 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 (A-E) root at 9.00 hrs.

   Hence, Team 1 will reach B at 9.30, and come to A at 10.00 hrs. After that, they will go to E at 10.30 hrs.

   Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations A, E, and C.

   Hence, the roots of team 4 to reach station 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 E from A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach E at 11.30 is (A-E-A-C-A-E), and at 12.00 hrs, team 4 will come back to station 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 E at 10.30 hrs, and they will reach station B at 11.30 hrs, which is only possible when they travel to B via 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 E to F since they can't choose E to 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-A

   

   Now the roots for team 2 going back to A is from 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

6. Passage

   A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.
   The underlying principle that they are working on is the following:
   Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day.

   Q6.CAT 2017

   If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is:

   • 45

   • 90

   • 180

   • 135

   Show solution

   There are ten cities. We need to find the minimum number of flights required to travel from any city to any city. Any two cities can be selected in 10C2 ways. Now for these two cities, a person will need minimum 4 flights. (1 to go from A to B, 1 to go from B to A. Similarly, 1 to return to A and 1 to return to B) Thus, minimum number of required flights = 45*4 = 180.

7. Q7.CAT 2017

   Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:

   • 54

   • 120

   • 96

   • 60

   Show solution

   
   From each hub, there will be flights to 7 cities. So total total number of flights originating or terminating at each hub = 7*4 = 28. For all three hubs, it would be 28*3 = 84

   There are three hubs in total. Each hub must also be interconnected. The total number of flights between any two hubs will be 4. For three hubs it will be 12.

   Hence, the required number will be 84 + 12 = 96.

8. Q8.CAT 2017

   Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:
   1. Both cities are in G1
   2. Between A and any city in G2
   3. Between B and any city in G3
   4. Between C and any city in G4
   Then the minimum number of direct flights that satisfies the underlying principle of the airline is:

   Answer: 40

   Show solution

   In G1, we have three cities namely A, B, C. Person living in any of these three cities should be able to travel to other city once in the morning nad one in the night.  Therefore, a total of 4 flights are required between a pair of cities.
   A ---> B (Morning flight)
   A ---> B (Evening flight)
   B ---> A (Morning flight)
   B ---> A (Evening flight)
   
   

   Number of flights between the cities of G1 = 3c2*4 = 12
   Between cities in A and any city in G2 = 3*4 = 12
   Between B and any city in G3 = 2*4 = 8
   Between C and any city in G4 = 2*4 = 8
   Total = 12*2 + 8*2 = 40

9. Q9.CAT 2017

   Suppose the 10 cities are divided into 4 distinct groups Gl, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that Gl consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:
   1. Both cities are in G1
   2. Between A and any city in G2
   3. Between B and any city in G3
   4. Between C and any city in G4
   However, due to operational difficulties at A, it was later decided that the only flights that would operate at A would be those to and from B. Cities in G2 would have to be assigned to G3 or to G4.
   What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose?

   Answer: 4

   Show solution

   The cities those were a part of G2 will be shifted to either G3 or G4 but that will not have any impact on the number of the total flights from G1. The only reduction which will take place due to the number of flights shutting down from A to C.
   Hence, the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose = 4 

   Alternate method : 

   Let us determine the number of flights under new conditions.
   Flights between A and B = 4
   Between B and any city in G3 = 4*4 = 16
   Between C and B = 4
   Between C and any city in G4 = 4*4 = 12
   Hence, total flights = 36
   Thus, reduction = 40 - 36 = 4.
   
   

10. Passage

    Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N). The roads from A to M, and from N to B, are both short and narrow. In each case, one car takes 6 minutes to cover the distance, and each additional car increases the travel time per car by 3 minutes because of congestion. (For example, if only two cars drive from A to M, each car takes 9 minutes.) On the road from A to N, one car takes 20 minutes, and each additional car increases the travel time per car by 1 minute. On the road from M to B, one car takes 20 minutes, and each additional car increases the travel time per car by 0.9 minute.
    The police department orders each car to take a particular route in such a manner that it is not possible for any car to reduce its travel time by not following the order, while the other cars are following the order.
    

    Q10.CAT 2017

    How many cars would be asked to take the route A-N-B, that is Akala-Nanur-Bakala route, by the police department?

    Answer: 2

    Show solution

    Since there are two routes i.e A-M-B and A-N-B and four cars, then 2 cars must be allowed to take each route. In case if one car tried to break rule, then its travel time will increase. Now assume that on route A-M-B three cars are allowed and on route A-N-B one car is allowed, then one car running on A-M-B can break the rule and reduce its travel time. Hence, two cars must be allowed on each route.

11. Q11.CAT 2017

    If all the cars follow the police order, what is the difference in travel time (in minutes) between a car which takes the route A-N-B and a car that takes the route A-M-B?

    • 1

    • 0.1

    • 0.2

    • 0.9

    Show solution

    Since, two cars are allowed on each route, The A-M part and N-B will take same time. The difference will be in travelling M-B part and A-N part, and that difference is 0.1 minute. In route M-B car will take 20+0.9 = 20.9 min and in route A-N car will take 20+1 = 21 min. So difference = 0.1 min

12. Q12.CAT 2017

    A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M-N-B. On the road from M to N, one car takes 7 minutes and each additional car increases the
    travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.
    How many cars would the police department order to take the A-M-N-B route so that it is not possible for any car to reduce its travel time by not following the order while the other cars follow the order? (Assume that the police department would never order all the cars to take the same route.)

    Answer: 2

    Show solution

    Case 1:  Let us assume 1 car takes AMB route, 3 cars take ANB route
    Then travel time of AMB will be A-M + M-B = 6 + 20 = 26
    Then travel time of ANB will be A-N + N-B = (20+2) + (6+3*2) = 34
    
    Now, one car(A) travelling on ANB broke the rule and decided to move on AMB route. then
    
    Case 2: Let us assume 2 cars take AMB route, 2 cars take ANB route
    Then travel time of AMB will be A-M + M-B =(6+3) + (20+0.9) = 29.9
    Then travel time of ANB will be A-N + N-B = (20+1) + (6+3) = 30
    
    Since the car A reduced its time from 34 to 29.9, the case-1 route is not optimal, hence case 1 is invalid.
    Now, one car(B) travelling on ANB broke the rule and decided to move on AMB route. then, case
    
    Case 3:  Let us assume 3 cars take AMB route, 1 car take ANB route

    Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9*2) = 33.8
    Then travel time of ANB will be A-N + N-B = (20) + (6) = 29

    Now, one car(C) travelling on AMB broke the rule and decided to move on AMNB route. then
    Case 4:  Let us assume 2 cars take AMB route, 1 car take AMNB route and other take ANB route. 

    Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9) = 32.9
    Then travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7) + (6+3) = 28
    Then travel time of ANB will be A-N + N-B = (20) + (6+3) = 29
    
    Since this car(C) reduced its time from 33.8 to 28, the route is not optimal, hence case 3 is invalid.
    Since this car(B) reduced its time from 30 to 28, the route is not optimal, hence case 2 is invalid.
    Now, one car(D) travelling on AMB broke the rule and decided to move on AMNB route. then

    Case 5:   Let us assume 1 car takes AMB route, 2 cars take AMNB route and other takes ANB route.

    Then, the portion A-M will be travelled by 3 cars, M-B by one car, M-N by 2 cars, A-N by 1 car and N-B by 3 cars. 

    Then, travel time of AMB will be A-M + M-B = (6+3*2) + (20) = 32
    Then, travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7+1) + (6+3*2) = 32
    Then, travel time of ANB will be A-N + N-B = (20) + (6+3*2) = 32

    It is clear that the car D reduced its time from 32.9 min to 32 min if it broke the rule. Hence, case 4 is invalid.
    In this arrangement of case 5, no car can improve their travel time by changing their path
    Hence, the optimal allocation will be to order 2 cars on A-M-N-B route.

CAT Routes and Networks previous year questions with solutions — AthenaPrep