Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Tests  >  GATE Computer Science Engineering(CSE) 2025 Mock Test Series  >  GATE Mock Test Computer Science Engineering (CSE) - 1 - Computer Science Engineering (CSE) MCQ

GATE Mock Test Computer Science Engineering (CSE) - 1 - Computer Science Engineering (CSE) MCQ


Test Description

30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - GATE Mock Test Computer Science Engineering (CSE) - 1

GATE Mock Test Computer Science Engineering (CSE) - 1 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) - 1 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 1 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) - 1 below.
Solutions of GATE Mock Test Computer Science Engineering (CSE) - 1 questions in English are available as part of our GATE Computer Science Engineering(CSE) 2025 Mock Test Series for Computer Science Engineering (CSE) & GATE Mock Test Computer Science Engineering (CSE) - 1 solutions in Hindi for GATE Computer Science Engineering(CSE) 2025 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free. Attempt GATE Mock Test Computer Science Engineering (CSE) - 1 | 65 questions in 180 minutes | Mock test for Computer Science Engineering (CSE) preparation | Free important questions MCQ to study GATE Computer Science Engineering(CSE) 2025 Mock Test Series for Computer Science Engineering (CSE) Exam | Download free PDF with solutions
GATE Mock Test Computer Science Engineering (CSE) - 1 - 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.

Find the difference between the profit of A and B together in 2015 and the profit of A and B together in 2017:

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

Profit of A and B together in 2015 = 6000 + 16000 = Rs. 22,000

Profit of A and B together in 2017 = 8000 + 20,000 = Rs. 28,000

Required difference = Rs. 28,000 − 22,000 = Rs. 6,000

∴ The difference between the profit of A and B together in 2015 and the profit of A and B together in 2017 is Rs. 6000

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 2

 Directions: Select the most appropriate word/phrase among the choices to complete the sentence.

You can make your quixotic experiments with someone else, I do not wish to be your _______

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 2

 'Guinea pig' is term used to refer any person or thing used in an experiment. As first part of sentence refers the experiments as quixotic (impractical or daring), so perfect word suited to this blank is 'guinea pig', which is used in labs for testing on biological experiments.

1 Crore+ students have signed up on EduRev. Have you? Download the App
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 3

Direction: Fill in the blank with the most appropriate article choosing from the options given below.
She doesn't want to study at _______ university.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 3

Article 'a' is used before a word beginning with the consonant sound of  'yu' but begins with a vowel letter. For example, a university, a union, a useful thing, etc.
Such a rule doesn't apply to 'an' and 'the'. 
'few' on the other hand, is not an article.
Therefore, given the general idea of any university here, article 'a' is to be used.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 4

If prices reduce by 20% and sales increase by 15%, what is the net effect on gross receipts?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 4

Let original price = p, and original sales = s. Therefore, original gross receipts = ps.
Let new price = 0.80p, and new sales = 1.15s. Therefore, new gross receipts = 0.92ps.
Gross receipts are only 92% of what they were. Hence, gross receipts decrease by 8%.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 5

In the figure given below, ABC is a right-angled triangle where ∠A = 90∘, AB = p cm and AC = q cm. On the three sides as diameters semicircles are drawn as shown in the figure. The area of the shaded portion, in square cm, is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 5

Given,

AB = p cm and AC = q cm

∵ Area of right-angled triangle  = 1/2 ×  Base × Height

Area of triangle 

 ABC = pq/2

Applying Pythagoras theorem in △ABC,

BC2 = AB2 + AC2

⇒ BC2 = (p2 + q2)

∵ Area of semi-circle = (π/8) × (diameter)2

Area of semi-circle with diameter AB = (π/8) × (p)2 = πp2/8 cm2

Area of semi-circle with diameter AC = (π/8) × (q)2 = πq2/8 cm2

Area of semi-circle with diameter BC = (π/8) × (p2 + q2) = π(p2 + q2)/8 cm2

∴ Area of shaded region 

= (pq/2) + (πp2/8) + (πq2/8) − [π(p2 + q2)/8]

=(pq/2)cm2

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 6

In a town, 48% people are educated, 51% people are young and 60% are servicemen. 24% are educated and young, 25% are young and servicemen, 27% are educated and servicemen and 5% have all the qualities. If the total number of persons in this town is 300, what is the ratio of those who have exactly two characteristics and those who have only one characteristic?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 6

 

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 7

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

What is the average number of people infected from COVID -19 in Nagpur and Nasik

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 7

Given,

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

From the data given in question, we have:

Required average = (7,50,000 + 6,00,000)/2

= 13,50,000/2

= 6,75,000

∴ The average number of people infected from COVID-19 in Nagpur and Nasik is 6,75,000.

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

Directions: Read the given context and answer the question that follows.

Most Reality TV shows centre on two common motivators: fame and money. The shows transform waitresses, hairdressers, investment bankers, counsellors and teachers, to name a few, from obscure figures to household names. A lucky few successfully parlay their fifteen minutes of fame into celebrity. The luckiest stars of Reality TV also reap huge financial rewards for acts, including eating large insects, marrying someone they barely know, and revealing their innermost thoughts to millions of people.

Which of the following options best supports the above paragraph?

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

The correct answer is the first option as it is stated in the passage that most Reality TV shows centre on two common motivators: fame and money.

Choices 2 and 4 are not supported by the passage because passage is about Reality TV stars and not Reality TV.

Choice 3 is incorrect because the paragraph states that some Reality TV stars manage to parlay their fifteen minutes of fame into celebrity

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 9

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.

Rose said, “I am very busy now"

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 9

Option 2: Rose said that she was very busy.

This option correctly changes the sentence from direct speech to indirect speech without changing the tense or meaning of the original sentence. It is the most appropriate and accurate representation of the given sentence.

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

The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below:

The distances covered during four laps of the journey are listed in the table below:

From the given data, we can conclude that the fuel consumed per kilometre was least during the lap

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

Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr. So, least fuel was consumed per litre in lap Q.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 11

Consider the given statements:

Statement A: All cyclic groups are abelian groups.

Statement B: The order of the cyclic group is the same as the order of its generator.

Which of these are true/false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 11

Abelian Group: Let {G=e, a, b} where e is identity. The operation 'o' is defined by the following composition table. Then(G, o) is called Abelian if it follows the following property:

  1. Closure Property
  2. Associativity
  3. Existence of Identity
  4. Existence of Inverse
  5. Commutativity

Cyclic Group: A group a is said to be cyclic if it contains an element 'a' such that every element of G can be represented as some integral power of 'a'. The element 'a' is then called a generator of G, and G is denoted by <a> (or [a]).

Theorem:

(i) All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic.

(ii) The order of a cyclic group is the same as the order of its generator. 

Thus it is clear that A and B both are true.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 12

The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp(t) =  .

It is given that C3 = 3 + j5. Then, C-3 is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 12

Here, C3 = 3 + j5
For real periodic signal,
C-k = Ck*
Thus, C-3 = 3 - j5

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 13

Consider a hard disk with 16 recording surfaces (0 − 15) having 16384  cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 13

 As we know,

Since starting on disk starts from <1200,9,40> so no. of sectors left on 9th  surface is 24.

So, on 9th  surface total storage of "12288 B" is possible.

Now, a part from 9th  surface, on cylinder no. 1200 only 6 surface is left.

To storage possible on these 6 surfaces are = 6 × 26 × 29→ Storage on each sector.

No. of sectors on each surface

= 196608 B

So, total on cylinder no. 1200, storage possible is:

⇒ 196608 + 12288

= 208896 B

So, since the file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So, we are left only with only 43615232 B.

Since in 1 cylinder, storage possible is = 24 × 26 × 29 B

= 524288/B

So, we need about = 43615232 B/524288 B

= 83.189 more cylinders

So, we'll need the [1284th Cylinder] to completely store the file. C02 after 1283rd  cylinder we will left with data which will need 189.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 14

The eigen values of the matrix A =   are

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 14

Sum of eigen values = Sum of diagonal elements (Trace) = 12
Product of eigen values = Value of determinant = 40
Now check by options, only (4) satisfies these conditions.
Hence, eigenvalues are 1, 2, 4 and 5.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 15

Assume the level of the root node in a tree is 0. We have a tree with nodes having values as 1,3,7,15,29. Then the maximum value of level number possible is _______.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 15

Given,

The values of a node of a tree are:

1,3,7,15,29

As we know,

In worst case, tree might be skewed tree and in that case, there will be 5 levels and root will be at 0. So, the levels can be like this.

1 - Level 0

3 - Level 1

7 - Level 2

15 - Level 3

29 - Level 4

The maximum value of level number possible is 4

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 16

Consider the following log sequence of two transactions on a bank account, with initial balance 12000, that transfer 2000 to a mortgage payment and then apply a 5% interest.

  1.  T1 start
  2.  T1 B old = 12000 new = 10000
  3.  T1 M old = 0 new = 2000
  4.  T1 commit
  5.  T2 start
  6.  T2 B old = 10000 new = 10500
  7.  T2 commit

Suppose the database system crashes just before log record 7 is written. When the system is restarted, which one statement is true of the recovery procedure?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 16

We must undo log record 6 to set B to 10000 and then redo log records 2 and 3 because system fails before commit operation. So, we need to undo active transactions (T2) and redo committed transactions (T1).
Note: Here we are not using checkpoints.

Checkpoint : Checkpoint is a mechanism where all the previous logs are removed from the system and stored permanently in a storage disk. Checkpoint declares a point before which the DBMS was in consistent state, and all the transactions were committed.

Recovery: When a system with concurrent transactions crashes and recovers, it behaves in the following manner:
 The recovery system reads the logs backwards from the end to the last checkpoint.
 It maintains two lists, an undo-list and a redo-list.
 If the recovery system sees a log with <Tn, start> and < Tn, Commit> or just <Tn, Commit >, it puts the transaction in the redo-list.
 If the recovery system sees a log with <Tn, start> but no commit or abort log found, it puts the transaction in undo-list.All the transactions in the undo-list are then undone and their logs are removed. All the transactions in the redo-list and their previous logs are removed and then redone before saving their logs.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 17

Which of the following choices given in the options is/are not shared by all the threads in a process?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 17

Multiple threads of the same process share other resources of process except register, stack and stack pointer. In particular, a process is generally considered to consist of a set of threads sharing an address space, heap, static data, code segments and file descriptors.

Thread of the same process doesn't share program counter (register), stack, registers

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 18

Left factor the grammar:

A → ad | a | ab | abc | b

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 18

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 19

What is the relation between DFA and NFA on the basis of computational power?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 19

On the basis of computational power, DFA and NFA are both equally powerful. Since we have an algorithm to convert NFA to DFA, and DFA is by default an NFA, both of them have equal computational power.

DFA is said to be a specific case of NFA and for every NFA that exists for a given language, an equivalent DFA also exists.

The computational power of other mechanisms:

  • Power of DFA = Power of NFA.
  • Power of NPDA (Non-deterministic Push-down Automata) > Power of DPDA (Deterministic Push-down Automata) i.e. language accepted by DPDA is a subset of the language accepted by NPDA.
  • Power of NTM (Non-deterministic Turing Machine), Multi-Tape TM and DTM (Deterministic TM) are the same since every NTM can be converted to corresponding DTM.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 20

The postfix expression for the infix expression A + B* (C + D)/F + D*E is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 20

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 21

For x > 1,(2x)2y = 4e2x − 2y, then (1 + loge⁡2x)2dy/dx is equal to:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 21

Given,

(2x)2y = 4e2x − 2y

logb⁡(Mp) = plogb⁡(M)

logb⁡(MN) = logb⁡(M) + logb⁡(N)

Taking log both sides, we get

ln⁡ (2x)2y = ln⁡ 4e2x − 2y

⇒ ln⁡ (2x)2y = ln ⁡4+ln ⁡(2x − 2y)

⇒ 2yln ⁡(2x) = ln ⁡22 + (2x−2y)

⇒ 2yln ⁡(2x) = 2ln ⁡2+(2x−2y)… (1) 

⇒ yln⁡ (2x) = ln⁡ 2+(x−y)

⇒ y(1+ln⁡ 2x) = x+ln ⁡2

∴ y = x + ln⁡ 2/1 + ln⁡ (2x)…(2)

As we know,

d/dx(ln⁡x) = 1/x

If f and g are both differentiable, then,

Differentiating equation (1) with respect to x, we get,

Putting equation (2) in equation (3), we get,

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 22

A router has the following (CIDR) entries in its routing table.

What does the router do, for each of the following IP addresses, if a packet with that address arrives?

(A) 135.46.63.10 (B) 192.53.40.7

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 22

Taking the first 22 bits of 135.46.63.10 as network address, we have 135.46.60.0. The bit pattern of 135.46.63.10 is 10000111.00101110.00111111.00001010.
When we perform the bit and operation with 22 leading bit 1s and 10 bit 0s, it is equivalent of making the last 10 bit zero.
We get the following network address bit pattern: 10000111.00101110.00111100.00000000.
The first two bytes are not changed. The 3rd type changes from 63 to 60 while the 4th byte become zero. Match with network address in the routing table.
The 2nd row matches. The router will forward the packet to Interface 1.
Taking the first 23 bits of the IP address 192.53.40.7 as network address, we have 192.53.40.0.
The packet will be forwarded to Router 1.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 23

Consider the following functional dependencies for relational schema R(PQRSTU) are given:

PQ → RS,ST→ U,R → T,U → S,PT → R and T → P

Which of the following is/are false?

Note: {X}+contains all the attributes that can be derived.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 23

As we know,

{PT}+ = {PTR}

Therefore, option (A) is false.

{ST}+ = {STUPR}

Since Q is not present in {ST}+, so, option (B) is false.

{QT}+ = {PQRSTU}

Therefore, option (C) is true.

{PQ}+ = {PQRSTU}

In option (D), S is not present.

∴ It is false.

Now,

You might feel that option (D) is also correct but in question, it is already defined that it should contain all are derived attributes.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 24

Which software is built to communicate directly on the hardware?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 24

An operating system is built to directly communicate with the hardware of the system. An OS instructs the hardware to perform intended operations and gives commands for the same.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 25

In a network two hosts, A and B are connected by an intermediate router R. Host A is connected to R through Link 1, and Host B is connected to R though Link 2. Each Link is 500 meters long and has a bandwidth of 1 Gbps. Queuing delay of R is 10 microseconds and the propagation speed 2.5 × 105 km/s. The processing delay is negligible. What is the total transfer delay of 2 KB packet from A to B in microseconds?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 25

Given,

Packet Size = 2 KB = 211 bytes = 214 bits

Bandwidth = 1 Gbps = 109 bps

Lenght of link 1= Lenght of link 2 = 500 m

Propagation speed = 2.5 × 105 km/s = 2.5 × 108 m/s

Queuing delay at R = 10 microseconds

Diagram:

As we know,

Total transfer delay = transmission delay from A to link 1+ propogation delay from A to R+ queuing delay at R+ transmission delay from R to link 2+ propagation delay from R to B

Now,

Ttrans  (from A to link 1)

 = Ttrans  (from R to link 2)

= 214 ÷ 109 bps

= 16.384 microseconds

Tprop  (from A to R)

= Tprop  (form R to B)

= (500 m) ÷ (2.5 × 108 m/s)

= 2 microseconds

Total transfer delay = 16.384+0.2+10+16.384+2

= 46.768 microseconds

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 26

Consider the instruction where operand field specifies a value of 200. At memory location 200, a value of 400 is stored. If the PC has value 300, what will be the effective address in indirect and relative addressing respectively?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 26

In indirect addressing:
Opcode 200

In relative addressing,
effective address = (200 + value of PC) = 200 + 300 = 500

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 27

Consider 4-block cache (initially empty) with the following main memory block references:
4,5,7,12,4,5,13,4,5,7
Identify the hit ratio for direct-mapped cache?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 27

Cache Memory is a special very high-speed memory. It is used to speed up and synchronize with a high-speed CPU. Cache memory is costlier than main memory or disk memory but more economical than CPU registers. Cache memory is an extremely fast memory type that acts as a buffer between RAM and the CPU. It holds frequently requested data and instructions so that they are immediately available to the CPU when needed.

The performance of cache memory is frequently measured in terms of a quantity called hit ratio.

{K MOD N = i}

4 − M;4 MOD 4 = 0

5 − M;5 MOD 4 = 1

7 − M;7 MOD 4 = 3

12 − M;12 MOD 4 = 0

4 − M;4 MOD 4 = 0

5 − H

13 − M;13 MOD 4 = 1

4 − H

5 − M;5 MOD 4 = 1

7 − H

⇒ H = 3/10

= 0.3

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 28

Let L1 be a recursive language. Let L2 and L3 be the languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 28

L1 → recursive

L2,L3 → recursively enumerable but not  recursive.

So L1 can be recursive enumerable.

RE - RE = RE

So L1 → L3 can be recursive enumerable.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 29

Consider a system having 'N' resources of the same type. X,Y, and Z are the three processes that shared the resource. The peak demand of X,Y, and Z are 5,11 and 9 respectively. What is/are the value of 'N' to ensures the system is in deadlock?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 29

Given,

Number of processes = 3

The available instance of resources = N

Max needs of process (X,Y,Z) = (5,11,9)

Now,

Deadlock can occur: needed (requested) resource > available resources

Max resource per process to be in deadlock = needed −1

Nmax = (5 − 1) + (11− 1) + (9 − 1) = 4 + 10 + 8

= 22

∴ ≤ 22 will leads to deadlock.

∴ 20,21,22 will leads to deadlock.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 30

Given A = {0, 1}, then the number of possible strings of length n is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 30

Number of possible strings = 2 x 2 x 2 x 2…..n times = 2n.

View more questions
55 docs|215 tests
Information about GATE Mock Test Computer Science Engineering (CSE) - 1 Page
In this test you can find the Exam questions for GATE Mock Test Computer Science Engineering (CSE) - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for GATE Mock Test Computer Science Engineering (CSE) - 1, EduRev gives you an ample number of Online tests for practice

Top Courses for Computer Science Engineering (CSE)

Download as PDF

Top Courses for Computer Science Engineering (CSE)