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Practice Test: Computer Science Engineering (CSE) - 6 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test - Practice Test: Computer Science Engineering (CSE) - 6

Practice Test: Computer Science Engineering (CSE) - 6 for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Practice Test: Computer Science Engineering (CSE) - 6 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 6 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) - 6 below.
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Practice Test: Computer Science Engineering (CSE) - 6 - Question 1

Select the pair that does not expresses a relationship similar to that expressed in the pair:

Wheel: spokes

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 1

The spokes are units which radiate of a wheel and make the wheel a complete entity. Similarly, fingers, tentacles and petals are integral units of hand, octopus and flower respectively. On the other hand, roots and leaves do not share this kind of relationship. Thus option 4 does not expresses a relationship similar to that expressed in the given pair.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 2

If a, b and c are three positive integers such that a and b are in the ratio 3:4 while b and c are in the ratio 2:1, then minimum integer value of a + b + c is _________ 


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 2

Let a = 3x and b = 4x

Similarly b = 2y and c = y

∴ 4x = 2y ⇒ y = 2x

∴ c = 2x

Now a + b + c = 3x + 4x + 2x = 9x

So, the minimum integer value = 9

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Practice Test: Computer Science Engineering (CSE) - 6 - Question 3

Reaching a place of appointment on Friday. I found that I was two days earlier than the scheduled day. If I had reached on the following Wednesday then how many days late would I have been?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 3

Friday → 2 days earlier

Therefore, scheduled day = Friday + 2 = Sunday

Sunday + 3 = Wednesday

Therefore, I would have been late by 3 days

Practice Test: Computer Science Engineering (CSE) - 6 - Question 4

Which one of the following options is the closest in meaning to the word 'mitigate'?

Practice Test: Computer Science Engineering (CSE) - 6 - Question 5

Choose the most appropriate word(s) from the options given below to complete the following sentence.

It was hoped at the time that that place would become the centre from which the civilization of Africa would proceed; but this ________ was not fulfilled.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 5

The sentence implies that it was hoped that that place would become the point from where the African civilization would proceed, but the belief that it would happen was not fulfilled.

Therefore, the correct word to fill in the blank is expectation as it means a strong belief that something will happen or be the case.

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

3, k, 2, 8, m, 3

The arithmetic mean of the list of numbers above is 4. If k and m are integers and k ≠ m, what is the median of the list?

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

Mean = 4

k ≠ m so k = m = 4 is out.

{k, m} = {1, 7} or {2, 6} or {3, 5}

for median of {3, k, 2, 8, m, 3}

{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}

Median: (3 + 3) /2 = 3

Practice Test: Computer Science Engineering (CSE) - 6 - Question 7

Consider a random walk on an infinite two-dimensional triangular lattice, a part of which is shown in the figure below.

If the probabilities of moving to any of the nearest neighbour sites are equal. What is the probability that the walker returns to the starting position at the end of exactly three steps?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 7

A person can take 1st step in any direction independently

Suppose he moves to A

Now to return to O ( initial point) in 2 steps he can move in 2 directions. Either B or f

Thus probability he will move to either B or F will be = 2/6

Suppose he moves to B

Now to return back to O

he has only 1 option

⇒ to move in BO

Probability he will move in BO direction  = 1/6

Total probability he will return

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

Twelve straight lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. A circle is now drawn in the same plane such that all the points of intersection of all the lines lie inside the circle. What is the number of non-overlapping regions into which the circle is divided?

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

The nth line drawn will add n more regions to the circle. 

1st line adds 1 region to the circle for a total of 2 regions. 

2nd line adds 2 more regions to the circle, bringing the total number of regions to 2 + 2 = 4. 

3rd line adds 3 more regions to the circle, bringing the total number of regions to 4 + 3 = 7. 

4th line adds 4 more regions to the circle, bringing the total number of regions to 7 + 4 = 11.

5th line adds 5 more regions to the circle, bringing the total number of regions to 11 + 5 = 16. 

6th line adds 6 more regions to the circle, bringing the total number of regions to 16 + 6 = 22. 

7th line adds 7 more regions to the circle, bringing the total number of regions to 22 + 7 = 29. 

8th line adds 8 more regions to the circle, bringing the total number of regions to 29 + 8 = 37. 

9th line adds 9 more regions to the circle, bringing the total number of regions to 37 + 9 = 46. 

10th line adds 10 more regions to the circle, bringing the total number of regions to 46 + 10 = 56. 

11th line adds 11 more regions to the circle, bringing the total number of regions to 56 + 11 = 67. 

12th line adds 12 more regions to the circle, bringing the total number of regions to 67 + 12 = 79. 

Practice Test: Computer Science Engineering (CSE) - 6 - Question 9

Electromagnetic radiation is an insidious culprit. Once upon a time, the major concern around electromagnetic radiation was due to high tension wires which carry huge amounts of electricity to cities. Now, we even carry sources of this radiation with us as cell phones, laptops, tablets and other wireless devices. While the most acute exposures to harmful levels of electromagnetic radiation are immediately realized as burns, the health effects due to chronic or occupational exposure may not manifest effects for months or years.

Which of the following can be the viable solution for electromagnetic radiation reduction?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 9

The correct answer is option 2 i.e. To implement hardware protocols to minimize risks and reduce electromagnetic radiation production significantly. 

The passage states about the electromagnetic radiations released from the devices and how they affect the individuals. Out of the given options, only option 2 states the possible solution which can be implemented to reduce electromagnetic radiation.

Practice Test: Computer Science Engineering (CSE) - 6 - Question 10

In a mock exam, there were 3 sections. Out of all students, 60 students cleared the cut off in section 1, 50 students cleared the cutoff in section 2 and 56 students cleared the cut off in section 3. 20 students cleared the cutoff in section 1 and section 2, 16 cleared cut off in section 2 & section 3, 26 cleared the cut off in section 1 & section 3. The number of students who cleared cutoff of the only one section was equal & was 24 for each section. How many students cleared cut off all the three sections?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 10

Let the number of students be X who cleared cut off in all sections.

The number of students who cleared the cutoff in section 1 and section 2, only = 20 – X

The number of students who cleared the cutoff in section 2 and section 3, only = 16 – X

The number of students who cleared the cutoff in section 1 and section 3, only = 26 – X

 

Now, consider section 1:

24 + 20 – x + x + 26 – x = 60

70 – x = 60

x = 10

Practice Test: Computer Science Engineering (CSE) - 6 - Question 11

The regular expression which represents the set of strings in which every 0 is immediately followed by at least two 1's is ____________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 11

If w is in L, then either

(a) w does not contain any 0, or

(b) it contains a 0 followed by 11. So, w can be written as w1w2.........wn, where each wi is either 0 or 011.

So, L is represented by the regular expression (1 + 011)*.

Therefore, option 2 is the correct answer.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 12

A and B are the only two host on a LAN which uses CSMA/CD protocol. A minimum time required by ‘A’ to detect collision is 600 μs. Find the time taken by the packet to travel from host A to host B.


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

Concept:

Contention period is minimum time a host must transmit for before it can be sure that no other host's packet has collided with its transmission. It takes minimum of RTT to detect collision.

Calculation:

Contention Period = RTT = 2 × Tp

600 = 2 × Tp

Tp = 300 μs

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 13

What is the minterm that equals 1 if x1 = x3 = 0 and x2 = x4 = x5 = 1, and equals 0 otherwise?


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

The minterm y1y2 ...yn is 1 if and only if each yi is 1, and this occurs if and only if xi = 1 when yi = xi and xi=0 when yi = xi.

Therefore, the minterm will be x1'x2x3'x4x5 i.e. 01011. 

The answer will be 11.

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

Consider a 1024 MB free partition and the following memory request.

R1 requests 120 MB

R2 requests 250 MB

R3 requests 480 MB

R4 requests 40 MB

Which of the following allocation technique will merge the partition into the original 1024 KB segment, when R4 finishes?

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

The buddy system allocates memory from a fixed-size segment consisting of physically contiguous pages. Memory is allocated from this segment using a power-of-2 allocator, which satisfies requests in units sized as a power of 2.

Practice Test: Computer Science Engineering (CSE) - 6 - Question 15

Consider a double-ended queue with elements 31, 17, 4, 22, 19, 8. What is the time complexity of deletion of last element i.e. 8 and insertion of new element 10 at the rear?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 15

In a double-ended queue, elements can be inserted and deleted from both the front and back of the queue.

Time complexity of all operations like insert element at front, insert at rear, delete element from front, and delete element from rear is O(1).

Practice Test: Computer Science Engineering (CSE) - 6 - Question 16

The column vector    is a simultaneous Eigen vector of   

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 16

For all values of a and b,     is an Eigen vector of A.

⇒ ab + b2 = 2a2

⇒ 2a2 - ab - b2 = 0

⇒ 2a2 - 2ab + ab - b2 = 0

⇒ 2a (a - b) + b (a - b) = 0

⇒ b = a, b = -2a

For b = a, (or) b = -2a, X is an eigen vector of matrix B.

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 17

Consider a processor that includes a base with indexing addressing mode. Suppose an instruction is encountered that employs this addressing mode and specifies a displacement of 1500. Base register contains the value 3456 and index register contains the value 4. What is the address of the operand?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 17

In indexing addressing mode, the address field references a main memory address, and the referenced register contains a positive displacement from that address.

Address of operand = base register + index register + displacement

Address of operand = 3456 + 4 + 1500 = 4960

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 18

The address of a class C host is to be split into subnets with a n-bit subnet number. The maximum number of hosts in each subnet is 14. Find the value of n.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 18

The number of available hosts on a subnet is 2h−2, where h is the number of bits used for the host portion of the address.

h2 = 14

h = 4

Practice Test: Computer Science Engineering (CSE) - 6 - Question 19

Which of the following options is INCORRECT?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 19

Option 4 is incorrect. It can be corrected as:

The cartesian product of two countable sets is countable.

Proof: There are three cases to consider:

Case 1: If both A and B are finite with |A|=m and |B|=n, then it is easy to show that |A×B|=mn and hence A×B is finite and so it is countable.

Case 2: If A is finite with |A|=n and B is countably infinite, there exist bijective functions f: A→{1,2,...,n} and g: B→N. We can define h: A×B→N for all (a,b)∈A×B.

Then clearly h is injective. So A×B is countable.

Case 3: If A and B are both countable, there exist bijective functions f: A→N and g: B→N and defining h: A×B→N gives us an injective function, and so A×B is countable. 

Practice Test: Computer Science Engineering (CSE) - 6 - Question 20

Consider a following resource allocation graph with multiple instance of each resource type.

Which of the following statement is true about above system?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 20

In resource allocation graph, a claim edge Pi → Rj indicates that process Pi may request resource Rj at some time in the future. An assignment edge Rj → Pi indicates that process Pi is holding resource Rj. The above system is deadlock-free although there is a cycle in a graph because there are more than one instance per resource type.

Practice Test: Computer Science Engineering (CSE) - 6 - Question 21

Which of the following is the correct way to allocate a memory for the following structure?

struct Student

{

 int sid;

 int age;

 char grade;

};

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 21

 

The sizeof operator returns the size of the struct in bytes. The size of a struct is not always the sum of the size of the individual members, hence use sizeof rather than a literal.

struct Student *sptr;

sptr = (struct Student*)malloc( sizeof(struct Student) );

Practice Test: Computer Science Engineering (CSE) - 6 - Question 22

 

Solve the integral   

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 22

We know that,

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 23

In TCP connection, When SYN segment is sent, the value of retransmission-timeout is set to 11 sec. and when the SYN + ACK segment is received at the sender side the time required for the segment to reach the destination (i.e. at the sender side) and be acknowledged is equal to 3.2 sec. Find the retransmission – timeout.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 23

Concept:

The measured round-trip time for a segment is the time required for the segment to reach the destination and be acknowledged, although the acknowledgment may include other segments. In TCP, there can be only one RTT measurement in progress at any time.

Calculation:

RTTM = 3.2

RTTS = 3.2

RTTD = RTTS/2 = 1.6

RTO = RTTS + 4 × RTTD = 9.6

Practice Test: Computer Science Engineering (CSE) - 6 - Question 24

Consider the following syntax directed definition:

The above SDD is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 24

The first rule defines the inherited attribute T’.val using F’.val and F appears to the left of T’ in the production. The second rule define T’1.val using the inherited attribute T’.val associated with head and F.val, where F appears to the left of T’ in the production

Practice Test: Computer Science Engineering (CSE) - 6 - Question 25

There are two boxes each containing two components. Each component is defective with probability ¼, independent of all other components. The probability that exactly one box contains exactly one defective component equals?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 25

Probability (1 faulty in box) 

The chances of box having other than 1 defective is 5/8

The probability exactly 1 one fault in exactly 1 box

Alternate:

The given situation is possible if

(1) 1 faulty among 4 components

(2) 3 faulty among 4 components

For case (1)

For case (2)

Practice Test: Computer Science Engineering (CSE) - 6 - Question 26

Which of the following statements is false?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 26

Out of all the options, option 3 is false.

It can be corrected as: It is desirable to maximize CPU utilization and throughput and to minimize turnaround time, waiting time, and response time. 

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 27

A data is sent to UDP along with a pair of socket address and the length of data. After receiving data, UDP adds the header and passes the user datagram to IP with the socket address. What is the maximum size of the data that can be encapsulated in a UDP datagram?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 27

In IPv4, the maximum length of packet size is 65,536 and length of IP header is 20 bytes. So, maximum data size (including UDP header) = 65,535 – 20 = 65515

Size of UDP header = 8 bytes

Maximum data size = 65,515 – 8 = 65507

Practice Test: Computer Science Engineering (CSE) - 6 - Question 28

What is the output of following C code:

#include <stdio.h>

int main(void) {

 char *p;

 p = "Programming";

 p++ ;

 ++p;

 --p;

 p--;

 printf( p );

 return 0;

}

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 28

 

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 29

f the number of balanced parenthesis possible with 'n' pair of parenthesis is 14, what is the value of 'n'?


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 29

Number of balanced parenthesis = Number of binary search trees

Number of binary search trees = 

When n = 4

Number of binary search trees =

Therefore, value of n = 4

*Answer can only contain numeric values
Practice Test: Computer Science Engineering (CSE) - 6 - Question 30

Consider the following statements:

S1: The identifying relationship is many-to-one from the weak entity set to the identifying entity set, and the participation of the weak entity set in the relationship is total.

S2: It is possible to have a weak entity set with more than one identifying entity set.

The number of correct statements are ______


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 6 - Question 30

Both the given statements are true.

It is possible to have a weak entity set with more than one identifying entity set. A particular weak entity would then be identified by a combination of entities, one from each identifying entity set. The primary key of the weak entity set would consist of the union of the primary keys of the identifying entity sets, plus the discriminator of the weak entity set. 

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