ISC Class 12 Computer Science Sample Paper 2025 with Solutions

PART I - 20 MARKS

Question 1

(i) The compliment of the Boolean expression Aꞌ • (B • Cꞌ + Bꞌ • C) [1]
(A) Aꞌ • (B+C+Bꞌ +C)
(B) A+ (B+Cꞌ) •(B+Cꞌ)
(C) A+(Bꞌ +C) • (B+Cꞌ)
(D) Aꞌ • (Bꞌ +Cꞌ +Bꞌ •C)

Answer: (C) A+(Bꞌ +C) • (B+Cꞌ)

Applying De Morgan's Law: (Aꞌ • (B • Cꞌ + Bꞌ • C))ꞌ = A + (B • Cꞌ + Bꞌ • C)ꞌ = A + (Bꞌ + C) • (B + Cꞌ).

Teacher's Note:
a) Use De Morgan's law and distributive laws of Boolean algebra to find complements.
b) Be careful with inner term negations during expansion.

(ii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option.
Assertion: Recursion utilises more memory as compared to iteration.
Reason: Time complexity of recursion is higher due to the overhead of maintaining the function call stack. [1]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.

Both statements are individually true, but the memory overhead is due to stack frames, which is distinct from time complexity reasons.

Teacher's Note:
a) Recursion stores activation records on the stack.
b) Verify if the reason directly explains the assertion statement before choosing option (A).

(iii) According to the Principle of duality, the Boolean equation (Aꞌ + B) • (1 + B) = Aꞌ + B will be equivalent to: [1]
(A) (A + Bꞌ) • (0 + B) = A + Bꞌ
(B) (Aꞌ • B) + (0 • B) = Aꞌ • B
(C) (Aꞌ • B) + (0 • B) = Aꞌ + B
(D) (Aꞌ + B) • (0 + B) = Aꞌ + B

Answer: (B) (Aꞌ • B) + (0 • B) = Aꞌ • B

Replacing AND (.) with OR (+), OR (+) with AND (.), 1 with 0, and 0 with 1 gives the dual equation.

Teacher'sNote:
a) The principle of duality swaps operators and identity elements.
b) Variables themselves are not complemented during dual formation unless already complemented.

(iv) Distributive law states that: [1]
(A) A + B • C = (A + B) • (A +C)
(B) A + ( A • B) = A
(C) A • (B + C) = (A • B) + (B • C)
(D) A + B • C = A • B + A • C

Answer: (A) A + B • C = (A + B) • (A +C)

This represents the distributive law of addition over multiplication.

Teacher's Note:
a) Boolean algebra has two forms of distributive laws.
b) Memorize standard Boolean laws thoroughly.

(v) The complement of the reduced expression of F(A,B) = Σ (0,1,2,3) is: [1]
(A) 1
(B) A • B
(C) 0
(D) Aꞌ + Bꞌ

Answer: (C) 0

F(A, B) covering all minterms equals 1, so its complement is 0.

Teacher's Note:
a) Summing all possible minterms for two variables yields 1.
b) The complement of 1 is always 0.

(vi) Study the given propositions and the statements marked Assertion and Reason that follow it. Choose the correct option on the basis of your analysis.
p = I am a triangle
q = I am a three-sided polygon
s1 = p → q
s2 = q → p
Assertion: s2 is converse of s1
Reason: Three-sided polygon must be a triangle. [1]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.

The converse of p → q is q → p by definition, making the assertion true, but the definition does not depend on the geometric reason.

Teacher's Note:
a) Converse swaps hypothesis and conclusion.
b) Understand propositional logic definitions completely.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and choose the correct option.
Assertion: In Java, the String class is used to create and manipulate strings, and it is immutable.
Reason : Immutability ensures that once a String object is created, its value cannot be changed. [1]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

String immutability precisely means its internal character sequence cannot be modified after creation.

Teacher's Note:
a) String objects in Java reside in the string constant pool.
b) Modifications create new objects rather than changing existing ones.

(viii) Consider the following statement written in class Circle where pi is its data member.
static final double pi = 3.142;
Which of the following statements are valid for pi?
I. It contains a common value for all objects class Circle.
II. Its value is non-changeable.
III. At a time two access modifiers, static and final, cannot be applied to a single data member pi. [1]
(A) I and II
(B) II and III
(C) I and III
(D) Only III

Answer: (A) I and II

Static makes it a class variable shared across objects, and final makes it a constant.

Teacher's Note:
a) Static and final can be combined to define class constants.
b) Statement III is false because multiple modifiers can coexist.

(ix) For Big O notation, state the difference between O(n) and O(n²). [1]

Answer:
O(n) represents linear time complexity typically associated with a single loop, whereas O(n²) represents quadratic time complexity typically associated with nested loops.

Teacher's Note:
a) Mention loop structures as practical indicators.
b) Keep answers concise and direct for 1-mark questions.

(x) A full adder needs five gates and those are 3 AND gates, 1 OR gate and 1 XOR gate. When a full adder is constructed using 2 half adders, it also requires 5 gates. State the names along with the quantity those gates. [1]

Answer:
2 AND gates, 1 OR gate and 2 XOR gates.

Teacher's Note:
a) Two half adders and an OR gate implement a full adder circuit.
b) Count the exact number of logical gates required in the composite circuit.

Question 2

(i) Convert the following infix notation to prefix form.
( A – B ) / C * ( D + E ) [2]

Answer:
*/-ABC+DE

Teacher's Note:
a) Parenthesize operations based on precedence.
b) Convert sub-expressions systematically from inside out.

(ii) A matrix M[-6….10, 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. [2]

Answer:
Address = B + W * ((J – LC) * M + (I – LR))
= 1025 + 4 * ((8 - 4) * 17 + (4 - (-6)))
= 1025 + 4 * (4 * 17 + 10)
= 1025 + 4 * (68 + 10)
= 1025 + 4 * 78
= 1025 + 312
= 1337

Teacher's Note:
a) Calculate row count as Upper Bound - Lower Bound + 1 = 10 - (-6) + 1 = 17.
b) Apply column-major address calculation formula carefully.

(iii) The following function getIt() is a part of some class. Assume x is a positive integer, f is the lower bound of arr[ ] and l is the upper bound of the arr[ ].
Answer the questions given below along with dry run/working.
public int getIt(int x,intarr[],int f,int l)
{
    if(f>l)
        return -1;
    int m=(f+l)/2;
    if(arr[m]<x)
        return getIt(x,m+1,l);
    else if(arr[m]>x)
        return getIt(x,f,m-1);
    else
        return m;
}

(a) What will the function getIt( ) return if arr[ ] = {10,20,30,40,50} and x=40? [2]

Answer:
3

Teacher's Note:
a) Trace the midpoints recursively until the target element is found.
b) Index 3 contains the value 40.

(b) What is function getIt( ) performing apart from recursion? [1]

Answer:
Function getIt() is performing binary search on arr[].

Teacher's Note:
a) Binary search divides the search space in half recursively.
b) The array must be sorted for this algorithm to work.

(iv) The following is a function of class Armstrong. This recursive function calculates and returns the sum of the cubes of all the digits of num, where num is an integer data member of the class Armstrong.
[A number is said to be Armstrong if the sum of the cubes of all its digits is equal to the original number].
There are some places in the code marked by ?1?, ?2?,?3? which may be replaced by a statement/expression so, that the function works properly.
public int sumOfPowers(int num)
{
    if (num == 0)
        return ?1?;
    int digit = ?2?;
    return (int) Math.pow(digit, 3) + ?3?;
}

(a) What is the expression or statement at ?1? [1]

Answer:
0

Teacher's Note:
a) Base case returns 0 when number reduces to 0.
b) This serves as the additive identity.

(b) What is the expression or statement at ?2? [1]

Answer:
num%10

Teacher's Note:
a) Modulo operator extracts the last digit.
b) Essential for processing digits individually.

(c) What is the expression or statement at ?3? [1]

Answer:
sumOfPowers(num/10)

Teacher's Note:
a) Recursive call reduces the number by removing the last digit.
b) Combines results via addition.

SECTION - A

Question 3

(i) A shopping mall announces a special discount on all its products as a festival offer only to those who satisfy any one of the following conditions.
• If he/she is an employee of the mall and has a service of more than 10 years.
OR
• A regular customer of the mall whose age is less than 65 years and should not be an employee of the mall.
OR
• If he/she is a senior citizen but not a regular customer of the mall.
The inputs are :
INPUTS
E - Employee of the mall
R - Regular customer of the mall
S - Service of the employee is more than 10 years
C - Senior citizen of 65 years or above
(In all the above cases, 1 indicates yes and 0 indicates no.)
Output: X - Denotes eligible for discount [1 indicates YES and 0 indicates NO in all cases]
Draw the truth table for the inputs and outputs given above and write the SOP expression for X ( E, R, S, C ). [5]

Answer:

X (E, R, S, C) = Σ (1, 3, 4, 6, 9, 10, 11, 14, 15)
= E'R'S'C + E'R'SC + E'RS'C' + E'RSC' + ER'S'C + ER'SC' + ER'SC + ERSC' + ERSC

Teacher's Note:
a) Formulate minterms corresponding to each true condition in the problem statement.
b) Verify all 16 rows against the given logical conditions carefully.

(ii) Reduce the above expression X ( E, R, S, C ) by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).
Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [5]

Answer:
Karnaugh map reduction yields two quads and a pair:
Quad 1 (m₁ + m₃ + m₉ + m₁₁) = R'C
Quad 2 (m₁₀ + m₁₁ + m₁₄ + m₁₅) = ES
Pair (m₄ + m₆) = E'RC'
Hence, F(E, R, S, C) = R'C + ES + E'RC'

[Figure: Logic gate diagram showing three AND gates for terms R'C, ES, and E'RC' connected to a 3-input OR gate producing output F(E,R,S,C)]

Teacher's Note:
a) Group adjacent 1s in powers of 2 (quads and pairs).
b) Ensure all 1s are covered efficiently with minimal product terms.

Question 4

(i) (a) Reduce the Boolean function F(P,Q,R,S) = (P+Q+R+S) • (P+Q+R+Sꞌ) • (P+Q+Rꞌ+S) • (P+Qꞌ+R+S) • (P+Qꞌ+R+Sꞌ) • (P+Qꞌ+Rꞌ+S) • (P+Qꞌ+Rꞌ+Sꞌ) • (Pꞌ+Q+R+S) • (Pꞌ+Q+R+Sꞌ) by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). [4]

Answer:
Plotting zeros for the given maxterms results in three quads:
Quad 1: (M₀ M₁ M₈ M₉) = Q + R
Quad 2: (M₄ M₅ M₆ M₇) = P + Q'
Quad 3: (M₀ M₂ M₄ M₆) = P + S
Hence, F(P, Q, R, S) = (Q + R) • (P + Q') • (P + S)

Teacher's Note:
a) Convert maxterm indices to K-map positions for POS reduction.
b) Combine adjacent zeros to form optimal quads.

(b) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs. [1]

Answer:

[Figure: Logic gate diagram showing three OR gates for terms (Q+R), (P+Q'), and (P+S) connected to a 3-input AND gate producing output F]

Teacher's Note:
a) POS expression uses OR gates followed by an AND gate.
b) Label inputs clearly.

(ii) From the given logic diagram :
[Figure: Logic diagram showing gate 1 as AND gate with inputs A and B, gate 2 as AND gate with inputs B' and C, gate 3 as NOR gate taking outputs of gate 1 and gate 2, followed by a NOT gate producing output X]
(a) Derive Boolean expression and draw the truth table for the derived expression. [4]

Answer:
Derived expression: [((A • B)' + B' • C)']'
Truth Table:

Teacher's Note:
a) Trace outputs step by step from individual gates.
b) Double inversion simplifies the expression.

(b) If A=1, B=0 and C=1 then find the value of X. [1]

Answer:
0

Teacher's Note:
a) Substitute values into the derived Boolean expression.
b) Evaluate according to operator precedence.

Question 5

(i) Draw the logic circuit to decode the following binary number (0001, 0101, 0111, 1000, 1010, 1100, 1110, 1111) to its hexadecimal equivalents. Also state the Hexadecimal equivalents of the given binary numbers. [5]

Answer:

[Figure: Decoder circuit diagram with vertical input lines A, B, C, D and their complements, connected to 8 multi-input AND gates corresponding to hex outputs 1, 5, 7, 8, A, C, E, F]

Hexadecimal equivalents:
(0001) = 1
(0101) = 5
(0111) = 7
(1000) = 8
(1010) = A
(1100) = C
(1110) = E
(1111) = F

Teacher's Note:
a) Use AND gates to decode specific minterms.
b) Map 4-bit binary patterns to their respective hex symbols.

(ii) Verify if the following proposition is valid using the truth table:
(X ∧ Y) => Z = (Y => Z) ∧ (X => Y) [3]

Answer:

INVALID (The truth table columns do not match identically in all rows).

Teacher's Note:
a) Construct truth tables for both sides of the logical equivalence.
b) If result columns differ, the proposition is invalid.

(iii) Answer the following questions related to the below image:
[Figure: NAND gate with inputs A and B producing an output]

(a) What is the output of the above gate if input A=0, B=1? [1]

Answer:
1

Teacher's Note:
a) NAND outputs 0 only when all inputs are 1.
b) Inputs 0 and 1 yield 1.

(b) What are the values of the inputs if output =1? [1]

Answer:
A=0, B=0 (or any input combination where not all inputs are 1)

Teacher's Note:
a) Multiple input combinations can produce 1 for a NAND gate.
b) State one valid input state clearly.

SECTION - B

Question 6 [10]

Answer:
import java.util.*;
class StringOp
{
    String str;
    String nstr;
    String msk;
    Scanner sc = new Scanner(System.in);
    StringOp()
    {
        str = "";
        nstr = "";
        msk = "";
    }
    void accept()
    {
        System.out.println("Enter the original word");
        str = sc.next() + sc.nextLine();
        System.out.println("Enter the mask string");
        msk = sc.next();
    }
    void form()
    {
        int l1 = str.length();
        for (int i = 0; i < l1; i++)
        {
            char c1 = str.charAt(i);
            if (msk.indexOf(c1) == -1)
                nstr = nstr + c1;
        }
    }
    void display()
    {
        System.out.println("Original string: " + str);
        System.out.println("Changed string: " + nstr);
    }
    public static void main()
    {
        StringOp ob = new StringOp();
        ob.accept();
        ob.form();
        ob.display();
    }
}

Teacher's Note:
a) Use indexOf() to check character existence efficiently.
b) Ensure all class data members and methods requested in the specification are implemented.

Question 7 [10]

Answer:
import java.util.*;
class Mixarray
{
    int arr[];
    int cap;
    static Scanner sc = new Scanner(System.in);
    Mixarray(int mm)
    {
        cap = mm;
        arr = new int[cap];
    }
    void input()
    {
        System.out.println("Enter the content of the array");
        for (int i = 0; i < cap; i++)
            arr[i] = sc.nextInt();
    }
    void display()
    {
        for (int i = 0; i < cap; i++)
            System.out.print(arr[i] + " ");
        System.out.println();
    }
    Mixarray mix(Mixarray P, Mixarray Q)
    {
        Mixarray res = new Mixarray(6);
        int k = 0;
        for (int i = 0; i < 3; i++)
            res.arr[k++] = P.arr[i];
        for (int i = 0; i < 3; i++)
            res.arr[k++] = Q.arr[i];
        return res;
    }
    public static void main()
    {
        System.out.println("Enter the capacity of both the array");
        int c1 = sc.nextInt();
        int c2 = sc.nextInt();
        Mixarray ob1 = new Mixarray(c1);
        Mixarray ob2 = new Mixarray(c2);
        System.out.println("Enter the content of 1st array");
        ob1.input();
        System.out.println("Enter the content of 2nd array");
        ob2.input();
        Mixarray r = new Mixarray(c1 + c2);
        Mixarray res = r.mix(ob1, ob2);
        System.out.println("Content of the combined array");
        res.display();
    }
}

Teacher's Note:
a) Object-oriented array blending requires returning an instance of the class.
b) Verify array bounds before copying elements.

Question 8 [10]

Answer:
import java.util.*;
class LCM
{
    int n1, n2;
    int large, sm;
    int l;
    static Scanner sc = new Scanner(System.in);
    void accept()
    {
        System.out.println("Enter 2 different integers:");
        n1 = sc.nextInt();
        n2 = sc.nextInt();
        if (n1 > n2)
        {
            large = n1;
            sm = n2;
        }
        else if (n2 > n1)
        {
            large = n2;
            sm = n1;
        }
    }
    int getLCM()
    {
        if (large != sm)
        {
            if (large > sm)
                large = large - sm;
            else if (large < sm)
                sm = sm - large;
            return getLCM();
        }
        else
            return (n1 * n2) / large;
    }
    void display()
    {
        l = getLCM();
        System.out.println("LCM of " + n1 + " and " + n2 + " = " + l);
    }
    public static void main()
    {
        LCM ob = new LCM();
        ob.accept();
        ob.display();
    }
}

Teacher's Note:
a) LCM can be computed recursively using the relation LCM(a, b) = (a * b) / GCD(a, b).
b) GCD is calculated here using repeated subtraction.

SECTION - C

Question 9

(i) Specify the class ReCycle giving details of the functions void pushfront(int) and int poprear( ). Assume that the other functions have been defined.
The main( ) function and algorithm need NOT be written. [4]

Answer:
class ReCycle
{
    void pushfront(int v)
    {
        if (front != 0)
            q[front--] = v;
        else
            System.out.println("FULL FROM FRONT");
    }
    int poprear()
    {
        if (front != rear)
            return (q[rear--]);
        else
            return -999;
    }
}

Teacher's Note:
a) Handle boundary conditions for underflow and overflow properly.
b) Match pointer decrement and increment logic with deque specifications.

(ii) Name the entity described above and state its principle. [1]

Answer:
Entity is deque (double-ended queue) and works on the principle of FIFO (or allows insertion and deletion at both ends).

Teacher's Note:
a) Deque supports operations at both front and rear.
b) State both the name and operational principle clearly.

Question 10 [5]

Answer:
class Compute extends Library
{
    private int d;
    private double f;
    public Compute(String name, String author, double p, int d)
    {
        super(name, author, p);
        this.d = d;
        f = 0.0;
    }
    public void fine()
    {
        int d1 = d - 7;
        if (d1 >= 1 && d1 <= 5)
            f = d1 * 2;
        else if (d1 >= 6 && d1 <= 10)
            f = d1 * 3;
        else
            f = d1 * 5;
    }
    public void show()
    {
        super.show();
        System.out.println("Fine = " + f);
        System.out.println("Total amount = " + ((0.02 * p * d) + f));
    }
}

Teacher's Note:
a) Use super keyword to invoke parameterized constructor and methods of the parent class.
b) Implement slab-based fine calculation accurately.

Question 11

(i) A linked list is formed from the objects of the class Node. The class structure of the Node is given below:
class Node
{
    int n;
    Node link;
}
Write an Algorithm OR a Method to search for a number from an existing linked list.
The method declaration is as follows:
void FindNode( Node str, int b ) [2]

Answer:
void FindNode(Node str, int b)
{
    Node temp = str;
    while (temp != null)
    {
        if (temp.n == b)
        {
            System.out.println(b + " is found");
            break;
        }
        temp = temp.link;
    }
    if (temp == null)
        System.out.println(b + " is not found");
}

Teacher's Note:
a) Traverse the linked list using a temporary pointer until null is reached.
b) Check node data fields for matching values.

(ii) Answer the following questions from the diagram of a Binary Tree given below:
[Figure: Binary tree with root A, left child B with child D and grandchild E; right child F with left child C and right child G with child H]
(a) Name the root of the left sub tree and its siblings. [1]

Answer:
Root: B, Sibling: F

Teacher's Note:
a) The left child of the root node A is B.
b) B and F share the same parent A, making them siblings.

(b) State the size and depth of the right sub tree. [1]

Answer:
Size: 4, Depth: 2

Teacher's Note:
a) Size is the total number of nodes in the right subtree (F, C, G, H).

Teacher's Note:
b) Depth counts edges from root F to deepest leaf H.

(c) Write the in-order traversal of the above tree structure. [1]

Answer:
E D B A C F G H

Teacher's Note:
a) In-order follows Left-Root-Right traversal order recursively.
b) Trace left subtree, root, then right subtree for each node.