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Practice Test: Computer Science Engineering (CSE) - 11 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Practice Test: Computer Science Engineering (CSE) - 11

Practice Test: Computer Science Engineering (CSE) - 11 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) - 11 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 11 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) - 11 below.
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Practice Test: Computer Science Engineering (CSE) - 11 - Question 1

If the product of two integers 53 and 22 in a specific number system is 1276, then what is the decimal equivalent of the number 4371 in the same number system?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 1
Since ‘6’ and ‘7’ have been used in the product of 53 and 22, the numbers have to be using base ‘8’ or ‘9’.

If we try with base ‘8’,-(53)8-= 8 × 5 + 3 = (43)10

-(22)8-=8 × 2 + 2 = (18)10

-43 × 18-= (774)10

whereas,-(1276)8 -=512×1+64 × 2+8 × 7+6 =(702)10

If we try with base ‘9’,-(53)9- = 9 × 5 + 3 =(48)10

-(22)9 - = 9 × 2 + 2 =(20)10

-48 × 20 -= (960)10

whereas,- (1276)9 - = 729 × 1 + 81 × 2 + 9 × 7+ 6 = (960)10

Converting-(4371)9 –

= 4 × (729) + 3 × (81) + 7 × (9) + 1 =3223.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 2

If a and b are rational numbers and , then a : b is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 2

After rationalization,

On comparison, we get

a = 2

b = 1

a:b = 2:1

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Practice Test: Computer Science Engineering (CSE) - 11 - Question 3

If the world ‘MAJORITY’ is encoded as ‘PKBNXSHQ’, how will the word ‘DAUGHTER’ be encoded?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 3

Divide the given word in 2 equal parts

M A J O R I T Y and each part is rewritten in reverse sequence

O J A M Y T I R.

Position of letters in the alphabet is 15/10/1/13 and 25/20/9/18 and position of letters in the given code PKBN/XSHQ is 16/11/2/14 and 24/19/8/17.

In comparison, you notice, the first half is obtained by adding ‘1’ and the second half is obtained by subtraction ‘1’.

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

Which device is generally used in applications like ATMs, hospitals, airline reservation etc.?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 4
  • A touch screen is a display that can recognize a touch to its surface area, either with a finger or a stylus.

  • Touch screens are generally used in applications like ATMs, hospitals, airline reservations, supermarkets, etc.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 5

Direction: In these questions, the relationship between different elements is shown in the statements. These statements are followed by two conclusions.

Statement: A > B, B ≥ C = D < E

Conclusions:

I. C < A

II. D ≤ B

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 5
A > B, B ≥ C = D < E: A > B, B ≥ C → A>C & D= C, B ≥ C → D ≤B

Conclusion:

I. C > A: True

II. D ≤ B: True

Practice Test: Computer Science Engineering (CSE) - 11 - Question 6

Direction: Study the following pie chart carefully to answer the question that follow:

Percentage of students enrolled in different streams in a college.

Total number of student = 3500

Percentage break-up of girls enrolled in these stream out of the total students

Total number of girls = 1500

What is the total number of boys enrolled in Management and IT together?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 6

The total number of boys enrolled in Management and IT together

= 3500 × 20 + 16100 - 1500 × 18 + 12100

= 35 × 36 - 15 × 30

= 1260 - 450 = 810

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

In the following question, select the odd letters from the given alternatives.

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

If we arrange alphabets in two rows as shown below, the letters are related as follows,

Thus JP are the odd letters.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 8

Who was the first Viceroy of Portuguese in India?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 8
  • Almeida was the first Viceroy of Portuguese in India .

  •  

    Almeida took the island of Diu in his possession in AD 1509.

  •  

    Albuquerque was the next Viceroy who won Goa in AD 1510.

  •  

    Later on, the Portuguese also took Bassein & Daman in their possession.

 

Practice Test: Computer Science Engineering (CSE) - 11 - Question 9

Direction: In the following question, the sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.

Moving to the city was eye-__________ for the straitlaced country girl.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 9

Clearing means an open space in a forest, especially one cleared for cultivation.

Freeing means release from confinement or slavery.

Saving means release from confinement or slavery.

Eye-opening is a phrase meaning something that surprises you and teaches you new facts about life, people, etc.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 10

Select the option in which the given figure is embedded.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 10

After carefully observing the figures given in the question, it is very clear that the question figure is embedded in answer figure (A). It is shown as given below

Hence, Option A is the correct response.

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

Serial communications involves interfacing of data among:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 11
It exhibits serial exchange between microprocessor and peripherals such as printers, external drives, scanners or mice. The interface has parallel-to-serial converter which serves as data transmitter and serial-to-parallel converter which serves as data receiver.
Practice Test: Computer Science Engineering (CSE) - 11 - Question 12

What is the wrong statement?

1) Complement of DFA that accepts those strings where after each ‘ a, always ‘ b ‘ appears, is the DFA that accepts those strings where after ‘ a ‘ never ‘ b ‘ appears.

2) In a language – no. of state in minimal NFA is n, then in worst case, maximum no. of state in equivalent DFA is 2n.

3) Minimal DFA always contains only one dead state.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 12
For strings where after each ‘ a , always ‘ b ‘ appears, language is-

L1= {ab,abab,abb,bbab,.........,bbb,bbbbbb........}

For string where after ‘ a ‘ never ‘ b ‘ appears ,language is-

L2={ε,a,b,aa,ba,,bbb,bbbb.}

Some strings are there on intersection of L1 & L2 (Like bbb, bbbb etc)

So L1 & L 2 are not complementary to each other .

In a language – no. of state in minimal NFA is n, then in worst case, maximum no. of state in equivalent DFA is 2n.

More than one dead state gets combined during minimization of DFA, so DFA contains only one dead state.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 13

The number of 4 line to-16 line decoders required to make an 8-line-to-256 line decoder is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 13
8-line to-256-line decoder using 4-line to-16-line decoders is shown below

Number of 4 x 16 MUX required

= 256 / 16 + 16/16 = 16 + 1 = 17

Practice Test: Computer Science Engineering (CSE) - 11 - Question 14

Which of the following is a hardware generated signal.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 14
A trap is a software-generated interrupt. An interrupt can be used to signal the completion of an I/O to obviate the need for device polling. A trap can be used to call operating system routines or to catch arithmetic errors.
Practice Test: Computer Science Engineering (CSE) - 11 - Question 15

In an array of 2N elements that is both 2-ordered and 3-ordered, what is the maximum number of positions that an element can be from its position if the Arrays were 1-ordered?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 15
An Arrays A [ 1 … n] is said to be k-ordered if

A [i - k] < A [ i ] < A [i + k]

for each i such that k < i < n – k.

For example, the Arrays 1 4 2 6 3 7 5 8 are 2- ordered.

Now come to question: Arrays are both 2 ordered and 3 ordered. i.e.

1 2 3 4 5 6 7 8

Now if the Arrays were 1-ordered then Arrays will be 1 2 3 4 5 6 7 8 maximum number of positions that an element can be from its position=1.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 16

Consider the following augmented grammar

S→aAb |eb

A→e |f

Find the number of states in LR(0) construction.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 16

No conflict in above DFA.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 17

Consider the following synchronous counter made up of JK, D and T flip-flops.

Find the modulus value of the counter.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 17
Consider characteristics equation of J-K flip-flop

Consider characteristic equation of D-Flip-Flop : QN+1=D

Consider characteristics equation of T-Flip-flop: Q_N+1=T⊕Q_N

Using equations (i), (ii) and (iii)

The number of used states = 5

∴ Modulus value of the counter = 5.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 18

The maximum value of “a” such that the matrix has three linearly independent real eigenvectors is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 18

The characteristic equation of A is

|A-XI| = 0

⇒ f(x) = x3 + 6x2 + 11x + 6 + 2a

= (x + 1)(x + 2)(x + 3)+2a = 0

f(x) cannot have all 3 real roots (if any) equal

for if f(x) = (x-k)3, then comparing coefficients, we get

6 = –3k, 3k2 = 11

No such k exists

A) Thus f(x) = 0 has repeated (2) roots α, α, β

or

B) f(x) = 0 has real roots (distance) α, β, δ

Now

At x1, f(x) has relative maximum

At x2, f(x) has relative minimum

The graph of f(x) will be as below

Case A. repeated roots (α, α, β)

Case B. distinct roots

Note that the graph of f(x) cannot be like the one given below

Thus in all possible cares we have

Practice Test: Computer Science Engineering (CSE) - 11 - Question 19

Match the following -

LIST- I

a. Token

b. Pattern

c. Lexeme

LIST – II

I) A rule describing the set of lexemes that can represent a particular token in the source program

II) A sequence of characters in the source program that is matched by the pattern for a token

III) A set of characters grouped together

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 19
Token is a set of characters grouped together like literal, id, num, const, if etc.

Pattern is a rule describing the set of lexemes that can represent a particular token in the source program. For example, if the token is related then its informal description of the pattern will be < or <= or = or > or >= or <> etc.

Lexeme is a sequence of characters in the source program that is matched by the pattern for a token. For example let the token be an identifier then its sample lexemes can be pi, count, a, b etc.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 20

The Servlet Response interface enables a servlet to formulate a response for a client using the method ___________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 20
void set Context type (String type)

- It’s used to send the content type of a response to be sent to a client, if the response has not been sent yet. The content type may include character encoding specifications.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 21

Let S = {1, 2, 3, . . . n} and G be a simple graph where every subset of S is a vertex in G. There will be an edge between two sub-set of S when they intersect in exactly 3 elements. What will be the degree of a vertex containing 4 elements?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 21
Let the vertex having 4 elements is {1, 2, 3, 4}. Now select any 3 elements from this, say X = {1, 2, 3 ……..}, the number of ways to do this is 4C3.

Now for remaining n-4 vertices there is a choice whether to be included or not. So using the product rule, we get 2 × 2 × 2 × ….. (n-4) times.

So, we got the degree of the vertex having 4 elements is 4C3 x 2n-4

Practice Test: Computer Science Engineering (CSE) - 11 - Question 22

The Breadth First Search algorithm has been implemented using the queue data structure. One possible order of visiting the nodes of the following graph if Q is starting node.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 22

Q →M

Q →N

Q →P

M→R

P →O

Using BFS Algorithm Order of visiting the nodes QMNPRO or QMNPOR

A) MNOPQR → If you try to run BFS, after M, you must traverse NQR (In some order) Here P is traversed before Q, which is wrong.

B) NQMPOR → This is also not BFS. P is traversed before O!

D) QMNPOR → Incorrect. Because R needs to be traversed before O.(Because M is ahead of N in the queue.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 23

A decimal number has 18 digits, The number of bits needed for its equivalent binary representation is(Approximately)?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 23
Maximum number using 18 digits = 1018-1

1018-1= 2n

Take log both side

log10(1018-1) = log10(2n)

~ log10(1018) = log10(2n)

18 = n log102

n = 18/ log102

n = 59.79

~ n = 60

Practice Test: Computer Science Engineering (CSE) - 11 - Question 24

Which one of the following is NOT a valid identity?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 24

(a) x ⊕ y = (xy + x′y ′)′

= (xy )′

= x ⊕ y, it is valid.

(b)

=

= Σm(1, 2, 4, 6)

x⊕(y+z)=x ̅(y+z)+x((y+z) ̅ )

=

= Σm(1, 2, 3, 4)

(x + y) ⊕ z ≠ x ⊕ (y + z)

So option (b) is invalid.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 25

Consider the following functions:

f(n)=2n

g(n)=n!

h(n)=nlogn

Which of the following statements about the asymptotic behaviour of f(n), g(n), and h(n) is true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 25

For asymptotic behaviour

h(n) = O(f(n))

g(n) = Ω(f(n))

A function having its upper bound denoted by O notation

Ω notation provides a lower bound of asymptote.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 26

Husband and wife apply for two vacancies that arise in a private company. Two friends apply for that position. If the probability of 1st friend's selection and 2nd friend's selection are 3/5 and 1/6 respectively,

Find the probability of both friends getting rejected.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 26

Probability of 1st friend's selection is 3/5

Probability of 1st friend's rejection is (1-3/5)=2/5

Probability of 2nd friend's selection is 1/6

Probability of 2nd friend's rejection is 1-1/6=5/6

probability of both friends get rejected is (2/5)×(5/6)=1/3

Practice Test: Computer Science Engineering (CSE) - 11 - Question 27

If A is 3 × 3 matrix, whose elements are given by aij = i2 – j2 where 1 ≤ i, j ≤ 3 then A–1 = _____

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 27

aij = i2 – j2∀ i, j

∴ A3×3 is singular

∴ |A| = 0.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 28

Consider the system which have 3 processor p1, p2, p3 the pack demand for each 5, 9,13 for the resource 'R' then what is the minimum no of resources require to insure deadlock free operation

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 28

P1 → 5 → 4

P2 → 9 → 8

P3 → 13 → 12

So P1 + P2 + P3 + 1 = 25.

Practice Test: Computer Science Engineering (CSE) - 11 - Question 29

Consider the languages L1={wwR∣w∈{0,1}∗}

L2={w#wR∣w∈{0,1}∗}, where # is a special symbol

L3={ww∣w∈(0,1}∗)

Which one of the following is TRUE?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 29
In formal language theory, a context-free language (CFL) is a language generated by some context-free grammar (CFG). Different CF grammars can generate the same CF language. It is important to distinguish properties of the language (intrinsic properties) from properties of a particular grammar (extrinsic properties). The set of all context-free languages is identical to the set of languages accepted by pushdown automata, which makes these languages amenable to parsing. Indeed, given a CFG, there is a direct way to produce a pushdown automaton for the grammar (and corresponding language), though going the other way (producing a grammar given an automaton) is not as direct.
Practice Test: Computer Science Engineering (CSE) - 11 - Question 30

Consider the following code with 2-processes.

If two processes are executing concurrently, which of the following execution orders never arise by them?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 11 - Question 30

order never arises.

Process P2 can execute D first then E. So option D. is correct.

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