MARKING SCHEME FRAGMENT-2 EXAMINATION(2019-20) …ahiyaraipur.thelps.edu.in/UploadedFiles/UpdateDirectory/Computer... · xii/cs(283)/fragment-2-exam. page 1 of 18 marking scheme fragment-2

XII/CS(283)/FRAGMENT-2-EXAM. PAGE 1 OF 18

MARKING SCHEME

FRAGMENT-2 EXAMINATION(2019-20)

CLASS-XII

SUBJECT: COMPUTER SCIENCE(283)

TIME: 3 Hrs M.MARKS:70 General Instructions:

(a) All questions are compulsory.

(b) Programming Language : C++

(c) Read the questions before writing answers.

1 a) What is the output of this program? Select the correct option.

#include <iostream.h>

void main()

{

intarr[] = {4, 5, 6, 7};

int *p = (arr + 1);

cout<<arr;

}

A. 4,5,6,7

B. 5

C. address of 4

D. 7

1

Answer: C. address of 4

1 Mark for correct option

b) Which of the following declaration(s) are illegal?

A. char *str[ ]={“hello”};

B. char *str = “hello”;

C. char str = “hello”;

D. char *str=[“hello”];

1

Answer: C. char str = “hello”;

D.char *str=[“hello”];

½ Mark for each correct option

c) Give difference between the following:

i) int *n=new int(5);

ii) int *n=new int[5];

1

Answer: In i) case n pointer address is initialized by value 5 where as

in ii) array of 5 integer pointers is created and represented by n.

1 Mark for correct difference


½ Mark for only explaining one part

d) Find the output of the following program :

#include<iostream.h>

void main ( )

{

int X[ ] = {10, 25, 30, 55, 100};

int *p = X ;

while ( *p < 110)

{

if (*p%3!=0)

*p = *p + 1 ;

else

*p = *p + 2 ;

p++;

}

for(int I = 4 ;I>= 1 ;I--)

{

cout<< X[I] << “*” ;

if ( I%3 = = 0) cout<<endl ;

}

cout<<X[0]*3<<endl ;

}

2

Answer: 101*56*

32*26*33

1 Mark for each correct output(line)

½ Mark is deducted if ‘*’ not printed

½ Mark deducted if output is displayed in same line

e) Find and write the output of the following C++ program code

Note: Assume all required header files are already being included in the

program.

void main( )

{

char P[ ]=”google”;

int points[ ]={100,200,101,201};

char *K=P;

int *G=points;

cout<<*K<<”<<”<<*G%4<<endl;

K=K+2;

G+=2;

cout<<K<<”::”<<*G<<endl;

3


*K=*K+1;

*G+=10;

cout<<K<<”==”<<*G;

}

Answer:

g<<0

ogle::101

pgle==111

1 Mark for each correct output(line)

½ Mark deducted if output is displayed in same line

f) Find and write the output of the following C++ program code :

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[C]=Modify(LIST[C]);

for (C=0; C<=3; C++)

{

cout<<*P<<":";

P++;

}

}

2

Answer:

40:55:46:61:

1 Mark for correct output

½ Mark deducted for output without ‘:’

2 a) Write the definition of a member function DeletePacket( ) only, of a class

QUEUE in C++, remove/delete Packet information from a dynamically

allocated QUEUE of Packets considering the following code is already written

as a part of the program.

struct Packet {

int PID;

char Address[20];

Packet *LINK;

4


};

class QUEUE {

Packet *Front, *Rear;

public:

QUEUE( )

{

Front=NULL;

Rear=NULL;

}

void AddPacket( );

void DeletePacket( );

void TRAVERSE( );

~QUEUE();

};

Answer:

void QUEUE ::DeletePacket( )

{

if(Front==NULL)

cout<<”empty”;

else

{

Packet *T=Front;

Front=Front->LINK;

delete T;

T=NULL;

}

}

1 Mark for checking if Queue is Empty

1 Mark for assigning Front to T

1 Mark for changing Front value

1 Mark for deleting T

Marks can be given for any other similar answer

b) Write functions in C++ to perform insert operation in a static circular Queue

containing Book‟s information (represented with the help of an array of

structure BOOK).

struct BOOK

{ long Accno; // Book Accession Number

char Title [20]; // Book Title

};

void insertCQ(BOOK CQ[ ],BOOK T,int&rear, int&front,int size)

{

if(front==0&&rear==size-1||front==rear+1)

3


cout<<”full”;

else if(rear==-1)

{

front=rear=0;

CQ[rear]=T;

}

else if(rear==size-1)

{

rear=0;

CQ[rear]=T;

}

else

CQ[++rear]=T;

}

3Marks for correct function definition

1 Mark for each for checking for overflow

½ Mark for each for checking for empty

1 Mark for reset rear to0 condition

½ Mark for inserting for normal case

o Note:Parameters can be globally declared

c) Write the definition of a member function PUSH( )in C++, to add a new book

in a dynamic stack of TEXTBOOK considering the following code is already

included in the program :

struct TEXTBOOK

{ char ISBN[20], TITLE[80];

TEXTBOOK *Next;

};

class STACK

{ TEXTBOOK *Top;

public:

STACK( ){Top=NULL;}

void PUSH();

void POP( );

~STACK( );

};

4

Answer:

void STACK::PUSH()

{

//create a Textbook


TEXTBOOK *R=new TEXTBOOK;

gets(R->ISBN);

gets(R->TITLE);

R->Next=NULL;

if(Top==NULL)

Top=R;

else

{

R->Next=Top;

Top=R;

}

}

1 Mark for creating a new node and assigning/entering appropriate

values in it

1 Mark for checking if Queue is Empty)

1 Mark for assigning Rear and Front as Temp - if Queue is Empty)

1 Mark for assigning Rear->Link as Front and Rear as Temp


d) Write the definition of insertQ(int ) and deleteQ( ) to insert and delete elements

for the statically allocated queue as per following code.

#include<iostream.h>

#define size 4

class QUEUE

{

int data[size];

int front,rear;

public: QUEUE( );

void insertQ( int);

void deleteQ( );

void display( );

};

4

Answer:

void QUEUE::insertQ(int n)

{

if(rear==size-1)

cout<<”full”

else if(rear==-1)

{

rear=front=0;

data[rear]=n;

}

else


data[++rear]=n;

}

void QUEUE::deleteQ()

{

if(front==-1)

cout<<”empty”;

elseif(front==rear)

front=rear=-1;

else

front++;

}

2 Marks for each correct function definition

½ Mark for each if condition(for full/empty)

1 ½ Mark for each insert/delete part

No deduction if :: is missing


3 a) A Queue is a linear list implemented in____ form.

A. LIFO

B. FIFO

C. SIFO

D. FISO

1

Answer: B. FIFO

1 Mark correct answer

b) Write two applications of Stack in computer System.

1

Answer:

To evaluate mathematical expression.

For execution of various nested functions.

½ Mark for each application

Any other applications can also be accepted

c) Write the definition of PUSH(int) and POP() to insert and delete element from a

statically allocated stack represented by the following code.

const int size=10;

class STACK

{

int STK[size];

int top;

public:

STACK(){top=-1;}

void PUSH(int);

void POP();

void traverse();

4


};

Answer:

void STACK::PUSH(int n)

{

if(top==size-1)

cout<<”full”;

else

STK[++top]=n;

}

void STACK::POP( )

{

if(top==-1)

cout<<”empty”

else

top--;

}

2 Marks for each correct function definition

1 Mark for each if condition(for full/empty)

1 Mark for each insert/delete part

No deduction if :: is missing


d) Convert the following infix expression to its equivalent postfix expression

showing stack contents for the conversion:

X – Y / (Z + U) * V

OR

Convert the following infix expression into postfix. show the stack status after

execution of each operation:

(TRUE OR FALSE )AND NOT FALSE OR FALSE

2

Answer:

Mark start and end of expression by enclosing in between [ ]

[X – Y / (Z + U) * V ]

Read

character

stack Postfix expression

[ [

X [ X

- [- X

Y [- XY

/ [-/ XY

( [-/( XY

Z [-/( XYZ


+ [-/(+ XYZ

U [-/(+ XYZU

) [-/ XYZU+

* [-* XYZU+/

V [-* XYZU+/V

] EMPTY XYZU+/V*-

Postfix expression: XYZU+/V*-

ORPart

Mark start and end of expression by enclosing in between [ ]

[(TRUE OR FALSE) AND NOT FALSE OR FALSE]

Read

character

stack Postfix expression

[ [

( [(

TRUE [( TRUE

OR [(OR TRUE

FALSE [(OR TRUE FALSE

) [ TRUE FALSE OR

AND [AND TRUE FALSE OR

NOT [AND NOT TRUE FALSE OR

FALSE [AND NOT TRUE FALSE OR FALSE

OR [OR TRUE FALSE OR FALSE NOT AND

FALSE [OR TRUE FALSE OR FALSE NOT AND FALSE

] Empty TRUE FALSE OR FALSE NOT AND FALSE

OR

Postfix expression : TRUE FALSE OR FALSE NOT AND FALSE OR

2 Marks for correct solution(any one)

1 Marks if only correct answer is written

½ Mark for each 4 steps shown correctly in sequence

e) Evaluate the following postfix expression and show the status of the stack after

every step .

12, 7, 3, – , /, 2, 1, 5, +, *, +

OR

TRUE, FALSE, NOT, AND, NOT,FALSE,OR

2

Answer:

12, 7, 3, – , /, 2, 1, 5, +, *, +

Steps Read

character

Action

performed

Stack Intermediate operations

1 12 PUSH 12

2 7 PUSH 12,7

3 3 PUSH 12,7,3

4 - POP 12,7 OP2=3


POP

PUSH

12

12,4

OP1=7

RESULT=OP1-OP2=4

5 / POP

POP

PUSH

12

Empty

3

OP2=4

OP1=12

RESULT=OP1/OP2=3

6 2 PUSH 3,2

7 1 PUSH 3,2,1

8 5 PUSH 3,2,1,5

9 + POP

POP

PUSH

3,2,1

3,2

3,2,6

OP2=5

OP1=1

RESULT=OP1+OP2=6

10 * POP

POP

PUSH

3,2

3

3,12

OP2=6

OP1=2

RESULT=OP1*OP2=12

11 + POP

POP

PUSH

3

EMPTY

15

OP2=12

OP1=3

RESULT=OP1+OP2=15

ANSWER:15

OR part

TRUE, FALSE, NOT, AND, NOT,FALSE,OR

Steps Read

character

Action

performed

Stack Intermediate operations

1 TRUE PUSH TRUE

2 FALSE PUSH TRUE, FALSE

3 NOT POP

PUSH

TRUE

TRUE,TRUE

OP=FALSE

RESULT =NOT

OP=TRUE

4 AND POP

POP

PUSH

TRUE

EMPTY

TRUE

OP2=TRUE

OP1=TRUE

RESULT=OP1 AND

OP2=TRUE

5 NOT POP

PUSH

EMPTY

FALSE

OP=TRUE

RESULT =NOT

OP=FALSE

6 FALSE PUSH FALSE,FALSE

7 OR POP

POP

PUSH

FALSE

EMPTY

FALSE

OP2=FALSE

OP1=FALSE

RESULT=OP1 OR

OP2=FALSE

ANSWER:FALSE

2 Marks for correct solution(any one)


½ Marks if only correct answer is written

½ Mark for each 2/3 steps shown correctly in sequence

Intermediate operation may not be written

4 a) According to Commutative law , which of the following options is correct

(a) A+B = B+A (b) A.B = B.A

(c) Both (a) & (b) (d) None of these

1

Answer: (c)

1 Mark for correct answer

½ mark for only writing (a) or (b)

b) Any Boolean expression that is either in the form of sum of minterms or

product of maxterms is said to be in _______________

(a) Standard form (b) Non standard form

(c) Canonical form (d) None of the above

1

Answer: (c) Canonical form


c) Given the truth table of a function F(X, Y, Z). Write the S-O-P and P-O-S

expression from the following truth table:

X Y Z F

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

1

Answer:

SOP:

F(X, Y, Z)=X‟Y‟Z+X‟YZ‟+X‟YZ+XYZ

POS:

F(X, Y, Z)=(X+Y+Z)(X‟+Y+Z)(X‟+Y+Z‟)(X‟+Y‟+Z)

½ Mark each for correct SOP and POS

d) State the principle of duality in Boolean Algebra and give the dual of the

Boolean expression

(X + Y) . (X‟ + Z‟) . (Y + Z)

2

Answer:

It states that from a given Boolean expression another expression can be formed

by carrying out following steps:


Change all AND’s with OR’s

Change all OR’s with AND’s

Change all 0 to 1 or vice-versa

Dual of (X + Y) . (X‟ + Z‟) . (Y + Z)

is (X.Y) +(X‟.Z‟)+(Y.Z)

1 Mark for defining principle of duality

1 Mark for dual of given expression

e) State and prove any one De Morgan‟s law with the help of Truth Table.

2

Answer:

De Morgan‟s Law: (X.Y)‟=X‟+Y‟

X Y X‟ Y‟ X.Y (X.Y)‟ X‟+Y‟

0 0 1 1 0 1 1

0 1 1 0 0 1 1

1 0 0 1 0 1 1

1 1 0 0 1 0 0

Values Of columns (X.Y)‟ and X‟+Y‟ are same hence (X.Y)‟=X‟+Y‟

OR

De Morgan‟s Law: (X+Y)‟=X‟.Y‟

X Y X‟ Y‟ X+Y (X+Y)‟ X‟.Y‟

0 0 1 1 0 1 1

0 1 1 0 1 0 0

1 0 0 1 1 0 0

1 1 0 0 1 0 0

Values Of columns (X+Y)‟ and X‟.Y‟ are same hence (X+Y)‟=X‟.Y‟

1 Mark for Writing any one of the De Morgan’s Law

1 Mark for showing proofing using Truth Table.

f) State and prove Absorption law algebraically. 2

Answer:

X+XY=X

L.H.S

= X+XY

=X(1+Y) [as 1+Y=1]

=X(1) [as 1.x=x]

=X =R.H.S

OR


X.(X+Y)=X

LHS

X.(X+Y)

=X.(X+Y)

=X.X+X.Y

=X+XY [X.X=X]

=X(1+Y) [1+Y=1]

=X(1) [1.X=X]

=X RHS

1 Mark for writing any one of the Absorption law

1 Mark for its Proofing

5 a) Find the complement of the following Boolean expression:

(a + b)(c' + a')(b + d)

1

Answer:

Answer:

(a‟.b‟)+(c.a‟)+(b‟.d‟)


½ Mark for any two correct term in the expression

b) Draw the logic circuit for : A‟B+AB‟ 2

Answer:


½ Mark for circuit for A’Bor AB’ only

c) Write the Boolean function/expression for the following logic circuits:

2


Answer:

(A.B+C).D‟


½ Mark only for partial correct answer.

d) Reduce the following Boolean expression using K-map:

F(A,B,C,D)=∑(0,1,3,4,5,6,7,8,9,10,12,13)

3

Answer:

AB CD

Octet : m0+m1 m4+m5+m12+m13+ m8+m9 reduces to C’

Quad 1: m1+m3 m5+m7 reduces to A‟D

Quad 2: m4+m5 m6+m7 reduces to A‟B

Pair :m8+m10 reduces to A‟BD‟

Therefore final reduced expression:

F(A,B,C,D)=C‟+A‟D+A‟B+A‟BD‟

½ Mark for placing all 1s at correct positions in K-Map

½ Mark for each grouping

(1 Mark for writing final expression in reduced/minimal form)

Note: Deduct ½ mark if wrong variable names are used

Marks can be given for any other solution following K-Map reduction method

C’D’ C’D CD CD’

00 01 11 10

A’B’

00

1

0

1

1

1

3

2

A’B

01

1

4

1

5

1

7

1

6

AB

11

1

12

1

13

15

14

A’B

10

1

8

1

9

11

1

10


e) F(W, X, Y,Z) = π (0,1, 2, 4, 5, 8, 9, 10,15) 3

WX YZ

Quad 1: M0.M1.M4.M5 reduces to W+Y

Quad 1: M0.M1.M8.M9reduces to X+Y

Quad 1: M0.M2.M8.M10reduces to X+Z

Single :M15: W‟+X‟+Y‟+Z‟

Therefore final reduced expression:

F(W,X,Y,Z)= (W+Y)( X+Y)( X+Z)( W‟+X‟+Y‟+Z‟)

½ Mark for placing all 1s at correct positions in K-Map

½ Mark for each grouping

(1 Mark for writing final expression in reduced/minimal form)

Note: Deduct ½ mark if wrong variable names are used

Marks can be given for Any other solution following K-Map reduction

method

Y+Z Y+Z’ Y’+Z’ Y’+Z

00 01 11 10

W+X

00

0

0

0

1 3

0

2

W+X’

01

0

4

0

5 7

6

W’+X’

11 12 13

0

15

14

W’+X

10

0

8

0

9

11

0

10

6 a) Give one difference between:

i. Cardinality and Degree.

ii. Candidate key and Alternate key.

2

Answer:

i. Cardinality is number of records or tuples in a Table and Degree is

number of fields or columns in a table.

ii. Candidate is(are) the attribute(s) in a table that is(are) eligible to


become primary key in a table whereas alternate key is the candidate

key(s) that did not become the primary key.

1 Marks for each correct part

½ Mark for each part for partially answer

Marks can be given for another answer for similar meaning

b) Consider the following tables Stationery.

Table: Stationery

S_ID StationeryName Company Price

BP01 Ball Pen Reynolds 10

PL02 Pencil Natraj 5

ER05 Eraser Natraj 3

PL01 Pencil Apsara 6

GP02 Gel Pen Reynolds 15

Write SQL statements for (i) to (vi) (6)

(i) To display the Stationery detail in descending order of their company name.

(ii) To display the Name and Price of Stationery whose Price is in the range 10

to 15.

(iii) To display the StationeryName,S_id for Stationery where company is

either "Reynolds" or „Apsara‟.

(iv) To increase the Price of all Stationery by 20 Rupees where Stationery name

contains letter „c‟.

(v) To delete the items for „Reynolds‟ where price is below 10.

(vi) To insert the record: C101,‟file‟,‟faber‟,NULL

Write the output of SQL queries for (vii) to (x) on the basis of original table.(2)

(vii) SELECT * FROM Stationery WHERE Stationery Name Like 'B%'

(viii) SELECT DISTINCT Company FROM Stationery;

(ix) SELECT Company, S_ID as “code” from Stationery where NOT

price<10;

(x) SELECT StationeryName, Price FROM Stationery where price>=10

order by price desc;

8

Answer:

(i) SELECT * from Stationery order by company;


½ Mark for answer without order by clause.

(ii) Select stationery Name,price from stationery where price between 10

and 15;

Or

Select stationery Name,price from stationery where price >= 10 and price<=15;



½ Mark for answer without order by clause.

(iii) SELECT StationaryName, S_id FROM Stationery WHERE company

IN(„REYNOLDS‟,‟APSARA‟)

Or

SELECT StationaryName, S_id FROM Stationery WHERE company

=„REYNOLDS‟ or company =‟APSARA‟;


½ Mark for answer without PROPER CONDITION.

(iv) UPDATE STATIONERY SET PRICE=PRICE+20 WHERE

StationaryName LIKE „%C%‟;


½ Mark for answer without proper condition.

(v) DELETE from STATIONERY WHERE PRICE BELOW<10 and

company=‟ Reynolds‟;


½ Mark for answer without proper condition.

(vi) INSERT INTO STATIONERY

VALUES(„C101‟,‟file‟,‟faber‟,NULL);

(vii)

S_ID StationaryName Company Price

BP01 Ball Pen Reynolds 10

½ Mark for correct output with or without headings

(viii)

Company

Reynolds

Natraj

Apsara


(ix)

Company code

Reynolds BP01

Reynolds GP02

½ Mark for correct output with headings

(x)

StationeryName Price

Gel Pen 15

Ball Pen 10



7 a) Give two differences between DDL and DML commands. 2

Answer:

DDL( Data definition Language commands) are used to create and modify

structure of database whereas DML (Data manipulation language commands)

are used for reecord maintenance like to insert ,delete,modify and access

records

DDL examples are Create table,altertable and DML examples are Update,select


1 mark for partial correct answer

½ mark for each correct difference point.

b) Which of the following SQL Command is used to display the structure of a

table?

(a) DROP TABLE (b) ALTER TABLE

(c ) CREATE TABLE (d) DESC

1

Answer: (d) DESC


c) RDBMS stands for _________________.

1

Answer: Relational Database Management


d) Which of the following is an example of open source RDBMS?

(a) Oracle (b) Microsoft SQL Server

(c ) My SQL (d) MS-Access

1

Answer: (c) MySQL


Documents

MARKING SCHEME FRAGMENT-2 EXAMINATION(2019-20) …ahiyaraipur.thelps.edu.in/UploadedFiles/UpdateDirectory/Computer... · xii/cs(283)/fragment-2-exam. page 1 of 18 marking scheme fragment-2