# 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.

**CAT 2022 LRDI Slot 1 | DI - Routes & Networks**

Given above is the schematic map of the metro lines in a city with rectangles denoting terminal stations (e.g. A), diamonds denoting junction stations (e.g. R) and small filled-up circles denoting other stations. Each train runs either in east-west or north-south direction, but not both. All trains stop for 2 minutes at each of the junction stations on the way and for 1 minute at each of the other stations. It takes 2 minutes to reach the next station for trains going in east-west direction and 3 minutes to reach the next station for trains going in north-south direction. From each terminal station, the first train starts at 6 am; the last trains leave the terminal stations at midnight. Otherwise, during the service hours, there are metro service every 15 minutes in the north-south lines and every 10 minutes in the east west lines. A train must rest for at least 15 minutes after completing a trip at the terminal station, before it can undertake the next trip in the reverse direction. (All questions are related to this metro service only. Assume that if someone reaches a station exactly at the time a train is supposed to leave, (s)he can catch that train.)

If Hari is ready to board a train at 8:05 am from station M, then when is the earliest that he can reach station N?

- A.
9:06 am

- B.
9:01 am

- C.
9:13 am

- D.
9:11 am

Answer: Option D

**Explanation** :

A train leaves M every 10 minutes starting at 6 am.

If Hari reaches M at 8:05, he can catch the train that leaves at 10 am.

Time taken by Hari to reach N:

There are total 20 stations from M till N, hence total travelling time = 20 × 2 = 40 minutes.

There are 2 junctions, hence stoppage time = 2 × 2 = 4 mins

There are 17 other stations, hence stoppage time = 17 × 1 = 17 mins

∴ Total time taken = 40 + 4 + 17 = 61 mins.

∴ Hari reaches N at 8:10 am + 61 minutes = 9:11 am.

Hence, option (d).

Workspace:

**CAT 2022 LRDI Slot 1 | DI - Routes & Networks**

If Priya is ready to board a train at 10:25 am from station T, then when is the earliest that she can reach station S?

- A.
11:12 am

- B.
11:07 am

- C.
11:28 am

- D.
11:22 am

Answer: Option A

**Explanation** :

Priya can go from T to S via R or V.

**Case 1:** T – R – S

The first train to depart T will be at (6 am + 9 + 2 + 2 =) 6:13 am.

∴ Now a train will depart T every 15 minutes, hence the train which Priya can catch for T → R is at 10:28.

Time taken to reach R = 12 + 3 = 15 mins.

∴ Priya reaches R at 10:43 am.

The first train to depart R will be at (6 am + 8 + 3 + 2 =) 6:13 am.

∴ Now a train will depart R every 10 minutes, hence the train which Priya can catch for R → S is at 10:43 itself.

Time taken to reach S = 20 + 9 = 29 mins.

∴ Priya reaches S at 11:12 am.

**Case 2:** T – S – R

The first train to depart T will be at (6 am + 8 + 3 + 2 =) 6:13 am.

∴ Now a train will depart T every 10 minutes, hence the train which Priya can catch for T → V is at 10:33 itself.

Now time taken from T → V is 29 mins and time taken from V → S is 15 mins.

Even if Priya does not have to wait at V she can S latest by 10:33 am + 29 + 15 = 11:17 am.

∴ Priya can reach S latest by 11:12 am.

Hence, option (a).

Workspace:

**CAT 2022 LRDI Slot 1 | DI - Routes & Networks**

Haripriya is expected to reach station S late. What is the latest time by which she must be ready to board at station S if she must reach station B before 1 am via station R?

- A.
11:35 pm

- B.
11:49 am

- C.
11:39 pm

- D.
11:43 pm

Answer: Option C

**Explanation** :

Workspace:

**CAT 2022 LRDI Slot 1 | DI - Routes & Networks**

What is the minimum number of trains that are required to provide the service on the AB line (considering both north and south directions)?

Answer: 8

**Explanation** :

A train leaves A for B and B to A every 15 minutes starting from 6 am.

Total time taken by a north-south train:

There are total 10 stations from M till N, hence total travelling time = 10 * 3 = 30 minutes.

There are 2 junctions, hence stoppage time = 2 * 2 = 4 mins

There are 7 other stations, hence stoppage time = 7 * 1 = 7 mins

∴ Total time taken = 30 + 4 + 7 = 41 mins.

The train which starts from A to B at 6 am reaches B at 6:41 am. It needs to rest for 15 minutes before it can start the journey back to A i.e., it can start it’s journey at 6:56 or after that.

As per schedule north-south trains leave every 15 minutes starting from 6 am. Hence, the train that starts from A to B at 6 am, can start its backward journey at 7 am.

∴ There have to be trains starting at 6 am, 6:15 am, 6:30 am and 6:45 am before trains coming from A can start their journey back.

Hence, we need at least 4 trains from B to A and similarly from A to B.

∴ We need at least 8 trains on the route AB.

Hence, 8.

Workspace:

**CAT 2022 LRDI Slot 1 | DI - Routes & Networks**

What is the minimum number of trains that are required to provide the service in this city?

Answer: 48

**Explanation** :

Consider the solution to previous question.

We need at least 8 trains on AB route as well as 8 trains on CD route.

∴ 16 trains on north-south routes.

Total time taken by a east-west train = 61 mins (calculated in first question of this set)

A train going from M to N at 6 am will reach N at 7:01 and will be ready for journey back at 7:16 am. A train starts from N every 10 mins. Hence, this train can start its journey at 7:20 am.

∴ There have to be trains starting at 6 am, 6:10 am, 6:20 am, 6:30 am, 6:40 am, 6:50 am, 7 am and 7:10 am before trains coming from M can start their journey back.

Hence, we need at least 8 trains from N to M and similarly from M to N.

∴ We need at least 16 trains on the route MN and 16 trains on PQ route

∴ 32 trains on east-west routes.

⇒ we need at least 16 + 32 = 48 trains.

Hence, 48.

Workspace:

**Answer the next 5 questions based on the information given below:**

Every day a widget supplier supplies widgets from the warehouse (W) to four locations – Ahmednagar (A), Bikrampore (B), Chitrachak (C), and Deccan Park (D). The daily demand for widgets in each location is uncertain and independent of each other. Demands and corresponding probability values (in parenthesis) are given against each location (A, B, C, and D) in the figure below. For example, there is a 40% chance that the demand in Ahmednagar will be 50 units and a 60% chance that the demand will be 70 units. The lines in the figure connecting the locations and warehouse represent two-way roads connecting those places with the distances (in km) shown beside the line. The distances in both the directions along a road are equal. For example, the road from Ahmednagar to Bikrampore and the road from Bikrampore to Ahmednagar are both 6 km long.

Every day the supplier gets the information about the demand values of the four locations and creates the travel route that starts from the warehouse and ends at a location after visiting all the locations exactly once. While making the route plan, the supplier goes to the locations in decreasing order of demand. If there is a tie for the choice of the next location, the supplier will go to the location closest to the current location. Also, while creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse. If both paths are available (direct and via warehouse), the supplier will choose the path with minimum distance.

**CAT 2022 LRDI Slot 2 | DI - Routes & Networks**

If the last location visited is Ahmednagar, then what is the total distance covered in the route (in km)?

[**Note**: There is an ambiguity in this question and hence was discarded by IIM Bangalore.]

Answer: 35

**Explanation** :

This questions was discarded by IIM Bangalore.

A cannot be the last city to be visited while satisfying all the conditions given in the caselet.

**Explanation**:

Demand

A – 50 (40%); 70 (60%)

B – 40 (30%); 60 (70%)

C – 70 (30%); 100 (70%)

D – 30 (40%); 50 (60%)

For Ahmednagar to be last, it should have the least demand of the 4 cities.

⇒ The only way Ahmednagar’s demand can be the least of the 4 cities is when its demand is 50.

Now, demand of all other cities should be greater than or equal to 50.

⇒ Demand at

B = 60

C = 70 or 100

D = 50

∴ Sequence of cities according to demand will be C → B → D → A

Distance travelled from

Warehouse → C = 12

C → B = 4

B → W → D = 12

D → W → A = 7 [shortest route from D to A is through Warehouse and not the direct route]

∴ Total distance travelled = 12 + 4 + 12 + 7 = 35.

**Ambiguity**: There is some ambiguity in this question. Once you reach B, demand at both A and D is same (i.e., 50). You would go the nearest of A and D which is A and hence A cannot be the last city to be visited then.

Hence, this question was discarded.

**Note**: The answer given by IIM-B in the cadidate response sheet was 35.

Workspace:

**CAT 2022 LRDI Slot 2 | DI - Routes & Networks**

If the total number of widgets delivered in a day is 250 units, then what is the total distance covered in the route (in km)?

Answer: 38

**Explanation** :

Demand

A – 50 (40%); 70 (60%)

B – 40 (30%); 60 (70%)

C – 70 (30%); 100 (70%)

D – 30 (40%); 50 (60%)

Maximum demand possible = 70 + 60 + 100 + 50 = 280

Actual demand is 250. This is possible only when demand at C is 70 instead of 100.

∴ Actual demands at various cities is:

A → 70

B → 60

C → 70

D → 50

Sequence of cities visited is: A → C → B → D

[A is closer to warehouse than C, hence first city to be visited will be A.]

∴ Total distance travelled = 5 + 17 + 4 + 12 = 38.

Hence, 38.

Workspace:

**CAT 2022 LRDI Slot 2 | DI - Routes & Networks**

What is the chance that the total number of widgets delivered in a day is 260 units and the route ends at Bikrampore?

- A.
7.56%

- B.
10.80%

- C.
17.64%

- D.
33.33%

Answer: Option A

**Explanation** :

Demand

A – 50 (40%); 70 (60%)

B – 40 (30%); 60 (70%)

C – 70 (30%); 100 (70%)

D – 30 (40%); 50 (60%)

For route to end at B, B should have least demand i.e., 40.

Total demand is 260, hence demand at other cities should be higher of the two values.

∴ Demand at A = 70 (60%)

Demand at B = 40 (30%)

Demand at C = 100 (70%)

Demand at D = 50 (60%)

∴ Required possibility = 60% × 30% × 70% × 60%

= 0.6 × 0.3 × 0.7 × 0.6 = 0.0756 = 7.56%

Hence, option (a).

Workspace:

**CAT 2022 LRDI Slot 2 | DI - Routes & Networks**

If the first location visited from the warehouse is Ahmednagar, then what is the chance that the total distance covered in the route is 40 km?

- A.
3.24%

- B.
5.4%

- C.
18%

- D.
30%

Answer: Option C

**Explanation** :

Demand

A – 50 (40%); 70 (60%)

B – 40 (30%); 60 (70%)

C – 70 (30%); 100 (70%)

D – 30 (40%); 50 (60%)

If first city visited is Ahmednagar, this is possible when A’s demand is highest. This is only possible when A’s demand is 70.

∴ Demand at C should be 70

Demand at B = 40 or 60

Demand at D = 30 or 50

∴ Sequence of cities can be

A → C → B → D: distance travelled = 38 kms

A → C → D → B: distance travelled = 40 kms

∴ Demand at D ≥ B

⇒ Demand at D = 50 (60%) and demand at B = 40 (30%)

⇒ Required possibility = 60% × 30% = 18%

Hence, option (c).

Workspace:

**CAT 2022 LRDI Slot 2 | DI - Routes & Networks**

If Ahmednagar is not the first location to be visited in a route and the total route distance is 29 km, then which of the following is a possible number of widgets delivered on that day?

- A.
210

- B.
250

- C.
200

- D.
220

Answer: Option A

**Explanation** :

Demand

A – 50 (40%); 70 (60%)

B – 40 (30%); 60 (70%)

C – 70 (30%); 100 (70%)

D – 30 (40%); 50 (60%)

If A is not the first city to be visited, the first city will have to be C.

Distance travelled from Warehouse to C = 12 kms.

∴ To visit the remaining 3 cities, distance travelled should be (29 – 12 =) 17 kms.

There are two possibilities for this.

Case 1: W → C → B → A → D

Here,

highest demand is from C i.e., 70 or 100

2^{nd} highest demand is from B i.e., 60

3^{rd} highest demand is from A i.e., 50

4^{th} highest demand is from D i.e., 30

Total widgets delivered can be 210 or 240

Case 2: W → C → D → A → B

Here,

highest demand is from C i.e., 70 or 100

2^{nd} highest demand is from D i.e., 50

3^{rd} highest demand is from A i.e., 50

4^{th} highest demand is from B i.e., 40

Total widgets delivered can be 210 or 240.

[**Note**: shortest route from A to D or vice-versa is through the warehouse.]

Hence, option (a).

Workspace:

**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:

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