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Practice Test: Computer Science Engineering (CSE) - 4 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Practice Test: Computer Science Engineering (CSE) - 4

Practice Test: Computer Science Engineering (CSE) - 4 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The Practice Test: Computer Science Engineering (CSE) - 4 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 4 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Computer Science Engineering (CSE) - 4 below.
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Practice Test: Computer Science Engineering (CSE) - 4 - Question 1

Identify the correct spelling of the word.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 1

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 2

This is the place that _______

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 2

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place. 

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Practice Test: Computer Science Engineering (CSE) - 4 - Question 3

She is brave. Her brother is more brave.

Select the most suitable sentence with respect to grammar and usage.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 3

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 4

When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 4

Let number is N

N = 65 k +29

Practice Test: Computer Science Engineering (CSE) - 4 - Question 5

I was ___ ___  for the bus and then I ___ sight of Craig passing by.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 5

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 6

 – digit number greater than 5000 are randomly formed from the digits 0, 2, 3, 5 and 7. The probability of forming a number divisible by 5 when the digits are repeated is ______


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 6

For a number to be greater than 5000, d1 should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d4 should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = 99/249 = 0.397

Practice Test: Computer Science Engineering (CSE) - 4 - Question 7

It is theoretically possible that bacteria developed on Venus early in its history and that some were carried to Earth by a meteorite. However, strains of bacteria from different planets would probably have substantial differences in protein structure that would persist over time, and no two bacterial strains on Earth are different enough to have arisen on different planets. So, even if bacteria did arrive on Earth from Venus, they must have died out. 

The argument is most vulnerable to which of the following criticisms?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 7

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable. 

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 8

A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 8

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

Net gain on the sale of all the articles = 

Net gain on the sale of all the articles = 6.66%

Practice Test: Computer Science Engineering (CSE) - 4 - Question 9

Which of the following inferences can be drawn from the above graph?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 9

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 10

DRQP is a small square of side a in the corner of a big square ABCD of side A. What is the ratio of the area of the quadrilateral PBRQ to that of the square ABCD, given A/a = 3?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 10

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A2/3

Area of ΔRCB = A2/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A2 = a2 + A2/3 + A2/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

Practice Test: Computer Science Engineering (CSE) - 4 - Question 11

Given 

Assume

Then the product of the eigen values of B is________

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 11

Practice Test: Computer Science Engineering (CSE) - 4 - Question 12

If 'k' is the number of states of the NFA, how many states will the DFA simulating NFA have?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 12

If 'k' is the number of states of the NFA, it has 2k subsets of states. Each subset corresponds to one of the possibilities that the DFA must remember, so the DFA simulating the NFA will have 2states.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 13

A complete graph is a graph in which each pair of graph vertices is connected by an edge. The chromatic number of a complete graph having 100 vertices is _______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 13

The chromatic number of a complete graph is the number of vertices it has. 

Therefore, the chromatic number of a complete graph having 100 vertices is 100.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 14

A processor has 128 distinct instructions. A 24-bit instruction word has an opcode, register, and operand. The number of bits available for the operand field is 7. The maximum possible value of the general-purpose register is ______.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 14

No. of bits required for 128 instructions = log2 128 = 7

No. of bits required for operand field = 7 

No. of bits required for register field = 24 – 7 – 7 = 10

Maximum no. of registers = 210 = 1024

i.e. from 0 to 1023. Hence, maximum value = 1023

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 15

The value of 


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 15

Let 

Practice Test: Computer Science Engineering (CSE) - 4 - Question 16

Which of the following expressions is equivalent to A.B+A′.B+A′.B′ ?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 16

The given expression can be solved as:

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 17

Consider the following C function.

int strg(char *str)

{

    static int temp=0;

    if(*str!=NULL)

    {

        temp++;

        strg(++str);

    }

    else

    {

        return temp;

    }

}​

What is the output of strg(abcabcbb)?


Practice Test: Computer Science Engineering (CSE) - 4 - Question 18

The probability of a shooter hitting the target is 1/3 and three shots at the bull’s eye are needed to win the game. What could be the least number of shots for the shooter to give him more than half chance of winning the game?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 18

1 – P (x) ≥ 50%

Practice Test: Computer Science Engineering (CSE) - 4 - Question 19

Suppose a circular queue of capacity n elements is implemented using an array. Circular queue uses REAR and FRONT as array index variables, respectively. Initially, REAR = FRONT = -1. The queue is initially full with 5 elements i.e. 1 2 3 4 5. After that 3 dequeue operations are performed. What is the condition to insert an element in to the above queue?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 19

In a circular queue, the new element is always inserted at the rear position. If queue is not full then, check if (rear == SIZE – 1 && front!= 0) if it is true then set rear=0 and insert new element.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 20

Which of the following is/are correct inorder traversal sequence(s) of binary search tree?

1. 5, 2, 6, 7, 9, 11, 1, 10

2. 10, 15, 16, 23, 38, 56, 89

3. 3, 7, 9, 16, 67, 88, 98

4. 7, 1, 8, 56, 34, 66, 45

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 20

The inorder traversal of the binary search tree gives the sequence in ascending order.

Out of the given sequences, only 2 and 3 are in ascending order.

Therefore, option 2 is the correct answer.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 21

Consider the propagation delay along the bus and through the ALU is 35 ns and 120 ns respectively. It takes 18 ns for a register to copy data from the bus. The total time required to transfer data from one register to another is ______ ns


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 21

Transfer time = propagation delay + copy time

= 35 + 18 

= 53 ns

Practice Test: Computer Science Engineering (CSE) - 4 - Question 22

Consider the following set of statements:

S1: Given a context-free language, there is a Turing machine which will always halt in the finite amount of time and give answer whether language is ambiguous or not.

S2: Given a CFG and input alphabet, whether CFG will generate all possible strings of input alphabet (∑*) is undecidable.

S3: Consider three decision problems P1, P2 and P3. It is known that P1 is decidable and P2 is undecidable. P3 is undecidable if P2 is reducible to P3.

Which of the given statements is true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 22

Statements S2 and S3 are true.

Statement S1 can be corrected as: Given a context-free language, there is no Turing machine which will always halt in the finite amount of time and give answer whether language is ambiguous or not.

Practice Test: Computer Science Engineering (CSE) - 4 - Question 23

Consider a relation A with n elements. What is the total number of relations which can be formed on A which are irreflexive?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 23

A relation R on a set S is irreflexive provided that no element is related to itself; in other words, xRx for no x in S.

An irreflexive relation contains (n2 - n) elements.

Therefore, number of irreflexive relations =

Practice Test: Computer Science Engineering (CSE) - 4 - Question 24

In the IPV4 addressing format, the 214 number of networks allowed under ______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 24

In class B, size of net id = 16 bits 

Size of host id = 16 bits

In network id first two bits are reserved for leading bits i.e. 10

Hence total number of networks possible = 216 – 2 = 214

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 25

What is the output of following C code?

#include <stdio.h>

#define MUL(x) (x * x) 

int main( )

{

int i=4;

int p,q;

p= MUL(i++);

q = MUL(++i);

printf("%d", p + q); 

return 0;

}


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 25

In C, when preprocessor sees the #define directive, it goes through the entire program in search of the macro templates; wherever it finds one, it replaces the macro template with the appropriate macro expansion. In this program wherever the preprocessor finds the phrase MUL(x) it expands it into the statement (x * x).

p = 4 x 5 = 20

now i = 6

q = 8 x 8 = 64

p + q = 84

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 26

Consider a disk pack with 8 surfaces, 128 tracks per surface, 128 sectors per track and 512 bytes per sector. The number of bits required to address the sector is ______.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 26

Number of sectors  = Number of surfaces*Number of tracks per surface*Number of sectors per track

= 2* 27 * 27 = 217

Therefore, number of bits required = 17

Practice Test: Computer Science Engineering (CSE) - 4 - Question 27

Consider the implementation of Dijkstra’s shortest path algorithm on a weighted graph G(V, E) with no negative edge. If this algorithm is implemented to find the shortest path between all pair of nodes, the complexity of the algorithm in a worst-case is ______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 27

Dijkstra’s algorithm solves the single-source shortest-paths problem on a weighted, directed graph G for the case in which all edge weights are nonnegative. The time complexity of Dijkstra’s algorithm is O(E log V).

One can implement Dijkstra’s shortest path algorithm to find all pair shortest path by running it for every vertex. The time complexity of this Dijkstra’s algorithm is O(V3 log V).

Practice Test: Computer Science Engineering (CSE) - 4 - Question 28

Consider the two cascaded 2-to-1 multiplexers as shown in the figure:

The minimal sum of products form of the output X is

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 29

In a TCP connection the size of the available buffer space in the receiver is 8 and senders window size is 2. The size of the congestion window is ________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 4 - Question 29

A receiver and network can dictate to the sender the size of the sender's window. If the network cannot deliver the data as fast as they are created by the sender, it must tell the sender to slow down. In addition to the receiver, the network is a second entity that determines the size of the sender's window.

Actual window size = minimum (rwnd, cwnd)

rwnd = size of the receiver window

cwnd = size of the congestion window

2 = minimum(8, cwnd)

cwnd = 2

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 4 - Question 30

Consider two weighted complete graph G1 and G2 on the vertex set V1, V2, V3,…, V5 such that weight of the edge (Vi, Vj) is min(i,j) for the first graph and max(i,j) for the second graph respectively. The difference between the weight of a minimum spanning tree of G1 and G2 is ______.


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