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

Practice Test: Computer Science Engineering (CSE) - 15 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) - 15 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 15 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) - 15 below.
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Practice Test: Computer Science Engineering (CSE) - 15 - Question 1

A 300-meter-long train passes a 450-meter-long platform in 5 sec. If a man is walking at a speed of 4 m/sec along the track and the train is 100 m away from him, how much time will it take to reach the man?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 1

The train can cover (300 + 450) m distance in 5 sec.

The speed of the train = 150 m/sec

Relative speed of the man and the train is 154 m/sec or 146 m/sec

To cover the distance of 100 m, in either of the case, it will take less than 1 second.

Hence, (C) is the correct option.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 2

A boat takes half the time moving a certain distance downstream than upstream. The ratio of the speed of the boat to that of the current is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 2
Let,

Boats speed = x kmph

Current speed = y kmph

Time taken in upstream = t hours

Therefore, (x - y) × 2t = (x + y) × t

⇒ 2x – 2y = x + y

⇒ x = 3y ⇒ x : y = 3 : 1

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Practice Test: Computer Science Engineering (CSE) - 15 - Question 3

Which of the following word is opposite to the word “connivance”?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 3

Connivance- willingness to allow or be secretly involved in an immoral or illegal act

Conspiracy- a secret plan by a group to do something unlawful or harmful.

Sufferance - capacity to endure pain, hardship, etc.; endurance

Complot - a plot involving several participants; conspiracy

Intolerance - unwillingness or refusal to tolerate or respect persons of a different social group, especially members of a minority group.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 4

Direction: A number is wrong in the following number series. Select the wrong number.

64, 130, 264, 536, 1076

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 4
The series is

64 × 2 + 2= 130

130 × 2 + 4= 264

264 × 2 + 6= 534

534 × 2 + 8 = 1076

Practice Test: Computer Science Engineering (CSE) - 15 - Question 5

Direction: Each statement has a blank followed by four options. Select the most appropriate word for the blank.

Guru was always able to maintain a _______ face when he said something silly, and that contrast made everyone laugh.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 5
The statement means to state that Guru had an ability to maintain a serious face while saying something silly. The biggest hint in the question is the word ‘contrast’. Hence, c is the correct answer.
Practice Test: Computer Science Engineering (CSE) - 15 - Question 6

A supermarket launched a scheme that if a customer purchases two CDs, one extra CD will be free and if he purchases 3 Mobile he will get one extra Mobile free. If the cost price of 3 CD and 4 Mobile be Rs.6700 and Rs.232500 respectively. If a customer purchase 2 CD and 3 Mobile as per scheme he availed 1 product free of each category, then at what price these product should be sold so that the agency can get overall profit of 17.5%

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

CP of 2 CD and 3 Mobile = 6700 + 232500 = 300000

Now, since we required 17.5% profit,

So, SP= 300000*117.5/100 = Rs. 352500.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 7

In the following question, two/three statements are given followed by four conclusions. You have to consider the statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follow from the given statements.

Statements:

I. Some cats are owls.

II. Some owls are elephant.

Conclusion:

I. Some cats are elephant.

II. All owls are elephant.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 7

Here two possibilities emerge based on statement I and II.

Conclusion I:- It cannot be said with certainty that some cats are elephants.

Conclusion II: - It cannot be said with certainty that all owls are elephants.

Hence, neither conclusion I nor II follows.

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

In the following question, a set of labelled sentences is given. Out of the four alternatives, select the most logical order of the sentences to form a coherent paragraph.

P) The over 11-km Mundka-Bahadurgarh section of Delhi Metro’s Green Line, connecting the capital of Haryana, was inaugurated by the Prime Minister.

Q) The rapidly growing industry in the area was waiting for Metro connectivity, said Mr.PM, adding that it would create new employment opportunities and benefit students.

R) Built at a cost of Rs 2,028 crore, the Mundka-Bahadurgarh elevated section, with seven stations, is the third Metro extension between Delhi and Haryana.

S)Mr. Prime Minister said Bahadurgarh was known as the Gateway of Haryana and would now be the Gateway of Development with the advent of the Metro.

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

The given passage talks about the plan of a metro, hence P should be the first sentence as it states about the initiation of the plan. R gives another basic information about the plan, i.e. its cost etc. Thus, R should be the second sentence. Sentence S must follow next because Q contains the abbreviation which is mentioned in sentence R. Hence, option A is correct.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 9

Study the information given below and answer the questions based on it.

Eight friends, A, B, C, D, E, F, G and H are sitting around a circle (not necessarily in the same order) facing the centre. B sits third to the left of F. E is an immediate neighbour of both B and H. Only one person sits between A and H. C and G are immediate neighbours of each other. Neither C nor G is an immediate neighbour of B. Only one person sits between C and D.

Who amongst the following is an immediate neighbour of both A and H?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 9

From the given seating arrangements, we can conclude as follows

Hence F is an immediate neighbour of both A and H.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 10

Direction: Study the line graph carefully and answer the given questions.

The graph shows that the total number of students attended the exams and the total number of students passed in the exam in different years of a school.

Note:

1. The total number of students enrolled in each year= The total number of students who appeared in the exam + The total number of students who didn’t appear in the exam

2. The total number of students who appeared in the exam= The total number of students passed in the exam + The total number of students who failed in the exam

What is the average number of students failed over the years?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 10

Total number of students failed over the years = 120 + 100 + 80 + 80 + 60 + 40 = 480

So, required average=480/6=80

Practice Test: Computer Science Engineering (CSE) - 15 - Question 11

S={1,2,...,n} of n unit-time tasks with deadlines {d1, d2, ..., dn} and penalty(non –ve) {w1, w2 ,..., wn} for missing the deadlines. Objective is to find a schedule for S that minimizes the total penalty. Execution of each task requires one unit of time. Given below is a set of tasks:

Tasks that are left out or that missed deadlines

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 11

The set S={1,2,3,4,5,6,7} of tasks is sorted on the penalty.

For t=1,2,..,n; let Nt(A) denote no of tasks in A whose di ≤ t i.e. deadline is t or earlier. Clearly, if Nt(A) > t for some t, then there is no way to make a schedule with no late tasks for set A, because there are more than t tasks to finish before time t.

● Next, we sort A on deadlines so that schedule is early tasks <2,4,1,3,7> followed bylate tasks <5,6>.

11-35

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

Consider the following statements regarding Eigenvalues:

S1: Eigenvalues of A and A-1 are the same.

S2: Eigenvalues of A and AT are the same.

S3: If A is a singular matrix, then at least one of its Eigenvalue is 0.

Which option is correct?

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

S1: The Eigenvalues of A 1 are the reciprocal of Eigenvalues of A, Hence False

S2: The Eigenvalues of AT are the same as of Eigenvalues of A, Hence True

S3: If A is singular matrix, then |A| = 0. As the determinant is given by the product of Eigenvalues, hence for the product to be 0, at least one of the Eigenvalue must be 0.

Hence S2 and S3 are correct.

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

Consider a new data structure which is named as A- tree. Consider the properties of new discovered A-tree:

• The nodes are inserted in such a way that all the nodes in the left subtree of a particular node are smaller than that node. Also, all the nodes in the right subtree of a particular node are larger than that node

• The height of the tree is balanced in such a way that the height of the left subtree and height of the right subtree differ by almost one.

Let the minimum number of nodes required to generate an A- tree of height 6 = 'a' and the maximum number of nodes which can be there in A- tree of height 6 = 'b' , find |a-b| = ________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 13
Let h(n) be the minimum no. of nodes required by AVL tree of height n.

for height (O) h(n) = 1 (root node)

height (l) h(n) = 2 height (2) h(2) 4

Also, h(n) = h(n-1) + h(N-2) + 1

for height 6.

h(3) = h(1) + h(2) + 1 = 7

h(4) = h(3) + h(2) + 1 = 12

h(5) = h(4) + h(3) + 1 = 20

h(6) = h(5) + h(4) + 1 = 20 + 12 + 1 = 33

value of a = 33

Also, no. of Nodes will be maximum when tree is a complete binary tree

No. of Nodes in a complete tree with height (n) = 2n+1 - 1

= 27 – 1 = 127

value of b = 127

|a-b| = 127 – 33 = 94

Practice Test: Computer Science Engineering (CSE) - 15 - Question 14

Which of the following statements about parser is/are CORRECT?

I. Canonical LR is more powerful than SLR

II. SLR is more powerful than LALR

III. SLR is more powerful than Canonical LR

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 14
Bottom up parsers in decreasing order of their power: CLR≫ LALR≫ SLR≫ LR (0)

The given statements:

I. Canonical LR is more powerful than SLR is CORRECT.

II. SLR is more powerful than LALR is INCORRECT

III. SLR is more powerful than Canonical LR is INCORRECT.

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

Which algorithm is mainly used in military applications to send and receive messages.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 15
Military applications use flooding algorithms as during war routers may get damaged or get exploded, in flooding algorithm message is forwarded through all available paths so it gives guarantee that message will surely reach the destination.
Practice Test: Computer Science Engineering (CSE) - 15 - Question 16

Given a disconnected graph on 10 vertices, what is the max number of edges it can have ________.


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 16
Maximum number of edges in a graph with 'n' vertices =

Minimum number of edges to be removed to make the graph disconnected = (n-1)

Hence overall maximum number of edges in a disconnected graph =

Hence for 10 vertices, (9 x 8) / 2 = 36

Practice Test: Computer Science Engineering (CSE) - 15 - Question 17

C What does the following fragment of c program print?

#include

int main()

{

static int GATE[]={100,200,300,400,500};

static int *ptr = {GATE+2,GATE,GATE+3,GATE+4,GATE+1}

int **p = ptr;

ptr++;

printf("%d%d", ptr - p, **ptr};

}

The output of the program is______

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 17
In order to simplify programs involving complex operations on pointers, we suggest you to draw proper diagrams in order to avoid silly mistakes. Let’s assume that integer is of 4 Bytes and Pointer size is also 4 Bytes.

Let’s assume array a Base address is 1000. Array name actually holds the array base address.

Array A

Let’s assume array p Base address is 2000.

Array P.

Double Pointer ptr Base Address is 3000.

Ptr

Now ptr is actually pointing to the first element of array p. ptr++ will make it point to the next element of array p. Its value will then change to 2004.

One of the Rule of Pointer Arithmetic is that When you subtract two pointers, as long as they point into the same array, the result is the number of elements separating them.

ptr is pointing to the second element and p is pointing to the first element so ptr-p will be equal to 1(Excluding the element to which ptr is pointing).

Now ptr = 2004 →*(2004) = 1000 →*(1000) →100.

Therefore, the final answer is 1100.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 18

Assume host A having an IP address of 125.32.16.5 with a subnet mask of 255.255.255.128. Also, assume another host B, having an IP address of 125.32.16.120 with a subnet mask of 255.255.255.192. Which of the following is correct?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 18
IP address of 'A' = 125.32.16.5

Subnet mask of 'A' = 255.255.255.128

Subnet ID of A according to 'A' = 125.32.16.0

IP address of B' = 125.32.16.120

Subnet ID of B according to A' = 125.32.16.0

Hence 'A' assumes 'B' to be on the same network

IP address of B' = 125.32.16,120

Subnet mask of 'B' = 255.255.255.192

Subnet ID of B according to 'B' = 125.32.16.64

Subnet ID of A according to 'B' = 125.32.16.0

Hence 'B' assumes 'A' to be on a different network.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 19

Consider the following statements:

S1: The number of relations which are both reflexive and asymmetric is 0

S2: The number of relations which are both symmetric and asymmetric is 0

Which option is correct :

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 19

S1: No relation is possible which is both reflexive and asymmetric because to be reflexive, all self-loops must be there but asymmetric doesn’t allow self-loops. Hence 0 relations.

S2: Only a single relation, i.e. empty relation is both symmetric and asymmetric.

So only S1 is correct.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 20

A system Jarvis requested for a specific site "abc.co", in return to this request it got the error as "LOOKUP FAILED". Now consider the following services:

1) SMTP

2) TCP

3) UDP

4) DNS

Which of the above service(s) are failed when we get the error as "LOOKUP FAILED"

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 20
Lookup failed means DNS has failed. The conversion from website name to its IP address can't be done.
Practice Test: Computer Science Engineering (CSE) - 15 - Question 21

Consider a binary tree in which every node has 3 attributes (Left child pointer, DATA, right child pointer). The tree has 3 levels. Now consider the following three cases:

Case 1: The tree is completely filled

Case 2: The last level of the tree is half-filled

Case 3: The last level of the tree is quarterly filled

If the total number of pointers in case 1 = a

similarly, the total number of pointers in case 2 = b

and, the total number of pointers in case 3 = c

Find the value of (b+c-a).

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 21

Case 1:

Total pointers, a = 14

Case 2:

Total pointers, b = 10

Case 3:

Total pointers, c = 8

(b+c-a) = 10 + 8 – 14 = 4

Practice Test: Computer Science Engineering (CSE) - 15 - Question 22

Consider a Demand Paging Environment Where the system requires 4 bytes to store integers and page size is of 256 bytes.

Assume the given code.

Int a [][] = new int a[200][250];

Int i=0, j=0;

While (i++ < 200)

{

while (j++< 250)

a [i][j]= 0;

}

What is the number of pages referenced from the system while executing the above code?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 22

Here, page size= 256 B

Integer Size= 4 B

No of integers on a single page= 256/4 = 64

Total no of integers to be stored in array= 200* 25 = 50000

Therefore, the total no of pages= no of integers/ integers per page

= 781.28

= 782 pages.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 23

Consider the following statements:

S1 : When in any application the number of insertion are very large but the number of deletion are very less. Then unordered array data structure give better performance than ordered array data structure.

S2 : When in any application number of insertion and number of deletion are same then binary search tree data structure give better performance than max heap data structure.

Which of the following is correct?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 23
S1 : Since number of insertions are high, hence in unordered array every insertion can be done at end which will take O(1) time, but in an unordered array, all the insertions will be done somewhere in there right positions which will take O(log n) time. Hence an unordered array will give better performance than ordered array data structures.

S2: All the operations in a binary search tree takes O(h) time where the height can go till O(n), which is not the case in max heap. Hence the statement is false:

Practice Test: Computer Science Engineering (CSE) - 15 - Question 24

A secondary index on a non candidate key

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 24
Secondary index may be from a field which is a candidate key and has a unique value in every record, or a non-key with duplicate values. Here the secondary key on a non candidate key will have duplicates.

Since the search key of the secondary index is not a candidate key and the file is unordered, only one record pointer per search key is not enough. There must be record pointer for each record. Hence dense index.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 25

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) - 15 - Question 25

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) - 15 - Question 26

The trace and determinant of a 2 × 2 matrix are known to be –2 and –35 respectively. Its eigenvalues are :

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 26

∴ Trace = λ1 + λ2 = -2

Determinant = λ1 λ2 = -35

Solving , we get λ1 = -7, and λ2 = 5

∴ a + d = -2

and ad – bc = -35

Alternately

Now,

= (λ – a) (λ – d) – bc = 0

λ2 – (a + d) λ – bc + ad = 0

λ2 + 2λ – 35 = 0

⇒ (λ + 7) (λ – 5) = 0

λ = -7 or λ = 5

Practice Test: Computer Science Engineering (CSE) - 15 - Question 27

Assume that the time required for the eight functional units, which operate in each of the eight cycles, are as follows

5ns, 8ns, 6ns, 10ns, 15ns, 12ns, 6ns, 8ns

Assume that pipelining adds 1ns of overhead. Find the speedup versus the single cycle datapath.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 27
Since the unpipelined machine executes all instructions in a single clock cycle, its average time per instruction is simply clock time. The clock is equal to the sum of the times for each step in the execution

Average instructions execution time = 5 + 8 + 6 + 10 + 15 + 12 + 6 + 8 = 70

The clock cycle time on the pipelined machine must be the largest time for any stage in the pipeline (15 ns) + the overhead of 1 ns for a total 16ns

Speed for the pipelining = Average instruction time unpipelined / An instruction time pipelined

= 70ns / 16ns

= 4.375 times.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 28

If the connection of the input 3×8 decoder is as shown in the figure, then the output F(A,B,C) can be expressed as

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 28

From the decoder circuit it is clear that F(A,B,C)= I1+I4+I6

Where I1 = C’A’B

I4 = CA’B’

I6 = CAB’

But till now the function is expressed as F(C,A,B) as C is the MSB in the given decoder. So rearranging the terms, we get -

I1 = A’BC’ = 2

I4 = A’B’C = 1

I6 = AB’C = 5

Therefore, F(A,B,C) = Σm(1,2,5)

Practice Test: Computer Science Engineering (CSE) - 15 - Question 29

Consider the disk with following characteristics. Computer Network

• Transfer speed of 107 bytes/sec

• 10,000 RPM

• Average seek time of 8ms

• Each disk operation has 2 ms overhead, Assume system bus has a maximum bandwidth of 133 Megabytes per second and network connection has bandwidth of 107 bytes/sec. HTML files have an average of 8000 bytes. Each HTML file occupied one sector.

Find the throughput to transfer HTML files? (Round to nearest integer). Assume throughput is the number of HTML files read per second?

ptions~

(a) 72

(b) 73

(c) 75

(d) 76


Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 29
Time taken to transfer one HTML file

= = 0.8 ms

Avg. rotational latency = ½ x 6ms = 3 ms

[10,000 RPM ⇒1 rotation = 6 ms]

∴ Avg time to read an HTML file

= 0.8 + 3 ms + 8 ms + 2 ms = 13.8 ms

Time taken to read one HTML file = 13.8ms

1 sec ⁡⇒ 1 sec = 72 files

∴ 72 HTML files can read per second.

Practice Test: Computer Science Engineering (CSE) - 15 - Question 30

Consider a B+ tree in which the maximum number of keys in a node is 8. What is the minimum number of keys present in a non-leaf node?

Correct Answer

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 15 - Question 30
maximum pointer = keys + 1 (order) = 9

minimum pointer = 9 / 2= 5

minimum keys = 4

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