GATE Exam  >  GATE Tests  >  Test : CSE Past Year Paper 2019 - GATE MCQ

Test : CSE Past Year Paper 2019 - GATE MCQ


Test Description

30 Questions MCQ Test - Test : CSE Past Year Paper 2019

Test : CSE Past Year Paper 2019 for GATE 2024 is part of GATE preparation. The Test : CSE Past Year Paper 2019 questions and answers have been prepared according to the GATE exam syllabus.The Test : CSE Past Year Paper 2019 MCQs are made for GATE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test : CSE Past Year Paper 2019 below.
Solutions of Test : CSE Past Year Paper 2019 questions in English are available as part of our course for GATE & Test : CSE Past Year Paper 2019 solutions in Hindi for GATE course. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free. Attempt Test : CSE Past Year Paper 2019 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study for GATE Exam | Download free PDF with solutions
Test : CSE Past Year Paper 2019 - Question 1

Two cars start at the same time from the same location and go in the same direction. The speed of the first car is 50 km/h and the speed of the second car is 60 km/h. The number of hours it takes for the distance between the two cars to be 20 km is _________.

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 1

Let us take two cars A and B
VA = 50 km/hr
V= V 60 km/hr




 

Test : CSE Past Year Paper 2019 - Question 2

2. The expenditure on the project _______ as follows: equipment Rs.20 lakhs, salaries Rs.12 lakhs, and contingency Rs.3 lakhs.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test : CSE Past Year Paper 2019 - Question 3

Ten friends planned to share equally the cost of buying a gift for their teacher. When two of them decided not to contribute, each of the other friends had to pay Rs 150 more. The cost of the gift was Rs __________.

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 3

Let us consider cost of gift = X
If 10 friends contributes then share = X/10
Two friends denied to contribute then remaining are 8 each of remaining to be given 150 more

Test : CSE Past Year Paper 2019 - Question 4

A court is to a judge as ________ is to a teacher.

Test : CSE Past Year Paper 2019 - Question 5

The search engine‟s business model ________ around the fulcrum of trust.

Test : CSE Past Year Paper 2019 - Question 6

Three of the five students allocated to a hostel put in special requests to the warden. Given the floor plan of the vacant rooms, select the allocation plan that will accommodate all their requests.
Request X: Due to pollen allergy, I want to avoid a wing next to the garden.
Request by Y: I want to live as far from the washrooms as possible, since I am very sensitive to 
smell. 
Request by Z: I believe in Vaastu and so want to stay in the South-west wing.
The shaded rooms are already occupied. WR is washroom.

Test : CSE Past Year Paper 2019 - Question 7

The police arrested four criminals –P, Q, R and S. The criminals knew each other. They made the following statements:
P says “Q committed the crime.”
Q says “S committed the crime.”
R says “I did not do it.”
S says “What Q said about me is false.”
Assume only one of the arrested four committed the crime and only one of the statement made above is true. Who committed the crime?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 7

Explanation: Given that only one person committed the crime and only one given statement is correct.

When P is correct and Q, R, S are false: it implies,Q committed crime, S did not commit crime, R committed crime, S committed crime.

This situation is impossible.

When Q is correct and P, R, S are false: it implies,Q did not commit crime, S committed crime, R committed crime, S committed crime.

This situation is impossible.


When R is correct and P, Q, S are false: it implies,Q did not commit crime, S did not commit crime, R did not commit crime, S committed crime.


This situation is impossible.


When S is correct and P, Q, R are false: it implies,Q did not commit crime, S did not commit crime, R committed crime, S did not commit crime.

This situation is possible means R has committed crime.

So, option (B) is correct.

Test : CSE Past Year Paper 2019 - Question 8

“A recent High Court Judgment has sought to dispel the ideal of begging as a disease which leads to its stigmatization and criminalization – and to regard it as a symptom. The underlying disease is the failure of the state to protect citizens who fall through the social security net.”
Which one of the following statements can be inferred from the given passage?

Test : CSE Past Year Paper 2019 - Question 9

In a college, there are three student clubs, Sixty students are only in the Drama club, 80 students are only in the Dance club, 30 students are only in the Maths club, 40 students are in both Drama and Dance clubs, 12 students are in both Dance and Maths clubs, 7 students are in both Drama and Maths clubs, and 2 students are in all the clubs. If 75% of the students in the college are not in any of these clubs, then the total number of students in the college is ________.

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 9

Total number of students in the all three clubs
= 60 + 38 + 80 + 5 + 2 + 10 + 30 = 225
Total number of students in the college = X

Given that 75% of students are not of any of these clubs.
Remaining 25% of X = 225
0.25X = 225 ⇒ X = 225/0.25 = 900

Test : CSE Past Year Paper 2019 - Question 10

In the given diagram, teachers are represented in the triangle, researchers in the circle and administrators in the rectangle. Out of the total number of the people, the percentage of administrators shall be in the rage of ___________.

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 10

Total number of people = 70 + 10 + 20 + 20 + 40 = 160
Total number of administrators = 10 + 20 + 20 = 50
% of administrators = 

Test : CSE Past Year Paper 2019 - Question 11

Which one of the following kinds of derivation is used by LR parsers?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 11

LR parser uses rightmost derivation in reverse order.

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 12

Consider the following C program:
#include <stdio.h>
int main(){
int arr[ ]={1,2,3,4,5,6,7,8,9,0,1,2,5},  *ip=arr+4;
printf("%d \n", ip[1]);
return 0;
}
The number that will be displayed on execution of the program is _________.


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 12

*ip=arr+4 will point element 5 in the array, since it will increment the base address of the array by 4.
ip[1] is equal to *(ip+1)
it will further increment the pointer to next element and hence it will print 6.

Test : CSE Past Year Paper 2019 - Question 13

For ∑ = {a,b} , let us consider the regular language L = {x | x a2+3k or  x = b10+12k, k ≥ 0}. 
Which one of the following can be a pumping length (the constant guaranteed by the pumping lemma) for L?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 13

The set x is consists of {a2 , a5 , a8 , a11........} here the pumping length is 3 i.e. to get the another string we can repeat the length 3.
The regular expression is aa(aaa)*
Or { b10 , b22 , b34 , b46 ...........}, here the pumping length is 12 
The regular expression is b10 (bbbbbbbbbbbb)*
Pumping length can‟t be 5
Pumping length can‟t be 9, because it will generate 18 which is not multiple of 12
Pumping length can‟t be 3, since it will generate 15, which is not multiple of 12
Possible pumping length can be 24 since every repetition will be multiple of 3 and 12.

Test : CSE Past Year Paper 2019 - Question 14

Let U = {1, 2, …, n}. Let A = {( x, X ) | x ∈ X, X ⊂ U}. Consider the following two statements on |A|.

Which of the above statements is/are TRUE?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 14

U {1, 2, ....n}
X U
i.e., X is subset of U
and number of subsets are = 2[ since n elements are there]
In case of nC0 n i.e., subset is φ
We cannot select any element
In case of mC1 we have only one choice
In case of nC2 we have selected two elements subset
So there exists two value of x & . t x ∈ x
And this goes on.

Test : CSE Past Year Paper 2019 - Question 15

A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the Tag and the Index fields respectively in the addresses generated by the processor?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 15

The division of fully associative cache is 
 Main memory address(32 bits)
Cache size is 16 kB
Block size is 16 Bytes , byte offset is log2 = 4bits
Number of blocks 16kb/16 = 1k = 210
Number of bits required to index is 10 bits.
Since fully associative is not having index field hence TAG field is 32 – 4 = 28 bits (D) is the Key.

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 16

Consider the grammar given below:
S → Aa
A → BD
B → b | ∈
D → d | ∈
Let a, b, d, and $ be indexed as follows:

Compute the FOLLOW set of the non-terminal B and write the index values for the symbols in the FOLLOW set in the descending order. (For example, if the FOLLOW set is {a, b. d, $}, then the Key should be 3210)


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 16

follow(B)  = first(D)
first(D) is { d , ∈ }
follow (B)  is {d} and since first(D) contains e
follow (B) is first (A)
first(A) is {a}
Index of a is 3 and d is 1
Hence 31.

Test : CSE Past Year Paper 2019 - Question 17

Let X be a square matrix. Consider the following two statements on X.
I. X is invertible
II. Determine of X is non-zero.
Which one of the following is TRUE?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 17

If | A | ≠ 0 then A is invertible matrix
If A -1 exists then | A | ≠ 0
∴ I and II are equivalent statements.

Test : CSE Past Year Paper 2019 - Question 18

The chip select logic for a certain DRAM chip in a memory- system design is shown below Assume that the memory system has 16 address lines denoted by A15 to A0. What is the range of addresses (in hexadecimal) of the memory system that can get enabled by the chip select (CS) signal? 

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 18

Chip select signal will be enable only of 

And rest of the bits 

Will be available from 00000000000 to 11111111111 Hence address 

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 19

Consider a sequence of 14 elements: A = {-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0]. The subsequence sum  Determine the maximum of S (i, j ) , where 0 ≤ i ≤ j < 14 . (Divide and conquer approach may be used.)


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 19

The maximum sum of the array
{6, 3, -1, -2, 13, 4, -9, -1, 4, 12 }
Which sum to 29.

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 20

An array of 25 distinct elements is to be sorted using quick sort. Assume that the pivot element is chosen uniformly at random. The probability that the pivot element gets placed in the worst possible location in the first round of partitioning (rounded off to 2 decimal places) is


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 20

The worst possible location for the pivot is either first place or the last place
Hence the probability is 2/25 = 0.08

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 21

The value of 351 mod 5 is _______.


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 21

351 mod 5 = ?

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 22

Consider the concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialization to 100.

The processes are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y–X is _________.


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 22

For maximum value:
First P2 & P1 read the D value as 100.
So, P2 update it first 
⇒ D = D - 50 = 100 - 50 = 50
After P2, P1 overwrite the value 50 by 120
⇒ D = D + 20 =100 + 20 =120
Then P3 read the value of D as 120 and adds 10 to it
⇒ D = 120 + 10 = 130
∴ The max value of D after all 3 processes are executed is 130(Y).
For minimum Value:
All processes read the D value as 100.
But, ensure that last update is from P2.
So that, the min value of D after all 3 processes are executed is 50 (X).
∴ Difference between X and Y is ( Y - X) = 130 - 50 = 80

Test : CSE Past Year Paper 2019 - Question 23

Compute 

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 23

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 24

Consider the following C program:
#include <stdio.h>
int jumble(int x, int y){
x=2*x+y;
return x; }
int main(){
int x=2, y=5;
y= jumble(y,x);
x= jumble(y,x);
printf("%d \n", x);
return 0;
}
The value printed by the program is _________


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 24

Initial value of x and y is given as 2 and 5
y= jumble(y,x);
In function
5 is stored in x and 1 stored in y
x = 2*x+y is equal to 2*5+2 = 12 x = 12
return value update y to 12
x= jumble(y,x);
now 12 is stored in x and 2 stored in y
x = 2*x+y is equal to 2*12+2 = 26
x = 26
return value of update x to 26 value printed is 26

Test : CSE Past Year Paper 2019 - Question 25

Let G be an arbitrary group. Consider the following relations on G:

Which of the above is/are equivalence relation/relations?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 25

Let us take an example of multiplication modulus 5

Let us take R2
For Reflexive

So Not reflexive, So R is not equivalence
Let us take R1
For Reflexive

So It is reflexive
If a = g-1bg

[Since multiplication is commutative]

For transitive

So transitive,  so R1 is equivalent relation
 

Test : CSE Past Year Paper 2019 - Question 26

Consider the following two statements about database transaction schedules:
I. Strict two-phase locking protocol generates conflict serializable schedules that are also recoverable.
II. Timestamp-ordering concurrency control protocol with Thomas‟ Write Rule can generate view serializable schedules that are not conflict serializable.
Which of the above statements is/are TRUE?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 26

Strict 2 phase locking protocol ensures conflict serializability as well as strict recoverability.
So statement I is correct.
Thomas write rule sometimes discard the operation in case of multiple write and it is similar to view serializability, focus on last write.
So statement-II is also correct.

*Multiple options can be correct
Test : CSE Past Year Paper 2019 - Question 27

Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 27

Option (B):
For labeled nodes,

For an undirected complete graph G.
Number of Hamiltonian cycles are 

3 cycles are;
ABCDA
ACBDA
ACDBA
Option (C):
For unlabelled nodes:
Every Hamilton cycle will be similar. So answer is 1.
Since in question it is not mentioned whether the graph is labeled or not. So both answers are accepted.

Test : CSE Past Year Paper 2019 - Question 28

Which of the following protocol pairs can be used to send and retrieve e-mails (in that order)?

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 28

SMTP and POP3 are application layer protocols which are responsible for email services 
SMTP → Sending or outgoing mail(P ush protocol)
specifically
POP3 → Retrieving or downloading mail(Pull protocol)

*Answer can only contain numeric values
Test : CSE Past Year Paper 2019 - Question 29

The following C program is executed on a Unix/Linux system:
# include < unistd.h>
int main ( )
{int i; for (i = 0; i < 10; i++) if (i % 2 = = 0) fork ( ); return 0;}
The total number of child processes created is ________.


Detailed Solution for Test : CSE Past Year Paper 2019 - Question 29

According to given logic, child creation is successful only for 5 times between 0 to 9 (As for loop is executed for 10 times)
∴ Number of child processes = 2n -1 = 25 -1 = 31

Test : CSE Past Year Paper 2019 - Question 30

Consider Z = X – Y, where X, Y and Z are all in sign-magnitude form. X and Y are each represented in n bits. To avoid overflow, the representation of Z would require a minimum of:​

Detailed Solution for Test : CSE Past Year Paper 2019 - Question 30

Ist method:
Take X=31, Y=-31, see both need 6 bits. But Z=X-Y=62 needs 7 bits in sign magnitude form. So, n+1 bits should be answer.
(or)
IInd method:
It is given that x, y, z are in sign magnitude form and represented using „n‟ number of bit.


110 → represent (-2) which is between overflow
If we represent 3 using 4 bits

Same result can be verified by taking 
7+7
15 + 15
31 + 31.......So on
So in general z needs n + 1 number of bits.

View more questions
Information about Test : CSE Past Year Paper 2019 Page
In this test you can find the Exam questions for Test : CSE Past Year Paper 2019 solved & explained in the simplest way possible. Besides giving Questions and answers for Test : CSE Past Year Paper 2019, EduRev gives you an ample number of Online tests for practice
Download as PDF