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

The fundamental principle of multiplication:

Let us suppose there are two tasks A and B such that task A can be done in m different ways following which the second task B can be done in n different ways. 

Then the number of ways to complete tasks A and B in succession respectively is given by: 

m × n ways

Here, we have to form a 5 letter code using the first 6 letters of the English alphabet such that no letter is letter repeated.

Number of ways to choose 1^st  letter for the code = 6

Number of ways to choose 2^nd  letter for the code = 5

Number of ways to choose 3^rd letter for the code = 4

Number of ways to choose 4^th  letter for the code = 3

Number of ways to choose 5^th  letter for the code = 2

The number of ways to form a 5 letter code using first 6 letters of english alphabet = 6 × 5 × 4 × 3 × 2

= 720

Hence, the correct option is (A).

The word 'Perpetual' means going on and on without any interruptions.

The synonyms of the word 'Perpetual' are "constant, changeless, stable".

From the synonym of the given word, we can say that the word 'Constant' has the same meaning.

The word 'Constant' means a situation that does not change.

Hence, the correct option is (A).

Given,

The length of a rectangle is decreased by 4 cm and the width is increased by 3 cm.

We know that:

Area of rectangle = (l × b) sq. unit

Area of square = (side)² sq. unit

Perimeter of rectangle = 2(l + b) units

Let the length and breadth of the rectangle be x and y.

Area of rectangle = (l × b) sq. unit

= xy cm²

New length and breadth = (x − 4) and (y + 3)

According to question,

⇒ (x − 4) = (y + 3)

x − y = 7…..(1)

Area of square = (x − 4) × (y + 3)

According to question,

Area of rectangle = Area of square

xy = (x − 4) × (y + 3)

⇒ 3x − 4y = 12…..(2)

On solving equations (1) and (2) we get,

x = 16 and y = 9

Perimeter of rectangle = 2(l + b) units

= 2(16 + 9)

= 2 × 25

= 50 cm

Perimeter of the original rectangle is 50 cm.

Hence, the correct option is (C).

For a better and bright future education is important. Making education compulsory for all children up to class XII will not affect the industries employing school dropouts. The only disadvantage that these industries face is they will not get cheap labour. Thus, Argument I is not effective.

Education is important for a better future. Making education compulsory for all children up to class XII will help the students to have a bright future ahead. Thus, Argument II is effective.

Hence, the correct option is (B).

I don't know him from Adam is a phrase used to refer to someone you have never met or known.

It is used to show the memory as old as Adam from the Bible to signify the unknown.

Example: While interrogation was going on, the culprit claimed that he did not know the victim from Adam.

Hence, the correct option is (C).

Contemplate is a verb which is generally followed by a gerund. Hence, the correct usage of contemplate is verb + ing form.

There are ¹⁰C₄ x 4! = 5040 sequences of 4 distinct digits, out of which there is only one sequence in which the lock opens.
Required probability = 1/5040

The smallest lot is 6,699 square feet and the largest lot is 9,004 square feet.
Required percentage = (6699/9004) × 100 = 0.744 × 100 ≈ 75%

Let the first pipe fill the reservoir in x hours.
Then, the second pipe can fill the reservoir in x – 10 hours.
Now, according to the question,

⇒ 12(x – 10 + x) = x² – 10x
⇒ 24x – 120 = x² – 10x
 ⇒ x² – 34x + 120 = 0
 ⇒ x² – 30x - 4x + 120 = 0
 ⇒ (x – 30)(x – 4) = 0
 ⇒ x = 30, 4
But, x ≠ 4 (∵ x – 10 should be positive)
∴ x = 30 hours
x – 10 = 20 hours

Let original price = p, and original sales = s.
Therefore, original gross receipts = ps. Let new price = 0.80p, and new sales = 1.15s.
Therefore, new gross receipts = 0.92ps. Gross receipts are only 92% of what they were.

We know that,

Probability of complementary event:

Let A be an event such that  is the complementary event then 
.

Probability of union of two events:

Let A and B be two events then the following relation holds:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Conditional Probability:

For two events the conditional probability is given by:

Now, it is given that P(A) = 0.6,P(B) = 0.5 and P(A ∩ B) = 0.4.

We can calculate the probability of complementary events as follows:

Therefore in the given case:

Using this we write:

= 0.4 + 0.5 − 0.1

= 0.8

Therefore, the first statement is wrong.

Using the formula for conditional probability we write:

Using the given data we will first determine the value of P(A∪B).

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= 0.6 + 0.5 − 0.4

= 0.7

Now, using the complementary event formula we get:

= 1 − 0.7

= 0.3

Substituting in the formula for the conditional formula:

 
= 0.3/0.4

= 0.75

Therefore, the second statement is also wrong.

Hence, the correct option is (D).

The algorithm to find the maximum of a given array is as follows:
Maximum(A, n)
{
Max = A[0];
for (i = 1 to n −1)
{
if Max< A[i]
then Max = A[i];
}
return Max;
}
The correct code is:
#include <bits/stdc++.h>
using namespace std;
int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low + 1; i<= high; i++)
{
if (arr[i] >max)
max = arr[i];
else
break;
}
return max;
}
int main()
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = sizeof(arr)/sizeof(arr[0]);
cout<< "The maximum element is "<< findMaximum(arr, 0, n-1);
return 0;
}
Output:
70
Hence, the correct option is (B).

After constructing the Huffman tree:

The total number of bits needs to send:

= 1 × 45 + 3 × 12 + 3 × 13 + 4 × 5 + 4 × 9 + 3 × 16

= 224
 
= 224/100

= 2.24

Hence, the correct answer is 2.24.

Relational Algebra and Relational Calculus both have same power. But SQL have more power compared to these. Tuple Relational Calculus is a non-procedural query language, unlike relational algebra. Tuple Calculus provides only the description of the query but it does not provide the methods to solve it. A Safe Expression is one that is guaranteed to yield a finite number of tuples as its results. Otherwise, it is called unsafe.

For example: {t| ~(t belongs R)} this is unsafe query.

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

The given alphabet contains only one symbol {a} and the given NFA accepts all strings with any number of occurrences of ‘a’. In other words, the NFA accepts {a⁺}. Therefore complement of the language accepted by automata is an empty string.

The Σ = {a} and the given NFA accepts the strings {a, aa, aaa, aaaa,........} i.e. the language accepted by the NFA can be represented by the regular expression: {a⁺}

So, the complement of language is {a^∗ − a⁺} = {ε}

Hence, the correct option is (B).

Assuming A and B are exhaustive events i.e. P(A U B) = 1
We have P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 1 + 0.5 - 1 = 0.5
Now, P(B|A) = P(A ∩ B)/P(A)
= 0.5/1
= 0.5

Turn around time of a process is the total time between submission of the process and its completion. LRTF (Longest Remaining Time First), means the process which has the remaining time largest, will run first and in case of the same remaining time, the lowest process will be given priority to run.

Let the processes be p0, p1 and p2. These processes will be executed in the following order.

Gantt chart is as follows:

First 4 sec, p2 will run, then remaining time p2 = 4, p1 = 4, p0 = 2.

Now, p1 will get chance to run for 1 sec, then remaining time p2 = 4, p1 = 3, p0 = 2.

Now, p2 will get chance to run for 1 sec, then remaining time p2 = 3,p1 = 3,p0 = 2.

By doing this way, you will get the above Gantt chart.

Scheduling table:

AT = Arrival Time, BT = Burst Time, CT = Completion Time, TAT = Turn Around Time As we know, turn around time is the total time between submission of the process and its completion. i.e turn around time = completion time-arrival time. i.e. TAT = CT − AT.
Turn around time of p0 = 12 (12 − 0)
Turn around time of p1 = 13 (13 − 0)
Turn around time of p2 = 14 (14 − 0)
Average turn around time is,

= 13 

=13
Hence, the correct answer is 13.

Union is the special data type used in C programming it allows the storage of different data types in the same memory location. Therefore the output of the program or size of union definition is 4. The union can be defined as a user-defined data type which is a collection of different variables of different data types in the same memory location. The union can also be defined as many members, but only one member can contain a value at a particular point in time.

Hence, the correct answer is 4.

Postfix notation is also called as 'suffix notation' and 'reverse polish'. Postfix notation is a useful form of intermediate code if the given language is expressions. Postfix notation is a linear representation of a syntax tree. In the postfix notation, any expression can be written unambiguously without parentheses. The ordinary (infix) way of writing the sum of x and y is with an operator in the middle: x^∗y. But in the postfix notation, we place the operator at the right end as xy *. In postfix notation, the operator follows the operand.

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

2 - phase locking protocol suffers from the deadlock is a true statement.

Timestamp protocol suffers from more aborts is a true statement (suffer from cascading rollback).

A block hole in a DFD is a data store with only inbound flows is a true statement.

Multivalued dependency among attributes is checked at 3NF level. It is a false statement (checked at 4NF level).

An entity-relationship diagram is a tool to represent the event model. It is a false statement (tool to represent data model).

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

Normalization is used to minimize redundancy from a set of relations. It is used to organize data effectively in the database. Normal forms are used to reduce redundancy from the database.

Third Normal form: A relation is said to be in third normal form if it is in 2NF & there must not exist any transition dependency.

Also, it must satisfy these properties:

1. For the function A → B, A should be a super key.

2. B should be a part of a key attribute or prime attribute i.e. B should be a part of a candidate key.

BCNF: A relation is said to be in Boyce-Codd normal form (BCNF) if it is in 3NF & must satisfy this property:

1. For the function A → B, A should be a super key.

Assume a relation R(ABC) with the following functional dependencies:

A(ABC)

{AB → C,C → A}

Candidate keys are AB and BC.

AB → C is in BCNF but C → A not in BCNF.

So, the statement which is false is "Relation with every attribute is prime always in BCNF".

Hence, the correct option is (C).

Required to address 8K of memory. We know that, 10 address bits allow the addressing of 1024 bytes (1K) of memory.

So, “1024” starts with “10”. Therefore, we need an additional 3 address bits to increase this by a factor of 8 (8 = 2 cubed), giving a total of 13 address bits.

Memory size = 8 K × 16 = 2¹³ × 16

So, address line  = 13, data line = 16

Hence, the correct option is (B).

AB to JK

A simplified version of the versatile J − K flip-flop. The outputs feedback to the enabling NAND gates. This is what gives the toggling action when J = K = 1.
While this implementation of the J − K flip-flop with four NAND gates works in principle, there are problems that arise with the timing. The timing pulse must be very short because a change in Q before the clock pulse goes off can drive the circuit into an oscillation called "racing". Modern ICs are so fast that this simple version of the J − K flip-flop is not practical.

Truth Table for input A:

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

Both statements S1 and S2 are true for TM. TM always halts in the final state if the given input string is valid, otherwise, it may halt in the non-final state or never halts.

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

In option (A) we have,

S → AB

A → aA | bA | ε

B → aBb ∣ ε

Here, L = (a + b) × a′bⁿ = (a + b) is regular.

In option (B) we have,

S → AaB

A → aA | bA | ε

B → aB | bB | ε

Here, L= set of all strings which contain a as substring.

In option (C) we have,

S → AB

A → aA ∣ ε

B → aBb ∣ ε

Here, L= {a^mbⁿ ∣ m > = n} is non regular but CFL.

Hence, the correct option is (C).

Given,

x = 0 

y = 0

Execution Sequences:

1. A → B → C → D

Thus, output is x = 6,y = 4

2. A → C → B → D

Thus, output is x = 6,y = 5

3. C → D → A → B

Thus, output is x = 1, y = 5

4. A → C → D → B

Thus, output is x = 6,y = 10

As we can see that outputs 1,2 and 3 are possible but not 4th.

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

Address line: An address line is basically referred to as the physical connection between a CPU/Chipset and memory. They specify which addresses to access in memory. When there are k address lines, then 2^k memory word can be accessed.

Data line: Data lines provide the information to be stored in memory. It represents the number of bits in the word.

Now,

Here, it is given that memory unit size  = 200K × 32

Where 96K represents the number of words.

16 represents the number of bits per word.

As, 128 < 200 ≤ 256 so, we have to take 128 i.e. 2⁸.

Number of words = 200K

= 2⁸ × 2¹⁰

= 2¹⁸

It means the address lines to access those words are 18.

Number of input-output data lines = number of bits in a word = 32

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

Given,

ρ(A) = 2

ρ(B) = 3

Rank: The rank of a matrix is a number equal to the order of the highest order non-vanishing minor, that can be formed from the matrix.

For matrix A, it is denoted by ρ(A).

The rank of a matrix is said to be r if, there is at least one non-zero minor of order r.

Every minor of matrix A having an order higher than r is zero.

Property of Rank of matrix:

ρ(AB) ≤ min[ρ(A), ρ(B)]
The rank of a matrix does not change by elementary transformation, we can calculate the rank by changing the matrix into Echelon form. In the Echelon form, the rank of a matrix is the number of non-zero rows of the matrix.
The rank of a matrix is zero if the matrix is null.
The rank of a skew-symmetric matrix cannot be one.
Using properties,

ρ(AB) ≤ min[ρ(A), ρ(B)]

ρ(AB) ≤ min(2,3)

⇒ ρ(AB) ≤ 2

∴ We don't know the dimension of A and B, we cannot predict the exact rank of AB but its maximum rank will be 2.

Hence, the correct option is (D).

You can do it in two ways:

By converting base 4 into decimal adding them and than converting them back to base 4.

= (2.3)₄ + (1.2)₄

Here you can minimize the values as:

You can directly convert it to base 4 now, (no need to divide).

= (10.1)₄

Alternate method:

As every number has base 4, adding them both and if we get an overflow we take the values as x mod 4 and extra value as carry.

 (2.3)

+(1.2)

 10.1

(For decimal values 3 + 2 = 5 and 5 mod 4 = 1 carry = 1)

(For units place 2 + 1 + 1 = 4 and 4 mod 4 = 0 carry = 1)

Hence, the correct option is (B).

Let ASCII value of A be x and therefore the ASCII value of E will be x+4.
char *p=c; //p is a pointer of type char that points to the starting address of character array.
printf("%s", p+p[3]-p[1]); //This will print the string from where updated pointer p points to the address (which is 104).
p = p+p[3]-p[1]
p = 100+'E'-'A'
p =100+(x+4)-(x)
p = 104 = starting address
printf("%s", p);
Therefore, output is 2022.
Hence, the correct option is (C).