ORDINARY FUNCTION MEMBER FUNCTIONsolved sample paper computer science-2014-15 by shivender kumar bhardwaj, pgt-computer sc, the air force school, subroto park 1 | p a g e ordinary

SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15

BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK

1 | P a g e

ORDINARY FUNCTION MEMBER FUNCTION 1. Must be declared outside a class,

generally before main () function.

1. Must be declared inside a class.

2. Globally declared functions can be

accessed or called by any other

function present in the program.

2. Can be accessed by the

members of the same class and

also by the objects of the class

(if declared in public section.

3. Cannot access members of the

same class

3. Can access members of the

same class

class square { int side; public:

void input();

-------------- ---------------

}; void Area(); void main() {

square A; A.input(); Area(); ______ ________

}

Member Functions are declared inside a class

An object is required to call a member function

Ordinary Functions do not require object

Ordinary Functions are declared outside any class



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b. Write the related library function name based upon the given information

in C++. (i) Get single character using keyboard. This function is available in

stdio.h file.

(ii) To check whether given character is alpha numeric character or not.

This function is available in ctype.h file. [1] Ans.

c. Rewrite the following C++ program after removing all the syntactical

errors (if any), underlining each correction. : [2]

include<iostream.h> #define PI=3.14

void main( )

{ float r;a; cout<<‟enter any radius‟;

cin>>r;

a=PI*pow(r,2); cout<<”Area=”<<a

}

d. Write the output from the following C++ program code: [2]

#include<iostream.h> #include<ctype.h>

void strcon(char s[])

{

for(int i=0,l=0;s[i]!='\0';i++,l++); for(int j=0; j<l; j++)

{

i. getchar()

ii. isalnum()

Ans.

#include<iostream.h>

#include<math.h>

#define PI 3.14

void main()

{

float r, a;

cout<<”enter any radius”;

cin>>r;

a = PI*pow(r,2);

cout<<”Area =”<<a;

}



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if (isupper(s[j]))

s[j]=tolower(s[j])+2; else if ( islower(s[j]))

s[j]=toupper(s[j])-2;

else

s[j]='@'; }

}

void main() {

char *c="Romeo Joliet";

strcon(c); cout<<"Text= "<<c<<endl;

c=c+3;

cout<<"New Text= "<<c<<endl;

c=c+5-2; cout<<"last Text= "<<c

}

Ans.

e. Find the output of the following C++ program: [3]


#include<conio.h>

#include<ctype.h> class Class

{

int Cno,total;

char section; public:

Class(int no=1)

{ Cno=no;

section='A';

total=30;

} void addmission(int c=20)

{

section++; total+=c;

}

void ClassShow() {

cout<<Cno<<":"<<section<<":"<<total<<endl;

}

} ;

Text = @tMKCM@lMJGCR

New Text = KCM@lMJGCR

Last Text = @lMJGCR



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void main()

{ Class C1(5),C2;

C1.addmission(25);

C1.ClassShow();

C2.addmission(); C1.addmission(30);

C2.ClassShow();

C1.ClassShow(); }

Ans.

5 : B : 55

1 : B : 50

5 : C : 85

f. Study the following C++ program and select the possible output(s) from it :

Find the maximum and minimum value of L. [2]

#include<stdlib.h>


#include<string.h>

void main()

{

randomize();

char P[]="C++PROGRAM";

long L;

for(int I=0;P[I]!='R';I++)

{

L=random (sizeof(L)) +5;

cout<<P[L]<<"-";

}

}

i) R-P-O-R-

ii) P-O-R-+-

iii) O-R-A-G-

iv) A-G-R-M-

Ans.

C + + P R O G R A M

0 1 2 3 4 5 6 7 8 9

POSSIBLE VALUES

Only Possible Output is (iii) O-R-A-G-

Minimum value of L = 5

Maximum Value of L = 8



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Q2.a. How encapsulation and abstraction are implemented in C++ language?

Explain with an example. [2]

b. Answer the questions (i) and (ii) after going through the following C++ class: [2]

class Stream

{ int StreamCode ; char Streamname[20];float fees;

public:

Stream( ) //Function 1

{ StreamCode=1;

strcpy (Streamname,"DELHI");

fees=1000; }

void display(float C) //Function 2

{

Ans. 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. We can say that

encapsulation is way of achieving abstraction.

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()

{

Cout<<L<<B;

}

};

Data and member functions are

together - Encapsulation

Only Input() and Output functions

are visible outside the class i.e. only

these 2 can be accessed outside the

class using objects of the same

class - Abstraction



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cout<<StreamCode<<":"<<Streamname<<":"<<fees

<<endl; }

~Stream( ) //Function 3

{

cout<<"End of Stream Object"<<endl; }

Stream (int SC,char S[ ],float F) ; //Function 4

}; i) In Object Oriented Programming, what are Function 1 and Function 4

combined together referred as? Write the definition of function 4.

ii) What is the difference between the following statements? Stream S(11,”Science”,8700);

Stream S=Stream(11,”Science”,8700);

Ans.

c. Define a class Customer with the following specifications. [4] Private Members :

Customer_no integer

Customer_name char (20) Qty integer

Price, TotalPrice, Discount, Netprice float

Member Functions:

Public members: A constructer to assign initial values of Customer_no as 111,

Customer_name as “Leena”, Quantity as 0 and Price, Discount and

Netprice as 0. Input( ) – to read data members(Customer_no, Customer_name,

Quantity and Price) call Caldiscount().

Caldiscount ( ) – To calculate Discount according to TotalPrice and NetPrice

TotalPrice = Price*Qty

TotalPrice >=50000 – Discount 25% of TotalPrice

i) Constructor Overloading

Stream (int Sc, char S[], float F)

{

StreamCode = Sc;

strcpy(Streamname, S);

fees = F;

}

ii) The constructors can be called explicitly or implicitly. The method

of calling the constructor implicitly is also called

the shorthand method.

Stream S(11,”Science,8700); - In this statement constructor

is called implicitly.

Stream S = Stream(11,”Science,8700); - Here constructor is

called explicitly.



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TotalPrice >=25000 and TotalPrice <50000 - Discount 15% of

TotalPrice TotalPrice <250000 - Discount 10% of TotalPrice

Netprice= TotalPrice-Discount

Show( ) – to display Customer details.

Ans.

d. Answer the questions (i) to (iv) based on the following code: [4]

class AC

{ char Model[10];

char Date_of_purchase[10];

char Company[20];

public( );

Class Customer

{ int Customer_no;

char Customer_name;

int Qty;

float Price, TotalPrice, Discount, Netprice;

public:

Customer()

{ Customer_no = 111;

strcpy( Customer_name, “Leena”);

Qty=0;

lPrice =0, Discount=0, Netprice=0;

}

Void Input()

{ cin>>Customer_no;

gets( Customer_name);

cin>>Qty>>TotalPrice;

Caldiscount();

}

Void Caldiscount()

{ TotalPrice = Price*Qty;

if ( TotalPrice >= 50000)

Discount = 0.25 * TotalPrice;

else if (TotalPrice >= 2 5000)


else


Netprice = TotalPrice – Discount;

}

void Show()

{ cout<<Customer_no<<” “<< Customer_name<<” “<<Qty<” “

<<Price<<” “<<TotalPrice<<” “<<Discount<<” “<<Netprice;

}

};



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AC( );

void entercardetail( ); void showcardetail( );

};

class Accessories : protected AC

{ protected:

char Stabilizer[30];

char AC_cover[20]; public:

float Price;

Accessories( ); void enteraccessoriesdetails( );

void showaccessoriesdetails( );

};

class Dealer : public Accessories {

int No_of_dealers;

char dealers_name[20]; int No_of_products;

public:

Dealer( ); void enterdetails( );

void showdetails( );

};

(i) How many bytes will be required by an object of class Dealer and class Accessories?

(ii) Which type of inheritance is illustrated in the above c++ code?

Write the base class and derived class name of class Accessories. (ii) Write names of all the members which are accessible from the

objects of class Dealer.

(iv) Write names of all the members accessible from member functions of class Dealer.

Ans.

i) Object of Dealer = 118 bytes and object of Accessories = 98 bytes

ii) Multilevel Inheritance, Base class = AC, Derived class = Dealer

iii) enterdetails(), showdetails(), price, enteraccessoriesdetails(),

showaccessoriesdetals()

iv) No_of_dealers, dealers_name, No_of _products, enterdetails(),

showdetails(), Stablizer, Ac_cover, price, enteraccessoriesdetails(),

showaccessoriesdetals(), entercardetail, showcardetail()

Q3a) An array T[-1..35][-2..15] is stored in the memory along the row with each

element occupying 4 bytes. Find out the base address and address of

element T[20][5], if an element T[2][2] is stored at the memory location

3000. Find the total number of elements stored in T and number of bytes

allocated to T. [3]



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Ans.

Since element T[2][2] is stored at the memory location 3000,

LOC (A[I][J]) = BASE(A) + W*[C*(I-LBI) + (J-LBJ)]

3000 = BASE(A) + 4 [ 18 * (2-(-1)) + (2 – (-2))]

3000 = BASE(A) + 4 [ 54 + 4]

3000 – 4*58 = BASE(A)

BASE (A) = 2768

To find address of element T[20][5],

LOC (A[I][J]) = BASE(A) + W*[C*(I-LBI) + (J-LBJ)]

LOC (T[20][5]) = 2768 + 4 [ 18 * (20-(-1)) + (5 – (-2))]

= 2768 + 4 * [18 * 21 + 7]

= 2768 + 1540

= 4308

Total number of elements stored in T = 37*18 = 666

Total number of bytes allocated to T = 666 * 4 = 2664

b. Write a function SORTSCORE() in C++ to sort an array of structure IPL in

descending order of score using selection sort . [3]

Note : Assume the following definition of structure IPL. struct IPL

{ int Score;

char Teamname[20];

}; Ans.

void SORTSCORE(IPL a[], int n)

{ int largest;

IPL temp;

for ( int K = 0; K < n-1; K++)

{ largest = K;

for ( int j = K+1; j< n; j++)

{ if ( a[j].score > a[largest].score)

{ largest = j;

}

}

temp = a[K];

a[K] = a[largest];

a[largest] = temp;

}

}

c. Write member functions to perform POP and PUSH operations in a

dynamically allocated stack containing the objects of the following structure:[4]

struct Game

{ char Gamename[30];

int numofplayer;

Game *next; };

Ans.



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struct Game

{

char Gamename[30];

int numofplayer;

Game *next;

};

class Stack

{

Game *Top;

public :

Stack()

{

Top = NULL;

}

void Push();

void Pop();

void display();

~Stack();

};

void Stack::Push()

{

Game *temp = new Game;

cout<<"Enter Data : ";

gets(temp->Gamename);

cin>>temp->numofplayer;

temp->next =Top;

Top = temp;

}

void Stack::Pop()

{

if ( Top != NULL)

{

Game *temp = Top;

cout<<Top->Gamename<<"

Deleted";

Top = Top->next;

delete temp;

}

else

cout<<"Stack is empty....";

}

d. Write a function in C++ to print the sum of all the non-negative elements

present on both the diagonal of a two dimensional array passed as the argument

to the function. [2]

void Diagonal(int A[][20], int N)

{ int K, Sdiag=0;

for(int K=0;K<N;K++)

if ( A[K][K] > 0)

Sdiag+=A[K][K];

for(int K=0;K<N;K++)

if ( A[N-K-1][K] > 0 )

Sdiag+=A[N-K-1][K];

cout<<”Sum of all the non-negative elements on both the diagonals= “<<Sdiag;

}



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e. Evaluate the following postfix expression. Show the status of stack after

execution of each operation separately:

2,13, + , 5, -,6,3,/,5,*,< [2]

ITEM SCANNED OPERATION STACK

2 PUSH 2 2

13 PUSH 13 2,13

+

POP 13 and 2

Evaluate 2 + 13 = 15

PUSH 15

15

5 PUSH 5 15, 5

-

POP 5 & 15

EVALUATE 15 – 5 = 10

PUSH 10

10

6 PUSH 6 10,6

3 PUSH 3 10, 6, 3

/

POP 3 & 6

EVALUATE 6/3= 2

PUSH 2

10,2

5 PUSH 5 10, 2, 5

*

POP 5 & 2

EVALUATE 2*5 = 10

PUSH 10

10, 10

<

POP 10 & 10

EVALUATE 10<10 = FALSE

PUSH FALSE

FALSE

RESULT = FALSE

Q4. a. Write the command to place the file pointer at the 10th and 4th record

starting position using seekp() or seekg() command. File object is „file‟ and

record name is „STUDENT‟. [1]

Ans. file.seekp( 9 * sizeof(STUDENT), ios::beg);

file.seekp( 3 * sizeof(STUDENT), ios::beg);

b. Write a function in C++ to count and display the no of three letter words in

the file “VOWEL.TXT”. [2]

Example:



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If the file contains:

A boy is playing there. I love to eat pizza. A plane is in the sky.

Then the output should be: 4


#include<string.h>

void wordcount()

{ char word[80];

int cnt=0;

ifstream f1;

f1.open(“VOWEL.TXT”);

while (f1>>word)

{

if(strlen(word) == 3)

{

cnt++;

cout<<word<<endl;

}

}

cout<<” number of three letter words = “<<cnt;

f1.close();

}

c. Given the binary file CAR.Dat, containing records of the following class CAR

type: [3]

class CAR

{

int C_No;

char C_Name[20];

float Milage;

public:

void enter( )

{

cin>> C_No ; gets(C_Name) ; cin >> Milage;

}

void display( )

{

cout<< C_No ; cout<<C_Name ; cout<< Milage;

}



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int RETURN_Milage( )

{

return Milage;

}

};

Write a function in C++, that would read contents from the file CAR.DAT and

display the details of car with mileage between 100 to 150.

void CARSearch()

{

fstream FIL;

FIL.open(“CAR.DAT ”,ios::binary|ios::in);

CAR C;

int Found = 0;

while (FIL.read((char*)&C,sizeof(C)))

{

if ((C.RETURN_Milage( ) > 100) && (C.RETURN_Milage( ) < 150) )

{

C.display();

Found++;

}

}

if (Found==0)

cout<<”Sorry! No car found with mileage between 100 to 150”;

FIL.close();

}

Q5. a. Define degree and cardinality. Based upon given table write degree and

cardinality. [2]

Ans. 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. Write SQL commands for the queries (i) to (iv) and output for (v) & (viii)

based on a table COMPANY and CUSTOMER [6]



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(i) To display those company name which are having prize less than 30000.

SELECT NAME FROM COMPANY

WHERE COMPANY.CID = CUSTOMER.CID AND PRICE < 30000;

ii) To display the name of the companies in reverse alphabetical order.

SELECT NAME FROM COMPANY

ORDER BY NAME DESC;

(iii) To increase the prize by 1000 for those customer whose name starts with „S‟

UPDATE CUSTOMER

SET PRICE = PRICE + 1000

WHERE NAME LIKE ‘S%’;

iv) To add one more column totalprice with decimal(10,2) to the table customer

ALTER TABLE CUSTOMER

ADD TOTALPRICE DECINAL(10,2);

v) SELECT COUNT(*) ,CITY FROM COMPANY GROUP BY CITY;

3 DELHI

2 MUMBAI



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1 MADRAS

(vi) SELECT MIN(PRICE), MAX(PRICE) FROM CUSTOMER WHERE QTY>10 ;

50000, 70000

(vii) SELECT AVG(QTY) FROM CUSTOMER WHERE NAME LIKE “%r%;

11

(viii) SELECT PRODUCTNAME,CITY, PRICE FROM COMPANY,CUSTOMER WHERE

COMPANY.CID=CUSTOMER.CID AND PRODUCTNAME=”MOBILE”;

MOBILE MUMBAI 70000

MOBILE MUMBAI 25000

Q6. a) State and define principle of Duality. Why is it so important in Boolean

algebra? [2]

Ans. 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 1's by 0's, and all 0's by 1'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) Write the equivalent boolean expression for the following logic circuit [2]

Ans. ((X’.Y)’ + (X.Y’)’)’



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F (a,b,c,d) =

(a+b+c+d).(a+b+c+d’).(a+b’+c+d).(a+b’+c’+d’).(a’+b+c+d).(a’

+b+c+d’).(a’+b’+c+d).(a’+b’+c+d’).(a’+b’+c’+d)

d) Obtain the minimal SOP form for the following Boolean expression using K-

Map.

F(A,B,C,D) = (0,2,3,5,7,8,10,11,13,15) [3]

Ans.

1

1 1 1

1 1

1 1

1 1



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Quad 1 = m0 +m2 +m8 +m10

= B’D’

Quad 2 = m3 +m7 +m15 +m11

= CD

Quad 1 = m5 +m7 +m15 +m13

= BD

Minimal SOP = B’D’ + CD + BD

Q7.a.Give any two advantage of using Optical Fibres. [1]

Ans. Two advantage of using Optical Fibres are

(i) Not affected by any kind of noise

(ii) High transmission capacity

b. Indian School, in Mumbai is starting up the network between its different

wings. There are Four Buildings named as SENIOR, JUNIOR, ADMIN and HOSTEL

as shown below.: [4]

The distance between various buildings is as follows:

ADMIN TO SENIOR 200m

ADMIN TO JUNIOR 150m

ADMIN TO HOSTEL 50m

SENIOR TO JUNIOR 250m

SENIOR TO HOSTEL 350m

JUNIOR TO HOSTEL 350m

Number of Computers in Each Building

SENIOR 130

JUNIOR 80

ADMIN 160

HOSTEL 50

(b1) Suggest the cable layout of connections between the buildings.

Ans. One of the layouts of connection is given below:

ADMIN

JUNIOR HOSTEL SENIOR



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b2) Suggest the most suitable place (i.e. building) to house the server of this

School, provide a suitable reason.

Ans. Server should be housed in ADMIN building because it has

maximum number of computers.

c. Identify the Domain name and URL from the following. [1]

http://www.income.in/home.aboutus.hml

Ans. Domain name - income.in

URL - http://www.income.in/home.aboutus.hml

d. What is Web Hosting? [1]

Ans. 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. What is the difference between packet & message switching? [1]

Ans. Packet Switching There is a tight upper limit on the block size. In

message switching there was no upper limit. A fixed size of packet is

specified. All the packets are stored in main memory in switching office.

In message switching packets are stored on disk. This increases the

performance as access time is reduced.

f. Define firewall. [1]

Ans. 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. Which protocol is used to creating a connection with a remote

machine? [1]

Ans. Telnet

It is an older internet utility that lets us log on to remote computer

system. It also facilitates for terminal emulation purpose.

Documents

ORDINARY FUNCTION MEMBER FUNCTIONsolved sample paper computer science-2014-15 by shivender kumar bhardwaj, pgt-computer sc, the air force school, subroto park 1 | p a g e ordinary