GATE Computer Science Engineering(CSE) 2027 Mock Test Engineering (CSE)

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).

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).

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).

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) = ⁿC_r

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).

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).

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.

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.

CP : MP = 2 : 3

Let CP = 200, MP = 300

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

⇒ x = 8.33%

Discount = 2x = 16.66%

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

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.

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).

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.

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).

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.

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).

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).

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).

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).

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).

Given,

T₁ → Π_{id, name (σage >13∧ age <16}(STU))

As we know,

Output:

Number of rows: 5

T₂→Π_{id, 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.

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).

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.

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).

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.

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).

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).

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).

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).

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.

We know that,

L= Size of frame

BW→ Bandwidth = 10 Mbps = 10 × 10⁶ b/s

T_p → Propagation time = 25.6μs = 25.6 × 10⁻⁶ s

T_t → Transmission time

As we know,

For CSMA/CD,

T_t ≥ 2 × T_p
T_t = L/BW

Now,
L/BW ≥ 2 × T_p

 L ≥ 2 × T_p × BW

L ≥ 2 × 25.6 × 10⁻⁶ × 10 × 10⁶

 L ≥ 512 bits

L ≥ 64 × 8 bits  (8 Bit = 1 byte)

L_min = 64 bytes

Therefore, the minimum size of a frame in the network is 64 bytes.

Hence, the correct option is (C).