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GATE Mock Test Computer Science Engineering (CSE) - 8 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - GATE Mock Test Computer Science Engineering (CSE) - 8

GATE Mock Test Computer Science Engineering (CSE) - 8 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The GATE Mock Test Computer Science Engineering (CSE) - 8 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 8 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Computer Science Engineering (CSE) - 8 below.
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GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 1

Direction: Study the following graph carefully and answer the question based on the information given below.

Percentage wise Distribution of Employees in Six Different Professions.

Total Number of Employees = 26800


What is the difference between the total number of employees in teaching and medical profession together and the number of employees in management profession?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 1

Given,

Total Number of Employees = 26800

Teaching = 15%

Medical = 27%

Management = 17%

Now,

Total percentage of employees in teaching and medical,

= 15% + 27%

= 42%

Now,

Difference between total employees in teaching and medical and total employees in management,

= (42% − 17%)  of 26800

= 25% of 26800

 
= 25/100 × 26800

= 6700

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 2

Direction: In the following question out of the four alternatives, choose the one which is best expresses the meaning of the given word.

Agitated

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 2

Agitate: Feeling or appearing troubled or nervous. e.g. "There's no point getting agitated".

Arouse: Excite or provoke (someone) to anger or strong emotions. e.g. "An ability to influence the audience and to arouse the masses".

Still: Means stable, static, stationary, constant, stagnant, etc. e.g. "The sheriff commanded him to stand still and drop the gun".

Glamour: an attractive or exciting quality that makes certain people or things seem appealing. e.g. "Gloomy forecasts about the economy".

Beginner: Means neophyte, learner, abecedarian, apprentice, fresher etc. e.g. "A person is beginner for learn a skill or take part in an activity".

Hence, the correct option is (C).

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GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 3

In the figure given below, ABCD is a square of side 4 cm. Quadrants of a circle of diameter 2 cm are removed from the four corners and a circle of diameter 2 cm is also removed. What is the area of the shaded region?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 3

Area of square = ( side )= (4)= 16 cm2

The sum of the area of four quadrants of a circle removed from the corners of the square is equal to the area of the circle of diameter 2 cm.
Radius of circle 
= Diameter/2
= 2/2 = 1 cm
Area of circle = π (radius )2 
= 22/7 cm2
∴ Area of shaded region = Area of square − 2 (Area of circle) 
= 16 − 44/7
= 68/7

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 4

Direction: Study the pie chart carefully to answer the following questions:

Pie chart represents the number of infected person from COVID - 19 in 4 cities of Maharashtra.

Total number of infected persons in Maharashtra = 30,00,000


The number of person infected from COVID-19 in Pune is what percent the number of persons infected from COVID-19 in Mumbai?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 4

Given,

Total number of Persons infected in Maharashtra =30,00,000

From the data given in the question,

= 450%
∴ The number of person infected from COVID-19 in Pune is 450% the number of persons infected from COVID-19 in Mumbai.
Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 5

Direction: In the following question, a sentence has been given with a part of it highlighted in bold. It is followed by alternatives that try to explain the meaning of the idiom/phrase given in bold. Choose the alternative which explains the meaning of the phrase correctly without altering the meaning of the sentence given in the question.

Nowadays, one gets good literary books once in a blue moon.     

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 5

The meaning of once in a blue moon is something that occurs rarely.

Example:

Peter only comes out for a drink once in a blue moon now that he has kids.

Hence, the correct option is (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 6

Find the sentence that has a mistake in grammar or usage. If you find no mistake, mark choice 4.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 6

The correct verb form is 'has broken'.

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 7

Directions: The following verbal analogy has two parts. One part is complete while the other one is incomplete. Complete the portion that is incomplete by selecting the right choice from the given options.

Son : Nuclear : : ______ : Extended

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 7

A son is part of a nuclear family, and a cousin is part of an extended family.

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 8

Directions: Choose the sentence that best combines the given sentences.

The federal government has diversity of jobs and geographic locations. The federal government offers flexibility in job opportunities that is unmatched in the private sector.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 8

The subordinator 'because' in choice 3 establishes the logical causal relationship between the subordinate and main clause. Choices 1 and 2 do not make sense. Choice 4 has faulty construction.

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 9

The first term of an AP is equal to the sum of the common ratio of a GP and the first term of the GP, which is equal to the common difference of the AP. If the sum of the first two terms of the GP is equal to the sum of the first 2 terms of the AP, then the ratio of the first term of the GP to the first term of the AP is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 9

We know that the three term of the AP are α - β, α, α + β and those of the GP are a, ar, ar2.
So, according to the quesition,

α - β = r + a = β

a + ar = 2α - β
a+ar = 2α - 2β + β
a(1 + r) = 2 (r + a) + r + a
a(1+ r) = 3 (r + a)
a + r = a(1 + r)/3
Now, the required ratio is   = a/(a + r) = (a × 3)/(a (1 + r)) = 3/(1 + r)
This is independent of the first term of the GP.

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 10

Ramu is planning to get wireless internet service at his house. Two service providers A and B offer different rates as shown in the table below.

If Ramu plans on using 25 hours of internet service per month, which of the following statements is true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 10

When used for 25 hours per month, provider A will cost $20 + 7.5 x $1 (for the hourly charge above the free hours).
This equals $27.50.
Provider B will cost $20 + 5 x $1.50 (for the hourly charge above the free hours).
This equals $20 + $7.50 = $27.50 as well.
So, choice (3) is the correct answer.

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 11

If G is the forest with 54 vertices and 17 connected components, G has _______ total number of edges.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 11

Here we are given a forest with 54 vertices and 17 components.

A component is itself a tree and since there are 17 components means that every component has a root, therefore we have 17 roots.

Each new vertex of the forest contributes to a single edge to a forest.

So, for remaining we can have m−n edges.

= 54 − 17

= 37 edges

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 12

The following problems are solved using Divide and Conquer. Match these problems to their respective recurrence relation:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 12

The correct match is: 

Hence, the correct option is (B).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 13

Let L be the language on A = {a,b,c} which consists of all words of form w = arbsct where r, s, t > 0. Which of the following is valid regular expression 'r' such that L = L(r)?

1. r = abc

2. r = aabbcc

3. r = aabcc

4. r = aabc


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 13

Statement 1:

If L = {w = arbsc∣ r,s,t ≥ 0}

Then L(r) = abc

Statement 2:

If L = {w = arbsc∣ r, s, t > 0}

Then L(r) = aabbcc

DFA for Language L:

Statement 3:

If L = {w = arbsc∣ s ≥ 0 and r, t > 0}

Then L(r) = aabcc

Statement 4:

If L = {w = arbsc∣ s,t ≥ 0 and r > 0}

Then L(r) = aabc

Hence, the correct answer is 2.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 14

Which of the following is/are the stateless protocol?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 14

As we know,

Stateless protocols are the type of network protocols in which the client sends the request to the server and the server responds back according to the current state.

It does not require the server to retain session information or status about each communicating partner for multiple requests.

Application Layer protocol: HTTP and DNS

Transport Layer protocol: TCP and UDP

Stateless protocol: HTTP, UDP, DNS

Therefore, only TCP is not a stateless transport layer protocol.

UDP is an unreliable connectionless-transport layer protocol used for its simplicity and efficiency in applications where error can be provided by the application layer.

UDP provided process-to-process communication.

Hence, the correct options are (B), (C) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 15

Consider the following DFA D.

The number of states in the minimization of D is ________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 15

Now, equivalence 0 ⟶ [q0q1q2q3] [q4]

Why q1q2q3 is in same set because 

on seeing q1 on a we get q4

q2,q3 on a also q4

on seeing q1 on b we get q2

q2 on b q1

q3 on b q2

Which arein the same set in equivalence 0 as fir a it is q4 which is in one set For b ot is q1,q2 which is also in sameset so only 3 states in q0 on as a it is going to q3which is defferent set from q4 so seperate q0 on 2nd equivalence also got same result.

The minimized DFA after combining the q1,q2 and q3 are given below.

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 16

Find the Eigen values of matrix:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 16

To find the Eigen values it satisfy the condition,

 |A − λI| = 0

= (2 − λ)((2 − λ)− 1) − 1((2 − λ) − 0) + 0

= (2 − λ)(4 + λ2 − 4λ − 1) − 2 + λ

= (2 − λ)(λ2 − 4λ + 3) − 2 + λ

= (-λ3 + 6λ2 − 11λ + 6) − 2 + λ

= −λ3 + 6λ2 − 10λ + 4

= −λ3 + 2λ2 + 4λ2 − 8λ − 2λ + 4

= −λ2(λ − 2) + 4λ(λ − 2) − 2(λ − 2)

= (λ − 2)(−λ2 + 4λ − 2)… (i)

By using Shri Dronacharya's formula for quadratic equation, we get,

 

So,

And from equation (i) we get,

⇒ x = 2

By solving the above equation, we get,

Hence, the correct option is (A).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 17

Consider a database that has the relation schema Random (A,B,C). An instance of the schema Random is as given below:


Tuple calculus expression for the above instance is given as:
{t . B∣ t ∈ r ∧ t[A] = 10 ∧ t[C] = 7}
What is the sum of the elements in the output row of the given expression?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 17

Given,

A = 10 and C = 7

{t . B∣ t ∈ r ∧ t[A] = 10 ∧ t[C] = 7}

Now,

Therefore, sum = 10 + 7 + 13 + 2
= 32
Hence, the correct answer is 32.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 18

Consider an array consisting of the following elements in unsorted order (placed randomly), but 60 as the first element.

60, 80, 15, 95, 7, 12, 35, 90, 55

Quicksort partition algorithm is applied by choosing the first element as the pivot element. How many total numbers of arrangements of array integers is possible preserving the effect of the first pass of partition algorithm.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 18

Given,

Elements in unsorted order:

60, 80, 15, 95, 7, 12, 35, 90, 55

We have to choose first element as pivot.

Here 60 is given as first element.

After first pass, the pivot element goes to it's exact location.

Here 60 goes to 6th place.

All the elements less than 60 go to left of 60 and all the elements greater than 60 go to right of 60.

After 1st  pass.


= 5! × 3!
= 720 possible arrangements.
Hence, the correct answer is 720.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 19

If a queue is implemented using two stacks. In enqueue operation, all the elements are pushed from the first stack to the second stack. In dequeue operation pop an element from 1st  stack.

Which of the following is/are true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 19

Enqueue:

While 1st  stack is not empty, push everything from 1st  to 2nd  stack.

Push an element to 1st  stack.

Push everything back to 1st  stack.

Here time complexity will be O(n).

Dequeue:

While 1st  stack is not empty then Pop an item from stack 1 and return it. Here time complexity will be O(1).

Hence, the correct options are (A) and (C).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 20

Consider the following statements:

Statement 1: An execution plan is a blueprint for evaluating a query, and is usually represented as a tree of relational operators.

Statement 2: Database management system will always be efficient as compared to file system.

Which of the statements is true or false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 20

Statement 1:

"An execution plan is a blueprint for evaluating a query and is usually represented as a tree of relational operators".

This statement is true. Since the execution plan of a query is the most efficient way of executing in that machine according to the optimizer.

Statement 2:

"A database management system will always be efficient as compared to a file system".

This statement is false. Consider the example for the operation/manipulation of text data that is not supported by DBMS then performing those operations/manipulations will be easier on the file system rather than with DBMS.

Hence, the correct options are (A) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 21

Consider the following transaction involving two bank accounts x and y. Read (x); x : = x − 50; write (x); read (y); y:  = y + 50; write (y). The constraint that the sum of the accounts x and y should remain constant is that of:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 21

Given,

Read (x); x : = x − 50; write (x); read (y);  y: = y + 50; write (y)

As we know,

Consistency in database systems refers to the requirement that any given database transaction must only change affected data in allowed ways, that is the sum of x and y must not change. Any data written to the database must be valid according to all defined rules, including constraints, cascades, triggers, and any combination thereof.

So, the constraint that the sum of the accounts x and y should remain constant is that of consistency.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 22

Horizontal micro-instructions have which of the following attributes?

  1. Short formats
  2. Limited ability to express parallel micro-operations
  3. Considerable encoding of the control information
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 22

Horizontal micro-instructions generally follow a sequential approach to specify the next microinstruction in a microprogram, similar to conventional machine language format. Each bit is identified specifically with a single control point, which indicates that the corresponding micro-operation is to be executed. Horizontal micro-instructions have the following attributes:

  1. Long formats
  2. Ability to express a high degree of parallelism
  3. Little encoding of the control information

Hence, the correct option is (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 23

Which are the very useful combinational circuits used in communication systems?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 23

Parity bit checker and Parity bit generator are very useful combinational circuits used in communication systems. A combinational circuit consists of logic gates whose outputs at any instant of time are determined directly from the present combination of inputs without regard to previous input. Examples of combinational circuits: Adder, Subtractor, Converter, and Encoder/Decoder.

It is a logic circuit that checks for possible errors in the transmission. This circuit can be an even parity checker or odd parity checker depending on the type of parity generated at the transmission end. When this circuit is used as an even parity checker, the number of input bits must always be even.

There are different types of parity generator/checker ICs are available with different input configurations such as 5-bit, 4 - bit, 9 - bit, 12 - bit, etc. One of the most commonly used and standard types of parity generator/checker IC is 74180.

Hence, the correct options are (A) and (B).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 24

Which of the following statement is/are true in the context of interpreters?

S1: Interpreters process the program according to the logical flow of control through the program.

S2: Interpreter translates and executes the error-free first instruction before it goes to the second.

S3: Interpreter processing time is less compared with the compiler.

S4: LISP and Prolog are interpreted languages.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 24

As we know,

An interpreted language is a programming language that is generally interpreted, without compiling a program into machine instructions. It is one where the instructions are not directly executed by the target machine, but instead, read and executed by some other program.

Given,

S1:True

Interpreters process the program according to the logical flow of control through the program.

S2: True

An interpreter translates and executes the error-free first instruction before it goes to the second.

Interpreter Translates and executes line by line.

S3: False

Interpreter processing time is less compared with the compiler.

The compiler processing time is lesser than Interpreter because the compiler scans and translates the whole program at a time.

S4: True

LISP and Prolog both are Interpreted languages.

Hence, the correct options are (A), (B) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 25

Any string of terminals that can be generated by the following CFG satisfy which of the given choices?

S → XY

X → aX | bX | a

Y → Ya | Yb | a

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 25

Consider the string aa,

S → XY

S → aY

S → aa

The string ‘aa' can be generated from the given grammar. This eliminates option (A) & (C).

Also,

S → XY

S → aY (X → a)

S → aYb (Y → Yb)

S → aab (Y → a)

Eliminate option (D) as string ends in b.

Hence, the correct option is (B).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 26

Disk requests come to a disk driver for cylinders in the order 11, 23, 21, 3, 41, x, 25, and 39, at a time when the disk drive is reading from cylinder 15. The seek time is 4 ms/cylinder. The total seek time, if the disk arm scheduling algorithm is first-come-first-served is 560 ms. What is/are the value of x?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 26

Given,

Seek time = t = 4 ms/cylinder.

Head = 15

Now,

Since it is a first-come-first-serve scheduling request will be served in the given sequence:

11, 23, 21, 3, 41, x, 25 and 39.

Diagram:

As we know,

Total seek time = Total head movement x t

Now,

560 = total head movement x 4

Total head movement = 140

⇒ (15 − 11) + (23 − 11) + (23 − 21) + (21 − 3) + (41 − 3) + |41 − x| + |25 − x| + (39 − 25) = 140

⇒ 4 + 12 + 2 + 18 + 38 + |41 − x| + |25 − x| + 14 = 140

⇒ 88 + |41 − x| + |25 − x| = 140

⇒ |41 − x| + |25 − x| = 52

Case 1:

⇒ 66 − 2x = 52

∴ x = 7

Case 2:

⇒ 2x − 66 = 52

∴ x = 59

Hence, the correct options are (C) and (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 27

Which of the following option is/are true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 27

Option (A): TRUE

Depth search explores edge (u, v), it is in the direction from u to v, then vis undiscovered until that time, for otherwise, the search would have explored this edge already in the direction from v to u. Thus, (u, v) becomes a tree edge. If the search explores (u, v) first in the direction from v to u, then (u, v) is a back edge.

Option (B): TRUE

An undirected graph may entail some ambiguity in the classification of edges since (u, v) and (v, u) are really the same edge. Hence, forward and cross edges never occur in a depth-first search of an undirected graph.

Option (C): FALSE

Pre-order traversal in a binary tree is similar to a depth-first traversal.

Option (D): TRUE

Suppose that a depth-first search produces a back edge (u, v) Then vertex v is an ancestor of vertex u in the depth-first forest. Thus, G contains a path from v to u, and the back edge (u, v) completes a cycle.

Hence, the correct options are (A), (B) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 28

Consider the matrix:

If the eigen values of A are 4 and 8, then __________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 28

Given,

Sum of eigen values = (A) = 2 + y

Product of eigen values = |A| = 2y − 3x

According to question,

⇒ 4 + 8 = 2 + y

⇒ 2 + y = 12

y = 10…(i)

⇒ 4 × 8 = 2y − 3x

⇒ 2y − 3x = 32…(ii)

Putting y = 10 in equation (ii),

⇒ 2y − 3x = 32

⇒ 20 − 3x = 32

⇒ 3x = −12

x = −4

So, the value of x and y are −4 and 10.

Hence, the correct option is (D).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 29

Consider the following circuit:

The minimum number of NAND gates (having 2 fan-in) require to implement the same circuit is ______. (Assume complements of each variable are already available).

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 29

Since there is a NOR-NOR realization that can be replaced with an OR-AND realization. So, a given circuit can also be drawn as:

From the above circuit diagram, we can write as

f = (A+B)(B+C).C’

f = ABC’ + BC'

f = ABC'

So, the required minimum number of NAND gates is 4.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 30

The variables of type auto, static and extern are all stored in:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 8 - Question 30

Auto variables are always local and are stored on the stack. the register modifier tells the compiler to do its best to keep the variable in a register if at all possible. Otherwise, it is stored on the stack. extern variables are stored in the data segment.
The variables of type auto, static and extern are all stored in:

Hence, the correct option is (B).

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