THE EMIRATES NATIONAL SCHOOL SHARJAH THIRD MODEL EXAMINATION COMPUTER SCIENCE ( Code : 083) ANSWER KEY

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1 THE EMIRATES NATIONAL SCHOOL SHARJAH THIRD MODEL EXAMINATION COMPUTER SCIENCE ( Code : 08) ANSWER KEY. MARK a) Function overloading is implemented in C++ programs when multiple functions are defined with the same name, but are differentiable by the number and/or type of arguments and these overloaded functions behave differently. Example: void area(int a, int b) cout<<a*b; void area(int a) cout<<a*a; void area(float a) cout<<(.4*a*a; void area(float b, float h ) cout<<0.5*b*h; b) i) getchar() ii) isalnum() c) class Music //error intmno; float Fees; public ://error void Register( ) cin>>mno>>fees; //error void Display( ) cout<<mno<<" : "<<Fees<<endl; void main() Music del; //error 4 del.register(); del.display(); d) $ 6$ $7$ 0$90 e) 6:0:0 6:4:00 f) Sizeof(L)=4 L=random(sizeof(L)) +5 L=random(4)+5 Random(4)=0,,, L=random(4)+5 = 5,6,7,8 P[L] can be O,G,R,A Loop is run four times Therefore possible value is O-R-A-G Minimum value of L =5 Maximum value of L= 8. a) Abstraction refers to the act of representing essential features without including the background details or explanations and encapsulation is wrapping up of data and function together in a single unit. Both these concepts are implemented in C++ in the form of a class. In a class data and function members are wrapped together. Only essential features are shown to the outside world in the form of member functions with public mode of accessibility. Example: class Rectangle int L, B; public: void Input() cin>>l>>b; void Output()

2 cout<<l<<b; b) i) Polymorphism /Constructor overloading Stream (int Sc, char S[], float F) StreamCode = Sc; strcpy(streamname, S); fees = F; ii) Stream S(, Science,8700); - In this statement object S is created by calling the parameterized constructor implicitly. Stream S = Stream(, Science,8700); - In this statement object S is created by calling the parameterized constructor explicitly ie. A temporary instance is created and the returned values are copied to the object S. c) class STATE long population,no_of_girls,tot_girls; char statename[0]; long CALC_PER() long per=no_of_girls/tot_girls*00; return per; public: STATE() strcpy(statename, NOT ALLOCATED ); void INSTATE() cout<< \nenter name of the state : ; cin.getline(statename,0); cout<< \nenter Population : ; cin>>population; cout<< \nenter Number of girls under 6 : ; cin>>no_of_girls; cout<< \nenter Total Number of girls under 6 : ; cin>>tot_girls; void OUTSTATE() long Percent; cout<< \nthe name of the state : <<statename; cout<< \npopulation : <<population; cout<< \nnumber of girls under 6 : <<no_of_girls; cout<< \ntotal Number of girls under 6 : <<tot_girls; Percent=CALC_PER(); cout<< Percentage <<Percent; d) i) Object of Dealer = 8 bytes and object of Accessories = 94 bytes ii) Multilevel Inheritance, Base class = AC, Derived class = Dealer iii) enterdetails(), showdetails(), price, enteraccessories(), showaccessories() iv) No_of_dealers, dealers_name, No_of _products, enterdetails(), showdetails(), Stablizer, Ac_cover, price, enteraccessories(), showaccessories(), entercar(), showcar(). a) Row major Address of A[i][j] = B + W[n(i-lr) + (j-lc) ] m= 5-(-)+=7, n=5-(-)+=8, W =4, lr =-, lc=-,i=,j= 000 = B+ 4 [ 8 * (-(-)) + ( (-))] 000 = B + 4 [ ] 000 4*58 = B 4 4

3 B = 768 To find address of element T[0][5], Address of A[i][j] = B + W[n(i-lr) + (j-lc) ] Address of (T[0][5]) = [ 8 * (0-(-)) + (5 (-))] = * [8 * + 7] = = 408 Total number of elements stored in T = (no.of rows * no. of columns)=7*8 = 666 Total number of bytes allocated to T = 666 * 4 = 664 bytes b) struct IPL int Score; char Teamname[0]; void SORTSCORE(IPL a[], int n) int big; IPL temp; for ( int i= 0; i < n-; i++) big = a[i]; P=i; for ( int j = i+; j< n; j++) if ( a[j].score > a[big].score) big =a[j] ; p=j; temp = a[i]; a[i] = a[p]; a[p] = temp; for( int i=0;i<n;i++) cout<<a[i]<< \t ; c) struct Game char Gamename[0]; int numofplayer; Game *next; class Queue Game *f,*r; public : Queue() f=r= NULL; void PUSH(); void POP(); void Queue::PUSH() char ch; Begin: Game *p; p=new Game; cout<<"enter Game name : "; cin>>p->gamename; cout<< Enter Number of Players ; cin>>p->numofplayer; 4

4 p->next =NULL; if (r==null) f=r=p; r ->next=p; r=p; cout<< \n Do you want to add ; cin>>ch; if (ch== Y ch== y ) goto Begin; void Queue::POP() if ( f==null) cout<< Queue is empty ; Game *p; p=f; cout<< The popped item <<endl; cout<<p->gamename<< \t <<p->numofplayer<<endl; f=f->next; delete p; d) int sum(int a[][0],int n) int sum =0; for(inti=0;i<n;i++) for(int j=0;j<n;j++) if(((i==j) (i+j==n-))&&a[i][j]>0) sum+=a[i][j]; cout<< sum <<sum; e) Stack elements Stack operation output / 5 * < push push push 5 push 6 push push 5 Ans : 0 4. a) a) file.seekg(0,ios::beg); b) 6 bytes b) void wordcount() char ch[80]; int count=0; ifstream f; f.open( VOWEL.TXT ); if(!f) cout << File not found ; while (f>>ch) if(strlen(ch) == ) count++;, 5 5,5 0 0,6 0,6, 0, 0,,5 0,0 0 (FALSE)

5 cout<< number of three letter words = <<count; f.close(); c) void Disp() int flag=0; CAR C; ifstream f("car.dat"); if(!f) cout<<"file does not exist"<<endl; while(f.read((char*)&c,sizeof(c))) if(c.return_milage()>=00 && C.RETURN_Milage()<=50) C.display(); flag=; f.close(); if(flag==0) cout<<"record does not exist"<<endl; 5. a) Degree is the number of attributes or columns present in a table. Cardinality is the number of tuples or rows present in a table. Degree of the given table = 4 Cardinality of the given table = 5 b) i) SELECT NAME FROM COMPANY WHERE COMPANY.CID = CUSTOMER.CID AND PRICE < 0000; ii) SELECT NAME FROM COMPANY ORDER BY NAME DESC; iii) UPDATE CUSTOMER SET PRICE = PRICE WHERE NAME LIKE S% ; iv) ALTER TABLE CUSTOMER ADD TOTALPRICE DECIMAL(0,); v) vi) vii) COUNT(*) CITY DELHI MUMBAI MADRAS MIN(PRICE) MAX(PRICE) / / AVG(QTY) / viii) PRODUCTNAME CITY PRICE MOBILE MUMBAI MOBILE MUMBAI 5000 /

6 6 a) The principle of duality pronounces that given an expression which is always valid in Boolean algebra, the dual expression is also always valid. The dual expression is found by replacing all + operations with (.), all (.) operation with (+) all 's by 0's, and all 0's by 's. The principle of duality is an important concept in Boolean algebra, particularly in proving various theorems. The principle of duality is used extensively in proving Boolean algebra theorem. Once we prove that an expression is valid, by the principle of duality, its dual is also valid. Hence, our effort in proving various theorems is reduced to half. b) ((X.Y) + (X.Y ) ) c) G(P,Q,R) = (P+Q+R)(P+Q +R)(P+Q +R )(P +Q +R) d) a) i)not affected by any kind of noise (ii) High transmission capacity F(w, x, y, z) = x z + xz + yz b) b) SENIOR ADMIN JUNIOR HOSTEL b) Server should be placed in ADMIN as it has the maximum number of computers,so that the total cable length can be made minimum reducing cabling cost and increasing efficiency b) ) SENIOR R JUNIOR R ADMIN HOSTEL R-Repeater Repeater should be placed between SENIOR and JUNIOR buildings to amplify the signals as the distance between these buildings is more than 70m. ) Switch/Hub should be placed in all the buildings to connect the computers within each building. (Draw the hub/switch in each building) SENIOR JUNIOR ADMIN HOSTEL b4) Radio Waves

7 c Domain name - income.in URL - d Web hosting is the service that makes our website available to be viewed by others on the Internet. A web host provides space on its server, so that other computers around the world can access our website by means of a network or modem. e Trojan Horse- It is a code hidden in a program such as games or spreadsheet that may look safe but can carry out malicious operations without the user s knowledge(like a false file listing command which destroys files instead of displaying the list) Virus A malicious program that requires a host. It can spread from computer to computer by attaching themselves to program files or hard drive or data files. f A system designed to prevent unauthorized access to or from a private network is called firewall. it can be implemented in both hardware and software or combination of both. g File Transfer Protocol

THE EMIRATES NATIONAL SCHOOL SHARJAH THIRD MODEL EXAMINATION 2015 COMPUTER SCIENCE ( Code : 083) CLASS : XII MAX MARKS: 70

THE EMIRATES NATIONAL SCHOOL SHARJAH THIRD MODEL EXAMINATION 2015 COMPUTER SCIENCE ( Code : 083) CLASS : XII MAX MARKS: 70 DATE :25/03/2015 TIME: 3Hrs Instructions: (i) All questions are compulsory. (ii)