KENDRIYA VIDYALAYA SANGATHAN, GUWAHATI REGION HALF-YEARLY EXAMINATION-2015 CLASS XI : COMPUTER SCIENCE Time Allotted : 3 hrs. Max.
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1 KENDRIYA VIDYALAYA SANGATHAN, GUWAHATI REGION HALF-YEARLY EXAMINATION-2015 CLASS XI : COMPUTER SCIENCE Time Allotted : 3 hrs. Max. Marks : 70 Q operator in C++, requires three operands. [1] Q.2 Peta Byte= bits [1] Q.3 Which of the following are hardware and software? [1] (i) Capacitor (ii) Internet Explorer (iii) Hard disk (iv) UNIX Q.4 Write names of two components of CPU. [1] Q.5 What is the difference between \n and \t in terms of functionality? [1] Q.6 Explain the following terms: (i) Disk Defragmentation [1] (ii) Cache memory [1] Q.7 Expand following terms and explain: (i)agp Ports [1] (ii)ps-2 Port [1] Q.8 Name the header file(s) that required for successful compilation of the following C++ code. void main( ) [1] int p=3,q=5,x; x=pow(p,q); cout<<x; Q.9 Declare a variable of type float and initialize it with the value 4.5. [1] Q.10 What is the difference between RAM and ROM? (Any Two) [1] 1
2 Q.11 What will be the output produced by the following code. [1] int main() int x; int y; x = 30; y = 2; cout << x * y + 9 / 3 << endl; return 0; Q.12 Name the header files to which the following belongs to : [2] i. getch( ) ii. isalnum( ) iii. sqrt( ) iv. random( ) Q.13 Write the equivalent c++ expressions: [2] (i) p=2(l+b) (ii) (iii) S=1/2mv 2 (iv) Q.14 Rewrite the following program after removing syntactical errors, underline each Correction. #include<iostream.h> [2] main() int Sum; for(i=0;i<2;i++) for(j=0;j<=3;i++) cout<<sum; Q.15 Explain Break and Continue statement in C++ with example. [2] Q.16 Define Prettyprinting [2] Q.17 What is the difference between Entry controlled loop and Exit controlled loop? Explain with example? [2] 2
3 Q.18 What do you mean by Robustness of a program [2] Q.19 Classify the following identifiers of C++ into valid and invalid category [2] (i) int_1 (ii) num 2 (iii) break (iv) num.1 Q.20 What will be result of following statements if a=5, b=5 initially [2] (i) ++a<=5 (ii) b++<=5 Q.21 What is difference between a compiler and an interpreter? [2] Q.22 What do you mean by Comments? Give an example [2] Q.23 What is the difference between Implicit Type Conversion and Explicit Type Conversion? Explain with suitable example. [2] Q.24 What is the difference between Unary and Binary Operator? Explain with suitable example. [2] Q.25 What is the output of the following C++ code? [ 2X2=4 ] i) ii) # include<iostream.h> void main ( ) int i=0; cout<<i++<< <<i++<< <<i++<<endl; cout<<++i<< <<++i<< <<++i<<endl # include<iostream.h> void main ( ) for(int i=20;i<=100;i+=10) j=i/2; cout<<j<< ; Q.26 Convert the following into its binary equivalent codes. [2] (i)(0.375) 10 = (?) 8 (ii)( ) 2 = (?) 8 Q.27 What is Operating System? Write different functions performed by the Operating System. [2] Q.28 Define types of Program maintenance [2] Q.29 Write a program that inputs three integers from the keyboard, and prints the smallest and largest of these numbers. [2] 3
4 Q.30 Write a program to print the following using for loop. [3] Q.31 What is debugging? Explain types of Errors with suitable example. [4] Q.32 Write a program in C++ to check whether a given number is prime or not. [4] Q.33 Define and explain stages of program development process [4] Q.34 Draw a flow chart/c++ Program to print Sum of Digits of a given number. [4] e.g. if n=587, then output should be 5+8+7=20 ********* 4
5 KENDRIYA VIDYALAYA SANGATHAN, GUWAHATI REGION HALF-YEARLY EXAMINATION-2015 SUBJECT: COMPUTER SCIENCE(Marking-Scheme) Time Allotted: 3 hrs. Max. Marks: 70 Ans.1 Ternary/conditional operator (1 mark for correct answer) Ans X1024X1024X1024X1024 X8 bits OR 2 10 X2 10 X2 10 X2 10 X2 10 X8 bits 1 marks for correct calculation. Ans.3 (i) H/W (ii) S/W (iii) H/W (iv) S/W (1 marks for correct calculation). Ans.4 ALU CU (½ mark for each correct component) Ans.5 \n is a Newline OR Linefeed non-graphic character, it feeds the current line and place the cursor to the next line. \t is a Horizontal tab non-graphic character, it places a tab size space from the current position of the cursor. 1 mark for any 2correct differences Ans.6 Disk Defragmentation:It is a process which collect free space at one end and occupied spaces at other end in a contiguous block. Occupied spaces Free spaces Cache Memory:It is a type of memory.it places between processor and RAM in order to speed up access to data and instruction. CPU RAM CACHE MEMORY 1 mark for correct definition of Disk de-fragmentation. 1 mark for correct definition of Cache Memory. Ans.7 (i)agp Ports: Accelerated Graphics Port It is used to connect graphics card that provides high speed video performance.
6 (ii)ps-2: Personal System/2 It is used to connect mouse and keyboard to computer. ½ mark for correct expansion. ½ mark for correct explanation. Ans.8 Iostream.h and math.h 1 mark for each correct answer. Ans.9 float f=4.5; ½ mark for declaration ½ mark for initialization Ans.10 RAM Random Access memory Volatile in nature Read and Write operations can take place ROM Read Only Memory Non-Volatile in nature Only read operation can take place 1 mark for any two correct difference,1/2 for each difference Ans mark for correct answer Ans.12 (i)getch() conio.h (ii)isalnum() ctype.h (iii)sqrt() math.h (iv)random() stdlib.h 1/2 Mark for each correct answer Ans.13 (i) p=2*(l+b); (ii) z=2*pow((p/q),2)) or 2*p/q*p/q; (iii) s=1/2*m*v*v; or s=1/2*m*pow(v,2); (iv) x=(-b+sqrt(b*b-4*a*c))/2*a; 1/2 mark for each correct answer Ans.14 #include<iostream.h> main() int Sum;
7 for(i=0;i<2;i++) //i is undeclared for(j=0;j<=3;j++) //j is undeclared cout<<sum; //S must be capital //Closing brace for main is missing Corrected Code: #include<iostream.h> void main() int Sum; int i,j; for(i=0;i<2;i++) for(j=0;j<=3;j++) cout<<sum; ½ mark for finding and removing each error( 4 errors award 2 marks) Ans.15 break makes the compiler to transfer the control out of the loop... for... if here; true/false condition break; the control is transferred to "here" continue makes the compiler to execute the next iteration of the loop for condition.. continue; ****;
8 **** will not b executed and next iteration will happen continue can be given inside an if condition 1 mark for description, 1 mark example. Ans.16 When program formatting is done to make a program more readable is known as prettyprinting 2 mark for each correct definition. Ans.17 In entry controlled condition checks first then body executes. In exit controlled first execute then check condition. For and while entry controlled and do while exit controlled. 1 mark for each correct answer with suitable example. Ans.18 Robustness is the ability of a program to recover following an error and to continue operating within its environment 2 mark for correct definition. Ans.19 i. valid ii. Invalid iii. Invalid iv. Invalid ½ mark for each correct answer. Ans.20 (i) false (ii) true 1 mark for each correct output. Ans.21 An interpreter reads the program line by line and if it is error free it converts source program (HLL) into object program(lll) line by line
9 Whereas compiler reads the complete program in one go and if it is error free it converts source program into object program in one go. 1 mark for interpreter definition and 1 mark for compiler definition. Ans.22 Comments are the words or group of words written within the program to make program more understandable for readers. The Comments provides explanatory notes to the readers. e.g. // This program calculates Area of Circle void main() code for calculating area of circle Ans.23 1 mark for definition and 1 mark for any correct example. Implicit Type Conversion:. The process of converting one data type to another data type automatically by the System (Compiler). Ex: int m; char c= a ; m=c; Explicit Type Conversion: It is also known as type casting. The process of converting one data type to another data type explicitly (by the user). Ex: float f=(float)(m+n/3); 1 /2 mark for each correct definition ½ mark for each correct example. Ans.24 Binary Operators:These are the operators which needs two operands to operate.ex:+,-,*,/,% Unary operator: These are the operators which needs one operand to operate.ex:+,-- Ans.25 1 mark of each correct definition. 1mark for each correct example. (i) 0 1 2
10 Ans mark for each correct line of output. (ii) marks for correct output. (i)(0.375) 10 = (0.3) 8 1 mark for correct conversion. (ii)( ) 2 = (26.24) 8 1 mark for correct conversion. Ans.27 Operating System ia a program which provides a platform on which application runs. Functions of Operating System: (i)process Management(Management of different Programs/Applications) (ii)memory Management(Management of storage required for Applications) (iii)resource Management(Management of peripheral devices) 1 mark for correct points 1 mark for writing point s detail. Ans.28 Program maintenance refers to the modification of a program. Types of program maintenance are (i) Corrective maintenance : Correcting unexpected errors during compilation (ii) Adaptive maintenance: Correcting errors due to change in environment, Government policies, rules etc. (iii) Preventive maintenance: Correcting possible errors before they actually occurs (iv) Perfective maintenance: Maintaining existing system with new features,new facilities, new technologies 1 mark for correct points Ans.29 1 mark for writing point s detail. #include<iostream.h> #include<conio.h> main() clrscr();
11 int number1,number2,number3,largest,smallest; cout << "Input three different integers: "; cin >> number1 >> number2 >> number3; largest = number1; if ( number2 > largest ) largest = number2; if ( number3 > largest ) largest = number3; smallest = number1; if ( number2 < smallest ) smallest = number2; if ( number3 < smallest ) smallest = number3; cout<< "\nlargest is " << largest << endl; cout<< "\nsmallest is " << smallest; getch(); ½ mark for correct header files ½ mark for correct input 1 ½ mark for correct logic ½ for correct output ANS:30 #include<iostream.h> #include<conio.h>
12 main() clrscr(); int i,j; for(i=1;i<=5;i++) cout<< \n ; for(j=1;j<=i;j++) cout<<i; getch(); ½ mark for correct header files 2½ mark for correct logic Ans.31 A bug ig known as an error.debugging is a process to find and remove error. Types of Error: ERROR Compile-Time Logical Run-Time Syntax Semantics (i)compile Time: Error arises at compilation time. (a)syntax Error: occur when rules of programming language are misused. Ex:cout<<a //missing of semicolon (b)semantics Error: occur when statement is not meaningful Ex:b+c=a; //statement is not meaningful (ii)run time(execution Time): Error arises at Run/Execution time. Ex:b=a/0; //divide by zero error
13 (iii)logical: arises when output of a program is not according to the requirement of that program. Ex:a=b+c; //want to calculate multiplication in place of addition 1 mark for correct definition of debugging ½ mark for correct definition of compile time error ½ mark for correct example ½ mark for correct definition of run time error ½ mark for correct example ½ mark for correct definition of logical time error ½ mark for correct example Ans.32 #include<iostream.h> #include<conio.h> main() clrscr(); int i,a,p=0; getch(); cout<< Enter a number ; cin>>a; for(i=2;i<=(a/2);i++) if(a%i==0) if(p==0) cout<< No.is not prime ; p=1; break; cout<< No. is Prime ; ½ mark for correct header files
14 ½ mark for correct input 2 ½ mark for correct logic ½ for correct output Ans 33 Stages of program development process 1. Crack the problem 2. Code the algorithm 3. Compile the program 4. Execute the program Short descriptions of above points. 2 marks for correct points and 2 marks for detail of points Ans 34 Start Input n S=0 If n>0 Print s r=n%10 s=s+r End n=n/10 4 marks for correct flow chart/program 2 marks for partially correct flow chart or program
15 KENDRIYA VIDYALAYA SANGATHAN, GUWAHATI REGION HALF-YEARLY EXAMINATION-2015 SUBJECT: COMPUTER SCIENCE(Blue-Print) Time Allotted: 3 hrs. Max. Marks: 70 S.No. UNIT VSA SA I SA II LA TOTAL (1mark) (2mark) (3 mark) (4mark) 1. Computer Fundamentals : 3(3) 6(3) 9(3) 0(0) 18(9) 2. Programming Methodology: 2(2) 10(5) 0(0) 8(2) 20(9) 3. Introduction to C++ : 9(9) 4(2) 3(1) 8(2) 24(14) 4. Programming in C++ :(Flow of Control only) 0(0) 0(0) 0(0) 8(2) 08(2) TOTAL 14(14) 20(10) 12(4) 24(6) 70(34)
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