Introduction
UGC NET Exam is a very popular exam in India for people interested in research. Previous Year Questions are an excellent option to learn about the exam pattern. By solving the PYQs, you will get a basic idea about your preparation.
Refer to Dec 2014 Paper II Part 2 here.
You can evaluate your weak areas and work on them to perform better in the examination. In this article, we have given the questions of UGC NET Dec 2014 Paper II. We have also explained every problem adequately to help you learn better.
Dec 2014 Paper II Part 1
1. Consider a set A = {1, 2, 3, …….., 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5 ?
(A) 533 (B) 599
(C) 467 (D) 66
Answer: C
From set A numbers {3,6,9,……..999} which are divisible by 3 are 999 / 3 (A)= 333 .
From set A numbers {5,10,……995,1000} which are divisible by 5 are 1000 / 5 (B)= 200.
From set A numbers {15, 30, …990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66.
So, numbers divisible by 3 or by 5 or by both 3 and 5:
(A ∨ B) = A + B – (A ∧ B)
(A ∨ B) = 333 + 200 – 67.
(A ∨ B) = 467.
2. A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph?
(A) 5 (B) n – 3
(C) 20 (D) 11
Answer: D
There are 2 vertices of degree 4, 1 vertex of degree 3, 1 vertex of degree 2 and vertex of degree one is unknown.
Let’s assume k is the nof vertex of degree one.
Total vertex = 2 + 1 + 1 + k = k + 4.
Number of edges = vertex – 1
i.e. k + 4 – 1
= k + 3.
Now apply handshaking lemma,
2 * 4 + 1 * 3 + 1 * 2 + 1 * K = 2 * (No of edges)
i.e. 13 + k = 2 * (k + 3)
k = 7.
Total vertex = 7 + 4 = 11.
3. Consider the Graph shown below :
This graph is a ...............
(A) Complete Graph (B) Bipartite Graph
(C) Hamiltonian Graph (D) All of the above
Answer: C
There is no edge between D and B, A and C and one more thing for the complete graph edge must be n * (n – 1) / 2 which is not valid in given graph.
If a graph is two colourable then it is bipartite but this graph is not bipartite.
According to Dirac’s Theorem, If each vertex has a degree greater than n / 2 then it is
4. A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout at least once ?
(A) 0.5 (B) 0.704
(C) 0.632121 (D) 0.68
Answer: C
Favourable outcome k = 1
Total outcome = 1000000 = 106
Probability of k = 1;
Probability = favourable outcome / total outcome
= 1 / 106 = 106
Probability of k != 1:
= 1 – Probability of k = 1;
1 – 106 = 0.999
Probability that k = 1 is never printed in all 106 printouts:
= 0.999*0.999*…….0.9999 (106 time)
= 0.999106
The probability that 1 is printed at least once:
= 1 the probability that 1 is never printed
= 10.99910^6
=0.6321
5. If we define the functions f, g and h that map R into R by :
f(x) = x4, g(x)=√(x2+1), h(x) = x2+72, then the value of the composite functions ho(gof) and (hog)of are given as
(A) x8 – 71 and x8 – 71
(B) x8 – 73 and x8 – 73
(C) x8 + 71 and x8 + 71
(D) x8 + 73 and x8 + 73
Answer: D
f(x) = x^{4}, g(x) = √(x^{2} + 1) and h(x) = x^{2} + 72
Now ho(gof) = ho√(x^{8} + 1)
= x^{8} + 72 + 1
= x^{8} + 73.
We have to find (hog)of:
(hog)of = (x^{2} + 1) + 72
= x^{2} + 73
= x^{8} + 73.
6. The BCD adder to add two decimal digits needs minimum of
(A) 6 full adders and 2 half adders
(B) 5 full adders and 3 half adders
(C) 4 full adders and 3 half adders
(D) 5 full adders and 2 half adders
Answer: D
7. The Excess3 decimal code is a selfcomplementing code because
(A) The binary sum of a code and its 9’s complement is equal to 9.
(B) It is a weighted code.
(C) Complement can be generated by inverting each bit pattern.
(D) The binary sum of a code and its 10’s complement is equal to 9.
Answer: A, C
The Excess3 decimal code is a selfcomplementing code because the binary sum of a code and its 9’s complement is equal to 9 and complement can be generated by inverting each bit pattern.
8. How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A * B) + (C * D/E) ?
(A) 4 PUSH and 3 POP instructions
(B) 5 PUSH and 4 POP instructions
(C) 6 PUSH and 2 POP instructions
(D) 5 PUSH and 3 POP instructions
Answer: B
Reverse polish notation is a system of formula notation without brackets or special punctuation.
To evaluate (A ∗ B) + (C ∗ D / E):
First avoid brackets and punctuation and convert it into postfix form i.e. AB+CDE/*+
Now push AB
On * pop AB and perform A * B. Now push back the result(say it X).
Push CDE.
On / pop DE and push back the result(say it Y).
On * Pop CY and perform * operation and push the result(say it z).
On + pop XZ and perform + operation and and push back the final answer.
Above computation include 5 PUSH and 4 POP instructions.
9. The range of representable normalised numbers in the floating point binary fractional representation in a 32bit word with 1bit sign, 8bit excess 128 biassed exponent and 23 bit mantissa is
(A) 2^{–128} to (1 – 2^{–23}) × 2^{127}
(B) (1 – 2^{–23}) × 2^{–127} to 2^{128}
(C) (1 – 2^{–23}) × 2^{–127} to 2^{23}
(D) 2^{–129} to (1 – 2^{–23}) × 2^{127}
Answer: D
The range of representable normalised numbers in the floating point binary fractional representation in a 32bit word :
bais = 2^{n1} = 2^{81 }= 128
Number 1 in normalised format = 0.0000000....23times = 2^{0128} = 2^{128}
Maximum Number 1111111..22 times = 0.111111...23 times = (1 2^{23} ) * (2^{255128}) = (1 2^{23} ) * (2^{127})
10. The size of the ROM required to build an 8bit adder/subtractor with mode control, carry input, carry output and two’s complement overflow output is given as
(A) 2^{16} × 8 (B) 2^{18} × 10
(C) 2^{16} × 10 (D) 2^{18} × 8
Answer: B
Total input to the rom decoder will be (8+8 ( two 8 bit number ) +1( mode ) +1( carry in))
So, the total number of words out of the decoder will be 2^18 . result will be 8 bit so 8 vertical lines + ( 1 for carry ) +1 ( for saying underflow) .
11. What will be the output of the following ‘C’ code ?
main ( )
{ int x = 128;
printf (“\n%d”, 1 + x++);
}
(A) 128
(B) 129
(C) 130
(D) 131
Answer: B
In the above C program There is post increment operation:
So, printf will print 1 + 128 as output.
++ will increment the x value i.e. x = 129.
12. What does the following expression means ?
char *(*(* a[N]) ( )) ( );
(A) a pointer to a function returning array of n pointers to function returning character pointers.
(B) a function return array of N pointers to functions returning pointers to characters
(C) an array of n pointers to function returning pointers to characters
(D) an array of n pointers to function returning pointers to functions returning pointers to characters.
Answer: A,B,C,D
a  a
a[N]  is an Nelement array
*a[N]  of pointers
(*a[N])()  to functions
*(*a[N])()  returning pointers
(*(*a[N])())()  to functions
*(*(*a[N])())()  returning pointers
char *(*(*a[N])())()  to char
13. Which of the following is not a member of class ?
(A) Static function (B) Friend function
(C) Const function (D) Virtual function
Answer: B
Friend function is not a member of the class. No such function exists.
14. When an array is passed as parameter to a function, which of the following statements is correct ?
(A) The function can change values in the original array.
(B) In C, parameters are passed by value, the function cannot change the original value in the array.
(C) It results in compilation error when the function tries to access the elements in the array.
(D) Results in a run time error when the function tries to access the elements in the array.
Answer: A
When an array is passed as parameter to a function then the function can change values in the original array.
15. Which of the following differentiates between overloaded functions and overridden functions ?
(A) Overloading is a dynamic or runtime binding and overridden is a static or compile time binding.
(B) Overloading is a static or compile time binding and overriding is dynamic or runtime binding.
(C) Redefining a function in a friend class is called overloading, while redefining a function in a derived class is called as overridden function.
(D) Redefining a function in a derived class is called function overloading, while redefining a function in a friend class is called function overriding.
Answer: B
Overloading is a static or compile time binding and overriding is dynamic or runtime binding.
Function Overloading (achieved at compile time)
It provides multiple definitions of the function by changing signature i.e changing number of parameters, change datatype of parameters, return type doesn’t play any role.
It can be done in base as well as derived class.
16. Division operation is ideally suited to handle queries of the type :
(A) customers who have no account in any of the branches in Delhi.
(B) customers who have an account at all branches in Delhi.
(C) customers who have an account in atleast one branch in Delhi.
(D) customers who have only joint account in any one branch in Delhi
Answer: B
Division operation is ideally suited to handle queries of the type:
 Customers who have an account at all branches in Delhi.

Students enrolled all courses.
etc.
17. Which of the following is true ?
I. Implementation of selfjoin is possible in SQL with table alias.
II. Outerjoin operation is basic operation in relational algebra.
III. Natural join and outer join operations are equivalent.
(A) I and II are correct. (B) II and III are correct.
(C) Only III is correct. (D) Only I is correct.
Answer: D
Implementation of selfjoin is possible in SQL with table alias. This statement is correct as selfjoin is usable in SQL.
Outerjoin operation is basic operation in relational algebra. The given statement is incorrect as Outerjoin is a complex operation.
Natural join and outer join operations are equivalent. This statement is incorrect as natural join is similar to inner join, where as outer join is different from inner join.
18. What kind of mechanism is to be taken into account for converting a weak entity set into strong entity set in entityrelationship diagram ?
(A) Generalisation (B) Aggregation
(C) Specialisation (D) Adding suitable attributes
Answer: D
Weak entity is an entity which does not have sufficient attributes to form a primary key.
Weak entity is represented by Double rectangle In an ER Diagram.
It must be in a 1:M relation.
A member of a weak entity is called a subordinate entity.
For converting a weak entity set into strong entity set in entityrelationship diagram by simply adding appropriate attributes.
19. The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies F = {AB®C, AB®D, C®A, D®B} is
(A) BoyceCodd Normal form (B) Third Normal form
(C) Second Normal form (D) First Normal form
Answer: B
Here C> A (prime attribute > prime attribute) and D>A (prime attribute > prime attribute) are partially dependent FDs . This type of FD must not be in BCNF and there is no problem with this type of FDs in 3NF.
20. Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A→B, B→C, AC→D}
(A) A (B) AE
(C) BE (D) CE
Answer: A
A+ = ABCD
But we can’t derive E from A, so We have to add E i A
I.e. AE+ = ABCDE.
21. Convert the following infix expression into its equivalent post fix expression (A + B^ D) / (E – F) + G
(A) ABD^ + EF – / G+ (B) ABD + ^EF – / G+
(C) ABD + ^EF / – G+ (D) ABD^ + EF / – G+
Answer: A
(A + B^ D) / (E – F) + G
= (A + B^ D)(E – F)/ + G
= (A + B^ D)(E – F)/G+
= A + BD^(E – F)/G+
= ABD^+EF/G+
So, option (A) is correct.
22. You have to sort a list L, consisting of a sorted list followed by a few ‘random’ elements. Which of the following sorting method would be most suitable for such a task?
(A) Bubble sort (B) Selection sort
(C) Quick sort (D) Insertion sort
Answer: D
For a sorted list with a few random numbers:
Bubble sort will take O(n2) time in the best case.
The selection sort will take O(n2) time in the best case.
Quicksort will take O(n2) time in this case because it is the worst case for quick sort.
Insertion sort will take O(n). So, option (D) is correct.
23. The directory can be viewed as .................... that translates filenames into their directory entries.
(A) Symbol table (B) Partition
(C) Swap space (D) Cache
Answer: A
The directory can be viewed as a symbol table that translates filenames into their directory entries. Symbol table is an important data structure created and maintained by compilers in order to store information about the occurrence of various entities such as variable names, function names, objects, classes, interfaces, etc.
24. Consider an array A[20, 10], assume 4 words per memory cell and the base address of array A is 100. What is the address of A[11, 5]? Assume rowmajor storage.
(A) 560 (B) 565
(C) 570 (D) 575
Answer: A
Element A[11][0] is stored at “Base Address + 11 * 10 * 4” which is “Base Address + 440” = 540. So A[11][5] is stored at 540 + 5*4 = 560.
25. A full binary tree with n leaves contains
(A) n nodes (B) log2 n nodes
(C) 2n –1 nodes (D) 2n nodes
Answer: C
A Binary Tree is full if every node has 0 or 2 children. Following are examples of full binary tree. We can also say a full binary tree is a binary tree in which all nodes except leaves have two children. A full binary tree with n leaves contains 2 * n – 1 nodes.