ISC Class 12 Computer SOLVED Question Paper 2023

ISC Computer Science Solved Question Paper 2023

ISC Computer Science Paper 2023 answer key was conducted by CISCE on March 20, 2023 and if you are looking for ISC Computer Question Paper 2023 Answer Key then you have clicked on the right link.

In this article you will get ISC Computer Solved Boards Paper 2023, also you will get solution of Section A, Section B and Section C.

Previous year solved paper are proved to be very helpful for all students as it helps students to score good marks in board exams because they are board question papers and helps a lot of students who are going to appear in board examinations.

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How to Score 100 in ISC Computer Science

ISC Computer Science is one of the easiest subject among other subjects for both Commerce and Science students and you can easily score 90+ marks. If you really want some tips to score good marks in ISC Computer Science, then below are the tips shared by a topper student who score 90+ marks in ISC Computer Science.

• Solve Boolean Algebra questions very carefully, you can easily grab 30 marks in these questions. Always remember to write Laws in solutions related to Boolean algebra.
  
  
• While writing the code of Inheritance program, remember to write extends keyword and the use of super keyword.
  
  
• Always provide documentation i.e. comments to your code.
  
  
• Use Buffer Reader Input instead of using Scanner Input this will add 1 more marks in your code.
  
  
• Don't attempt Recursion program question if you don't have good understanding of this concept.
  
  
• Pay close attention while solving Output questions otherwise your silly mistake can make you lose some marks.   

ISC Computer Science Paper 2023 Solved

Part - I (20 Marks)

  

(i) According to De Morgan’s law (a + b + c’)’ will be equal to: 

 (a) a’ + b’ + c’ 

 (b) a’ + b’ + c 

 (c) a’ . b’ . c’ 

 (d) a’ . b’ . c 

(i) Option (d) is Correct,

(ii) The dual of (X’ + 1) . (Y’ + 0) = Y’ is: 

(a) X . 0 + Y . 1 = Y 

(b) X’ . 1 + Y’ . 0 = Y’ 

 (c) X’ . 0 + Y’ . 1 = Y’ 

(d) (X’ + 0) + (Y’ + 1) = Y’

(ii) Option (c) is correct

(iii) The reduced expression of the Boolean function F(P, Q) = P’ + PQ is: 

(a) P’ + Q 

(b) P 

 (c) P’ 

(d) P + Q

(iii) Option (a) is correct

(iv) If (~p => ~q) then its contra positive will be: 

(a) p => q 

(b) q => p 

 (c) ~q => p 

(d) ~p => q

(iv) Option (b) is correct

(v) The keyword that allows multi-level inheritance in Java programming is: 

(a) implements 

(b) super 

(c) extends 

(d) this

(v) Option (c) is correct

(vi) Write the min term of F(A, B, C, D) when A = 1, B = 0, C = 0 and D = 1.

(vi) A.B’.C’.D (Find any min term we have to find the SOP expression for variables. Here A=1, B=0, C=0, D=1 so to find the min term we have to change 0 to 1. To achieve this use the complements of B and C) 

(vii) Verify if (A + A’)’ is a Tautology, Contradiction, or a Contingency

(vii) (A + A’)’ = A’.A’’ = A’.A = 0 Hence, it is a contradiction. The truth table of the above expression is:

(viii) State any one purpose of using the keyword this in Java programming.

(viii) this keyword returns the address of the current object during the execution

(ix) Mention any two properties of the data members of an Interface.

(ix)  Two properties are:

(a) Data members of an Interface are always declared with final keyword. 

(b) Data members of an Interface cannot be changed in the due course of the program.

(x) What is the importance of the reference part in a Linked List?

(x) Reference part in a Linked List holds the address of the next node in the Linked List.

 

Question 2 

(i) Convert the following infix notation to prefix notation. 

     (A – B)/C*(D + E)

(i) Prefix of the (A–B)/C*(D+E) : 

 (A–B)/C*(D+E) 

 =(–AB)/C*(+DE) 

 =(/–ABC)*(+DE) 

 =*/–ABC+DE

(ii) A matrix M[–6...10m 4...15] is stored in the memory with each element requiring 4 bytes of storage. If the base address is 1025, find the address of M[4][8] when the matrix is stored in Column Major Wise.

(ii) A = B + W( (I – I0) + (J – J0) × R) 

B = 1025 

W = 4 

I, J = 4, 8 

I0, J0 = –6, 4 

R = 10 + 6 + 1 = 17 

A = 1025 + 4 × ((4 + 6) + (8 – 4) × 17) 

= 1025 + 4 × (10 + 68) 

= 1025 + 312

= 1337

(iii) With reference to the code given below, answer the questions that follow along with dry run/working. 

boolean num(int x) 

{ int a=1; 

for(int c=x; c>0’c/=10) a*=10; 

return(x*x%a)=x; }

(a) What will the function num() return when the value of x = 25?

(b) What is the method num() performing? 

(iii) 

(a) X = 25 A = 1

C = 25 ® a = a * 10 = 10 ® c = 25 10 = 2 

C = 2 ® a = a * 10 = 100 ® c = 2 10 = 0 

X * X = 25 * 25 = 625 ® 625%100 = 25 ® true

(b) Checking whether the number x is Automorphic Number or not.

      Checking whether the number x is present in its square or not.

(iv) The following function task() is a part of some class. Assume ‘m’ and ‘n’ are positive integers, greater than 0. Answer the questions given below along with dry run/working. 

int task(int m, int n) 

{if(m==n) 

return m; 

else if(m>n) 

return task(m–n,n); 

else return task(m,n–m); } 

(a) What will the function task() return when the value of m = 30 and n = 45? 2 

(b) What function does task() perform, apart from recursion?

(iv) 

(a) task(30, 45) → task(30, 5) →  task(25, 5) → task(20, 5) → task(15, 5) → task(10, 5) → task(5, 5) → 5

(b) Returning the HCF of two numbers, m and n.

ISC Computer Paper 2023 Solved Section A

SECTION - A

Question 3 

(i) Given the Boolean function F(A, B, C, D) = S(2, 3, 6, 7, 8, 10, 12, 14, 15). 

 (a) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). 

 (b) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. 

 

(ii) Given the Boolean function F(A, B, C, D) = p(0, 1, 2, 4, 5, 8, 10, 11, 14, 15). 

 (a) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs). 

 (b) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.

(i) F(A, B, C, D) = ∑(2, 3, 6, 7, 8, 10, 12, 14, 15)

(a) 

Quad 1(2, 3, 6, 7) = A’.C 

Quad 2(6, 7, 14, 15) = B.C

Quad 3(8, 10, 12, 14) = A.D’ 

F(A, B, C, D) = A’C + BC + AD’

(b) 

(ii) F(A, B, C, D) = ∏(2, 3, 6, 7, 8, 10, 12, 14, 15) 

(a) 

Quad 1(0, 1, 4, 5) = A + C

Quad 2(10, 11, 14, 15) = A’ + C’ 

Quad 3(0, 2, 8, 10) = B + D 

F(A, B, C, D) = (A + C) . (A’ + C’) . (B + D)

(b) 

Question 4 

(i) A shopping mall allows customers to shop using cash or credit card or any nationalised bank. It awards’ bonus points to their customers on the basis of criteria given below:  

 The customer is an employee of the shopping mall and makes the payment using a credit card. 

 OR 

 The customer shops items which carry bonus points and makes the payment using a credit card with a shopping amount of less than 10,000.

OR 

 The customer is not an employee of the shopping mall and makes the payment not through a credit card but in cash for the shopping amount above 10,000.

(In all the above cases, I indicates yes and 0 indicated no.) 

Output: X[1 indicates bonus point awarded, 0 indicates bonus point not awarded for all cases] 

Draw the truth table for the inputs and outputs given above and write the POS expression for X(C, A, E, I). 

(ii) Differentiate between half adder and full adder. Write the Boolean expression and draw the logic circuit diagram for the SUM and CARRY of a full adder.

(iii) Verify the following expression by using the truth table: 

       (A ⊕ B)’ = (A ⊕ B) 

4. (i) 

X(C, A, E, I) = (C + A + E + I).(C + A + E + I’).(C + A + E’ + I).(C + A + E’ + I’).(C + A’ + E’ + I).

(C + A’ + E’ + I’).(C’ + A + E + I).(C’ + A’ + E + I).(C’ + A’ + E + I’)

= ∏(0, 1, 2, 3, 6, 7, 8, 12, 13)

(ii) 

Boolean expression of Full Adder:
SUM = A ⊕ B ⊕ C
CARRY = (A ⊕ B ).C + AB

(iii) 

Hence, proved (A ⊙B)’ = A ⊕ B

Question 5 

(i) What is an encoder? How is it different from a decoder? Draw the logic circuit for a 4 : 1 multiplexer and explain its working. 

(ii) Form the logic diagram given below, write the Boolean expression for (1) and (2). Also, derive the Boolean expression (F) and simplify it. 

(iii) Convert the following cardinal expression to its canonical form: F(P, Q, R) = p(0, 1, 3, 4)

5. (i) Encoder is sequential circuit that converts decimal/octal/hexadecimal digit to its equivalent binary form. Decoder is a circuit that does the opposite of Encoder i.e. Binary to the other form.

Logic circuit of 4 : 1 multiplexer

(ii) 

Gate (1) = (x. y)’ 

Gate (2) = (y + z)’ 

 F(x, y, z) = (x. y)’ + (y + z)’

(iii) F(P, Q, R) = (P + Q + R).(P + Q + R’).(P + Q’ + R’).(P’ + Q + R)

ISC Computer Paper 2023 Solved Section B

SECTION - B

Question 6 

Design a class NumDude to check if a given number is a Dudeney number or not. 

(A Dudeney number is a positive integer that is a perfect cube, such that the sum of its digits is equal to the cube root of the number.) Example: 5832 = (5 + 8 + 3 + 2)³  = (18)³  = 5832.

Specify the class NumDude giving details of the constructor(), void input(), int sumDigits(int) and void is Dude(). Define a main() function to create an object and call the functions accordingly to enable the task.

6.

import java.util.Scanner; 

class NumDude { 

int num; 

public NumDude() { 

num=0; } 

public void input() { 

Scanner sc=new Scanner(System.in); 

System.out.println(“Enter a no”); 

num=sc.nextInt(); } 

public int sumDigits(int x) {

if(x==0) 

return 0; 

else return x%10+sumDigits(x/10); } 

public void isDude() { 

int n=sumDigits(num); 

if(n*n*n==num) 

System.out.println(“Dudeney number”); 

else 

System.out.println(“Not a Dudeney number”); } 

public static void main(String args[]) { 

NumDude obj=new NumDude(); 

obj.input(); obj.isDude(); 

} 

}

Question 7 

A class Trans is defined to find the transpose of a square matrix. A transpose of a matrix is obtained by interchanging the elements of the rows and columns. 

Example: If size of the matrix = 3, then

Specify the class Trans giving details of the constructor(), void fillarray(), void transpose() and void display(). Define a main() function to create an object and call the functions accordingly to enable the task.

7. 

import java.util.Scanner;

class Trans

{

int arr[][];

int m;

public Trans(int mm)

{

m=mm;

arr=new int[m][m];

}

public void fillarray()

{

Scanner sc=new Scanner(System.in);

System.out.println(“Enter the array elements”);

for(int i=0;i<m;i++)

{

for(int j=0;j<m;j++)

{

arr[i][j]=sc.nextInt();

}

}

}

public void transpose()

{

System.out.println(“TRANSPOSE”);

for(int i=0;i<m;i++)

{

for(int j=0;j<m;j++)

{

System.out.print(arr[j][i]+”\t”);

}

System.out.println();

}

}

public void display()

{

System.out.println(“ORIGINAL ARRAY”);

for(int i=0;i<m;i++)

{

for(int j=0;j<m;j++)

{

System.out.print(arr[i][j]+”\t”);

}

System.out.println();

}

transpose();

}

public static void main(String args[])

{

Scanner sc=new Scanner(System.in);

System.out.println(“Enter the size”);

int mm=sc.nextInt();

Trans obj=new Trans(mm);

obj.fillarray();

obj.display();

}

}

Question 8 

A class SortAlpha has been defined to sort the words in the sentence in alphabetical order. 

Example: 

Input: THE SKY IS BLUE 

Output: BLUE IS SKY THE

Specify the class SortAlpha giving details of the constructor(), void acceptsent(), void sort(SortAlpha) and void display(). Define a main() function to create an object and call the functions accordingly to enable the task.

8. 

import java.util.*;

class SortAlpha

{

String sent;

int n;

public SortAlpha()

{

sent=””;

n=0;

}

public void acceptsent()

{

Scanner sc=new Scanner(System.in);

System.out.println(“Enter a sentence”);

sent=sc.nextLine().toUpperCase();

}

public void sort(SortAlpha P)

{

String arr[]=P.sent.split(“\\s”);

int len=arr.length;

for(int i=0;i<len-1;i++)

{

 for(int j=0;j<len-1-i;j++)

 {

 if(arr[j].compareTo(arr[j+1])>0)

 {

 String x=arr[j];

 arr[j]=arr[j+1];

 arr[j+1]=x;

 }

 }

 }

 for(int i=0;i<len;i++)

 sent+=arr[i]+” “;

 }

 public void display()

 {

 System.out.println(sent);

 }

 public static void main(String args[])

 {

 SortAlpha obj=new SortAlpha();

 SortAlpha tem=new SortAlpha();

 obj.acceptsent();

 tem.sort(obj);

 obj.display();

 tem.display();

 }

 }

ISC Computer Paper 2023 Solved Section C

SECTION - C

Question 9 

A double ended queue is a linear data structure which enables the user to add and remove integers from either ends i.e., from front or rear.

(i) Specify the class deQueue giving details of the functions void addFront(int) and int popFront(). Assume that the other functions have been defined.

(ii) Differentiate between a stack and a queue.

9. (i) 

class deQueue 

{ 

int Qrr[]; 

int lim, front, rear;  

void addFront(int a) 

{ 

if(front==0) 

System.out.println(“OVERFLOW FROM FRONT”); 

else 

Qrr[--front]=a; 

} 

int popFront( ) 

{ 

if(front==rear) 

return -999; 

else 

return 

Qrr[front++]; 

} 

}

(ii) Stack follows LIFO order where as Queue follows FIFO order

Question 10 

A super class Demand has been defined to store the details of the demands for a product. Define a subclass Supply which contains the production and supply details of the products.

Assume that the super class Demand has been defined. Using the concept of inheritance, specify the class Supply giving the details of the constructor(...), double calculation() and void display().

10. 

class Supply extends Demand 

{ 

int pproduced; 

double prate; 

public Supply(String a, String b, int c, int d, double e) 

{ 

super(a,b,c); 

produced=d; 

prate=e;

} 

public double calculation( )

{ 

return

Math.abs(prate*pdemand – prate*pproduce);

}

public void display( ) 

{ 

super.display( ); 

System.out.println(calculation( ));

}

}

Question 11 

(i) A linked list is formed from the objects of the class given below: 

Class Node 

{

double sal; 

Node next;

} 

Write and Algorithm OR a Method to add a node at the end of an existing linked list. The method declaration is as follows: 

void addNode(Node ptr, double ss)

(ii) Answer the following questions from the diagram of a Binary Tree given below: 

(a) Write the pre-order traversal of the above tree structure. 

(b) Name the parent of the nodes D and B. 

(c) State the level of nodes E and F when the root is at level 0.

(ii)

(a) A F D G B H E 

(b) F and A 

(c) Level of E = 3 and level of F =1

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