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

Practice Test: Computer Science Engineering (CSE) - 2 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) - 2 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Practice Test: Computer Science Engineering (CSE) - 2 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) - 2 below.
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Practice Test: Computer Science Engineering (CSE) - 2 - Question 1

A vendor sells his articles at a certain profit percentage. If he sells his articles at ¼th of his actual selling price then he incurs a loss of 60%. What is his actual profit percentage?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 1

Let the cost price = 100 Rs.

From the options:

If profit % = 60%

Then SP = 160 Rs.

New SP = 160/4 = 40 Rs.

Then,

Percentage loss= (100-40)/100 = 60%

Hence Verified

ALTERNATE:-

Let the selling price is 100

New selling price 100/4 = 25

He suffers a loss of 60%.

CP*0.4 = 25

CP = 25/0.4 = 62.5

Actual Profit = 100 - 62.5 = 37.5

profit% = 37.5/62.5 *100 = 60%

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

Find wrong number in series: 23, 29, 31, 33, 41, 43, 47

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

the given series is prime numbers from 23

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Practice Test: Computer Science Engineering (CSE) - 2 - Question 3

Criteria for selecting candidate for internship programme

The candidate:

(1) can preferably start the internship between 18th Oct'17 and 17th Nov'17

(2) are preferably available for duration of 6 months

(3) have computer skills and interest in designing

(4) have already graduated or are currently in any year of study

(5) knows to deal with customers

Nick is a high school student and wants to do an internship as his summer project. He is a very vibrant boy and goes well with people. Is he the right candidate for the internship?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 3
He does is looking for summer internships and November is not summer time. Hence, option B is the correct option.
Practice Test: Computer Science Engineering (CSE) - 2 - Question 4

Direction: In the given question, one statement with a blank along with four words is given. Two of the given words can fit into the given blank. Five options with various combinations of these words are given. Choose the combination of the words that best fits into the blank.

Comedian Vasu Primlani takes a hilarious ______ at the new trend of renaming Indian cities.

a) Jabs

b) Jokes

c) Satires

d) Gags

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 4
The meaning of the words as follows:

a. Jab as a noun refers to a quick, sharp blow, especially with the fist.

b. A joke is a thing that someone says to cause amusement or laughter, especially a story with a funny punchline.

c. Satire is the use of humour, irony, exaggeration, or ridicule to expose and criticize people's stupidity or vices, particularly in the context of contemporary politics and other topical issues.

d. A gag is a joke or an amusing story, especially one forming part of a comedian's act, or in a film.

‘Hilarious satires’ and ‘hilarious jabs’ seem inappropriate and ambiguous. Among all the options, ‘jokes’ and ‘gags’ fit perfectly in the blank. Therefore, option D is the apt answer.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 5

Which of the following is MOST OPPOSITE in meaning to Locus?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 5
Locus (noun) -a particular position or place where something occurs or is situated.
Practice Test: Computer Science Engineering (CSE) - 2 - Question 6

The ratio between the speed of a bus and train is 15 : 27, respectively. Also, a car covered a distance of 720 km in 9 h. The speed of the bus is three- fourth of the speed of the car. How much distance will the train cover at 7 h?

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

speed of car = distance covered / time taken

720 / 9 = 80 km/h

Now, speed of bus = 3 / 4 x 80 = 60 km/h

Speed of train = 27 / 15 x 60 = 108 km/h

Distance covered by train in 7 h = 108 x 7 = 756 km

Practice Test: Computer Science Engineering (CSE) - 2 - Question 7

The Union Sports Ministry has approved five lakh rupees from the National Welfare Fund for Sportspersons for Kaur Singh who is suffering from heart disease. Kaur Singh is associated with which of the following sports?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 7

Kaur Singh, former heavyweight Boxer is struggling with the treatment for heart disease and admitted at a private hospital in Mohali. Under such circumstances, the Union Sports Ministry has approved five lakh rupees from the National Welfare Fund for Sportspersons for Kaur Singh.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 8

Direction: In the following table data is given about an electronic shop. Some data is given and some data is hidden. Study the given data carefully and answer the related questions given below.

Selling price of T.V is what percent of Marked price of laptop?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 8

Selling price of T.V = 5000 x 100 / 10 x 90 / 100 = 45000

Marked price of laptop = 45000 + 15000 = 60000

Required % = 45000/60000 x 100 = 75%

Practice Test: Computer Science Engineering (CSE) - 2 - Question 9

Direction: In the given question below there are three statements followed by two conclusions numbered I and II. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Statements:

All oils are sands

Some clays are oils

All clays are rocks

Conclusions:

I. At Least some clays are sands

II. Some oil is not rock

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 9

Statements

All oils are sands

Some clays are oils

All clays are rocks

Combining all three statements, we get

Conclusions

I. At least some clays are sands

II. Some oil is not rock

Conclusion I

Some clays are oils + All oils are sands = Some clays are sands

Hence, Thus, the conclusion I follow.

Conclusion II

Some clays are oils → conversion → some oils are clays + All clays are rocks = Some oils are rock. Hence, conclusion II does not follow.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 10

Direction: In the given question, the 1st part of the sentence is given. The rest of the sentence/passage is split into four parts and named A, B, C and D. These four parts are not given in their proper order. Read the sentence and find out which of the four combinations is correct.

1) According to Indian Express, on Sunday, six men reached a South Delhi businessman's house in Malviya Nagar in a Tata Safari car bearing the Haryana government’s sticker fixed on the windscreen.

(A) One of the guys first took away all the cell phones of the businessman’s family members alleging that they had come on government duty to investigate a tax evasion charge.

(B) The plan could have been successfully executed but one of the family members found their behaviour suspicious.

(C) When the family member raised an alarm about the same, about 150 people from the same locality gathered outside the trader's house and the con men were beaten up and interrogated before they were handed over to the police.

(D) They “searched” the house, moving from room to room, and collected Rs 20 lakh in cash that they kept in their cars.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 10

The first sentence 1 introduced a situation, and the rest of the sentences should be arranged in the order that they happened. Clearly A is the first in sequence as it uses the word 'first' indicating the first step of these six men. The next step is described in D. Now, the twist comes in sentence C as the suspicions rise. People cannot get suspicious of something unless something actually happened. So, we place A first, followed by D and B. This leaves C for the last, which describes the result of those suspicions. Hence, the correct answer is d.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 11

Consider the following conditions:

1. tp < Δt

2. Δt < T

3. tp > Δt

4. Δt > T

Where, tp = pulse width,

Δt = propagation delay and

T = clock time period.

The race around condition in the Flip-Flop can be avoided if conditions

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 11
  • Let the inputs to JK flip flop are J = K = 1 and Q = 0. After the propagation delay Δt, the output will change to Q = 1.
  • Now, we have J = K = 1 and Q = 1 and after another time interval of Δt, the output will change back to Q = 0.
  • Hence, for the duration tp of the clock pulse, the output will oscillate back and forth between 0 and 1.
  • At the end of the clock pulse, the value of Q is uncertain. This situation is referred to as the race around the condition.


Therefore, the race around the condition can be avoided by keeping the propagation delay as tp < Δt < T. 

Note:

For JK flip-flop if J, K, and Clock are equal to 1 the state of flip-flop keeps on toggling which leads to uncertainty in determining the output of the flip-flop. This problem is called Race around the condition.

This can be eliminated by using the following methods.

  • Increasing the delay of flip-flop
  • Use of edge-triggered flip-flop
  • Use of master-slave JK flip flop
Practice Test: Computer Science Engineering (CSE) - 2 - Question 12

Consider the following schedules involving three transaction:

S1 : W2(x), W1(x), R3(x), W2(y), R3(y), R3(z), R2(x), R1(y)

S2 : R2(z), W2(x), W2(y), R1(x), R3(x), R2(z), R3(y), W1(x)

Which of the above schedules are conflict serializable?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 12
A schedule is serializable if the precedence graph does not contain any cycle.

It contains cycles so it is not serializable.

It contains no cycle so is serializable.

So option (C) is correct.

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

Consider the grammar defined by the following production rules:

S → A * C

A → B + A | B

B → A – B | A

A → id

B → id

C → id

Which of the following is true?

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

+ and – are both right associative and + has higher precedence than*. So the correct option is (c).

Practice Test: Computer Science Engineering (CSE) - 2 - Question 14

The output waveforms of a counter circuit shown below:

The counter is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 14
The waveform is of the ring counter because in one clock period only one flip-flop output is active and the same is repeating.
Practice Test: Computer Science Engineering (CSE) - 2 - Question 15

For k number of users, how many keys are needed using private key cryptography and public key cryptography schemes respectively?

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

In symmetric ⇒ Asymmetric ⇒ 2k

Explanation for asymmetric:

Every user has a (public key, private key) pair.

1 user ⇒ 2 keys

k user ⇒ 2k keys

Hence option (C) is the right answer.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 16

A binary operation Ηon a set of integers is defined as x⊙y=x2+y2+2xy. Which one of the following statements is true aboutΗ?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 16
1.Check for commutative

(x⊙y)=(y⊙x)

x2 + y2 + 2xy = y2 + x2 + 2yx

Since LHS = RHS

So ⊙ is commutative.

2. Check for associative:

(x⊙y)⊙z=x⊙(y⊙z)

(x2 + y2 + 2xy) Η z = x Η ( y2 + z2 + 2yz)

x4 + y. + 4x2y2 + 2x2y2 + 4xy3 + 4x3y + z2 + 2x2z + 2y2z + 4xyz ≠ x2 + y4 + z4 + 4y2z2 + 2y2z2 + 4y3z + 4yz3 + 2xy2 + 4xyz

So A is not associative.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 17

A token bucket scheme is used for traffic shaping. A new token is put into the bucket every 10 μsec. Assume each token picks one packet which contains 2 bytes of data. It is observed that the initial capacity of the bucket is 10 Mbits. The computer can transmit at the full speed of 8 Mbps for ________ (in seconds upto 2 decimal places)

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 17
We know,

C + R × S = M × S

Where, C = Initial capacity

R = Token rate

M = Output rate

S = Bursty traffic

Given, C = 10 Mbits

Token arrives at an interval of 10 μsec.

Then in 1 sec 100000 token arrives.

So, 100000 × 16 bit

= 16 × 105 bits/sec

= 1.6 Mbps

Hence R = 1.6 Mbps

Substituting we get

S = = 1.5625

Practice Test: Computer Science Engineering (CSE) - 2 - Question 18

The incorrect match (when n > 1) is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 18
MD model has n-control units and since in multiprocessing, it requires n ALU units also.
Practice Test: Computer Science Engineering (CSE) - 2 - Question 19

An Internet Service Provider (ISP) has a block 219.50.0.0/16. There are 3 groups, Group1 has 128 customers and each requires 64 IP addresses. Group2 has 64 customers and each requires 256 IP addresses. Group3 has X customers and each requires Y IP addresses. After successfully completing customer requests only 38K IP address left with (ISP)What can be the possible value of X and Y respectively.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 19

38 × 210 = [216 – [Group1 + Group2 + Group3]]

38 × 210 = [216 – [27 × 26 + 26 × 28 + 2m × 2n]]

2x × 2y = 216 – 213 – 214 – 38 × 210

= 210[26 – 23 – 24 – 38]

= 210[64 – 8 – 16 – 38] = 210 × 21

2x × 2y = 211

Option (d), which 26 × 25 = 211

Practice Test: Computer Science Engineering (CSE) - 2 - Question 20

Let R be a relation on the set A = {1, 2, 3}, such that R = {(1,1), (2,2), (3,3), (1,2)}. Now consider the following relations.

I. {(1,1), (2,2), (3,3)}

II. {(1,1), (2,2)}

III. {(1,1), (2,2), (3,3), (1,2)}

Which of the above correctly represent(s) the smallest reflexive closure of R?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 20
The common mistake here is to pick both I and III as the answer, which is wrong. Why? Because first of all in order for a relation T (say) to be reflexive closure of a relation R, then firstly T must be a superset of R, and only then we think about reflexivity, So I is actually a subset of R and thus cannot be the reflexive closure of R. II does not contain (3,3) and therefore is not reflexive, So the correct answer is choice (c).
Practice Test: Computer Science Engineering (CSE) - 2 - Question 21

The sorting algorithm which requires least number of swaps in the worst case is

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 21

Insertion Sort ⇒ O(n2)

Selection Sort ⇒ O(n)

Bubble Sort ⇒ O(n2)

Quick Sort ⇒ O(n2)

Amongst all these algorithms, selection sort requires the least number of swaps.

Selection sort, which in many respects has incredibly poor performance (not adaptive; quadratic performance re: num of comparisons), actually requires the minimum number of swaps. It is based on the concept of doing a significant number of comparisons before moving each element directly to its eventual sorted resting place. At most the algorithm requires N swaps; once you swap an element into place, you never touch it again.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 22

We are given an array A in which every element is either 0 or 1. The time complexity of the most efficient algorithm which sorts A in descending order is equal to

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 22
The simplest way is to scan the entire array once, and maintain a count of the number of 0's (zero_count) and 1's in the array (one_count) - for every 0 encountered, increment the zero_count and similarly do the same for the 1's also.

And then overwrite the array by first filling the array with 1's - the number of 1's being equal to one_count value, and do the same for 0's also.

All this will take O(n) time and (A) is the answer.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 23

A 4-way set associative cache has lines of 32 byte and a total cache size of 16 KB. Which of the following main memory blocks is mapped onto the set ‘13’ of the cache memory when 16 MB of main memory is used?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 23

Number of lines =

Number of sets =

Cache representation

Physical address size = 16 MB = 24 bit

24 = x + 7 + 5

x = 12 bit

Now, to map physical memory at 13th set bit (b11 b10 b9 b8 b7 b6 b,) MM address should be 13.

In option (d)

Hence option (d) is correct.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 24

Consider the following relations, SQL query and given instances of relations: (where keys are underlined)

Student (snum, sname)

Enroll (snum, cname)

SELECT S.name FROM Student S WHERE

S.snum NOT IN (SELECT E.snum FROM Enroll E)

Number of tuples returned by the SQL query is ________.

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 24

SELECT S.Name FROM Student S WHERE

S.snum NOT IN (SELECT E.snum FROM Enroll E)

It return snum of all students who is enrolled in any course.

Query return the sname of a student who is not enrolled in any course.

SQL does not eliminates duplicate so relation given by SQL query is

Total 2 tuple returns.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 25

Consider two languages, L1and L2:

L1= {an| n > = 0} and L2 = {bn | n > = 0}

Which of the following correctly represents L1⋅ L2, where ‘⋅’ is the concatenation operation?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 25

L1 . L2 will be equal to {an bm| m , n > = 0}, which is same as {an bm| m = n, n > –1}.

Practice Test: Computer Science Engineering (CSE) - 2 - Question 26

Number of binary trees formed with 5 nodes are

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - 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) - 2 - Question 27

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

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 27



For a 2 × 1 MUX is shown above, the output function F is expressed as:

F = S̅I+ S1 I1

i.e. when S1 = 0, I0 is transmitted to the output.

And when S1 = 1, I1 is transmitted to the output.

Application:

The output of the 1st MUX will be:

E = B̅C + BC̅ 

The final output will be:

f = E̅.0 + E.A = E.A

f = (B̅C + BC̅) A

f = AB̅C + ABC̅ 

Practice Test: Computer Science Engineering (CSE) - 2 - Question 28

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

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 28
According to the 4-colour theorem states that 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) - 2 - Question 29

Consider the following statements about simple connected undirected graph having more than 2 vertices:

1) At least two vertices have the same degree

2) At least three vertices have the same degree

Which one of the above statements hold true?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 29
As the graph is simple, there is no self loop or parallel edge. As the graph is connected, no vertex can have 0 degree and hence, the degree range from 1 to n-1 which implies that the degree of at least two vertices must be the same.
Practice Test: Computer Science Engineering (CSE) - 2 - Question 30

Given a combinational circuit below:

(2 X 1 multiplexer are being used)

What will be the sum minterms at output f?

Detailed Solution for Practice Test: Computer Science Engineering (CSE) - 2 - Question 30

f = ( C’ + BC ) D + A D’

f = C’ D + BCD + AD’

Min-terms that represent f are

C’ D - 0 0 0 1, 0 1 0 1, 1 0 0 1, 1 1 0 1

B C D - 0 1 1 1, 1 1 1 1

A D’ - 1 0 0 0, 1 0 1 0, 1 1 0 0, 1 1 1 0

f = min-terms (1 , 5 , 7 , 8 , 9 , 10, 12 , 13 , 14 ,15)

Now

1 + 5 + 7 + 8 + 9 + 10 + 12 + 13 + 14 + 15 = 94.

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