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

Practice Test: Computer Science Engineering (CSE) - 8 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) - 8 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 8 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) - 8 below.
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Practice Test: Computer Science Engineering (CSE) - 8 - Question 1

Direction: Study the following pie-chart and tables carefully and answer the questions given below.

Total Number of email received by the organization = 90000

Ratio of Read emails to Unread emails received by the organization

Q. What is the ratio of the number of emails read in January to those unread in the month of April in the organization?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 1

Number of read emails in the month of January = (90000*17/100*8/15)
Number of unread emails in the month of April = (90000* 8/100 * 5/12)
Ratio = 68:25

Practice Test: Computer Science Engineering (CSE) - 8 - Question 2

answer which can fill both the blanks of both the sentences.

I. Most of our employees get _____ abroad at some stage

II. The aircraft and its crew were ______ missing.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 2

Post means to send somebody to a place for a period of time as part of their job and to announce something publicly or officially.
Hence, option D is correct.

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Practice Test: Computer Science Engineering (CSE) - 8 - Question 3

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

Scurrilous

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 3

The meanings of the given words are as follows:
Scurrilous— abusive
Coarse— rough or harsh in texture
Sophisticated— developed to a high degree of complexity.
Insolent— showing a rude and arrogant lack of respect
Complimentary— expressing a compliment; praising or approving
Clearly, option D is the antonym of the given word.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 4

Statements:
All pillows are beds.
No fruit is pillow.
Some foods are fruits.
Conclusions:
I. At least some foods are pillows.
II. Some beds are definitely fruits.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 4

Consider the following least possible Venn diagram,

Conclusions:
I. At least some foods are pillows → it’s possible but not definite, hence false.
II. Some beds are definitely fruits → it’s possible but not definite, hence false.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 5

Direction: Study the following information carefully and answer the questions given below:

P, Q, R, S, T, U, V and F are sitting around a circle facing the centre. U is third to the right of Q, who is third to the right of F. P is third to the left of F. R is fourth to the left of P. T is third to the right of S. S is not a neighbour of P.Four of the following five are similar in a certain way based on their positions in the seating arrangement and so form a-group. Which of the following does not belong to that group?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 5


Option D does not belong the group.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 6

Following bar graphs shows time taken by different pipes to fill that particular percentage of tank which is described in tabular data. For Ex- Pipe 1 fills the 10 % of tank in 6 mins.
Now, read the following data and answer the question carefully:


A Tank has three pipes; two of them are used to fil the tank while another one used to empty the tank namely: pipe 3, pipe 5 to fill the tank and a third pipe for making the tank empty. When all three pipes are open, 7/18th part of the tank is filled in 1 hours. How much time will the third pipe take to empty the completely filled tank 10 times the bigger than the previous one.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 6

part of work in one hour=
(P3+P5+Third pipE. X 1 = 7/18th part of work
⇒ 7/18th part of work = 1 hr
⇒ Total work done = 18/7 hr by all three pipes
Pipe 3 does 50% of work in 90 mins
100 % of work = 90/50 * 100 = 180 mins = 3 hr
Pipe 5 does 50% of work in 60 mins
100 % of work = 60/50 * 100 = 120 mins = 2 hr
Time Tw Efficiency
Pipe -3 = 3 hr Total work/3 = 6
Pipe-5 = 2 hr Total work/2 = 9
All three = 18/7 hr Total work/(18/7)= 7
For Total work = LCM of (3,2,18/7) = 18
Efficiency of pipe 3 + Efficiency of pipe 5 + Efficiency of third pipe = total efficiency
6+9+X= 7
X = 7-15 = -8 (Negative sign shows the negative work herE.
Efficiency X Time = Total work
8 X Time = 18 * 10 ( as tank is 10 times bigger than the previous onE.
Time = 180/8 = 22.5 hrs

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

In a school, 12th class consists of 30% male students of which 30% male students failed in the class. Total 82% students passed in 12th examination out of 900 students. Calculate the total number of female passed students?

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

Short Trick:
Total students = 900
Boys : Girls = 30 % : 70% = 3 : 7
So, exactly Boys : Girls = 270 : 630
Total failed students = 18% of 900 = 162
Failed boys = 30% of 270 = 81
Failed girls = 162 - 81 = 81
Passed female students = 630 - 81 =549
Basic Method:
⇒ Let us consider total number of students be 100x
⇒ ∴ Males students = 30x , female students = 70x
⇒ According to condition given in the problem, of this 30x male students, 70 % male students pass the exam.
ie. Passed males = 

⇒ Total passed students = 

⇒ ∴ female students passed = 82x – 21x = 61x
⇒ ∴ Girls passed = 61% of total
∴ Answer is 

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

Directions: Rearrange the following six sentences (A), (B), (C), (D), (E) and (F) in the proper sequence to form a meaningful paragraph: then answer the questions given below them.

A). Indian Government remain concerned about potential spillover effects from the unconventional monetary policies of the advanced economies, which could cause disruptive volatility of exchange rates, asset prices and capital flows.
B). The challenges are related to high public debt and unemployment, poverty and inequality, lower investment and trade, negative real interest rates along with signs of prolonged low inflation in advanced economies.
C). In this context, emerging markets and developing countries (EMDCs) continue to be major drivers of global growth.
D). The global recovery continues, although growth remains fragile, with considerable divergences across countries and regions.
E). It is important to strengthen the framework of international financial cooperation, including through instruments such as swap-lines, to mitigate the negative impacts of monetary policy divergence in reserve currency issuing countries.
F). Structural reforms, domestic adjustment and promotion of innovation are important for sustainable growth and provide a strong and sustainable contribution to the world economy.
Which of the following will be the Second sentence?

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

Refer to the last question of the series.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 9

17 25 34 98 121 339

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 9

The pattern followed here is,
⇒ 17 + 23 = 17 + 8 = 25
⇒ 25 + 32 = 25 + 9 = 34
⇒ 34 + 43 = 34 + 64 = 98
⇒ 98 + 52 = 98 + 25 = 123
⇒ 123 + 63 = 123 + 216 = 339

Practice Test: Computer Science Engineering (CSE) - 8 - Question 10

There are 3 red balls and 5 green balls in a bag. If two balls are picked at random from the bag then find the probability of getting a red ball?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 10

Required probability
=3C1*5C1/8C2
= 3*5/28
= 15/28

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 8 - Question 11

The size of a proper subgroup of a finite group G is 31. The size of the smallest possible group G is ____.


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

The order of a subgroup divides the order of a group. Here the order of the subgroup is 31 therefore the smallest possible size of group is 62 which is evenly divisible by 31.
Tags: 1 Mark

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 8 - Question 12

Find the number of seven digit integers with sum of the digits equal to 11 and formed by using the digits 1, 2 and 3 only.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 12

Total possibility with sum = 11 and 7 digits
3,3,1,1,1,1,1
3,2,2,1,1,1,1
2,2,2,2,1,1,1

The number of 7 digit integers = 21+105+35 = 161

Practice Test: Computer Science Engineering (CSE) - 8 - Question 13

Consider the following statements:
i. Selection sort performs minimum number of swaps.
ii. Insertion sort performs worst in case of sorted array.
iii. Floyd Warshall uses dynamic programming to calculate all pairs of shortest paths

Q. Which of the statements are true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 13

Selection sort uses minimum number of swaps. In each pass, there can be at most 1 swap. So there can be at most n swaps in worst case.
In case of insertion sort, if every element is less than every element to its left, the running time of insertion sort is Θ(n2). Thus insertion sort is worst for reverse sorted data. Floyd Warshall algorithm uses dynamic programming for all pairs of shortest path computation.

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

Consider the following recursive function find.
int find (int A[], int n)
{
int sum=0;
if(n==0) return0;
sum = find(A, n-1)
if (A[n-1]<0) sum=sum+1;
return sum;
}
Q. What is the worst case running time above function find (A[], n) when array A has 0 to n-1 elements?

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

Recurrence relation for the function find:
F(n)=0; if n=0
=F(n-1)+1 ;n>0
Time complexity of F(n)=O(n)

Practice Test: Computer Science Engineering (CSE) - 8 - Question 15

Let A be a collection of objects. An efficient method for converting it into a set is first to sort the objects of A. Then, walk through the sorted sequence and remove all duplicates. What is the running time of this method?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 15

This takes O(nlogn) time to sort and n time to remove the duplicates. Overall, therefore, this is an O(nlogn) time method.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 16

The following function test takes argument a queue and uses stack to perform some function:
void test(Queue *A)
{
Stack T;
while (!isEmpty(A))
push(&T, deQueue(A));
while (!isEmpty(&T))
enQueue(A, pop(&T));
}
Q. What does the function test do?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 16

The first while loop copies the data of the queue into the stack and the next while loop copies into the queue but stack follows LIFO. Hence, the last element will be enqueued first and then so. The order of the queue is reversed.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 17

Consider the following code
int DO(char *gate)
{
char *gate1 = gate;
char *gate2 = gate + strlen (gate) – 1;
while (gate1 < =gate2)
{
if (*gate1 ++! = *gate2 --)
Return 0;
}
return 1;
}
Q. What is the functionality of above function Do ()?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 17

Here two pointers are used gate 1 and gate 2. One pointer points to beginning and other to the end. Loop is a set up that compares the characters pointed by these two pointers. If the characters do not match then it’s not a palindrome. It returns 1 for both even and odd palindrome.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 18

{
printf(“%d%d%d%d%d”, ‘ \n ‘ ,printf(“\0”),printf(“\n”),’ \0 ‘ , ‘ \b ‘ );
}
What will be printed by this above code?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 18

ASCII value of ‘ \n ‘ is 10.that is printed.
printf(“\0”) // it prints nothing ,only returns zero. This returned zero value is printed.
printf(“\n” // it prints nothing ,only returns one ,This returned 1 is printed
‘ \0 ‘. //ASCII value of this null character is zero.that is printed .
‘ \b ‘ // ASCII value of this backspace escape sequence is 8,that is printed.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 19

Given Ts : Transfer time
B : Number of bytes to be transferred
N : Number of bytes on a track
R : Rotational speed.
Q. Which of the following expression gives the total aveage access time?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 19

Average access time is

Practice Test: Computer Science Engineering (CSE) - 8 - Question 20

Cache with 4 Blocks of 4 bytes. Give the number of cache hits and cache miss respectively

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 20

Block address =Byte Address/ Block size
Cache index=Block address% No. of Cache blocks
Tag=block address/No. of cache blocks

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 8 - Question 21

We have three stations P, Q, R connected in serial manner. P is connected to Q through a 3Gbps fibre optic link and length is 500Km. Q is connected to R through 60Mbps link and length is 15Km.All the links are full duplex in nature. A file is sent from station A to C. Packet size is 1KB.We use sliding window protocol such that SWS=RWS. Find the optimal SWS packets.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 21

First we have to calculate RTT.
RTT= Transmission period + 2*propagation delay
= (8192/3*109) +(8192/60 *106) + 2*(5*105/2*108)(15*103/2*108)
= 5.29 ms.
Number of packets to be sent =5.29 *60 Mbps =31.74≈32 packets.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 8 - Question 22

Consider a 3-bit number A and 2 bit number B are given to a multiplier. The output of multipier is realized using AND gate and one bit full adders. If minimum number of AND gates required are X and one bit full adders required are Y, then X + Y =


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 22


Number of AND gates required X=6
Number of one bit full adders required Y=3
X+Y=6+3=9

Practice Test: Computer Science Engineering (CSE) - 8 - Question 23

The output F of the 4-to-1 MUX shown in figure is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 23

F = xy(1) + xy'(1) + x'y(0) + x'y'(0)
=xy + xy'
= x

Practice Test: Computer Science Engineering (CSE) - 8 - Question 24

Let L= {|M accepts some string} where M is a Turing machine. Find the language L?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 24

Simulate M on all strings of length atmost n for n steps and keep increasing n. We accept if the computation of M accepts some string. Language generated by a grammar is recursively enumerable hence it is turing recognizable. So it is also recursively enumerabl language.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 25

Let A, B, C and D are problems. Consider the following polynomial reductions to known about the problem B.
(i) A≤B (A is reducible to B)
(ii) C≤ D
(iii) B≤ D
Find the correct statement from the following.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 25

A≤B
C≤ D
B≤ D
A. A is decidable ⇒ B need not be decidable.
B. D is undecidable ⇒ B need not be undecidable.
C. C is decidable ⇒ D need not be decidable.
D. A is undecidable ⇒ B is undecidable.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 26

Consider the relation scheme of the relation SCHEDULE shown below. What is the highest Normal form of this relation?
SCHEDULE (Stud_ID, Class, Stud_Name, Stud_Major, Class_Time, Building_Room, Instructor)
Assume the following functional dependencies
Stud_ID →Stud_Name
Stud_ID →Stud_Major
Class →Class_Time
Class →Building_Room
Class →Instructor

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 26

As per the given data there are some patial dependencies on the key,
For example : Class →Building_Room or Stud_ID → Stud_Major
Therefore, the highest form of the relation is 1NF

Practice Test: Computer Science Engineering (CSE) - 8 - Question 27

In a database system, unique time stamps are assigned to each transaction using Lamport’s logical clock. Let TS(T1) and TS(T2) be the timestamps of transactions T1 and T2 respectively. Besides, T1 holds a lock on the resource R, and T2 has requested a conflicting lock on the same resource R. The following algorithm is used to prevent deadlocks in the database system assuming that a killed transaction is restarted with the same timestamp.
if TS(T2) < TS(T1) then
T1 is killed
else T2 waits.
Assume any transactions that is not killed terminates eventually. Which of the following is TRUE about the database system that uses the above algorithm to prevent deadlocks?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 27

Given,
if TS(T2) <TS(T1) then
T1 is killed
else T2 waits.

  • T1 holds a lock on the resource R
  • T2 has requested a conflicting lock on the same resource R

According to algo, TS(T2) <TS(T1) then T1 is killed else T2 will wait. So in both cases neither deadlock will happen nor starvation.
Therefore, option A is correct

Practice Test: Computer Science Engineering (CSE) - 8 - Question 28

Consider the following statements
(i) SQL’s aggregation can’t be expressed in relational algebra
(ii) SQL’s grouping construct can be expressed in relation algebra.
Which of the following is correct?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 28

Only (i) and (ii) is correct. Grouping construct can’t be expressed in relational algebra.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 29

Which of the following SQL query find all the tuples having temperature greater than that of Paris.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 29

Inner query returns the temperature of the city Paris and this will be compared with temperature of all tuples in outer query.

Practice Test: Computer Science Engineering (CSE) - 8 - Question 30

Consider the following assembly code
MOV R1,b ; R1b
MOV R2,c ;
MUL R1,R2 ;
MOV t1,R1 ;
MOV R1,a ;
MOV R2,t1 ;
ADD R1,R2 ;
MOV t2,R1 ;
MOV R1,t2 ;
MOV d,R1 ;
Find the correct expression which is equivalent to the above code.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 8 - Question 30

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