KENDRIYA VIDYALAYA SANGATHAN ERNAKULAM REGION FIRST COMMON PRE BOARD EXAMINATION MARKING SCHEME. Code No. 083

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1 KENDRIYA VIDYALAYA SANGATHAN ERNAKULAM REGION FIRST COMMON PRE BOARD EXAMINATION MARKING SCHEME CLASS:XII SUB:COMPUTER SCIENCE Code No. 083 TIME:3 Hrs MaxMarks:70 (a) Call by value : In this method, actual arguments of a calling function gets duplicated as formal arguments & are made available to the called function. As a result whatever change is made by called function in these arguments, are not reflected back in actual arguments. ½ mark Ex. void SWAP(float a, float b) { a=a+b; b=a-b; a=a-b; cout<<"\n << a="<<a<<" and b="<<b; ½ mark Call by reference : In this function call, the reference of actual arguments are provided to the called function i.e. memory location of actual arguments. As a result, any change made in these arguments by called function is reflected back to the actual arguments. ½ mark Ex. void SWAP(float &a, float &b) { a=a+b; b=a-b; a=a-b; In the function s argument we simply specify & operator along with the argument(s) which are being called by reference. ½ mark (b) i) conio.h ii) stdio.h ½ mark each (c) #include<iostream.h> #include<string.h> typedef char string[40]; void main( ) { string S= Australia ; int/long L=strlen(S); cout<< String <<S<< has <<L<< Characters <<endl; (½ Mark for each correction) OR ( Mark for identifying at least three errors, without suggesting correction) d) Ans: 7#9#5# 30#50#70#40# 200#200# ( mark for each line) e) Ans: sbojerets (2 marks for the correct answer) f) Ans: Option ii) is the possible output- The variable max will have the values from 0 to 3 only so Mynum can have values 20 to 23. ( mark for correct answer mark for justification) 2 a) CONSTRUCTOR OVERLOADING : When more than one constructor appears inside a single class, it is said to be constructor overloading i.e. if we have two or more constructors

2 inside a class, it is said to be an overloaded constructor. For ex. class ABC { private: int x; float y; public: ABC( ) //default constructor { x=5; y=0.0; ABC(int p, float q) //parameterized constructor { x=p; y=q; ABC(ABC &t) //Copy constructor { x=t.p; y=t.q; void INABC( ); void OUTABC( ); ; Here in the above written example, we see three constructors one after another. Since all of them share the same class name and are different in their type and number of arguments, therefore supports overloading mechanism of OOP. ( Mark for the correct explanation) ( Mark for the suitable example) b) i) Answer : In object oriented programming function 2 is referred as a parameterized constructor. When we create a object of class and pass required parameter during creating then this is invoked/called. Example of calling : ROOM r(30); // calling of parameterized constructor. (½ Marks for each correct answer and ½ marks for calling statement) ii) Answer : In Object Oriented Programming function 4 is referred as a destructor. It is invoked/called automatically when ever the scope of the object is end. Example : We doesn t called it explicitly it is called when the scope of object it end like this : void main () { ROOM r; { ROOM r2(30); // call destructor function for object r2 ( Mark for each correct answer through explanation OR example) c) Ans. class TAXPAYER {private: char Name[30],PanNo[30]; float Taxabincm; double TotTax; void CompTax() 2

3 { if(taxabincm >500000) TotTax= Taxabincm*0.5; else if(taxabincm>300000) TotTax= Taxabincm*0.; Else if(taxabincm>60000) TotTax= Taxabincm*0.05; else TotTax=0.0; public: TAXPAYER(char nm[], char pan[], float tax, double ttax) //parameterized constructor { strcpy(name,nm); strcpy(panno,pan); Taxabincm=tax; TotTax=ttax; void INTAX() { gets(name); cin>>panno>>taxabincm; CompTax(); void OUTAX() { cout<<name<< \n <<PanNo<< \n <<Taxabincm<< \n <<TotTax<<endl; ; (½ Mark for correct syntax of class header) (½ Mark for correct declaration of data members and member function) ( Mark for checking all conditions and calculating total tax in CompTax( )) (½ Mark for correct default constructor( ) with proper invocation ) (½ Mark for correct parameterized constructor) (½ Mark for correct OUTTAX ()for displaying the data on screen ) (½ Mark for correct INTAX() for getting data from user) d) i) PGENROL( ), PGSHOW( ) ii) None iii) Mname[25], year, age & marks iv) 3 bytes ( Mark for correct answer) 3 a) Solution: void accept(int a[ ],int size) { for (int i=0;i<size;i++) { if (a[i]%2= =0) a[i]=a[i]/2; else a[i]=a[i]*2; cout<<a[i]<<, ; ( Mark for correct Function Header with proper Arguments) ( Mark for correct loop) ( Mark for replacing array elements with proper values) b) Solution: Given Data: Arr[5][20] W=4 B=? R=5 C=20 L r = 0 L c = 0 Address of Arr[3][2] =? 3

4 Address of Arr[5][2] = 500. Address of an element (I,J) in row major = B+W(C(I-L r )+(J-L c )) Therefore, 500 = B+4(20(5-0)+(2-0)) 500 = B+4(20*5+2) 500 = B+4* = B+408 B = B=092 Address of Arr[3][2] =092+4(20*3+2) =092+4(62) = =340. ( Mark for calculating correct Base Address) ( Mark for writing correct formula/correct substituted values, for row major properly, for calculating Address of Arr[3][2]) ( Mark for calculating correct Address of Arr[3][2]) c) Solution: NODE *top=null; //declaring global pointer & initializing it with NULL void PUSH( ) { NODE *p=new NODE; //creating new dynamic list to go on to stack cout<< \nenter Name : ; gets(p->name); cout<< \nenter Fees : ; cin>>p->fees; p->next=null: if(top= = NULL) top=p; else { p->next=top; top=p cout<< \nlist inserted on the top of stack successfully ; getch( ); void POP( ) { if(top= = NULL) cout<< \nstack Empty ; else { NODE *temp=top; top=top->next; delete temp; cout<< \nlist deleted from top of stack successfully ; getch( ); ( Mark for correct function header) (½ Marks for creating new node and assigning value to it) (½ Mark for linking new node at last of link list) d) Solution: void accept(int a[ ][ ],int size) { cout<<"middle Row:"; 4

5 for (int i=0;i<size;i++) for(int j=0;j<size;j++) if (i= = size/2) cout<<a[i][j]<< \t ; cout<<"\n Middle Column:"; for (i=0;i<size;i++) for(j=0;j<size;j++) if(j= =size/2) cout<<a[i][j]<< \t ; (½ Mark for correct function header) (½ Mark for correct loop) (½ Mark for space printings) (½ Mark for printing rows elements& column elements) e) SOLUTION. (by tabular method): Steps INPUT ACTION STACK 30 Push # Push #30,6 3 4 Push #30,6,4 4 + Pop 4,6 & Push 6+4=0 #30,0 5 / Pop 0,30 & Push 30/0=3 #3 6 4 Push #3,4 7 + Pop 4,3 & Push 3+4=7 #7 8 4 Push # 7,4 9 * Pop 4,7 & Push 7*4=68 #68 Ans. 68 (For correct logic give ½ marks. 4 a) Ans. : For correct stack drawing give ½ marks. For correct answer give marks. ) INOUT.seekp(-*sizeof(school),ios::cur); OR INOUT.seekg(-*sizeof(school),ios::cur); ( Mark for correct answer) b) Ans.: void countvowel( ) { int c=0; char ch; ifstream fin( STRINGS.TXT ); while(!fin.eof( )) { fin.get(ch); OR fin>>ch; if(!fin) 5

6 c) Ans.: break; switch(ch) { case A : case a : case E : case e : case I : case i : case O : case o : case U : case u :c++; cout<< \ntotal vowels in the data file is <<c; fin.close( ); (½ Mark for opening file in the correct mode) (½ Mark for reading the content from the file and the loop) (½ Mark for correct comparison) (½ Mark for initialization and increment of the counter(variable)) void DELREC(long m) { TV ob; ifstream fin("tv.dat",ios::binary); ofstream fout("temp.dat",ios::app ios::binary); int flag=0; while(!fin.eof()) //for searching record { fin.read((char*)&ob,sizeof(tv)); if(!fin) break; if(ob.retmodel( )= =m) { flag=; break; if(!flag) { cout<<"\nrecord does not exist"; getch(); else //for deleting record { fin.seekg(0); while(!fin.eof()) { fin.read((char*)&ob,sizeof(tv)); if(!fin) break; if(ob.retmodel( )= =m) fout.write((char*)&ob,sizeof(ob)); remove("tv.dat"); rename("temp.dat","tv.dat"); cout<<"\nrecord DELETED SUCCESSFULLY..."; getch(); 6

7 fin.close(); fout.close(); (½ Mark for opening file in the correct mode) (½ Mark for the reading content from file) ( Mark for the correct loop) (½ Mark for writing content to the file) (½ Mark for deleting the record ) 5 a) Answer : DOMAIN : A domain is a pool of values from which the actual values appearing in a given column are drawn. A domain is said to be atomic if element of the domain are considered to be indivisible units. In other word we can say that a legal value range of a field is called its domain. Mean all the value which are allowed in a attribute is create its domain. Tuple : A row in a relation is called tuple. ( Markfor correct definition for Domain) ( Markfor correct definition for Tuple) b) i) Ans: SELECT * FROM Client WHERE City = Delhi ; ( ½ Mark for correct SELECT) ( ½ Mark for correct WHERE clause) ii) SELECT * FROM Product WHERE Price >=50 AND Price <=00; OR SELECT * FROM Product WHERE Price BETWEEN 50 AND 00; ( ½ Mark for correct SELECT) ( ½ Mark for correct WHERE clause) iii) SELECT ClientName, City, ProductName, Price, Client.P_ID FROM Client, Product WHERE Client.P_ID = Product.P_ID; ( ½ Mark for correct SELECT) ( ½ Mark for correct WHERE clause) iv) UPDATE Product SET Price = Price +0; ( ½ Mark for correct SELECT) ( ½ Mark for correct WHERE clause) v) DISTINCT City Bangalore Delhi Mumbai (½ Mark for correct output) OR ( ½ Mark for mentioning Address is not a Column in the Table Client OR mentioning ERROR) vi) Manufacturer MAX(Price) MIN(Price) Count(*) ABC LAK XYZ (½ Mark for correct output) 7

8 vii) ClientName Manufacturer Cosmetic Shop ABC Total Health ABC Live Life XYZ Pretty Woman XYZ Dreams LAK (½ Mark for correct output) OR ( ½ Mark for mentioning ManufactureName and Prod_Id are not valid Column in the respective Tables) viii) ProductName Price * 4 Talcom Powder 60 Face Wash 80 Bath Soap 220 Shampoo 480 Face Wash 380 (½ Mark for correct output) 6 a) Ans.: Absorption law states that : a) x+xy=x b) x(x+y)=x Verification: x+xy=x LHS=x+xy by distributive law =x(+y) since +y= =x. since.x=x =x = RHS, hence verified OR (using truth table) x y xy x+xy Here column x and x+xy are identical, hence proved. ( Mark for correct verification) b) Ans. Add extra column for min term and write min terms for the rows which have output (G) as :- A B C G Min Term A B C 0 0 A B C AB C ABC Now sum all the min terms to get the SOP as: Therefore SOP of F(A,B,C)= A B C +A B C+AB C +ABC ( Marks for correct Table) ( Mark for correct SOP form) 8

9 c) Ans: X Y Z (2 Marks for correct logic circuit ½ Marks if some part is correct) d) K-MAP Ans: A B A B AB AB C D C D CD CD F= C.D + A.C + B.D (½ Mark for drawing correct K-Map ) (½ Mark for plotting s correctly) ( Mark for correct grouping) ( Mark for correct Answer) 7 a) Packet Switching: It refers to protocols in which messages are broken up into small packets before they are sent. Each packet is transmitted individually across the net. Each packet has header information which enables to route the packet to its destination. At the destination the packets are reassembled into the original message. Message Switching: In this technique, first the source computer transfers data to the buffer of switching office computer. Further it looks for a free link to another switching office, and then the data are transferred to this link. (½ Mark for correct definition of message switching and packet switching ) (½ Marks for difference ) b) i) PHP Hypertext Preprocessor ii) SMSC Short Message Service Center (½ Mark for each correct expansion) c) INFRARED : Infrared technology allows computing devices to communicate via short-range wireless signals. The infrared transmission technology used in computers is similar to that used in consumer product remote control units. This ray transmits digital data bi-directionally through the air and can propagate throughout a room, but will not penetrate walls. d) SPAM :- It refers to electronic junk mail or junk newsgroup postings. Some people define it as any unsolicited . e) Proprietary software: It refers to any computer software that has restrictions on any combination of the usage, modification, copying or distributing modified versions of the software. Proprietary software usually can be distributed at no cost or for a fee. Proprietary software may also be called closed-source software. In other words it is neither open nor freely available. f) Web hosting: It is a way of hosting web-server application on a computer system through which electronic content on the internet is readily available to any web-browser client. 9

10 g) Answer : (i) Suitable cable layout is II 2 II 4 20m 25m 30m II II 3 ( Mark for correct diagram) (ii) As the rule, th server should be placed in the building with maximum number of computers. Thus, we suggest that the server should be placed in Wing II 3. (½ Mark for mention of rule and ½Marks for mention of correct answer Wing II 3) (iii) II 2 II 4 20m Hub/switc 25m 30m Hub/switc II II 3 Hub/switc Hub/switc ( Mark for the correct device) (iv) Dial up Network Broad Band Cable(Twisted paid or Fiber optical or Coaxial cable) ISDN Radio Wave (½ Mark for each correct device or cable) (½ Mark for the any equivalent correct technology) 0

KENDRIYA VIDYALAYA SANGATHAN ERNAKULAM REGION FIRST COMMON PRE BOARD EXAMINATION

KENDRIYA VIDYALAYA SANGATHAN ERNAKULAM REGION FIRST COMMON PRE BOARD EXAMINATION CLASS:XII SUB:COMPUTER SCIENCE Code No. 083 Instructions: (i) All questions are compulsory. (ii) Programming language: C++