CBSE Question Paper 2018 class 12 Computer Science

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CBSE Question Paper 2018 class 12 Computer Science conducted by Central Board of Secondary Education, New Delhi in the month of March 2018. CBSE previous year question papers with solution are available in myCBSEguide mobile app and cbse guide website. The Best CBSE App for students and teachers is myCBSEguide which provides complete study material and practice papers to CBSE schools in India and abroad.

CBSE Question Paper 2018 class 12 Computer Science

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Class 12 Computer Science list of chapters

Review of Python
Concept of Object Oriented Programming
Classes in Python
Inheritance
Linear List Manipulation
Stacks & Queues in list
Data File Handling
Exception Handling & Green Functions
Databases Concepts and SQL
Structure Query Language
Boolean Algebra
Boolean Functions & Reduce Forms
Application of Boolean Logic
Networking Concepts (Part 1)
Networking Concepts (Part 2)
Networking Protocols
Mobile Telecommunication Technologies, Network Security and Internet Services

CBSE Question Paper 2018 class 12 Computer Science

General Instructions :

SECTION A refers to programming language C++.
SECTION B refers to programming language Python.
SECTION C is compulsory for all.
Answer either SECTION A or SECTION B.
It is compulsory to mention on the page 1 in the answer book whether you are attempting SECTION A or SECTION B.
All questions are compulsory within each section.

SECTION A
[Only for candidates, who opted for C++]

1. Write the type of C++ tokens (keywords and user-defined identifiers) from the following : (2)
  1. else
  2. Long
  3. 4Queue
  4. _count
  Ans.
  1. keyword
  2. Identifier
  3. None
  4. Identifier
    NOTE: Ignore (iii)
2. The following C++ code during compilation reports errors as follows :
  Error: ‘ofstream’ not declared
  Error: ‘strupr’ not declared
  Error: ‘strcat’ not declared
  Error: ‘FIN’ not declared
  Write the names of the correct header files, which must be included to compile the code successfully : (1)
  void main()
  {
  ofstream FIN(“WISH.TXT”);
  char TEXT2[]=”good day”;
  char TEXT1[]=”John!”;
  strupr(TEXT2);
  strcat(TEXT1, TEXT2);
  FIN<<TEXT1<<endl;
  }
  Ans.
  1. fstream
  2. string
3. Rewrite the following C++ code after removing any/all syntactical errors with each correction underlined. (2)
  Note : Assume all required header files are already included in the program.
  Typedef Count int;
  void main()
  {
  Count C;
  cout<<“Enter the count:”;
  cin>>C;
  for (K = 1; K<=C; K++)
  cout<< C “*” K <<endl;
  }
  Ans. typedef int Count ; //Error 1, Error 2
  void main()
  {
  Count C;
  int K; //OR Count K; //Error 3
  cout<<“Enter the count:”;
  cin>>C;
  for (K = 1; K<=C; K++)
  // OR for ( int K = 1; K<=C; K++) //Error 3
  // OR for ( Count K = 1; K<=C; K++) //Error 3
  cout<< C << “*” << K <<endl; //Error 4
  // OR cout<< C * K <<endl; //Error 4
  }
4. Find and write the output of the following C++ program code : (3)
  Note : Assume all required header files are already included in the program.
  void Revert(int &Num, int Last=2)
  {
  Last=(Last%2==0)?Last+1:Last-1;
  for(int C=1; C<=Last; C++)
  Num+=C;
  }
  void main()
  {
  int A=20,B=4;
  Revert(A,B);
  cout<<A<<“&”<<B<<endl;
  B–;
  Revert(A,B);
  cout<<A<<“#”<<B<<endl;
  Revert(B);
  cout<<A<<“#”<<B<<endl;
  }
  Ans.35&4
  38#3
  38#9
5. Find and write the output of the following C++ program code : (2)
  Note : Assume all required header files are already included in the program.
  #define Modify(N) N*3+10
  void main()
  {
  int LIST[ ]={10,15,12,17};
  int *P=LIST, C;
  for(C=3; C>=0; C–)
  LIST[I]=Modify(LIST[I]);
  for (C=0; C<=3; C++)
  {
  cout<<*P<<“:”;
  P++;
  }
  }
  Ans. Considering LIST[I] being replaced with LIST[C] 40:55:46:61:
6. Look at the following C++ code and find the possible output(s) from the options (i) to (iv) following it. Also, write the highest and lowest values that can be assigned in the array A. (2)
  Note :
  - Assume all the required header files are already being included in the code.
  - The function random(n) generates an integer between 0 and n – 1.
    void main()
    {
    randomize();
    int A[4], C;
    for(C=0; C<4; C++)
    A[C]=random(C+1)+10;
    for(C=3; C>=0; C–)
    cout<<A[C]<<“@”;
    }
    (i) (ii)
    13@10@11@10@ 15$14$12$10$
    (iii) (iv)
    12@11@13@10@ 12@11@10@10@
    Ans. (i) and (iv)
    A_Min = 10
    A_Max = 13

Which function(s) out of the following can be considered as overloaded function(s) in the same program ? Also, write the reason for not considering the other(s) as overloaded function(s). (2)
void Execute(char A,int B); // Function 1
void Execute(int A,char B); // Function 2
void Execute(int P=10); // Function 3
void Execute(); // Function 4
int Execute(int A); // Function 5
void Execute(int &K); // Function 6
Ans. Option [i] Functions 1,2,3 are overloaded
Reason: Function 4,5,6 would give ambiguity for Function 3
OR Any equivalent valid reasonOROption [ii] Functions 1,2,4,5 are overloaded
Reason: Function 3 and 6 not considered in this case because it would give redeclaration error for Function 5
OR Any equivalent valid reasonOROption [iii] Functions 1,2,4,6 are overloaded
Reason: Function 3 and 5 not considered in this case because it would give redeclaration error for Function 6
OR Any equivalent valid reason
Observe the following C++ code and answer the questions (i) and (ii).
Note : Assume all necessary files are included.
class FIRST
{
int Num1;
public:
void Display() //Member Function 1
{
cout<<Num1<<endl;
}
};
class SECOND: public FIRST
{
int Num2;
public:
void Display() //Member Function 2
{
cout<<Num2<<endl;
}
};
void main()
{
SECOND S;
_______________ //Statement 1
_______________ //Statement 2
}
1. Which Object Oriented Programming feature is illustrated by the definitions of classes FIRST and SECOND ? (1)
  Ans. InheritanceOREncapsulationORData AbstractionORData Hiding
2. Write Statement 1 and Statement 2 to execute Member Function 1 and Member Function 2 respectively using the object S. (1)
  Ans. S.FIRST::Display() //Statement 1
  S.Display() //Statement 2ORS.SECOND::Display() //Statement 2

Write the definition of a class CONTAINER in C++ with the following description : (4)
Private Members
– Radius, Height // float
– Type // int (1 for Cone,2 for Cylinder)
– Volume // float
– CalVolume() // Member function to calculate
// volume as per the Type

Type	Formula to Calculate Volume
1	3.14RadiusHeight
2	3.14RadiusHeight/3

Public Members
– GetValues() // A function to allow user to enter value
// of Radius, Height and Type. Also, call
// function CalVolume() from it
– ShowAll() // A function to display Radius, Height,
// Type and Volume of Container
Ans. class CONTAINER
{
float Radius, Height;
int Type;
float Volume;
void CalVolume();
public:
void GetValues();
void ShowAll();
};
void CONTAINER::GetValues()
{
cin>>Radius>>Height>>Type ;
CalVolume();
} void CONTAINER::ShowAll()
{
cout<<Radius<<Height<<Type<<Volume<<endl;
}

void CONTAINER::CalVolume()
{
if (Type == 1)
Volume=3.14*Radius*Height;
else if (Type == 2)
Volume=3.14*Radius*Height/3;
}

void CONTAINER::CalVolume()
{
switch (Type)
{
case 1:
Volume =3.14*Radius*Height;
break;
case 2:
Volume=3.14*Radius*Height/3;
}
}

Answer the questions (i) to (iv) based on the following : (4)
class Teacher
{
int TCode;
protected:
char Name[20];
public:
Teacher();
void Enter(); void Show();
};
class Course
{
int ID;
protected:
Char Title[30];
public:
Course();
void Initiate();
void Display();
};
class Schedule : public Course, private Teacher
{
int DD,MM,YYYY;
public:
Schedule();
void Start();
void View();
};
void main()
{
Schedule S;
}
1. Which type of Inheritance out of the following is illustrated in the above example?
  Single Level Inheritance, Multilevel Inheritance, Multiple Inheritance
  Ans. Multiple Inheritance
2. Write the names of all the members, which are directly accessible by the member function View() of class Schedule.
  Ans. Start(), DD, MM, YYYY
  Display(), Initiate(), Title
  Enter(), Show(), Name
  View() // Optional
3. Write the names of all the members, which are directly accessible by the object S of class Schedule declared in the main() function.
  Ans. View(), Start()
  Display(), Initiate()
4. What will be the order of execution of the constructors, when the object S of class Schedule is declared inside the main() function ?
  Ans. Course(), Teacher(), Schedule()

1. Write the definition of a function SumEO(int VALUES[], int N) in C++, which should display the sum of even values and sum of odd values of the array separately. (2)
  Example : If the array VALUES contains
  20 20 22 21 53
  Then the functions should display the output as :
  Sum of even values = 42 (i.e., 20+22)
  Sum of odd values = 99 (i.e., 25+21+53)
  Ans. void SumEO(int VALUES[], int N)
  {
  int SE = 0, SO = 0;
  for (int I=0;I<N;I++)
  {
  if(VALUES[I] %2 == 0)
  SE += VALUES[I];
  else
  SO += VALUES[I];
  }
  cout<< “Sum of even values = ” << SE<<endl;
  cout<< “Sum of odd values = ” << SO<<endl;
  }OR
  Any other correct alternative code in C++
2. Write a definition for a function UpperHalf(int Mat[4][4]) in C++, which displays the elements in the same way as per the example shown below. (3)
  For example, if the content of the array Mat is as follows :
  25 24 23 22
  20 19 18 17
  15 14 13 12
  10 9 8 7
  The function should display the content in the following format :
  25 24 23 22
  20 19 18
  15 14
  10
  Ans. void UpperHalf(int Mat[4][4])
  {
  for (int I=0;I<4;I++)
  {
  for (int J=0;J<4-I;J++)
  cout<<MAT[I][J]<< ” ” ;
  cout<<endl;
  }
  }OR
  void UpperHalf(int Mat[4][4])
  {
  for (int I=0;I<4;I++)
  {
  for (int J=0;J<4;J++)
  if ((I+J)<=3)
  cout<<MAT[I][J]<< ” ” ;
  cout<<endl;
  }
  }
  OR
  Any other correct alternative code in C++
3. Let us assume Data[20][15] is a two-dimensional array, which is stored in the memory along the row with each of its elements occupying 2 bytes. Find the address of the element Data[10][5], if the element Data[15][10] is stored at the memory location 15000. (3)
  Ans. LOC(Data[10][5]) = LOC(Data[15][10])+2(15*(10-15)+(5-10))
  = 15000 + 2((-75) + (-5))
  = 15000 + 2(-80)
  = 15000 – 160
  = 14840ORLOC(Data[I][J]) = Base(Data)+W*(NC*(I-LBR)+(J-LBC))
  Taking LBR=0, LBC=0
  LOC(Data[15][10]) = Base(Data)+2*(15*15+10)
  15000 = Base(Data)+2*(15*15+10)
  Base(Data) = 15000 – 2*(235)
  Base(Data) = 15000 – 470
  Base(Data) = 14530
  LOC(Data[10][5])= 14530 + 2*(10*15+5)
  = 14530 + 2*(155)
  = 14530 + 310
  = 14840ORLOC(Data[I][J]) = Base(Data)+W*(NC*(I-LBR)+(J-LBC))
  Taking LBR=1, LBC=1
  LOC(Data[15][10]) = Base(Data)+2*(15*14+9)
  15000 = Base(Data)+2*(15*14+9)
  Base(Data) = 15000 – 2*(219)
  Base(Data) = 15000 – 438
  Base(Data) = 14562
  LOC(Data[10][5])= 14562 + 2*(15*9+4)
  = 14562 + 2*(139)
  = 14562 + 278
  = 14840
4. Write the definition of a member function AddPacket() for a class QUEUE in C++, to remove/delete a Packet from a dynamically allocated QUEUE of Packets considering the following code is already written as a part of the program. (4)
  struct Packet
  {
  int PID;
  char Address[20];
  Packet *LINK;
  };
  class QUEUE
  {
  Packet *Front, *Rear;
  public:
  QUEUE(){Front=NULL;Rear=NULL;}
  void AddPacket();
  void DeletePacket();
  ~QUEUE();
  };
  Ans. void QUEUE::AddPacket()
  {
  if(Front != NULL)
  {
  Packet *T;
  T=Front;
  cout<<Front->PID<<Front->Address<<” removed”<<endl;
  //OR cout<<T->PID<<T->Address<<” removed”<<endl;
  Front = Front->LINK;
  delete T;
  if (Front==NULL)
  Rear=NULL;
  }
  else
  cout<< “Queue Empty”<<endl;
  }ORAny other equivalent code in C++
5. Convert the following Infix expression to its equivalent Postfix expression, showing the stack contents for each step of conversion : (2)
  U * V + (W – Z) / X
  Ans. ((U * V) + ((W – Z) / X))
  INFIX STACK POSFIX
  U U
  * * U
  V * UV
  ) UV*
  + + UV*
  W UV*W
  – + – UV*W
  Z + – UV*WZ
  ) + UV*WZ-
  / +/ UV*WZ-
  X +/ UV*WZ-X
  ) + UV*WZ-X/
  ) UV*WZ-X/+
  OR
  U * V + (W – Z) / X
  INFIX STACK POSFIX
  U U
  * * U
  V * UV
  + + UV*
  ( +( UV*
  W +( UV*W
  – + (- UV*W
  Z + (- UV*WZ
  ) + UV*WZ-
  / +/ UV*WZ-
  X +/ UV*WZ-X
  UV*WZ-X/+

INFIX	STACK	POSFIX
U		U
*	*	U
V	*	UV
)		UV*
+	+	UV*
W		UV*W
–	+ –	UV*W
Z	+ –	UV*WZ
)	+	UV*WZ-
/	+/	UV*WZ-
X	+/	UV*WZ-X
)	+	UV*WZ-X/
)		UV*WZ-X/+

INFIX	STACK	POSFIX
U		U
*	*	U
V	*	UV
+	+	UV*
(	+(	UV*
W	+(	UV*W
–	+ (-	UV*W
Z	+ (-	UV*WZ
)	+	UV*WZ-
/	+/	UV*WZ-
X	+/	UV*WZ-X
		UV*WZ-X/+

A text file named MATTER.TXT contains some text, which needs to be displayed such that every next character is separated by a symbol ‘#’.
Write a function definition for HashDisplay() in C++ that would display the entire content of the file MATTER.TXT in the desired format. (3)
Example :
If the file MATTER.TXT has the following content stored in it :
THE WORLD IS ROUND
The function HashDisplay() should display the following content :
T#H#E# #W#O#R#L#D# #I#S# #R#O#U#N#D#
Ans. void HashDisplay()
{
char ch;
ifstream F(“MATTER.TXT” );

while(F.get(ch))
cout<<ch<<‘#’;
F.close(); //IGNORE
}ORAny other correct function definition
Write a definition for a function TotalTeachers( ) in C++ to read each object of a binary file SCHOOLS.DAT, find the total number of teachers, whose data is stored in the file and display the same. Assume that the file SCHOOLS.DAT is created with the help of objects of class SCHOOLS, which is defined below: (2)
class SCHOOLS
{
int SCode; //School Code
char SName[20]; //School Name
int NOT; //Number of Teachers in the school
public:
void Display()
{cout<<SCode<<“#”<<SName<<“#”<<NOT<<endl;}
int RNOT(){return NOT;}
};
Ans. void TotalTeachers()
{
ifstream F;
F.open(“SCHOOLS.DAT”,ios::binary);
int Count=0;
SCHOOLS S;
while(F.read((char*)&S,sizeof(S)))
Count += S.RNOT();
cout<<“Total number of teachers :”<<Count<<endl;
F.close(); //IGNORE
}ORvoid TotalTeachers()
{
ifstream F;
F.open(“SCHOOLS.DAT”,ios::binary);
SCHOOLS S;
while(F.read((char*)&S,sizeof(S)))
cout<<S.RNOT()<<endl;//OR S.Display();
F.close(); //IGNORE
}ORAny other correct function definition

Find the output of the following C++ code considering that the binary file SCHOOLS.DAT exists on the hard disk with the following records of 10 schools of the class SCHOOLS as declared in the previous question (4 b). (1)

SCode	SName	NOT
1001	Brains School	100
1003	Child Life School	115
1002	Care Share School	300
1006	Educate for Life School	50
1005	Guru Shishya Sadan	195
1004	Holy Education School	140
1010	Play School	95
1008	Innovate Excel School	300
1011	Premier Education School	200
1012	Uplifted Minds School	100

void main()
{
fstream SFIN;
SFIN.open(“SCHOOLS.DAT”,ios::binary | ios::in);
SCHOOLS S;
SFIN.seekg(5*sizeof(S));
SFIN.read((char*)&S, sizeof(S));
S.Display();
cout<<“Record :”<<SFIN.tellg()/sizeof(S) + 1<<endl;
SFIN.close();
}
Ans. 1004#Holy Education School#140
Record :7

SECTION B
[Only for candidates, who opted for Python]

1. Differentiate between Syntax Error and Run-Time Error. Also, write a suitable example in Python to illustrate both. (2)
  Ans. Syntax error : An error of language resulting from code that does not conform to the syntax of the programming language.
  Example
  a = 0
  while a < 10 # : is missing as per syntax
  a = a + 1
  print a
  Runtime error : A runtime error is an error that causes abnormal termination of program during running time..
  Example
  A=10
  B= int(raw_input(“Value:”))
  print A/B
  # If B entered by user is 0, it will be run-time error
2. Name the Python Library modules which need to be imported to invoke the following functions : (1)
  1. sin()
  2. search()
  Ans.
  1. math
  2. re
3. Rewrite the following code in Python after removing all syntax error(s). Underline each correction done in the code. (2)
  Val = int(rawinput(“Value:”))
  Adder = 0
  for C in range(1,Val,3)
  Adder+=C
  if C%2=0:
  Print C*10
  Else:
  print C*
  print Adder
  Ans. Val = int(raw_input(“Value:”)) # Error 1
  Adder = 0
  for C in range(1,Val,3) : # Error 2
  Adder+=C
  if C%2==0: # Error 3
  print C*10 # Error 4
  else: # Error 5
  print C # Error 6
  print AdderORCorrections mentioned as follows:
  raw_input in place of rawinput
  : to be placed in for
  == in place of =
  print in place of Print
  else in place of Else
  C* is invalid, replaced by a suitable integer or C
4. Find and write the output of the following Python code : (2)
  Data = [“P”,20,”R”,10,”S”,30] Times = 0
  Alpha = ” ”
  Add = 0
  for C in range(1,6,2):
  Times = Times + C
  Alpha = Alpha + Data[C-1]+”$”
  Add = Add + Data[C] print Times, Add, Alpha
  Ans.1 20 P$
  4 30 P$R$
  9 60 P$R$S$
5. Find and write the output of the following Python code : (3)
  class GRAPH:
  def __init__(self,A=50,B=100):
  self.P1=A
  self.P2=B
  def Up(self,B):
  self.P2 = self.P2 – B
  def Down(self,B):
  self.P2 = self.P2 + 2*B
  def Left(self,A):
  self.P1 = self.P1 – A
  def Right(self,A):
  self.P1 = self.P1 + 2*A
  def Target(self):
  print “(“,self.P1.”:”,self.P2,”)”
  G1=GRAPH(200,150)
  G2=GRAPH()
  G3=GRAPH(100)
  G1.Left(10)
  G2.Up(25)
  G3.Down(75)
  G1.Up(30)
  G3.Right(15)
  G1.Target()
  G2.Target()
  G3.Target()
  Ans. ( 190 : 120 )
  ( 50 : 75 )
  ( 130 : 250 )
6. What possible output(s) are expected to be displayed on screen at the time of execution of the program from the following code ? Also specify the maximum values that can be assigned to each of the variables BEGIN and LAST. (2)
  import random
  POINTS=[20,40,10,30,15];
  POINTS=[30,50,20,40,45];
  BEGIN=random.randint(1,3)
  LAST=random.randint(2,4)
  for C in range(BEGIN,LAST+1):
  print POINTS[C],”#”,
  (i) 20#50#30# (ii) 20#40#45#
  (iii) 50#20#40# (iv) 30#50#20
  Ans. (ii) 20#40#45# and (iii) 50#20#40#
  Max value for BEGIN 3
  Max value for LAST 4
1. What is the advantage of super( ) function in inheritance ? Illustrate the same with the help of an example in Python. (2)
  Ans. In Python, super() function is used to call the methods of base class which have been extended in derived class.
  class person(object):
  def __init__(self,name,age):
  self.name=name
  self.age=age
  def display(self):
  print self,name, self.Age
  class student(person):
  def __init__(self,name,age,rollno,marks):
  super(student,self)._init_(self, name, age)
  self.rollno=rollno
  self.marks=marks
  def getRoll(self):
  print self.rollno, self.marks
2. class Vehicle: #Line 1 (2)
  Type = ‘Car’ #Line 2
  def __init__(self, name): #Line 3
  self.Name = name #Line 4
  def Show(self): #Line 5
  print self.Name,Vehicle.Type #Line 6
  V1=Vehicle(“BMW”) #Line 7
  V1.Show() #Line 8
  Vehicle.Type=”Bus” #Line 9
  V2=Vehicle(“VOLVO”) #Line 10
  V2.Show() #Line 11
  1. What is the difference between the variable in Line 2 and Line 4 in the above Python code ?
    Ans. The variable in Line 2 is a class attribute. This belongs to the class itself. These attributes will be shared by all the instances.
    The variable in Line 4 is an instance attribute. Each instance creates a separate copy of these variables.
  2. Write the output of the above Python code.
    Ans. BMW Car
    VOLVO Bus
3. Define a class CONTAINER in Python with the following specifications : (4)
  Instance Attributes
  – Radius,Height # Radius and Height of Container
  – Type # Type of Container
  – Volume # Volume of Container
  Methods
  – CalVolume() # To calculate volume
  # as per the Type of container
  # With the formula as given below :
  Type Formula to Calculate Volume
  1 3.14 * Radius * Height
  3 3.14 * Radius * Height/3
  – GetValue() # To allow user to enter values of
  # Radius, Height and Type.
  # Also, this method should call
  # CalVolume() to calculate Volume
  – ShowContainer() # To display Radius, Height, Type
  # Volume of the Container
  Ans. class CONTAINER: # class CONTAINER():/class CONTAINER(Object):
  def __init__(self): # def __init__(self,R,H,T,V):
  self.Radius=0 # self.Radius=R
  self.Height=0 # self.Height=H
  self.Type =0 # self.Type=T
  self.Volume=0 # self.Volume=V
  def CalVolume(self):
  if self.Type == 1:
  self.Volume = 3.14 * self.Radius * self.Height
  elif self.Type ==3:
  self.Volume = 3.14 * self.Radius * self.Height /3
  def GetValue(self):
  self.Radius = input(“Enter Radius”)
  self.Height = input(“Enter Height”)
  self.Type = input(“Enter type”)
  self.CalVolume() # OR CalVolume(self)
  def ShowContainer(self):
  print self.Radius
  print self.Height
  print self.Type
  print self.Volume
4. Answer the questions (i) to (iv) based on the following : (4)
  Class Top1(object):
  def __init__(self,tx): #Line 1
  self.X=tx #Line 2
  def ChangeX(self,tx):
  self.X=self.X+tx
  def ShowX(self):
  print self.XClass Top2(object):
  def __init__(self,ty): #Line 3
  self.Y=ty #Line 4
  def ChangeY(self,ty):
  self.Y=self.Y+ty
  def ShowY(self):
  print self.Y,class Bottom(Top1,Top2):
  def __init__(self,tz): #Line 5
  self.Z=tz #Line 6
  Top2.__init__(self,2*tz): #Line 7
  Top1.__init__(self,3*tz): #Line 8
  def ChangeZ(self,tz):
  self.Z=self.Z+tz
  self.ChangeY(2*tz)
  self.ChangeX(3*tz)
  def ShowZ(self):
  print self.Z,
  self.ShowY()
  self.ShowX()
  B=Bottom(1)
  B.ChangeZ(2)
  B.ShowZ()
  1. Write the type of the inheritance illustrated in the above.
    Ans. Multiple Inhertiance
  2. Find and write the output of the above code.
    Ans. 3 6 9OR“Error” / “No Output”
  3. What are the methods shown in Line 1, Line 3 and Line 5 known as ?
    Ans. Constructors
  4. What is the difference between the statements shown in Line 6 and Line 7 ?
    Ans. Initializing the member of child class in Line 6 and calling the parent class constructor in Line 7
1. Consider the following randomly ordered numbers stored in a list : (3)
  786, 234, 526, 132, 345, 467
  Show the content of the list after the First, Second and Third pass of the bubble sort method used for arranging in ascending order ?
  Note : Show the status of all the elements after each pass very clearly underlining the changes.
  Ans.
  I Pass 234 526 132 345 467 786
  II Pass 234 132 345 467 526 786
  III Pass 132 234 345 467 526 786
2. Write the definition of a method ZeroEnding(SCORES) to add all those values in the list of SCORES, which are ending with zero (0) and display the sum. (3)
  For example :
  If the SCORES contain [200, 456, 300, 100, 234, 678] The sum should be displayed as 600
  Ans. def ZeroEnding(SCORES):
  s=0
  for i in SCORES:
  if i%10==0:
  s=s+i
  print s
3. Write AddClient(Client) and DeleteClient(Client) methods in Python to add a new Client and delete a Client from a List of Client Names, considering them to act as insert and delete operations of the queue data structure. (4)
  Ans. def AddClient(Client):
  C=raw_input(“Client name: “)
  Client.append(C)
  def DeleteClient(Client):
  if (Client==[]):
  print “Queue empty”
  else:
  print Client[0],”Deleted”
  del Client[0] # OR Client.pop(0)ORclass queue:
  Client=[] def AddClient(self):
  a=raw_input(“Client name: “)
  queue.Client.append(a)
  def DeleteClient(self):
  if (queue.Client==[]):
  print “Queue empty”
  else:
  print queue.Client[0],”Deleted”
  del queue.Client[0]
4. Write a definition of a method COUNTNOW(PLACES) to find and display those place names, in which there are more than 5 characters. (2)
  For example :
  If the list PLACES contains
  [“DELHI”,”LONDON”,”PARIS”,”NEW YORK”,”DUBAI”] The following should get displayed :
  LONDON
  NEW YORK
  Ans. def COUNTNOW(PLACES):
  for P in PLACES:
  if len(P)>5:
  print P
5. Evaluate the following Postfix notation of expression : (2)
  22,11,/,5,10,*,+,12,-
  Ans.
  Elements Stack Contents
  22 22
  11 22, 11
  / 2
  5 2, 5
  10 2, 5, 10
  * 2, 50
  + 52
  12 52, 12
  – 40
  OR
  Any other way of stepwise evaluation
1. Write a statement in Python to open a text file STORY.TXT so that new contents can be added at the end of it. (1)
  Ans.file= open(“STORY.TXT”,”a”) OR file.open(“STORY.TXT”,”a”)
2. Write a method in Python to read lines from a text file INDIA.TXT, to find and display the occurrence of the word ‘‘India’’. (2)
  For example :
  If the content of the file is
  ‘‘India is the fastest growing economy.
  India is looking for more investments around the globe.
  The whole world is looking at India as a great market.
  Most of the Indians can foresee the heights that India is capable of reaching.’’
  The output should be 4.
  Ans. def display1():
  c=0
  file=open(‘INDIA.TXT’,’r’)
  c=0
  for LINE in file:
  Words = LINE.split()
  for W in Words:
  if W==”India”:
  c=c+1
  print c
  file.close()
  OR
  def display():
  c=0
  file=open(‘INDIA.TXT’,’r’)
  lines = file.read() # lines = file.readline()
  while lines:
  words = lines.split()
  for w in words:
  if w==”India”:
  c=c+1
  lines = file.read() # lines = file.readline()
  print c
  file.close()
3. Considering the following definition of class MULTIPLEX, write a method in Python to search and display all the contents in a pickled file CINEMA.DAT, where MTYPE is matching with the value ‘Comedy’. (3)
  class MULTIPLEX :
  def __init__(self,mno,mname,mtype):
  self.MNO = mno
  self.MNAME = mname
  self.MTYPE = mtype
  def Show(self):
  print self.MNO:”*”,self.MNAME,”$”,self.MTYPE
  Ans.def Search():
  file=open(‘CINEMA.DAT’,’rb’)
  try:
  while True:
  M=pickle.load(file)
  if M.MTYPE==”Comedy”:
  M.Show()
  except EOFError:
  pass
  file.close()

Type	Formula to Calculate Volume
1	3.14 * Radius * Height
3	3.14 * Radius * Height/3

Elements	Stack Contents
22	22
11	22, 11
/	2
5	2, 5
10	2, 5, 10
*	2, 50
+	52
12	52, 12
–	40

SECTION C
[For all the candidates]

Observe the following tables VIDEO and MEMBER carefully and write the name of the RDBMS operation out of (i) SELECTION (ii) PROJECTION (iii) UNION (iv) CARTESIAN PRODUCT, which has been used to produce the output as shown below. Also, find the Degree and Cardinality of the final result. (2)
TABLE : VIDEO

VNO	VNAME	TYPE
F101	The Last Battle	Fiction
C101	Angles and Devils	comedy
A102	Daredevils	Adventure

TABLE : MEMBER

VNO	MNAME
M101	Namish Gupta
M102	Sana Sheikh
M103	Lara James

TABLE : FINAL RESULT

VNO	VNAME	TYPE	MNO	MNAME
F101	The Last Battle Fiction	Fiction	M101	Namish Gupta
F101	The Last Battle Fiction	Fiction	M102	Sana Sheikh
F101	The Last Battle Fiction	Fiction	M103	Lara James
C101	Angels and Devils	Comedy	M101	Namish Gupta
C101	Angels and Devils	Comedy	M102	Sana Sheikh
C101	Angels and Devils	Comedy	M103	Lara James
A102	Daredevils	Adventure	M101	Namish Gupta
A102	Daredevils	Adventure	M102	Sana Sheikh
A102	Daredevils	Adventure	M103	Lara James

Ans. CARTESIAN PRODUCT
OR Option (iv)
DEGREE = 5
CARDINALITY = 9

Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables. (6)
TABLE : ACCOUNT
ANO ANAME ADDRESS
101 Nirja Singh Bangalore
102 Rohan Gupta Chennai
103 Ali Reza Hyderabad
104 Rishabh jain Chennai
105 Simran Kaur Chandigarh
TABLE : TRANSACT
TRNO ANO AMOUNT TYPE DOT
T001 101 2500 Withdraw 2017-12-21
T002 103 3000 Deposit 2017-06-01
T003 102 2000 Withdraw 2017-05-12
T004 103 1000 Deposit 2017-10-22
T005 101 12000 Deposit 2017-11-06
1. To display details of all transactions of TYPE Deposit from Table TRANSACT.
  Ans. SELECT * FROM TRANSACT WHERE TYPE = ‘ Deposit ‘ ;
2. To display the ANO and AMOUNT of all Deposits and Withdrawals done in the month of October 2017 from table TRANSACT.
  Ans. SELECT ANO,AMOUNT FROM TRANSACT
  WHERE DOT >= ‘2017-10-01 ‘ AND DOT <= ‘ 2017-10-31 ‘ ;ORSELECT ANO,AMOUNT FROM TRANSACT
  WHERE DOT BETWEEN ‘2017-10-01’ AND ‘2017-10-31 ‘ ;
3. To display the last date of transaction (DOT) from the table TRANSACT for the Accounts having ANO as 103.
  Ans. SELECT MAX(DOT) FROM TRANSACT WHERE ANO = 103;
4. To display all ANO, ANAME and DOT of those persons from tables ACCOUNT and TRANSACT who have done transactions less than or equal to 3000.
  Ans. SELECT ACCOUNT.ANO,ANAME,DOT FROM ACCOUNT,TRANSACT WHERE
  ACCOUNT.ANO=TRANSACT.ANO AND AMOUNT <=3000;ORSELECT A.ANO,ANAME,DOT FROM ACCOUNT A,TRANSACT T WHERE
  A.ANO=T.ANO AND AMOUNT <=3000;
5. SELECT ANO, ANAME FROM ACCOUNT
  WHERE ADDRESS NOT IN (‘CHENNAI’, ‘BANGALORE’);
  Ans.
  ANO ANAME
  103 Ali Reza
  105 Simran Kaur
  OR
  ANO ANAME
  101 Nirja Singh
  102 Rohan Gupta
  103 Ali Reza
  104 Rishabh Jain
  105 Simran Kaur
6. SELECT DISTINCT ANO FROM TRANSACT;
  Ans.DISTINCT ANO
  101
  102
  103
7. SELECT ANO, COUNT(*), MIN(AMOUNT) FROM TRANSACT
  GROUP BY ANO HAVING COUNT(*)> 1;
  Ans.
  ANO COUNT (*) MIN (ANOUNT)
  101 2 2500
  103 2 1000
8. SELECT COUNT(*), SUM(AMOUNT) FROM TRANSACT
  WHERE DOT <= ‘2017-06-01’;
  Ans.
  COUNT (*) SUM (AMOUNT)
  2 5000

ANO	ANAME	ADDRESS
101	Nirja Singh	Bangalore
102	Rohan Gupta	Chennai
103	Ali Reza	Hyderabad
104	Rishabh jain	Chennai
105	Simran Kaur	Chandigarh

TRNO	ANO	AMOUNT	TYPE	DOT
T001	101	2500	Withdraw	2017-12-21
T002	103	3000	Deposit	2017-06-01
T003	102	2000	Withdraw	2017-05-12
T004	103	1000	Deposit	2017-10-22
T005	101	12000	Deposit	2017-11-06

ANO	ANAME
103	Ali Reza
105	Simran Kaur

COUNT (*)	SUM (AMOUNT)
2	5000

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(i)	(ii)
13@10@11@10@	15$14$12$10$
(iii)	(iv)
12@11@13@10@	12@11@10@10@

(i) 20#50#30#	(ii) 20#40#45#
(iii) 50#20#40#	(iv) 30#50#20

I Pass	234	526	132	345	467	786
II Pass	234	132	345	467	526	786
III Pass	132	234	345	467	526	786