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

GATE Mock Test Computer Science Engineering (CSE) - 6 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) - 6 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 6 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) - 6 below.
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GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 1

How many different salads can be made from onion, cucumber, tomato, and carrot?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 1

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).

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

Direction: In the following question, a sentence is given with blank to be filled in with appropriate word. Some alternatives are suggested for each question. Choose the correct alternative from the given alternatives.

The child was so spoiled by her indulgent parents that she became ________ when she did not receive all of their attention.

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

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).

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GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 3

Direction: Study the following pie chart carefully & answer the question given below it.

The pie chart given below shows the break-up of the cost of construction of a house (in degrees). Assuming that the total cost of construction is Rs. 60000, answer the question given below.


The amount spent on labour exceeds the amount spent on supervision by:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 3

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 4

What will be the value of expression log⁡ cot⁡9 + log⁡ cot⁡45o + log cot⁡81o ?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 4

Given,

The given logarithmic expression is:

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

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⁡ 45o = 1

log⁡1 = 0

By applying the identity of log.

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

= log⁡(cot ⁡9o × cot ⁡45o × cot 81o)

= log⁡(cot⁡ 9o × cot⁡ 45o × cot ⁡81o)

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

= log⁡( cot ⁡9o × cot ⁡45o × tan⁡ 9o)

= log⁡1

= 0

Hence, the correct option is (A).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 5

From the given four options choose the correct sentence:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 5

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).

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

Directions: Select from the options given below that would best replace the underlined part of the sentence.

A corporation created by the federal government during the Great Depression, the Tennessee Valley Authority (TVA) is responsible for flood control, must generate electric power, and soil conservation.

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

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.

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 7

What is the chance that a leap year, selected at random, will contain 53 Sundays?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 7

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

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 8

In a group, the ratio of the number of boys to the number of girls is P. The ratio of the number of girls to the number of boys is Q, then P + Q is always

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 8

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

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

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) - 6 - Question 9

CP : MP = 2x : 3x

⇒ Profit = x

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

Let CP = 200, SP = 300

⇒ x = 8.33%

Discount = 2x = 16.66%

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 10

Directions: In the question given below, a statement is followed by three courses of action labelled (A), (B) and (C). A course of action is a step or administrative decision to be taken for improvement, follow-up or further action in regard to the problem, policy, etc. On the basis of the information given in the statement, you have to assume everything in the statement to be true and then decide which of the suggested courses of action logically follow(s) for pursuing.

Statement: Many political activists have decided to stage demonstrations and block traffic movement in the city during peak hours to protest against the steep rise in prices of essential commodities.
Courses of action:
(A) The Govt. should immediately ban all forms of agitations in the country.
(B) The police authority of the city should deploy additional forces all over the city to help traffic movement in the city.
(C) The state administration should carry out preventive arrests of the known criminals staying in the city.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 10

(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.

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 11

In a binary tree with n nodes, every node has an odd number of descendants. Every node is considered to be its own descendant. What is the number of nodes in the tree that has exactly one child?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 11

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 12

Consider the below given table in which regular expression is mapped to unique token:


What will be the output when the string "pppqppqprpqp" is scanned by the lexical analyzer if the analyzer outputs the token that matches the longest possible prefix if it is mandatory to use all the tokens at least once?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 12

Given,

String: pppqppqprpqp

Now,


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

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 13

Consider the following five disk access requests of the form (request-id, cylinder number) that are present in the disk scheduler queue at a given time.

(P, 155), (Q, 85), (R, 110), (S, 30), (T, 115)

Assume the head is positioned at cylinder 100. The scheduler follows Shortest Seek Time First scheduling to service the requests.

Which one of the following statements is/are true?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 13

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 14

Consider the relation given below and find the maximum normal form applicable to them:

  1. R(A, B) with productions {A→B}
  2. R(A, B) with productions {B→A}
  3. R(A, B) with productions  {A→B}, {B→A}
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 14

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 15

Consider the following grammar G:

S → aXa ⎪ bXb ⎪ a ⎪ b

X → aX ⎪ bX ⎪ ε

Identify the language generated by the above G?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 15

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 16

If the system,

2x − y + 3z = 2

x + y + 2z = 2

5x − y + az = b

Has infinitely many solutions, then the values of a and b, respectively, are:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 16

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 17

Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:

These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 illisecond, then the value of Z is ________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 17

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.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 18

Consider a person wants to make change for Rs. 1873 and he has infinite supply of denomination i.e. 1, 2, 5, 10, 20, 50, 100, 500, 1000 valued coins. What is the minimum number of coins needed to make the change?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 18

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.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 19

If T1 and T2 are average access time of upper level memory M1 and lower level memory M2 in a 2- level memory hierarchy and H is the hit rate in M1, then the overall average access time is given by _______, assuming that in case of a miss in M1, a block is first copied from M2 to M1 and then accessed from M1.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 19

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 20

Which of the given schedules is/are recoverable?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 20

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 Tj is ready a value written by Tj, the Ti must commit after Tj 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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 21

Which of the following statements are true?

S1: Inserting n elements into empty binary heap takes O(n) time.

S2: The most efficient algorithm take O(m+n) time for finding number of connected components in undirected graph of n vertex and m edges.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 21

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 22

By which of the following, spatial locality of reference is achieved?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 22

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 23

A message consisting of 2400 bits is to be passed over the Internet. The message is passed to the transport layer which appends a 150-bits header, followed by the network layer which uses a 120-bits header. Network layer packets are transmitted via two networks, each of which uses a 26-bits header. The destination network only accepts up to 900-bits long. The number of bits, including headers delivered to the destination network, is:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 23

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 24

Which of the following languages is/are undecidable?

Note: M indicates the encoding of the Turing machine M?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 24

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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 25

Consider the following segment table in segmentation scheme:

What happens if the logical address requested is segment Id 2 and offset 1000?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 25

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 26

Which of the following is/are true about Peterson's synchronization mechanism?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 26

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 27

Consider a control unit design in which 42 control signals are to be generated and the system is supporting 4−flag conditions. If the 64−bit control word is stored in control memory, then which of the following is true about control memory size and maximum degree of parallelism in both horizontal and vertical micro-programmed control unit. (For, vertical micro-programmed control unit, the same number of locations should be taken as horizontal).

Note:

HμPC denotes Horizontal micro-programmed control unit.

VμPC denotes Vertical micro-programmed control unit.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 27

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 2k 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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 28

The eigenvalues of a Hermitian matrix are _________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 28

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

  • The necessary and sufficient condition for a matrix A to be a Hermitian is that A = Aθ.
  • 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,………λn are eigenvalues of the matrix A of order n, then
  • λ1 + λ2 +……..λn =  trace of A.
  • λ1 × λ2 × …….λn =  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 AT 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).

GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 29

Consider the following format of 32 bit floating-point number:


(S) Sign: 1 bit
(E) Exponent: 8 bits
(M) Mantissa: 23 bits
The mantissa is normalized and has an implied "1" on the left of the point. Normalized form of mantissa is 1. M. The exponent is formatted using excess- 127 notation, with an implied base of 2. What will be the decimal value of the following 32 bit floating point number stored in the above-mentioned format?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 29

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: 1.1111011= 1.960937510

Exponent: 10000011= 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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 30

Total number of comparisons needed to find maximum and minimum of 200 element is __________.


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 6 - Question 30

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.

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Top Courses for Computer Science Engineering (CSE)

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Top Courses for Computer Science Engineering (CSE)