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CBSE Class 10 Solved Question Paper 2024 : Computer Science (Kendriya Vidyalaya (KV) No. 1, Jalahalli West, Bengaluru)

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March 2024 Scheme of Valuation Subject : Computer Science Code : 41 Qn. No. Version : English Marks PART A 1. C) Fourth 2. A) Sector 3. D) 4 4. C) Base 1 1 1 1 5. D) No error 1 6. B) Flowchart 1 7. A) Null character 1 8. D) \n 1 9. C) do-while 10. B) 2 index numbers 11. A) compiler 12. D) int 1 1 1 1 13. B) Calibri 1 14. D) .xls 15. B) Cell 16. Mantissa 17. Cout 18. Flow of control 19. Size 20. Global 1 1 1 1 1 1 1 PART B 21. 1) Weather forecasting. 2) Animated graphics. 3) Fluid dynamic calculation. 4) Nuclear energy research. 5) Space science. 6) Weapon and missile design. 7) Petroleum exploration. 22. 1) CRT 2) LCD 3) TFT 4) LED Any 2 uses 1 m for each Any 2 2 2 P.T.O. 41 -2- Qn. No. 23. 24. Marks Linker : is a system software which links the modules or program segments together so that they can be executed properly with proper reference. 1 Loader : is a system software which accepts the object program produced by the compiler or assembler and prepares these programs for execution. 1 If t test condition statement 1; else statement 2; 1 Example If (marks > 35) cout << pass ; else cout << fail ; 25. 26. 27. Any one example 1 Module : is a logically self contained unit which can be tested and executed independently. 1 Modularity : is a technique adopted to divide a complex problem into a number of self contained independent program segment. 1 Modifiers : are the one who alter the meaning of the basic data types. 1 Types of modifiers : 1) Signed 2) Unsigned 3) Long 4) Short 1 Any one Function : is a named group of statements developed to solve a sub-problem. 1 Types of functions 1 1) Library functions 2) User - defined function 28. Shortcuts : Save ctrl + S 1 Print ctrl + P 1 PART C 29. Types of keyboards 1) Standard keyboard 2) Ergonomics keyboard 3) Virtual keyboard 4) Compact keyboard 30. Binary number system : Data in a computer is represented in binary digits 0 and 1 The base of the binary system is 2 The positional value of a digit is expressed as power of 2 Atleast 2 features for each type any 3 type keyboard 1 3=3 2 Example Weight bits Value 22 1 4 21 0 2 20 1 1 1 41 -3- Qn. No. Marks 31. Tasks of process management component : Control access to shared resource like file, I/O, CPU Control execution of applications Any 3 - 1 m for each Create, execute and delete process Cancel or resume process Schedule a process Synchronize, communicate and deadlock handling for process 1 3=3 32. Advantages of modular programming : Reusability Debugging is easier Any 3 advantages with explanation one mark for each Building library Portability 1 3=3 Relational operators 33. Operator Use == Checks if the value of two operands is equal or not Checks if the value of two operand is not equal != Example a==b a!=b > Checks value of left operand is greater than the value of right operand a>b < Checks value of left operand is less than the value of right operand a #include <conio.h> void main ( ) { 1 int a, b, temp; cout<< Enter two numbers ; cin>>a>>b; cout<< Before interchange a = <<a<< and b = <<b; 1 temp = a; a = b; b = temp; cout<< After Interchange a = <<a<< and b = <<b; getch( ); 1 } 1 mark for header files 1 mark for declaration and I/O statements 1 mark for logic 41 -4- Qn. No. Marks 35. Types of arrays : 1) One dimensional array. 2) Two-dimensional array. 3) Multi-dimensional array 36. The members of a structure can be accessed using dot operator. The dot operator is called as member access operator The syntax for dot operator 1 Structure-variable-name.member.name It consists of 3 parts Part 1 name of the structure variable Part 2 .dot operator Part 3 member name 1 Ex : struct student { int roll no; char name [5]; }; student.rollno; student.name; 1 1 m for each type with explanation 1 3=3 PART D 37. Characteristics of computer : 1) Speed 2) Memory 3) Storage any 5 characteristics with explanation 4) Accuracy 1 m for each 5) Versatality 6) Automation 7) Diligence 1 5=5 38. Characteristics of an Algorithm : 1) Input 2) Definite 1 m for each characteristics with explanation 3) Effective 4) Terminate 5) Output 1 5=5 41 -5- Qn. No. Marks Unary operators : 39. ! Logical Not is used to reverse the logical state of its operand !a & Address of used to give the address of the operand &a ~ One s complement used to convert 1 to 0 and 0 to 1 Pointer difference used along with the operand to represent the pointer data type int a + Unary plus used to represent a signed positive operand +a Unary negation used to represent a signed negative number a Increment used to increment an operand a++ ++ Decrement used to decrement an operand 1 m for each operator with use and example. Any 5 operators. 1 5=5 a C++ program to print all the even numbers from 1 to 10 using do-while #include < iostream.h> 40. #include <conio.h> 1m void main ( ) { int i; i = 2; 1m 5 do { cout<<set w (5) <<i; i = i + 2; } 3m while (i < = 10) getch ( ) ; } 41 -6- Qn. No. Marks #include <iostream.h> 41. #include <conio.h> 1m void main ( ) { 1m int a[5] [5], b[5] [5], s [5] [5], i, j, m1, n1, m2, n2; clrscr( ); cout<< Enter the order of the first matrix ; cin>>m1>>n1; cout<< Enter order of the second matrix ; cin>>m2>>n2; cout<< Enter the elements of the first matrix ; for (i = 0; i < m1; i++) 1m for (j = 0; j < n1; j++) cin>>a [i] [j]; cout<< Enter elements of second matrix: ; for (i = 0; i < m2 ; i++) 5 for (j = 0; j < n2 ; j++) cin>>b[i] [j]; If ((m1==m2)&&(n1==n2)) { for(i = 0; i<m1; i++) 1m for (j = 0; j < n1; j ++) s[i] [j] = a[i] [j] + b [i] [j]; cout<< The resultant matrix is : ; for (i = 0; i < m1; i++) 1m { for (j = 0; j < n1; j++) cout<<setw(5)<<s[i] [j]; cout<<endl; } } else cout<< matrices are not compatible ;. } 41 -7- Qn. No. Marks 42. Steps to multiply numbers in cells Click a cell in which you want the calculation to be placed Type the equal sign = Click the first cell that you want to multiply & then type Click the cell you want to multiply Press enter 43. Statistical functions : 1) Average 2) count 3) AveragE Any 5 functions with explanation 1 m for each 4) counta 5) rank 6) large 7) small 1 5=5 44. Formatting tags : 1) 2) 3) 4) 5) <a> 6) <img> 1 5=5 1 m for each step Any 5 tags explanation with example 1 m for each tag 5 PART E FACE(16) 45. F 15 1111 A 10 1010 C 12 1100 E 14 1110 1m 001111101011001110 1 7 5 3 1 6 (2) 1m 5 Face(16) = 175316(8) 1 85 + 7 84 + 5 83 + 3 82 + 1 81 + 6 80 = 1 32768 + 7 4096 + 5 512 + 3 64 + 1 8 + 6 1 = 32768 + 28672 + 2560 + 192 + 8 + 6 = 64206(10) FACE(16) = 175316(8) = 64206(10) 1m 2m 41 -8- Qn. No. 46. Marks For explanation 2 Marks For program example 3 Marks #include <iostream.h> #include <conio.h> void main( ) { 1m void swap(int&, int&); int a, b; clrscr( ); cout<< Enter two members ; cin>>a>>b; cout<< Before calling a function a = <<a<< b= <<b; swap(a, b); cout<< After calling a function a= <<a<< b= <<b; getch( ); 3m 5 } void swap (int & x, int & y) { int temp; temp = x ; x = y; y = temp; } 47. 1m For explanation of for loop 1 Mark For syntax explanation 2 Marks for (initialization; condition; increment/decrement) 5 for example 2 Marks #include <iostream.h> #include <conio.h> void main( ) 5 { int n, i; cout<< Enter the limit ; cin>>n; for(i = 1 ; i < n ; i ++) cout<<i; getch( ); } Any suitable example

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