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

GATE Mock Test Computer Science Engineering (CSE) - 10 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) - 10 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 10 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) - 10 below.
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GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 1

Direction: Study the bar graph carefully to answer the following question.

The bar graph represents the profit earned by two individuals A and B over 5 years.

In which year, the profit percent increase of B is the highest compared to previous year?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 1

We can see that the profit increasing in year 2015 and 2017 only.

We know that,

Profit percent increase is given by:

In 2015, the profit is increasing as compared to the previous year but in 2017 profit percent increase is highest.

∴ The profit percent increase of B is the highest compared to the previous year in 2017.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 2

Direction: A sentence has been given in Direct/Indirect speech. Out of the four given alternatives, select the one which best expresses the same sentence in Indirect/Direct speech.

The traveller said, "Can you tell me the way to the nearest inn"?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 2

The given sentence is in Direct Speech.

The given sentence is in the Direct Speech because it quotes actual words. Therefore, it has to be changed into Indirect Speech.

The rule is that during conversion from direct to indirect pronouns change their forms. Example-“My” changes to “her”.

In the case of interrogative sentences, Indirect Speech is introduced by verbs like 'asked', 'inquired' etc.

Also, the actual words have been spoken by the speaker in the past, therefore narrating it in the present will require a change in the tense.

Therefore, the sentence in Indirect Speech will be an option (B) i.e. "The traveller asked if he could tell him the way to the nearest inn".

Hence, the correct option is (B).

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GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 3

The area of the region bounded externally by a square of side 2a cm and internally by the circle touching the four sides of the square is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 3

In the figure shown below, the shaded region is the region bounded externally by the square and internally by the circle,

Given,

Side of square =2a cm

Area of square =( side )2 = 4a2 cm2

∵ The circle touches the four sides of the square.

As we know,

Radius of a circle (r) = d/2

Diameter of circle = 2a cm
Area of a circle = π × r2
⇒ π × d/2 × d/2
⇒ π/4 × (d)2
= (π/4) × 4a2

= πa2 cm2

∴ Area of shaded region  =  Area of square − Area of circle = (4−π) a2 cm2

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 4

Direction: Read the data carefully and answer the following question.

The pie chart shows the percentage of total employees working in-5 companiesA,B,C,D and E. The total number of employees is 25000.

Find the average number of employees working in companies B,C and E:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 4

Average = 

Number of employees in company B = 

=5,000

Number of employees in company C = 

=3,750

Number of employees in company E = 

=5,500

Now, Average number of employees in B,C and E.

=4,750

∴ Average number of employees in B,C and E is 4,750.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 5

Direction: Given below idioms/phrases followed by four alternative meanings to each. Choose the response which is the most appropriate expression and mark your response.

Get the jitters

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 5

Let's look at the meaning of the given idiom:

Get the jitters: To experience a temporary state of nervous anxiety or anticipation.

Example: 

Lo Jill always gets the jitters before exams.

​Thus, from the explanation given above, we find that the first option is the correct choice.

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 6

There are 60 students in a class. The students are divided into three groups A, B and C of 15, 20 and 25 students, respectively. The groups A and C are combined to form group D. What is the average weight of the students in group D?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 6

There is no indication of weights of the students in the question. Therefore, it is not possible to find the relation between the weights of students in groups A, B, C, and D.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 7

The probability that a man will live 10 more years is 1/4 and the probability that his wife will live 10 more years is 1/3. The probability that neither of them will be alive in 10 years is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 7

Probability that a man will live 10 more years = 1/4
Probability that man will not live 10 more years = 1 - 1/4 = 3/4
Probability that his wife will live 10 more years = 1/3
Probability that his wife will not live 10 more years = 1 - 1/3 = 2/3
Then, probability that neither will be alive in 10 years = (3/4) x (2/3) = 1/2

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

Voting is the privilege for which wars have been fought, protests have been organized, and editorials have been written. "No taxation without representation," was a battle cry of the American Revolution. Women struggled for suffrage, as did many minorities. Eighteen year olds clamored for the right to vote, saying that if they were old enough to fight in the war, then they should be allowed to vote. Yet Americans have a deplorable voting history, and many will tell you that they have never voted.

Which of the following words is the best synonym for the word 'privilege' as used in the passage?

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

The context mentions that wars have been fought for 'voting' and 'privilege' means 'a special right'.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 9

Two trains start at the same time from two stations A and B towards each other.
They arrive at B and A respectively in 5 hours and 20 hours after they passed each other.

If the speed of the train that started from A is 56 kmph, then what is the speed of the second train (in kmph)?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 9

Let the speeds of two trains starting from station A and B be S1 and S2 respectively and the time taken by them after meeting be t1 and t2 respectively.
Then, we can use this equation to get the answer: 

S2 = 28 kmph

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

A man starts at a certain point, walks one km east, then two km north, one km east, one km north, one km east and finally one km north to arrive at the destination. What is the shortest distance (in km) from the starting point to the destination?

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

Let us assume that man starts from point O and reaches his destination at point A.

The shortest distance between O and his destination is OA = = 5 km

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 11

A certain processor uses a fully associative cache of size 16 kB. The cache block size Q.1
is 16 bytes. Assume that the main memory is byte addressable and uses a 32-bit
address. How many bits are required for the Tag and the Index fields respectively in
the addresses generated by the processor?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 11

Cache memory size = 16 kB
Block size = 16 B
Main memory address = 32 bit
Number of lines (N) = 16 K/16
⇒ 214/24
⇒ 210
Fully associative cache memory (N-way)
So, number of sets (S) = N
P-way = 210/210 = 1
∴ Address format:   

So, TAG = 28 bit
Index = 0 bit (No address)

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 12

 Let G be an undirected complete graph, on n vertices, where n > 2. Then, the number
of different Hamiltonian cycles in G is equal to

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 12

In a complete graph we can traverse the n vertices in any order and return to the starting vertex and form a Hamiltonian cycle. The number of such cycles will be n! However, since circular rotations will have to ignored. Since for example K4 with vertices {1, 2, 3, 4}, the cycle 1-2-3-4 is same as 2-3-4-1 is same as 3-4-1-2 etc. we now get only (n – 1)! distinct Hamiltonian cycles. Further, the cycle 1-2-3-4 and 1-4-3-2 are also same (clockwise and anticlockwise). So ignoring this orientation also we finally get (n-1)!/2 distinct Hamiltonian cycles which is option (d).

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 13

In 16-bit 2’s complement representation, the decimal number –28 is: 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 13

+28 ⇒ 0000 0000 0001 1100
–28 ⇒ 1111 1111 1110 0100 (2’s complement form)

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 14

Consider the following two statements about database transaction schedules:
I. Strict two-phase locking protocol generates conflict serializable schedules that are also recoverable.
II Timestamp-ordering concurrency control protocol with Thomas’ Write Rule can generate view serializable schedules that are not conflict serializable.
Which of the above statements is/are TRUE? 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 14

I. Strict 2PL guaranteed conflict serializable because of 2PL condition and also strict
recoverable.
II. Thomas Write time stamp ordering ensures serializable. Thomas write rule time
stamp ordering allowed to execute schedule which is view equal serial schedule
based on time stamp ordering

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 15

Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation
schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas
Y and Z where Y = (PR) and Z = (QRS).
Consider the two statements given below:
I. Both Y and Z are in BCNF
II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 15

X(PQRS) {QR → S, R → P, S → Q} decomposed into
Y(PR)                                                      Z(QRS)
{R → P}                                                  {QR → S, S → Q}
Candidate key : R                                  Candidate key : QR, RS
Relation Y in BCNF                               Relation Z in 3NF but not BCNF
Common attribute between Y and Z relations is R which is key for relation Y.
So that given decomposition is lossless join decomposition.
R → P in Y
QR →S
S →Q
both are in Z and dependency preserving decomposition.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 16

The difference between the eigen values of matrix  is


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 16

Characteristic equation is


Difference between the eigen values of the matrix = 12 - (-8) = 20

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 17

For what value of x, the given matrixbecomes singular?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 17

For a singular matrix, its determinant is always zero.
i.e. Determinant = 0
i.e.
4(0 – 3) – x(2  0 – 6) + 0(6 – 0) = 0
⇒ 4(–3) – x(– 6) + 0 = 0
– 12 + 6x = 0
– 12 + 6x = 0
6x = 12
x = 2

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 18

The range of signed decimal numbers that can be represented by 6-bit 1's complement number is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 18

The 1's complement number is a 6-bit number. We have to find the maximum and minimum number which can be represented by it.
When the number is negative, then the maximum value can be 111111, i.e. -(26 - 1) = (-31)10.
Here, MSB is for sign.
Maximum positive number is 0 11111, i.e. + (26- 1) = (+ 31)10.
Hence, the range of signed decimal numbers that can be represented by 6-bit 1's complement number is: -31 to +31.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 19

Which of the following is used for the purpose of link editing?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 19

Loader is a program which performs the functions of:
(1) Loading
(2) Link editing
Link editing is used to create a single program from several files of relocatable machine code.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 20

Which of the following notations satisfies the given condition?

f(n) ≤ c*g(n), where f(n) and g(n) are functions, and c is a positive constant.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 20

In O - notation, two functions f(n) and g(n) are such that f(n) is O(g(n)), if and only if there is positive constant c such that f(n) ≤ c*g(n).
O - notation is useful, when there exist only upper bound on the time complexity.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 21

Which of the following tables views the constraints in SQL?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 21

Owner constraints table, administrator constraints table and constraint type do not exist in SQL. User constraints table contains the information about each constraint in the database along with the constraint specific information.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 22

In which addressing mode is the operand given explicitly in the instruction?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 22

In immediate addressing mode, data is present in the address field of the instruction. It is designed in one address instruction format. In this mode, range of constants are limited by the size of address field.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 23

What is the post order traversal of the following?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 23

To find post order traversal,

  • scan left child of the tree
  • scan right child of the tree
  • scan the root

The postorder traversal will give the result as ACEDBHIGF.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 24

Which of the following networks cannot be an APIPA (Automatic Private IP Address)?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 24

The IP address range for APIPA is 169.254.0.1 - 169.254.255.254, with the subnet mask of 255.255.0.0.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 25

Consider a system with `m` resources of the same type being shared by `n` processes. Resources can be requested and released by processes only one at a time. The system is deadlock free if and only if

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 25

Suppose N = ∑  all Needsi, A = ∑ all Allocationsi, M = ∑ all maximum Needsi.
Assume that there is no deadlock in the given system. If there is a deadlock state, then A = m because there is only one kind of resource and it can either be requested or released at a time.
From the condition 1 and 2 we get, N + A = M < m + n. So, we get N < n. It shows that at least one process i that Needs = 0. From condition 1, one process can release at least 1 resource. Thus, there are n -1 processes sharing m resources. It is always true that no process will wait permanently. Hence, the given system is deadlock free.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 26

When a process is rolled out of memory, it loses its ability to use the CPU (atleast for a while). Describe another situation, where a process loses its ability to use the CPU, but it does not get rolled out

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 26

When an interrupt occurs the process loses the CPU, but regains it as soon as the handler completes. The process is never rolled out of memory.
Note that when timer runout occurs, the process is returned to the ready queue, and it may later be rolled out of memory. When the process blocks, it is moved to a waiting queue, where it may also be rolled out at some point.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 27

Directions: Consider the following statements.

  1. Boolean expressions and logic networks correspond to labelled acyclic digraphs.
  2. Optimal boolean expressions may not correspond to simplest networks.
  3. Choosing essential blocks first in a Karnaugh map and then, greedily choosing the largest remaining blocks to cover may not give an optimal expression.

Which of these statement(s) is/are correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 27
  • Boolean expressions and logic networks correspond to labelled acyclic digraphs. Correct
  • Optimal boolean expressions may not correspond to simplest networks. Correct
  • Choosing essential blocks first in a Karnaugh map and then, greedily choosing the largest remaining blocks to cover may not give an optimal expression. Correct

So, option (4) is correct.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 28

Which of the following statements is/are true with reference to the way of describing XML data?

  1. XML uses DTD to describe the data.
  2. XML uses XSL to describe the data.
  3. XML uses a description node to describe the data.
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 28

XML uses Document Type Definition (DTD) to describe the data. DTD is a specified document defining and constraining definition and XML uses a description node to describe the data.
Extensible Stylesheet Language (XSL) is used to transform and render the XML document.
Statements (1) and (3) are correct.
So, option (4) is correct.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 29

Recursive enumerable languages are not closed under

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 29

Recursively enumerable languages are also known as Type 0 grammars. They are not closed under set difference.
Here, T - P (where T and P are two sets) may or may not be recursively enumerable.
If T is recursively enumerable, then the complement of T is recursively enumerable if and only if T is also recursive.

GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 30

Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data is stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk respectively are:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 10 - Question 30

Number of surfaces = 16
Tracks = 16 × 128 × 256 = 24 × 27 × 28
Number of bits required by a sector = 219
So, 19 lines are required to address all the sectors.
Bytes = 219 × 512 B = 219 × 29 = 228
Bytes = 256 MB

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