# ISC Computer Science 2015 Class-12 Previous Year Question Papers Solved

**ISC Computer Science 2015** 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 2015 Class-12** Solved Previous Year Question Paper you can get the idea of solving.

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**ISC Computer Science 2015** 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 2015** Class-12 Previous Year Question Paper Solved** **

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

**Question 1.**

(a) Simplify: (A + C). (A + A.D) + A.C + C [2]
(b) Draw a logic circuit for (A + B).(C + D). C [2]
(c) Verify the following proposition with the help of a truth table: [2]
P ∨ (~P ∧ Q) = P ∨ Q

(d) State De Morgan’s law and verify it, using a truth table. [2]
(e) Answer the questions related to the circuit given below: [2]

(i) Give the output if, X = 1 and Y = 0

(ii) Name the basic gate represented by the above diagram.

**Answer 1:**

(a) (A + C). (A + A.D) + A.C + C

= (A + C).(1 + D).A + C(1 + A) [∵ 1 + D = 1]
= (A + C).A + C [∵ 1 + A = 1]
= A.A + A.C + C [∵ A.A = A]
= A + C(A + 1)

= A + C [∵ A + 1 = 1]

(b) Logic circuit for (A + B). (C + D). C

**(d) The two theorems suggested by De Morgan which are extremely useful in Boolean Algebra are as following:**

Theorem 1

NAND = Bubbled OR

Table showing verification of the De Morgans’s first theorem

Theorem 2

NOR = Bubbled AND

**Table showing verification of the De Morgans’s second theorem**

(i) when X = 1 and Y = 0

the output will be 0.

(ii) NOR gate

**Question 2.**

(a) Define computational complexity. Calculate the complexity using Big ‘O’ notation for the following code segment: [2]
for(int k=0; k < n; k++)

s+ = k;

**(b) Convert the following infix notation into postfix form: [2]**

X + (Y – Z) + ((W+E)*F)/J

(c) Differentiate between this keyword and the super keyword. [2]
(d) The array D [-2…10][3…8] contains double type elements. If the base address is 4110, find the address of D [4] [5], when the array is stored in Column Major Wise. [2]
(e) State any two characteristics of a Binary tree. [2]
**Answer 2:**

(a) Computational complexity theory focuses on classifying computational problems according to their difficulty. A problem is considered difficult if large amounts of resources (time, storage, communications, circuit gates, processors, etc.) are required to solve it, regardless of the algorithm used. A subset of computational complexity theory, known as analysis of algorithms, provides theoretical estimates for the resources needed for specific algorithms. These estimates are frequently expressed in Big O (or Big-Oh) notation, written as functions of the size of the input passed to the algorithm. For example, binary search is estimated to require a number of steps proportional to the logarithm of the length of the list being searched. In Big O notation, the estimate is written as 0(log (n)), where n = length of the list. O(N)

O(N) describes an algorithm whose performance will grow linearly and in direct proportion to the size of the input data set.

(b) = X + (Y – Z) + ((W + E) * F)/J

= X + (YZ-) + ((WE+) * F)/J

= X + (YZ-) + ((WE + F*))/J

= X + (YZ-) + (WE + F*)/J

= X + (YZ-) + (WEF*+)/J

= X + (YZ -) + (WEF * + J/)

= XYZ- + WEF* + J/+

(c) This refers to an object, that too, a current object or current class whereas super refers directly to its immediate above superclass. private variables cannot be accessed in an inherited class by using this, whereas by using “super” we can easily access the private variables in its superclass.

(d) D [-2… 10] [3 …8]
BaseAddress B = 4110

I = 4, J = 5

D [4] [5] = ?

W = 8 bytes

M = 10

**Columnwise**

D[I, J] = B + ((J – 1) * M + (I – 1)) * W

D [4] [5] = 4110 + ((5 – 1) + 10 + (4 – 1)) * 8

= 4110+ (40 + 3) × 8

= 4110 + 344

= 4454

**(e) A binary tree is an important type of structure which occurs very often.**

- It is characterized by the fact that any node can have at most two branches, i. e., there is no node with degree greater than two.
- For binary trees, we distinguish between the subtree on the left and on the right.
- A binary tree may have zero nodes.

**Question 3.**

(a) The following function is a part of some class. Assume ‘x’ and ‘y’ are positive integers, greater than 0. Answer the given questions along with dry run/working.

void someFun(int x, int y) { if(x>1) { if(x%y == 0) { System.out.print(y+ ""); someFun(x/y, y); } else someFun(x, y+1); } }

(i) What will be returned by someFun(24, 2)? [2] (ii) What will be returned by someFun(84, 2)? [2] (iii) State in one line what does the function someFun() do, apart from recursion? [1] (b) The following is a function of some class which checks if a positive integer is an Armstrong number by returning true or false. (A number is said to be Armstrong of the sum of the cubes of all its digits is equal to the original number.) The function does not use modulus (%) operator to extract digit. There are some places in the code marked by ?1?, ?2?, ?3?, ?4?, ?5? which may be replaced by a statement/expression so that the function works properly.

boolean ArmstrongNum(int N) { int sum = ?1?; int num= N; while(num>0) { int f=num/10; int s= ?2?; int digit = num-s; sum+=?3?; num= ?4?; } if(?5?) return true; else return false; }

(i) What is the statement or expression at ?1? [1]
(ii) What is the statement or expression at ?2? [1]
(iii) What is the statement or expression at ?3? [1]
(iv) What is the statement or expression at ?4? [1]
(v) What is the statement or expression at ?5? [1]
**Answer 3:**

(a)

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