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GATE Mock 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 - GATE Mock Test Computer Science Engineering (CSE) - 2

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

If (a  + b) : (b + c) : (c + a) = 7 : 6 : 5 and a + b + c = 27, then what will be the value of 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 1

Given,

(a + b) : (b + c) : (c + a) = 7 : 6 : 5

a + b + c = 27

(a + b) : (b + c) : (c + a) = 7k : 6k : 5k

⇒ a + b + b + c + c + a = 7k + 6k + 5k

⇒ 2(a + b + c) = 18k

⇒ (a + b + c) = 9k

⇒ c = 9k − 7k

= 2k

⇒ a = 9k − 6k

= 3k

⇒ b = 9k − 5k

= 4k

⇒ a : b : c = 4 : 3 : 6

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 2

Direction: The following sentence consists an bold word(s) followed by four options. Select the option that is nearest in meaning to the bold word and mark your response accordingly.

The inherent danger in the problem is that it would lead to many more problems.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 2

The correct answer is inbuilt:

Inherent: Existing in something as a permanent, essential, or characteristic attribute.

Inbuilt: Existing as an original or essential part of something or someone.

​​​Let's look at the meanings of the other given options:

  • outward- of, on, or from the outside
  • difficult- needing much effort or skill to accomplish, deal with, or understand
  • hallow- honor as holy

Thus, from the given meanings, we find that inherent and inbuilt are synonyms.

Hence, the correct option is (D).

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GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 3

Direction: Study the following graph carefully and answer the question based on the information given below.

Percentage distribution of teachers who teach six different subjects


If the percentage of Mathematics teachers is increased by 50% and percentage of Hindi teachers is decreased by 25%, what will be the total number of Mathematics and Hindi teachers together?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 3

Given,

Total number of teachers =1800

Now,

Percentage of Mathematics teacher =10%

After increased by 50% it becomes,

⇒10 + 5 = 15%

Similarly,

Percentage of Hindi teacher = 8%

After increased by 25% it becomes,

⇒ 8 − 2 = 6%

Now, percentage of total teachers becomes,

⇒ 15 + 6 = 21%

Therefore,

⇒ 21% of 1800 = 21/100 × 1800

= 21 × 18

= 378

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 4

In how many ways can an interview panel of 3 members be formed from 2 engineers, 3 psychologists and 4 managers if at least 1 psychologist must be included?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 4

The ways of arranging n different things = n!

The ways of arranging n things, having r same things and rest all are different 
 = n!/r!

The number of ways of arranging the n arranged thing and m arranged things together = n! × m!

The number of ways for selecting r from a group of n(n > r) = nCr

The selection can be done in the ways:

Case 1: 1 psychologist out of 3 and 2 other professionals out of  6(2 engineers +4 managers)

Aumber of ways = 3C1 × 6C2 = 3 × 15 = 45

Case 2:2 psychologists out of 3 and 1 other professional out of 6 (2 engineers +4 managers)

Number of ways = 3C2 × 6C1 = 3 × 6 = 18

Case 3:3 psychologist out of 3 and no other professional out of 6 (2 engineers +4 managers)

Number of ways = 3C3 = 1

Thus, total number ways N =45 + 18 + 1 = 64

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 5

Direction: In the following question find out the alternative which will replace the question mark.

Ornithologist : Bird :: Archaeologist : ?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 5

As Ornithologist is a specialist of Birds similarly Archaeologist is a specialist of Archaeology. 

An island or isle is any piece of subcontinental land that is surrounded by water.

The mediator assists and guides the parties toward their own resolution. The mediator does not decide the outcome but helps the parties understand and focus on the important issues needed to reach a resolution.

Aquatic means relating to water; living in or near water or taking place in water.

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 6

If the LCM of the first 100 natural numbers is N, then the LCM of the first 105 natural number will be:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 6

If we look at the numbers 100 < N ≤ 105, we see only 101 and 103 are not having their factors in N (because they are primes). So, the new LCM will be 101  x103 x N.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 7

Which of the following is an antonym of the word PROFESSIONAL?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 7

Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 8

The ratio of cost price and marked price of a cell phone is 2 : 3 and ratio of profit percentage and discount percentage is 3 : 2. What is the discount percentage?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 8

CP : MP = 2 : 3

Let CP = 200, MP = 300

(%) profit : (%) discount = 3 : 2

⇒ x = 8.33%

Discount = 2x = 16.66%

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 9

Directions: Choose the sentence that best combines the given sentences.

The airport is called the Glynco Jetport. The airline reservations and travel systems refer to the location as Brunswick, Georgia.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 9

Option 3 is the most logical subordinating phrase, showing a contrast. The other choices are not only illogical but also grammatically incorrect.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 10

Leo participated in a race of motor bikes. Flags of different colours are lowered to give different signals. How many different signals can be made here, using any number of flags from 5 flags of different colours?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 10

Signals can be made by using at a time one, two, three, four and five flags.
∴ Total number of signals that can be made = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

= 5 + (5 × 4) + (5 × 4 × 3) + (5 × 4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= 5 + 20 + 60 + 120 + 120
= 325
So, 325 signals can be made using five flags.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 11

In a trunk there are 3 types of ice cream of flavors vanilla, chocolate and blueberry. The probability of selecting one vanilla ice-cream is 12 and probability of selecting one blueberry ice-cream is 27. The total number of chocolate ice-cream is 6. Find the number of ice creams in the trunk:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 11

According to the question,

Let the ice cream of vanilla flavor be a.

Let the ice cream of blueberry ice cream be b.

Total number of ice cream = 6 + a + b

Probability (Selecting one vanilla ice cream) = 

⇒ Probability (Selecting one vanilla ice cream)  = 
 

Probability (Selecting one blue berry ice cream) = 

Solving equation (i) and (ii), we get

⇒a = 14 and b = 8

Total number of ice cream = 6 + 14 + 8

= 28

∴ Total number of ice cream in the trunk is 28.

Hence, the correct option is (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 12

Disk requests come to a disk driver for cylinders in the order 176, 79, 34, 60, 92, 11, 41 and 114. The initial head is positioned at 50 and the direction is moving from left to right. What is the total head movement (in the number of cylinders) incurred while servicing these requests, if LOOK scheduling is used?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 12

The total head movement,
= (60 − 50) + (79 − 60) + (92 − 79) + (114 − 92) + (176 − 114) + (176 − 41) +( 41 − 34) + (34 − 11)
= 291
Hence, the correct answer is 291.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 13

 Let R1(a,b,c) and R2(x,y,z) be two relations in which a is the foreign key of R1 that refers to the primary key of R2.

Consider the following four options:

Which of the following is/are correct about the referential integrity constraint?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 13

A foreign key in a table refers to the primary key of another table. It contains subset of entries of primary key of other table. It specifies the referential integrity constraint.

Given,

Two relations:

R1(a,b,c) and R2(x,y,z)

a is the foreign key of R1 that refers to the primary key of R2.

Option (A): Insert into R1

It will cause a violation. As a is the foreign key referring to the primary key of R2. If we insert values in R1, then that value must also be present in the primary key of R2 and which is not necessary as while inserting in R1 we are not aware of R2.

Option (B): Insert into R2

It will not cause a violation. Because the inserting element in R2 will not effect R1. Values present in R2's primary key may or may not present in the foreign key of R1.

Option (C): Delete from R1

It has no effect on the relation. It will not cause a violation. Values that are deleted from R1 need not be deleted from R2. As R1 contains the foreign key.

Option (D): Delete from R2

It will cause a violation. If we delete tuples from R2 relation, then if that tuple is present in R1 which is referring to the primary key of R2 must also be deleted.

Hence, the correct options are (A) and (D).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 14

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (=109 bits-per-second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 14

As sequence number field of TCP is 32 bits.

So, there are total 232 unique sequence number are possible (from 0 to 232−1 ), which is limit of TCP data.

But if you want to send data more than 232 bytes in TCP, then you need to repeat this procedure after sending 232 bytes of data or unique sequence numbers. This concept is known as wrap around which allow sending unlimited data using TCP.

Therefore, question is asking for wrap around time which is equal to pass all unique sequences first, i.e., 232, TCP assigns 1 sequence number to each byte of data.

As we know,

= 34.35 seconds

= 34 (in seconds)

Hence, the correct answer is 34.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 15

Which of the following option is true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 15

Context-sensitive Language: The language that can be defined by context-sensitive grammar is called CSL. Properties of CSL are: Union, intersection and concatenation of two context-sensitive languages is context-sensitive. The complement of a context-sensitive language is context-sensitive. CSL closed under complement so the complement of CSL is CSL.

RE and CFL are not closed under complement.

Complement of non-CFL can be CFL i.e. {ww = CSL} complement = CFL.

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 16

For the matrix  , one of the normalized eigen vectors is given as:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 16

Characteristic equation:

|A − λI| = 0

⇒ (5 − λ)(3 − λ) − 3 = 0;

⇒ λ2 − 8λ + 15 − 3 = 0

⇒ λ2 − 8λ + 12 = 0

⇒ λ = 2,λ = 6

Eigenvector for λ = 2,

(A − 2I) × x = 0

At, λ = 2

  

⇒ 3x1 + 3x2 = 0;

⇒ x1 = −x2;

The eigenvector will be  

So, the required vector is 

Hence, the correct option is (B).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 17

Which of the following statements are true about round-robin scheduling?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 17

We know that:

The round-robin scheduling algorithm is designed for time sharing systems. It is similar to FCFS but preemption is added to enable the system to switch between processes.

Now,

Some points about round- robin scheduling:

  • To implement round robin, we keep the queue as a FIFO queue of processes.
  • Round robin scheduling is preemptive.
  • If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units. Each process must wait no longer than (n−1) × q time units until its next time quantum.
  • Performance of the round robin algorithm depends on the size of time quantum.
  • If time quantum is too large, then round robin behaves like FCFS.
  • If time quantum is too small, then this approach is called processor sharing.

Hence, the correct options are (A) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 18

Consider a counting semaphore S. The operation P(S) performs the operation S = S − 1 and operation V(S) performs the operation V = V + 1. During program execution, 13 P and 5 V operation is performed in the some order. Find the number of processes in a blocked state with initial counting semaphore value 1.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 18

Given,

V(S): Signal will increment the semaphore variable, that is, S + +.

P(S): Signal will decrement the semaphore variable., that is, S − −.

Initial counting semaphore = x = 1

Signal operation =11 V

Wait operation =17 P

n process in blocked state.

Final counting semaphore (F) = −n

We know that:

n = x + 13P + 5V

Now,

n = 1 + 13( − 1) + 5(+1)

∴ n = −7

∴ Number of process in blocked state is 7.

Hence, the correct option is (A).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 19

Consider the following policies for preventing deadlock in a system with mutually exclusive resources.

Which of the below policies can be used for preventing deadlock?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 19

A deadlock will not occur if one of the following four conditions doesn’t occur:

  • Mutual exclusion: it prevents simultaneous access to a shared resource.
  • Hold and wait: Process is holding a resource that may be required by other processes.
  • Circular wait: It means there are processes that are waiting for other processes to finish.
  • No pre-emption: If a process that is holding some resources requests another resource and that resource cannot be allocated to it, then it must release all resources that are currently allocated to it.

Option (A): Processes should acquire all their resources at the beginning of execution. If any resource is not available, all resources acquired so far are released. If this is allowed, then hold and wait will never occur. So, the deadlock will not occur.

Option (B): The resources are numbered uniquely, and processes are allowed to request for resources only in increasing resource numbers. It violates circular wait.

Option (C): The resources are numbered uniquely, and processes are allowed to request for resources only in decreasing resource numbers. It also violates circular wait.

Option (D): The resources are numbered uniquely. A process is allowed to request only for a resource with a resource number larger than its currently held resources. It also violates circular wait.

Hence, the correct options are (A), (B), (C) and (D).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 20

Consider the following relational schema for students (STU):

The following query is made on the database.
T→ Πid, name (σage >13∧age<16(STU))
T→ Πid, name (σmarks >55∧ marks <66(STU))
T → T2 − T1
The total no. of rows in T is ________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 20

Given,

T1 → Πid, name (σage >13∧ age <16(STU))

As we know,

Output:

Number of rows: 5

T2→Πid, name (σmarks >55∧ marks <66(STU))

Output:

Number of rows: 3

T→T2−T1

Output:

T contains 1 row.

Hence, the correct answer is 1.

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 21

What is the minimum and maximum number of link field updations required respectively to insert a new node in the double linked list?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 21

Three cases:

(i) Insertion at the beginning.

p → new node

q → first node

∴ p → R link = q

p → L link = NULL

q → L link = p

∴3 updations.

(ii) Insertions at end (q is the last node).

p → R link = NULL

p → L link = q

p → R link = p

∴ 3 updations.

(iii) Insertions in middle of q and r.

p → R link = r

p → L link = q

q → R link = p

r → L link = p

∴ 4 updations.

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 22

Which of the following is hardware generated signal?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 22

A trap is a software-generated interrupt. An interrupt can be used to signal the completion of an I/O to obviate the need for device polling. A trap can be used to call operating system routines or to catch arithmetic errors. A trap is a synchronous interrupt triggered by an exception in a user process to execute functionality. Exception conditions like invalid memory access, division by zero, or a breakpoint can trigger a trap in an OS. A trap changes the mode of an OS to a kernel routine.

Hence, the correct option is (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 23

Consider a two dimensional array A[1…10][3….8] that is column index ranges from 3 to 8 and row index ranges from 1 to 10. Now suppose element are arranged in column major order. If base address of A is 1000 then address of element A[5][7] will be _________ (Assume size of each element is 1).


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 23

Given,

Number of rows = 10

Number of columns = 6

Starting of row indexing (n) = 1

Starting of column indexing (m) = 3

In the case of column major order,

Address of A[i][j] = ((j − m) ×  no. of rows + (i − n) ×  size of element + base address of A.

So, address of A[i][j] = ((7 − 3) × 10 + (5 − 1)) + 1000

= 1044

Hence, the correct answer is 1044.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 24

For the digital circuit shown in below figure, the output F is found to be '1' when P is logic '0'.

Choose the correct answer from the below given options for the function F:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 24

Expression for output F is given by:

 

(always whether Q is '0' or '1')

Therefore, F is independent to Q and R.

Hence, the correct options are (A) and (C).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 25

Consider X = 11111011 and Y = 00001010 be two 8 bit two's complement number. What is the value of their product in two's complement?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 25

Since 8 bit number is in 2's complement, weight of the last bit will be negative:

(11111011)2

=  −1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20

= −5

(00001010)2

= −0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20

= 10

Now, we get

(11111011)2 × (00001010)2

= −5 × 10

= −50

⇒ 11001110 = −1 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20

⇒ 11001110 = −128 + 64 + 8 + 4 + 2

= −50

So, the value of their product in two's complement is 11001110.

Hence, the correct answer is 11001110.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 26

Which of the following is/are true about conflict miss?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 26

Conflict misses: Even when the empty place is available, the block is trying to occupy an already filled line leads to conflict miss.

In direct-mapped, memory reference (x) should be placed in x mod C where C is number blocks in the cache. In set-associative mapped, memory reference (x) should be placed in x mod S where S is number sets in the cache. So, conflict miss occurs in the case of set-associative or direct-mapped block placement strategies.

In a fully associative mapping, the block number of main memory is equal to the tag size, therefore is no such constraint to put memory reference in a specific set or lines of a cache and so no conflict miss.

Hence, the correct options are (B) and (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 27

Consider the following problems L(G) denotes the language generated by a grammar G.L(M) denotes the language accepted by a machine M.

(I) For an unrestricted grammar G and a string w, where w = L(G).

(II) Given a Turing Machine M, whether L(M) is regular.

(III) Given two grammars G1 and G2 whether L(G1) = L(G2).

(IV) Given an NFA N, whether there is a deterministic PDA P such that N and P accept the same language.

Which one of the following statements is correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 27

Given,

Statement (I): Undecidable

For an unrestricted grammar G and a string w, where w = L(G).

Membership problem for RE → undecidable.

Statement (II): Undecidable

Given a turing Machine M, whether L(M) is regular.

Regularity problem for RE → undecidable.

Statement (III): Undecidable

Given two grammars G1 and G2 whether L(G1) = L(G2).

Equivalence problem for RE→ undecidable.

Statement (IV): Decidable

Given an NFA N, whether there is a deterministic PDA P such that N and P accept the same language.

Since DPDA P exists for every NFA N and is equivalent to it, this problem is trivially decidable.

Hence, the correct options are (A), (B) and (C).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 28

Suppose that a robot is placed on the cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i + 1,j) or (i,j + 1).

How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 28

At each move, robot can move either 1 unit right, or 1 unit up, and there will be 20 such moves required to reach (10,10) from (0,0). So we have to divide these 20 moves, numbered from 1 to 20, into 2 groups:

  1. Right group
  2. Up group

Right group contains those moves in which we move right, and up group contains those moves in which we move up.

Each group contains 10 elements each. So basically, we have to divide 20 things into 2 groups of 1010 things each, i.e., we need to find all possible arrangements of {r, r, r, r, r, r, r, r, r, r, u, u, u, u, u, u, u, u, u, u} where r represents right move and u represents up move. The arrangements can can be done in 
 ways

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 29

Each flip-flop in a 4-bit ripple counter introduces a maximum delay of 40 n sec. The maximum clock frequency is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 29

Synchronous counter: In this all the flip-flops work in sync with clock pulse as well as each other. Here clock pulse is applied to every flip flop simultaneously.

Asynchronous counter: In this clock pulse is applied only to the initial flip flop whose value would be considered as LSB. Instead of the clock pulse, the output of the first flip-flop acts as a clock pulse to the next flip-flop, whose output is used as a clock to the next in-line flip-flop and so on.

Ripple Counter:

  • Ripple counter is an asynchronous counter.
  • It got its name because the clock pulse ripples through the circuit.
  • An n−MOD ripple counter contains n number of flip-flops and the circuit can count up to 2 n values before it resets itself to the initial value.

Maximum clock frequency is given by:

 Now,

n = 4

⇒ delay (d) = 40 ns

Hence, the correct option is (B).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 30

Consider the regular language L = (111 + 11111). The minimum number of states in any DFA accepting this languages is:


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 2 - Question 30

Given language L = (111 + 11111)

Strings, that belongs to the language

L = {ϵ,111,11111,111111,11111111,111111111,111111111,…

Due to concatenation with (111) if we have three consecutive string lengths in L, then all higher string lengths will be in L.

We have strings of length 8,9,10 in L and so all higher length strings are also in L.

So, required DFA will be:

So, there are 5 states are final states and 4 states are non-final states, total number of states are 9 states.
Hence, the correct answer is 9.

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