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

## 65 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
 1 Crore+ students have signed up on EduRev. Have you?
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.

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

= (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

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

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

Opcode 200

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

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.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 31

What is the probability that the below-given f(x,y,z) will return 1?

Where f(x,y,z) is a boolean function (answer upto 2 decimal places).

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 31

Given,

De Morgan's Law

We know that,

NAND Gate Truth Table:

⇒ Probability of getting 1 = 3/4

= 0.75

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 32

The above DFA accepts the set of all strings over {0, 1} , so incorrect option is

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 33

The most appropriate matching for the following pairs is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 33

In Indirect  addressing mode the instruction does not have the address of the data to be operated on, but the instruction points where the address is stored(it is indirectly specifying the address of memory location where the data is stored or to be stored)

In Immediate addressing mode, the data is to be used is immediately given in instruction itself; so it deals with constant data.

In Auto decrement addressing mode. Before determining the effective address, the value in the base register is decremented by the size of the data item which is to be accessed.

Within a loop, this addressing mode can be used to step backwards through all the elements of an array or vector.

The correct match is:

Z: Auto decrement addressing → Loops.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 34

The bit sequence 0010 is serially entered (right-most bit first) into a 4-bit parallel out shift register that is initially clear. What is the Q output after two clock pulses?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 34

The operation of 4-bit parallel out shift register is as follows.
Let's assume that all the flip-flops have just been RESET and that all the outputs are at logic level 0, i.e. no parallel data output.
If the data input pin is connected to a logic ''1'', then on the first clock pulse, the output will be set HIGH to logic 1 with all the other outputs still remaining LOW at logic 0. Assume that the DATA input pin has returned LOW again to logic 0 giving us one data pulse or 0-1-0. The second clock pulse will change the output to logic 0 and the output is HIGH to logic 1 as its input has the logic 1 level on it from Q. The logic 1 has now moved or been shifted one place along the register to the right as it is now at Q. The output Q after two clock pulses is then 1000.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 35

Consider a paging system with 48 bit virtual address space. Each address defers to a byte in memory. Suppose the size of the page is 16 kB and the main memory size is 16 GB. Each page table entry contains frame number and 2 protection bits, 1 valid bit and 1 dirty bit. The minimum size of page table _____________ (in GB).

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 35

Given,

Page table size = 16 kB = 214

Main memory size = 234

As we know,

#Frame

= 234/214

= 220

Each page table entry has 20 + 4 = 24 bits.

Minimum page table size = # page × page table entry size

= 248/214 × 3 B

= 234 × 3 B

= 48 GB

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 36

If

and xy ≠ 0, then  is equal to

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 36

So, z is a homogeneous function of degree 1.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 37

A multilevel page table is preferred in comparison to a single-level page table for translating virtual address to physical address because it

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 37

Single level page table is not preferred since it requires the number of page table entries to be equal to the number of virtual memory addresses, but a multi-level page table has smaller number of entries to reduce the size of page table needed.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 38

A =  and I = , the value of A3 is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 38

The characteristic polynomial of A is:

By Cayley-Hamilton theorem, (Every square matrix satisfied its characteristic equation)
5A + A2 + 6I = 0
A2 = -5A - 6I .... (1)
A3 = -5A2 - 6A
A3 = -5(-5A - 6I) - 6A (from (1))
A3 = 25A + 30I - 6A
A3 = 19A + 30I

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 39

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

What is/are the sum of product form of the output X?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 39

Given,

Input given at 0 is R′ in MUX 2

As we know,

For a 2:1 multiplexer, if S is enabled and I0,I1 are the input, output = S′I0 + SI1

Here, output of MUX 1 is input to MUX 2.

Output of MUX 1 = P′Q + PR

Output of MUX 2 = Q′R + Q(P′Q + PR)

Output of MUX 2 = Q′R + P′Q + PQR

∴Q′R + P′Q + PQR

= Q′R + Q(P′ + PR)

= Q′R + Q(P′ + R)

= Q′R + P′Q + QR

So, the sum of product form of the output X are Q′R + P′Q + PQR and Q′R + P′Q + QR.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 40

If a > 0 and discriminant of ax2 + 2bx + c is negative, then

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 40

Since discriminant of ax2 + 2bx + c is negative ⇒ b2 - ac < 0, a > 0 ....(1)

Applying the transformation R1 = xR1, we get

R1 = R1 + R2 gives

R1 = R1 - R3 gives

= - ve (from equation (1))

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 41

Which of the following option is/are false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 41

Option (A): Most frequent operation on the symbol table is insert operation.

This is false statement since the most frequent operation on the symbol table is the lookup operation.

Option (B): Symbol table data is filled by the lexical analyzer and syntax analyzer.

This is a true statement since symbol table data is filled by the lexical analyzer and syntax analyzer.

Option (C): Efficient implementation of symbol table among, an ordered list, unordered list, and binary search tree is ordered list.

This is false statement since the efficient implementation of a symbol table among, an ordered list ( insert - O(n), lookup- O(logn) ), unordered list ( insert - O(1), look up - O(n) ), and binary search tree ( insert - O(logn), lookup - O(logn)) is a binary search tree.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 42

Consider the table R with attributes A, B and C. The functional dependencies that hold on R are : A → B, C → AB. Which of the following statements is/are true?

1. The decomposition of R into R1(C, A) and R2(A, B) is lossless.
2. The decomposition of R into R1(A, B) and R2(B, C) is lossy.
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 42

Decomposition of R into R1(C, A) and R2(A, B) is lossless.
Because C → A, A → B, so C → AB can be derived and there is no loss.
Decomposition of R into R1(A, B) and R2(B, C) is lossy.
Because A → B, C → B are derived but we can't derive C → AB, so it is lossy.
So, option (3) is correct.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 43

How many 4 to 1 multiplexers are required to implement 16 to 1 multiplexer?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 43

As we know,

A 16 × 1 multiplexer has 16 inputs so if we use 4 × 1 multiplexers:

At first stage: 4 multiplexers are needed for 16 inputs.

At the second stage: the output of these 4 multiplexers is connected to inputs of 1 multiplexer. Total 4 × 1 multiplexers needed are 4+1 = 5.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 44

Consider the relation schema as follows.

Employee (employee-name, sheet, city)
Works (employee-name, company-name, salary)
Company (Company-name, city)
Manages (Employee-name, manager-name)

Write a SQL query that gives the names of the employee who works in the same city and having salary greater than or equal to 17000 and less than or equal to 35500.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 44

SELECT employee-name FROM employee, works, company GROUP BY city having salary > = 17000 and 35500 salary ≤ 35500
This SQL statement will select the name of all the employees those who are having salary greater than or equal to 17000 and and less than or equal to 35500 from the relations employee, works and company in the same city.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 45

If f(x) = tan−1⁡[sin⁡x/(1+cos⁡x)], then what is first term derivative of f(x) ?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 45

Given,

f(x) = tan−1⁡[sin⁡x/(1+cos⁡x)]

As we know,

For first term derivative, we have to put x = 1.

The first derivative of f(x) = 1/2

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 46

The equivalent infix expression and value for the postfix form 1 2 + 3 * 4 5 * – will be

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 46

Given postfix expression : 1 2 + 3 * 4 5 * –
⇒ (1 + 2) 3 * 4 5 * –
⇒ ((1 + 2) * 3) 4 5 * –
⇒ ((1 + 2) * 3) (4 * 5) –
⇒ ((1 + 2) * 3) – (4 * 5)
So, the equivalent infix expression is (1 + 2) * 3 – (4 * 5) and it's value is -11.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 47

Consider the given 8 bit explicit normalized mantissa used for a floating data representation:

a1 a2 a3 a4 a5 a6 a7 a8

The difference value (in decimals) between maximum and minimum mantissa is __________ [Round off to 5 decimal places].

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 47

Given,

a1 a2 a3 a4 a5 a6 a7 a8

As we know,

Mantissa = M

Normalized Mmin

= 0.10000000

= 0.5

Normalized Mmax

= 0.11111111

=1 − 2−8

= 0.99609

The difference between Mmax and Mmin is:

= 0.99609 − 0.5

= 0.49609

So, the difference value (in decimals) between maximum and minimum mantissa is 0.49609.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 48

A → BC
C → + B {print ('+'); }C| ∈
B → D * B {print ('*')i}|D
D → (A)|id {print (id.value); }

For an input '5 + 6 * 7', this translation scheme prints

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 48

5 + 6 * 7
= id + id * id
= D + D * D
print 5 6 7 = D + D * B
print 5 6 7 *[∵ B →D * B,C→ ε] = D + B C
print 5 6 7 *+ = B C = A

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 49

Which of the following languages is regular?

1. L= {(bba(ba)na(n−1)∣ n > 0)}

2. L= {(anbn)∣ n < 1000)}

3. L= {(anbk)∣ n is odd or k is even }

4. L= {(wxwr∣ w, x ∈ (0 + 1))}

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 49

Statement 1 is not regular because it requires comparison between the number of ba′s and the number of a′s, i.e. if there are n the number of ba′s so we should have n−1 number of a′s.

Statement 2 is finite, so, it is regular.

In statement 3 let us use pumping lemma. For m ≥ 3,ay value of "aa" or "bb" will always pump the string. So, statement 3 is also regular.

In statement 4, string is regular as it can be expressed with regular expression 0(0 + 1)+0 + 1(0 + 1)+1.

So, the languages which are regular are:

L= {(anbn)∣ n < 1000)}

L= {(anbk)∣ n is odd or k is even }

L= {(wxwr∣ w, x ∈ (0 + 1))}

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 50

Consider the following two statements:

(A) Business intelligence and data warehousing is used for forecasting and data mining.
(B) Business intelligence and data warehousing is used for analysis of large volumes of sales data.

Which of the following options is correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 50
•  Business intelligence and data warehousing is used for forecasting and data mining. (Correct)
•  Business intelligence and data warehousing is used for analysis of large volumes of sales data. (Correct)

So, option (2) is correct.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 51

Which of the following file is an output of the compiler or assembler?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 51

An object file is a computer file containing object code, that is, the machine code output of the compiler or assembler. The object code is usually relocatable, and not usually directly executable. There are various formats for object files, and the same machine code can be packaged in different object file formats. An object file may also work as a shared library.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 52

Match the following.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 52

SYSDATE returns the current date and time set for the operating system on which the database resides. The datatype of the returned value is DATE.
SYSDATE + 60 will return the date 60 days from the today's date.
SMALLINT is a datatype used to declare the variables. It is used to store the small whole numbers ranging from -215 (-32,768) to 215-1 (32,767). So, it contains integer values up to 16 digits.
Data type 'Date' is used to get the calendar date.
The function Numeric (3,1) will return a 3-digit decimal number having 1 digit after the decimal place.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 53

Consider the following languages:

I. L1 = { < M >∣ M takes at least 2016 steps on some input }

II. L2 = {< M >∣ M takes at least 2016 steps on all inputs }

III. L3 = {< M >∣ M accepts ϵ}

Where for each Turing machine M, denotes a specific encoding of M.

Which one of the following is true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 53

Statement I:

L1 = {< M >∣ M takes at least 2016 steps on some input },

This language is decidable, because counting number of steps can always be decided. We can find TM whether it takes more than 2016 steps on some input string which has length up to 2016. If it processes 2016 steps till end, then will take at least 2016 steps. But if these are not taking 2016 steps, it means these have taken at most 2015 steps which means string is not processed

completely. Also, if Turing machine reaches the halt state before reaching the end of string of input, it can be decided. In all cases, Turing machine M is recursive.

Statement II:

L2 = {< M >∣ M takes at least 2016 steps on all inputs }

For this, consider M makes less than 2016 steps for some input and we can just give it all possible inputs of length less than 2016 steps and check if it can reach a halt state within 2016 steps. All the possible strings make a finite set. Thus, accepted by Turing machine. So, it is recursive language.

Statement III:

L3 = {< M >∣ M accepts ϵ},

It is based on the concept of rice theorem i.e. any non - trivial property of the language recognizable by a Turing machine is undecidable. As this language contains ϵ which is non - trivial property of recursive enumerable language and hence undecidable. So, it is not recursive.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 54

Consider the NFA M shown below.

Let the language accepted by M be L. Let L1 be the language accepted by the NFA M1 obtained by changing the accepting state of M to a non-accepting state and by changing the non-accepting states of M to accepting states.

Which of the following statements is true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 54

In the case of DFA, L(M) = L(M) but in the case of nfa this is not true. Infact L(M) and L (M) have no connection.
To find L1 = L(M), we have to look at M and directly find its language.
Clearly, λ = L(M), since q0 is accepting it (0 + 1) (0 + 1)* ∈ L(M), since q2 is accepting it.
∴ L(M) = L1 = l + (0 + 1) (0 + 1)*
∴ L1 = (0 + 1)* = {0, 1}*

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 55

Find the value of determinant:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 55

Properties of the determinant of a matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say  A, |A| = |A|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by −1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Now,

Applying C2 → C2 + C3;

Taking common (a + b + c) from column 2 , we get,

As we can see that the first and the second column of the given matrix are equal.

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 56

An I/O device transfers 10 MB/s of data into the memory of a processor over the I/O bus, which has a total data transfer capacity of 100 MB/s. The 10 MB/s of data is transferred as 2500 independent pages of 4 KB each. If the processor operates at 200 MA, then it takes 1000 cycles to initiate a DMA transaction and 1500 cycles to respond to the device's interrupt when the DMA transfer completes.

What percentage of processor time is spent handling the data transfer without DMA?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 56

Without DMA, the processor must copy the data into memory as the I/O device, if it sends over the bus. Because the device transfers 10 MB/s over the I/O bus, which has a capacity of 100 MB/s, 10% of each second is spent transferring data over the bus. Assuring the processor is busy handling data during the time each page is being transferred over the bus, 10% of the processor time is spent copying data into memory.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 57

Two cards are drawn randomly from a pack of 52 cards. What is the probability of getting one spade card and one diamond card?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 57

Given,

Pack of 52 cards

Two cards are drawn randomly from a pack of 52 cards.

We know that,

Total number of spade cards = 13

Total number of diamond cards = 13

Formula used:

P = Favorable outcomes/Total outcomes

Now,

Total outcomes = 52C

= 52!/(52 − 2)!2!

= (52 × 51)/2

= 1326

Favourable outcomes = 13C1 × 13C1

= 13 × 13

= 169

∴ The Required Probability

= 169/1326

= 13/102

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 58

Consider the given schemes:

Branch scheme = (Branch name, assets, branch city)
Customer scheme = (Customer name, street, Customer city)
Deposit scheme = (Branch name, account name, customer name, balance)
Borrow scheme = (Branch name, loan number, customer name, amount)
Client scheme = (Customer name, banker name)

Using the relational algebra the query that finds customers who have a balance of over 1000 is

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 58

The SQL statement is:
Select customer name from deposit where (balance > 1000)
∴ Relational algebra query is:
π customer name (σ balance > 1000(Deposits))
where π refers to select and σ specifies the condition that balance should not be greater than 1000.

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 59

Which of the following represent the relation for Heapify Procedure (MAX-HEAPIFY)?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 59

The running time of MAX-HEAPIFY on a subtree of size n rooted at a given node "i" is the 1 time to fix up the relationships among the elements A[i], A[LEFT. i], and A[RIGHT. i], plus the time to run MAX-HEAPIFY on a subtree rooted at one of the children of node "i" (assuming that the recursive call occurs).

The children's subtrees each have size at most 2n = 3, the worst case occurs when the bottom level of the tree is exactly half full and therefore we can describe the running time of MAX-HEAPIFY by the recurrence
T(n) = T(2n/3)+1.

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 60

Consider a 4-way set associative cache (initially empty) with total 16 cache blocks. The main memory consists of 256 blocks and the request for memory blocks is in the following order:
0, 255, 1, 4, 3, 8, 133, 159, 216, 129, 63, 8, 48, 32, 73, 92, 155.

Which one of the following memory block will NOT be in cache if LRU replacement policy is used?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 60

Total 3 sets are there in the cache and each set contains 3 blocks. The LRU replacement is shown here. Access to block 48 replaces block 216, not block 159 because of LRU strategy. Acccess to block 73 replaces block 1.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 61

Which of the following option is/are false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 61

Lexeme: A lexeme is any sequence of characters, defining a token.

As we know,

Option (A): A lexeme is any sequence of binary character.

Option (B): Lexical analysis and parsing are put as two different phases so as to make design simple, improve efficiency and enhance portability.

False options are:

Option (C): DFA does not allow to have an isolated state is false because DFA allows having an isolated state.

Option (D): A symbol table can be implemented by using a linked list and binary search trees is false because a symbol table can be implemented in one of the following ways: Linear (sorted or unsorted) list.

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

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 62

If the functions W, X, Y, Z are as: W =

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 62

From above, we can conclude that W = Z, X =

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 63

The eigen values of the matrix  are:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 63

Eigen values are the roots of the characteristic equation.

It is calculated by |A − λ|∣ = 0 where λ =  root or eigen value.

Given,

Now,

The characteristic equation is, |A − λ|∣ = 0

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

GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 64

Consider the following sets:
S1 : Set of all recursively enumerable languages over the alphabet {0, 1}. :
S2 : Set of all syntactically valid C programs. :
S3 : Set of all languages over the alphabet {0, 1}. :
S4 : Set of all non-regular languages over the alphabet {0, 1}. :
Which of the above sets are uncountable?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 64

S1: The set LRE is known to be countably infinite since it corresponds with set of turing
machines.
S2: Since syntactically valid C programs surely run on Turing machines, this set is also :
a subset of set of Turing machines, which is countable.
S3: Set of all languages = 2∑* which is known to be uncountable. Σ* countably infinite
⇒ 2∑* is uncountable.
S4: Set of all non-regular languages includes set : L NOT RE which is uncountable infinite
and hence is uncountable.
So, S3 and S4 are uncountable.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 65

Consider a classful addressing scheme in which the percentage % of IP address covered by Class A is x and by Class C is y. What is the value of x + y ?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 1 - Question 65

As we know,

Number of IP address possible = 232

So,

x = 50

y = 12.5
Therefore, x + y = 50 + 12.5
= 62.5

Hence, the correct answer is 62.5.

## GATE Computer Science Engineering(CSE) 2025 Mock Test Series

152 docs|216 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

### Up next

 Test | 65 ques
 Test | 65 ques
 Test | 65 ques
 Test | 65 ques
 Test | 65 ques

## GATE Computer Science Engineering(CSE) 2025 Mock Test Series

152 docs|216 tests

### Up next

 Test | 65 ques
 Test | 65 ques
 Test | 65 ques
 Test | 65 ques
 Test | 65 ques