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

Practice Test: Computer Science Engineering (CSE) - 12 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) - 12 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 12 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) - 12 below.
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Practice Test: Computer Science Engineering (CSE) - 12 - Question 1

Santosh’s car gives 5 km more per litre of diesel when driven on the highway in comparison to city drive. On a recent trip, Santosh drove 30 km on the highway and 130 km in the city consuming a total of 15 litres of diesel in the process. How many km/litre does Santosh’s car run in the city?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 1
Let the mileage of Santosh's car be n km/litre of diesel when driven in the city and (n+5) km/litre when driven on the highway. Translating the given information in to an equation, we can write

⇒ 3n2 + 17n – 130 = 0

Which gives n = 10 or n = -13/3

Hence we can say that Santosh's car runs 10 km/litre in the city.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 2

Choose the option which is opposite in meaning to RELINQUISH.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 2
‘Relinquish’ refers to surrender or give up whereas ‘possess’ refers to have or own something.
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Practice Test: Computer Science Engineering (CSE) - 12 - Question 3

Choose the best alternative which can be substituted for the given sentence”: “A person difficult to please”

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 3
‘Fastidious’ is someone who is excessively particular demanding or hard to please.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 4

Two solutions of alcohol A and B were mixed to obtain 20 litres of new solution C. Before they were mixed, the first solution A contained 1.6 litres of alcohol while the second solution B contained 1.2 litres of alcohol. Before mixing if the percentage of alcohol in the first solution A was twice that in the second B, what was the volume of the first solution A before mixing?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 4
Let the volume of the first A in the mixture be "x′′ "Itres, the volume of the second B in the mixture must be (20 − x)

% of alcohalin A=(1.6 / x) × 100

% of alcohalin B=(1.2 / 20 − x) × 100

(1.6 / x) × 100 = 2 × (1.2 / 20 − x) ×1 00

16 − 0.8x = 1.2x

x = 8 litres

Practice Test: Computer Science Engineering (CSE) - 12 - Question 5

In this question, out of the four alternatives, choose the option which is closest in meaning to FURORE.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 5
‘Furore’ refers to an outbreak of public anger or excitement.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 6

Find the pair of letters which will come in blank spaces marked as ‘?’.

V W X Y E D C B R S T ? ?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 6

A careful look at the alphabets in the given series shows a pattern

VWXY – EDCB RST? etc

∨ ↔ E; W ↔ D; X ↔C; Y ↔ B i.e. 5th from end corresponds to a* from the beginning etc. It is a cluster of 4 consecutive letters which is taken at a time, "Therefore, the next 4 letters will be RSTU-IHGF etc.

Hence the last 2 letters will be U and I i.e. option (b).

Practice Test: Computer Science Engineering (CSE) - 12 - Question 7

We are given a square of side 22 cm. A circle of maximum possible diameter is inscribed in this square. If a point is chosen at random inside the square, then the probability that it will lie inside the circle is ____________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 7

Required probability = Area of the circle / Area of the square

Biggest possible circle that can be inscribed in the given square would, be touching all the four sides of the square internally implying that the diameter of this circle = side of the square = 22cm

Required probability = = 0.785

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

Based on the given statements, select the most appropriate option to solve the question.

Sheetal wants to sell her bicycle at either a profit of K% or a loss of K%. What is the value of K?

Statement 1: Difference between the amount Sheetal gets in the 2 cases is ₹ 2560

Statement 2: If Sheetal’s profit is Rs K, her profit in percentage is 7.5%.

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

Let us assume k = K / 100 and the cost price = C

Based on SI, we can write C × (1 + K / 100) − C × (1 − K / 100) = 2560

i.e. 2CK / 100 = 2560 or Ck = 1280 which does not give the value of k or K. Hence Statement 1 is NOT sufficient.

Based onS2, C × 0.075 = K which gives C = 40K / 3 = 4000k / 3 which will NOT give the value of k or K.

When we combine the information given in both the statements, we will be able to find C as well as k or K. Hence option (c) is the correct option.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 12 - Question 9

Triangles ABC and CDE have a common vertex C with side AB of triangle ABC being parallel to side DE of triangle CDE. If length of side AB = 4 cm and length of side DE = 10 cm and perpendicular distance between sides AB and DE is 9.8 cm, then the sum of areas of triangle ABC and triangle CDE is _________ cm2.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 9
The answer is in between 40 and 41

Given AB‖DE

⇒ ∠B = ∠D (Alternate angles)

and ∠A = ∠E (Alternate angles)

△ABC − ΔEDC (AAA similarity)

⇒ h1 / h2 = AB / DE = 410 = 25

and h1 + h2 = 9.8cm (given)

h1 = 2.8cm and h2 = 7cm

Area of ΔABC = 12 × 4 × 2.8 = 5.6cm2

Area of ΔEDC = 12 × 10 × 7 = 35cm2

∴ Sum of areas of ΔABC and ΔEDC = 40.6cm2

Practice Test: Computer Science Engineering (CSE) - 12 - Question 10

Interior angles of a polygon with 8 sides are in arithmetic progression. If the smallest interior angle measures 100° then find the largest interior angle?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 10

If 100 is the smallest interior angle of the polygon i.e. Octagon, this gives the largest exterior angle = 180 100= 80

The sum of exterior angles of a convex polygon = 360

since the interior angles form an increasing arithmetic progression, the exterior angles will form a decreasing arithmetic progression, let us say with common difference 'd "

360 = 80 + (80 + d) + (80 + 2d)……8 terms

= 82[2 × 80 + (8 − 1)d] = 360 which gives d = −10

Using d=−10, we get the smallest exterior angle =80 + (8−1) × (−10) = 10 leading to the largest interior angle =180 − 10 = 170

Note: formula used for sum of terms of an AP is given by

Sn = n2[2a + (n − 1)d] where 'a' is the first term and 'd' is the common difference.

Alternatively,

Sum of all the interior angles of a polygon = (n − 2)180

⇒ a + (a + d) +……8 terms = (8 − 2)180∘

⇒ 82(100 × 2 + 7d) = 1080

⇒ d = 10

∴ Largest angle = a + 7d = 100 + 7(10) = 170

Practice Test: Computer Science Engineering (CSE) - 12 - Question 11

Consider the following problems.

1. longest common subsequence

2. Optimal Binary search tree

3. Fractional knapsack problem

4. Matrix chain multiplication

Which of the above problems can be solved using dynamic programming?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 11
Longest common subsequence, longest increasing subsequence, sum of subsets, optimal BST, matrix chain multiplication, Travelling salesperson, Balanced partition, Fibonacci sequence, Multistage graph problems , 0/1 knapsack are solved by using dynamic programming.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 12

Suppose that the minimum spanning tree of the following edge weighted graph contains the edges with weights x, y and z.

What is the maximum value of x+y+z?

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

x ≤ 11

y ≤ 6

z ≤ 8

Practice Test: Computer Science Engineering (CSE) - 12 - Question 13

Which of the following standard algorithm is not an greedy algorithm?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 13
Bellman ford shortest path algorithm is example of dynamic programming.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 14

For the sequence

500, 535, 512, 721, 436, 611, 624, 632, 643

Lexicographic sort gives time complexity of:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 14
T(n) = O(m × n) = O(3 × 9) = O(27)

Where m are the number of digits on each element of sequence

n : number of elements in sequence.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 15

Consider the following SDT

G→A-T{G.x=A.x-T.x}

A→Ea{A.x=E.x2}

EEb{E1.x=1+E2.x}

E→ξ{E.x=-1}

T→Eb(T.x=Ex-2}

If above SDT uses L- attributed definition then what is the value of an attribute x at root after evaluation for an input string “ba-bb”.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 15

In the above dependency graph, 2 is the value at G after the evaluation.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 16

Match the following:

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 16
Active web documents are the programs that run on the client side. The best example of active web documents is animated graphics on the screen which helps in interaction with clients.

While for static documents as the name suggests, contents are static.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 17

While opening a TCP connection, the initial sequence number is to be derived using a time-of-day (ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counters increments once per millisecond. The maximum packet lifetime is given to be 64s.

Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 17

The maximum packet lifetime is given to be 64 seconds in the question.

Thus, a sequence number increments after every 64 seconds.

So, minimum permissible rate = 1 / 64 = 0.015 per second.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 18

Both hosts and routers are TCP/IP protocol software. However, routers do not use protocol from all layers. The layer for which protocol software is not needed by a router is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 18
The features that stood out during the research, which led to making the TCP/IP reference model were:
  • Support for a flexible architecture. Adding more machines to a network was easy.
  • The network was robust, and connections remained intact until the source and destination machines were functioning.

The overall idea was to allow one application on one computer to talk to(send data packets) another application running on a different computer.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 19

If the instruction length in a system is 32 bits and memory has 32 words. Then find the 1 address instructions if there are 42 address instructions’.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 19
Number of 1 address instructions =(25 – 4) × 25 = 896.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 20

Consider execution of 100 instructions on a 5 stage pipeline. Let P be the probability of an instruction being a branch. What must be the value of P such that speed up is atleast 4?

(Assume each stage takes 1 cycle to perform it’s task and branch is predicted on fourth stage of the pipeline)

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 20

12P ≤ 1

= 0.08

Practice Test: Computer Science Engineering (CSE) - 12 - Question 21

What is the output of the following C code?

#include

#include

void main()

{

int index;

for(index=1; index<=5;i++)

{

printf("%d",index);

if(i == 3)

continue;

}

}

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 21
Here continue is the last line. So, it will not skip any number.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 22

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;

}

What is the functionality of the above function Do ()?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 22
Here two pointers are used: gate 1 and gate 2. One pointer points to the 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 palindromes.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 23

Consider the following code for inorder traversal of a binary tree.

Void traversal (node *root)

{

if (root ==0)

{

Print f(“empty tree”);

}

else

{

itraversal (root left);

printf("%d\n”, root data);

Iiraversal (root right);

}

}

Which of the following is true about the above snippet of code.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 23
“Empty tree” will be printed (N+1) times in addition to inorder traversal. It is so because, if root is zero, the control has to return, but it instead just prints “empty tree” (N+1) times, because for a binary tree with N nodes, there are (N+1) empty nodes.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 24

R{a,b,c} and S{x,y,z} are two relations in which x is a foreign key referring to the primary key of relation R.

Consider the following operations. Which among these operations may violate the referential integrity constraint?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 24
x is a foreign key referring to the primary key of the relation R. If we delete anything from R, it may so happen that there is a value in S linked to the deleted value. Or if we insert something in S, it may not be present in relation R.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 25

Suppose that we have the following three tuples in a legal instance of a relation schema S with three attributes ABC (listed in order): (1,2,3), (4,2,3), and (5,3,3). Which of the following dependencies can you infer does not hold over schema S?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 25
BC→ A does not hold over S (look at the tuples (1,2,3) and (4,2,3)). The other tuples hold over S.
Practice Test: Computer Science Engineering (CSE) - 12 - Question 26

Number of binary trees formed with 5 nodes are

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 26
Possible number of binary trees formed when n number of nodes are given is 2nCn / n+1 so in this case the number of nodes are 5 so

Number of binary trees possible are = 10C5 / 6 = 10!/ 5! * 5! * 6 → 42

Hence option D is the correct answer

Total number of different Binary tree of size

5 : 14 + 5 + 4 + 5 + 14 = 42

Practice Test: Computer Science Engineering (CSE) - 12 - Question 27

The Boolean function f implemented in the figure using two input multiplexers is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 27
f = E.A

E =

Practice Test: Computer Science Engineering (CSE) - 12 - Question 28

Consider a complete undirected graph A with 6 vertices. All the vertices are labelled. What is the number of distinct cycles of length 4 in this graph?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 28
We can pick 4 vertices from 6 in 6C4 ways i.e. 15. Now, there can be 3 distinct cycles from 4 vertices

A--------B

| |

| |

| |

C--------D

Distinct cycles = ABC , ACD , BCD

Hence 15×3 = 45 is the answer.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 29

The number of colours required to properly colour the vertices of every planar graph is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 29
According to the 4-colour theorem, the vertices of every planar graph can be coloured with at most 4 colours so that no two adjacent vertices receive the same colour.

Hence, Option (c) 4 is the correct choice.

Practice Test: Computer Science Engineering (CSE) - 12 - Question 30

While opening a TCP connection, the initial sequence number is to be derived using a time-of-day (ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counters increments once per millisecond. The maximum packet lifetime is given to be 64s.

Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 12 - Question 30

The maximum packet lifetime is given to be 64 seconds in the question.

Thus, a sequence number increments after every 64 seconds.

So, minimum permissible rate = 1 / 64 = 0.015 per second.

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