ICSE Class 10 Computer Specimen Paper 2023 Solved PDF - Overview
Key Highlights of the Computer Specimen Paper 2023 are as follows.
| Board | CISCE |
| Class | ICSE Class 10 |
| Subject | Computer Applications |
| Topic | ICSE Class 10 Computer Specimen Paper 2023 Solved |
| Exam Date | 23 March 2023 |
| Session | 2022 - 23 |
| Official Website | cisce.org |
Table of Content
• ICSE Class 10 Computer Specimen Paper 2023 Pattern discuss
• ICSE Class 10 Computer Specimen Paper 2023 PDF
• ICSE Class 10 Computer Specimen Paper 2023 PDF download
• ICSE Class 10 Computer Specimen Paper 2023 Solutions
• ICSE Class 10 Computer Specimen Paper 2023 Solved PDF
• ICSE Class 10 Computer Specimen Paper 2023 Solved PDF Download
ICSE Class 10 Computer Specimen Paper 2023 Pattern discuss
Below you will find ICSE class 10 Computer Application Specimen Paper 2023 pdf and its solutions. Observe carefully the pattern of the paper so that you don't face any proper in the exam. It is going to be a 100 Marks paper and time allowed for this paper is 2 hours. Answers to this paper must be written on the booklet provided separately. You will not be allowed to write for the first 15 minutes. This time is to be spent in reading the question paper.
This Paper is divided into two sections, Section A and Section B. Attempt all the questions from Section A and any four from Section B. In Section A, there are 2 questions. Under question number 1, there are 20 questions each of 1 Marks. Basically it consists of mcq questions. Under question number 2, there are 10 questions each of 2 Marks. So total marks in Section A is 40 marks. In Section B, there are total 6 questions out of which you have to attempt 4 questions each of 15 Marks. So total marks in Section B is 60 Marks.
ICSE Class 10 Computer Specimen Paper 2023 PDF
ICSE Class 10 Computer Specimen Paper 2023 PDF download
📌 ICSE Class 10 2023 Time Table PDF
ICSE Class 10 Computer Specimen Paper 2023 Solutions
Maximum Marks: 100
Time Allowed: Two hours.
Answers to this paper must be written on the paper provided separately.
You will not be allowed to write for the first 15 minutes.
This time is to be spent in reading the question paper.
This Paper is divided into two sections.
Attempt all the questions from Section A and any four from Section B.
Question 1:
(i) Wrapping up of data and methods together as one unit is termed as:
(a) Inheritance
(b) Polymorphism
(c) Encapsulation
(d) Abstraction
Answer: (c) Encapsulation
(ii) The data type which is specified that the method does not return a value is:
(a) Void
(b) void
(c) VOID
(d) boolean
Answer: (b) void
(iii) The logical operator which is a unary operator:
(a) &&
(b) ||
(c) !
(d) >>
Answer: (c) !
(iv) The Scanner class is a ________ class.
(a) Primitive
(b) Derived
(c) Wrapper
(d) Super class
Answer: (c) Wrapper
(v) Math.pow(625, 1 / 2) + Math.sqrt(144)
(a) 17.0
(b) 13.0
(c) 37.0
(d) 13
Answer: (b) 13.0
(vi) The correct if statement for the following ternary operation statement is:
System.out.println(n % 2 == 0? "true" : "false");
(a) if(n % 2 == 0)
return true;
else
return false;
(b) if(n % 2 == 0)
return "true";
else
return "false";
(c) if(n % 2 == 0)
System.out.println("true");
else
System.out.println("false");
(d) if(n % 2 == 0)
return false;
else
return false;
Answer: (c)
if(n % 2 == 0)
System.out.println("true");
else
System.out.println("false");
(vii) Multiple branching statement of Java is:
(a) For
(b) while
(c) do while
(d) switch
Answer: (d) switch
(viii) The number of bytes occupied by the constant 45 are:
(a) Four bytes
(b) Two bytes
(c) Eight bytes
(d) One byte
Answer: (a) Four bytes
(ix) do while loop is an:
(a) Entry-controlled loop
(b) Infinite loop
(c) Exit-controlled loop
(d) Finite loop
Answer: (c) Exit-controlled loop
(x) for(k = 1; k <= 2; k++){
for(m = 1; m <= 4; m++){
System.out.println(m * 2);
}
}
How many times the inner loop is executed?
(a) 4 times
(b) 8 times
(c) 2 times
(d) 16 times
Answer: (b) 8 times
(xi) A method with the same name as of the class and with arguments and no return data type is termed as:
(a) parameterized constructor
(b) default constructor
(c) non-parameterized constructor
(d) wrapper class method
Answer: (a) parameterized constructor
(xii) int res = 'A'; What is the value of res?
(a) A
(b) 66
(c) 65
(d) 97
Answer: (c) 65
(xiii) The style of expressing single line comment is:
(a) /* comment */
(b) * comment
(c) //comment
(d) /* comment
Answer: (c) //comment
(xiv) The method to check if a character is an alphabet or not is:
(a) isLetter(char)
(b) isAlpha(char)
(c) isUpperCase(char)
(d) isLowerCase(char)
Answer: (a) isLetter(char)
(xv) The output of Double.parseDouble("71.25") + 0.75 is:
(a) 72
(b) 72.0
(c) 71.0
(d) 71.75
Answer: (a) 72
(xvi) The method to convert a string to uppercase is:
(a) toUpperCase(char)
(b) toUPPERCASE(String)
(c) toUpperCase(String)
(d) touppercase(String)
Answer: (c) toUpperCase(String)
(xvii) The output of the method "DETERMINATION".substring(2, 6) is:
(a) "TERM"
(b) term
(c) "Term"
(d) "TERMI"
Answer:(a) "TERM"
(xviii) The array int x[10] occupies:
(a) 10 bytes
(b) 40 bytes
(c) 20 bytes
(d) 80 bytes
Answer: (b) 40 bytes
(xix) The element in x[4] of the array {3, 5, 7, 12, 16, 18, 20, 35, 42, 89} is:
(a) 16
(b) 12
(c) 7
(d) 18
Answer: (a) 16
(xx) Name the type of error that occurs for the following statement:
System.out.println(Math.sqrt(24 - 25));
(a) Syntax error
(b) Run-time error
(c) Logical error
(d) No error
Answer: (c) Logical error
Question 2:
(i) Evaluate the expression:
z += a++ + --b + ++a + --b;
where a = 10, b = 5, z = 10
Answer:
z = z + (a++ + --b + ++a + --b)
z = 10 + (10 + 4 + 12 + 3)
z = 10 + 29
z = 39
(ii) Write Java expression for: |x^2 + xy|
Answer:
Math.abs(x * x + x * y)
(iii) Rewrite the following using ternary operator:
if(x > y)
c = 'A';
else
c = 'a';
Answer:
c = (x > y)? 'A' : 'a';
(iv) Rewrite the following while loop using for loop:
int x = 5;
while(x <= 5){
x++;
}
System.out.println(x);
Answer:
int x;
for(x = 5; x <= 5; x++);
System.out.println(x);
(v) How many times the following loop will get executed? What is the output of the same?
int counter = 1;
do{
System.out.println(counter);
}while(counter++ < 5);
Answer:
The loop is executed 5 times.
Output:
1
2
3
4
5
(vi) "MISSISSIPPI".replace('S', 't').toLowerCase()
Answer: mittittippi
(vii) "REDUCE".compareTo("REVOLT") + "ANTARTICA".lastIndexOf('A')
Answer:
-18 + 8 = -10
(viii) Define boxing with an example.
Answer:
Boxing is the process of conversion from primitive data type to its corresponding object wrapper class.
For example:
int a = 5;
Integer b = a;
(ix) Consider the following program and answer the following questions given below:
class Sample{
int a, b;
Sample(int x, int y){
a = x;
b = y;
}
void calculate(){
int z;
z = a + b;
System.out.println(z);
}
}
(a) Name the global variables.
Answer: a and b are global variables.
(b) What are the method variables?
Answer: x, y and z are method variables.
(x) Consider the following array and answer the questions given below:
int x[] = {23, 45, 67, 12, 45, 89, 24, 12, 9, 7};
(a) What is the size of the array?
Answer :
Length-wise, the size is 10.
Memory-wise, the size is 40 bytes.
(b) What is the position of 89?
Answer: 5
Before going down I will suggest you to first try out the problems on your own and then see the solution.
Section - B
Question 3
Define a class with the following specifications:
Class name: Employee
Member variables:
eno - employee number
ename - name of the employee
age - age of the employee
basic - basic salary
[Declare the variables using appropriate data types]
Member methods:
void accept() - accept the details using Scanner class
void calculate() - to calculate the net salary as per the given specifications:
net = basic + hra + da - pf
hra = 18.5% of basic
da = 17.45% of basic
pf = 8.10% of basic
If the age of the employee is above 50, he/she gets an additional allowance of Rs. 5000.
void print() - to print the details as per the following format:
eno ename age basic net
void main() - to create an object of the class and invoke the methods.
Code:
import java.util.Scanner;
class Employee {
int eno;
String ename;
int age;
double basic;
double net;
public void accept() {
try (Scanner in = new Scanner(System.in)) {
System.out.print("Employee number: ");
eno = Integer.parseInt(in.nextLine());
System.out.print("Employee name: ");
ename = in.nextLine();
System.out.print("Age: ");
age = Integer.parseInt(in.nextLine());
System.out.print("Basic salary: ");
basic = Double.parseDouble(in.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
}
public void calculate() {
double hra = 0.185 * basic;
double da = 0.1745 * basic;
double pf = 0.0810 * basic;
net = basic + hra + da - pf;
if (age > 50) {
net += 5000;
}
}
public void print() {
System.out.println("eno\tename\tage\tbasic\tnet");
System.out.println("\n" + eno + "\t" + ename + "\t" + age + "\t" + basic + "\t" + net);
}
public static void main(String[] args) {
Employee obj = new Employee();
obj.accept();
obj.calculate();
obj.print();
}
}
Output -
Employee number: 101
Employee name: shekhar
Age: 25
Basic salary: 10000
eno ename age basic net
101 shekhar 25 10000.0 12785.0
Question 4
Define a class to overload the method print as follows: void print() - to print the format: 1 2 3 4 5 6 7 8 9 10 boolean print(int n) - to check whether the number is a Dudeney number. A number is dudeney if the cube of the sum of the digits is equal to the number itself. Example: 512 (5 + 1 + 2)^3 = 8^3 = 512. void print(int a, char ch) - If ch = s or S print the square of the number else if ch = c or C print the cube of the number.
Code:
import java.util.Scanner;
class Overload {
public void print() {
int v = 1;
for (int i = 1; i <= 4; i++) {
for (int j = 1; j <= i; j++)
System.out.print(v++ + "");
System.out.println();
}
}
public boolean print(int n) {
int sum = 0;
for (int i = n; i != 0; i /= 10)
sum += i % 10;
int cube = (int) Math.pow(sum, 3);
if (n == cube)
return true;
return false;
}
public void print(int a, char ch) {
if (ch == 'S' || ch == 's')
System.out.println("Square = " + (a * a));
else if (ch == 'C' || ch == 'c')
System.out.println("Cube = " + (a * a * a));
}
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
System.out.print("Enter a number: ");
int n = Integer.parseInt(in.nextLine());
System.out.print("Enter character 'c' or 's' ");
char c = in.next().charAt(0);
Overload obj = new Overload();
obj.print();
boolean r=obj.print(n);
System.out.println(r);
obj.print(n,c);
} catch (NumberFormatException e) {
e.printStackTrace();
}
}
}Output-
Enter a number: 512
Enter character 'c' or 's' s
1
23
456
78910
true
Square = 262144Question 5
Define a class to accept 10 integers and arrange them in descending order using bubble sort. Print the original array and the sorted array.
Code:
import java.util.Scanner;
class Bubble {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
int a[] = new int[10];
System.out.println("Enter 10 integers:");
for (int i = 0; i < a.length; i++){
a[i] = Integer.parseInt(in.nextLine());
}
System.out.println("Original array:");
for (int i = 0; i < a.length; i++)
System.out.print(a[i] + " ");
System.out.println();
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a.length - 1 - i; j++) {
if (a[j] < a[j + 1]) {
int temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
System.out.println();
System.out.println("Sorted array:");
for (int i = 0; i < a.length; i++)
System.out.print(a[i] + " ");
} catch (NumberFormatException e) {
e.printStackTrace();
}
System.out.println();
}
}Output-
Enter 10 integers:
1
2
3
5
4
6
7
9
8
0
Original array:
1 2 3 5 4 6 7 9 8 0
Sorted array:
9 8 7 6 5 4 3 2 1 0
Question 6
Define a class to accept values into a double data type array of size 20 and print the range of the array. Range is the difference between the largest and the smallest elements of the array.
Code:
import java.util.Scanner;
class Range {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
double a[] = new double[20];
double large = 0.0;
double small = 0.0;
System.out.println("Enter 20 numbers:");
for (int i = 0; i < a.length; i++) {
a[i] = Double.parseDouble(in.nextLine());
if (i == 0) {
large = a[i];
small = a[i];
} else {
if (large < a[i])
large = a[i];
if (small > a[i])
small = a[i];
}
}
double range = large - small;
System.out.println("Range = " + range);
} catch (NumberFormatException e) {
e.printStackTrace();
}
}
}Output -
Enter 20 numbers:1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Range = 19.0Question 7
Define a class to accept a string and print the same in reverse. Also print the number of vowels in the string.
Eg: s = “BEAUTIFUL” Output: “LUFITUAEB” No. of vowels = 5
Code:
import java.util.Scanner;
class reverse {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
System.out.print("Enter the string:");
String s = in.nextLine();
int v = 0;
String r = "";
for (int i = s.length() - 1; i >= 0; i--) {
char ch = s.charAt(i);
r += ch;
if ("AEIOUaeiou".indexOf(ch) >= 0)
v++;
}
System.out.println(r);
System.out.println("No. of vowels = " + v);
}
}
}Output:
Enter the string:BEAUTIFUL
LUFITUAEB
No. of vowels = 5Question 8
Define a class to accept the names of 10 students in an array and check for the existence of the given name in the array using linear search. If found, print the position of the name. If not found, print an appropriate message. Also print the names which begins with the word “SRI”.
Code:
import java.util.Scanner;
class Search {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
String a[] = new String[10];
int count = 0;
int pos = 0;
System.out.println("Enter 10 names:");
for (int i = 0; i < a.length; i++)
a[i] = in.nextLine();
System.out.print("Enter name to be searched: ");
String key = in.nextLine();
boolean found = false;
for (int i=0; i < a.length; i++) {
if (key.equalsIgnoreCase(a[i]) && !found) {
found = true;
pos = i;
}
if (a[i].startsWith("SRI"))
count++;
}
if (found)
System.out.println("Found at position: " + (pos + 1));
else
System.out.println("Not found!");
System.out.println("No. of names beginning with SRI = " + count);
}
}
}Output-
Enter 10 names:
aman
bhavan
digbijoy
elina
fraz
gajni
harash
ilina
jay
kartik
Enter name to be searched: kartik
Found at position: 10No. of names beginning with SRI = 0ICSE Class 10 Computer Specimen Paper 2023 Solved PDF
ICSE Class 10 Computer Specimen Paper 2023 Solved PDF Download
Recommended:
Post a Comment
If you have any doubts, please let me know