Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Tests  >  GATE Computer Science Engineering(CSE) 2025 Mock Test Series  >  Practice Test: Computer Science Engineering (CSE) - 13 - Computer Science Engineering (CSE) MCQ

Practice Test: Computer Science Engineering (CSE) - 13 - Computer Science Engineering (CSE) MCQ


Test Description

30 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Practice Test: Computer Science Engineering (CSE) - 13

Practice Test: Computer Science Engineering (CSE) - 13 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) - 13 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 13 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) - 13 below.
Solutions of Practice Test: Computer Science Engineering (CSE) - 13 questions in English are available as part of our GATE Computer Science Engineering(CSE) 2025 Mock Test Series for Computer Science Engineering (CSE) & Practice Test: Computer Science Engineering (CSE) - 13 solutions in Hindi for GATE Computer Science Engineering(CSE) 2025 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free. Attempt Practice Test: Computer Science Engineering (CSE) - 13 | 65 questions in 180 minutes | Mock test for Computer Science Engineering (CSE) preparation | Free important questions MCQ to study GATE Computer Science Engineering(CSE) 2025 Mock Test Series for Computer Science Engineering (CSE) Exam | Download free PDF with solutions
Practice Test: Computer Science Engineering (CSE) - 13 - Question 1

Direction: Two delivery boys Ram and Amar use their scooters to deliver pizza to different places in city Delhi. The line graph given below represents the distance travelled by them in months from January to June.

With the help of given data answer the following questions:

The distance travelled by Ram in the months of Feb and May is what percent less/more than the distance travelled by Amar in Feb and May?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 1

The distance travelled by Ram in February = 950 km

Distance travelled by him in month of May = 1200

Total distance travelled by Ram in Feb and May = 950+1200 = 2150 kms

The distance travelled by Amar in February = 1200 km

Distance travelled by him in month of May = 800

Total distance travelled by Amar in Feb and May = 1200+800 = 2000 kms

The distance travelled by Ram in Feb and May is 7.5% more than distance travelled by Amar in Feb and May.

Hence, option (c) is correct.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 2

Which of the following refers to ‘To fall asleep from exhaustion?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 2

‘Flake out’ means to fall asleep from exhaustion.

Fall back on - begin to use something held in reserve

Fritter away - To waste or squander time or money.

Ferret out - To discover something after searching.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Practice Test: Computer Science Engineering (CSE) - 13 - Question 3

Direction: Study the information carefully to answer the questions that follow.

In Vimla public school, students can choose among three sports Cricket, Football and Hockey. Out of 1000 students, each one is required to play at least one of the three sports. 40 students play all three sports. 380 students play only Hockey, 170 students play only Cricket, 60 students play exactly two sports Hockey and cricket and 240 students play only Football.

What is the number of students who play exactly two sports?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 3

It is given that the total number of students are = 1000

Only Hockey = 380

Only Cricket = 170

Only Football = 240

Hockey + Cricket + Football = 40 {all the three sports}

Now, let X be the number of students who play Cricket and Football.

And Y is the number of students who play football and hockey.

Now total number of students,

170 +60 +40 +380+ 240+ X + Y= 1000

X + Y = 110

And 60 students play only hockey and cricket, and that number we get after removing students who play all three sports from students who play hockey and cricket.

X + Y + 60 = 110+ 60 = 170

There are 170 students who play exactly two sports.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 4

He lost the game _________ he has improved a lot

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 4
Nonetheless can go at the end of a sentence. This sentence can also be written as: He lost the game; nonetheless, he has improved a lot.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 5

Rs. 200 are divided among Arun, Bholu and Chetan such that Arun's share is Rs. 30 more than Bholu and Rs. 20 less than Chetan's. What is Bholu's share?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 5
It is given that Rs. 200 divided among Arun, Bholu and Chetan

Let Bholu’s share = X Rs.

Arun’s Share = X + 30 Rs

Chetan’s Share = X + 30 + 20 = X+50

Now, Arun+ Bholu+ Chetan = (X + 30) + (X) + (X + 50) = 200

3X = 200 – 80

X = 120/3

X = 40

Bholu’s share is 40 Rs.

Hence, (b) is the correct option.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 6

The average of 5 consecutive integers starting with 'm' is n. What is the average of 6 consecutive integers starting with (m+2)?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 6

According to the question let m = 1

∴ 5 consecutive integers are = 1, 2, 3, 4, 5

n = 15/5 = 3

6 consecutive integers starting with (m+2) are = 3, 4, 5, 6, 7, 8

Now check from option to put n = 3

Option : (a)

(Satisfied)

Hence, the correct option is (a).

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

Ramesh sold a television for Rs 18500 at a loss of 7.5%. If he wants to earn 16% profit by selling the television then at what price he should sell the television?

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

Cost price of the television = = Rs. 20000

To earn 16% profit

Selling price = = Rs. 23200

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

A sum of money at compound rate of interest amounts to Rs. 750 in 1st year and to Rs. 1000 in 2nd year. Find the sum:

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

Amount after 1st year serves as principal for 2nd year. So,

Dividing (?) by (??) we get,

P=Rs.562.50

Practice Test: Computer Science Engineering (CSE) - 13 - Question 9

Present average age of Kanika and Shweta is ‘x’ years. Ratio of the age of Kanika one year ago to the age of Monika four year hence will be 1: 2. Present average age of Kanika, Monika and Shweta is 22 years. Find the value of ‘x’ if the present age of Kanika is 16 years

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 9

Present age of Kanika = 16 years

Age of Kanika one year ago =16-1=15 years

Age of Monika after four years = 15×2=30 years

Present age of Monika = 30-4=26 years=26 years

e present ages of Kanika, Monika and Shweta =22×3=66 years

Present age of Shweta =66-16-26-24 years

Present average age of Kanika and Shweta =x=(24 + 16)/2=20 years

So, the value of ‘x’ is 20 years

So option (d) is the correct answer.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 10

The HCF and LCM of two positive numbers are 5 and 585. If two numbers are in the ratio 9: 13, then find the sum of numbers.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 10

Let the two numbers be 9X and 13X

So, (9x) × (13x)= 5 × 585

117x2 = 25 x 117

x2=25

X = 5

Required sum =9 × 5 + 13 × 5 = 45 + 65 = 110

So option (a) is the correct answer.

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

From the types of sort given below choose the sort from which we get best average behavior

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 11
We can expect the best average behavior from Quick sort because it uses the concept of partitioning . Hence option b is correct
Practice Test: Computer Science Engineering (CSE) - 13 - Question 12

Consider an array consisting of the following elements in unsorted order (placed randomly), but 60 as the first element.

60, 80, 15, 95, 7, 12, 35, 90, 55

Quicksort partition algorithm is applied by choosing the first element as the pivot element. How many total number of arrangements of array integers is possible preserving the effect of first pass of partition algorithm.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 12
We have to choose the first element as pivot.

Here 60 is given as the first element.

After the first pass, the pivot element goes to it’s exact location.

Here 60 goes to 6th place.

All the elements less than 60 go to the left of 60 and all the elements greater than 60 go to the right of 60.

After 1st pass

⇒ 5! × 3!

⇒720 possible arrangements.

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

The grammar which is equivalent to -

E → E+/ES+/+

S → */S*

After eliminating the left recursion is-

E' → S+E'/+E'

E' → EPSILON

S → *S'

S' → *S'/EPSILON

E' → S+E'/+E'

S → *S'

S' → *S'/EPSILON

E' → +E'/EPSILON

S → *S'

S' → *S'/EPSILON

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

The way of removing left recursion is-

A→ AB / C is replaced by A → CA'

A' → BA' / epsilon

So the answer is - E → +E'

E' → S+E'/+E'

E' → EPSILON

S → *S'

S' → *S'/EPSILON

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

Consider the code segment

int i, j, x, y, m, n;

n = 20;

for (i = 0, i < n; i++)

{

for (j = 0; j < n; j++)

{

if (i % 2)

{

x + = ((4*j) + 5*i);

}

}

}

m = x + y;

Which one of the following is false?

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

int i,j,x,y,m,n;

N = 20;

for(i = 0; i < n; i++)

{

for(j = 0;j < n; j++)

{

if(i%2)

{

x + = 4*j[4<

y + = (7+4*j);

}

}

}

M = x + y;

Practice Test: Computer Science Engineering (CSE) - 13 - Question 15

One piece of data is being sent from one host to another host through the HyperText Transfer protocol. What port number will be attached to destination port number?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 15
HTTP uses port number 80 by default.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 16

A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 16

Given 8 – bit delimiter pattern of 01111110.

Output Bit string after stuffing is 01111100101

StuffedBit

Now, Input String is 0111110101.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 17

Consider an Ethernet network employing CSMA/CD, where Bw = 10 Mbps, length of the cable (L) is 200m and v = 2 x 108 m/sec. Calculate the maximum frame size in bytes.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 17
In CSMA/CD, Transmission time (Tt) = 2 × Tp

Therefore, maximum frame size = 2 × L × Bw / v

= 20 bits = 2.5 bytes.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 18

A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is a _________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 18

With the above-given data, after allocating 2 units of tape to each process, with 1 available unit any of the 3 processes can be satisfied in such a way that no deadlock will be there. So the answer is 7 tape units.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 19

If F= {A →BC, CD → E, E →C, D → AEH, ABH →BD, DH → BC} then canonical cover over F is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 19

F= {A →BC, CD → E, E →C, D → AEH, ABH →BD, DH → BC} then a nonredundant cover for F { A → BC, E →C, D → AEH, ABH →BD }. Then FD ABH → BD can be decomposed into FDs ABH → B and ABH →D

Now since, the FD A →B and AH → D. We also notice that AH →B is redundant since the FD A → B is already in F

That gives us the canonical cover as {A → B, A→C, E → C, D →A, D → E, D →H, AH → D}.

Practice Test: Computer Science Engineering (CSE) - 13 - Question 20

One to many relationship between X(one side) and Y(many side) entities in an E-R diagram is converted as

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 20
The E-R model can be represented with two relations X and Y therefore Y side includes the primary key of X side as foreign key.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 21

In ER models, which statement is not valid?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 21
It is seen that in ER model, entity is present in ER model while relational table is available in Relational Model. From above options we see that option (A) and option (B) are true as ER model supports multi valve and composite attributes, with this we see that option (C) is false. In case of option (D) we see that this option is true as an entry in the relational table has exactly one NULL value.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 22

Consider the following SQL query

SELECT studentid, studentname

FROM student

WHERE birthyear <=ALL (SELECT birth year from student)

Identify the answer returned by the above query?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 22
Given SQL query prints the oldest student information.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 23

A person is having 10 flowers in which 3 are of the same type A and 2 are of the same type B. He wants to make a garland out of all the flowers he is having. How many types of Garland are possible?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 23
3 are of type A and 2 are of type B so the possible garlands are = 15120
Practice Test: Computer Science Engineering (CSE) - 13 - Question 24

From the letters of the word ‘INSTITUTION’, let the number of words in which all I’s come together possible from the permutation of the letters is X and the number of words in which vowels come at even position possible from the permutation of the letters is Y, then what is X - Y?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 24
If all I's should be together then we make it a single unit '(III) TTTNNSUO' so now we've to arrange 9 units in which there are 3 T's and 2N 's so . There are 11 places in which position 2,4,6,8,10 are even and we need to place a vowel at an even position.

Therefore X - Y = 29040

Practice Test: Computer Science Engineering (CSE) - 13 - Question 25

If we run Dijkstra algorithm on vertex S to the following graph, then which of the following is shortest path distance from S to t

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 25
When we run the dijkstra algorithm on vertex S, it will explore all edges which are reachable from S and then we will choose an edge which has minimum weight. So < s, t > edge having minimum weight is 6.
Practice Test: Computer Science Engineering (CSE) - 13 - Question 26

Consider the following elements:

5 , 12 , 3 , 15 , 4 , 6 , 10

These elements are inserted one-by-one into two separate heaps, one min-heap and another max heap. Then these two heaps are stored in two different arrays starting from index How many elements have the same index value in both the arrays _______.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 26
Min-Heap after all insertions

Max-Heap after all insertions

Clearly, no element has the same index.

Hence, answer = 0

Practice Test: Computer Science Engineering (CSE) - 13 - Question 27

Consider a relation R related to 7 entities E1,E2…..E7 respectively, each of them has 2 attributes (A,B) (C,D) (E,F) (G,H) (I,J) (K,L) (M,N) with first attribute as candidate key in all entities. All entities are related to R with one-one mapping. What is the number of candidate keys possible for R :

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 27
For one to one mapping b/w a relation & entity set

No. of candidate keys at R = {A.C} i.e. 2

Therefore for a relation related to 7 entities

No. of candidate keys = 7 {A, C, E, G, I, K, M}

Practice Test: Computer Science Engineering (CSE) - 13 - Question 28

Consider a case in which A and B are been considered as non zero +ve integers, then find out the outcome of following Pascal program

While A < > B do

if A < B then

A : = A – B

else

B : = B – A;

Write(A);

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 28

We need to apply Optional Elimination

Taking A = 6 AND B = 4

Case it’s GCD when GCD = 2

Now dry run the code

1 → A = 2, B = 4

2→ A = 2, B = 2

Prints A which is 2 which means it’s a GCD

Practice Test: Computer Science Engineering (CSE) - 13 - Question 29

Consider the following elements in a heap {17, 9, 6, 3, 12, 2, 15, 14, 20, 13, 21}. The minimum number of exchanges to convert it into a min heap be P. Then the minimum number of elements that can be arranged in a BST of height P is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 29

From the given heap, it is obvious that 3 and 9 need to be exchanged for 17 and 3, but there are also exchanges needed for 2 and 6. So, 2 should be at the root as it is the minimum. Since the minimum number of exchanges is asked, it is clear that if exchanges are carried out in this manner, 2 – 6, 2 – 17, 6 – 17 and 3 – 9 separately, then the number of exchanges required = 4. But in other way, exchanges would be like:

3 – 9, 3 – 17, 9 – 17, (3 is at root now) and on the right side:

2 – 6, 2 – 3. Thus 5 exchanges. So P = 4.

But we have to find the minimum number of exchanges.

Now the elements in BST with height 4 is 4+1(root) = 5

Practice Test: Computer Science Engineering (CSE) - 13 - Question 30

Consider the following statement:

S1: Deletion in referencing relation causes deletion anomaly.

S2: Referential integrity constraints are specified between exactly two relations in a schema.

S3: Count(*) also counts the tuples which consist of NULL in a relation.

S4: Every relation is possible to decompose into BCNF with dependency preserving and lossless join decomposition.

Which option is correct?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 13 - Question 30

S1: Referencing relation is the relation referring to a primary key of some other relation. Deletion in that table will not cause any violation. Hence, no deletion anomaly hence this is true.

So, this statement is true.

S2: This statement is true, since foreign key can be created within a relation

C is a foreign key referring to the primary key "A'' of the same relation R.

S3: S3 is true since count (*) will count the no. of tuples in a relation, on the basis of ID value.

S4: It is true. Since it is possible to decompose every relation into BCNF with lossless join decomposition property but it may or may not preserve the dependencies of the relation.

View more questions
55 docs|215 tests
Information about Practice Test: Computer Science Engineering (CSE) - 13 Page
In this test you can find the Exam questions for Practice Test: Computer Science Engineering (CSE) - 13 solved & explained in the simplest way possible. Besides giving Questions and answers for Practice Test: Computer Science Engineering (CSE) - 13, EduRev gives you an ample number of Online tests for practice

Top Courses for Computer Science Engineering (CSE)

Download as PDF

Top Courses for Computer Science Engineering (CSE)