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

GATE Mock Test Computer Science Engineering (CSE) - 7 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) - 7 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The GATE Mock Test Computer Science Engineering (CSE) - 7 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) - 7 below.
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GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 1

How many 5 letter codes can be formed using the first 6 letters of the English alphabet if no letter can be repeated?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 1

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 1st  letter for the code = 6

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

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

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

Number of ways to choose 5th  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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 2

Direction: Choose the correct synonym (word with similar meaning) of the head word form the four options given and mark the right option.

Perpetual

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 2

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).

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GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 3

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 3

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 4

Direction: In the following question, there is a statement first, and two arguments, I and II, are given below it. You have to consider which of the argument/arguments are influential in the statement by considering the given arguments. Choose the correct answer from the given alternatives.

Statement: Should education be made compulsory for all children up to class XII?

Argument I: No, this will affect the industries employing school dropouts.

Argument II: Yes, school education forms the basis for higher education and a better future.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 4

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 5

Direction: In the following question, out of the four alternatives, choose the alternative which best expresses the meaning of Idiom/Phrase.

I don't know him from Adam

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 5

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).

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

Directions: Choose the most appropriate word(s) from the options given below to complete the following sentence.

I contemplated ________ Singapore for my vacation, but decided against it.

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

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

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

The number lock of a suitcase has 4 wheels, each labelled with ten digits, i.e. from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the sequence to open the suitcase?

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

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

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 8

Directions: The table lists the size of building lots in the Orange Grove subdivision and the people who are planning to build on those lots. For each lot, installation of utilities costs $12,516. The city charges impact fees of $3,879 per lot. There are also development fees of 16.15 cents per square foot of land.

What approximate percentage is the area of the smallest lot listed, as compared to the area of the largest lot?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 8

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%

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 9

A reservoir will be filled in 12 hours if two pipes function simultaneously. The second pipe fills the reservoir 10 hours faster than the first. How many hours will the second pipe take to fill the reservoir?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 9

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

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 10

If prices are reduced by 20% and sales are increased by 15%, what is the net effect on gross receipts?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 10

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.

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 11

If A and B are two events such that P(A) = 0.6,P(B) = 0.5 and P(A∩B) = 0.4, then consider the following statements:

Which of the statements is/are correct

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 11

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 12

Complete the following program to find the maximum of given array:
Maximum(A, n)
{
Max = A[0];
for (i = 1 to n −1)
{
if (1)
then {2}
}
return Max;
}
Choose the correct expression for 1 and 2:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 12

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 13

Consider the following frequency of characters appear in a document:


What is the total number of bits needed to send?


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 13

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.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 14

Consider the following way of retrieving the data from the database:

  1. TRC
  2. DRC
  3. SQL
  4. RA

Which of the following has same power regarding queries?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 14

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 15

Assume ∑ = {a} and ε is the empty string:


What is the complement of the language accepted by the NFA shown above?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 15

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).

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 16

Let P(E) denote the probability of the event E. Given, P(A) = 1 and P(B) = 1/2. The value of P(B|A) is


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 16

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

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 17

Consider three processes (process id 0,1,2 respectively) with compute time bursts 2,4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is:


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 17

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, pwill 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.

*Answer can only contain numeric values
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 18

What is output of the given code?
#include <stdio.h>
union abc {
char a,b,c,d,e,f,g,h;
int i;
}abc;
main()
{
printf( "%d", sizeof( abc ));
}


Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 18

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.

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 19

What is postfix notation also known as?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 19

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: xy. 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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 20

Which of the following statement is wrong?

  1. 2 - phase locking protocol suffers from deadlock.
  2. Timestamp protocol suffers from more aborts.
  3. A block hole in a DFD is a data store with only inbound flows.
  4. Multivalued dependency among attributes is checked at 3NF level.
  5. An entity-relationship diagram is a tool to represent the event model.
Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 20

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 21

Which of the following statement is false?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 21

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 22

How many separate address and data lines are needed for a memory of 8 K × 16 ?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 22

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 23

An AB Flip-flop is designed using JK FF which of the following can be Boolean function for the input of JK Flip-Flop, consider the truth table of AB Flip-flop as given below:

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 23

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 24

Consider the following statements about a TM that accepts a language L:

S1: If the string in the language L, then TM halts in final state.

S2: If the string not in the language L, then TM may halt in non-final state or never halt.

If the string given as an input to the TM, then which of the above statements are correct?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 24

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 25

Which of the following CFG generates a language where the language is CFL but not regular?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 25

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= {ambn ∣ m > = n} is non regular but CFL.

Hence, the correct option is (C).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 26

Consider the following program:

int x = 0;

int y = 0;

par begin

begin

x = 1;

y = y + x;

end

begin

y = 4;

x = x + 5;

end

par end

What will be the final values of x and y after completion of the above concurrent program.
1.  x = 1,y = 5
2.  x = 6,y = 10
3.  x = 6,y = 5
4.  x = 1, y = 4

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 26

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).

*Multiple options can be correct
GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 27

Consider a memory unit of size 200K × 32, where the first component represents the number of words and that the second component represents the number of bits per word. What will be the number of address lines and input-output data lines?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 27

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 2k 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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 28

If Rank (A) = 2 and Rank (B) = 3, then Rank (AB) is __________.

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 28

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 29

If (2.3)base 4 + (1.2)base 4 = (y)base 4, what is the value of y?

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 29

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).

GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 30

What does the following fragment of C program print?

char c [ ] = "GATE2022";

char * p = c;

printf("%s", p + p [3] - p [1]);

Detailed Solution for GATE Mock Test Computer Science Engineering (CSE) - 7 - Question 30

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).

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Top Courses for Computer Science Engineering (CSE)

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Top Courses for Computer Science Engineering (CSE)