2014 computer science_question_paper

XII COMPUTER SCIENCE
CBSE Board - 2014
1(a) What is the difference between call by reference and call by value with respect to memory allocation? Give a
suitable example to illustrate using C++ code.
2
Ans.
Call By Value Call by reference
 Call by value is used to create a temporary copy
of the data which is transferred from the actual
parameter in the final parameter.
 Call by reference is used to share the same
memory location for actual and formal
parameters
 The changes done in the function in formal
parameter are not reflected back in the calling
environment.
 The changes done in the function are reflected
back in the calling environment.
Example:
void compute (int A, int & B)
{
A++;
B++;
cout<<“The function on display gives “;
cout<<“A = “<<A<<“&”<<“B=”<<B<<endl;
}
void main( )
{
int I=50, J=25;
cout<<“Initial of function call “<<endl;
cout<<“I=”<<I<<“&”<<“J=”<<J<<endl;
compute(I,J); cout<<“After the call of the function”<<endl;
cout<<“I=”<<I<<“&”<<“J=”<<J<<endl;
getch( );
}
(b) Observe the following C++ code and write the name(s) of the header file(s), which will be essentially required
to run it in a C++ compiler:
void main()
{
char CH, STR[20];
cin>>STR;
CH=toupper(STR[0]);
cout<<STR<<"starts with"<<CH<<endl;
}
1
Ans. (i) iostream.h
(ii) ctype.h
(c) Rewrite the following C++ code after removing all the syntax error(s), if present in the code. Make sure that
you underline each correction done by you in the code.
Important Note:
 Assume that all the required header files are already included, which are essential to run this code.
 The corrections made by you do not change the logic of the program.
typedef char[80] STR;
void main()
{
Txt STR;
gets(Txt);
cout<<Txt[0]<<'t<<Txt[2];
2
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cout<<Txt<<endline;
}
Ans. typedef char STR[80];
void main()
{
STR Txt;
gets(Txt);
cout<<Txt[0]<<'t'<<Txt[2];
cout<<Txt<<endl;
}
(d) Obtain the output from the following C++ program as expected to appear on the screen after its execution.
Important Note:
All the desired header files are already included in the code, which are required to run the code.
void main()
{
char *Text="AJANTA";
int *P, Num[]={1, 5, 7, 9};
P=Num;
cout<<*P<<Text<<endl;
Text++;
P++;
cout<<*P<<Text<<endl;
}
2
Ans. 1AJANTA
5JANTA
(e) Obtain the output from the following C++ program, which will appear on the screen after its execution.
Important Note:
All the desired header files are already included in the code, which are required to run the code.
class Game
{
int Level, Score;
char Type;
public:
Game(char GType='P')
{
Level=1;
Score=0;
Type=GType;
}
void Play(int GS);
void Change();
void Show()
{
cout<<Type<<"@"<<Level<<endl;
cout<<Score<<endl;
}
};
void main()
{
Game A('G'), B;
B.Show();
A.Play(11);
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A.Change();
B.Play(25);
A.Show();
B.Show();
}
void Game::Change()
{
Type=(Type=='P')?'G':'P';
}
void Game::Play(int GS)
{
Score+=GS;
if(Score>=30)
Level=3;
else if(Score>=20)
Level=2;
else
Level=1;
}
Ans: P@1
0
P@1
11
P@2
25
(f) Read the following C++ code carefully and find out, which out of the given options (i) to (iv) are the expected
correct output(s) of it. Also, write the maximum and minimum value that can be assigned to the variable Taker
used in the code:
void main()
{
int GuessMe[4]={100, 50, 200, 20};
int Taker=random(2)+2;
for(int Chance=0;Chance<Taker;Chance++)
cout<<GuessMe[Chance]<<"#";
}
(i) 100#
(ii) 50#200#
(iii) 100#50#200#
(iv) 100#50
2
Ans. 100#50#
Note: that if you run the following program, again and again, the same random numbers will be generated. That
is, TurboC always seeds the random number generator with the same starting number.
Therefore, the function "randomize()" may be used to seed the random number generator with a number which
is developed from the system clock, which of course, is always changing.
So, Taker always have 2 as random is generating 2 on every execution of code.
2(a) What is function overloading? Write an example using C++ to illustrate the concept of function overloading. 2
Ans. A function name having several definitions that are differentiable by the number or types of their arguments is
known as function overloading.
Example :
#include <iostream.h>
class printData
{
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public:
void print(int i)
{
cout << "Printing int: " << i << endl;
}
void print(double f)
{
cout << "Printing float: " << f << endl;
}
void print(char* c)
{
cout << "Printing character: " << c << endl;
}
};
int main(void)
{
printData pd;
// Call print to print integer
pd.print(5);
// Call print to print float
pd.print(500.263);
// Call print to print character
pd.print("Hello C++");
return 0;
}
When the above code is compiled and executed, it produces following result:
Printing int: 5
Printing float: 500.263
Printing character: Hello C++
(b) Answer the questions (i) and (ii) after going through the following class:
class Hospital
{
int Pno, Dno;
public:
Hospital(int PN); //Function 1
Hospital(); //Function 2
Hospital(Hospital &H); //Function 3
void In(); //Function 4
void Disp(); //Function 5
};
void main()
{
Hospital H(20); //Statement 1
}
(i) Which of the functions out of Function 1, 2, 3, 4 or 5 will get executed when the Statement 1 is executed in
the above code?
(ii) Write a statement to declare a new object G with reference to already existing object H using Function 3.
2
Ans. (i) Function 1 will get executed when the statement 1 is executed in the above code.
(ii) Hospital G(H);
(c) Define a class Tourist in C++ with the following specification:
Data Members
• CN0 - to store Cab No
• Ctype - to store a chahracter 'A', 'B' or 'C' as City Type
4
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• PerKM - to store per Kilo Meter charges
• Distance - to store Distance travelled (in Km)
Member Functions
• A constructor function to initialize CType as 'A' and CNo as '0000'
• A function CityCharges() to assign PerKM as per the following table:
CType Per KM
A 20
B 18
C 15
• A function RegisterCab() to allow administrator to enter the values for CNo and CType. Also, this function
should call CityCharges() to assign PerKM Charges.
• A function Display() to allow user to enter the value of Distance and display CNo, CType, PerKM,
PerKM*Distance (as Amount) on scree.
Ans. class Tourist
{
int CNO;
char Ctype;
int PerKM;
int Distance;
public:
Tourist()
{
CType='A';
CNO=0000;
}
void CityCharges()
{
if(CType=='A')
{
PerKM=20;
}
else if(CType=='B')
{
PerKM=18;
}
else if(CType=='C')
{
PerKM=15;
}
}
void RegisterCab()
{
cout<<"enter Cab No=";
cin>>CNO;
cout<<endl<<"enter city type=";
cin>>CType;
CityCharges();
}
void Display()
{
cout<<"enter distence in KM=";
cin>>Distance;
cout<<"Cab No ="<<CNO<<endl;
cout<<"city Type="<<CType<<endl;
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cout<<"Kilo meter charges="<<PerKM<<endl;
cout<<"Amount is="<<PerKM*Distance;
}
};
(d) Consider the following C++ code and answer the questions from (i) to (iv):
class University
{
long Id;
char City[20];
protected:
char Country[20];
public:
University();
void Register();
void Display();
};
class Department: private University
{
long DCode[10];
char HOD[20];
protected:
double Budget;
public:
Department();
void Enter();
void Show();
};
class Student: public Department
{
long RollNo;
char Name[20];
public:
Student();
void Enroll();
void View();
};
(i) Which type of inheritance is shown in the above example?
(ii) Write the names of those member functions, which are directly accessed from the objects of class Student.
(iii) Write the names of those data members, which can be directly accessible from the member functions of
class Student.
(iv) Is it possible to directly call function Display() of class University from an object of class Department?
(Answer as Yes or No).
4
Ans. (i) Multilevel
(ii) void Enroll() and void view()
(iii) RollNo and Name[20]
(iv) No
3(a) Write code for a function void EvenOdd(int T[], int c) in C++, to add 1 in all the odd values and 2 in all the even
values of the array T.
Example : If the original content of the array T is
T[0] T[1] T[2] T[3] T[4]
35 12 16 69 26
The modified content will be:
3
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T[0] T[1] T[2] T[3] T[4]
36 14 18 70 28
Ans. void EvenOdd(int T[], int c)
{
for(int i=0;i<c;++i)
{
if(T[i]%2==0)
T[i]+=2;
else
T[i]+=1;
}
}
(b) An array A[20][30] is stored along the row in the memory with each element requiring 4 bytes of storage. If the
base address of array A is 32000, find out the location of A[15][10]. Also, find the total number of elements
present in this array.
3
Ans. Base Address (i.e., address of A[0][0]) B = 32000
Element size w = 4 bytes
Rows, Columns in Array A R, C = 20, 30
Location of array A I, J = 15, 20
Address of A[I][J] = B + w * (I * C + J)
Address of A[15][20] = 32000 + 4 * (15 * 30 + 20)
= 33880
Total Number of elements present in array A is R * C i.e., 20 * 30 = 600
(c) Write a user-defined function AddEnd2(int A[][4], int N, int M) in C++ to find and display the sum of all the
values, which are ending with 2 (i.e., units place is 2).
for example if the content of array is:
22 16 12
19 5 2
The output should be
36
2
Ans. #include <conio.h>
#include <iostream.h>
void AddEnd2(int A[][3], int N, int M)
{
int num,sum=0;
for(int i=0; i<N; i++)
{
for(int j=0; j<M; j++)
{
num=A[i][j];
int last=num%10;
if(last==2)
sum=sum+num;
}
}
cout<<sum;
}
void main()
{
int arr[2][3] ={{22, 16, 12}, {19, 5, 2}};
clrscr();
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AddEnd2(arr,2,3);
}
(d) Evaluate the following postfix expression. Show the status of stack after execution of each operation
separately:
T, F, NOT, AND, T, OR, F, AND
2
Ans.
token scanned from
postfix expression
Stack status ( bold letter shows the top of the
stack) after processing the scanned token
Operation performed
True True Push True
False True, False Push False
Not True, True Op1=pop() i.e. False
Push(NOT False) i.e. NOT False=True
And True Op2=pop() i.e. True
Op1=pop() i.e. True
Push(Op2 AND Op1) i.e. True AND
True=True
True True, True Push True
Or True Op2=pop() i.e. True
Op1=pop() i.e. True
Push(Op2 OR Op1) i.e. True OR
True=True
False True, False push False
And False Op2=pop() i.e. False
Op1=pop() i.e. True
Push(Op2 AND Op1) i.e. False AND
True=False
NULL Final result False Pop True and return False
(e) Write a function PUSHBOOK() in C++ to perform insert operation on a Dynamic Stack, which contains Book_no
and Book_Title. Consider the following definition of NODE, while writing your C++ code.
struct NODE
{
int Book_No;
char Book_Title[20];
NODE *Next;
};
4
Ans. struct NODE
{
int Book_No;
char Book_Title[20];
NODE *Next;
}*top, *newptr, *save;
void PUSHBOOK(NODE*);
void PUSHBOOK(NODE *np)
{
if(top==NULL)
top=np;
else
{
save=top;
top=np;
np->Next=save;
}
}
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4(a) Fill in the blanks marked as Statement 1 and Statement 2, in the program segment given below with
appropriate functions for the required task.
class Agency
{
int ANo; //Agent Code
char AName[20]; //Agent Name
char Mobile[12]; //Agent Mobile
public:
void Enter(); //Function to enter details of agent
void Disp(); //Function to display details of agent
int RAno()
{
return ANo;
}
void UpdateMobile() //Function to update Mobile
{
cout<<"Updated Mobile";
gets(Mobile);
}
};
void AgentUpdate()
{
fstream F;
F.open("AGENT.DAT",ios::binary|ios::in|ios::out);
int Updt=0;
int UAno;
cout<<"Ano (Agent No - to update mobile):";
cin>>UAno;
Agency A;
while(!Updt && F.read((char*)&A,sizeof(A)))
{
if(A.RAno()==UAno)
{
//Statement 1 : To call the function to update Mobile No.
______________________________________;
//Statement 2: To reposition file pointer to re-write the updated object back in the file
______________________________________________;
F.write((char*)&A, sizeof(A));
Updt++;
}
}
if (Updt)
cout<<"Mobile Updated for Agent"<<UAno<<endl;
else
cout<<"Agent not in the Agency"<<endl;
F.close();
}
1
Ans. Statement 1: int Pos= F.tellg()
Statement 2: F.seekp(Pos-sizeof(A));
(b) Write a function AECount in C++, which should read each character of a text file NOTES.TXT, should count and
display the occurrence of alphabets A and E (including small cases a and e too).
2
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Example:
If the file content is as follows:
CBSE enhanced its
CCE guideline further.
The AECount() function should display the output as
A:1
E:7
Ans. void AECount()
{
ifstream fin;
fin.open(“NOTES.TXT”, ios::in);
char word[50];
int c1=0,c2=0;
while(!fin.eof())
{
fin>>word;
if(strcmp(word,"A")==0||strcmp(word,"a")==0)
c1++;
if(strcmp(word,"E")==0||strcmp(word,"e")==0)
c2++;
}
cout<<"A :"<<c1;
cout<<"E :"<<c2;
fin.close();
}
(c) Assuming the class TOYS as declared below, write a function in C++ to read the objects of TOYS from binary file
TOYS.DAT and display those details of those TOYS, which are meant for childern of AgeRange "5 to 8".
class TOYS
{
int ToyCode;
char ToyName[10];
char AgeRange;
public:
void Enter()
{
cin>>ToyCode;
gets(ToyName);
gets(AgeRange);
}
void Display()
{
cout<<ToyCode<<":"<<ToyName<<endl;
cout<<AgeRange<<endl;
}
char* WhatAge()
{
return AgeRange;
}
};
3
Ans. void DisplayAgeRange()
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{
TOYS C;
fstream fin;
fin.open(“TOYS.DAT”, ios:: binary | ios::in);
while(fin.read((char*)&C, sizeof(C)))
{
if(C.WhatAge()==5|| C.WhatAge()==6|| C.WhatAge()==7|| C.WhatAge()==8)
C.Display();
}
fin.close();
}
5(a) Explain the concept of Cartesian product between two tables, with the help of appropriate example.
NOTE: Answer the questions (b) and (c) on the basis of the following tables SHOPPE and ACCESSORIES.
Table: SHOPPE
Id SName Area
S001 ABC Computronics CP
S002 All Infotech Media GK II
S003 Tech Shoppe CP
S004 Geek Tenco Soft Nehru Place
S005 Hitech Tech Store Nehru Place
TABLE: ACCESSORIES
No Name Price Id
A01 Mother Board 12000 S01
A02 Hard Disk 5000 S01
A03 Keyboard 500 S02
A04 Mouse 300 S01
A05 Mother Board 13000 S02
A06 Keyboard 400 S03
A07 LCD 6000 S04
T08 LCD 5500 S05
T09 Mouse 350 S05
T10 Hard Disk 4500 S03
Q. No. b iv SHOPPE and ACCESSORIES table ID values are totally different, answer prepared assuming values
are same in both the tables in Id attribute.
Q. No. c iv is wrong ACCESSORIES table does not have any attribute SNO , answer prepared assuming attribute
name ‘Id’ in query.
2
Ans. When you join two or more tables without any condition, it is called Cartesian product or Cross Join.
Example –
SELECT * FROM SHOPPE, ACCESSORIES;
(b) Write the SQL queries:
(i) To display Name and Price of all the Accessories in ascending order of their Price.
(ii) To display Id and SName of all Shoppe located in Nehru Place.
(iii) To display Minimum and Maximum Price of each Name of Accessories.
(iv) To display Name, Price of all Accessories and their respective SName where they are available.
4
Ans. (i) SELECT Name, Price FROM ACCESSORIES ORDER BY Price;
(ii) SELECT Id, SName FROM SHOPPE WHERE Area=’Nehru Place’;
(iii) SELECT Name, MAX(Price), MIN(Price) FROM ACCESSORIES ;
(iv) SELECT Name,Price,SName FROM ACCESSORIES A, SHOPPE S WHERE A.Id=S.Id;
(c) Write the output of the following SQL commands:
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(i) SELECT DISTINCT NAME FROM ACCESSORIES WHERE PRICE>=5000;
(ii) SELECT AREA , COUNT(*) FROM SHOPPE GROUP BY AREA;
(iii) SELECT COUNT(DISTINCT AREA) FROM SHOPPE;
(iv) SELECT NAME, PRICE*0.05 DISCOUNT FROM ACCESSORIES WHERE SNO IN (‘S02’, ‘S03’);
2
Ans. (i) Name
Mother Board
Hard Disk
LCD
(ii) AREA COUNT
CP 2
GK II 1
Nehru Place 2
(iii) COUNT
3
(iv) NAME PRICE
Keyboard 25.00
Mother Board 650.00
Keyboard 20.00
Hard Disk 225.00
6(a) Name the law shown below and verify it using a truth table.
X+ X’.Y=X+Y
2
Ans.
X Y X’ X’.Y X+X’.Y X+Y
0 0 1 0 0 0
0 1 1 1 1 1
1 0 0 0 1 1
1 1 0 0 1 1
Prove algebraically that X + X’Y = X + Y.
L.H.S. = X + X’Y
= X.1 + X’Y (X . 1 = X property of 0 and 1)
= X(1 + Y) + X’Y (1 + Y = 1 property of 0 and 1)
= X + XY + X’Y
= X + Y(X + X’)
= X + Y.1 (X + X’ =1 complementarity law)
= X + Y (Y . 1 = Y property of 0 and 1)
= R.H.S. Hence proved.
(b) Obtain the Boolean Expression for the logic circuit shown below: 2
Ans. A’.B+(C+D’)’
(c) Write the Product of Sum from of the function F(X, Y, Z) for the following truth table representation of F:
X Y Z F
0 0 0 1
0 0 1 0
1
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0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
Ans.
X Y Z F MAX TERMS
0 0 0 1 X+Y+Z
0 0 1 0 X+Y+Z’
0 1 0 0 X+Y’+Z
0 1 1 1 X+Y’+Z’
1 0 0 0 X’+Y+Z
1 0 1 0 X’+Y+Z’
1 1 0 1 X’+Y’+Z
1 1 1 1 X’+Y’+Z’
Now by multiplying Maxterms for the output 0s, we get the desired product of sums expression which is
(X + Y + Z’)(X + Y’ + Z’)(X’+Y+Z)(X’ + Y + Z’)
(d) Obtain the minimal form for the following Boolean expression using K-Map :
F (A,B,C,D) = Σ (1, 3, 4, 5, 6, 7, 12, 13)
3
Ans.
A’B’C’D + A’B’CD + A’BC’D’ +A’BC’D+A’BCD +A’BCD’+ABC’D’ + ABC’D
A’B’D(C’+C) + A’BC’(D+D’) + A’BC(D+D’) +ABC’(D’+D)
A’B’D+A’BC’+A’BC+ABC’
A’B’D+A’B(C’+C) +ABC’
A’B’D+A’B+ABC’
A’D(B’+B) +ABC’
A’D+ABC’
D(A’+A)+BC’
D+BC’
7(a) Write two characteristics of Wi-Fi. 1
Ans. 1. It allows an electronic device to exchange data or connect to the internet wirelessly using microwaves.
2. Network range of wi-fi is very less than other network technologies like wired LAN.
(b) What is the difference between E-mail and Chat? 1
Ans.  Chat is a type of software while Email is a protocol
 Chat requires the permission of both parties while Email does not
 Chat is typically software dependent while Email is not
 Chat needs accounts on the same provider while Email does not
(c) Expand the following:
• GSM
• GPRS
1
Ans. GSM – Global System for Mobile Communications
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GPRS –General Packet Radio Service
(d) Which type of network (out of LAN, PAN and MAN) is formed, when you connect two mobiles using Bluetooth
to transfer a video?
1
Ans. PAN
(e) Tech Up Corporation (TUC) is a professional consultancy company. The company is planning to set up their new
offices in Indian with its hub at Hyderabad. As a network adviser, you have to understand their requirement
and suggest to them the best available solutions. Their queries are mentioned as (i) to (iv) below.
Physical Locations of the blocks of TUC
Block to Block distances (in Mtrs.)
Block (From) Block (To) Distance
Human Resources Conference 60
Human Resources Finance 120
Conference Finance 60
Expected Number of Computers to be installed in each block
Block Computers
Human Resources 125
Finance 25
Conference 60
4
(i) What will the most appropriate block, where TUC should plan to install their server? 1
Ans. Human Resources will the most appropriate block, where TUC should plan to install their server.
(ii) Draw a block to block cable layout to connect all the buildings in the most appropriate manner for efficient
communication.
1
Ans.
(iii) What will be the best possible connectivity out of the following, you will suggest to connect the new setup of
offices in Bangalore with its London base office?
• Infrared
1
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• Satellite Link
• Ethernet Cable
Ans. Ethernet Cable
(iv) Which of the following devices will be suggested by you to connect each computer in each of the buildings?
• Gateway
• Switch
• Modem
1
Ans. Switch
(f) Write names of any two popular Open Source Software, which are used as Operating Systems. 1
Ans. Linux
OpenSolaris
(g) Write any two important characteristics of cloud computing. 1
Ans. Reduction of costs – unlike on-site hosting the price of deploying applications in the cloud can be less due to
lower hardware costs from more effective use of physical resources.
Choice of applications – This allows flexibility for cloud users to experiment and choose the best option for their
needs. Cloud computing also allows a business to use, access and pay only for what they use, with a fast
implementation time.
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2014 computer science_question_paper

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2014 computer science_question_paper