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Capgemini Previous Year Questions with Study Materials

Prasun Das| August 30, 2022 at 10:00 AM | 11 minutes Read

Capgemini is a prominent global leader in consulting, digital transformation, technology, and other engineering services. In the rapidly evolving world of cloud, digital, and platforms, the group is at the forefront of innovation in order to address the full spectrum of customer prospects.

The Capgemini interview kicks off with the following syllabus (the following syllabus is based on previous years patterns only and if format changes it will be a slight changes and most of the questions will be coming from previous year questions only).


These are the sections that appeared in the Capgemini test in the previous year-

Pseudo Code



  1.  Understanding of programming logic (tested through Pseudocode Questions)
  2. C
  3. C++
  4. Java
  5. Data Structures 
  6. Object-Oriented Programming (OOPS).
  7. DBMS
  8. Computer Networks
  9. OS


English Communication Test



  1. Reading Comprehension
  2. Para-Jumbles
  3. Direct and Indirect Speech
  4. Active and Passive Voice
  5. Sentence correction and completion


Behavioral Test



  1. This round (also referred to as the Adaptive Employee Personality Test or ADEPT-15) is a personality psychometric which is employed to assess work-related behavior. 


Game Test



  1. Deductive Logical Thinking
  2. Inductive Logical Thinking
  3. Grid Challenge
  4. Motion Challenge


According to the new syllabus for 2022, we predict the exam will be in the following format:




  1. Pseudo Code MCQs
  2. Data Structure Algorithm MCQs
  3. English communication MCQ
  4. Computer Fundamentals MCQs
  5. Database Management System Basics
  6. Cloud Computing Basics
  7. Computer Networking Basics
  8. Cognitive Games MCQs



In this blog we will focus on pseudo code DSA and computer fundamentals and English only.





Sample Questions


PSEUDO CODE:


Q1.Determine the output of the following pseudocode.



Integer arr[]={10, 20, 30, 40, 5}
Integer a, s
Set s = 0Set a = arr[1] + arr[2]
Print a


A) 25

B) 5

C) 50

D) 40


Ans: C


Explanation:

There is an array of integers arr[]={10,20,30,40,5}. There are two variables a and b declared. . The value initialized for s is 0. On the next line adding the 1st index value 20 and 2nd index value 30 arr[1] + arr[2]( 20+30), the answer is 50 will be stored in a. Finally, printing the updated values of a which now equals to 50.





Q2.Which of the following series will be printed by the given pseudocode?



Integer i, j, k, n
Set j=1, k=1for(each i from1to5)
  print k
  j=j+1
  k=k+j
endfor


A) 1 3 6 10 15

B) 1 2 3 4 5

C) 2 4 6 8 10

D) 1 1 2 3 5


Ans: A 


Explanation:

There are four variables i, j,k, and n declared. The value initialized for j is 1 and k is 1.

For loop, i value starts from 1 loop will run till the i<5, In the first iteration i value, is 1, printing k value is 1. Next line j value will be incremented by 1 (1+1) =>2. On the next line adding k and j (1+2), then the answer is 3.

2nd iteration i value will be incremented by 1, i=2. Print k, the updated k value is now 3. on the next line j value will be incremented by 1 (2+1) =>3. On the next line adding k and j (3+3) , then the answer is 6.

3rd iteration i value will be incremented by 1, i=3. Print k, the updated k value is now 6. on the next line j value will be incremented by 1 (3+1) =>4. On the next line adding k and j (6+4) , then the answer is 10.

4th iteration i value will be incremented by 1, i=4. Print k, the updated k value is now 10. on the next line j value will be incremented by 1 (4+1) =>5. On the next line adding k and j (10+5) , then the answer is 15.

5th iteration i value will be incremented by 1, i=5. Print k, the updated k value is now 15. Next line j value will be incremented by 1 (5+1) =>6. On the next line adding k and j (15+6), then the answer is 21.

Here, the for loop condition comes out to be false, hence it comes out of the for loop.

The output of Pseudocode is 1 3 6 10 15.



Q3. Find the output of the following pseudo-code:



Integer x,y,z;
x=0
y=1
x = y = z = 8
Print x


Options:

A) 0

B) 8

C) 1

D) None of the above


Correct Answer: Option B

Explanation: In this question, the value of x is initialized as 0 and y as 1 in the beginning. Later the value 8 gets assigned to the variable z and then the value of z is assigned to the variable y and the value of y is assigned to the variable x simultaneously. Finally, the value of x is updated as 8.


Q3.Find the output of the following pseudo-code:



 Integer value, n
 Setvalue = 1, n = 45while(valuelessthan equal to n)
 value = value << 1endloop
 Print value


Options:

A) 64

B) 32

C) 45

D) None of the above


Correct Answer: Option A

Explanation: Here, the left shift operation pushes the values towards the left once; when 1 is left-shifted the value that we will obtain will always be 2 to the power something. When one is converted in the beginning and not shifted the value will be 2^0 which is 1. The next iteration pushes one place towards the left, therefore the value now will be 2^1 which equals to 2; this goes on happening until the value stored is greater than 45. The value which is greater than 45 in 2 powers is 64. Now the loop gets terminated and the last value stored in the variable is 64 and the same will be printed as output.



Q4.Find the output of the following pseudo-code:



 Integer c, d
 Set c = 15, d = 12
 d = c – 1Print c //line
 c = d + (c – 2)
 if(c < 40)
   Goto line
 end if


Options:

A) 14 26 38

B) 27 39

C) 15 27 39

D) None of the above


Correct Answer: Option C

Explanation: c and a is initialized, and d as well; line 5 we are re-initializing c in every iteration of goto; the goto loop terminates only when the value of c is greater than 40. Now the last value stored in c which breaks the if condition is 51 and all the numbers 15 27 and 39 respectively will be printed according to the algebraic expression in line number 5.



Q5. Find the output of the following pseudo-code if x= 4 and y=5:



 Integer fun(intx, inty)
   if(x > 1)
     fun(x – 2, y + 2)
   end ifprinty
 End function fun()


Options:

A) 4 5 6

B) 7 6 5

C) 9 7 5

D) None of the above


Correct Answer: Option C

Explanation: the first reverse recursion would print 9 and returns to a previous function call, next it prints 7 and returns to the very first function call finally it prints a 5 and completes the execution.



Q6.What will be the output of the given pseudocode?



Integer x, y, z, a
Set x = 2, y = 1, z = 5
a = (x AND y) OR (z + 1)
Print a


Options:

A) 5

B) 3

C) 2

D) 1


Correct Answer: Option D

Explanation: AND gate works when both the conditions are true so the first (x and y) condition will be taken as true as both the inputs are true. OR gate works when any of its conditions are true, where it has already got any one of its input as true so without further checking the other input it will directly assign the value as 1.


By this point you should be able to work up some solutions itself. Here are some practice questions based on pseudo code and concept of C language.



Q7.  Find the output of the following pseudo code




  1. 2 2 3 4
  2. 2 3 4 5
  3. 2 2 2 2
  4. None of the above


Answer: Option A



Q8.Find the output of the following pseudo code




  1. 129, a
  2. 97 , X
  3. 32, a
  4. 32, X


Answer: Option A





Computer Fundamentals and Data Structures:


Q9. What is the functionality of the following code?




  1. Inserting a node at the very beginning of the list
  2. Deleting a node at the very beginning of the list
  3. Inserting a node at the very end of the list
  4. Deleting a node at the very end of the list


Answer: Option C




Q10.Recursion uses more memory space than iteration. Which of the following options is/are the valid reason for the same?

A. It uses the stack instead of a queue

B. Every recursion call has to be stored

Choose the correct answer from the following options.

A) Only A

B) Both A and B

C) Neither A nor B

D) Only B


Correct Answer: Option A

Explanation: We know that the Recursive Functions use the stack as the memory space technique. And note that the call has to be present to store the returned values.


Q11. Each node of the graph is represented as a _______?

A) Vertex

B) Root

C) Path

D) Edge


Correct Answer: Option A

Explanation: Every node cannot be a root node and path; the edge is related to the directions of the graph.



Q12. Identify the layer which provinces service to the user from OSI model



  1. Session Layer
  2. Application Layer
  3. Presentation Layer
  4. Physical Layer 

Answer: Option A

Reference: https://www.pephub.tech/blogs/computer-networking-cheatsheet-last-minute-notes-3a2ae987




Q13. Consider a source computer(S) which is transmitting a file of size 106 bits to a destination computer(D)over a network of two routers (R1 and R2) and three links(L1, L2, and L3). L1connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Now, let each link be of length 100 km. Assume that the signals travel over each link at a speed of 108 meters per second. Also assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Hence, find the total sum of transmission and propagation delays in transmitting the file from S to D?




  1. 1005 ms
  2. 1010 ms
  3. 3000 ms
  4. 3003 ms



Answer: Option A


Explanation: 

The propagation delay to travel from S to R1 = (Distance) / (Link Speed) = 10^5/10^8 = 1ms;

Total propagation delay to travel from S to D = 3*1 ms = 3ms.


Total Transmission delay for 1 packet = 3 * (Number of Bits) / Bandwidth = 3*(1000/10^6) = 3ms.


The first packet will take 6ms to reach D. While the first packet was reaching D, the other packets must have been processing simultaneously. So D will receive remaining packets 1 packet per 1 ms from R2. So logically the remaining 999 packets will take 999 ms.

Thus, total time will be equal to 999 + 6 = 1005 ms




Q14.Which of the following IP addresses can be used as (a) loop-back addresses?



  1. 0.0.0.0
  2. 127.0.0.1
  3. 255.255.255.255
  4. 0.255.255.255


Answer: (b) 127.0.0.1


Explanation: A loopback address is a special IP address whose IP address is exclusively between 127.0.0.1 to 127.255.255.255. It is reserved for loopback. Note that it doesn't require a physical connection to a network.


Q15. Choose among the following which is related to services provided by the cloud



  1. Reliablity
  2. Sourcing
  3. Ownership
  4. AaaS


Answer: B Souring


Cloud sourcing is a typical arrangement in which a company pays a third-party cloud hosting provider to deliver and support IT services that could have been provided in-house.

Salesforce is a good example as their entire suit of CRM applications are offered via the cloud.


16. What does “elasticity” in cloud computing refers to?



  1. Ability to scale down
  2. Parallel
  3. Ability to scale up
  4. Both A and C are correct


Answer: Option D

Cloud Elasticity is the property of a cloud to grow or shrink capacity for CPU, memory, and storage resources to adapt to the changing demands of an organization.


What is scalability and elasticity?

Scalability is defined as the ability of the system to accommodate larger loads just by adding resources either making the hardware stronger (scale up) or adding some additional nodes (scale out). Elasticity is the ability to fit the resources needed to cope with loads dynamically usually in relation to scale out.



Q17. Paas Stands for?



  1. Platform as a Service
  2. Platform as a software
  3. Parallel as a Service
  4. None

Answer: Option A


18.If you are using a Depth-first search (DFS) for traversing an unweighted graph, then which of the following will happen?

1. It produces the minimum spanning tree

2. It produces all pair shortest path tree


Choose the correct answer from the given options:

A) Both 1 and 2 are true

B) Both 1 and 2 are false

C) Only 2 is true

D) Only 1 is true


Ans: D

Explanation:

Depth-first search (DFS) for traversing an unweighted graph, will produce the minimum spanning tree.

Only a Depth-first search (DFS) for traversing a weighted graph, will produce the all pair shortest-path tree.




English communication

Q19. 



  1. I get a lot of invitations from friends.
  2. I have an open mind towards what other people might say.


Options:



  1. Agree to Statement 1
  2. Slightly Agree to Statement 2
  3. Slightly Agree to Statement 1
  4. Agree to Statement 2


Answer: Option B



Q20. 



  1. I like to live a fast-paced life.
  2. I like to have a balanced pace life


Options:



  1. Agree to Statement 2
  2. Slightly Agree to Statement 1
  3. Slightly Agree to Statement 2
  4. Agree to Statement 1


Answer: Option B




Q21.



  1. I try not to be involved in more than one thing.
  2. I try to do multiple things at a time


Options:



  1. Agree to Statement 1
  2. Agree to Statement 2
  3. Slightly Agree to Statement 2
  4. Slightly Agree to Statement 1


Answer: Option B


Q22. 



  1. When talking with someone, I encourage them to speak.
  2. I smile at everyone I see.


Options:



  1. Agree to Statement 1
  2. Agree to Statement 2
  3. Slightly Agree to Statement 1
  4. Slightly Agree to Statement 2


Answer: Option A


Q23. Mark the option which best expresses the sentence in Passive voice.

The watchman opened the school building's entrance.


A.The school building entrance was opened by the guard.

B.The school building’s entrance was opened by the guard.

C.The guard has opened the school building’s entrance.

D.The guard opens the school building’s entrance.


Answer: option B


Q24. Choose the option which is synonymous with the meaning of the word: Angry

A. Furious

B. Pleased

C. Precipitous

D. Hasty


Answer: A


Explanation: Angry means feeling or showing strong annoyance, displeasure or hostility. 

Meaning of Furious is extremely angry, pleased means feeling or showing pleasure and satisfaction, Precipitous means steep or high and the meaning of Hasty is acting with excessive speed. So looking at the meaning of different words “Furious” is the correct answer.



Q25. Choose the option which best expresses the below-given sentence in Active/Passive voice

Thrice a month, Rajesh cleans the room.

A. Thrice a month, the room is cleaned by Rajesh

B. Thrice a month, the room is being cleaned by Rajesh

C. Thrice a month, the room was cleaned by Rajesh

D. None of the above

Answer: A

Explanation: The given sentence is in active voice, we need to convert it into passive voice.

“Thrice a month, Rajesh cleans the room” is in the Simple present tense. Now to convert it to passive voice, the past participle form of the verb has to be used. The object of the active verb becomes the subject of the passive verb.


Q26. Choose the correct preposition and fill in the blanks:

The father said to his child, “You must be back ____________ eight o’clock.”

A. in

B. on

C. by

D. to

Answer: C

Explanation: Preposition ‘by’ is used to denote: not later than the time mentioned; before.


Q27. Choose the correct preposition and fill in the blanks:

Yesterday, he worked out at the gym from 8PM ________9PM.

A. since

B. for

C. till

D. before

Answer: C

Explanation: Till is used to denote time “up to”.


Q28. Ratan Tata _____ his car from office to home everyday.



  1. Drives
  2. Driving
  3. Drove 
  4. None of the above.


Answer: A




Study Resources

Picking up and learning computer fundamentals and cloud computing right before the interview can be a bit tricky. Below are some resources which would help:


Cloud Computing:

Basics cheatsheet: https://www.studocu.com/in/document/rajasthan-technical-university-kota/cloud-computing/cloud-computing-cheat-sheet/26618706

Cloud computing architecture: https://www.javatpoint.com/cloud-computing-architecture

Types of cloud: https://www.javatpoint.com/types-of-cloud

Cloud Service Models: https://www.javatpoint.com/cloud-service-models

Basics of Virtualisation : https://www.javatpoint.com/virtualization-in-cloud-computing


Computer Fundamentals:

Computer Network cheatsheet: https://www.pephub.tech/blogs/computer-networking-cheatsheet-last-minute-notes-3a2ae987

OS CheatSheet: https://www.pephub.tech/blogs/operating-system-cheatsheet-last-minute-notes-ded81d0b


DBMS and SQL:

DBMS PART 1: https://www.pephub.tech/blogs/database-management-system-last-minute-notes-part-1-61539f94

DBMS PART 2: https://www.pephub.tech/blogs/database-management-system-last-minute-notes-part-2-5117edcf

SQL Cheatsheet : https://www.pephub.tech/blogs/sql-cheatsheet-f93f9475

SQL Interview Questions: https://www.pephub.tech/blogs/sql-interview-questions-faa364a5

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