Introduction
UGC NET Exam is a very popular exam in India for people interested in the research domain. NET(National Eligibility Test) is conducted by NTA (NATIONAL TESTING AGENCY) to identify qualified candidates for Assistant Professor positions in colleges and universities, as well as Junior Research Fellowships (JRF). This article will discuss the UGC NET Exam of Aug 2016 PaperII(ReTest)  Part 1 solutions from questions 1 to 25. You can find the solutions to Q26 to 50 in Aug 2016 PaperII(ReTest)  Part 2.

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

p
Answer: D
Explanation: [~(~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

Let us assume that you construct ordered tree to represent the compound proposition (~(p˄q))↔(~p˅~q).Then, the prefix expression and postfix expression determined using this ordered tree are given as ........... and ............. respectively.
 ↔~˄pq˅ ~~pq, pq˄~p~q~˅↔
 ↔~˄pq˅ ~p~q, pq˄~p~q~˅↔
 ↔~˄pq˅ ~~pq, pq˄~p~ ~q˅↔

↔~˄pq˅ ~p~q, pq˄~p~~q˅↔
Answer: B
Explanation:.(~(p ∧ q))↔(~ p ∨ ~ q)
It is clearly specifying that
↔ is a root
(~(p ∧ q)) is left subtree
(~p ∨ ~q) is right subtree
We have the following compound proposition (~ (p ∧ q)) ↔ (~ p ∨ ~ q):
Now we'll make an ordered tree, we will make it from ordered tree i.e. ↔ ~ ∧ p q ∨ ~ p ~q
And for postorder, from the ordered tree it is p q ∧ ~ p ~ q ~ ∨ ↔.

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

A – B ≤ AB ≤ AB ≤ A + B
Answer: D
Explanation:.A ⊕ B = A + B  2A ∩ B
A ∪ B = A + B  A ∩ B
A – B = A  A ∩ B
so,
A – B < A ⊕ B < A ∪ B < A + B

What is the probability that a randomly selected bit string of length 10 is a palindrome?
 1/64
 1/32
 1/8

¼
Answer: B
Explanation: In A palindrome digits of a number are reversed, the number remains the same.
Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,
Here, string length is 10. If we consider the first 5 numbers as random choices like 0 or 1.
Other Remaining 5 numbers are fixed. Total number of possibilities are 2 10 . But we are only considering first 5 possibilities. The probability is 2 5 .
Step2: The probability 2 5 /2 10
= 1/2 5
= 1/32

Given the following graphs :
Which of the following is correct? G1 contains Euler circuit and G2 does not contain Euler circuit.
 G1 does not contain Euler circuit and G2 contains Euler circuit.
 Both G1 and G2 do not contain Euler circuit.

Both G1 and G2 contain Euler circuit.
Answer: C
Explanation:
Odd length cycle is not present in Euler Circuit.

The octal number 326.4 is equivalent to
 (214.2)_{10} and (D6.8)_{16}
 (212.5)_{10} and (D6.8)_{16}
 (214.5)_{10} and (D6.8)_{16}

(214.5)_{10} and (D6.4)_{16 }
Answer: C
Explanation: (326.4)8 = 82 * 3 + 81 * 2 + 80 * 6 . 81 * 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}.

Which of the following is the most efficient to perform arithmetic operations on the numbers?
 Signmagnitude
 1’s complement
 2’s complement

9’s complement
Answer: C
Explanation: Zero has a single representation in 2's complement, however it has two representations in Signmagnitude, 1's complement, and 9's complement (i.e., both positive zero and negative zero).When we are doing arithmetic operations such as addition or subtraction using 1's complement(or 9's complement) then we have to add an extra carry bit, to the result to get the correct answer. 2's complement doesn't require such extra calculation.

The Karnaugh map for a Boolean function is given as
The simplified Boolean equation for the above Karnaugh Map is AB + CD + AB’ + AD
 AB + AC + AD + BCD
 AB + AD + BC + ACD

AB + AC + BC + BCD
Answer: B
Explanation:
By grouping we will simply get AB + AC + AD + BCD

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

NOR
Answer: A
Explanation:
F is EXCLUSIVEOR between X and Y.

Match the following:
a. Controlled Inverter i. a circuit that can add 3 bits
b. Full adder ii. a circuit that can add two binary numbers
c. Half adder iii. a circuit that transmits a binary word or its 1’s complement
d. Binary adder iv. a logic circuit that adds 2 bits
Codes :
a b c d
(A) iii ii iv I
(B) ii iv i iii
(C) iii iv i ii
(D) iii i iv ii
Answer: D
Explanation:
Controlled Inverter : A circuit that send a binary word or its1’s complement.
Half adder : A logic circuit that adds 2 bits(ai,bi).
Full adder : A circuit capable of adding three bits,which adds two data bits(ai,bi) and also the carry bit(ci).
Binary adder : A circuit which can add two binary numbers together (A and B). Collection of Full adders form a binary adder

Given i= 0, j = 1, k = – 1
x = 0.5, y = 0.0
What is the output of given ‘C’ expression ?
x * 3 & & 3  j  k 1
 0
 1

2
Answer: C
Explanation: x * 3 && 3  j  k
= 1.5 && 3  j  k
= 1  j k
= 1

The following ‘C’ statement :
int * f[ ]( );
declares : A function returning a pointer to an array of integers.
 Array of functions returning pointers to integers.
 A function returning an array of pointers to integers.

An illegal statement.
Answer: B
Explanation: int *f [ ] ( ); It declares an array of functions that returning pointers to integers.

If a function is friend of a class, which one of the following is wrong ?
 A function can only be declared a friend by a class itself.
 Friend functions are not members of a class, they are associated with it.
 Friend functions are members of a class.

It can have access to all members of the class, even private ones.
Answer: C
Explanation: check here

In C++, polymorphism requires:
 Inheritance only
 Virtual functions only
 References only

Inheritance, Virtual functions and references
Answer: D
Explanation: check here

A function template in C++ provides ............ level of generalization.
 4
 3
 2

1
Answer: C
Explanation: In c++, a function template allows two degrees of generality. The c++ programming language has a feature called templates that allows functions and classes to work with generic types. This lets a function or class to work with a variety of data types without having to rewrite it.

DBMS provides the facility of accessing data from a database through
 DDL
 DML
 DBA

Schema
Answer: B
Solution;
The "Data Manipulation Language (DML)" is a family of syntax elements similar to computer programming language used for selecting, inserting, deleting and updating data in database. Performing readonly queries on data is sometimes also considered as the component of DML.

Relational database schema normalization is NOT for:
 reducing the number of joins required to satisfy a query.
 eliminating uncontrolled redundancy of data stored in the database.
 eliminating number of anomalies that could otherwise occur with inserts and deletes.

ensuring that functional dependencies are enforced.
Answer: A
Solution
Database normalisation is a technique of organizing data in the database. Normalisation is a systematic approach of decomposing tables to eliminate data redundancy and undesirable characteristics like insertion, update and deletion anomalies. It is a multistep process that puts data into tabular form by removing duplicated data from the relation tables.
Normalisation is used for mainly two purposes eliminating redundant data ensuring data dependencies make sense i. E. Data is stored logically normalisation doesn't reduce joins to be performed to execute a given query.
Also read anomalies in database

Consider the following statements regarding relational database model:
 NULL values can be used to opt a tuple out of enforcement of a foreign key.
 Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF.

The difference between the project operator (P) in relational algebra and the SELECT keyword in SQL is that if the resulting table/set has more than one occurrences of the same tuple, then P will return only one of them, while SQL SELECT will return all.
One can determine that: (a) and (b) are true.
 (a) and (c) are true.
 (b) and (c) are true.

(a), (b) and (c) are true.
Answer: D
Solution:
(a) null values can be used to opt a tuple out of the enforcement of a foreign key in a relational database schema. Correct
(b) assume table t only contains one candidate key. If q is found in 3nf, it is also found in bcnf. Correct
(c) the difference between the project operator () in relational algebra and the select keyword in sql is that if the resulting table/set contains many occurrences of the same tuple, the project operator () will only return one of them, but sql select will return them all.Correct

Consider the following EntityRelationship (ER) diagram and three possible relationship sets (I, II and III) for this ER diagram:
If different symbols stand for different values (e.g., t_{1} is definitely not equal to t_{2}), then which of the above could not be the relationship set for the ER diagram ? I only
 I and II only
 II only

I, II and III
Answer: A
Solution:
The S and T relations have a onetoone relationship, which is broken in set I. As a result, set I could not be the ER's relationship set. Manytomany relationships exist between S and P, S and Q, T and P, and T and Q, among others. Also, P and Q have a manytomany relationship. None of the relationships in the presented sets II and III are broken. As a result, they depict the correct relationship as defined by the ER diagram.The S and T relations have a onetoone relationship, which is broken in set I. As a result, set I could not be the ER's relationship set. Manytomany relationships exist between S and P, S and Q, T and P, and T and Q, among others. Also, P and Q have a manytomany relationship. None of the relationships in the presented sets II and III are broken. As a result, they depict the correct relationship as defined by the ER diagram.

Consider a database table R with attributes A and B. Which of the following SQL queries is illegal ?
 SELECT A FROM R;
 SELECT A, COUNT(*) FROM R;
 SELECT A, COUNT(*) FROM R GROUP BY A;

SELECT A, B, COUNT(*) FROM R GROUP BY A, B;
Answer: B
Solution: The COUNT() function calculates the number of rows that meet a set of criteria. The "GROUP BY" phrase normally specifies the criteria. COUNT(*) returns the total number of rows in the table without any criterion in query 2. As a result, the second query is inaccurate.

Consider an implementation of unsorted single linked list. Suppose it has its representation with a head and a tail pointer (i.e. pointers to the first and last nodes of the linked list). Given the representation, which of the following operation cannot be implemented in O(1) time ?
 Insertion at the front of the linked list.
 Insertion at the end of the linked list.
 Deletion of the front node of the linked list.

Deletion of the last node of the linked list.
Answer: D
Solution:
Insertion at the front of the linked list will take O(1) time, just take newNode.next = first and first = newNode.
Insertion at the end of the linked list will take O(1) time, just take last.next = newNode and last = newNode
Deletion of the front node of the linked list will take O(1) time, just make fisrt = first.next
Deletion of the last node of the linked list will take O(n) time, since you need to traverse from first node to second last node to make it last and delete the current last node.

Consider an undirected graph G where selfloops are not allowed. The vertex set of G is {(i, j)  1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a, b) and (c, d) if a – c ≤ 1 or b–d ≤ 1. The number of edges in this graph is
 726
 796
 506

616
Answer: C
Explanation: The vertex set of Graph G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}.
There is an edge between the points (a, b) and (c, d) if ca  <= 1 and
db <= 1.
The total number of vertex can be 12*12 and the vertices are (1, 1),
(1, 2),(1,3).....(1, 12) (2, 1), (2, 2),(2,3)...
Total number of edges in the graph?
The number of edges in the graph equals the number of pairs of vertices that meet the above criteria. The vertex pairs (1, 1) and (1, 2), for example, satisfy the above condition.
There may be an advantage to (1, 2), (2, 1), and (3, 1) for (1, 1). (2, 2). As noted in the question, there can be a selfloop.
The same goes for (12, 12), (1, 12), and (1, 12). (12, 1)
For vertex (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3),(1, 3)
Same is count (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11)
For vertex (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1),(2, 3), (3, 1), (3, 2), (3, 3)
Same is count for remaining vertices.
For all pairs (i, j) there can total 8 vertices connected to them if i and j are not in {1, 12}
There are total 100 vertices without a 1 or 12. So total 800 edges.
For vertices with 1, total edges = (Edges where 1 is first part) + (Edges where 1 is second part and not first part) = (3 + 5*10 + 3) + (5*10) edges
Same is count for vertices with 12
The total number of edges:= [(3 + 5*10 + 3) + (5*10) ] + [(3 + 5*10+5*10 + 3)] +800
= 800 + 106 + 106
= 1012
Since Two edges from v1 to v2 and v2 to v1 should be counted as one because the graph is undirected.
So, the total number of undirected edges = 1012/2 = 506.

The runtime for traversing all the nodes of a binary search tree with n nodes and printing them in an order is
 O(lg n)
 O(n lg n)
 O(n)

O(n^{2})
Answer: C
Solution:click here

Consider the following statements :
S_{1}: A queue can be implemented using two stacks.
S_{2}: A stack can be implemented using two queues.
Which of the following is correct ? S_{1} is correct and S_{2} is not correct.
 S_{1} is not correct and S_{2} is correct.
 Both S_{1} and S_{2} are correct.

Both S_{1} and S_{2} are not correct.
Answer: C
Solution
A queue can be created with a minimum of two stacks. A stack can be made up of at least two queues. Both assertions are correct.

Given the following prefix expression:
* + 3 + 3 ↑ 3 + 3 3 3
What is the value of the prefix expression? 2178
 2199
 2205

2232
Answer: C
Solution:
* + 3 + 3 ↑ 3 + 3 3 3
= * + 3 + 3 ↑ 3 6 3
= * + 3 + 3 729 3
= * + 3 732 3
= * 735 3
= 2205
Read about Bitwise Operators in C here.
FAQs
What is the UGC NET exam for?
The National Eligibility Test (NET) is a test administered by the University Grants Commission (UGC) to determine whether Indian nationals are eligible for Assistant Professorships, Junior Research Fellowships, or both in Indian universities and colleges.
What is UGC NET exam eligibility?
If you hold a Master's Degree or equivalent in Humanities (including languages), Social Science, Computer Science and Applications, Electronic Science, etc., you are eligible to take the UGC NET test.
Is PhD compulsory for net?
To teach in central institutions, one must hold a PhD or have completed the UGC NET qualification exam, according to UGC regulations. The UGC, on the other hand, intends to scrap the regulation requiring experts to be hired and replace it with specific titles such as professor of practice and associate professor of practise.
How many papers are there in the UGC NET exam?
There are two papers, and the candidates get 3 hours for both papers. There are 150 questions in UGC NET combining both papers.
What is the full form of UGC NET?
UGC stands for University Grants Commission, and NET stands for National Eligibility Test.