PREVIOUS YEAR SOLVED PAPERS - August 2016 Paper 2
1. The Boolean function [~(~p ∧ q) ∧ ~( ~p ∧ ~q)] ∨ (p ∧ r)] is equal to the Boolean function:
- Option : D
- Explanation : We have Boolean function: [~(~p ∧ q) ∧ ~( ~p ∧ ~q)] ∨ (p ∧ r)] = [(p ∨ ~q) ∧ (p ∨ q ) ∨ (p ∧ r)] = [p ∨ (p ∧ q) ∨ (p ∧ ~q) ∨(p ∧ r)] = p[1 ∨ q ∨ ~q ∨ r] = p So, option (D) is correct.
- Option : B
- Explanation : We have compound proposition (~ (p ∧ q)) ↔ (~ p ∨ ~ q): Now we will construct ordered tree:
We are asked to determine pre-order (i.e. parent-node left-node right-node), we will drive it from ordered tree i.e. ↔ ~ ∧ p q ∨ ~ p ~q And post-order(i.e. left-node right-node parent-node) from the ordered tree it is p q ∧ ~ p ~ q ~ ∨ ↔. So, option (B) is correct.
- Option : D
- Explanation : We will draw venn diagram for all set: |A – B|, |A ⊕ B|, |A| + |B| and |A ∪ B|
So, option (D) is correct. Alternative way -
|A – B| = |A| - |A ∩ B|
|A ⊕ B| = |A| + |B| - 2|A ∩ B|
|A ∪ B| = |A| + |B| - |A ∩ B|
Therefore,
|A – B| < |A ⊕ B| < |A ∪ B| < |A| + |B|
4. What is the probability that a randomly selected bit string of length 10 is a palindrome?
- Option : B
- Explanation : In the given question we have a palindrome: in even length palindrome half length is fixed and rest is repeated. So, in 10 bit palindrome, we have 5 position to be filled with 2 choices each-
i.e. 25 choices for first half and 25 choices for second half.
Probability = favorable outcome / total outcome
= 25/ 210
= 1 / 25
= 1 / 32.
So, option (B) is correct.
5. Given the following graphs:
Which of the following is correct?
- Option : C
- Explanation : Euler circuit does not contain odd length cycle. Refer: Eulerian path and circuit for undirected graph None of the above graph is Eulerian. So, option (C) is correct.
