# DI - Routes & Networks - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic DI - Routes & Networks. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**Answer the following question based on the information given below.**

The figure below shows the street map for a certain region with the street intersections marked from a through l. A person standing at an intersection can see along straight lines to other intersections that are in her line of sight and all other people standing at these intersections. For example, a person standing at intersection g can see all people standing at intersections b, c, e, f, h, and k. In particular, the person standing at intersection g can see the person standing at intersection e irrespective of whether there is a person standing at intersection f.

Six people U, V, W, X, Y, and Z, are standing at different intersections. No two people are standing at the same intersection.

The following additional facts are known.

- X, U, and Z are standing at the three corners of a triangle formed by three street segments.
- X can see only U and Z.
- Y can see only U and W.
- U sees V standing in the next intersection behind Z.
- W cannot see V or Z. 6. No one among the six is standing at intersection d.

**CAT 2019 LRDI Slot 1 | DI - Routes & Networks**

Who is standing at intersection a?

- A.
No one

- B.
W

- C.
Y

- D.
V

Answer: Option A

**Explanation** :

From (1), X, U, Z are at b,c, g or at b, f, g in some order. Thus, X, U or Z is definitely at g.

Let X is at g.

Case (i): U and Z at b and f.

From (4), U has to be at b, Z at f and V at j. From (2), no one is at c, e, k and h. As Y sees both U and W, Y must be at a and W at i. But then W sees V, which contradicts (5).

Thus, this case is not valid.

Case (ii): U and Z at b and c.

From (4), U has to be at c, Z at b and V at a. From (2), no one is at c, e,f, k and h. But then Y must be at I, j or l. But in that case Y cannot see U. Thus, this case is not valid.

Therefore, X cannot be at g.

Let Z is at g.

From (4), U is at c or f.

Case (i) U is at c and hence V is at k and X is at a. Again there is no place for V. This case is invalid.

Case (ii) U is at f, X is at b and V at h. V will be at e as he sees U. But then he will be able to see Z also. So this case is also invalid.

Thus, U is at g. Therefore, Z is at f, V at e and X at b. So, from (2) no one will be at a, c and j. From (3) and (5), it can be concluded that Y is at k and W at l. Thus, we have

No one is standing at intersection a.

Hence option 1.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Routes & Networks**

Who can V see?

- A.
Z only

- B.
U only

- C.
U and Z only

- D.
U, W and Z only

Answer: Option C

**Explanation** :

Consider the solution to first question of this set.

V can see only U and Z.

Hence option 3.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Routes & Networks**

What is the minimum number of street segments that X must cross to reach Y?

- A.
2

- B.
3

- C.
1

- D.
4

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

X must reach Y via g so that he would cross minimum street segments. i.e., he would cross 2 street segments.

Hence option 1.

Workspace:

**CAT 2019 LRDI Slot 1 | DI - Routes & Networks**

Should a new person stand at intersection d, who among the six would she see?

- A.
V and X only

- B.
W and X only

- C.
U and Z only

- D.
U and W only

Answer: Option B

**Explanation** :

Consider the solution to first question of this set.

A new person standing at d would see W and X only.

Hence option 2.

Workspace:

**Answer the following question based on the information given below.**

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.

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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:

- A.
45

- B.
90

- C.
180

- D.
135

Answer: Option C

**Explanation** :

Since there are a total of 10 cities we can have a combination of 2 cities in 10C2 or 45 ways. Now, if for example we have 2 cities City 1 and City 2, then the cities can be connected in the following 4 ways:

Morning flight from city 1 to city 2

Morning flight from city 2 to city 1

Evening flight from city 1 to city 2

Evening flight from city 2 to city 1

So the minimum number of direct flights to connect all cities is 45 × 4 or 180 ways.

Hence, option 3.

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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:

- A.
54

- B.
120

- C.
96

- D.
60

Answer: Option C

**Explanation** :

Let us suppose that City 1, City 2 and City 3 are the hubs and City 4, City 5... upto City 10 are 7 of the other 10 cities.

Now City 1, City 2 and City 3 connect with each other in 4 possible ways (as mentioned in the answer to the previous question).

Now 2 out of 3 cities can be a chosen in 3C2 or 3 ways. So total of no ways City 1, City 2 and City 3 connect with each other is 3 × 4 or 12 ways.

Now City 1 will connect with each of City 4, City 5, City 6 ..... City 10 in 4 possible ways (as explained in the previous questions answer).

So, total number of flights between City 1 and the cities 4 to 10 is 28.

Similarly there will be 28 flights each for City 2 and City 3 that will connect it with the 7 cities. So total minimum number of flights between 2 cities will be 12 + 28 + 28 + 28 = 96.

Hence, option 3.

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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:

- Both cities are in G1
- Between A and any city in G2
- Between B and any city in G3
- Between C and any city in G4

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

Answer: 40

**Explanation** :

Now A, B and C are in G1. As there is no restriction of flights that can connect with each other in G1, the total number of flights connecting 2 cities in G1 is ^{3}C_{2} × 4

= 12

A can connect with any of the 3 cities in G1 in ^{3}C_{1} × 4 = 12 ways

B can connect with any of the 2 cities in G2 in ^{2}C_{1} × 4 = 8 ways

C can connect with any of the 3 cities in G3 in ^{2}C_{1} × 4 = 8 ways

Total minimum number of direct flights that satisfy the underlying principle is 12 + 12 + 8 + 8 = 40

Answer: 40

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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:

- Both cities are in G1
- Between A and any city in G2
- Between B and any city in G3
- 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

**Explanation** :

Given the above conditions, we know that in G1, A would connect to B and B would connect with C. B would connect with any of the 2 cities in G3 and C would connect with any of the 2 cities in G4.

Now the 2 cities from G2 that were connected to A earlier can now connects to one of the cities in G3 or G4. There is no flight connection possible now between A and C in G1.

So the total number of flight connections as compared to the number of flight connections in the previous question will be less by number of flights connecting A and C i.e., 4.

Answer: 4

Workspace:

**Answer the following question based on the information given below.**

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, 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.

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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

**Explanation** :

The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.

2 cars taking A-M-B route – 9 + 20.9 = 29.9 minutes.

2 cars taking A-N-B route – 21 + 9 = 30 minutes.

Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes. So 2 cars would take the A-N-B route.

Answer: 2

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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?

- A.
1

- B.
0.1

- C.
0.2

- D.
0.9

Answer: Option B

**Explanation** :

The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.

2 cars taking A-M-B route – 9 + 20.9 = 29.9 minutes.

2 cars taking A-N-B route – 21 + 9 = 30 minutes.

Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.

Minimum travel time would be when 2 cars are assigned to each of the 2 routes by the police department.

The difference in travel time when the cars follow the police order is 30-29.9 or 0.1 minutes.

Hence, option 2.

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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

**Explanation** :

The police department would ask 2 cars to take the A-N-B route and 2 cars to take the A-M-B route, because 2 cars taking each of these routes would minimize the time for each of the 4 cars.

2 cars taking A-M-B route – 9 + 20.9 = 29.9 minutes.

2 cars taking A-N-B route – 21 + 9 = 30 minutes.

Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.

The police department would order two cars to take the A-M-N-B route

Answer: 2

Workspace:

**CAT 2017 LRDI Slot 1 | DI - Routes & Networks**

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.If all the cars follow the police order, what is the minimum travel time (in minutes) from A to B? (Assume that the police department would never order all the cars to take the same route.)

- A.
26

- B.
32

- C.
29.9

- D.
30

Answer: Option B

**Explanation** :

2 cars taking A-M-B route – 9 + 20.9 = 29.9 minutes.

2 cars taking A-N-B route – 21 + 9 = 30 minutes.

Increasing to 3 cars on the A-M-B would increase travel time of each car by 4.8 minutes and increasing to 3 cars on the A-N-B route would increase travel time of each car by 4 minutes.

The minimum travel time from A to B is 32 minutes.

Hence, option 2.

Workspace:

**Answer the following question based on the information given below.**

A low-cost airline company connects ten Indian cities, A to J. The table below gives the distance between a pair of airports and the corresponding price charged by the company. Travel is permitted only from a departure airport to an arrival airport. The customers do not travel by a route where they have to stop at more than two intermediate airports.

**CAT 2007 LRDI | DI - Routes & Networks**

What is the lowest price, in rupees, a passenger has to pay for travelling by the shortest route from A to J?

- A.
2275

- B.
2850

- C.
2890

- D.
2930

- E.
3340

Answer: Option D

**Explanation** :

Possible routes from A to J are as shown in the table .

The shortest distance is by the route A-C-F-J.

The price is 1350 + 430 + 1150 = Rs. 2930

Hence, option 4.

Workspace:

**CAT 2007 LRDI | DI - Routes & Networks**

The company plans to introduce a direct flight between A and J. The market research results indicate that all its existing passengers travelling between A and J will use this direct flight if it is priced 5% below the minimum price that they pay at present. What should the company charge approximately, in rupees, for this direct flight?

- A.
1991

- B.
2161

- C.
2707

- D.
2745

- E.
2783

Answer: Option B

**Explanation** :

The current market price paid by the customers is Rs. 2275 (A-H-J).

Therefore, the company should charge (2275 × 0.95) = Rs. 2161.25

Hence, option 2.

Workspace:

**CAT 2007 LRDI | DI - Routes & Networks**

If the airports C, D and H are closed down owing to security reasons, what would be the minimum price, in rupees, to be paid by a passenger travelling from A to J?

- A.
2275

- B.
2615

- C.
2850

- D.
2945

- E.
3190

Answer: Option C

**Explanation** :

If C, D and H are closed, the cheapest route will be A-F-J and it will cost Rs. 2850.

Hence, option 3.

Workspace:

**CAT 2007 LRDI | DI - Routes & Networks**

If the prices include a margin of 10% over the total cost that the company incurs, what is the minimum cost per kilometer that the company incurs in flying from A to J?

- A.
0.77

- B.
0.88

- C.
0.99

- D.
1.06

- E.
1.08

Answer: Option B

**Explanation** :

The minimum cost per km that the company incurs would correspond to the minimum price per km route.

By observation from the table, minimum price per kilometre is for the route AHJ and is equal to

2275/2350 = 0.97

Minimum cost per kilometre = 0.97/1.1 = 0.88

Hence, option 2.

Workspace:

**CAT 2007 LRDI | DI - Routes & Networks**

If the prices include a margin of 15% over the total cost that the company incurs, which among the following is the distance to be covered in flying from A to J that minimizes the total cost per kilometer for the company?

- A.
2170

- B.
2180

- C.
2315

- D.
2350

- E.
2390

Answer: Option D

**Explanation** :

Even if the margin for the prices changes the minimum cost per km would correspond to the same route namely A-H-J.

∴ From the table, the distance for the travel = 2350 km

Hence, option 4.

Workspace:

**Answer the following question based on the information given below.**

A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated by the number adjacent to the arrow representing the street.

Motorists travelling from point S to point T would obviously take the route for which the total cost of travelling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the traffic gets evenly distributed among all the least cost routes.

The government can control the flow of traffic only by levying appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs. 14 (i.e. Rs. 9 + Rs. 5) plus the toll charged at junction A.

**CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that all motorists travelling from S to T pay the same amount (fuel costs and toll combined) regardless of the route they choose and the street from B to C is under repairs (and hence unusable), then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
2, 5, 3, 2

- B.
0, 5, 3, 1

- C.
1, 5, 3, 2

- D.
2, 3, 5, 1

- E.
1, 3, 5, 1

Answer: Option B

**Explanation** :

Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.

Since the cost of travel including toll on routes S-A-T, S-D-T, S-B-A-T and S-D-C-T is the same,

∴ 14 + a = 13 + d = 9 + a + b = 10 + c + d

Thus, b = 5, d – a = 1, c = 3

If a = 0, d = 1, If a = 1, d = 2 and if a = 2, d = 3

Hence, both options 2 and 3 satisfy the given criteria.

Note: The question makers took care of this inconsistency while calculating scores.

Workspace:

**CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that no traffic flows on the street from D to T, while equal amount of traffic flows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
1, 5, 3, 3

- B.
1, 4, 4, 3

- C.
1, 5, 4, 2

- D.
0, 5, 2, 3

- E.
0, 5, 2, 2

Answer: Option E

**Explanation** :

Since the cost of travel including toll on routes S-A-T, S-B-C-T, S-B-A-T and S-D-C-T is the same,

∴ 14 + a = 7 + b + c = 9 + a + b = 10 + c + d

∴ b = 5, d = 2, c – a = 2

Only option 5 satisfies these criteria.

Hence, option 5.

Workspace:

**CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that all routes from S to T get the same amount of traffic, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
0, 5, 2, 2

- B.
0, 5, 4, 1

- C.
1, 5, 3, 3

- D.
1, 5, 3, 2

- E.
1, 5, 4, 2

Answer: Option D

**Explanation** :

Since the cost of travel including toll on all routes is the same.

∴ 14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d

∴ b = 5, d = 2, c = 3 and a = 1

Hence, option 4.

Workspace:

**CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that the traffic at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
0, 5, 4, 1

- B.
0, 5, 2, 2

- C.
1, 5, 3, 3

- D.
1, 5, 3, 2

- E.
0, 4, 3, 2

Answer: Option A

**Explanation** :

If we make the cost of travelling on all the routes equal, traffic along S-B will be twice that along S-A.

But we want traffic along S-A, S-B and S-D to be the same.

As routes lead to C from both B and D, we can increase the toll at C so that the cost of travelling along S-B-C-T and S-D-C-T is more than that along the other three routes.

Now, 14 + a = 9 + a + b = 13 + d

∴ a = 0, b = 5 and d =1

Also, 7 + b + c > 14 and 10 + d + c > 14

∴ c > 3

Hence, option 1.

Workspace:

**CAT 2006 LRDI | DI - Routes & Networks**

The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 per cent of the total traffic passes through junction B. The cost incurred by the commuter travelling from point S to point T under this policy will be:

- A.
Rs. 7

- B.
Rs. 9

- C.
Rs. 10

- D.
Rs. 13

- E.
Rs. 14

Answer: Option C

**Explanation** :

If toll charges at all junctions are made 0, 100% traffic will pass through S-B-C-T. This is not possible.

If toll charges at A and B are made 0, then 100% traffic will pass through S-B-A-T. This is also not possible.

If toll charges at C and D are made 0, that at B are made Rs. 3, then the traffic will get equally divided between S-D-C-T and S-B-C-T.

Thus, the cost incurred will be Rs. 10.

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

Shown below is a layout of major streets in a city.

Two days (Thursday and Friday) are left for campaigning before a major election, and the city administration has received requests from five political parties for taking out their processions along the following routes.

Congress: A-C-D-E

BJP: A-B-D-E

SP: A-B-C-E

BSP: B-C-E

CPM: A-C-D

Street B-D cannot be used for a political procession on Thursday due to a religious procession. The district administration has a policy of not allowing more than one procession to pass along the same street on the same day. However, the administration must allow all parties to take out their procession during these two days.

**CAT 2003 LRDI - Retake | DI - Routes & Networks**

Congress procession can be allowed:

- A.
Only on Thursday.

- B.
Only on Friday.

- C.
On either day.

- D.
Only if the religious procession is cancelled.

Answer: Option A

**Explanation** :

∵ Route BD cannot be used on Thursday.

∴ BJP has to take out its procession on Friday.

∵ Route DE is common to BJP and Congress.

∴ Congress should take out its procession on Thursday.

∵ Route AB is common to SP and BJP.

∴ SP should take out its procession on Thursday.

∴ BSP and CPM have to take out their processions on Friday.

∴ Congress procession can be allowed only on Thursday.

Hence, option 1.

Workspace:

**CAT 2003 LRDI - Retake | DI - Routes & Networks**

Which of the following is not true?

- A.
Congress and SP can take out their processions on the same day.

- B.
The CPM procession cannot be allowed on Thursday.

- C.
The BJP procession can only take place on Friday.

- D.
Congress and BSP can take out their processions on the same day.

Answer: Option D

**Explanation** :

∵ Route BD cannot be used on Thursday.

∴ BJP has to take out its procession on Friday.

∵ Route DE is common to BJP and Congress.

∴ Congress should take out its procession on Thursday.

∵ Route AB is common to SP and BJP.

∴ SP should take out its procession on Thursday.

∴ BSP and CPM have to take out their processions on Friday.

From the explanation we get that,

Congress takes out a procession on Thursday and BSP on Friday.

∴ Only option 4 is not true.

Hence, option 4.

Workspace:

**Answer the following question based on the information given below.**

The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is 400, at Jyotishmati is 400, at Panchal is 700, and at Vidisha is 200. Each arrow indicates the direction of material flow through the pipeline. The flow from Vaishali to Jyotishmati is 300. The quantity of material flow is such that the demands at all these locations are exactly met. The capacity of each pipeline is 1000.

**CAT 2001 LRDI | DI - Routes & Networks**

The quantity moved from Avanti to Vidisha is

- A.
200

- B.
800

- C.
700

- D.
1000

Answer: Option D

**Explanation** :

Demand at Panchal = 700

∴ Quantity of material flowing in the pipeline from Jyotishmati to Panchal = 700

Also, Flow from Vaishali to Jyotishmati + Flow from Vidisha to Jyotishmati = Demand at Jyotishmati + Demand at Panchal

∴ 300 + Flow from Vidisha to Jyotishmati = 400 + 700 = 1100

∴ Flow from Vidisha to Jyotishmati = 1100 − 300 = 800

Now, Flow from Avanti to Vaishali = Demand at Vaishali + Flow from Vaishali to Jyotishmati

∴ Flow from Avanti to Vaishali = 300 + 400 = 700

Also, Flow from Avanti to Vidisha = Demand at Vidisha + Flow from Vidisha to Jyotishmati

∴ Flow from Avanti to Vidisha = 200 + 800 = 1000

The diagram above shows the flow of the material in the pipelines and the demand at the respective points.

∴ Quantity of material moved from Avanti to Vidisha = 1000

Hence, option 4.

Workspace: