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1 GATE 2021 Computer Science solved Question Paper

GATE 2021 Computer Science solved Question Paper

GATE 2021 Computer Science Exam has been successfully conducted on 13 February 2021. There were two question papers.

First GATE 2021 CSE Solved Question Paper was conducted in the morning shift f9:30 am12:312:30 pm. THE second GATE 2021 CSE Solved Question Paper was conducted in the second shift f2:002:00 pm:00 5:00 pm today. In this post, we will discuss the questions asked in GATE CSE 2021 Question Paper which has been conducted in Morning Shift and GATE 2021 CSE Question Paper Answer Key. These questions are Memory Based.

The official Answer Key of the GATE 2021 CSE Questions Paper is expected, and the response sheet is expected to upload at Gate 2021 Official Website.

GATE 2021 CSE Solved Questions Paper Subject Wise Score

Weightage of Each subject in GATE 2021 CSE Question Paper First is as shown in following table. You can check the GATE 2021 CSE Syllabus.

GATE 2021 CSE Solved Questions Subject Wise

These questions are asked in GATE 2021 Computer Science Questions Paper are based on memory. GATE 2021 Computer Science Subject wise Questions and GATE 2021 CSE Question Paper Answer Key are as follow –

GATE 2021 Data Structure and Algorithm Questions

Data Structure and Algorithm based questions asked in the GATE 2021 Computer Science Questions paper are as follow

Question 1.Consider the following sequence of operations and Find the value of S + Q =?

Answer: 86

Explanation: Output of the Last statement of Stack Implementation will be 62. 62 will be Pop from stack and Output of last R = Dequeue() will be 24 so S+ R will be = 62+24 =86

Note – Students can Read this Stack Tutorial to understand the concept of Stack Data Structure.

Questions 2. Which among the following take minimum numbers of comparison to sorting the array given below

23,32,44,44,54,63,72,89

(a)Selection Sort

(b) Insertion Sort

(d) Merge Sort

Answer: B

Question 3. What is the minimum weight of the Minimum Spanning Tree for the undirected Graph Given Below?

Answer: 1.6

Explanation: Minimum Spanning Tree of a Graph is a tree that consists of all the vertices of the graph and doesn’t have any cycle. When we find the minimum spanning tree of this graph using Kruskal’s or Prim’s algorithm, then there will be eight edges in the tree. Seven edges are having a weight 0.1 and one edge having weight 0.9 . So total weight of MST will be 0.1 *7+ 0.9=1.6.

Note – To Understand the Concept of MST students can read this Minimum Spanning Tree Tutorial.

Question 4 : In Binary Search Tree what is time to find an element which is smaller than maximum element.

(a) O(N)

(b) O(1)

(c)O(N log N)

(d) O( log N)

Answer : B is right option

Questions 6.Recurrence Relation based Question

Answer: Option a is correct

Gate 2021 CSE Automata Theory Questions

Questions 7. 4 DFAs diagram were given in the options and we have to find the DFA which accept the string W {0,1}* ending with substring 011.

Answer: DFA is given below the DFA shown in following picture will always accpet the string ending with 011 .

for example 011, 1011,0011,11011 etc.

Note – Students can practice previous Theory of Automata Question (TOC) GATE Questions

Question 8. If L1 is a regular language and L2 is a context Free the which among the following not necessarily will be the context free language

(a) L1 U L2

(b) L1-L2

(d) L1∩L2

Answer : B, L1-L2 will not be necessarily context free language.

GATE 2021 Computer Organization and Architecture Questions

Question 9. Main Memory Size is 232 KB. Direct Cache Mapped of 32 KB. Block Size is 64 Bytes. What is the Length of tag Field.

Answer :17 bits are required for TAG field

See the solution given below since main main memory size is 2 Pow(32) Byte and Memory is byte addressable so Address will be of 32 bits.

Since block size is 64 Byte so 6 bits are need to represent word field.

Total number of block or line will be 2 Pow(9) so 9 bits are required to represent Block field.

Remaining ( 32- (6+9)=17) 17 bits are used for TAG field.

Note – Students can read this cache mapping techniques tutorial to understand the concepts used in this question.

Note – Students can read this cache mapping techniques tutorial to understand the concepts used in this question

Questions 10. 5 stages Pipeline delay 150,140,130,160,150 nanosecond register delay of 5 nanosecond. What is time taken to execute 100 instruction on pipeline.

Answer:17160 ns

Solution is given below

Note – Students can practice some more COA Previous Year GATE Questions

GATE 2021 C Programming Questions

Question 11. Which C Library Function always invoke system function call

(a) Malloc

(b) Strlen

(d) Sleep

Answer : (a,c,d)

GATE 2021 Engineering Mathematics Questions

Questions 12. What is the largest Eigen value of the following matrix.

Answer : Largest Eigen value will be 3.

Questions 13.A Simple Planar graph having 8 vertices and 5 faces the what are the total number of edges.

Answer: Number of Edges will be 11 , Use simple planar graph formula R= E-V+2. Where R is the number of faces, E number of edges and V number of vertices in the graph.

5 = E -8+2

E=11

Questions 14.Solvethefollowingquestion

Answer :0.25

Question 15:

$gate 2021 cse question paper$

Question 16: If H is a Sub Group of G . | G | = 6

1 <= | H | < = 6

Then which among the following must be cyclic.

(A) G but not H

(B) H but not G

( D) Neither G nor H

Answer : B is right option

Question 17 : Consider A and B as given below

Which one among the following is correct

(a) Only A is tautology

(b) Only B is tautology

(d) None

Answer : a is the right option

A is tautology B is not

GATE 2021 Operating System Questions

Question 18. Consider three process A, B, C all arrived at time Zero. If burst time of process A , B and C is 10 , 20 and 16 ms respectively then what is the minimum average waiting time for non preemptive cpu scheduling algorithm.

Answer : 12 ms is minimum average waiting time.

Explanation : This questions will be solved using Non Preemptive SJF CPU scheduling. Because using non preemptive minimum waiting time will be obtained only through non preemptive SJF.

When process will be execute using Non Preemptive SJF then completion order of process will be A, C, B and their waiting time for A, C, B will be 0,10,26 respectively. So average waiting time will be (10+26=36/3=12 ms)

Question 19. Find the correct options among the following

(a) Paging incurred memory overhead

(b) Multilevel paging is necessary for pages with different size

( c ) Page size has no impact on internal fragmentation.

(d) Paging help solve the issue of external fragmentation

Answer : (a,d) are true.

Note – Students can learn the Operating System Tutorials

Gate 2021 Database Management System Questions

Question 20.A relation have 1200 tuples with 2 attributes A and B. A values range from 6 to 20. B value Range from 1 to 20 then what will be the estimated number of tuple returns as result of the following query.

Answer : ?

Question 21.Consider the following two schedules

Which among the following is correct option?

(a) Both S1 and S2 are Conflict Serializability Schedule.

(b) Only S1 is Conflict Serialiazble.

(d) Neither S1 and nor S2 is Conflict Serializable.

Answer: C, S2 is conflict serializable.

Explanation : Since we know that if we get a serial schedule by swapping the non conflicting operations in the given schedule then given schedule will be Conflict Serializable.

So in S1 we did not get serial schedule after swapping the non conflicting operations

But in Schedule S2 we get a serail schedule <T2, T1> so Schedule S2 will be conflict serializable.

To understand the concept in more detail students are requested to read this Conflict Serializability in DBMS Tutorial.

GATE 2021 Digital Electronics Questions

Question 22. Floating Point representation Based

Answer : -7.75

Question 23. Consider the number 210 in base 3 then what is corresponding Hexadecimal Number.

Answer : 15 is the right answer.

First we find 210 in base 3 will be equal to 21 in base 10. It means this 21 so When we convert 21 in Hexadecimal then We get 15.

Question 24:

GATE 2021 Compiler Design Questions

Question 25: Which among the following statements are true.

S1 : All SLR(1) Grammar are unambiguous but there are some unambiguous Grammar which are not SLR(1).

S2: There Exist a Parser with O(n ³) time complexity which can parse all CFG Grammar.

(a) ONLY S1 is true.

(b) ONLY S2 is true

(d) None is true

Answer: C

GATE 2021 Computer Networks Questions

Question 26 : Which among the following is true

I: ARP Request is Broadcasting.

II: ARP Reply is Broadcasting

(a) Only I is true.

(b) Only II is true.

(d) Neither I nor II is true.

Answer : Only I is true. ARP request message to find MAC Address corresponding to IP Address is true.

Note – Students can practice some more previous year Computer Networks GATE Questions.

General Aptitude Based Questions GATE 2021

Question 27. Consider the following Sentences

S1: Everybody in the class is prepared for the exam

S2: Rahul Invited Danish to his home because he enjoys playing chess

(a) S1 is grammatically correct and S2 is unambiguous

(b) S1 is grammatically correct and S2 is ambiguous

(d) S1 is grammatically incorrect and S2 is unambiguous

Answer : B

Question 28. In a class room ratio of boys and girls is 7 : 3. How many students are there in the class ?

(a) 21

(b) 49

(d) 15

Answer : C

Explanation

When there will be 35 Boys in the class and 15 girls then ratio of boys and girls will be 7: 3 and total students in the class will be 35+15=50.

Question 29.Consider the following statements and conclusions

Statements

S1: All Pathogens are micro organism

S2: All Bacteria are micro organism

Conclusions

I. Some pathogens are Bacteria

II. All pathogens are not bacteria

a. Both I and II are true

b. Only Conclusion I is true.

c. Only Conclusion II is true.

d. Neither conclusion I is true and nor II is true.

Answer : D

Explanation

By considering Above Statements Following Situation may be possible as shown in picture given below. In this situation neither Conlcusion I is followed nor Conclusion II is follow so Option D is right.

Question 30. For the following Question

Answer : (d) 10

Question 31.

Answer : C , is the right option.

Questions 32. Which among the following is not a Convex Polygon

Answer: B is the right answer.

Questions 33. There are five bags and each have set of 10 distinct chocolate , 1 chocolate is picked from each bag. What is the probability of having same chocolates.

Answer – 0.6976

Questions 34 : Given a M * N Rectangle of size 6 * 1. A cylinder is made by folding the shortest side of rectangle and it is closed from the both side. A cube of maximum possible area is made by cutting square from rectangle. Then what is ratio of volume of the cylinder to the volume of cube.

Answer: 1.

Conclusion and Summary

As per the feedback received from candidate participated in GATE 2021 difficulty level of GATE 2021 CSE Question Paper was high as compare to GATE 2020 CSE Question paper.

Here in this post we have discussed some questions asked in GATE 2021 CSE Question Paper conducted on 13 February 2021. These questions are memory based. GATE 2021 CSE Question Paper Answer Key and Response sheet are expected to release on 18 Feb 2021.

If you have any query or confusion is answer of any question please ask in comment section. We will try to solve your query.

I hope this post will be beneficial for the Candidate who are preparing for GATE CSE and UGC NET CS exam.

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