Oswal 36 Sample Question Papers ISC Class 12 Computer Science Solutions
Section-A
Answer 1.
(iii) (c) dr.substring(0,4)+”.”+dr.substring(11)
substring(0,4) returns characters from index 0 to 3 that is “Dr.” and substring(11) returns characters from index 11 to the end that is “Chatterjee”.
(iv) (b) Inner loop
In bubble sort the largest element is pushed towards the end in each pass, hence the inner loops becomes smaller after every pass.
Answer 2.
| n | i | condition | k = 1 |
| 6 | 3 | true | 1 × 6 × 3 = 18 |
| 5 | 2 | true | 18 × 5 × 2 = 180 |
| 4 | 1 | true | 180 × 4 × 1 = 720 |
| 3 | 0 | false |
= 720
PART-II
Section-A
Answer 3.
Answer 4.
(i) Truth table for the inputs and outputs :
| S | P | C | T | X | |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 2 |
| 0 | 0 | 1 | 0 | 0 | 2 |
| 0 | 0 | 1 | 1 | 1 | 3 |
| 0 | 1 | 0 | 0 | 1 | 4 |
| 0 | 1 | 0 | 1 | 1 | 5 |
| 0 | 1 | 1 | 0 | 1 | 6 |
| 0 | 1 | 1 | 1 | 1 | 7 |
| 1 | 0 | 0 | 0 | 0 | 8 |
| 1 | 0 | 0 | 1 | 0 | 9 |
| 1 | 0 | 1 | 0 | 1 | 10 |
| 1 | 0 | 1 | 1 | 1 | 11 |
| 1 | 1 | 0 | 0 | 1 | 12 |
| 1 | 1 | 0 | 1 | 1 | 13 |
| 1 | 1 | 1 | 0 | 1 | 14 |
| 1 | 1 | 1 | 1 | 1 | 15 |
Answer 5.
| F3 | F2 | F1 | F0 | |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 |
| 2 | 0 | 0 | 1 | 0 |
| 3 | 0 | 0 | 1 | 1 |
| 4 | 0 | 1 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 |
| 6 | 0 | 1 | 1 | 0 |
| 7 | 0 | 1 | 1 | 1 |
| 8 | 1 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 |
| A | B | C | A ⊕ B ⊕ C | A ⊙ B ⊙ C | A ⊙ B ⊙ C |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 | 1 |
(iii)
| Proposition | wff |
| A proposition is an elementary atomic sentence that returns either true or false. | wff’s are well formed formula which is a proposition that is satisfiable or valid. |
Section-B
Answer 6.
import java.util.*;
public class Emirp
{
int n, rev, f;
Emirp(int nn)
{
n = nn;
rev = 0;
f = 2;
}
int isprime(int x)
{
if(n = = x) return 1;
else if(n % x = = 0 || n = = 1)
return 0;
else
return isprime(x + 1);
}
void isEmirp()
{
int x = n;
while(x != 0)
{
rev = rev * 10 + x % 10; x = x / 10;
}
int ans1 = isprime(f);
n = rev;
f = 2;
int ans2 = isprime(f);
if(ans1 = = 1 && ans2 = = 1)
System.out.println(n+“ is an Emirp no.”);
else
System.out.println(n+“is not an Emirp no.”);
}
public static void main()
{
Scanner sc = new Scanner System.in);
System.out.println(“Enter a no.”);
int x = sc.nextInt();
Emirp obj = new Emirp(x);
Obj.isEmirp();
}
}
Answer 7.
import java.util.*;
public class Exchange
{
String sent, rev;
int size;
Exchange()
{
sent = “”;
rev = “”;
}
void readSentence()
{
Scanner sc = new Scanner(System.in);
System.out.println(“Enter a string”);
sent = sc.nextLine();
size = sent.length();
}
void exfirstlast()
{
int p = 0;
char ch;
String b;
for (int i = 0; i < size; i++)
{
ch = sent.charAt(i);
if(ch = = ‘ ’ || ch = = ‘.’)
{
b = sent.substring(p, i);
if(b.length() != 1)
{
rev += b.charAt(b.length() – 1);
rev += b.substring (1, b.length() – 1);
rev += b.charAt(0);
}
else { rev += b; rev = rev + “ ”; }
}
}
}
void display()
{
System.out.println(“\nInput :” +sent);
System.out.println(“\nOutput :” +rev);
}
public static void main()
{
Exchange ob = new Exchange();
obj.readSentence();
obj.exfirstlast();
obj.display();
}
}
Answer 8.
import.java.io.*;
class Combine
{
int com[],size;
public Combine(int nn)
{
size=nn;
com=new int[size];
}
void inputarray()throws IOException
{
BufferedReader br=new BufferedReader (new InputStreamReader(System.in));
int i;
for(i=0;i<size;i++)
{
System.out.println(“Enter a no.”);
com[i]=Integer.parselnt(br.readLine());
}
}
void sort()
{
int i,j,t;
for(i=0;i<size;i++)
{
for(j=i+1;j<size;j++)
{
if(com[i]>com[j])
{
t=com[i];
com[i]=com[j];
com[j]=t;
}
}
}
}
void mix(Combine A, Combine B)
{
int i,j;
for(i=0,j=0;i<A.size;i++,j++)
{
com[j]=A.com[i];
}
for(i=0;i<B.size;i++,j++)
{
com[j]=B.com[i];
}
}
void display()
{
int i;
for(i=0;i<size;i++)
{
System.out.println(com[i]);
}
}
public static void main(String args[])throws IOException
{
Combine c1=new Combine(5);
Combine c2=new Combine(3);
Combine c3=new Combine(8);
c1.inputarray();
c2.inputarray();
c3.mix(c1,c2);
c3.sort();
c3.display();
}
}
Section-C
Answer 9.
(i) class Link
{
int lnk[]=new int[100];
int begin,end,max;
public Link(int mm)
{
max=mm;
begin=end=0;
}
void addlink(int v)
{
if(end= =0)
{
end=begin=1;
lnk[end]=v;
}
else if(end= =max)
{
System.out.println(‘List is Full’);
}
else
{
lnk[++end]=v;
}
}
int dellink()
{
int a;
if(begin= =0)
{
System.out.println(“Empty...”);
return(-99);
}
else if(begin= =end)
{
a=lnk[begin];
begin=end=0;
return a;
}
else
{
a=lnk[begin];
begin++;
return(a);
}
}
void display()
{
int i;
for(i=begin;i<=end;i++)
{
System.out.println(lnk[i]);
}
}
}
}
(ii) It is a queue.
Answer 10.
class Stock
{
protected String item;
protected int qty;
protected double rate, amt;
public Stock(String it, int q, double r)
{
item = it;
qty = q;
rate = r;
amt = qty * rate;
}
public void display()
{
System.out.println(item + “ “ + qty + “ “ + rate + “ “ + amt);
}
class Purchase extends Stock
{
int pqty;
double prate;
public Purchase(String it1, int q1, double r1, double pr, int pq)
{
super(it1, q1, r1);
pqty = pq;
prate = pr;
}
void update()
{
qty += pqty;
if(rate != prate)
{
rate = prate;
}
amt = qty * rate;
}
public void display()
{
System.out.println(“Details Before Updation “);
super.display();
System.out.println(“Details After Updation “);
update();
super.display();
}
}
}
Answer 11.
(i) Method
void search (node start, string b)
{
boolean found = false;
node tmp = start;
while (tmp ! = nul)
{
if ((tmp. n). equals (b) = = true)
{
System out.println (“Item found” + tmp. p);
System.out,println (tmp.n);
found = true;
break;
}
tmp = tmp. next;
if (found = = false)
}
System.out.print (“Item not found”);
}
Algorithm
Step 1 : Input b, the string to search
Step 2 : Let tmp = start, a temporary node created.
Step 3 : If (tmp.n = b) then print “found”, tmp.p, tmp.n, Goto End Else Goto steple
Step 4 : tmp = tmp-next, move tmp to the next node.
Step 5 : if tmp ! = null , then Goto step 3 else Goto step 6
Step 6 : Print “Not Found”
Step 7 : End
(ii) (a) D, E, G
(b) C
(c) B, D, C, E, A, F, G
ISC 36 Sample Question Papers
All Subjects Combined for Class 12 Exam 2023
ISC 36 Sample Question Papers
All Subjects Combined for Class 12 Exam 2023
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