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# Class 12 ISC - Java Question Paper 2019

### PART I

###### (Attempt all questions)

__Question 1__

#### (a) Name and draw the logic gate represented by the following truth table, where A and B are inputs and X is the output.

#### Solution

#####
(a) XOR gate

#### (b) Write the canonical POS expression of: F(P, Q) = Π(0, 2)

#### Solution

##### (b) (A + B).(A’ + B)

#### (c) Find the dual of: X.Y+X.Y’ = X + 0

#### Solution

##### (c) (X + Y).(X + Y’) = X.1

#### (d) If F(A, B, C) = A’.B’.C’ + A’.B.C’ then find F’ using De Morgan’s Law.

#### Solution

#####
(d) (AB’C + ABC’) (∵ x’ + y’ = x’.y’)

= (AB’C’)’. (A’.B.C’)’ (∵(x’.y’) = x’ + y’)

= ((A’B’)’ + (C)’. ((A’B)’ + (C’)’)

= ((A’)’ + (B’)’) + C).((A’)’ + (B)’ + C)

= (A + B + C).(A + B’ + C)

= AA + AB’ + AC + BA + BB’ + BC + CA + CB’ + CC (∵x.x = x)

= A + AB’ + C + AC + AB + B’C + BC

#### (e)
If A = “It is cloudy” and B = “It is raining”, then write the proposition for

(i) Contrapositive

(ii) Converse

#### Solution

#####
(e) (i) If it is not raining then it is not cloudy.

(ii) If it is raining then it is cloudy

__Question 2__

#### (a)
What is an Interface? How is it different from a class?

#### Solution

##### (a) An interface is just like Java Class, but it only has static constants and abstract method. Java uses Interface to implement multiple inheritances. An interface is syntactically similar to a class, but it lacks in field declaration and the methods inside an interface do not have any implementation. A class can be instantiated but an interface not.

#### (b) A matrix ARR[-4 ….. 6, 3 ……. 8] is stored in the memory with each element requiring 4 bytes of storage. If the base address is 1430, find the address of ARR[3] [6] when the matrix is stored in Row Major Wise.

#### Solution

#####
(b) Given, Address of ARR[3] [6] = 1430

According to row-major,

Address = B + w(n(p – L1) + (q – L2)

B = 1430, w = 4 bytes

p = 3, q = 6,U1 = 6, U2 = 8, L1 = -4, L2 = 3

Number of columns n = U2 – L2 + 1

⇒ n = 8 – 3 + 1 = 6

Address of ARR[3][6] = 1430 + 4(6 (3 – (-4))+ (6 – 3))

= 1430 + 4(6 × 7 + 3)

= 1430 + 180

= 1610

#### (c)
Convert the following infix notation to postfix form:

(A + B * C) – (E * F / H) + J

#### Solution

#####
(c) (A + B * C) – (E * F / H) + J

= (A + (BC*)) – ((EF*) / H) + J

= (ABC *+) – (EF * H/) + J

= (ABC *+ EF * H/-) + J

= ABC*+ EF * H / – J*

#### (d)
Compare the two complexities O(n^{2}) and O(2^{n}) and state which is better and why.

#### Solution

#####
(d) O(n^{2}) is better than O(2^{n})

Let us see with the help of an example.

Suppose n = 8

n^{2} = 8^{2} = 64

2^{n} = 2^{8} = 256

Therefore, the complexity in 2^{n} is higher than n^{2}. If n increases, 2^{n} increases much more than n^{2}. Therefore, the time complexity is O(n^{2}) is better than O(2^{n}).

#### (e) State the difference between internal nodes and external nodes of a binary tree structure.

#### Solution

##### (e) Internal nodes are not leaf nodes whereas external nodes are leaf nodes.

__Question 3__

The following function Mystery( ) is a part of some class. What will the function Mystery( ) return when the value of num=43629, x=3 and y=4 respectively? Show the dry run/working.

#####
Given n = 43629, x = 3, y = 4

int Mystery (int num, int x, int y) { if(num < 10) return num; else { int z = num % 10; if(z%2 == 0) return z*x + Mystery (num/10, x, y); else return z*y + Mystery(num/10, x, y); } }

#### Solution

#####
Given n = 43629, x = 3, y = 4

Step 5 returns 4

Step 4 returns 12 + 4 = 16

Step 3 returns 18 + 16 = 34

Step 2 returns 6 + 34 = 40

Step 1 returns 36 + 40 = 76

### PART II

### SECTION A

###### (Attempt any two questions from this Section.)

__Question 4__

####
(a) Given the Boolean function F(A, B, C, D) = Σ (0, 2, 3, 4, 5, 8, 10, 11, 12, 13).

(i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs).

(ii) Draw the logic gate diagram for the reduced expression using only NAND gates. Assume that the variables and their complements are available as inputs.

(b) Given the Boolean function : F(P, Q, R, S) = π (0, 1, 2, 8, 9, 11, 13, 15).

(i) Reduce the above expression by using a 4-variable Karnaugh map, showing the various groups (i.e, octal, quads and pairs).

(ii) Draw the logic gate diagram for the reduced expression using only NOR gates. Assume that the variables and their complements are available as inputs.

#### Solution

#####
(a) (i)

(ii)

(b) (i)

(ii)

__Question 5__

####
(a) How is a decoder different from a multiplexer? Write the truth table and draw the logic circuit diagram for a 3 to 8 decoder and explain its working.

(b) From the logic circuit diagram given below, derive the Boolean expression and simplify it to show that it represents a logic gate. Name and draw the logic gate.

(c) Using a truth table, state whether the following proposition is a Tautology, Contradiction or Contingency:

~(P =>Q)<=>(~P∨Q)

#### Solution

#####
(a) A multiplexer is a device which consists of multiple input channels through a single line while decoder consists of multiple inputs passing through multiple outputs.
Multiplexer converts inputs from unary codes (initial) to binary codes while decoder converts binary codes to inputs.

(b)

Expression : F = (X+Y). XZ

= XZ +XYZ

= XZ

It represents AND gate.

(c)

The proposition is a contradiction

__Question 6__

####
(a) The owner of a company pays the bonus to his salesmen as per the criteria are given below:

If the salesman works overtime for more than 4 hours but does not work on off days/ holidays.

OR

If the salesman works when festival sales are on and updates showroom arrangements.

OR

If the salesman works on an off day/holiday when the festival sales are on.

The inputs are
O - Works overtime for more than 4 hours

F - Festival sales are on

H - Working on an off day/holiday

U - Updates showroom arrangements

(In all the above cases 1 indicates yes and 0 indicates no)

Output: X[1 indicates yes, 0 indicates no for all cases]

Draw the truth table for the inputs and outputs given above and write the POS expression for X(O, F, H, U).

(b) What is a half adder? Write the truth table and derive an SOP expression for sum and carry for a half adder.

(c) Simplify the following expression, using Boolean laws:

(X + Z).(X.Y + Y.Z’) + X.Z + Y

#### Solution

#####
(a)

(b) The half adder is an example of a simple, functional digital circuit built from two logic gates. The half adder adds two one-bit binary numbers (AB). The output is the sum of the two bits (S) and the carry (C).
The truth table for a half-adder is:

The simplified sum of products (SOP) expressions is:

S = x’y + xy’, C = xy

(c) (X + Z).(X.Y + Y.Z’) + X.Z + Y

= (X + Z).(Y.Z’) + X.Z + Y (a.b = 0)

= X.Y.Z’ + Z.Z’ + X.Z + Y (a.a’ = 0) and (a.c = 0)

= 0.Z’ + 0 + 0 + Y

= Y

### SECTION B

###### (Attempt all questions from this Section.)

__Question 7__

######
Design a class ArmNum to check if a given number is an Armstrong number or not.
[A number is said to be Armstrong if sum of its digits raised to the power of length of the number is equal to the number]
Example:
371 = 3^{3} + 7^{3} + 1^{3}
1634 = 1^{4} + 6^{4} + 3^{4} + 4^{4}
54748 = 5^{5} + 4^{5} + 7^{5} + 4^{5} + 8^{5}
Thus, 371, 1634 and 54748 are all examples of Armstrong numbers.
Some of the members of the class are given below:
Class name: ArmNum
Data members/instance variables:
n: to store the number
l: to store the length of the number
Methods/Member functions:
ArmNum (int nn): parameterized constructor to initialize the data member n = nn
int sum_pow(int i): returns the sum of each digit raised to the power of the length of the number using recursive technique eg., 34 will return 3^{2} + 4^{2} (as the length of the number is 2)
void isArmstrong(): checks whether the given number is an Armstrong number by invoking the function sum_pow () and displays the result with an appropriate message.
Specify the class ArmNum giving details of the constructor( ), int sum_pow(int) and void isArmstrong( ). Define a main() function to create an object and call the functions accordingly to enable the task.

^{3}+ 7

^{3}+ 1

^{3}1634 = 1

^{4}+ 6

^{4}+ 3

^{4}+ 4

^{4}54748 = 5

^{5}+ 4

^{5}+ 7

^{5}+ 4

^{5}+ 8

^{5}Thus, 371, 1634 and 54748 are all examples of Armstrong numbers. Some of the members of the class are given below: Class name: ArmNum Data members/instance variables: n: to store the number l: to store the length of the number Methods/Member functions: ArmNum (int nn): parameterized constructor to initialize the data member n = nn int sum_pow(int i): returns the sum of each digit raised to the power of the length of the number using recursive technique eg., 34 will return 3

^{2}+ 4

^{2}(as the length of the number is 2) void isArmstrong(): checks whether the given number is an Armstrong number by invoking the function sum_pow () and displays the result with an appropriate message. Specify the class ArmNum giving details of the constructor( ), int sum_pow(int) and void isArmstrong( ). Define a main() function to create an object and call the functions accordingly to enable the task.

###### Solution:

import java.util.*; class ArmNum { int n; int l; ArmNum(int nn) { n=nn; l=0; } int sum_pow(int i) { if(i!=0) { return (int)Math.pow(i%10,l)+sum_pow(i/10); } else { return 0; } } void isArmstrong() { int n1=n; while(n1!=0) { ++l; n1=n1/10; } int sum=sum_pow(n); if (sum==n) { System.out.println("Armstrong no."); } else { System.out.println("Not an Armstrong no."); } } public static void main(String args[]) { Scanner sc=new Scanner(System.in); System.out.println("Enter a number:"); int no=sc.nextInt(); ArmNum obj=new ArmNum(no); obj.isArmstrong(); } }

__Question 8__

######
Design a class MatRev to reverse each element of a matrix.
Example:
Some of the members of the class are given below:
Class name: MatRev
Data members/instance variables:
arr[][] : to store integer elements
m: to store the number of rows
n: to store the number of columns
Member functions/methods:
MatRev(int mm, int nn): parameterized constructor to initialise the data members m = mm and n = nn
void fillarray(): to enter elements in the array
int reverse(int x): returns the reverse of the number x
void revMat(MatRev P): reverses each element of the array of the parameterized object and stores it in the array of the current object
void show(): displays the array elements in matrix form
Define the class MatRev giving details of the constructor ( ), void fillarray (), int reverse(int), void revMat(MatRev) and void show(). Define the main () function to create objects and call the functions accordingly to enable the task.

###### Solution:

import java.util.*; class MatRev{ int arr[][]; int m; int n; public MatRev(int mm, int nn) { m=mm; n = nn; arr=new int[m][n]; } public void fillArray( ){ Scanner sc = new Scanner(System.in); for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { System.out.println("Enter the "+i+j+" Element: "); arr[i][j] = sc.nextInt(); } } } public int reverse(int x) { int r = 0; while(x> 0) { r=r* 10+x % 10; x =x/ 10; } return r; } public void revMat(MatRev P) { for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { this.arr[i] [j] = reverse(P.arr[i] [j]); } } } public void show() { for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { System.out.print(arr[i][j] + "\t"); } System.out.println(); } } public static void main(String args[]) { Scanner sc = new Scanner(System.in); System.out.print("Enter number of rows::"); int x = sc.nextInt(); System.out.print("Enter number of columns::"); int y = sc.nextInt(); MatRev obj1 = new MatRev(x, y); MatRev obj2 = new MatRev(x, y); obj1.fillArray(); obj2.revMat(obj1); System.out.println("Original Matrix is::"); obj1.show(); System.out.println("Matrix with reversed elements"); obj2.show(); } }

__Question 9__

######
A class Rearrange has been defined to modify a word by bringing all the vowels in the word at the beginning followed by the consonants.
Example:
ORIGINAL becomes OIIARGNL
Some of the members of the class are given below:
Class name: Rearrange
Data Member/instance variable:
wrd: to store a word
newwrd: to store the rearranged word
Member functions/methods:
Rearrange(): default constructor
void readword(): to accept the word in UPPER case
vow freq_vow_con(): finds the frequency of vowels and consonants in the word and displays them with an appropriate message
void arrange(): rearranges the word by bringing the vowels at the beginning followed by consonants
void display(): displays the original word along with the rearranged word
Specify the class Rearrange, giving the details of the constructor(), void readword(), void freq _vow_con(), void arrange() and void display(). Define the main() function to create an object and call the functions accordingly to enable the task.

###### Solution:

import java.util.*; class Rearrange { String wrd; String newwrd; public Rearrange() { wrd = ""; newwrd = ""; } public void readword(){ Scanner sc = new Scanner(System.in); System.out.print("Enter the word in Upper Case:"); wrd = sc.next(); } public void freq_vow_con() { int vc=0,cc=0; int l=wrd.length(); for(int i=0;i< l;i++) { char ch=wrd.charAt(i); if(ch=='A' || ch =='E' || ch =='I' || ch== 'O' || ch=='U') { vc++; } else { cc++; } } System.out.println("Frequency of vowels: "+vc); System.out.println("Frequency of consonants: "+cc); } public void arrange() { String v =""; String c = ""; for(int i = 0; i < wrd. length(); i++) { char ch = wrd.charAt(i); if(ch=='A' || ch =='E' || ch =='I' || ch== 'O' || ch=='U') { v+=ch; } else { c+=ch; } } newwrd = v + c; } public void display() { System.out.println("Original word:" + wrd); System.out.println("Rearranged word:" + newwrd); } public static void main(String args[]) { Rearrange obj = new Rearrange(); obj.readword(); obj.freq_vow_con(); obj.arrange(); obj.display(); } }

__Question 10__

######
A superclass Record contains names and marks of the students in two different single dimensional arrays. Define a subclass Highest to display the names of the students obtaining the highest mark
The details of the members of both classes are given below:
Class name: Record
Data member/instance variable:
n[] : array to store names
m[]: array to store marks
size: to store the number of students
Member functions/methods:
Record(int cap): parameterized constructor to initialize the data member
size = cap
void readarray() : to enter elements in both the arrays
void display() : displays the array elements
Class name: Highest
Data member/instance variable:
ind: to store the index
Member functions/methods:
Highest(…): parameterized constructor to initialize the data members of both the classes
void find(): finds the index of the student obtaining the highest mark and assign it to ‘ind’
void display(): displays the array elements along with the names and marks of the students who have obtained the highest mark
Assume that the superclass Record has been defined. Using the concept of inheritance, specify the class Highest giving the details of the constructor(…), void find() and void display().
The superclass, main function and algorithm need NOT be written.

###### Solution:

import java.util.*; class Record { String n[]; int m[]; int size; public Record(int cap) { size=cap; n=new String[size]; m=new int[size]; } public void readarray() { Scanner sc = new Scanner(System.in); System.out.println("Enter names : "); for(int i=0;i < size;i++) { n[i]= sc.next(); } System.out.println("Enter marks : "); for(int i= 0;i < size;i++) { m[i]= sc.nextInt(); } } public void display() { for(int i=0;i < size;i++) { System.out.print(n[i] + " "); } System.out.println(); for(int i=0;i < size;i++) { System.out.print(m[i] + " "); } System.out.println(); } } class Highest extends Record { int ind; Highest(int cap) { super(cap); ind=-1; } void find() { readarray(); int hm=m[0]; for (int i=0;ihm) { hm=m[i]; ind=i; } } } public void display() { super.display(); for(int i=0;i < size;i++) { if(m[i] == m[ind]) { System.out.println(" Highest obtained by " +n[i] + " marks " +m[i]); } } } } class rec { public static void main(String args[]) { Highest obj = new Highest(5); obj.find(); obj.display(); } }

__Question 11__

######
A linear data structure enables the user to add an address from rear end and remove address from front. Define a class Diary with the following details :
Class name: Diary
Data members/instance variables:
Q[]: array to store the addresses
size: stores the maximum capacity of the array
start: to point the index of the front end
end: to point the index of the rear end
Member functions:
Diary(int max): constructor to initialize the data member size = max, start=0 and end=0
void pushadd(String n): to add the address in the diary from the rear end if possible, otherwise display the message “NO SPACE”
String popadd(): removes and returns the address from the front end of the diary if any, else returns “?????”
void show (): displays all the addresses in the diary
(a) Specify the class Diary giving details of the functions void pushadd(String) and String popadd(). Assume that the other functions have been defined.
The main function and algorithm need NOT be written.
(b) Name the entity used in the above data structure arrangement.

###### Solution:

public class Diary { String Q[]; int size,start,end; public Diary(int max){ size=max; Q=new String[size]; start=end=0; } void pushadd(String n) { if(end == size-1) { System.out.println("NO SPACE"); } else if (start == 0) { start=end=1; Q[end]=n; } else { Q[++end]=n; } } String popadd() { String n; if(end == 0) { return "?????"; } else if(start == end) { n=Q[start]; start=end=0; } else { n=Q[start++]; } return n; } }

__Question 12__