ISC Computer Science 2019 Class-12 Previous Year Question Paper Solved for practice. Step by step Solutions with Questions of Part-1 and Part-2 (Section-A,B and C). By the practice of Computer Science 2019 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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## ISC Computer Science 2019 Class-12 Previous Year Question Paper Solved

Part-I

Section-A of Part-II

Section-B of Part-II

Section-C of Part-II

Maximum Marks: 70
Time allowed: 3 hours

• Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.
• Answer all questions in Part-I (compulsory) and six questions from Part-II, choosing two questions from Section-A, two from Section-B and two from Section-C.
• All working, including rough work, should be done on the same sheet as the rest of the answer.
• The intended marks for questions or parts of questions are given in brackets [ ].

Part – I (20 Marks)

### ISC Computer Science 2019 Class-12 Previous Year Question Paper Solved

While answering questions in this Part, indicate briefly your working and reasoning, wherever required.

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. 

 A B X 0 0 0 0 1 1 1 0 1 1 1 0

(b) Write the canonical POS expression of: F(P, Q) = Π(0, 2) 
(c) Find the dual of: X.Y+X.Y’ = X + 0 
(d) If F(A, B, C) = A’.B’.C’ + A’.B.C’ then find F’ using De Morgan’s Law. 
(e) If A = “It is cloudy” and B = “It is raining”, then write the proposition for 
(i) Contrapositive
(ii) Converse
(a) XORgate (b) (A + B).(A’ + B)
(c) (X + Y).(X + Y’) = X.1
(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) (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? 
(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  when the matrix is stored in Row Major Wise. 
(c) Convert the following infix notation to postfix form: 
(A + B * C) – (E * F / H) + J
(d) Compare the two complexities O(n2) and O(2n) and state which is better and why. 
(e) State the difference between internal nodes and external nodes of a binary tree structure. 
(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) Given, Address of ARR  = 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 = 1430 + 4(6 (3 – (-4))+ (6 – 3))
= 1430 + 4(6 × 7 + 3)
= 1430 + 180
= 1610

(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) O(n2) is better than O(2n)
Let us see with the help of an example.
Suppose n = 8
n2 = 82 = 64
2n = 28 = 256
Therefore, the complexity in 2n is higher than n2. If n increases, 2n increases much more than n2. Therefore, the time complexity is O(n2) is better than O(2n).

(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. 

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);
}
}

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 (50 Marks)

Answer six questions in this part, choosing two questions from Section A, two from Section B and two from Section C.

Section – A

### Previous Year Question Paper Solved ISC Computer Science 2019 Class-12

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. 
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