Coding Made Easy: cbse-net july 2016 solved paper

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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

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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

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.

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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

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

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

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