COMMON PRE-BOARD EXAMINATION COMPUTER SCIENCE

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Diane Goodwin
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1 COMMON PRE-BOARD EXAMINATION COMPUTER SCIENCE SET 1 Subject Code: 083 CLASS XII MARKING SCHEME Time Allowed: 3 hours Maximum Marks: 70 Q.No. Answer Marks 1. a Macro User defined function 2 Macros are built on the #define User defined functions are pre-processor. Macros are compiled before they executed. executed before the source code is compiled. #define SQUARE(x) x * x void SQUARE(x) int Y=x * x; cout<<y; b <iostream.h> //cin, cout 1 <stdio.h> //getchar(), putchar() c int NEXTPLANE (char PLANE[][20], int CALL[], int DUE[] ) 2 int COUNT ; cout<< PLANE [0]; COUNT=0; while (!strcmp( PLANE[COUNT],"ZZZ")) PLANE [COUNT] =PLANE [COUNT+1]; CALL [COUNT] =CALL [COUNT+1]; DUE [COUNT] =DUE [COUNT+1]; COUNT=COUNT+1; return COUNT; d 2 BAJJAB#-1 e Try.#0#0#0 3 it is.#1#4#5 5&4 Page 1 of 9

2 f Option (ii) and (iv) 2 (ii) (iv) TVM@ 2. a Abstraction refers to the act of representing essential featues without 2 including the background details or explanation. You are driving a car. You only know the essential features to drive eg. Gear handling, streeing handling, use of clutch breaks etc. But while driving do you get in to internal details of a car like wiring, motor working etc? What is happening is hidden from you. This is abstraction wher you only the essential things. b i) constructor overloading 2 Travel T1; // call for function 1 Travel T2(5); //call for function 3 ii) Function 4 is a destructor function, which get executed when an object goes out of its scope. Function 4 will not take any arguments. c class BODYMASSINDEX 4 char Name[25]; float Height; float Weight; float BMI; public: BODYMASSINDEX(); void INPUT(); void CALCULATEBMI(); void DISPLAY(); ; BODYMASSINDEX: :BODYMASSINDEX() Height=0.0; Weight=0.0; BMI=0.0; void BODYMASSINDEX: :INPUT() cin>>name; cin>>height; cin>>weight; void BODYMASSINDEX: : CALCULATEBMI() BMI= Weight/( Height*Height); void BODYMASSINDEX: :DISPLAY() Page 2 of 9

3 if(bmi>30) cout<< Extremely Fat ; else if(bmi>25 && BMI<=30) cout<< Fat ; else if(bmi>22 && BMI<=25) cout<< Over Weight ; else if(bmi>=20 && BMI<=22) cout<< Normal Weight ; else cout<< Under Weight ; void main() BODYMASSINDEX BI; BI.INPUT(); BI. CALCULATEBMI(); BI.DISPLAY(); d i) NO 4 ii) 69 bytes iii) void getfruitproducts(); void putfruitproducts(); void getproduct(); void putproduct(); iv) Hierarchical Inheritance 3. a TRANSFER( int CODE1[], char COUNTRY1[][25], int CODE2[], char 3 COUNTRY2[][25], int CODE3[], char COUNTRY[][25], int M, int N) int LIC1=0,LIC2=0,LIC3=0; while(lic1<m && LIC2<N) if(code1[lic1]<=code2[lic2]) CODE3[LIC3]=CODE1[LIC1]; strcpy(country3[lic3],country1[code1]); LIC3++; LIC1++; else CODE3[LIC3]=CODE2[LIC2]; strcpy(country3[lic3],country2[code1]); LIC3++; LIC2++; if(lic1>n-1) Page 3 of 9

4 while(lic2<m) CODE3[LIC3]=CODE2[LIC2]; strcpy(country3[lic3],country2[code1]); LIC3++; LIC2++; if(lic2>m-1) while(lic1<n) CODE3[LIC3]=CODE1[LIC1]; strcpy(country3[lic3],country1[code1]); LIC3++; LIC1++; b Column Major: 3 MAT[i][j]=B + W[(i-0)+nr(j-0)] rr= 20 W=2 MAT[2][4]=B +2[(2-0)+20(4-0)] 1000 = B +164 B= B=836 MAT[15][10]= [(15-0)+20(10-0)] = = Total number of elements: 240 c Void AddEnd3(int X[][5],int R, int C) 2 int sum=0; for(int i=0;i<r; i++) for(int j = 0j<C;j++) if(x[i][j]%10==3) sum+=x[i][j]; cout<<sum; d ((P-Q)/(R*(S-T)+U)) 2 Element Action Stack status Output Page 4 of 9

5 ( Push ( ( Push (( P Print (( P - Push ((- P Q Print ((- PQ ) Pop - ( PQ- Remove ( / Push (/ PQ- ( Push (/( PQ- R Print (/( PQ-R * Push (/(* PQ-R ( Push (/(*( PQ-R S Print (/(*( PQ-RS - Push (/(*(- PQ-RS T Print (/(*(- PQ-RST ) Pop (/(* PQ-RST- Remove ( + Pop * (/(+ PQ-RST-* Push + U Print (/(+ PQ-RST-*U ) Pop + (/ PQ-RST-*U+ Remove ( ) Pop / Empty PQ-RST-*U+/ Remove ( e void QUEUE::DELETE() 4 Node *ptr; if (Front==NULL) cout<< Queue underflow ); else if(front ===Rear) ptr=front; front=rear=null; delete ptr; else ptr=front; Front=Front->Link; delete ptr; 4. a Statement 1 : File.tellg(); 1 Statement 2: File.seekp(-Filepos,ios::cur); b void CountYouMe() 2 char word[25]; ifstream Fin( MESSAGE.TXT ); Fin.getline(word,25, ); while(!fin.eof()) Page 5 of 9

6 if(strcmp(word, you )==0 strcmp(word, me )==0) count++; Fin.getline(word,25, ); cout<< Count of you/me in file : <<count; Fin.close(); c void searchlaptop() 3 ifstream Fin; Fin.open( LAPTOP.DAT,ios::binary); LAPTOP L; long mn; cout<< Enter the model Number ; cin>>mn; Fin.read((char *)&L,sizeof(L)); while(!fin.eof()) if(mn==l. ReturnModelNo ( )) L. StockDisplay ( ) ; goto end; Fin.read((char *)&L,sizeof(L)); cout<< Model not found ; end: getch(); Fin.close(); 5. a Intersection operation will select tuples which are common in both the tables. Example: RS Name Age Class Nisha Nishant 14 9 YRS Name Age Class Nishant 14 9 Niki 13 8 RS YRS Page 6 of 9

7 Name Age Class Nishant 14 9 b i SELECT Model, Size of engine, Amount FROM CAR, PRICE WHERE 1 CAR.Registration = PRICE.Registration; ii SELECT Registration, Make FROM PRICE WHERE Amount BETWEEN AND 23000; iii SELECT * FROM CAR ORDER BY Model asc; 1 iv SELECT Make, MAX(Amount), MIN(Amount) FROM PRICE GROUP 1 BY Make; c i COUNT(*) COUNT(DISTINCT Make) ½ 7 3 ii Model ½ Zafira Jetta Vectra iii Make COUNT(*) ½ Opel 1 Volkswagen 3 Renault 2 iv Model Size of engine ½ Vectra 1.8 Zafira 2.0 Polo 1.2 Megane 1.4 Clio a X + YZ = (X + Y)(X + Z) 2 Distributive law (X + Y)(X + Z)= XX + XZ+XY+YZ X+XZ+XY+YZ (XX=X Indempotance law) X+XY+XZ+YZ X(1+Y)+Z(X+Y) (1+Y = 1 property of 0 and 1) X.1+Z(X+Y) X+XZ+YZ X(1+Z)+YZ (1+Z=1 property of 0 and 1) X.1+YZ (X.1 =X property of 0 and 1) X+YZ b A B C X Page 7 of 9

8 c (X+Y ) (X +Y ) 1 d F(P,Q,R,S,T) = (1, 3, 4, 5, 6, 7, 12, 13) 3 R S R S RS RS 1 1 P Q P Q PQ PQ P Q +QR +P S 7. a Internet Mail Access Protocol 1 Subscriber Identity Module b i) A hub 1 ii) A switch iii) A proxy server iv) A bridge c Virtual keyboard to enter username and password 1 Encrypted smart cards to access the account Biometric sensors to access the account d Mobile Telephone Laptop 1 Advantage Internet can be Internet can be accessed anywhere accessed wherever Wi-Fi available Disadvantage Difficult to read the Difficult to carry internet pages everywhere e i) Switch 1 ii) Admin block, since it has maximum number of computers 1 iii) Star Topology Page 8 of 9

9 Customer Care Block 1 Admin Block HRD Block Conference Block iv) Satellite 1 f Open Source Software Free Software 1 Source code is freely availabe to Source code is not freely available the customer g Cloud storage Cloud computing 1 Cloud storage is a model of Cloud computing is is typically data storage where the digital data defined as a type of computing that is stored in logical pools, the relies on sharing computing physical storage spans multiple resources rather than having local servers (and often locations), and servers or personal devices to the physical environment is handle applications. typically owned and managed by a hosting company Page 9 of 9

KENDRIYA VIDYALAYA SANGATHAN (KOLKATA REGION) Second Pre Board Examination ( ) COMPUTER SCIENCE (Theory) Class-XII Marking Scheme

KENDRIYA VIDYALAYA SANGATHAN (KOLKATA REGION) Second Pre Board Examination (2014-15) COMPUTER SCIENCE (Theory) Class-XII Marking Scheme Ques. 1 a) [1] Automatic Type Conversion Type casting It is also