GATE Mock Test Computer Science Engineering (CSE) - 8 - Computer Science Engineering (CSE) MCQ

Given,

Total Number of Employees = 26800

Teaching = 15%

Medical = 27%

Management = 17%

Now,

Total percentage of employees in teaching and medical,

= 15% + 27%

= 42%

Now,

Difference between total employees in teaching and medical and total employees in management,

= (42% − 17%)  of 26800

= 25% of 26800

 
= 25/100 × 26800

= 6700

Hence, the correct option is (C).

Agitate: Feeling or appearing troubled or nervous. e.g. "There's no point getting agitated".

Arouse: Excite or provoke (someone) to anger or strong emotions. e.g. "An ability to influence the audience and to arouse the masses".

Still: Means stable, static, stationary, constant, stagnant, etc. e.g. "The sheriff commanded him to stand still and drop the gun".

Glamour: an attractive or exciting quality that makes certain people or things seem appealing. e.g. "Gloomy forecasts about the economy".

Beginner: Means neophyte, learner, abecedarian, apprentice, fresher etc. e.g. "A person is beginner for learn a skill or take part in an activity".

Hence, the correct option is (C).

Area of square = ( side )² = (4)² = 16 cm²

The sum of the area of four quadrants of a circle removed from the corners of the square is equal to the area of the circle of diameter 2 cm.
Radius of circle 
= Diameter/2
= 2/2 = 1 cm
Area of circle = π (radius )² 
= 22/7 cm²
∴ Area of shaded region = Area of square − 2 (Area of circle) 
= 16 − 44/7
= 68/7

Hence, the correct option is (C).

Given,

Total number of Persons infected in Maharashtra =30,00,000

From the data given in the question,

= 450%
∴ The number of person infected from COVID-19 in Pune is 450% the number of persons infected from COVID-19 in Mumbai.
Hence, the correct option is (B).

The meaning of once in a blue moon is something that occurs rarely.

Example:

Peter only comes out for a drink once in a blue moon now that he has kids.

Hence, the correct option is (D).

The correct verb form is 'has broken'.

A son is part of a nuclear family, and a cousin is part of an extended family.

The subordinator 'because' in choice 3 establishes the logical causal relationship between the subordinate and main clause. Choices 1 and 2 do not make sense. Choice 4 has faulty construction.

We know that the three term of the AP are α - β, α, α + β and those of the GP are a, ar, ar2.
So, according to the quesition,

α - β = r + a = β

a + ar = 2α - β
a+ar = 2α - 2β + β
a(1 + r) = 2 (r + a) + r + a
a(1+ r) = 3 (r + a)
a + r = a(1 + r)/3
Now, the required ratio is   = a/(a + r) = (a × 3)/(a (1 + r)) = 3/(1 + r)
This is independent of the first term of the GP.

When used for 25 hours per month, provider A will cost $20 + 7.5 x $1 (for the hourly charge above the free hours).
This equals $27.50.
Provider B will cost $20 + 5 x $1.50 (for the hourly charge above the free hours).
This equals $20 + $7.50 = $27.50 as well.
So, choice (3) is the correct answer.

Here we are given a forest with 54 vertices and 17 components.

A component is itself a tree and since there are 17 components means that every component has a root, therefore we have 17 roots.

Each new vertex of the forest contributes to a single edge to a forest.

So, for remaining we can have m−n edges.

= 54 − 17

= 37 edges

Hence, the correct option is (B).

The correct match is: 

Hence, the correct option is (B).

Statement 1:

If L = {w = a^rb^sc^t ∣ r,s,t ≥ 0}

Then L(r) = a^∗b^∗c^∗

Statement 2:

If L = {w = a^rb^sc^t ∣ r, s, t > 0}

Then L(r) = aa^∗bb^∗cc

DFA for Language L:

Statement 3:

If L = {w = a^rb^sc^t ∣ s ≥ 0 and r, t > 0}

Then L(r) = aa^∗b^∗cc^∗

Statement 4:

If L = {w = a^rb^sc^t ∣ s,t ≥ 0 and r > 0}

Then L(r) = aa^∗b^∗c^∗

Hence, the correct answer is 2.

As we know,

Stateless protocols are the type of network protocols in which the client sends the request to the server and the server responds back according to the current state.

It does not require the server to retain session information or status about each communicating partner for multiple requests.

Application Layer protocol: HTTP and DNS

Transport Layer protocol: TCP and UDP

Stateless protocol: HTTP, UDP, DNS

Therefore, only TCP is not a stateless transport layer protocol.

UDP is an unreliable connectionless-transport layer protocol used for its simplicity and efficiency in applications where error can be provided by the application layer.

UDP provided process-to-process communication.

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

Now, equivalence 0 ⟶ [q₀q₁q₂q₃] [q₄]

Why q₁q₂q₃ is in same set because 

on seeing q₁ on a we get q₄

q₂,q₃ on a also q₄

on seeing q₁ on b we get q₂

q₂ on b q₁

q₃ on b q₂

Which arein the same set in equivalence 0 as fir a it is q₄ which is in one set For b ot is q₁,q₂ which is also in sameset so only 3 states in q₀ on as a it is going to q₃which is defferent set from q₄ so seperate q₀ on 2^nd equivalence also got same result.

The minimized DFA after combining the q₁,q₂ and q₃ are given below.

Hence, the correct option is (A).

To find the Eigen values it satisfy the condition,

 |A − λI| = 0

= (2 − λ)((2 − λ)² − 1) − 1((2 − λ) − 0) + 0

= (2 − λ)(4 + λ² − 4λ − 1) − 2 + λ

= (2 − λ)(λ² − 4λ + 3) − 2 + λ

= (-λ³ + 6λ² − 11λ + 6) − 2 + λ

= −λ³ + 6λ² − 10λ + 4

= −λ³ + 2λ² + 4λ² − 8λ − 2λ + 4

= −λ²(λ − 2) + 4λ(λ − 2) − 2(λ − 2)

= (λ − 2)(−λ² + 4λ − 2)… (i)

By using Shri Dronacharya's formula for quadratic equation, we get,

So,

And from equation (i) we get,

⇒ x = 2

By solving the above equation, we get,

Hence, the correct option is (A).

Given,

A = 10 and C = 7

{t . B∣ t ∈ r ∧ t[A] = 10 ∧ t[C] = 7}

Now,

Therefore, sum = 10 + 7 + 13 + 2
= 32
Hence, the correct answer is 32.

Given,

Elements in unsorted order:

60, 80, 15, 95, 7, 12, 35, 90, 55

We have to choose first element as pivot.

Here 60 is given as first element.

After first pass, the pivot element goes to it's exact location.

Here 60 goes to 6^th place.

All the elements less than 60 go to left of 60 and all the elements greater than 60 go to right of 60.

After 1^st  pass.

= 5! × 3!
= 720 possible arrangements.
Hence, the correct answer is 720.

Enqueue:

While 1^st  stack is not empty, push everything from 1^st  to 2^nd  stack.

Push an element to 1^st  stack.

Push everything back to 1^st  stack.

Here time complexity will be O(n).

Dequeue:

While 1^st  stack is not empty then Pop an item from stack 1 and return it. Here time complexity will be O(1).

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

Statement 1:

"An execution plan is a blueprint for evaluating a query and is usually represented as a tree of relational operators".

This statement is true. Since the execution plan of a query is the most efficient way of executing in that machine according to the optimizer.

Statement 2:

"A database management system will always be efficient as compared to a file system".

This statement is false. Consider the example for the operation/manipulation of text data that is not supported by DBMS then performing those operations/manipulations will be easier on the file system rather than with DBMS.

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

Given,

Read (x); x : = x − 50; write (x); read (y);  y: = y + 50; write (y)

As we know,

Consistency in database systems refers to the requirement that any given database transaction must only change affected data in allowed ways, that is the sum of x and y must not change. Any data written to the database must be valid according to all defined rules, including constraints, cascades, triggers, and any combination thereof.

So, the constraint that the sum of the accounts x and y should remain constant is that of consistency.

Hence, the correct option is (B).

Horizontal micro-instructions generally follow a sequential approach to specify the next microinstruction in a microprogram, similar to conventional machine language format. Each bit is identified specifically with a single control point, which indicates that the corresponding micro-operation is to be executed. Horizontal micro-instructions have the following attributes:

1. Long formats
2. Ability to express a high degree of parallelism
3. Little encoding of the control information

Hence, the correct option is (D).

Parity bit checker and Parity bit generator are very useful combinational circuits used in communication systems. A combinational circuit consists of logic gates whose outputs at any instant of time are determined directly from the present combination of inputs without regard to previous input. Examples of combinational circuits: Adder, Subtractor, Converter, and Encoder/Decoder.

It is a logic circuit that checks for possible errors in the transmission. This circuit can be an even parity checker or odd parity checker depending on the type of parity generated at the transmission end. When this circuit is used as an even parity checker, the number of input bits must always be even.

There are different types of parity generator/checker ICs are available with different input configurations such as 5-bit, 4 - bit, 9 - bit, 12 - bit, etc. One of the most commonly used and standard types of parity generator/checker IC is 74180.

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

As we know,

An interpreted language is a programming language that is generally interpreted, without compiling a program into machine instructions. It is one where the instructions are not directly executed by the target machine, but instead, read and executed by some other program.

Given,

S1:True

Interpreters process the program according to the logical flow of control through the program.

S2: True

An interpreter translates and executes the error-free first instruction before it goes to the second.

Interpreter Translates and executes line by line.

S3: False

Interpreter processing time is less compared with the compiler.

The compiler processing time is lesser than Interpreter because the compiler scans and translates the whole program at a time.

S4: True

LISP and Prolog both are Interpreted languages.

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

Consider the string aa,

S → XY

S → aY

S → aa

The string ‘aa' can be generated from the given grammar. This eliminates option (A) & (C).

Also,

S → XY

S → aY (X → a)

S → aYb (Y → Yb)

S → aab (Y → a)

Eliminate option (D) as string ends in b.

Hence, the correct option is (B).

Given,

Seek time = t = 4 ms/cylinder.

Head = 15

Now,

Since it is a first-come-first-serve scheduling request will be served in the given sequence:

11, 23, 21, 3, 41, x, 25 and 39.

Diagram:

As we know,

Total seek time = Total head movement x t

Now,

560 = total head movement x 4

Total head movement = 140

⇒ (15 − 11) + (23 − 11) + (23 − 21) + (21 − 3) + (41 − 3) + |41 − x| + |25 − x| + (39 − 25) = 140

⇒ 4 + 12 + 2 + 18 + 38 + |41 − x| + |25 − x| + 14 = 140

⇒ 88 + |41 − x| + |25 − x| = 140

⇒ |41 − x| + |25 − x| = 52

Case 1:

⇒ 66 − 2x = 52

∴ x = 7

Case 2:

⇒ 2x − 66 = 52

∴ x = 59

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

Option (A): TRUE

Depth search explores edge (u, v), it is in the direction from u to v, then vis undiscovered until that time, for otherwise, the search would have explored this edge already in the direction from v to u. Thus, (u, v) becomes a tree edge. If the search explores (u, v) first in the direction from v to u, then (u, v) is a back edge.

Option (B): TRUE

An undirected graph may entail some ambiguity in the classification of edges since (u, v) and (v, u) are really the same edge. Hence, forward and cross edges never occur in a depth-first search of an undirected graph.

Option (C): FALSE

Pre-order traversal in a binary tree is similar to a depth-first traversal.

Option (D): TRUE

Suppose that a depth-first search produces a back edge (u, v) Then vertex v is an ancestor of vertex u in the depth-first forest. Thus, G contains a path from v to u, and the back edge (u, v) completes a cycle.

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

Given,

Sum of eigen values = (A) = 2 + y

Product of eigen values = |A| = 2y − 3x

According to question,

⇒ 4 + 8 = 2 + y

⇒ 2 + y = 12

y = 10…(i)

⇒ 4 × 8 = 2y − 3x

⇒ 2y − 3x = 32…(ii)

Putting y = 10 in equation (ii),

⇒ 2y − 3x = 32

⇒ 20 − 3x = 32

⇒ 3x = −12

x = −4

So, the value of x and y are −4 and 10.

Hence, the correct option is (D).

Since there is a NOR-NOR realization that can be replaced with an OR-AND realization. So, a given circuit can also be drawn as:

From the above circuit diagram, we can write as

f = (A+B)(B+C).C’

f = ABC’ + BC'

f = ABC'

So, the required minimum number of NAND gates is 4.

Hence, the correct option is (B).

Auto variables are always local and are stored on the stack. the register modifier tells the compiler to do its best to keep the variable in a register if at all possible. Otherwise, it is stored on the stack. extern variables are stored in the data segment.
The variables of type auto, static and extern are all stored in:

Hence, the correct option is (B).