Oswal 36 Sample Question Papers ISC Class 12 Computer Science Solutions

PART-I

(i) (d) X+Y

(ii) (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.

(iii) (c) P+Q+R

(iv) (a) a tautology

Explanation :

A+A′ = 1 and that is a tautology

(v) (d) A′+B+C+D

Explanation :

In maxterm 0 represents the literal itself and 1 represents the complement of the literal.

(vi) (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.

(vii) A statement block is used to organize a sequence of statements as a single statement group.

(viii) False. Java does not allow array overruns at run time.

(ix)

1. Reusability
2. Transitive nature

(x) “Overflow” means attempt to INSERT when the list is full and no free space is available.

(i) F = (A + B ′) . (B + CD) ′

$$=(\text{A + B'}).(\overline{\text{B +CD}})\\= (\text{A +} \overline{\text{B}}).\lbrack\overline{\text{B}}.\overline{\text{CD}}\rbrack\\\text{(using De Morgan’s Law)}\\=(\text{A +}\bar{\text{B}})\lbrack\overline{\text{B}}.(\overline{\text{C}} + \overline{\text{D}})\rbrack\\=(\text{A+}\overline{\text{B}})(\overline{\text{B}}.\overline{\text{C}}) + (\overline{\text{B}}.\overline{\text{D}})\\\text{(using Distributive Law)}\\=\text{A}.\bar{\text{B}}.\bar{\text{C}} + \text{A}.\bar{\text{B}}.\bar{\text{D}} +\\ \bar{\text{B}}.\bar{\text{B}}.\bar{\text{C}} +\bar{\text{B}}.\bar{\text{B}}.\bar{\text{D}}\\= \text{A} .\bar{\text{B}}.\bar{\text{C}} + \text{A}.\bar{\text{B}}.\bar{\text{D}} +\bar{\text{B}}.\bar{\text{C}}+ \bar{\text{B}}.\bar{\text{D}}\\=\bar{\text{B}}.\bar{\text{C}}(\text{A +1}) + \bar{\text{B}}.\bar{\text{D}}(\text{A+1})\\= \bar{\text{B}}.\bar{\text{C}} + \bar{\text{B}}.\bar{\text{D}}\\=\bar{\text{B}}(\bar{\text{C}} + \bar{\text{D}})$$

(ii) BA = 1012, ES = 2, R = 10, I = 7, J = 3, LR = X, LC = 1

Address of B [I, J] = BA + ES [(I – LR) + R (J – LC)]

B [7, 3] = 1012 + 2 [(7 – X) + 10 (3 – 1)]

1060 = 1012 + 2 [(7 – X) + 10 (2)]

1060 = 1012 + 2 [(7 – X) + 20)]

1060 – 1012 = 2 [(7 – X) + 20)]

48 = 2 [(7 – X) + 20)]

(7 – X) = 48/2 – 20

7 – X = 4

X = 7 – 4 = 3.

(iii) (a) What does the method recursive (86, 1) return?

count=0

recursive(86,1) → 8*6=48, count=1

recursive(48,1) → 4*8=32, count=2

recursive(32,1) → 3*2=6, count=2

return 2

(b) What does the method recursive (341, 1) return?

count=0

recursive(341,1) → 3*4*2=24, count=1

recursive(24,1) → 2*4=8, count 1

return 1

(iv) 1. mul*=m%10;

2. m==0

3. mul=1;

4. count++;

Explanation :

Section-A

(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 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

X (S, P, C, T) = Σ (1, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15)

$$\bar{\text{S}}\bar{\text{P}}\bar{\text{C}}\text{T} + \bar{\text{S}}\bar{\text{P}}\text{CT +} \bar{\text{S}}\text{P}\bar{\text{C}}\bar{\text{T}} +\bar{\text{S}}\text{P}\bar{\text{C}}\text{T} +\\\bar{\text{S}}\text{P}\text{C}\bar{\text{T}} +\bar{\text{S}}\text{PCT}+ \text{S}\bar{\text{P}}\text{C}\bar{\text{T}} +\text{S}\bar{\text{P}}\text{C}\text{T} +\\\text{SP}\bar{\text{C}}\bar{\text{T}} +\text{SP}\bar{\text{C}}\text{T} + \text{SPC}\bar{\text{T}} + \text{SPCT}$$

(ii) (a) Gate 1 = (A.B)′ = A′ +B′

Gate 2 = (B′.C)′ = B+C′

Gate 3 = (A.B)′+(B′.C)′ OR (A′+B′)+(B+C′) = 1

Gate 4 = ((A.B)′+(B′.C)′)′ = (A.B)′′.(B′.C)′′ = A.B.B′.C = 0

(b) Σ( 0, 2, 7, 6)

$$=\overline{\Sigma(1,3,4,5)}$$

= Π(1, 3, 4, 5)

(i) (a) F (A, B, C, D) = Σ (4, 6, 7, 10, 11, 12, 14, 15)

Quad 1 = m6 + m7 + m14 + m15 = BC

Quad 2 = m10 + m11 + m14 + m15 = AC

$$\text{Quad 3 = }m4 + m6 + m12 + m14 =\text{B}\bar{\text{D}}$$

(b) So the expession is:

(ii) X (S,P,C,T) = Σ(1,3,4,5,6,7,10,11,12,13,14,15)

 C′.T′ C′.T C.T C.T′ S′.P′ 1 1 S′.P 1 1 1 1 S.P 1 1 1 1 S.P′ 1 1

Octet(m4,m5,m7,m6,m12,m13,15,m14) = P

X (S,P,C,T) = P + S′.T +S.C

(i) Decimal to binary encoder is used to convert decimal numbers into its equivalent binary form.

 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

$$\text{(ii)\space} \text{A}⊕\text{B} = \text{A}\bar{\text{B}} + \bar{\text{A}}.\text{B}\space\text{A}⊙\text{B}\\=\overline{\text{A}⊙\text{B}}$$

 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

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();

}

}

import java.util.*;

public class Exchange

{
String sent, rev;

int size;

Exchange()

{
sent = “”;

rev = “”;

}

{

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.exfirstlast();

obj.display();

}

}

import java.util.*;

class MatrixSum

{
int matA[][];

int matB[][];

int m, n;

MatrixSum(int mm , int nn)

{
m =mm;

n =nn;

matA=new int [m][n];

matB=new int [m][n];
}

void input()
{
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 value for matA”);

matA[i][j]=sc.nextInt();

System.out.println(“Enter the value for matB”);

matB[i][j]=sc.nextInt();

}
}

void sum()
{
for(int i=0; i<m; i++)
{
for(int j =0; j<n; j++)
{
System.out.print((matA[i][j]+matB[i][j])+”\t”);
}
System.out.println( );
}
}

public static void main(String ar[])

{
MatSum ob=new MatSum(3,4);

ob.input();

ob.sum();
}
}

Section-C

{
int lnk[]=new int[100];

int begin,end,max;

{
max=mm;

begin=end=0;
}

{
if(end= =0)

{
end=begin=1;

lnk[end]=v;
}

else if(end= =max)
{
System.out.println(‘List is Full’);
}

else
{
lnk[++end]=v;
}
}

{
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.

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();
}
}
}

(i) The dominant term is the term that grows the fastest. For example, n² grows faster than n, so if we have something like g(n) = n² + 5n + 6, it will be big O(n²).

(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 2024

ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2024