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

Practice Test: Computer Science Engineering (CSE) - 14 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) - 14 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 14 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) - 14 below.
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Practice Test: Computer Science Engineering (CSE) - 14 - Question 1

Pipe A can fill a tank three times as fast as pipe B. If together two pipes can fill the tank in 48 min, the slower pipe alone will be able to fill the tank in:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 1

A = 3B

Ratio of efficiency, A : B = 3 : 1

Ratio of times, A : B = 1 : 3 Total capacity = Total efficiency × Total time = 4 × 48 = 192 unit Time taken by slower pipe

B = Total Capacity / Efficiency of B = 192 / 1 = 192 min

Practice Test: Computer Science Engineering (CSE) - 14 - Question 2

In the following question, out of the four alternatives, select the word opposite in meaning to the given word.

Gratuitous

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 2
The word “gratuitous” means given or done free of charge. Thus, the word “costly” would be the correct antonym of the given word.

Gratis means without charge; free.

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Practice Test: Computer Science Engineering (CSE) - 14 - Question 3

Find the area bounded between parabola and the line y2 = x2y = 2

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 3
y = 2 y2 = x

⇒ x = 22 = 4

: Parabola and line intersect at the point (4,2) ∴ Area =

⇒ A = = 8/3 sq.units

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

Direction: The bar graph shows the number of employees working under the six different Departments (A, B, C, D, E, F) of a certain company. Study the diagram and answer the following questions.

If departments F and D are merged to create a new department G, then which department will have the least number of employees?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 4
If departments F and D are merged to create a new department G, then

Employees in department A = 25

Employees in department B = 6

Employees in department C = 10

Employees in department E = 15

Employees in department G = 8

∴ Department B has the least number of employees.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 5

In the following question, a 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.

Confusion prevails in madrasas in Uttar Pradesh over the distribution of free NCERT textbooks at the academic session ____________ from August.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 5
The answer is ‘has begun’ because we use Present Perfect Tense, if the action is important and not the time of action or an action that has recently finished.
Practice Test: Computer Science Engineering (CSE) - 14 - Question 6

A sum of Rs. 400 amounts to Rs. 480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 6

We know that, S.I = P x R x T / 100

A = S.I + P

480 = 5.1+400

⇒S.I = 480 − 400 = 80

⇒S.I = P × R × T100

⇒80 = 400 × R × 4100

⇒R = 5%

Now rate is increased by 2% So, new rate is 7% New S.I=400 × 7 × 4 / 100 = Rs.112

New Amount = S.I + P = 112 + 400 = Rs.512

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

The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.

P: The Information and Broadcasting Ministry plans to conduct an independent study to gauge the

impact of government advertisements on people.

Q: The advertisements are carried on various platforms, including print and visual media.

R: The Directorate of Advertising and Visual Publicity (DAVP) is the nodal agency of the government

for advertising on behalf of the various ministries.

S: The initiative comes ahead of the Lok Sabha election in 2019 for which the government is expected

to reach out to the people and highlight the works done by it in the past 4 years.

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

The paragraph talks about the plans and advertisements of The Information and Broadcasting Ministry, which is given in sentence P. P is the first statement. The word ‘initiative’ given in the sentence S is talking about the plans. Hence, S must follow P. Now, the introduction of the advertising agency is given in the sentence R, which must be the next statement. Thus, the sequence after rearrangement is PSRQ and option B is the correct answer.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 8

In the following question, some parts of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If the sentence is free from error, select 'No error'.

The gold foil used liberal (1)/ in Thanjavur paintings serves (2)/ many objectives that makes the painting more attractive. (3)/ No error

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 8

The error is in part (1) of the sentence. Change ‘liberal’ to ‘liberally’ because in this sentence it is in adjective form while the proper usage of liberal is in its adverb form i.e. ‘liberally’ as it qualifies the gold foil here.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 9

A,B and C can do a job in 6 days, 12 days and 15 days respectively. C works till 1/8 of the work is completed and then leaves. Rest of the work is done by A and B together. Time taken to finish the remaining work by A \& B together is how much?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 9

Remaining work = 1 - 1/8 = 7 / 8

(A+B)′s1 day′s work =

∴ Time taken in doing 7/8 part of work = ⅞ x 4 = 7/2

= days

Practice Test: Computer Science Engineering (CSE) - 14 - Question 10

A invests 1/3 part of the capital for 1/6 of the time, B invests 1/4 part of the capital for 1/2 of the time and C invests the rest of the capital for the rest of the time. Out of a profit of Rs. 23000, B’s share is?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 10

Ratio of their investment,

=

=

=

= 4 : 9 : 10

B's share = 23000 x 9/23 = Rs. 9000

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

Identify the correct translation into logical notation of the following assertion.

“You can not ride the roller coaster if you are under 4 feet tall unless you are older than 16 years old.”

Where q = “You can ride the roller coaster”

r = “You are under 4 feet tall”

s = “You are older than 16 years old”

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 11
A à B can be represented like B if A.

The third option seems the most appropriate as ~Q is on the right-hand side.

(r∧= s)→-q . If you try to interpret this mathematical statement, you will get that this is the most appropriate.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 12

The traditional computer system stores the data in the form of binary. Arjun is a professor who teaches binary systems to the students. He observes that if ternary (radix 3) is used instead of binary, space can be utilised more efficiently. If n be the no of bits in string when data is stored in the form of binary, then ternary system will need

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 12
Far a n bit string in binary, maximum data which can be

represented = 2n

No of bits required to stare it intaternary format= log3⁡2n = nlog3⁡2

Practice Test: Computer Science Engineering (CSE) - 14 - Question 13

Consider a processor that includes a base register with indexing addressing mode. Suppose an instruction is encountered that employs this addressing mode and specifies a displacement of 7B2, in hexadecimal. Currently the base and index register contain the decimal numbers 48022 and 8, respectively. What is the address of the operand (In Decimal)?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 13
Address of operand= base register value + Index Value + Displacement

= 48022 + 8+ 1970

= 50000

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

A non-pipelined processor has a clock rate of 2.5 GHz and an average CPI (cycles per instruction) of 4. An upgrade to the processor introduces a five-stage pipeline. However, due to internal pipeline delays, such as latch delay, the clock rate of the new processor has to be reduced to 2 GHz.

What is the speedup achieved for executing 100 instructions and MIPS of the upgraded processor?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 14
For a non-pipelined processor,

Time for execution= no of cycle per instruction × cycle time* no of instruction

= 5 × 100 × 0.4 nsec

= 200 nsec

For a pipelined processor,

Time for execution= [no of stages + (no of instruction -1)]× cycle time

= [5 + 99] × 0.5 nsec

= 104 × 0.5 nsec

= 52 nsec

Speed Up= 200/ 52

= 3.84

For upgraded processor, clock frequency= 2GHz

Cycle time= 1/2 nsec = 0.5 nsec

In pipelined processor, the average time of executing an instruction is 1 cycle

Therefore no of instructions in 1 sec= 1/0.5 × 109 instructions

=2000 MIPS

Practice Test: Computer Science Engineering (CSE) - 14 - Question 15

Consider a relation R(A B C) with attribute size of A as 8 bytes. Disk block size is 512 bytes and block pointer is 8 bytes. The best choice for degree (maximum value) for B+ tree, if B+ tree was used for creating indexing on R(A B C) is _________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 15
Let p be degree of B + tree internal node

(p - 1) keys + p Block pointers should fit in a block ie (p-1) keys + p Block pointers size ⇐ 512

(2p - 1) × 8 ⇐ 512

p ⇐ 65/2

p = 32

If you take p = 33 node size becomes 520 bytes so not possible to fit in a block hence the correct answer is 32.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 16

Find the number of states in minimal FA for the following language :

L = {(a+b) × | number of a's = (0 mod 4) and (0 mod 8)}


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 16

Let set A = number of a's = 0 mod 4

and set B = number of a's = 0 mod 8

clearly , B is a subset of A

Hence A ∩ B = B

So the number of states will be determined only by set B

Hence , the number of a's = 0 mod 8 requires 8 states.

Hence answer = 8

Practice Test: Computer Science Engineering (CSE) - 14 - Question 17

Find the number of M-N equations for the language, L = [anbm | m,n 0]

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 17

Number of M-N equations = number of states in DFA

The DFA for given language is :

Hence M-N equations = 4

Practice Test: Computer Science Engineering (CSE) - 14 - Question 18

Suppose a processor uses a round robin scheduling algorithm to schedule the process. Previously, it was using a time slice of 2 units while scheduling the process. An update is made to the system and the time slice is now changed to 4 units. Then the turnaround time of the process will

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 18
For the process with small burst times, turnaround time will decrease but for the process with large burst time, it will increase. Hence we can say it varies according to the burst time of the process.
Practice Test: Computer Science Engineering (CSE) - 14 - Question 19

Which of the following is/are true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 19

1. Option A is false, because semaphores are the solution used to avoid busy waiting.

2. Option B is false, Since Priority inversion causes the problem of LiveLock and priority Inheritance is the solution for Live Lock.

3. Option C is true.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 20

To realise a 128x1 MUX , how many 4x1 MUX are required _____________


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 20

128/4 = 32

32/4 = 8

8/4 = 2

2/4 = 0.5 ≅ 1

Total = 32 + 8 + 2 + 1 = 43

Practice Test: Computer Science Engineering (CSE) - 14 - Question 21

Consider the following Relational Schema:

Sailors (sid: integer, sname: string, rating: integer, age: real)

Boats (bid: integer, bname: string, color: string)

Reserves (sid: integer, bid: integer, day: date)

Consider the Following Statements:

S1:SELECT S.sname

FROM Sailors S

WHERE NOT EXISTS (SELECT *

FROM Sailors S2

WHERE S2.age < 21 AND S.rating <= S2.rating )

S2: SELECT S.sname

FROM Sailors S

WHERE S.rating > ANY (SELECT S2.rating

FROM Sailors S2

WHERE S2.age < 21 )

Which of the following is true regarding S1 and S2?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 21

S1: it will generate the name of sailors whose rating is more than the rating of every sailor whose age is less than 21.

S2: it will generate the name whose rating is more than the rating of the same sailor with age less than 21.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 22

Consider a binary tree with the following more conditions :

Condition 1: The elements present in the left sub-tree of a specific node are smaller than that node and elements present in the right sub-tree of a specific node are greater than that node.

Condition 2: The tree is always balanced i.e. the height of the left subtree and right sub-tree is approximately the same. (maximum differ by 1)

What is the worst-case time complexity to run a membership test for an element on this tree?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 22
The tree is nothing but a balanced BST where the worst case time to find an element = O(logn)
Practice Test: Computer Science Engineering (CSE) - 14 - Question 23

Consider a lower triangular matrix. When this lower triangular matrix is stored in array format then only the elements a[i][j] with i≥j are stored in array i.e only the elements present in lower triangular matrix are stored. Hence less size is consumed to store the array. Consider a lower triangular matrix as [25----100 , 25----100] with base address as 1000 and size of each element in matrix is 10, If the array is stored in column major order then find the address of the element a [80][45] stored in array ________________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 23

In general , if the array is a [lb1---ub1 , lb2----ub2]

Base address = BA

Size of element = c

Number of rows = nr = ub1- lb1 + 1

Number of columns = nc = ub2 - lb2 + 1

and lower triangular matrix stored in column major order then

=

So,

Practice Test: Computer Science Engineering (CSE) - 14 - Question 24

Suppose there are certain number of peoples who’s usage are ALOHA dependent and they are producing 100 requests/sec. Consider time to be slotted in 50msec units. After considering all of the above cases find out the chances of win on the 1st attempt?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 24

Number of slots per second = 1/50msec = 20

Channel load = = 100/20 = 5

We know, Poisson Distribution

or the success on the first attempt

K=0

P0 = e-G = e-5

= 6.73 x 10-3

Practice Test: Computer Science Engineering (CSE) - 14 - Question 25

You have a class B network 172. 16. 0. 0. You use 11 bits for subnetting. Which of the following is a correct range of IP addresses that belong to the same network?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 25

If we use 11 bits for subnetting, we have 5 networking bits and 11 node bits

From the subnetting formulas above:

M = 5, N=11

The first range = 255.255.X.1 to 255.255.Y.254 where X = 2(N-8) + 1 and Y = 2(N-8-1) – 2

For the next ranges, just add 2(N-8) at each end of the range

The first range = 255.255.9.1 to 255.255.14.254 = 2(N-8)= 23=8

The second range = 255.255.17.1 to 255.255.22.254 so on and so forth.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 26

Consider the following relations, SQL query and given instances of relations: (where keys are underlined)

Student (snum, sname)

Enroll (snum, cname)

SELECT S.name FROM Student S WHERE

S.snum NOT IN (SELECT E.snum FROM Enroll E)

Number of tuples returned by the SQL query is ________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 26

SELECT S.Name FROM Student S WHERE

S.snum NOT IN (SELECT E.snum FROM Enroll E)

It return snum of all students who is enrolled in any course.

Query return the sname of a student who is not enrolled in any course.

SQL does not eliminates duplicate so relation given by SQL query is

Total 2 tuple returns.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 27

Consider two languages, L1and L2:

L1= {an| n > = 0} and L2 = {bn | n > = 0}

Which of the following correctly represents L1⋅ L2, where ‘⋅’ is the concatenation operation?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 27
L1 . L2 will be equal to {an bm| m , n > = 0}, which is same as {an bm| m = n, n > –1}.
Practice Test: Computer Science Engineering (CSE) - 14 - Question 28

Consider the following regular expressions over the alphabet {0, 1}.

I. 1* 0(0 + 1)*

II. (0 + 1)* 01*

III. 1* 0(1 + 0)* + (0 + 1)* 01*

Which of the above regular expressions are equivalent?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 28
I and II are quite easy to understand - both denote strings containing at least one 0. III is actually the union of I and II, but III is again denoting strings containing at least one 0, as both regular expressions (i.e. I and II) are equivalent, so union won't make any difference.

Hence all 3 regular expressions are equivalent.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 29

A 4-bit preset table UP counter has preset input 0111. The preset operation takes place as soon as the counter becomes maximum, i.e. 1111. The modulus of this counter is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 29

Preset → 8 clock pulses, it is repeating

So mod 8.

Practice Test: Computer Science Engineering (CSE) - 14 - Question 30

Let R be a recursive language. Consider the following operations on languages.

I. Union

II. Intersection

III. Complement

Let X be the number of operations under which R is closed. Then the value of X is ________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 14 - Question 30
Recursive languages are closed under all the above operations.
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