ordinary function member functionsolved sample paper computer science-2014-15 by shivender kumar...
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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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
2 | P a g e
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;
}
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
3 | P a g e
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<iostream.h>
#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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
4 | P a g e
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<iostream.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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
5 | P a g e
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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
6 | P a g e
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.
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
7 | P a g e
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)
Discount = 0.15 * TotalPrice;
else
Discount = 0.10 * TotalPrice;
Netprice = TotalPrice – Discount;
}
void Show()
{ cout<<Customer_no<<” “<< Customer_name<<” “<<Qty<” “
<<Price<<” “<<TotalPrice<<” “<<Discount<<” “<<Netprice;
}
};
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
8 | P a g e
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]
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
9 | P a g e
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.
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
10 | P a g e
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;
}
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
11 | P a g e
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:
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
12 | P a g e
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<iostream.h>
#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;
}
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
13 | P a g e
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]
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
14 | P a g e
(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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
15 | P a g e
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’)’)’
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
16 | P a g e
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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
17 | P a g e
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
SOLVED SAMPLE PAPER COMPUTER SCIENCE-2014-15
BY SHIVENDER KUMAR BHARDWAJ, PGT-COMPUTER SC, THE AIR FORCE SCHOOL, SUBROTO PARK
18 | P a g e
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.