ICSE Class 10 Computer Applications Board Exam Question Paper 2022 with Solutions

ICSE Class 10 Computer Applications Board Exam Question Paper 2022 with Solutions

SECTION A

Question 1.
Choose the correct answers to the questions from the given options. (Do not copy the question. Write the correct answer only.) [10]

(i) Return data type of isLetter(char) is... [1 Mark]
(a) Boolean
(b) boolean
(c) bool
(d) char

Answer: (b) boolean

The Character.isLetter(char) method returns a primitive boolean value (true or false).

Teacher's Note:
a) Character methods in Java like isLetter() and isDigit() return the primitive boolean data type.
b) Students often confuse primitive boolean with the wrapper class Boolean.

(ii) Method that converts a character to uppercase is....... [1 Mark]
(a) toUpper()
(b) ToUpperCase()
(c) toUppercase()
(d) toUpperCase(char)

Answer: (d) toUpperCase(char)

Character.toUpperCase(char ch) is the correct standard library method to convert a character to uppercase.

Teacher's Note:
a) Note the exact camelCase naming convention used in Java for Character methods.
b) Be careful with string methods like toUpperCase() versus character methods.

(iii) Give output of the following String methods: [1 Mark]
“SUCESS".indexOf(‘S’)+“SUCCESS". lastIndexOft‘S’)
(a) 0
(b) 5
(c) 6
(d) -5

Answer: (c) 6

"SUCESS".indexOf(‘S’) returns 0. "SUCCESS".lastIndexOf(‘S’) returns 6. Therefore, 0 + 6 = 6.

Teacher's Note:
a) indexOf() returns the index of the first occurrence, while lastIndexOf() returns the index of the last occurrence.
b) Indexing in Java strings always begins at 0.

(iv) Corresponding wrapper class of float data type is....... [1 Mark]
(a) FLOAT
(b) float
(c) Float
(d) Floating

Answer: (c) Float

The wrapper class for primitive type float in Java is Float with a capital F.

Teacher's Note:
a) All numeric wrapper classes in Java start with a capital letter corresponding to their primitive counterparts.
b) Watch out for case sensitivity in Java class names.

(v) …....class is used to convert a primitive data type to its corresponding object. [1 Mark]
(a) String
(b) Wrapper
(c) System
(d) Math

Answer: (b) Wrapper

Wrapper classes encapsulate primitive values into objects.

Teacher's Note:
a) Wrapper classes exist in the java.lang package for all eight primitive types.
b) This mechanism is fundamental for collections and object-oriented data handling.

(vi) Give the output of the following code: System.out.println("Good”.Concat(“Day”)); [1 Mark]
(a) GoodDay
(b) Good Day
(c) Goodday
(d) goodDay

Answer: (a) GoodDay

The concat() method joins two strings without inserting any extra spaces.

Teacher's Note:
a) concat() simply appends the argument string to the end of the invoking string.
b) Check for exact spacing when evaluating string concatenation.

(vii) A single dimensional array contains N elements. What will be the last subscript? [1 Mark]
(a) N
(b) N-1
(c) N -2
(d) N+1

Answer: (b) N-1

Since array indexing in Java starts from 0, an array of size N has valid subscripts from 0 to N-1.

Teacher's Note:
a) Off-by-one errors are common when traversing arrays up to N instead of N-1.
b) Always remember that the first element is at index 0.

(viii) The access modifier that gives least accessibility is: [1 Mark]
(a) private
(b) public
(c) protected
(d) package

Answer: (a) private

Private members are accessible only within the class in which they are declared.

Teacher's Note:
a) Access hierarchy from most restrictive to least restrictive is private, default (package), protected, public.
b) Private enforces strict encapsulation in object-oriented programming.

(ix) Give the output of the following code :
String A =“56.0”, B = “94.0”;
double C = Double.parseDouble(A);
double D = Double.parseDouble(B);
System.out.println((C+D)); [1 Mark]
(a) 100
(b) 150.0
(c) 100.0
(d) 150

Answer: (b) 150.0

C becomes 56.0, D becomes 94.0. Their sum is 150.0, which prints as 150.0 because C and D are double variables.

Teacher's Note:
a) Double.parseDouble() converts a string representation of a decimal number into a primitive double.
b) Arithmetic addition on double values results in a double type output.

(x) What will be the output of the following code?
System, out. println(“Lucknow". substring (0,4)); [1 Mark]
(a) Lucknow
(b) Luckn
(c) Luck
(d) luck

Answer: (c) Luck

substring(beginIndex, endIndex) extracts characters from beginIndex up to endIndex - 1. Here indices 0, 1, 2, 3 give "Luck".

Teacher's Note:
a) The end index in the substring method is exclusive.
b) Length of the substring is always equal to endIndex minus beginIndex.

SECTION B

(Attempt any four questions from this Section.)

Question 2.
Define a class to perform binary search on a list of integers given below, to search for an element input by the user, if it is found display the element along with is position, otherwise display the message “Search element not found”. [10]
2, 5, 7, 10, 15, 20, 29, 30, 46, 50

Answer:

import java.util.Scanner;
public class BinarySearchDemo {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int arr[] = {2, 5, 7, 10, 15, 20, 29, 30, 46, 50};
        System.out.println("Enter element to search:");
        int srch = sc.nextInt();
        int lb = 0, ub = arr.length - 1, pos = -1;
        while(lb <= ub) {
            int mid = (lb + ub) / 2;
            if(arr[mid] == srch) {
                pos = mid;
                break;
            }
            else if(arr[mid] < srch) {
                lb = mid + 1;
            }
            else {
                ub = mid - 1;
            }
        }
        if(pos != -1) {
            System.out.println("Element found at index: " + pos);
        }
        else {
            System.out.println("Search element not found");
        }
    }
}

Teacher's Note:
a) Binary search requires the input array to be sorted in ascending order.
b) Ensure the loop condition is lb <= ub to avoid missing edge elements.

Question 3.
Define a class to declare a character array of size ten. accept the characters into the array and display the characters with highest and lowest ASCII (American Standard Code for Information Interchange) value. [10]
EXAMPLE :
INPUT:
‘R’, ‘z’, ‘q’, ‘A’, ‘N’, ‘p’, ‘m’, ‘U’, ‘Q’, ‘F’
OUTPUT :
Character with highest ASCII value = z
Character with lowest ASCII value = A

Answer:

import java.util.Scanner;
public class AsciiMinMax {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        char arr[] = new char[10];
        System.out.println("Enter 10 characters:");
        for(int i = 0; i < 10; i++) {
            arr[i] = sc.next().charAt(0);
        }
        char max = arr[0];
        char min = arr[0];
        for(int i = 1; i < 10; i++) {
            if(arr[i] > max) {
                max = arr[i];
            }
            if(arr[i] < min) {
                min = arr[i];
            }
        }
        System.out.println("Character with highest ASCII value = " + max);
        System.out.println("Character with lowest ASCII value = " + min);
    }
}

Teacher's Note:
a) Characters in Java can be compared directly using relational operators since they have underlying integer ASCII values.
b) Initializing max and min to arr[0] is a standard and robust approach for finding extremes.

Question 4.
Define a class to declare an array of size twenty of double datatype, accept the elements into the array and perform the following : [10]
• Calculate and print the product of all the elements.
• Print the square of each element of the array.

Answer:

import java.util.Scanner;
public class DoubleArrayOps {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        double arr[] = new double[20];
        double product = 1.0;
        System.out.println("Enter 20 double elements:");
        for(int i = 0; i < 20; i++) {
            arr[i] = sc.nextDouble();
            product = product * arr[i];
        }
        System.out.println("Product of all elements = " + product);
        System.out.println("Square of each element:");
        for(int i = 0; i < 20; i++) {
            System.out.println(arr[i] + " squared = " + (arr[i] * arr[i]));
        }
    }
}

Teacher's Note:
a) Initialize the product accumulator to 1.0, not 0, otherwise the final product will always be zero.
b) Use double datatype consistently for all floating-point calculations.

Question 5.
Define a class to accept a string, and print the characters with the uppercase and lowercase reversed, but all the other characters should remain the same as before. [10]
EXAMPLE: INPUT : WelCoMe_2022
           OUTPUT : wELcOmE_2022

Answer:

import java.util.Scanner;
public class ReverseCaseString {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a string:");
        String s = sc.nextLine();
        String res = "";
        for(int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if(Character.isUpperCase(ch)) {
                res = res + Character.toLowerCase(ch);
            }
            else if(Character.isLowerCase(ch)) {
                res = res + Character.toUpperCase(ch);
            }
            else {
                res = res + ch;
            }
        }
        System.out.println("OUTPUT : " + res);
    }
}

Teacher's Note:
a) Character wrapper class methods like isUpperCase() and toLowerCase() make case manipulation straightforward and clean.
b) Non-alphabetic characters such as digits and underscores must be appended without modification.

Question 6.
Define a class to declare an array to accept and store ten words. Display only those words which begin with the letter ‘A’ or ‘a’ and also end with the letter ‘A’ or ‘a’.
EXAMPLE :
Input : Hari, Anita, Akash, Amrita, Alina, Devi Rishab, John, Farha, AMITHA
Output: Anita
           Amrita
           Alina
           AMITHA

Answer:

import java.util.Scanner;
public class FilterWords {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String words[] = new String[10];
        System.out.println("Enter 10 words:");
        for(int i = 0; i < 10; i++) {
            words[i] = sc.next();
        }
        System.out.println("Output:");
        for(int i = 0; i < 10; i++) {
            char first = words[i].charAt(0);
            char last = words[i].charAt(words[i].length() - 1);
            if((first == 'A' || first == 'a') && (last == 'A' || last == 'a')) {
                System.out.println(words[i]);
            }
        }
    }
}

Teacher's Note:
a) Use charAt(0) for the first character and charAt(length() - 1) for the last character of each word.
b) Check both uppercase and lowercase 'A' conditions using logical OR and AND operators.

Question 7.
Define a class to accept two strings of same length and form a new word in such a way that, the first character of the first word is followed by the first character of the second word and so on. [10]
Example : Input string 1 – BALL
           Input string 2 – WORD
OUTPUT : BWAOLRLD

Answer:

import java.util.Scanner;
public class InterleaveStrings {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter first string:");
        String s1 = sc.next();
        System.out.println("Enter second string:");
        String s2 = sc.next();
        String res = "";
        if(s1.length() == s2.length()) {
            for(int i = 0; i < s1.length(); i++) {
                res = res + s1.charAt(i) + s2.charAt(i);
            }
            System.out.println("OUTPUT : " + res);
        }
        else {
            System.out.println("Strings must be of the same length.");
        }
    }
}

Teacher's Note:
a) A single loop running up to the length of the string efficiently alternates characters from both strings.
b) Validating that both strings are of the same length demonstrates good programming practice.