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Practice Test: Computer Science Engineering (CSE) - 7 - Computer Science Engineering (CSE) MCQ


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65 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Practice Test: Computer Science Engineering (CSE) - 7

Practice Test: Computer Science Engineering (CSE) - 7 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) - 7 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 7 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) - 7 below.
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Practice Test: Computer Science Engineering (CSE) - 7 - Question 1

His reactions to unpleasant situations tended to _________ everyone’s nerves.

The word that best fills the blank in the above sentence is 

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 1

The sentence implies that his reactions to unpleasant situations annoyed everyone. The correct word to fill in the blank is aggravate as it means 'to annoy or exasperate'.

The meaning of the other words is:

Deviate: depart from an established course 

Revue: a light theatrical entertainment consisting of a series of short sketches, songs, and dances, typically dealing satirically with topical issues

Meander: wander at random

Practice Test: Computer Science Engineering (CSE) - 7 - Question 2

If XY + Z = X(Y + Z) which of the following must be true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 2

XY + Z = X(Y + Z)

⇒ XY + Z = XY + XZ

⇒ Z = XZ ⇒ Z - XZ = 0 ⇒ Z (1 - X) = 0

Either Z = 0 (in this case X can take any value) OR X = 1 (in this case Z can take any value).

X = 1 or Z = 0

Practice Test: Computer Science Engineering (CSE) - 7 - Question 3

What is the number missing from the table?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 3

XY + Z = X(Y + Z)

⇒ XY + Z = XY + XZ

⇒ Z = XZ ⇒ Z - XZ = 0 ⇒ Z (1 - X) = 0

Either Z = 0 (in this case X can take any value) OR X = 1 (in this case Z can take any value).

X = 1 or Z = 0

5 × 3 + 1 = 16

16 × 3 + 1 = 49

9 × 3 + 2 = 29

29 × 3 + 2 = 89

15 × 3 + 3 = 48

48 × 3 + 3 = 147

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

ARCHIPELAGO : ISLAND

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

An ‘archipelago’ is ‘a group of islands’, i.e. it is it is formed of several ‘islands’. So the answer should be the option where the group of the latter word forms the former. Option 1 is incorrect as a group links forms a chain and not the other way round. Similarly, option C and D can be disqualified. An encyclopedia is ‘a book or set of books giving information on many subjects’. The correct answer is option B as many beads form a necklace.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 5

There are three basket of fruits. First basket has twice the number of fruits in the second basket. Third basket has 3/4 th of the fruits in the first. The average of the fruits in all the basket is 30. The number of fruits in the first basket is


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 5

Let the number of fruits in first basket is x

Fruits in 2nd basket = x/2

Fruits in 3rd basket = 3/4x

 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 6

Until now only injectable vaccines against Swine Flu have been available. They have been primarily used by older adults who are at risk for complications from Swine Flu. A new vaccine administered in an oral form has proven effective in preventing Swine Flu in children. Since children are significantly more likely than adults to contract and spread Swine Flu, making the new vaccine widely available for children will greatly reduce the spread of Swine Flu across the population.          

Which of the following, if true, most strengthens the argument?    

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 6

The passage points out that the oral form of vaccine will allow Swine Flu to be greatly reduced among the children who are more susceptible to catching the Flu (as compared to adults). Undoubtedly, the non painful method of oral vaccination is more likely to encourage the parents to get their children vaccinated who are skeptical of injections that cause pain. Thus, among all the given options, 4 strengthens the given argument. Options 1, 2 and 3 are completely irrelevant and 5 would weaken the argument as it cites the probable negative effects of the new vaccine which will definitely not encourage parents to get their children vaccinated.

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

Four friends Rishabh, Keshav, Lavish and Hemang are out for shopping. Rishabh has less money than three times the amount that Keshav has. Lavish has more money than Keshav. Hemang has an amount equal to the difference of amounts with Keshav and Lavish. Rishabh has three times the money with Hemang. Each of them has to buy at least one shorts, or one sleeper, or one sleeveless t-shirt, or one goggle that is priced 200, 400, 600, and 1000 a piece, respectively. Lavish borrows 300 from Rishabh and buys a goggle. Keshav buys a Sleeveless t-shirt after borrowing 100 from Rishabh and is left with no money. Rishabh buys three shorts. What is the costliest item that Hemang could buy with his own money?

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

Let, Rishabh has 'P' amount of money, Keshav has 'Q' amount of money, Lavish has 'R' amount of money, and Hemang has 'S' amount of money.

Therefore, according to the question

P < 3 × Q

R > Q,

S = R - Q

P = 3 × S

Also, Lavish borrows 300 from Rishabh and buys a goggle and Keshav buys a Sleeveless t-shirt after borrowing 100 from Rishabh and is left with no money. Implies that, 

Q = 600 - 100 (Rishabh) = 500

R ≥ 1000 - 300 (Rishabh) ≥ 700

Also, Rishabh buys three shorts means he has a minimum of 1000 (100 + 300 + 3 × 200 = 1000) and the maximum is less than 3Q.

Therefore, 1000 ≤ P < 3Q

1000 ≤ P < 3 × 500

1000 ≤ P < 1500

Since Rishabh has three times the money with Hemang

1000 ≤ 3 × S  < 1500

333 ≤  S  < 500

Hence, Hemang could buy a sleeper with his own money.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 8

I wouldn’t _______ with a soldier who was wearing a metal _______, awarded for a display of _______.

The words that best fill the blanks in the above sentence are:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 8

The sentence implies that I would not poke a soldier who was wearing a metal medal which was awarded for the display of determination.

The meaning of the words is:

Meddle: interfere in something that is not one's concern

Medal: a metal disc typically of the size of a large coin and bearing an inscription or design, made to commemorate an event or awarded as a distinction to someone such as a soldier or athlete

Mettle: a person's ability to cope well with difficulties; spirit and resilience

Therefore, the correct sequence of words to fill in the blanks is option 2.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 9

A paper sheet is in the shape of a right angle triangle and cut along a line parallel to hypotenuse leaving a smaller triangle. There was 25% reduction in the length of the hypotenuse of the triangle. If area of triangle initially was 28 cm2 then area of smaller triangle will be ______ cm2.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 9

Let PQR is the initial triangle and SQT is the final triangle.

ΔPQR is similar to ΔSQT

∵ ST = 0.75 PR

∴ SQ = 0.75 PQ

And QT = 0.75 QR

Practice Test: Computer Science Engineering (CSE) - 7 - Question 10

A drinks machine offers three selections - Tea, Coffee or Random but the machine has been wired up wrongly so that each button does not give what it claims. If each drink costs 50p, how much minimum money do you have to put into the machine to work out which button gives which selection?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 10

For the first drink press, the button labelled Random. We know this is NOT the Random button so if it dispenses Tea it is the Tea button likewise if it dispenses Coffee it is the Coffee button. From here it is simple to work out the other buttons knowing that they can't be what they say they are.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 11

Suffix of a string is any number of trailing symbols. What is the number of possible suffixes for the string of length 10?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 11

For a given string of length n, the number of possible suffixes = n+1

For example: In a string abcd, the suffixes are: ε, d, cd, bcd, abcd. Therefore, 5 suffixes are there in a string of length 4.

Hence, for a string of length 10, the number of suffixes is 11.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 12

If =   ,  find k so that A2=kA−2I where I is an identity matrix.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 12

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements:

4k = 4 ⇒ k = 1

Practice Test: Computer Science Engineering (CSE) - 7 - Question 13

Consider sender’s sliding window size in Go - Back - n ARQ is equal to sender’s window size in Selective Repeat ARQ. Which of the following expression represents the relation between Go - Back - n and selective repeat ARQ in terms of n? (Where n is number of bits that are used to represent frame sequence number)

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 13

Sender’s sliding window size in Go – Back – n ARQ = 2n – 1

Sender’s sliding window size in Selective Repeat ARQ = 2n-1

2n – 1 = 2n-1

2n – 2n-1 = 1

2n – 2n/2 = 1

2n+1 – 2n = 2

Practice Test: Computer Science Engineering (CSE) - 7 - Question 14

Consider the schema a r(A, B, C, D, E) with the following functional dependencies:

A → BC

CD → E

B → D

E → A

The decomposition of the schema into r1(A, B, C) and r2(A, D, E) is _________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 14

R1 and R2 form a lossless decomposition of R if at least one of the following functional dependencies is in F:

• R1 ∩ R→ R1

• R1 ∩ R2 → R2

In the given schema r(A, B, C, D, E), the candidate key is A because A+ = {A, B, C, D, E}

Practice Test: Computer Science Engineering (CSE) - 7 - Question 15

Consider an array which stores a maximum of 100 elements. For case 1, the user inserts 1 element into an array. For case 2, the user inserts 100 elements into an array. What is the time complexity for both the cases?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 15

To store 1 element, 100 memory spaces are required.

To store 2 elements, 100 memory spaces are required.

.

.

.

To store n element, 100 memory spaces are required.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 16

Evaluate 

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 16

Practice Test: Computer Science Engineering (CSE) - 7 - Question 17

Which of the following expressions is valid for "There is an Indian who does not like watching Television" where

c(x): x likes watching television

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 17

All Indians like watching Television: Ɐxc(x)

Negating the above expression

∼(Ɐxc(x))

Ǝx∼c(x)

Practice Test: Computer Science Engineering (CSE) - 7 - Question 18

A frame consists of m data (i.e., message) bits and r redundant (i.e. check) bits. Column I consists of different type of codes and Column II consists of the way the redundant bits are calculated. Select the most appropriate option.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 18

The correct answer is option 3 i.e. A-3, B-1, C-2.

In a block code, the r check bits are computed solely as a function of the m data bits with which they are associated, as though the m bits were looked up in a large table to find their corresponding r check bits.

In a systematic code, the m data bits are sent directly, along with the check bits, rather than being encoded themselves before they are sent.

In a linear code, the r check bits are computed as a linear function of the m data bits. Exclusive OR (XOR) or modulo 2 addition is a popular choice.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 19

Consider the following set of statements:

S1: If L1 and L2 are regular languages, then L1 - Lis also a regular language.

S2: Context free languages are closed under concatenation.

S3: Context free languages are closed under intersection.

S4: Regular language is closed under infinite union of regular languages.

Which of the given statements are true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 19

Only statements S1 and S2 are true.

Context free languages are not closed under intersection. Also, regular languages are not closed under infinite union of regular languages.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 20

Mean of 12 observation was found to be 28. Later on, it was detected that an observation 62 was misread as 26. Then the correct mean of the observations is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 20

Given, mean of 12 observation was found to be 28.

Mean = sum of observations/number of observations.

Sum of observations = 12 × 28 = 336

Now, it was detected that an observation 62 was misread as 26.

∴ Correct mean = 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 21

Which of the following activation record unit points to non-local data stored in other activation records?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 21

Temporary values: stores the values that arise in the evaluation of an expression.

Machine status: holds the information about the status of a machine just before the function call.

Access link: points to non-local data stored in other activation records.

Control link: points to activation record of a caller.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 22

Which of the following function is used to assign a name to a socket?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 22

bind is used to bind a name to socket.

int bind(int sockfd, const struct sockaddr *addr, socklen_t addrlen);

When a socket is created with socket(2), it exists in a name space (address family) but has no address assigned to it. bind() assigns the address specified by addr to the socket referred to by the file descriptor sockfd. addrlen specifies the size, in bytes, of the address structure pointed to by addr.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 23

Let A = {1,2,3,4}. What is the number of irreflexive relations which can be formed on the set A?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 23

Note: The number of irreflexive relations on set A when |A| = n is 

Here |A| = 4

Therefore, total number of irreflexive relations = 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 24

Consider an interpolation search which is an improvement over binary search where the values in a sorted array are uniformly distributed. In interpolation search construction of new data points take place at different locations according to the value of the key being searched. Find the time complexity of interpolation search.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 24

To narrow down the search space binary search uses mid   while interpolation search uses mid =

On average the interpolation search makes about log (log n) comparisons if the elements are uniformly distributed, where n is the number of elements to be searched.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 25

A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is _______.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 25

M for men, W for women, E for employed & U for unemployed

P(M) = 0.5, P(W) = 0.5

P(U/M) = 0.2 & P(U/W) = 0.5

∴ Total probability is

P(U) = P(M). P(U/M) + P(W) ⋅ P(U/W)

⇒ P(U) = 0.35

∴ P(E) = 1 – P(U) = 1-0.35 = 0.65

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 26

There are 17 engineers and 3 official languages. Every pair of engineers communicates in one of the official languages. What is the number of engineers communicating in the same language pairwise?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 26

Since we need to find the number of engineers communicating in the same language pairwise, the number of pairs that can be formed = 9.

By pigeonhole principle, there are ceil (9/3)  = 3  ​engineers communicating in the same language pairwise.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 27

Which of the following statement is correct about WiFi?

I. WiFi systems are full duplex.

II. WiFi uses access control protocol - CSMA/CA.

III. Channel bandwidth of high rate wifi is 25 MHz

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 27

IEEE 802.11 wireless LANs use a media access control protocol called Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA).

All WiFi networks are contention-based TDD systems, where the access point and the mobile stations all use the same channel. Because of the shared media operation, all WiFi networks are half duplex.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 28

Which of the following indexes is created on a non-key field and defined on an ordered data file?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 28

The correct answer is option 1 i.e. clustering index.

A clustering index is defined on an ordered data file. The data file is ordered on a non-key field. 

The primary index is defined on an ordered data file. The data file is ordered on a primary key field. 

A secondary index is defined on an unordered data file. It can be either a candidate key or some non-key field.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 29

Consider the following languages over the input alphabet {a,b}:

A: {ambn | m>n}

B: {xcy | x,y Є (a,b)*}

C: {aibj | i = 2j}

D: {wwwR | w Є (a,b)*}

The number of context free languages is ____.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 29

Out of the given languages, only D i.e. {wwwR | w Є (a,b)*} is not a context free language.

Push down automata for all the other languages can be drawn.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 30

Consider a simple paging system with 1024 MB of physical memory and the page size of 4KB with a logical address space of 128 pages. Find the number of bits used to represent physical address, frame number and page number. 

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 30

Physical memory = 1024 MB = 1024 × 1024 × 1024 bytes = 210 × 210 × 210 = 230

Hence, 30 bits are required to represent physical address.

Page size = 4 KB = 4 × 210 = 212

Number of frames =

Hence, 18 bits are required to represent frame number.

128 pages i.e. 27 pages being accessed. Hence, 7 bit represent page number.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 31

Apply the Dijkstra’s shortest path algorithm on the following graph.

The correct path from vertex 1 to 6 is ______.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 32

Consider the 4 to 1 multiplexer with two select lines S1 and S0 given below.

The minimal sum of products form of the Boolean expression for the output F of the multiplexer is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 32

The correct answer is option 2.

The output is given by:

F = P'Q' + P'QR + PQR'

= P'[Q' + QR] + PQR'

= P'[Q' + Q][Q' + R] + PQR'

= P'[Q' + R] + PQR'

= P'Q' + P'R + PQR'

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 33

Consider a disk system with 100 cylinders. The requests to access the cylinders occur in the following sequence:

16, 24, 7, 9, 25, 11, 8, 87

Assuming that the head is currently at cylinder 58, what is the time taken (in ms) to satisfy all requests if first come first serve policy is used?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 33

(58 – 16) + (24 - 16) + (24 - 7) + (9 - 7) + (25 - 9) + (25 - 11) + (11 - 8) + (87 - 8)

= 42 + 8 + 17 + 2 + 16 + 14 + 3 + 79

= 181 ms

Practice Test: Computer Science Engineering (CSE) - 7 - Question 34

Which of the following instruction cannot be represented in 1 - address instruction format?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 34

LOAD and STORE are 1 - address instruction. LOAD is used to load data from memory to accumulator. STORE is used to store the accumulator result in memory.

e.g. LOAD X   AC = M[X]

e.g. STORE YM[Y] = AC​

MOV is a 2 – address instruction. MOVE is used to move contents of a register to memory and vice versa or from one register to another.

e.g. MOV R1, X          R1 = M[X]

Practice Test: Computer Science Engineering (CSE) - 7 - Question 35

Let   then

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 35

The given matrix is upper triangular matrix. Hence Eigen values = Diagonal elements = 1,4, 9

M has distinct Eigen values, hence eigen vectors are linearly independent.

So, M is diagonalizable

The eigen values of M2 = 12, 42, 92

= 1,76,25

M2 also having distinct eigen values.

Hence both M and M2 are diagonalizable.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 36

Which of the control signal specifies that the ALU operation should be determined from the function bits?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 36

Control signal ALUOp determines either specifies the ALU operation to be performed or specifies that the operation should be determined from the function bits.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 37

Consider a multicore system in which an application has 40% serial component and 60% parallel component. After executing this application on a system with three processing cores, the maximum potential gain achieved is ________. (Calculate value up to two decimal places)


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 37

In parallel computing, Amdahl's law is mainly used to predict the theoretical maximum speed-up for program processing using multiple processors.

If S is the portion of the application that must be performed serially on a system with N processing cores, the formula appears as follows:

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 38

Consider a LAN with five nodes. Time is divided into fixed-size slots. A node can begin its transmission only at the beginning of a slot. A collision occurs if more than two nodes transmits in the same slot. The probabilities of generation of a frame in a time slot by these LAN are 0.2, 0.3, 0.25, 0.16 and 0.09, respectively. The probability of transmission in the first slot without any collision by any of these five stations is _________. (Compute the answer up to 2 decimal places.)


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 38

Probability of transmission on the first slot by 1st node:

= 0.2 × (1 – 0.3) × (1 – 0.25) × (1 – 0.16) × (1 – 0.09)

= 0.2 × 0.7 × 0. 75 × 0.84 × 0.91

 

Probability of transmission on the first slot by 2nd node:

= (1 - 0.2) × 0.3 × (1 – 0.25) × (1 – 0.16) × (1 – 0.09)

= 0.8 × 0.3 × 0. 75 × 0.84 × 0.91

 

Probability of transmission on the first slot by 3rd node:

= (1 - 0.2) × (1 - 0.3) × 0.25 × (1 – 0.16) × (1 – 0.09)

= 0.8 × 0.7 × 0.25 × 0.84 × 0.91

 

Probability of transmission on the first slot by 4th node:

= (1 - 0.2) × (1 - 0.3) × (1 - 0.25) × 0.16 × (1 – 0.09)

= 0.8 × 0.7 × 0.75 × 0.16 × 0.91

 

Probability of transmission on the first slot by 5th node:

= (1 - 0.2) × (1 - 0.3) × (1 - 0.25) × (1 - 0.16) × 0.09

= 0.8 × 0.7 × 0.75 × 0.84 × 0.09

 

Probability = 0.080262 + 0.137592 + 0.107016 + 0.061152 + 0.031752 = 0.4177

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 39

Consider the DFA given below. What is the number of states obtained after minimizing the given DFA?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 39

Since the state q2 is not reachable from any of the states, we do not consider it in the transition table. The  modified transition diagram is as follows:

0 equi sets: [q0, q1, q3], [q4]

1 equi sets: [q0], [q1, q3], [q4]

2 equi sets: [q0], [q1, q3]. [q4]

Therefore, the number of states in minimized DFA is 3.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 40

An operating system uses Shortest Job first (SJF) process scheduling algorithm. Consider the arrival times and execution times (in milliseconds) for the following processes:

What is the average response time (in milliseconds)?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 40

Gantt chart:

 

 

Response time is the amount of time from when a request was submitted until the first response is produced.

Response time = submission time – Arrival time

Response time of process P1 = 8 – 0 = 8

Response time of process P2 = 0 – 0 = 0

Response time of process P3 = 5 – 1 = 4

Response time of process P4 = 16 – 3 = 13

Average response time = 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 41

Consider an unrolled linked list with n elements. This list stores multiple elements in each node. What is the worst case time complexity to find the kth element if the number of nodes and the number of elements in each node are equal?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 41

An unrolled linked list stores multiple elements in each node.

Here, the number of elements in each node is equal to the number of nodes in a list. Hence, there are total √n elements in each node and total √n  nodes in a list.

To find the kth element, traverse the list up to    node. If an element is present in the last node, total  √n nodes should be traversed. Hence, time complexity to search an element is 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 42

Consider the following query:

select distinct emp_id

from employee 

where project = ’Alpha’ and joining_year = 2009 and

emp_id not in (select emp_id

from employee

where project= ’Beta’ and joining_year= 2009);

What does the given query perform?

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 43

Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 43

Transmission time (Tx) for 1 frame = L/B

= 20 × 10^3/ 10 × 10^6

= 2 ms

Transmission time for 35 frame = 70 ms

Round trip time = 90 ms

The sender cannot receive acknowledgement before 90 ms.

After sending 35 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 90 – 70 = 20 ms

Practice Test: Computer Science Engineering (CSE) - 7 - Question 44

What is the generating function for the sequence <2, 0, 2, 0, 2, 0...>?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 44

The generating function for the sequence 

Adding (1) and (2)

Practice Test: Computer Science Engineering (CSE) - 7 - Question 45

The expression (P ⇒ Q) ˄ (R ⇒ Q) is equivalent to

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 45

The correct answer is option 2 i.e. (P ˅ R) ⇒ Q.

The truth table for the given expressions is:

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 46

Consider six memory partitions of size 200 KB, 400 KB, 600 KB, 500 KB, 300 KB, and 250 KB, where KB refers to kilobyte. These partitions need to be allotted to four processes of sizes 357 KB, 210 KB, 468 KB and 491 KB in that order. If the best fit algorithm is used, which partitions are NOT allotted to any process?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 46

Best fit allocates the smallest block among those that are large enough for the new process. So the memory blocks are allocated in below order.

Fixed partitions
(Only one process can reside in one partition)
Hence partitions of size 200 KB and 300 KB will not be allotted to any process.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 47

What is the output of the following code?

int main()

{

            float k = 5.375;

            char *m;

            int i;

            m = (char*)&k;

            for(4; 7; 2)

                        printf("%d ", m[4]);

            return 0;

}

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 47

m[4] is similar to some address[4] which returns some garbage value. Here in condition of for loop, there is a non-zero value, thus it becomes infinite for loop.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 48

The keys 25, 78, 15, 109, 50, and 6 are inserted into an initially empty hash table of length 7 using quadratic probing with hash function h(k)= k mod 7. Find the maximum number of iteration (i) to solve collision.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 48

If slot hash(x) % S is full, then use (hash(x) + 1*1) % S for next iteration use (hash(x) + 2*2)

For ith iteration use (hash(x) + i*i)

h(25) = 25 mod 7 = 4

h(78) = 78 mod 7 = 1

h(15) = 15 mod 7 = 1 → 15 + 1 × 1 mod 7 = 2

h(109) = 109 mod 7 = 4 → 109 + 1 × 1 mod 7 = 5

h(50) = 50 mod 7 = 1 → 50 + 1 × 1 mod 7 = 2 → 50 + 2 × 2 mod 7 = 5 → 50 + 3 × 3 mod 7 = 3

h(6) = 6 mod 7 = 6

Practice Test: Computer Science Engineering (CSE) - 7 - Question 49

Classless Inter-domain Routing (CIDR) receives a packet with address 182.25.110.68. The router’s routing table has the following entries:

The identifier of the output interface on which this packet will be forwarded is ______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 49

182.25.110.68 -> 182.00011001.110.68

182.20.0.0/ 13 -> 182.00011000.0.0

Bitwise AND result -> 182.00011000.0.0 -> 182.24.0.0

182.25.110.68 -> 182.00011001.110.68

182.22.0.0/ 12 -> 182.00010000.0.0

Bitwise AND result -> 182.00010000.0.0 -> 182.16.0.0

182.25.110.68 -> 182.00011001.110.68

182.24.0.0/ 14 -> 182.00011000.0.0

Bitwise AND result -> 182.00011000.0.0 -> 182.24.0.0

Practice Test: Computer Science Engineering (CSE) - 7 - Question 50

Let 'i' represents the cache line number, 'j' represents main memory block number and 'm' represents the number of lines in the cache. Which of the following is the correct expression for direct mapping?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 50

Direct mapping maps each block of main memory into only one possible cache line. The mapping is expressed as:

i = j modulo m

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 51

Consider the grammar with the following translation rules and E as the start symbol.

E -> E @ T                 E1.trans = E2.trans + T.trans

E -> T                        E.trans  = T.trans

T -> T $ F                  T1.trans = T2.trans * F.trans

T -> F                        T.trans  = F.trans

F -> INTLITERAL       F.trans  = INTLITERAL.value

F -> (E)                       F.trans  = E.trans​

Compute E.value for the root of the parse tree for the expression:4 $ (7 @ 8)


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 51

Syntax-directed translation refers to a method of compiler implementation where the source language translation is completely driven by the parser.

A common method of syntax-directed translation is translating a string into a sequence of actions by attaching one such action to each rule of a grammar.

4 $ (7 @ 8)

4 *(7+8)

4*15

60

Practice Test: Computer Science Engineering (CSE) - 7 - Question 52

Consider the following infix expression. While converting this infix expression to postfix expression, what will be the stack content (from bottom to top) after scanning operand 5? 

((13 × 4) + 16) ÷ 4 – ((8 × 5) ÷ 10)

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 52

 

Practice Test: Computer Science Engineering (CSE) - 7 - Question 53

Consider the following C-fragment:

#include<stdio.h> 

int (*foo())[3]

{

   int a[3] = {1,2,3};

   printf("%d", *a);

   return &a;

}

int main()

{

   int (*p)[3];

   p = foo();

   printf("%d", *(*p + 2));

   return 0;

}

What is the output of the given program?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 53

The function foo() holds an array which returns the address of the first element of the array a.

Also, p holds the address of the first element of the array a. Thus, *p returns the address of the first element of the array a and **p returns the value of the first element of the array a.

Therefore, (*p + 2) will point to the third location of the array. *(*p+2) will now return the value of the third element of the array.

Hence, the output will be 13.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 54

Consider the following schedules:

Which of the given schedules is/are conflict serializable?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 54

The wait-for-graph for the given schedules is as follows:

 

Since there is no cycle in S1, S1 is conflict serializable.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 55

In HTTP request message, head of the request is consist of

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 55

A start-line describes the requests to be implemented, or its status of whether successful or a failure.

An optional set of HTTP headers specifies the request, or describes the body included in the message.

The start-line and HTTP headers of the HTTP message are collectively known as the head of the requests, whereas its payload is known as the body.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 56

What is the number of perfect matching in a complete graph K6?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 56

The number of perfect matching in a complete graph K2n is given by 

Therefore, the number of perfect matching in a complete graph K6 is 

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 57

Consider the following strings: 

S1 = PQPTSR

S2 = QPRQPS

Find the length of the longest common subsequence by using dynamic programming.


Practice Test: Computer Science Engineering (CSE) - 7 - Question 58

What is the output of the function fun(-1+2419)?

int fun(int n)

{

    if (n == 0)

       return 0;

    int temp = n % 10 + fun(n / 10);

    if(temp>9)

        return fun(temp);

    else

        return temp;

}

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 58

fun(-1+2419) → fun(2418)

fun(2418) returns value 15 which is greater than 9. Hence it will call fun(15)

Practice Test: Computer Science Engineering (CSE) - 7 - Question 59

Which of the following is the drawbacks of the I/O mode of transfer?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 59

The disadvantages of I/O mode of transfer are:

  1. The I/O transfer rate is limited by the speed with which the processor can test and service a device.
  2. The processor is tied up in managing an I/O transfer; a number of instructions must be executed for each I/O transfer.

Programmed I/O is simple to implement and does not require any special software or hardware.

Programmed I/O is probably the most common I/O technique because it is very cheap and easy to implement, and in general does not introduce any unforeseen hazards. 

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 60

The minimum positive integer p such that 8Y (mod 23) = 1 then Y is ______.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 60

Fermat's little theorem states that if p is a prime number, then for any integer a, the number ap − a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as ap ≡ a (mod p)

If a is not divisible by p, Fermat's little theorem is equivalent to

ap-1 ≡ 1 (mod p)

Here a = 8 and p = 23

Y = 23 – 1 = 22

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 61

Consider the following grammar:

S0-> AS|PB|SB|S

S → AS|QB|SB

A → RS|XS|a

B → TS|VV|US|XS|a

X → a

Y → b

V → b

P → AS

Q → a

R → XA

T → SY

U → XA

What is the number of steps required to derive aaaaaaaa?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 61

The above grammar is in CNF.

Therefore, the number of steps required in the derivation of an n-length string from the given CNF is 2n-1.

'aaaaaaaa' is a string of lenght 8, therefore, to derive it, the number of steps required = 15.

Note: It can be easily proved by derivating the string from the given grammar.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 7 - Question 62

Consider the following equation:

(247)8 = (1132)x

What will be the value of 'x' so that the equation is satisfied?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 62

On converting (247)8 into binary, we get (010100111)2.

(247)= (010100111)2 = (167)10

(167)10 = (1132)5

Therefore, x = 5

Practice Test: Computer Science Engineering (CSE) - 7 - Question 63

Consider the following recursive function:

int sum(n int) 

{

    if n == 1 

    {

        return 1

    }

    return sum(n) + Sum(n-1)

}

What is the time complexity of the above code?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 63

Recurrence relation for above code is

T(1) = 1

T(n) = 1 + T(n-1), when n > 1

1 + T(n-1) 

= 1 + (1 + T(n-2)) 

= 2 + T(n-2)

= 2 + (1 + T(n-3)) 

= 3 + T(n-3) …

= k + T(n-k) …

= n - 1 + T(1) 

= n - 1 + 1

= Θ(n)

Practice Test: Computer Science Engineering (CSE) - 7 - Question 64

A typical SQL query has the form

select A1, A2,..., An 

from r1, r2,...,rm 

where P;

Each Ai represents an attribute, and each ri a relation. P is a predicate.

What is the relational algebra expression for the given query?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 64

The correct answer is option 3 i.e.

The select operation is represented by   while the project operation is represented by σ.

Practice Test: Computer Science Engineering (CSE) - 7 - Question 65

Consider the following grammar:

S → AB 

B → +S 

B → ε

A → int C 

A → (S) 

C → *A 

C → ε

Find the follow of A.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 7 - Question 65

First(S) = {int, (}

First(A) = {int, (}

First(B) = { ε, +}

First(C) = { ε, *}​

Follow(S) = {), $}

Follow (A) = {), $, +}

Follow (B) = { ), $}

Follow (C) = {), $, +}

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