GATE Mock Test Computer Science Engineering (CSE) - 3 - Computer Science Engineering (CSE) MCQ

Given,

Diameter of semicircle =

Radius =  

Total no. of chords =6

We know that,

Since the chords are equal in length, they will subtend equal angles at the center. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.

Formula used:

Area of sector = 

Area of triangle = 
 
Now,

The angle subtended by each chord =180% no. of chord

 
=

=30∘

Area of sector  AOB=

Area of triangle  AOB=

 ∴ Area of shaded region =

=7 cm²

∴ Area of shaded region is 7 cm².

Hence, the correct option is (A).

The correct logical order is BDCA.

The passage tells about book and the opening sentence will be B introducing the topic with 'the book'. D follows B adding further about the content of the book. D is followed by C which tells the speaker's opinion about the book which he/she proposes to lend a copy to the listener as mentioned in A.

Hence, the correct option is (C).

Given,

Boys =100

Girls =80

Total boys pass =48%

Total girls pass =30%

We know that,

Failed percentage 

Now,

Total number of students =100+80

=180

According to the question, we have

The number of boys who passed the examination =100 ×

=48

The number of boys who failed the examination =100−48

=52

The number of girls who passed the examination =  

=24

The number of girls who failed the examination =80−24

=56

The total number of students who failed the examination =52+56

=108

Now,

The percentage of students who failed the examination = 

=60%

∴ The percentage of students who failed the examination is 60%.

Hence, the correct option is (B).

G → T (place value of 'T' = 20)

A → Z (place value of 'Z' = 26)

R → I (place value of 'I' = 9)

M → N (place value of 'N' = 14)

E → V (place value of 'V' = 22)

N → M (place value of 'M' = 13)

T → G (place value of 'G' = 7)

So, GARMENT is coded as 202691422137.

Similarly,

I → R (place value of 'R' = 18)

N → M (place value of 'M' = 13)

D → W (place value of 'W' = 23)

U → F (place value of 'F' = 6)

L → O (place value of 'O' = 15)

G → T (place value of 'T' = 20)

E → V (place value of 'V' = 22)

So, INDULGE is coded as 1813236152022.

Hence, the correct option is (D).

1. Product rule: The log of a product equals the sum of two logs.

⇒ log_a⁡(mn) = log_a⁡m+log_a⁡n

2. Quotient rule: The log of a quotient equals the difference of two logs.

⇒ log_a⁡  = log_a⁡m−log_a⁡n

3. Power rule: In the log of a power the exponent becomes a coefficient.

⇒ log_a⁡mⁿ = n log_a⁡m

4. Change of base rule:

⇒ log_m⁡n = 

If m = n;

⇒log_m⁡m = 

5. log_m⁡n = 

Now,

Here, we have to find the value of log₃⁡log_3⁡ 

⇒ log₃⁡log₃ = log₃⁡log₃ 

= log₃⁡log₃ 

= log₃⁡log₃ ⁡

= log₃⁡log₃  

From power rule;

=log₃⁡[×log₃⁡3][∵log_a⁡(m)n=n×log_a⁡(m)]

=log₃⁡  (∵logm⁡m=1)

=log₃⁡  = log₃⁡3−log₃⁡4

=1−log₃⁡4=1−log₃⁡22

=1−2log₃⁡(2)

Hence, the correct option is (C).

'Contentious' means 'causing or likely to cause an argument' and hence controversial.

The code for RENT is %+×$.

Look at the chart to see all of the land he owns:
The total amount of land he owns is, 8023 + 9004 + 8269 + 6774 = 32070 square feet.

Nimble means quick and light in motion; sluggish means slow or inactive.

The ratio of speeds before and after is 10 : 11. The ratio of time will be 11 : 10.
or,
He takes 1 unit less time now which is given as 2 minutes. The time taken will be 22 minutes at his usual speed.

Let a₁ be the number of games played until day 1, and so on, a_i be the no games played until i.

Consider a sequence like a₁,a₂,…a₃₀ where 1 ≤ a_i ≤ 45,∀_ai.

Add 14 to each elements of the sequence we get a new sequence a₁ + 14, a₂+14,…a₃₀+14 where 15 ≤ a_i + 14 ≤ 59, ∀a_i.

Now we have two sequences:

1. a₁,a₂,…,a₃₀

2. a₁ + 14, a₂+14,…,a₃₀+14

Having 60 elements in total with each elements taking a value ≤ 59.

So, according to pigeon hole principle there must be at least two elements taking a same value ≤ 59 i.e., a_i = a_j +14 for some i and j.

Therefor, there exists at least a period such as a_j to a_i, in which 14 matches are played.

Hence, the correct option is (D).

Given,

S₁: A graph with distinct edge weights can have more than one shortest path.

Two path (a − d − c) (a − b − c)

S₂: Adding a number on every edge of a graph may change the shortest path between two vertices.

As we know,

S₂: Consider the graph as shown below:

There are two shortest paths between a and c i.e. (a − b − c) and (a − c). Add 1 to every edge.

Now, only (a−c) is the shortest path thus changed.

Hence, the correct option is (C).

The effect of the partition algorithm when a pivot is P: After the Partition algorithm, P will go to its correct position and all the elements less than or equal to P will go to one side and all the elements greater than P will go to the other side.

We have to choose the first element as a pivot:

Here 60 is given as the first element.

After the first pass, the pivot element goes to its 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.

Option 3 : Regular, Not regular 

We know that, 

If A is any square matrix of order n, we can form the matrix [A−λ], where I is the n^th order unit matrix. The determinant of this matrix equated to zero i.e. ∣A−λI∣ = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the equation AX = λX.

Properties of Eigenvalues: 

The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A.

The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A.

Now,

⇒(1−λ)(2−λ) = 0

⇒λ=1,2

Given,

Eigenvectors are:  

For λ = 1, let the Eigenvector is 

⇒ a = 0

For λ = 2, let the Eigenvector is  
  

Now, a + b = 

Hence, the correct option is (B).

Given,

Physical address space (PAS)  = 32 bits

Page size = 4 KB

Page table size = 72 MB

As we know,

Number of frames = 

Now,

Number of frames  =  = 2²⁰

Translation =20 bits

A byte is a unit of digital information that most commonly consists of eight bits. So, we have

Entry of page table = 2 + 1 + 1 + 20 = 24 bits

=3 byte

Number of pages  =  
=

=24 MB

Virtual address = number of pages × page size = 24 × 2²⁰ × 4 KB

Virtual address =96×2³⁰  B

Length of virtual address space =log_2⁡ 96 × 2³⁰

=37 bits

Hence, the correct answer is 37.

The correct code is:

#include<stdio.h>
int main()
{
int a[9]={56,2,18,56,2,11,2,18,12};
int i, j, n=9, count = 0;
for(i=0; i< n; i++)
{
for(j=i+1; j< n; j++)
{
if(a[i] == a[j])
{
count++;
break;
}
}
}
printf("%d", count);
return 0;
}
In the above C code, variable count keeps track of duplicate elements. The second loop starts from i +1 because there is a need to search for duplicate elements in the next subsequent elements, from the current element. The body of the inner loop check for a duplicate element. If a duplicate element is found then increment the duplicate count. Also, terminate the inner loop if a duplicate element is found.
56 appears 2 times.
2 appears 3 times.
18 appears 2 times.
11 appears 1 time.
12 appears 1 time.
Hence, the correct answer is 4.

Given,

Option (A): int arr[] = {1, 2, 3, 4, 5, 6, 7};

This is a valid array declaration statement.

Option (B): int arr[10] = {0};

This is a valid array declaration statement. All the elements in arr will be initialized with 0.

Option (C): int arr[];

This is not a valid array declaration statement. Since we have to mention the size of the array.

Option (D): int arr[10] = {1,2,3};

This is a valid array declaration statement, the 0,1,2 will contains 1,2,3 respectively and all other indexes will be initialized with 0.

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

Given,

In an ER model, suppose relation R is a 1 : M relationship from entity set Y to entity set X.

For X,

Minimum participation = 1

Maximum participation = 1

So, every entity in X is associated with exactly one entity in Y.

For Y,

Minimum participation = 0

Maximum participation = M

So, the maximum participation of Y in the relationship is M.

As we know,

In question X and Y are sets, the cardinality of a set is instances of an entity.

Instances of X > Instances of Y.

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

For example,

|X| = y,|Y| = x

In X(KLMN), we have

N → K, M → N, K → L

{M}⁺ = {MNKL}

Therefore, M is a key.

Maximum tuple possible = |Y| = 2
From above example:
Since |X| = 300 and |Y| = 20
∴ Maximum tuple possible = 20
Hence, the correct option is (A).

We know that,

Cache memory is the smallest and fastest memory available in the computer system. Data of main memory that are used frequently are stored in the cache for easy accessibility.

The DRAM which forms the main memory is slow devices. This reduces the speed of execution. To speed up the process, high-speed memories such as SRAMs must be used.

In the memory system, a small section of SRAM is added along with the main memory referred to as cache memory. The program which is to be executed is loaded in the main memory but the part of the program and data that work at a particular time is accessed from cache memory.

Diagram:

Hence, the correct option is (B)

A frequency divider is a circuit that takes an input signal of a frequency fin and generates an output signal of a frequency given by:

f_out = f_in/n

Where, 'n' is an integer.

Option (A):

S = 1 and R = 0

∴ Y = 1 (flip-flop set)

S = 0 and R = 1

∴ Y = 0 (flip-flop reset)

So, it is a MOD-2 circuit and will act as a frequency divider circuit with:

f_out = f_in/2

When Q = 0 and  = 1,D = 1

After clock pulse application, Y=1

Now, when Q = 1 and   = 0, D = 0

After clock pulse, Y = 0 

So, it is also a MOD-2 circuit and will act as a frequency divider circuit with:

 f_out  = f_in/2

Option (C):

Here, J = 1 and K = 1

So, Y will toggle its states.

Therefore, it is also MOD-2 and will act as a frequency divider circuit with:

 f_out  = f_in/2

Option (D):

Here, T = 0

So, Q = Hold state and will remain the same.

Therefore, f_out  = 0

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

Option (A):

Mealy and Moore's machines are finite state machines that accept the same language. Both machines have the same power.

Option (B):

DPDA (deterministic pushdown automata) accepts deterministic context-free languages and NPDA stands for non-deterministic pushdown automata.

L = pⁿqⁿr^m ∪ p^mqⁿrⁿ n ∣ m,n > 0

Language L is not accepted by deterministic pushdown automaton (DPDA) but it is accepted by non-deterministic pushdown automaton. That's why NPDA is more powerful than DPDA.

Option (C):

• Mealy and Moore's machines are finite-state transducers (fst).
• The finite accepter plays a central role in the study of formal language but in other areas such as digital design, transducers are more important.
• A finite-state transducer has a finite set Q of internal states and operates in a discrete time frame with translation from one state to another made in the interval between two instances t_n and t_n+1.

Option (D):

• Finite state automata are a machine that accepts some input.
• It accepts regular languages. Finite state machines are not language translators.

As we know,

Every language accepted by DPDA is accepted by NPDA but vice versa is not true.

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

Given,

Statement (1): "For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine".

It is a true statement. A non-deterministic Turing machine can be simulated by a deterministic Turing machine with exponential time.

Statement (2): "Turing recognizable languages are closed under union and complementation".

Turing recognizable languages are "not" closed under complementation. For any Turing recognizable language, the turing machine "T' recognizing 'L', may not terminate on inputs x ∉ L.

So, this is false.

Statement (3): "Turing decidable languages are closed under intersection and complementation".

Turing decidable languages are closed under union and complementation. It is easy to construct a turning machine to decide L₁ ∪ L₂, L₁ if L₁, L₂ are Turing decidable.

So, this is a true statement.

Statement (4): "Turing recognizable languages are closed under union and intersection".

It is a true statement.

Hence, the correct option is (C).

Given,

Statement S1: "Using a larger block size in a fixed partition scheme may increase internal fragmentation".

As the size of the block in the fixed partition scheme increases, the chances of internal fragmentation also increase. Because with a larger block size, if a smaller process arrives; the remaining block will be wasted. This is because one block can be allocated to only one process at a time.

Statement S2: "Variable partition and segmentation suffers from external fragmentation".

Variable partition scheme suffers from external fragmentation; segmentation is one of the variable partition schemes.

Segmentation is similar to dynamic partitioning, only this time process can have several segments, while in dynamic partitioning a partition is accommodating the entire process.

A different memory management approach known as dynamic partitions (also called a variable partition) creates partitions dynamically to meet the requirements of each requesting process.

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

Now,

Circuit for Johnson counter:

If Q₀ is MSB and Q₃ is LSB:

Count sequence of 8 → 12 → 14 → 15 → 7 → 3 → 1 → 0 → 8 → (repeat).
If Q₃ is MSB and Q₀ is LSB:

Count sequence of 8 → 0 → 1 → 3 → 7 → 15 → 14 → 12 → 8 → (repeat).

As we know,
Since the least significant bits and the most significant bit is not signified.
Hence, the correct options are (B) and (D).

Given,

Using the formula,

On comparing both matrix, we get

⇒ 3b = 1

b = 1/3

⇒ 2a − 0.1b = 0

⇒ 2a = 0.1b

a = 0.1b/2

Putting value of b:

 a = 1/60 

Therefore, 
 a = 1/60, 
 b = 1/3 

⇒ a + b = 1/60 + 1/3

a + b = 7/20

Hence, the correct option is (B).

We know that,

A MUX is a combinational circuit that accepts data from multiple input lines and allows the data at the output to pass through a single line.

Select lines decide which input comes at the output.

For an N : 1 MUX

Number of inputs = N

Number of select lines = log₂⁡ N

Number of output = 1

A multiplexer with n-data select inputs can implement any function of n-variables and some function of (n + 1) variables.

Example:

Let us consider a 3 variable function.

 
 f = Σm(0,3,7)

The function can be implemented by connecting the given minterm to +V_cc and the rest of the minterm to the ground (0 V).

Similarly, any 3 variable functions can be implemented using 8 : 1 MUX.

Let us consider a 4 variable function.

Reducing (1):

Let us consider another function.

Any 4 variable functions, in which the output does not depend on the 4^th  variable, can be implemented using 8 : 1 MUX.

f₁ can be implemented using 8 : 1 MUX.
f₂ cannot be implemented using 8 : 1 MUX.

Hence, the correct option is (D).

Max heap: Root node value should be greater than child nodes.

Whenever we insert the new element in the heap, we will insert it at the last level of the heap.

After inserting the element if the max heap doesn't follow the property then we will apply heapify algorithm until we get the max heap.

Whenever insertion will be done in heap, it will always be inserted in the last level from left to right. So, we insert '1' and '7' as a child of node 5 now, we perform heapify algorithm until the heap property will satisfied and then we get the heap whose level order traversal is 10, 8, 7, 3, 2, 1, 5.

Initially heap has 10, 8, 5, 3, 2.

After insertion of 1:

No need to heapify as 5 is greater than 1.

After insertion of 7:

Heapify 5 as 7 is greater than 5.

No need to heapify any further as 10 is greater than 7:

Hence, the correct option is (A).