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

**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

-: Select Your Topics :-

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)**

**Answer all questions.**

**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. [1]

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) [1]

(c) Find the dual of: X.Y+X.Y’ = X + 0 [1]

(d) If F(A, B, C) = A’.B’.C’ + A’.B.C’ then find F’ using De Morgan’s Law. [1]

(e) If A = “It is cloudy” and B = “It is raining”, then write the proposition for [1]

(i) Contrapositive

(ii) Converse

**Answer 1:**

(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? [2]

(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. [2]

(c) Convert the following infix notation to postfix form: [2]

(A + B * C) – (E * F / H) + J

(d) Compare the two complexities O(n2) and O(2n) and state which is better and why. [2]

(e) State the difference between internal nodes and external nodes of a binary tree structure. [2]

**Answer 2:**

(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[3] [6] = 1430

According to row-major,

Address = B + w(n(p – L_{1}) + (q – L_{2})

B = 1430, w = 4 bytes

p = 3, q = 6,U_{1} = 6, U_{2} = 8, L_{1} = -4, L_{2} = 3

Number of columns n = U_{2} – L_{2} + 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) (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(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) 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. [5]

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); } }

**Answer 3:**

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**

**Answer any two questions.**

### 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). [4]

(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. [1]

(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). [4]

(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. [1]

**Answer 4:**

(a)

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