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

GATE Mock Test Computer Science Engineering (CSE) - 9 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) - 9 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 9 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) - 9 below.
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GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 1

If X2 + 4Y2 = 4XY, find the value of X3 : Y3 :

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 1

Given,

X2 + 4Y2 = 4XY

⇒ X2 + 4Y2 − 4XY = 0

⇒ (X − 2Y)2 = 0

⇒ X − 2Y = 0

⇒ X = 2Y
⇒ X/Y = 2/1

On cubing both the sides,

⇒ X3 : Y3 = 8 : 1

Hence, the correct option is (C).

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

Direction: Fill in the blank with the most appropriate option.

The advisor has not yet returned the student lists, but when ____________ they will be put on the bulletin board.

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

In the given sentence, there is a conjunction 'but' and the second clause does not contain a verb. Therefore, the answer must contain a verb. Further, the second clause also does not contain a subject. Thus, the answer must also include a subject that matches the subject in the first sentence, that is, the answer must contain a pronoun that matches with 'the advisor' which must be a third-person singular pronoun. Also, since the tense of the first clause is present tense, the answer must also be in the present tense and since the action in both the clauses is the same, the verb must be a form of the verb 'do'.

Option (A) is incorrect. In 'he will', 'will' is in the simple future tense, and hence, is incorrect. Further, 'will' is also not a form of 'do'.

Option (B) is incorrect. In 'they do', although 'do' is present, 'they' is a third-person plural pronoun. Thus, it is incorrect

Option (C) is correct. 'he does' contains a third person pronoun 'he'. Further, the verb 'does' is the simple present form of 'do' for third-person singular pronouns. Thus, it is the correct answer.

Option (D) is incorrect. 'they are' contains a third person plural pronoun 'they', making it incorrect. Further, the verb 'are' is also not a form of the verb 'do'. Thus, it is incorrect.

Hence, the correct option is (C).

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GATE Mock Test Computer Science Engineering (CSE) - 9 - 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.


What is the total numbers of teachers who teach Chemistry, English and Biology?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 3

Given,

Total number of teachers = 1800

Chemistry = 23%

English = 27%

Biology = 12%

Now,

(Chemistry + English + Biology) % of Total number of teachers = (23 + 27 + 12)% of 1800

= 62% of 1800
= 62/100 × 1800

= 1116

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 4

In how many ways can the letters of the word SOFTWARE be arranged so that all the vowels be together?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 4

Given,

Word = SOFTWARE

Arrangement of n letter of a word.

Case 1: If there were no repeating letters = n!

Case 2: If there were "r" letter of one kind = n!/r!

Keeping all the vowel at one place and consider it as one vowel = SFTWR(OAE)

Here, n = 6 and number of vowel = 3

All letter can be arranged in 6! ways and vowel can be arranged in 3! ways.

Total number of ways to arrange the word,

= 6! × 3!

= 6 × 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1

= 4320

∴ Total number of ways to arrange the word = 4320

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 5

Directions: Choose the option that best expresses the meaning of the idiom which is underlined.

He is a plain, simple and sincere man. He will always call a spade a spade.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 5

If you 'call a spade a spade', you speak honestly and directly about a subject even if it offends people.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 6

Pick the odd one out.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 6

Swimming is done in a large water body. All the other words relate to activities involving smaller volume of water.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 7

Pick the odd one out.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 7

Break, hiatus and pause, all mean a temporary halt. End is the final halt.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 8

SBI lent 1 lakh to farmer Kishan Singh at 6% per annum, at simple interest for a period of 10 years. Meanwhile, the bank offered a discount in the rate of interest, according to the Govt. policy for farmers. Thus, the rate of interest is decreased to 4% p.a. In this way, Kishan Singh had to pay a total amount 1.48 lakh. After how many years did Kishan Singh got the discount in the rate of interest?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 8

Suppose that 6% per annum be the rate of interest for x years, then the 4% per annum be the rate of interest for (10 - x) years.
Interest paid by Kishan Singh = Rs. 48,000
Now,

48 = 6x + 40 - 4x
2x = 8
x = 4 years
It means Kishan Singh availed the discount after 4 years.
Hence, option (3) is correct.

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

After striking the floor, a Tennis ball rebounds to 4/5th of the height from which it has fallen. What will be the total distance (in m) that it travels before coming to rest if it has been gently dropped from a height of 90 metres?

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

The first drop is 90 metres. After this, the ball will rise by 72 metres and falls by 72 metres, now this process will be continue in the form of an infinite GP with common ratio 0.8 and first term 72.
The required answer will be got by 


= 90 + 2(360)
= 810 m

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 10

Four cities are connected by a road network, as shown in the figure. In how many ways can you start from any city and come back to it without travelling on the same road more than once?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 10

Suppose that a man starts from a city. The condition is that there should be no repetition of path during his journey. So, two paths are restricted for one pass only.
Thus, the total number of possible cases = 12.
Travelling with each node in one direction only will give 6 possible routes. There are another 6 routes, when travelling in opposite direction from the first one. So, the total possible ways are 12 only.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 11

Consider the following expression:

  1. A ∧ B
  2. A ∨ B
  3. ¬(A ∨ B)
  4. TRUE
  5. FALSE

The number of expression above that are logically implied by (A ∧ B) ∧ (¬A → B) is ___________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 11

Important Points:

. ≡ ∧

+ ≡ ∨

′ ≡¬

We know that,

(A ∧ B) ∧ (¬A → B) ≡ A ⋅ B ⋅ (A′ + B) ≡ A ⋅ B ≡ A ∧ B

Statement 1: A ∧ B ≡ A . B

(A ∧ B) → (A ∧ B) ≡ ¬(A ∧ B) ∨ (A ∧ B) ≡ X′ + X ≡ 1 ≡ TRUE

Statement 2: A ∨ B ≡ A + B

(A ∧ B) → (A ∨ B) ≡ (A ⋅ B)′ + (A + B) ≡ (A′ + B′) + (A + B) ≡ A′ + A + B′ + B ≡ 1 ≡  TRUE

Statement 3: ¬(A∨ B) ≡ A′ . B′(A ∧ B) → (¬(A∨B)) ≡ (A . B)′+ A′ ⋅ B′ ≡ (A′ + B′) + A′ . B′

≡ A′ + B′ ≡ (A . B)′ ≡ FALSE (if A =TRUE and B =TRUE)

Statement 4: TRUE

(A ∧ B) → TRUE ≡ ¬(A ∧ B) ∨ 1 ≡ 1 ≡ TRUE

Statement 5: FALSE

(A ∧ B) →  FALSE  ≡¬(A ∧ B) ∨ 0 ≡ ¬(A ∧ B) ≡  FALSE (if A = TRUE and B = TRUE )

So, the statements I, II and IV are logically implied by (A ∧ B) ∧ (¬A → B).
Hence, the correct option is (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 12

When an I/O device completes an I/O operation, which of the following is/are not correct regarding the sequences of hardware events?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 12

When an I/O device completes an I/O operation, the following sequence of hardware events occurs:

  1. The device issues an interrupt signal to the processor.
  2. The processor finishes the execution of the current instruction before responding to the interrupt.
  3. The processor test for an interrupt and sends an acknowledgment signal to the device that issued the interrupt.
  4. The processor now needs to save information needed to resume the current program at the point of an interrupt.
  5. information saved to be is the status of the processor and the location of the next instruction to be executed.
  6. The processor now loads the program counter with the entry location of the interrupt handling program.
  7. The interrupt handler next processes the interrupt.
  8. After interrupt processing, saved register values are retrieved from the stack and restored and the next instruction will be executed.

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

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 13

In a leaky bucket system if the output rate is 5 KB/sec and input burst of 50 KB/sec for 10 sec and 10 KB/sec for 60 sec then bucket size in KB is ___________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 13

Given,

Input burst = 50 KB/sec for 10 sec and 10 KB/sec for 60 sec

Output rate = 5 KB/sec

Total time = 60 sec

Total Length = 500 KB + 500 KB

= 1000 KB.

Will, it is in 60 sec,

Now, output rate = 5 KB/sec

It is in 60 sec.

⇒ 5 × 60 = 300 KB

In 60 sec we can transmit 300 KB.

We should have a bucket of size = 1000 KB − 300 KB

= 700 KB

Hence, the correct answer is 700.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 14

The output Y of the logic circuit given below is:


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 14

XOR GATE:

Symbol:

Truth Table:

Output Equation: 
 Y = A ⊕ B =
As we know,
1. If B is always high, the output is the inverted value of the other input A, i.e. 
2. The output is low when both the inputs are the same.
3. The output is high when both the inputs are different.

 

Hence, the correct answer is 1.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 15

Let L be the language represented by the regular expression Σ0011Σ where Σ = {0, 1}. What is the minimum number of states in a DFA that recognizes  (complement of L )?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 15

If minimal DFA for any regular language L has n states then minimal DFA for the complement of L will also have n states.

DFA for language represented by the regular expression Σ0011Σ where, Σ = {0,1} is as following below:

The complement of the above DFA will have the same number of states but there will be non-final states to final states and final states to non-final states.

The minimum number of states in the minimal DFA that recognizes 
 
 is 5. The given regular expression matches with all strings that contain substring 0011. The complement should match with all strings except the strings with 0011 as a substring.

Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 16

The factorized form of the following determinant is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 16

Given,

Applying R2 → R- R1

Applying R→ R3 - R1

Applying R3 → R3 - R1

Now, expanding from a11.

= (m − l)(n − l)(n + l − m − l)

= (m − l)(n − l)(n − m)

Hence, the correct option is (B).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 17

Consider a two dimensional array X[−10…5,7…15] in which staring location is at 250. If every data of a given array takes 4 byte of space and store in column major order, then what will be the location of A[2][10]?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 17

We know that,

Range of given array:

[−10…5,7……15]

Old location (Given):

A[2][10]

A[−10][7] = 250

Number of rows = 5 − ( − 10) + 1

= 16

Number of columns = 15 − 7 + 1

= 9

Array size:

A16×9

New range:

[0…15,0…8]

New location:

A[2 + 10][10 − 7] = A[12][3]

Size of data = 4

A[0][0] = 250

Formula:

A[n][m] = A[0][0] + ( Row size  × m + n) ×  Size of cell

Now,

A[12][3] = 250 + (16 × 3 + 12) × 4

A[12][3] = 250 + 60 × 4

A[12][3] = 490

Old location of A[ − 10][7] =  New location of A[12][3] = 490.

Hence, the correct answer is 490.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 18

Consider 3 processes whose {X,Y,Z} whose arrival and burst time are given below

Which of the following is/are true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 18

Sequence: Y, Z, X

Gantt chart:

Processs Table:

Average waiting time 
= 8/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 13/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 11/3
Sequence: X, Y, Z
Gantt chart:

Average waiting time 
 = 13/3
{Y, Z, X} is an optimal non-preemptive scheduling sequence that gives 8/3 average waiting time.
Formula used:
TAT=CT-AT
WT=TAT-BT

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

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 19

A disk drive has 200 cylinders, numbered 0 to 199. Disk requests come to the disk driver for cylinders 82, 170, 43, 140, 24, 16 and 190. The driver is currently serving a request at a cylinder 50. How much seek time is needed for the FCFS algorithm?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 19

FCFS is the simplest disk scheduling algorithm. As the name suggests, this algorithm entertains requests in the order they arrive in the disk queue. The algorithm looks very fair and there is no starvation (all requests are serviced sequentially) but generally, it does not provide the fastest service.

Algorithm:

1. Let Request array represents an array storing indexes of tracks that have been requested in ascending order of their time of arrival. ‘head’ is the position of disk head.

2. Let us one by one take the tracks in default order and calculate the absolute distance of the track from the head.

3. Increment the total seek count with this distance.

4. Currently serviced track position now becomes the new head position.

5. Go to step 2 until all tracks in the request array have not been serviced.

Given,

A disk drive has cylinders = 200

They are numbered 0 to 199.

Disk requests come to the disk driver for cylinders 82, 170, 43, 140, 24, 16 and 190.

The driver is currently serving a request at a cylinder 50.

So, total seek time:

= (82 − 50) + (170 − 82) + (170 − 43) + (140 − 43) + (140 − 24) + (24 − 16) + (190 − 16)

= 642

Hence, the correct answer is 642.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 20

A combinational digital circuit design is attempted using only NAND logic. The circuit has four inputs A,B,C,D, and one output Z. The output Z is required to be 1 (high) whenever either A = B = 1, or C = D = 1, or both.

Which of the following conclusions is/are correct based on the given data?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 20

As per design demand, the truth table is prepared as shown below:

Two level NAND implementation has the delay = 10 + 10
= 20 ns.
Hence, the correct options are (C) and (D).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 21

For a carry look ahead binary adder for n(n > 5) bit binary numbers A and B, then consider the following SOP expression for generating carry:

C5 = G5 + PaGb + PcP5G3 + P5P4P3Gd + P5PeP3P2G1 + P5P4P3P2P1G0 + P5P4P3P2P1P0Cf

Where Pi is the propagation function and Gi is the generating function and have usual meaning.

i.e. Gi = Ai × Bi

Pi = Ai ⊕ Bi

C−1 is the input carry.

What will be correct values for the subscripts a, b, c, d, e, f?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 21

In a carry look-ahead adder if we ignore the input size of a gate, then to add two n bit numbers using CLA the total no of AND and OR gate required. In a look-ahead carry generator, the carry generate function Gi and the carry propagate function Pi for inputs Ai and Bi are given by:

Pi = Ai ⊕ Bi and Gi = Ai × Bi

In the carry generator function first thing is that either the carry Ci is generated by the current bits i.e. by Ai and Bi therefore by Gi or it is generated on lower level and propagated to the higher level.

Therefore,

C5 = G5 + P5G4 + P5P4G3 + P5P4P3G2 + P5P4P3P2G1 + P5P4P3P2P1G0 + P5P4P3P2P1P0C− 1

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 22

The main advantage of Static RAM over Dynamic RAM is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 22

The main advantage of Static RAM over Dynamic RAM is that SRAM is faster than DRAM which means it is faster in operation. SRAM can be used to create a speed-sensitive cache. SRAM only has medium power consumption. SRAM has a shorter cycle time since it does not require pausing between accesses.

Hence, the correct option is (D).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 23

If we implement a queue with a linked list by keeping track of a front pointer and a rear pointer then which of these pointers will change during insertion into an empty queue?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 23

If the queue is implemented with a linked list, keeping track of a front pointer, Only rear pointers will change during an insertion into a non-empty queue.

A queue data structure can be used to implement the least recently used (LRU) page fault algorithm and Quicksort algorithm.

When the queue is empty both the pointers are initialized to NULL front ptp = rear ptp = NULL

Now, when a new node is inserted into the linked list both the front and a rear pointer will point to the new node.

After that, whenever a new node will be inserted the rear ptp will change and point to the new node.

As we know that ptp is a pointer.

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

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 24

ICMP error message should not be sent after receiving:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 24

ICMP error message should not be sent after receiving when:

  • After another ICMP error message.
  • After an IGMP message of any kind.
  • After any fragment other than the first fragment from a fragmented IP datagram.

ICMP error message should be sent after receiving when:

  • If DF ( do not fragment) bit is set and a router needs to fragment the datagram to send it further then the ICMP error message must be sent to the sender.
  • If TTL is exceeded then also it should report to the sender by sending the ICMP error message.

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

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 25

Consider 4 processes, P0,P1,P2 and P3 shown in table:


The completion order or the processes under the policies SRTF and LRTF are: (Conflict for same burst time resolved in favour of lowest process id) is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 25

Shortest remaining time, also known as shortest remaining time first (SRTF), is a scheduling method that is a pre-emptive version of the shortest job next schedule. In this scheduling algorithm, the process with the smallest amount of time remaining until completion is selected to execute. Since the currently executing process is the one with the shortest amount of time remaining by definition, and since that time should only reduce as execution progresses, processes will always run until they are complete or a new process is added that requires a smaller amount of time.
SRTF: Gantt chart:

Here, we know that
At time 0, P1 is the only process, P1 runs for 15 time units.
At time 15, P2 arrives, but P1 has the shortest remaining time. So, P1 continues for 5 more time units.
At time 20,P2 is the only process. So, it runs for 10 time units. at time 30, P3 is the shortest remaining time process. So, it runs for 10 time units.
At time 40,P2 runs as it is the only process. P2 runs for 5 time units. At time 45, P3 arrives, but P2 has the shortest remaining time. So, P2 continues for 10 more time units. P2 completes its execution at time 55 .
So, order is P0, P2, P1 and P3.
LRTF: Gantt chart:


So, order is P0, P1, P2 and P3.
Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 26

CREATE TABLE Student (stu_id NUMERIC NOT NULL, Name VARCHAR(20) , dept_name VARCHAR (20) UNIQUE (Name));

INSERT INTO Student VALUES(1001, Arun, CSE);

INSERT INTO Student VALUES(1002,Rita,ECE );

INSERT INTO Student VALUES(1005,Rita,EE );

What will be the result of the query?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 26

The not null specification prohibits the insertion of a null value for the attribute.

The unique specification says that no two tuples in the relation have the same values on all the listed attributes (Here name is the listed attribute). Rita is repeated and here it is given UNIQUE(Name).

So, INSERT INTO Student VALUES(1005,Rita,EE ); will generate error.

Hence, the correct option is (D).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 27

What is the number of terminal string of length ≤ 6 generated by the CFG shown below?

S → 0A1

A → BB1/B

B → A/0/1


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 27

To count the no of strings we have to check for every length string:

For 0 length string − No string possible

For 1 length string − No string possible

For 2 length string − No string possible

For 3 length string − S → 0A1 → 0 B1 → {001,011}

For 4 length string − No string possible

For 5 length string − S → 0 A1 → 0BB11 → {00011, 00111, 01011, 01111}

For 6 length string − No string possible

Hence, the correct answer is 6.

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 28

Differentiate the following expression:

 

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 28

Given,

Differentiating with respect to x, we get

Hence, the correct option is (C).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 29

The output of logic circuit given below represents ___________ gate.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 29

Output expression Q is equivalent to NAND gate.

NAND GATE:

Symbol:

As we know,
1. If A is always high, the output is the inverted value of the other input B,i.e. 
2. The output is low only when both the inputs are high.
3. It is a universal gate.
Hence, the correct option is (B).

GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 30

Consider the three problems:

  1. Equivalence
  2. Ambiguity
  3. Regularity

What is the correct option regarding these given problems?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 9 - Question 30

For regular grammar and CFL grammar:

1. Equivalence: It can be checked by developed product automate for XOR function. If XOR = 0 ⇒ equal else not. Two context-free grammars are defined as being structurally equivalent if they generate the same sentences and assign similar parse trees (differing only in the labeling of the nodes) to each. It is argued that this type of equivalence is more significant than weak equivalence, which requires only that the same sentences be generated.

2. Ambiguity: Ambiguity is decidable for regular languages & grammar since they are deterministic in nature. A context-free language is called ambiguous if there is no unambiguous grammar to define that language and it is also called inherently ambiguous.

3. Regularity: It is the trivial problem. For context-free language. It is not closed in any of those problems.

So, Regular grammar is closed in all three problems given but CFL is not closed in anyone.

Hence, the correct option is (C).

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