ISC Computer Science 2018 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 2018 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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## ISC Computer Science 2018 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)Answer all questions.

ISC Computer Science 2018 Class-12 Previous Year Question Paper Solved

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

Question 1.
(a) State the Commutative law and prove it with the help of a truth table.  (b) Convert the following expression into its canonical POS form:  F(X, Y, Z) = (X + Y) . (Y’ + Z)
(c) Find the dual of  (A’ + B) . (1 + B’) = A’+B
(d) Verify the following proposition with the help of a truth table:  (P∧Q)∨(P∧~Q) = P
(e) If F(A, B, C) = A'(BC’ + B’C), then find F’.  Answer 1:
(a) Commutative law states that the interchanging of the order of operands in a Boolean equation does not change its result.
Using OR operator → A + B = B + A
Using AND operator → A * B = B * A
Truth Table for Commutative Law A + B = B + A

(b) F(X, Y, Z) = (X + Y’).(Y’ + Z)
By De-Morgan’s theorem, we have
((X + Y’).(Y’ + Z))’ = (X + Y’)’ + (Y’+ Z)’
=X’.Y” + Y”.Z’
= X’.Y + Y.Z’
= Y.(X’ + Z’)
Again applying De-Morgan’s theorem, we have
(Y.(X’ + Z’))’ = (X’ + Z’)’

(c) In Principle of Duality;
replace (+) by (.)
replace (.) by (+)
replace 1 b 0
Taking L.H.S. = (A’ + B).(1 + B’)
= (A’.B) + (0.B’)
=A’.B
Again applying the principle of duality, we have = A + B’ Question 2.
(a) What are the Wrapper classes? Give any two examples.  (b) A matrix A[m] [m] is stored in the memory with each element requiring 4 bytes of storage. If the base address at A  is 1500 and the address of A is 1608, determine the order of the matrix when it is stored in Column Major Wise.  (c) Convert the following infix notation to postfix form:  A + (B – C*(D/E) * F)
(d) Define Big ‘O’ notation. State the two factors which determine the complexity of an algorithm.  (e) What is exceptional handling? Also, state the purpose of finally block in a try-catch statement.  Answer 2:
(a) A Wrapper class is a class whose object wraps or contains primitive data types. When we create an object to a wrapper class, it contains a field and in this field, we can store primitive data types. In other words, we can wrap a primitive value into a wrapper class object. (b) Base Address, B = 1500
W = 4 Bytes
i = 4, j = 5, r = 1, c = 1
Formula for Column Major Wise
A = B + W[m(j – c) + (I – r)] 1608 = 1500 + 4[m(5 – 1) + (4 – 1)] 1608 – 1500 = 16m + 12
16m = 96
m = 6 (d) Big ‘O’ notation is a particular tool for assessing algorithm efficiency and is used to describe the performance or complexity of an algorithm. While analyzing an algorithm, we mostly consider time complexity and space complexity.
The time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the input. Similarly, the Space complexity of an algorithm quantifies the amount of space or memory taken by an algorithm to run as a function of the length of the input.

(e) The exception handling in java is one of the powerful mechanism to handle the runtime errors so that the normal flow of the application can be maintained.
Java finally block is a block that is used to execute important code such as closing connection, stream etc.
Java finally block is always executed whether an exception is handled or not.
Java finally block follows try or catch block.

Question 3.
The following is a function of some class which checks if a positive integer is a Palindrome number by returning true or false. (A number is said to be palindrome if the reverse of the number is equal to the original number.) The function does not use the 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 PalindromeNum(int N)
{
int rev=?1?;
int num=N;
while(num>0)
{
int f=num/10;
int s= ?2?;
int digit = num-?3?;
rev= ?4? + digit;
num/= ?5?;
}
if(rev==N)
return true;
else
return false;
}```
1. What is the statement or expression at ?1?
2. What is the statement or expression at ?2?
3. What is the statement or expression at ?3?
4. What is the statement or expression at ?4?
5. What is the statement or expression at ?5?

1. 0
2. 0
3. 1
4. rev*10
5. 10

### 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 2018 Class-12

Question 4.
(a) Given the Boolean function F(A, B, C, D) = Σ (0, 2, 4, 8, 9, 10, 12, 13).
(i) Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.eoctal, quads and pairs).  (ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.  (b) Given the Boolean function: F(A, B, C, D) = π(3, 4, 5, 6, 7, 10, 11, 14, 15).
(i) Reduce the above expression by using the 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).  (ii) Draw the logic gate diagram for the reduced expression. Assume that the variables and their complements are available as inputs.  Answer 4:
(a) (i) F(A, B, C, D) = Σ (0, 2, 4, 8, 9, 10, 12, 13) Product at cell
(0) = A’.B’C’.D’
(4) = A’.B.C’.D’
(12) = A.B.C’.D’
(8) = A.B’.C’.D’ (common is C’.D’)
Product at
(12) = A.B.C’.D’
(8) = A.B’.C’.D’
(9) = A.B’.C’.D
(13) = A.B.C’.D (common is A.C’)
Product at
(2) = A’.B’.C.D’
(10) = A.B’.C.D’ (common is B’.C.D’)
Reduced expression is CD’ + AC’ + B’.C.D’
(ii) C’.D’ + A.C’ + B’.C.D’
⇒ A.C’ + B’.C.D’ + C’.D’
Logic Gate Read Next 👇 Click on Page Number Given Below 👇