Coding Made Easy

Friday, December 9, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q:: 32 The content of the accumulator after the execution of the following 8085 assembly

language program, is :

MVI A, 42H

MVI B, 05H

UGC: ADD B

DCR B

JNZ UGC

ADI 25H

HLT

(1) 82 H (2) 78 H

(3) 76 H (4) 47 H

Answer:: (3)

Explanation::

Content of accumulator register (A):- 01000010

Content of register (B) :- 00000101

A : 01000010

+B : 00000101

: 01000111

B = B - 1

B = 00000100

A : 01000111

+B : 00000100

: 01001011

B = B - 1

B = 00000011

A : 01001011

+B : 00000011

: 01001110

B = B - 1

B = 00000010

A : 01001110

+B : 00000010

: 01010000

B = B - 1

B = 00000100

A : 01010000

+B : 00000001

: 01010001

B = B - 1

B = 00000000

Now the loop exits as contents of B register becomes 00000000.

Add 25H to A register

A:- 01010001

+00100101

01110110

So, Contents of accumulator (A) becomes 01110110 i.e. 76H.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

ugc net computer science question papers,ugc net computer science december 2014 question papers,
ugc net computer science june 2014 question papers,
ugc net computer science december 2014 question papers,
ugc net computer science june 2013 question papers,
ugc net computer science december 2013 question papers,
ugc net computer science june 2012 question papers,
ugc net computer science december 2012 question papers,
ugc net computer science june 2011 question papers,
ugc net computer science december 2011 question papers,
ugc net computer science june 2010 question papers,
ugc net computer science december 2010 question papers,
ugc net computer science june 2009 question papers,
ugc net computer science december 2009 question papers,
ugc net computer science june 2008 quest

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q::11 Given i = 0, j = 1, k = –1, x = 0.5, y = 0.0

What is the output of the following expression in C language ?

x * y < i + j || k

(1) – 1 (2) 0

(3) 1 (4) 2

Answer:: (3)

Explanation::

= x * y < i + j || k

= 0.5 * 0.0 < 0 + 1 || -1

= 0.0 < 0 + 1 || -1

= 0.0 < 1 || -1

= 1 || -1

= 1

Here is operator precedence table for you::

Category	Operator	Associativity
Postfix	() [] -> . ++ - -	Left to right
Unary	+ - ! ~ ++ - - (type)* & sizeof	Right to left
Multiplicative	* / %	Left to right
Additive	+ -	Left to right
Shift	<< >>	Left to right
Relational	< <= > >=	Left to right
Equality	== !=	Left to right
Bitwise AND	&	Left to right
Bitwise XOR	^	Left to right
Bitwise OR	\|	Left to right
Logical AND	&&	Left to right
Logical OR	\|\|	Left to right
Conditional	?:	Right to left
Assignment	= += -= *= /= %=>>= <<= &= ^= \|=	Right to left
Comma	,	Left to right

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q::9 The simplified form of a boolean equation (AB'+AB'C+AC)(A'C'+B') is :

(1) AB' (2) AB'C
(3) A'B (4) ABC

Answer:: (1)

Explanation::

=( AB' + AB'C + AC ) ( A'C' + B' )

=( AB'( 1 + C ) + AC ) ( A'C' + B' ) [ x + 1 = 1 ]

=( AB' + AC ) ( A'C' + B' )

=( AA'B'C' + AB'B' + AA'CC' + AB'C )

=( 0 + AB' + 0 + AB'C ) [ x . x' = 0 ]

= AB'( 1 + C )

= AB'

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

Sunday, December 4, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q:: 6 Which of the following logic expressions is incorrect ?
(1) 1 ⊕ 0 = 1
(2) 1 ⊕ 1 ⊕ 1 = 1
(3) 1 ⊕ 1 ⊕ 0 = 1
(4) 1 ⊕ 1 = 0

Answer:: (3)

Explanation::

The symbol ⊕ stands for X-OR gate. X-OR gives output as 1 when only one of the two inputs are supplied as 1 otherwise 0.

A B A⊕B
0 0 0
0 1 1
1 0 1
1 1 0

So, on inspecting option (3) produce 1 ⊕ 1 ⊕ 0 = 0 ⊕ 0 = 0.
Hence it is incorrect. All others are correct.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q:: A clique in a simple undirected graph is a complete subgraph that is not contained in any
larger complete subgraph. How many cliques are there in the graph shown below ?

(1) 2 (2) 4
(3) 5 (4) 6

Answer:: Marks to All

Explanation::

So, there are total 8 cliques in all and no option matches.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Q::4 There are three cards in a box. Both sides of one card are black, both sides of one card are
red, and the third card has one black side and one red side. We pick a card at random and
observe only one side.
What is the probability that the opposite side is the same color as the one side we
observed ?
(1) 3/4 (2) 2/3
(3) 1/2 (4) 1/3

Answer:: (2)

Explanation::

Total number of cases = 3 i.e. (BB,RR,BR) Preferred number of cases =2 (BB,RR) since 2 cards have the same colour both sides. Probability = 2/3.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

UGC-NET Computer Science July 2016 Solved Paper 2

Q::2 The number of different spanning trees in complete graph, K₄ and bipartite graph, K_2,2 have ______ and _______ respectively.

(1) 14, 14 (2) 16, 14

(3) 16, 4 (4) 14, 4

Answer:: (3)

Explanation::

Let 'n' be total number of vertices in complete graph K_n then total different spanning trees are n^n-2

n = 4 as given then total different spanning trees are 4^4-2= 4²= 16.

The number of labelled spanning trees in a bipartite graph K_m,n is given by m^n-1*n^m-1.
As given in question number of spanning trees in bipartite graph K_2,2 = 2^2-1 * 2^2-1 = 4.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

Complete Result of candidates qualified UGC NET July-August 2016

Click Here to Download pdf

Friday, September 2, 2016

CBSE NET July 2016 Official Answer key

http://cbsenet.nic.in/cbsenet/KeyChallange/Login.aspx

Sunday, August 14, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

Q::15 What is the value returned by the function f given below when n=100 ?
int f (int n)
{
if (n==0) then
return n;
else
return n + f(n-2);
}

(A)2550 (B)2556

(C)5220 (D)5520

Answer: (A)

Explanation:

f(100)-->(100 + f(98)-->(98 + f(96)-->96 + f(94)........... 2 + f(0) <-- 0

So, it will be an A.P.

2 + 4 +6 +8 + ... + 98 + 100 = n(n+1) {Sum of even numbers upto 'n'}

=50 * 51 = 2550.

Hence, Option A is correct.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

Monday, August 1, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

Q:2 Suppose that R1 and R2 are reflexive relations on a set A.

Which of the following statements is correct ?

(A) R1∩R2 is Reflexive and R1∪R2 is irreflexive

(B) R1∩R2 is irReflexive and R1∪R2 is reflexive

(D) Both R1∩R2 and R1 ∪R2 are irreflexive

Answer: C

Explanation:

Suppose if both R1 and R2 are Reflexive, Symmetric and Transitive i.e. R1 and R2 are both equivalence relations then:

R1∩R2 is Reflexive, Symmetric and Transitive.

R1∪R2 is Reflexive, Symmetric but not Transitive.

Discrete maths solved questions ugc net,Discrete maths solved questions ugc-net,Discrete maths solved questions cbse net,Discrete maths solved questions cbse-net

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

Friday, July 29, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

Q:1 How many different equivalence relations with exactly three different equivalence classes are there on a set with five elements?

(A)10
(B)15
(C)25
(D)30

Ans: C

Number of equivalence classes with 5 elements with three elements in each class

could be 2,2,1 and 3,1,1.

3,1,1 could be chosen in (5C3*2C1*1C1)/2! = 10

2,2,1 could be chosen in (5C2*3C2*1C1)/2! = 15

So, total 10 + 15 = 25.

UGC-NET Computer Science july 2016 solved paper,UGC-NET Computer Science july 2016 paper 2,cbse-net july 2016 solved paper

Monday, April 11, 2016

CBSE-UGC NET RESULT - December 2015

http://cbseresults.nic.in/UGC/net_dec2015.htm

NATIONAL ELIGIBILITY TEST (NET)

On behalf of UGC, the Central Board of Secondary Education announces holding of the National Eligibility Test (NET) on 10^thJuly, 2016 (SUNDAY) for determining the eligibility of Indian nationals for the Eligibility for Assistant Professor only or Junior Research Fellowship & Eligibility for Assistant Professor Both in Indian universities and colleges. CBSE will conduct NET in 83 subjects at 88 selected NET Examination Cities spread across the country.

The candidates who qualify for the award of Junior Research Fellowship are eligible to pursue research in the subject of their post-graduation or in a related subject and are also eligible for Assistant Professor. The universities, institutions, IITs and other national organizations may select the JRF awardees for whole time research work in accordance with the procedure prescribed by them. The award of JRF and Eligibility for Assistant Professor both OR Eligibility for Assistant Professor only will depend on the performance of the candidate in all three papers of NET. However, the candidates qualifying exclusively for Assistant Professor will not be considered for award of JRF.

Tuesday, April 5, 2016

UGC NET JULY 2016 Notification

E-mail: net@cbse.gov.in

Website: www.cbsenet.nic.in

Tele. 7042399520, 7042399521,

7042399524, 7042399525,

7042399526, 7042399528,

7042399529.

Fax. 0120-2427772

CENTRAL BOARD OF SECONDARY EDUCATION

(An Autonomous Organisation under the Union Ministry of Human Resource Development Govt. of India)

H-149, Sector-63, Noida, District Gautam Budh Nagar – 201 309(UP).

_________________________________________________________________________

CBSE/NET/2015-16 04/04/2016

UGC NATIONAL ELIGIBILITY TEST (NET)

SHORT NOTIFICATION

It is notified that CBSE will conduct the next UGC-NET for Junior Research

Fellowship & Eligibility for Assistant Professor on 10th July 2016 (Sunday). The candidates who

desire to appear in the test may see the detailed notification available on the website

www.cbsenet.nic.in from 12th April, 2016. The candidates are required to apply online from

12.04.2016. The last date for applying online is 12st May 2016 and fee can be paid up to 13

May, 2016.

th

EXECUTIVE DIRECTOR

Download Complete notification here

ugc net july 2016,cbse net july 2016,ugc net july 2016 notification,cbse net july 2016 notification,ugc net latest notification

Saturday, April 2, 2016

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q::7 Number of binary trees formed with 5
nodes are
(A) 32 (B) 36
(C) 120 (D) 42

Answer:(D)

Explanation:

In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively-defined objects. They are named after the Belgian mathematician Eugène Charles Catalan (1814–1894).

Using zero-based numbering, the nth Catalan number is given directly in terms of binomial coefficients by

C_n = \frac{1}{n+1}{2n\choose n} = \frac{(2n)!}{(n+1)!\,n!} = \prod\limits_{k=2}^{n}\frac{n+k}{k} \qquad\mbox{ for }n\ge 0.

The first Catalan numbers for n = 0, 1, 2, 3, … are

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, …

Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases C₃ = 5 and C₄ = 14.

Lattice of the 14 Dyck words of length 8 - ( and ) interpreted as up anddown

C_n is the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's (see also Dyck language). For example, the following are the Dyck words of length 6:

XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY.

Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, C_n counts the number of expressions containing n-pairs of parentheses which are correctly matched:

((())) ()(()) ()()() (())() (()())

C_n is the number of different ways n + 1 factors can be completely parenthesized (or the number of ways of associating n applications of a binary operator). For n = 3, for example, we have the following five different parenthesizations of four factors:

((ab)c)d (a(bc))d (ab)(cd) a((bc)d) a(b(cd))

The associahedron of order 4 with the C₄=14 full binary trees with 5 leaves

Successive applications of a binary operator can be represented in terms of a full binary tree. (A rooted binary tree is full if every vertex has either two children or no children.) It follows that C_n is the number of full binary trees with n + 1 leaves:

C_n is the number of non-isomorphic ordered trees with n vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed left-to-right order.)
C_n is the number of monotonic lattice paths along the edges of a grid with n × n square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".

The following diagrams show the case n = 4:

This can be succinctly represented by listing the Catalan elements by column height:

[0,0,0,0][0,0,0,1][0,0,0,2][0,0,1,1]

[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2][0,0,2,2][0,0,1,3]

[0,0,2,3][0,1,1,3] [0,1,2,2][0,1,2,3]

The triangles correspond to nodes of the binary trees.

C_n is the number of different ways a convex polygon with n + 2 sides can be cut into triangles by connecting vertices with straight lines(a form of Polygon triangulation). The following hexagons illustrate the case n = 4:

C_n is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences. These are the permutations that avoid the pattern 231.
C_n is the number of permutations of {1, ..., n} that avoid the pattern 123 (or any of the other patterns of length 3); that is, the number of permutations with no three-term increasing sub-sequence. For n = 3, these permutations are 132, 213, 231, 312 and 321. For n = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
C_n is the number of non-crossing partitions of the set {1, ..., n}. A fortiori, C_n never exceeds the nth Bell number. C_n is also the number of non-crossing partitions of the set {1, ..., 2n} in which every block is of size 2. The conjunction of these two facts may be used in a proof by mathematical induction that all of the free cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices.
C_n is the number of ways to tile a stairstep shape of height n with n rectangles. The following figure illustrates the case n = 4:

C_n is the number of rooted binary trees with n internal nodes (n + 1 leaves or external nodes). Illustrated in following Figure are the trees corresponding to n = 0,1,2 and 3. There are 1, 1, 2, and 5 respectively. Here, we consider as binary trees those in which each node has zero or two children, and the internal nodes are those that have children.

Coding Made Easy

Friday, December 9, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Sunday, December 4, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Complete Result of candidates qualified UGC NET July-August 2016

Friday, September 2, 2016

CBSE NET July 2016 Official Answer key

Sunday, August 14, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

Monday, August 1, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Friday, July 29, 2016

UGC-NET Computer Science July 2016 Solved Paper 2

UGC-NET Computer Science July 2016 Solved Paper 2

Monday, April 11, 2016

CBSE-UGC NET RESULT - December 2015

NATIONAL ELIGIBILITY TEST (NET)

Tuesday, April 5, 2016

UGC NET JULY 2016 Notification

Saturday, April 2, 2016

UGC-NET Computer Science Data Structures Questions with Explanation

Applications in combinatorics

Coding Acceleration Program

Search This Blog