ISC Class 11 Computer Science Evaluation II Sample Paper 2026-27 with Solutions

This is a full ISC Class 11 Computer Science Evaluation II paper for 2026-27, followed by a complete, checked answer key. Part I tests Boolean algebra, propositional logic and Java output questions. Part II covers number-system conversions, a full adder and a scholarship truth table, three Java class-design programs (circular prime, matrix boundary and string encryption), and three short Python programs.

Attempt it in 3 hours, then compare your answers. All truth tables, conversions and programs are worked out in full below.

COMPUTER SCIENCE – EVALUATION II (2026-27)

Class XI | Maximum Marks: 100 | Time Allowed: Three Hours

Answer all questions from Part I (compulsory) and six questions from Part II, choosing two questions from each section. The intended marks for questions or parts of questions are given in brackets [ ].

PART I – 20 Marks (Answer all questions)

Question 1 [10 × 1]

(i) Which Boolean expression represents the XNOR gate?

• (a) A ⊕ B
• (b) (A ⊕ B)'
• (c) A + B
• (d) A.B

(ii) Which logical connective is used to represent 'implies'?

• (a) ∧
• (b) ∨
• (c) →
• (d) ↔

(iii) The dual of the Boolean expression A + (B.C) is:

• (a) A.(B + C)
• (b) A + B + C
• (c) A.B.C
• (d) A + (B + C)

(iv) The complement of F = A + B.C is:

• (a) A'.(B' + C')
• (b) A'.B'.C'
• (c) A' + B' + C'
• (d) A'.(B.C)

(v) Which law is represented by P + 0 = P?

• (a) Identity law
• (b) Complement law
• (c) Commutative law
• (d) Involution law

(vi) What is the output of the following code?

String state[] = {"GOA", "KERALA", "GUJARAT", "ODISHA", "PUNJAB"};
System.out.println(state[2].length() == state[4].length());
System.out.print(state[1].substring(2, 5));

• (a) true RAL
• (b) false RAL
• (c) true KER
• (d) false KER

(vii) Which of the following is a contradiction?

• (a) X + X'
• (b) X.X'
• (c) X + Y
• (d) X

(viii) Assertion: Every contradiction is always false. Reason: The complement of a contradiction is a tautology.

• (a) Both true, R explains A
• (b) Both true, R does not explain A
• (c) A true, R false
• (d) A false, R true

(ix) Assertion: A truth table is used to determine the output of a logical proposition. Reason: A truth table lists all possible truth values.

• (a) Both true, R explains A
• (b) Both true, R does not explain A
• (c) A true, R false
• (d) A false, R true

(x) If P → Q, then its inverse is:

• (a) Q → P
• (b) P' → Q'
• (c) Q' → P'
• (d) P ↔ Q

Question 2

(a) Complete the palindrome method by writing the correct code at ?1? to ?5?. [5]

boolean palindrome(int n)
{
    int rev = 0, temp = n;
    while (?1?)
    {
        int d = ?2?;
        rev = rev * 10 + ?3?;
        temp = ?4?;
    }
    if (?5?) return true;
    else return false;
}

(b) Find the output of the following method when a = {{2,4,6},{1,5,3},{7,8,9}}. [2]

public static void display(int a[][])
{
    int s = 0;
    for (int i = 0; i < a.length; i++)
        s = s + a[i][i];
    System.out.println(s);
}

(c) Construct the truth table for (P ∨ Q) → (P ∧ Q). [3]

PART II – 50 Marks

SECTION A (Attempt any two questions)

Question 3 – Perform the following [10]

• (a) (110101)₂ + (101011)₂
• (b) (111101)₂ − (100111)₂
• (c) (245.625)₁₀ to binary, octal and hexadecimal
• (d) (7254.34)₈ to binary, hexadecimal and decimal
• (e) (9AF)₁₆ + (4D6)₁₆

Question 4

(a) Explain the full adder with its truth table and logic diagram. [5]

(b) A college scholarship is granted to a student if any one of the following conditions is satisfied: [5]

• The student has scored 90% or above and belongs to the Economically Weaker Section (EWS), OR
• The student has represented the school at the national level in sports, OR
• The student has scored 90% or above and has won a state-level science competition.

Inputs: M – scored 90% or above, E – belongs to EWS, N – represented the school at the national level in sports, S – won a state-level science competition. Output: F – 1 if eligible, 0 otherwise. Draw the truth table and write the Sum of Products expression for F(M, E, N, S).

Question 5

(a) Verify using a truth table: (P → Q)' = P.Q' [5]

(b) Determine whether (P ∨ Q') → Q is a tautology, contradiction or contingency. [5]

SECTION B (Attempt any two questions)

Question 6 – CircularPrime [10]

A circular prime is a number that remains prime under all cyclic rotations of its digits. For example, 197 is a circular prime because 197, 971 and 719 are all prime. Design a class CircularPrime:

• Data member: int num
• CircularPrime(int n): parameterised constructor to initialise num
• boolean isPrime(int n): returns true if n is prime
• int countDigits(int n): returns the number of digits in n
• int rotate(int n): moves the first digit to the last position and returns the rotated number
• void checkCircular(): checks all rotations of num and displays whether it is a circular prime

Also define main() to create an object and call the methods.

Question 7 – MatrixBoundary [10]

The boundary elements of a square matrix are those in its first row, last row, first column and last column. Design a class MatrixBoundary:

• Data members: int arr[][], int M (order of the matrix)
• MatrixBoundary(int mm): initialises M and allocates arr
• void fillArray(): accepts M × M elements
• void displayMatrix(): displays the matrix
• int boundarySum(): returns the sum of the boundary elements, counting each corner only once
• void displayBoundary(): displays the boundary elements and their sum using boundarySum()

Question 8 – Encrypt [10]

A sentence is encrypted by replacing every vowel with '*'. Design a class Encrypt:

• Data members: String text, String encrypted, int count (vowels replaced)
• Encrypt(): default constructor
• void accept(): accepts a sentence
• boolean isVowel(char ch): returns true if ch is a vowel
• void convert(): builds encrypted by replacing vowels with '*' and updates count
• void display(): displays the original sentence, the encrypted sentence and count

SECTION C – Python (Attempt any two questions)

Question 9 [5]: Write a Python program to accept the interest rate, principal amount and time from the user and print the simple interest.

Question 10 [5]: Write a Python program to accept three numbers and swap them without using a temporary variable.

Question 11 [5]: Write a Python program to input the name and basic salary of an employee and display the name, gross salary and net salary, where DA = 30% of basic, HRA = 18% of basic, PF = 12.5% of basic, gross = basic + DA + HRA and net = gross − PF.

ANSWER KEY WITH SOLUTIONS

Question 1 – MCQ answers

• (i) (b) (A ⊕ B)': XNOR is the complement of XOR.
• (ii) (c) →
• (iii) (a) A.(B + C): swap + with . (and 0 with 1).
• (iv) (a) A'.(B' + C'): by De Morgan's law, (A + BC)' = A'.(BC)' = A'.(B' + C').
• (v) (a) Identity law
• (vi) (b) false RAL: "GUJARAT" has 7 letters and "PUNJAB" has 6, so the comparison is false. "KERALA".substring(2, 5) gives the characters at indices 2, 3 and 4, which is "RAL".
• (vii) (b) X.X': it is 0 for every value of X.
• (viii) (b): both are true, but R talks about the complement. It does not explain why a contradiction is always false.
• (ix) (a): because a truth table lists every possible combination of truth values, it can show the output of the proposition in every case.
• (x) (b) P' → Q': the inverse negates both parts. (Q → P is the converse, and Q' → P' is the contrapositive.)

Question 2

(a) ?1? temp > 0 | ?2? temp % 10 | ?3? d | ?4? temp / 10 | ?5? rev == n

(b) 16: the method adds the main diagonal, 2 + 5 + 9 = 16.

(c)

P  Q | P∨Q  P∧Q | (P∨Q)→(P∧Q)
T  T |  T    T   |      T
T  F |  T    F   |      F
F  T |  T    F   |      F
F  F |  F    F   |      T

Question 3 – Conversions

(a) 110101 (53) + 101011 (43) = (1100000)₂ (96)

(b) 111101 (61) − 100111 (39) = (10110)₂ (22)

(c) 245 ÷ 2 repeatedly gives 11110101. For the fraction, .625 × 2 = 1.25 → 1, .25 × 2 = 0.5 → 0, .5 × 2 = 1.0 → 1. So (245.625)₁₀ = (11110101.101)₂.

• Octal: group the bits in threes, 011 110 101 . 101 → (365.5)₈
• Hexadecimal: group the bits in fours, 1111 0101 . 1010 → (F5.A)₁₆

(d) Replace each octal digit with 3 bits: 7 = 111, 2 = 010, 5 = 101, 4 = 100, 3 = 011, 4 = 100. So (7254.34)₈ = (111010101100.011100)₂.

• Hexadecimal: 1110 1010 1100 . 0111 → (EAC.7)₁₆
• Decimal: 7×512 + 2×64 + 5×8 + 4 + 3/8 + 4/64 = 3756 + 0.4375 = (3756.4375)₁₀

(e) F + 6 = 21 = 15 + carry 1 → 5. A + D + 1 = 24 = 16 + 8 → 8 carry 1. 9 + 4 + 1 = 14 → E. Result: (E85)₁₆.

Question 4

(a) A full adder is a combinational circuit that adds three bits: A, B and a carry-in (Cin). It produces a Sum and a Carry-out (Cout).

A  B  Cin | Sum  Cout
0  0  0   |  0    0
0  0  1   |  1    0
0  1  0   |  1    0
0  1  1   |  0    1
1  0  0   |  1    0
1  0  1   |  0    1
1  1  0   |  0    1
1  1  1   |  1    1

Sum = A ⊕ B ⊕ Cin and Cout = A.B + B.Cin + A.Cin, which can also be written as A.B + Cin.(A ⊕ B). The logic diagram is shown at the top of this post.

(b) F = 1 when (M AND E) OR N OR (M AND S).

Canonical SOP: F(M,E,N,S) = Σ(2, 3, 6, 7, 9, 10, 11, 12, 13, 14, 15). Written out as minterms: M'E'NS' + M'E'NS + M'ENS' + M'ENS + ME'N'S + ME'NS' + ME'NS + MEN'S' + MEN'S + MENS' + MENS. This simplifies to F = ME + N + MS.

Question 5

(a)

P  Q | P→Q | (P→Q)' | Q' | P.Q'
T  T |  T  |   F    | F  |  F
T  F |  F  |   T    | T  |  T
F  T |  T  |   F    | F  |  F
F  F |  T  |   F    | T  |  F

The (P→Q)' and P.Q' columns are identical, so the identity is verified.

(b)

P  Q | Q' | P∨Q' | (P∨Q')→Q
T  T | F  |  T   |    T
T  F | T  |  T   |    F
F  T | F  |  F   |    T
F  F | T  |  T   |    F

The result column has both T and F, so the proposition is a contingency.

Question 6 – CircularPrime

import java.util.Scanner;

class CircularPrime
{
    int num;

    CircularPrime(int n)
    {
        num = n;
    }

    boolean isPrime(int n)
    {
        if (n < 2)
            return false;
        for (int i = 2; i * i <= n; i++)
            if (n % i == 0)
                return false;
        return true;
    }

    int countDigits(int n)
    {
        int c = 0;
        do
        {
            c++;
            n /= 10;
        } while (n > 0);
        return c;
    }

    int rotate(int n)
    {
        int p = (int) Math.pow(10, countDigits(n) - 1);
        int first = n / p;          // leading digit
        int rest = n % p;           // remaining digits
        return rest * 10 + first;
    }

    void checkCircular()
    {
        int digits = countDigits(num);
        int n = num;
        boolean flag = true;
        for (int i = 1; i <= digits; i++)
        {
            System.out.println(n);
            if (!isPrime(n))
            {
                flag = false;
                break;
            }
            n = rotate(n);
        }
        if (flag)
            System.out.println(num + " is a Circular Prime");
        else
            System.out.println(num + " is not a Circular Prime");
    }

    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        CircularPrime ob = new CircularPrime(sc.nextInt());
        ob.checkCircular();
    }
}

The number of digits is counted once, before rotating. This matters for numbers with a 0, such as 101: the rotation 011 becomes 11, which has fewer digits.

Question 7 – MatrixBoundary

import java.util.Scanner;

class MatrixBoundary
{
    int arr[][];
    int M;

    MatrixBoundary(int mm)
    {
        M = mm;
        arr = new int[M][M];
    }

    void fillArray()
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter " + (M * M) + " elements:");
        for (int i = 0; i < M; i++)
            for (int j = 0; j < M; j++)
                arr[i][j] = sc.nextInt();
    }

    void displayMatrix()
    {
        for (int i = 0; i < M; i++)
        {
            for (int j = 0; j < M; j++)
                System.out.print(arr[i][j] + "\t");
            System.out.println();
        }
    }

    int boundarySum()
    {
        int sum = 0;
        for (int i = 0; i < M; i++)
            for (int j = 0; j < M; j++)
                if (i == 0 || j == 0 || i == M - 1 || j == M - 1)
                    sum += arr[i][j];      // each cell is visited once, so corners are not repeated
        return sum;
    }

    void displayBoundary()
    {
        System.out.println("Boundary elements:");
        for (int i = 0; i < M; i++)
        {
            for (int j = 0; j < M; j++)
            {
                if (i == 0 || j == 0 || i == M - 1 || j == M - 1)
                    System.out.print(arr[i][j] + "\t");
                else
                    System.out.print("\t");
            }
            System.out.println();
        }
        System.out.println("Boundary sum = " + boundarySum());
    }

    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter the order of the matrix: ");
        MatrixBoundary ob = new MatrixBoundary(sc.nextInt());
        ob.fillArray();
        System.out.println("Original matrix:");
        ob.displayMatrix();
        ob.displayBoundary();
    }
}

Checking every cell with one condition guarantees that each corner is added only once, and it also works for a 1 × 1 matrix.

Question 8 – Encrypt

import java.util.Scanner;

class Encrypt
{
    String text, encrypted;
    int count;

    Encrypt()
    {
        text = "";
        encrypted = "";
        count = 0;
    }

    void accept()
    {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a sentence: ");
        text = sc.nextLine();
    }

    boolean isVowel(char ch)
    {
        return "AEIOUaeiou".indexOf(ch) != -1;
    }

    void convert()
    {
        encrypted = "";
        count = 0;
        for (int i = 0; i < text.length(); i++)
        {
            char ch = text.charAt(i);
            if (isVowel(ch))
            {
                encrypted += '*';
                count++;
            }
            else
                encrypted += ch;
        }
    }

    void display()
    {
        System.out.println("Original sentence : " + text);
        System.out.println("Encrypted sentence: " + encrypted);
        System.out.println("Vowels replaced   : " + count);
    }

    public static void main(String args[])
    {
        Encrypt ob = new Encrypt();
        ob.accept();
        ob.convert();
        ob.display();
    }
}

Questions 9 to 11 – Python

p = float(input("Enter principal amount: "))
r = float(input("Enter rate of interest: "))
t = float(input("Enter time in years: "))
si = (p * r * t) / 100
print("Simple Interest =", si)

a = int(input("Enter first number: "))
b = int(input("Enter second number: "))
c = int(input("Enter third number: "))
print("Before swapping: a =", a, " b =", b, " c =", c)
a, b, c = b, c, a          # tuple packing/unpacking, no temporary variable
print("After swapping:  a =", a, " b =", b, " c =", c)

If the examiner expects arithmetic instead of tuple assignment, two numbers can be swapped with a = a + b; b = a - b; a = a - b. Repeat the same steps with b and c to rotate all three numbers.

name = input("Enter employee name: ")
basic = float(input("Enter basic salary: "))
da = 0.30 * basic
hra = 0.18 * basic
pf = 0.125 * basic
gross = basic + da + hra
net = gross - pf
print("Name         :", name)
print("Gross salary :", gross)
print("Net salary   :", net)