GATE Mock Test Computer Science Engineering (CSE) - 6 - Computer Science Engineering (CSE) MCQ

The number of ways in which r distinct objects can be selected from a group of n distinct objects, is:

n! = 1 × 2 × 3 × …× n

0! = 1

There are a total of 4 ingredients.

A salad can be made by using any combination of all or some of these.

The required number of possibilities is:

(Number of salads with 1 ingredient) + (Number of salads with 2 ingredients) + (Number of salads with 3 ingredients) + (Numer of salads with 4 ingredients)

= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄

= 4 + 6 + 4 + 1

= 15

Hence, the correct option is (C).

As we know,

Sullen: Bad-tempered and sulky.

Discrete: Careful and prudent in one's speech or actions, especially in order to keep something confidential or to avoid embarrassment.

Suspicious: Having or showing a cautious distrust of someone or something.

Elated: Make (someone) ecstatically happy.

So, the correct sentence is "The child was so spoiled by her indulgent parents that she became sullen when she did not receive all of their attention."

Hence, the correct option is (C).

Given,

Total cost = Rs. 60000

The break-up of the cost of construction of a house (in degrees) for labor = 90

The break-up of the cost of construction of a house (in degrees) for supervision = 54

The difference in the angle of labour and supervision = 90 − 54

= 36

=2160000/360

= Rs. 6000

Hence, the correct option is (D).

Given,

The given logarithmic expression is:

log⁡ cot ⁡9^o + log ⁡cot ⁡45^o + log cot ⁡81^o

Formula Used:

Basic concept of trigonometric ratio and identities

We know that,

log⁡ m + log ⁡n = log⁡(m × n)

tan⁡θ = cot⁡(90 − θ)

tan⁡ θ × cot⁡ θ = 1

cot⁡ 45^o = 1

log⁡1 = 0

By applying the identity of log.

∴ log⁡ cot⁡9^o + log ⁡cot⁡45^o + log⁡ cot⁡81^o

= log⁡(cot ⁡9^o × cot ⁡45^o × cot 81^o)

= log⁡(cot⁡ 9^o × cot⁡ 45^o × cot ⁡81^o)

= log⁡(cot ⁡9^o × cot ⁡45^o × cot ⁡(90 − 9)^o)

= log⁡( cot ⁡9^o × cot ⁡45^o × tan⁡ 9^o)

= log⁡1

= 0

Hence, the correct option is (A).

The correct answer is: 'She was going to the party.'

'I got up at 5 am every day' is incorrect.

The given sentence is a habitual action.

Habitual actions are always in the Present Tense.

'James goes to the market yesterday' is incorrect.

The given sentence is in the past tense which is evident with the use of the word 'yesterday'.

Structure:

Subject + Past tense + Object

Therefore, the verb 'went' would be used.

'She was going to the party' is the correct answer.

The given sentence is in the past continuous tense and the structure for a sentence in the past continuous tense is:

Subject + was/were +V¹+ ing + Object.

'We will seen him after two days' is the incorrect answer.

The given sentence is in the simple future tense and the structure for a sentence in the simple future tense is:

Subject + will/shall +V¹ form of the verb + Object.

Hence, the correct option is (C).

The three underlined elements make a coordinated series. To clearly express their relationship with each other, they need to abide by one consistent grammatical construction.
In choice 1, the verb 'must generate' breaks the parallelism.
In choice 2, the word 'for' breaks the parallelism.
In choice 3, the series changes construction, adopting a different type of parallel construction.
Hence, choice 4 will be the correct one here.

There are 52 complete weeks in a calendar year i.e. 52 x 7 = 364 days
Number of days in a leap year = 366
∴ Probability of 53 Sundays = 2/7

We know that (x + 1/x) is always greater than or equal to 2.
Now, let x be the number of boys and y be the number of girls, then
x/y = P and y/x = Q
So, P + Q = (x/y) + (y/x) ≥ 2

CP : MP = 2x : 3x

⇒ Profit = x

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

Let CP = 200, SP = 300

⇒ x = 8.33%

Discount = 2x = 16.66%

(A) is not feasible in a democracy. (C) is not followed because the problem is not concerned with criminals. (B) is the only course that authorities can resort to.

It is mentioned that each node has odd number of descendants, including node itself, so all nodes must have even number of descendants - 0, 2, 4 and so on, which means each node should have either 0 or 2 children. So, there will be no node with 1 child. Hence, 0 is answer.
Following are a few examples.

Such a binary tree is full binary tree (a binary tree where every node has 0 or 2 children).

Given,

String: pppqppqprpqp

Now,

The output is 3123.
Hence, the correct option is (D).

In the Shortest Seek Time First (SSTF) disk scheduling algorithm, the I/O request which requires the least disk arm movement, irrespective of the direction, from the current position is selected.

Now,

Given the disk, the head is currently at cylinder number 100.

Using SSTF it would next serve the request-id R with cylinder number 110 because it requires the least amount of arm movement. After R being served, T would be served and so on, as shown below:

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

Given relation,

(i) R(A,B) with productions {A → B}

(ii) R(A,B) with productions {B → A}

(iii) R(A,B) with productions {A → B,B → A}

As we know,

One of the more desirable normal forms that we can obtain is Boyce-Codd normal form (BCNF). It eliminates all redundancy that can be discovered based on functional dependencies.

• BCNF is the advance version of 3 NF. It is stricter than 3 NF.
• A table is in BCNF if every functional dependency X → Y, X is the super key of the table.
• For BCNF, the table should be in 3 NF, and for every FD, LHS is super key.

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

Given,

The grammar G is:

S → aXa ⎪ bXb ⎪ a ⎪ b

X → aX ⎪ bX ⎪ ε

Now we have,

L(G) = {a, b, aaa, aba, bab, bbb,......}

L(G) = All non-empty strings that start and end with the same symbol.

∴ L(G) is regular but not finite.

Hence, the correct option is (A).

Given linear system is:

2x − y + 3z = 2

x + y + 2z = 2

5x − y + az = b

Then augmented matrix form is written below;

For rank (A) < n = 3

'a' must be = 8

For rank [A ∣ B] < 3, b = 6

Therefore, a = 8 and b = 6

Hence, the correct option is (B).

Process Table 1:

Average waiting time =1
 
The value of Z is 2.
Assume that: P₄ ≥ P₃
Gantt chart:

Process Table 2:

Given average waiting time =1

5/4 ≠ 1
∴ P₄ > P₃ is not possible
∴ Z = 2
Hence, the correct answer is 2.

Find the largest denomination which is smaller than given amount i.e. 1000.

New value =1873−1000 = 873

Find the largest denomination which is smaller than given amount i.e. 500.

New value = 873 − 500 = 373

Find the largest denomination which is smaller than given amount i.e. 100.

New value = 373 − 100 = 273

Find the largest denomination which is smaller than given amount i.e. 100.

New value = 273 − 100 = 173

Find the largest denomination which is smaller than given amount i.e. 100.

New value = 173 − 100 = 73

Find the largest denomination which is smaller than given amount i.e. 50.

New value = 73 − 50 = 23

Find the largest denomination which is smaller than given amount i.e. 20.

New value = 23 − 20 = 3

Find the largest denomination which is smaller than given amount i.e. 2.

New value = 3 − 2 = 1

Find the largest denomination which is smaller than given amount i.e. 1.

New value = 1 − 1 = 0

So, coins required 1, 2, 20, 50, 100, 100, 100, 500, 1000.

Hence, the correct answer is 9.

We know that,

First, a page is accessed in upper-level memory, if it is present there then it is called a hit. But if the page is not found in upper-level memory, then it is called a miss, for this page is first copied from lower memory to upper-level memory. During miss, it involves first access time for upper-level memory then access time for lower-level memory.

Now,

T₁ is the memory access time for memory M₁.

T₂ is the memory access time for memory M₂.

H is the hit rate in M₁.

So, 1 − H is the hit rate in M₂.

In serial manner:

Average access time will be = H₁ T + (1 − H) × (T₁ + T₂) = T₁ + (1 − H)T₂

Because the first page is accessed in memory M₁ and it takes T₁ time. If page not found there, then it will be:

Accessed in M₂ and takes T₂ time with miss rate (1 − H).

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

Given,

S₁ : r₁(x), r₂(z), r₁(z), r₃(x), r₃(y), w₁(x), c₁, w₃(y), c₃, r₂(y), w₂(y)w₂(z), c,

S₂ : r₁(x), r₂(z), r₁(z), r₃(x), r₃(y), w₁(x), w₃(y), r₂(y), w₂(z), w₂(y), c₁, c₂, c₃

As we know,

In S₂ : w₃(y) is first and r₂(y) appears second.

So, c₂ should appear after c₃.

In recoverable [If T_j is ready a value written by T_j, the T_i must commit after T_j commits].

S₃ : r₁(x), r₂(z), r₃(x), r₁(z), r₂(y), r₃(y), w₁(x), c₁, w₂(z), w₃(y), w₂(y)c3, c₂

S₄ : r₁(x), r₂(z), r₃(x), r₁(z), r₂(y), r₃(y), w₁(x), w₂(z), w₃(y), w₂(y), c₁, c3, c₂

Schedules S₁, S₃ and S₄ are recoverable.

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

Given,

S₁: True

Inserting n elements in an empty binary heap takes O(n) time because the build heap method takes O(n).

S₂: True

BFS algorithm takes O(m + n) time to find the number of connected components in undirected graphs of n vertices and m edges.

Hence, the correct option is (C).

The spatial locality of reference refers that if an item is referenced, items whose addresses are closed by tending to be referenced soon. We are achieving this with the help of data moved in form of blocks. As it will not only contain the currently referenced word but also the near ones too. Spatial locality means instruction or data near to the current memory location that is being fetched, may be needed soon in the near future. This is slightly different from the temporal locality.

Hence, the correct option is (B).

Given,

Message size = 2400 bits

The header of the transport layer = 150 bits

Segment size at transport layer = (Header + Message size)

= 150 + 2400

= 2550 bits

Maximum, transmission unit at destination network = 900 bits

So, at a time, data supported by the destination network = 900 − 26

= 876 bits

Therefore, 2550 bits are divided into packets having a maximum of 876 bits. 

So,

Length of packet 1 = 874 bits

Length of packet 2 = 874 bits

Length of packet 3 = 802 bits

Total = 2550 bits.

Network layer packets are transmitted via two networks, each of which uses a 26-bits header.

So, the number of bits, including headers delivered to the destination network is:

Packet 1 (Data + Header) size = 26 + 26 + 874

= 926 bits

Packet 2 (Data + Header) size = 26 + 26 + 874

= 926 bits

Packet 3 (Data + Header) size = 26 + 26 + 802

= 854 bits

Total = 2706 bits.

Hence, the correct option is (A).

The decidability problem of Turing Machines can directly be determined using Rice's theorem.

L₁ is undecidable. According to Rice's theorem, emptiness problem of Turing machine is undecidable.

L₂ is decidable. This is because here we have to check whether the Turing machine reaches a particular step 'q' on a given input in finite steps or not. This is a decidable problem.

L₃ is undecidable. There is no algorithm to computationally determine whether a Turing machine accepts a recursive language or not. Some Turing machines may accept recursive languages, while other may not.

L₄ is also undecidable. According to Rice's theorem, membership problem of Turing machine is undecidable.

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

Segmentation is a process of dividing each process into variable size segments, where each segment performs related functions. Segment table stores details about segments. Base register of a segment contains the base address value i.e. smallest physical address from where each segment starts. Limit register contains the last range value of a segment.

Now,

For a segment address to be in the range of the segment, it must be:

Base address ← segment address ← (Base address + limit register value)

Since the offset value is exceeding the value of the limit of segment Id 2, the trap will be generated.

Hence, the correct option is (B).

Peterson's algorithm (or Peterson's solution) is a concurrent programming algorithm for mutual exclusion that allows two or more processes to share a single-use resource without conflict, using only shared memory for communication. It was formulated by Gary L. Peterson in 1981. While Peterson's original formulation worked with only two processes, the algorithm can be generalized for more than two.

Peterson's synchronization mechanism:

• It is a busy waiting solution.
• Mutual exclusion and progress are guaranteed.
• It doesn't use semaphore variable, but it uses to integer variables (turn and interest) to achieve synchronization.

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

As we know that,

In a horizontal micro-programmed control unit, one bit per control signal is required and in a vertical micro- programmed control unit, control signal is encoded for k bits as 2^k signals.

Given,

Case 1:

Horizontal micro- programmed control unit (HμPC):

 
Size of control memory in words =2²⁰
But each word contain 64 bits,
Control memory size in byte = 
=2²³
=8 MB
Maximum degree of parallelism = control signals =42
Case 2:
Vertical micro programmed control unit (VμPC):
The number of locations is the same, So the next part will remain the same as of 20 bits.

Control word is of 20 + 6 + 2 = 28 bits.
Control memory size in bytes = 
=2²⁰×3
=3 MB
Maximum degree of parallelism in vertical micro programmed control unit is always 1.
Hence, the correct options are (A) and (D).

Hermitian matrix: A square matrix A = (a_ij)_n×n is said to be Hermitian if 
a_ij =  for all i,j.

• The necessary and sufficient condition for a matrix A to be a Hermitian is that A = A^θ.
• The diagonal element of a Hermitian matrix is purely real.

Example:   is a hermitian matrix.

• The eigenvalue of a real symmetric (or Hermitian) matrix is always real and the eigenvalues of a real skew-symmetric (or skew Hermitian) matrix are either zero or purely imaginary.

Properties of eigenvalues:

• For the lower and upper triangular matrix, the diagonal elements of the matrix are eigenvalues.
• If λ₁, λ₂,………λ_n are eigenvalues of the matrix A of order n, then
• λ₁ + λ₂ +……..λ_n =  trace of A.
• λ₁ × λ₂ × …….λ_n =  det of A.
• 0 is an eigenvalue of matrix A if and only if A is singular.
• If all the eigenvalues of A are non-zero then A is non-singular.
• The eigenvalue of A and A^T is the same.
• If λ is the eigenvalue of an orthogonal matrix then 1/λ is also another eigenvalue of the same matrix A.

Hence, the correct option is (C).

Given,

The following format of 32 bit floating-point number:

(S) Sign: 1 bit

(E) Exponent: 8 bits

(M) Mantissa: 23 bits

As we know that,

Mantissa: 1.1111011₂ = 1.9609375₁₀

Exponent: 10000011₂ = 131₁₀

Biased Exponent = 131 − 127 = 4

∵ Sign bit is 1.

It is a negative number.

= −1.9609375 × 2⁴

= −31.375

Hence, the correct option is (A).

Given,

n = 200

Using divide and conquer paradigm approach Min-Max algorithm:

As we know,

Minimum number of comparisons = 3n/2 − 2

= 300 − 2

= 298

Hence, the correct answer is 298.