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CAT Exam  >  CAT Tests  >  Logical Reasoning (LR) and Data Interpretation (DI)  >  Routes & Network - CAT MCQ

Routes & Network - CAT MCQ


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9 Questions MCQ Test Logical Reasoning (LR) and Data Interpretation (DI) - Routes & Network

Routes & Network for CAT 2024 is part of Logical Reasoning (LR) and Data Interpretation (DI) preparation. The Routes & Network questions and answers have been prepared according to the CAT exam syllabus.The Routes & Network MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Routes & Network below.
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Routes & Network - Question 1
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Which one among the following stations is visited the largest number of times?

Detailed Solution for Routes & Network - Question 1

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.

Routes & Network - Question 2
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How many times the teams pass through station B in a day?

Detailed Solution for Routes & Network - Question 2

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 Erfrom A at 10.00 hrs, at that time team 4 can't choose the same route. Hence, 1 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 Fat 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.

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Routes & Network - Question 3
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Which team patrols the street connecting Stations D and E at 10:15 hrs?

Detailed Solution for Routes & Network - Question 3

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

Routes & Network - Question 4
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How many times does team 4 Pass through station E in a day?
Detailed Solution for Routes & Network - Question 4

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

Routes & Network - Question 5
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How many teams pass through Station C in a day?
Detailed Solution for Routes & Network - Question 5

 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, ar 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-Α-Ε).
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 Fat 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

Routes & Network - Question 6
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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:
Detailed Solution for Routes & Network - Question 6

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.

Routes & Network - Question 7
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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:
Detailed Solution for Routes & Network - Question 7

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.

Routes & Network - Question 8
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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:
Detailed Solution for Routes & Network - Question 8

Then the minimum number of direct flights that satisfies the underlying principle of the airline is:
Ans: 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

Routes & Network - Question 9
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What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose?
Detailed Solution for Routes & Network - Question 9

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.

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