ISC 2016 Sample Paper Computer Application Paper II (Practical) [Solved]

Click To Download Solved Paper in PDF

COMPUTER SCIENCE
Paper – 2
(PRACTICAL)
(Three hours)
Maximum Marks: 30
(Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start
writing during this time)
----------------------------------------------------------------------------------------------------------------------------
The total time to be spent on the Planning session and Examination session is Three hours.
Planning session: 90 minutes
Examination session: 90 minutes
Note: Candidates are to be permitted to proceed to the Examination Session only after the
90 minutes of the Planning Session are over.

----------------------------------------------------------------------------------------------------------------------------

This paper consists of three problems from which candidates are required to attempt any one problem.
Candidates are expected to do the following:
1. Write an algorithm for the selected problem.
(Algorithm should be expressed clearly using any standard scheme such as pseudo code or
in steps which are simple enough to be obviously computable) [3]
2. Write a program in JAVA language. The program should follow the algorithm and should
be logically and syntactically correct. [5]
3. Document the program using mnemonic names / comments, identifying and clearly
describing the choice of data types and meaning of variables. [2]
4. Code / Type the program on the computer and get a print out ( Hard Copy ). Typically, this
should be a program that compiles and runs correctly. [2]
5. Test run the program on the computer using the given sample data and get a print out of the
output in the format specified in the problem. [5]
6. Viva-Voce on the Selected Problem. [3] 

In addition to the above, the practical file of the candidate containing the practical work related
to programming assignments done during the year is to be evaluated as follows:

• Programming assignments done throughout the year (by the Teacher) [5]

• Programming assignments done throughout the year (by the Visiting Examiner) [5]

Attempt any one of the following

Question 1:

An Evil number is a positive whole number which has even number of 1’s in its binary equivalent.
Example: Binary equivalent of 9 is 1001, which contains even number of 1’s.
A few evil numbers are 3, 5, 6, 9....
Design a program to accept a positive whole number and find the binary equivalent of the number and
count the number of 1’s in it and display whether it is a Evil number or not with an appropriate
message.
Output the result in format given below:
Example 1
INPUT : 15
BINARY EQUIVALENT : 1111
NO. OF 1’s : 4
OUTPUT : EVIL NUMBER
Example 2
INPUT : 26
BINARY EQUIVALENT : 11010
NO. OF 1’s : 3
OUTPUT : NOT AN EVIL NUMBER

import java.util.*;
public class Evil_Number
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the Number to be checked");
int  n=sc.nextInt();
int ar[];
int temp=n;
int f=0;
while(true)
{
if(Math.pow(2,f)f++;
else
break;
}

ar=new int [f];
int i=0;
while(temp>0)
{
ar[i++]=temp%2;
temp=temp/2;
}
int c=0;
for( i=0;iif(ar[i]==1)
c++;

System.out.print("Binary Equivalent : ");

for(i=0;iSystem.out.print(ar[i]);

if(c%2==0)
System.out.println("\nEvil Number");
else
System.out.println("\nNot An Evil Number");
}
}

Question 2

The encryption of alphabets are to be done as follows:

A = 1

B = 2

C = 3

.

.

.

Z = 26

The potential of a word is found by adding the encrypted value of the alphabets.

Example: KITE

Potential = 11 + 9 + 20 + 5 = 45

Accept a sentence which is terminated by either “ . ” , “ ? ” or “ ! ”. Each word of sentence is separated

by single space. Decode the words according to their potential and arrange them in ascending order.

Output the result in format given below:

Example 1

INPUT : THE SKY IS THE LIMIT.

POTENTIAL : THE = 33

SKY = 55

IS = 28

THE = 33

LIMIT = 63

OUTPUT : IS THE THE SKY LIMIT

Example 2

INPUT : LOOK BEFORE YOU LEAP.

POTENTIAL : LOOK = 53

BEFORE = 51

YOU = 61

LEAP = 34

OUTPUT : LEAP BEFORE LOOK YOU

import java.util.*;
public class Potential{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
String s;
do{
System.out.println("Enter the Sentence ending with . ? or !");
s=sc.nextLine();
if(s.endsWith(".") || s.endsWith("?") || s.endsWith("!"))
break;
}
while(true);
s=s.substring(0,s.length()-1);// This will remove the full-Stop
StringTokenizer str=new StringTokenizer(s);
String ar[]=new String[str.countTokens()];
int pot[]=new int [ar.length];
int i=0;
while(str.hasMoreTokens())
ar[i++]=str.nextToken();

//calculating potential

for( i=0;i{
int p=0;
for(int k=0;kp+=((int)ar[i].charAt(k))-64;

pot[i]=p;
}

System.out.println("Potential");
for( i=0;iSystem.out.println(ar[i]+":\t"+pot[i]);
//arranging
for(i=0;i{
for(int k=0;kif(pot[k]>pot[k+1])
{
int t=pot[k];
pot[k]=pot[k+1];
pot[k+1]=t;

String s1=ar[k];
ar[k]=ar[k+1];
ar[k+1]=s1;
}
}

System.out.println("OUTPUT: ");
for(i=0;iSystem.out.print(ar[i]+" ");
}
}

Question 3

Given a square matrix M [ ] [ ] of order ‘n’. The maximum value possible for ‘n’ is 10. Accept three
different characters from the keyboard and fill the array according to the instruction given below.
Fill the upper and lower elements formed by the intersection of the diagonals by character 1.
Fill the left and right elements formed by the intersection of the diagonals by character 2.
Fill both the diagonals by character 3.
Output the result in format given below:
Example 1
ENTER SIZE : 4
INPUT : FIRST CHARACTER ‘*’
SECOND CHARACTER ‘?’
THIRD CHARACTER ‘#’
OUTPUT :
# * * #
? # # ?
? # # ?
# * * #
Example 2
ENTER SIZE : 5
INPUT : FIRST CHARACTER ‘$’
SECOND CHARACTER ‘!’
THIRD CHARACTER ‘@’
OUTPUT :
@ $ $ $ @
! @ $ @ !
! ! @ ! !
! @ $ @ !
@ $ $ $ @
Example 3
ENTER SIZE : 65
OUTPUT : SIZE OUT OF RANGE

import java.util.*;
public class FillArray
{
public static void main(String args[])
{
Scanner sc=new Scanner (System.in);
int n;
System.out.println("Enter the size.(Maximum Limit : 10)");
n=sc.nextInt();

if(n>10)
{System.out.println("OUT OF RANGE");
System.exit(0);
}

System.out.print("INPUT: First Character : ");
char ch=sc.next().charAt(0);
System.out.print("\n      Second Character : ");
char ch1=sc.next().charAt(0);
System.out.print("\n      Third Cahracter : ");
char ch2=sc.next().charAt(0);

char ar[][]=new char[n][n];

for(int i=1;i{
ar[0][i]=ch;
ar[n-1][i]=ch;
ar[i][0]=ch1;
ar[i][n-1]=ch1;
}

for(int i=0;i{
for(int k=0;kif(i+k==n-1)
ar[i][k]=ch2;

ar[i][i]=ch2;
}

System.out.println("Array:");

for(int i=0;i{
for(int k=0;kSystem.out.print(ar[i][k]);
System.out.println();
}
}
}

Click To Download Solved Paper in PDF

Click To Download Solved Paper in Text File