UGC NET CS 2016 Aug – II

Question 1

The Boolean function [~(~p ∧ q) ∧ ~( ~p ∧ ~q)] ∨ (p ∧ r)] is equal to the Boolean function:

• A. q
• B. p ∧ r
• C. p ∨ q
• D. p

Answer: (D)

UGC NET CS 2016 Aug – II Propositional and First Order Logic.
Discuss it

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.

Question 2

Let us assume that you construct ordered tree to represent the compound proposition (~ (p ∧ q)) ↔ (~ p ∨ ~ q). Then, the prefix expression and post-fix expression determined using this ordered tree are pgiven as ____ and _____ respectively.

• A. ↔~∧pq∨ ~ ~ pq, pq∧~p~q~∨↔
• B. ↔~∧pq∨ ~ p~q, pq∧~p~q~∨↔
• C. ↔~∧pq∨ ~ ~ pq, pq∧~p~~q∨↔
• D. ↔~∧pq∨ ~ p~ q, pq∧~p~ ~q∨↔

Answer: (B)

UGC NET CS 2016 Aug – II Propositional and First Order Logic.
Discuss it

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.

Question 3

Let A and B be sets in a finite universal set U. Given the following: |A – B|, |A ⊕ B|, |A| + |B| and |A ∪ B| Which of the following is in order of increasing size ?

• A. |A – B| < |A ⊕ B| < |A| + |B| < |A ∪ B|
  
• B. |A ⊕ B| < |A – B| < |A ∪ B| < |A| + |B|
  
• C. |A ⊕ B| < |A| + |B| < |A – B| < |A ∪ B|
  
• D. |A – B| < |A ⊕ B| < |A ∪ B| < |A| + |B|

Answer: (D)

UGC NET CS 2016 Aug – II Propositional and First Order Logic.
Discuss it

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|

Question 4

What is the probability that a randomly selected bit string of length 10 is a palindrome?

• A. 1/64
• B. 1/32
• C. 1/8
• D. 1/4

Answer: (B)

UGC NET CS 2016 Aug – II Probability
Discuss it

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.

Question 5

Given the following graphs: Which of the following is correct?

• A. G₁ contains Euler circuit and G₂ does not contain Euler circuit.
• B. G₁ does not contain Euler circuit and G₂ contains Euler circuit.
• C. Both G₁ and G₂ do not contain Euler circuit.
• D. Both G₁ and G₂ contain Euler circuit.

Answer: (C)

UGC NET CS 2016 Aug – II Graph Theory
Discuss it

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.

Question 6

The octal number 326.4 is equivalent to

• A. (214.2)₁₀ and (D6.8)₎₁₆
• B. (212.5)₁₀ and (D6.8)₎₁₆
• C. (214.5)₁₀ and (D6.8)₎₁₆
• D. (214.5)₁₀ and (D6.4)₎₁₆

Answer: (C)

UGC NET CS 2016 Aug – II Digital Logic & Number representation Number Representation
Discuss it

To solve this question we will apply traditional approach: (326.4)₈ = 8² * 3 + 8¹ * 2 + 8⁰ * 6 . 8^-1 * 4 = ( 214.5)₁₀ is decimal representation. For hexadecimal representation group binary sequence of (214.5) = (011010110.100)2 into group of 4. i.e. 0 1101 0110. 1000 (0 can be padded after decimal) this is equivalent to- (D6.8)₁₆. So, option (C) is correct.

Question 7

Which of the following is the most efficient to perform arithmetic operations on the numbers?

• A. Sign-magnitude
• B. 1’s complement
• C. 2’s complement
• D. 9’s complement

Answer: (C)

UGC NET CS 2016 Aug – II Digital Logic & Number representation Number Representation
Discuss it

Sign's magnitude is only for sign convention (MSB is 1 then no is negative and if 0 then no is positive). Main difference is that while adding numbers using 1′ s complement, we first do binary addition, then add in an end-around carry value. But, 2′ s complement has only one value for zero, and doesn’t require carry values. 9’s complement of a decimal number is the subtraction of it’s each digits from 9. Like 1’s complement, 9’s complement is used to subtract a number using addition. 2's complement representation is unambiguous for 0 (i.e., only positive 0), but Sign-magnitude, 1’s complement and 9’s complement are ambiguous representation for 0 (i.e., both positive and negative 0). So, option (C) is correct.

Question 8

The Karnaugh map for a Boolean function is given as The simplified Boolean equation for the above Karnaugh Map is

• A. AB + CD + A`B + AD
• B. AB + AC + AD + BCD
• C. AB + AD + BC + ACD
• D. AB + AC + BC + BCD

Answer: (B)

UGC NET CS 2016 Aug – II Digital Logic & Number representation Logic functions and Minimization
Discuss it

By grouping we will simply get AB + AC + AD + BCD. So, option (B) is correct.

Question 9

Which of the following logic operations is performed by the following given combinational circuit?

• A. EXCLUSIVE-OR
• B. EXCLUSIVE-NOR
• C. NAND
• D. NOR

Answer: (A)

UGC NET CS 2016 Aug – II Digital Logic & Number representation Combinational Circuits
Discuss it

F is EXCLUSIVE-OR between X and Y. So, option (A) is correct.

Question 10

Match the following:

• A. (1)
• B. (2)
• C. (3)
• D. (4)

Answer: (D)

UGC NET CS 2016 Aug – II Digital Logic & Number representation Combinational Circuits
Discuss it

Controlled Inverter is a circuit that transmits a binary word or its 1’s complement. Full adder is a circuit that can add 3 bits. Half adder is a logic circuit that adds 2 bits. Binary adder is a circuit that can add two binary numbers. So, option (D) is correct.