KE DRIYA VIDYALAYA SA GATHA, CHE AI REGIO. COMMO PRE-BOARD EXAMI ATIO COMPUTER SCIE CE (CLASS-XII) MARKI G SCHEME

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KE DRIYA VIDYALAYA SA GATHA, CHE AI REGIO. COMMO PRE-BOARD EXAMI ATIO 2008-09. COMPUTER SCIE CE (CLASS-XII) MARKI G SCHEME 1 1. (a) (i) Arrays bring together a group of items of the same data type whereas structures bring together a group of related data items of any data type. (ii) Structures are declared with Keyword struct. By default, all members are public in a structure, on the other hand, all members are private by default in classes. (b) (i) #include<ctype.h> (ii) #include<string.h> (c) Error 1: Reference Variable b must be initialized. Error 2: Illegal operator >> for cout #include<iostream.h> void large(int &a, int &b); int main() int a=17; int x=2; int &b=x; large(b,a); cout<<a<<b; void large(int &a, int &b) if(a>b) a = -1; b = -1; (d) or *-* bys (3) (e) Result = 40 (2) (f) 20 400 420 (2) 20 20 420 2. (a) Inheritance is the capability of one class of things to inherit capabilities or properties from another class. When a subclass inherits only from one base class, it is known as single inheritance. X Y

2 When a subclass inherits from multiple base classes, it is known as multiple inheritance. X Y Z (b) B=7 C=5 A=3 B=7 C=6 B=2 C=7 (c) #include<iostream.h> #include<stdio.h> #include<string.h> class QUARTERS private: char name_of_society[20]; int house_number; int number_of_members; char flat[10]; float income; public: void input() cout<< \nenter Name of Society: ; gets(name_of_society); cout<< \nenter house number: ; cin>>house_number; cout<< \nenter number of members: ; cin>>number_of_members; cout<< \nenter Income: ; cin>>income; void alloc_flat() if(income>=50000) strcpy(flat, A Type ); (2)

if((income>=25000)&&(income<50000)) strcpy(flat, B Type ); if(income<25000) strcpy(flat, C Type ); cout<< \n\nthe allocated flat is : <<flat; void show() cout<< \nname of Society : <<name_of_society; cout<< \nhouse Number : <<house_number; cout<< \nnumber of members: <<number_of_members; cout<< \nflat Type : <<flat; cout<< \nincome : <<income; ; 3 (d) (i) Multilevel Inheritance (ii) 76 bytes (iii) Data members : No_of_players, Venue (iv) Member functions: initialize(), olympicsentry(),olympicsdisp(), outdoorentry(), outdoordisp(), hockeyentry(), hockeydisp() 3(a) void Swap(int A[],int size) int i,j,tmp,mid=size/2; if(size%2==0) j=mid; j=mid+1; for(i=0;i<mid;i++,j++) tmp=a[i]; A[i]=A[j] * 2; A[j]=tmp; (b) MAT[I][J]=B+W((I-L1)+M(J-L2)) MAT[2][4]=B+2((2-1)+20(4-1)) 1000=B+2(1+20X3) 1000=B+122 B=1000-122 B=878 MAT[15][10]=978+2((15-1)+20(10-1)) =978+2(14+20X9)

(c) struct NODE int data; NODE *link; ; =978+388 =1366 4 Node * del_q(node *front, int &val) NODE * temp; clrscr(); if(front==null) cout<< Queue Empty ; val=-1; temp=front; front=front->link; val=temp->data; temp->link = NULL; delete temp; return (front); (d) void sum(int A[5][5],int m,int n) int i,j,sumrow[5],sumcol[5]; for(i=0;i<m;i++) sumrow[i]=0; for(j=0;j<n;j++) sumrow[i]+=a[i][j]; for(j=0;j<n;j++) sumcol[j]=0; for(i=0;i<m;i++) sumcol[j]+=a[i][j]; cout<< \nrow sum=\n ; for(i=0;i<m;i++) cout<<sumrow[i]<< \n ;

for(i=0;i<n;i++) cout<<sumcol[i]<< \n ; 5 (e) Operation Stack Status Output ( ( True ( True && (&& False (&& True False ) True False && True False &&!! True False && (! ( True False && False!( True False && False! ( True False && False True! ( True False && False True )! True False && False True True False && False True! Empty True False && False True! 4.(a) tellg()- used in random file access. It is used to return the current file pointer in terms of bytes. seekg()- used to place the file pointer at the required position. Both comes under ifstream ½ mark for each correct definition (b) void calculate() fstream f; clrscr(); f.open( TOUR.DAT,ios::in); char arr[80]; char ch; int i=0,sum=0,n=0; while(f) f.get(ch); arr[i]=ch; i++; if(ch== ) i--; i=0; n++; cout<< Total no. blanks: <<n; (b) void write()

PRODUCTS P; fstream f; f.open( PROD.DAT,ios::app ios::out ios::binary); int n,i; clrscr(); cout<< \nenter how many records to be added: ; cin>> n; for(i=0;i<n;i++) P.getproducts(); f.write((char*)&p,sizeof(p)); f.close(); void read() PRODUCTS P; fstream f; f.open( PROD.DAT,ios::in ios::binary); char prodname[20]; while(f) f.read((char*)&p,sizeof(p)); strcpy(prodname,p.getproduct()); if(strcmp(prodname, Ponds Powder )==0) cout<< \n ; P.showproducts(); f.close(); 6 5. (a) A view is a virtual table that does not really exist but is instead derived from one or more underlying base table(s). No stored file is created to store the contents of a view rather its definition is only stored. If we drop a table in which view exists the view is automatically deleted. Yes, we can create View of a View. (½ ) (b) [6X1=6] (i) SELECT Title FROM LIBRARY WHERE Price BETWEEN 100 AND 300; (ii) SELECT Title, Author FROM LIBRARY WHERE Subject = Prog ; (iii) SELECT * FROM LIBRARY WHERE Price >300 ORDER BY Quantity; (iv) UPDATE LIBRARY SET Price = Price + 315; (v) 213 (vi) 4370

(vii) 3 (viii) SUBJECT MAX(PRICE) COUNT(*) ------------- -------------------- --------------- DBMS 230 1 DS 320 1 Prog 515 3 7 6. (a) To Prove : X(X+Y)=X = X.X+X.Y = X+XY = X(1+Y) = X.1 = X = RHS (b) X.Y + Z Z + X.Y [ X+Y=Y+X ] (Z+X)(Z+Y ) [ X+YZ=(X+Y)(X+Z)] Figure : [1] (c) G(U,V,W)= (U+V +W)(U+V +W )(U +V +W)(U +V +W ) (d) Z W Z W ZW ZW X Y 1 1 1 [4X½=2] K-Map X Y 1 1 1 1 XY 1 1 XY 1 1 1 There are 4 groups : 1 Octet, 2 Quads, 1 Pair The Octet (m1+m3+m5+m7+m9+m11+m13+m15 reduces to W The Quad1(m0+m1+m4+m5) reduces to X Z The quad2(m4+m5+m6+m7) reduces to X Y The pair(m0+m11) reduces to XY Z Therefore final expression is : F(X,Y,Z,W)=W+X Z +X Y+XY Z 7. (a) A gate way is a device that connects dissimilar networks. (b) (i) File Transfer Protocol (ii) Network File Server (iii) Hyper Text Transfer Protocol

(iv) Uniform Resource Locator (c ) Cookies are messages that a web server transmits to a web browser so that the web browser can keep track of the user s activity on a specific web site. (d) (a) Total cable length required for this layout = 75 mts 8 GV1 GV3 GV2 GV4 (b) To give dedicated bandwidth, the computers in each building should be connected via switches as switches offer dedicated bandwidths. (c) By installing routers in each building, shared internet access can be made possible. (d) (i) Satellite as it can connect offices across globe. (ii) WAN ( Wide Area Network )