UGC NET Computer Science Solved Paper II November 2017 - Part 1

1.       If the time is now 4 O'clock, what will be the time after 101 hours from now?

(1) 9 O'clock

(2) 8 O'clock

(3) 5 O'clock

(4) 4 O'clock

Answer: 1
Explanation:
After 24 hours , time will again be 4 O'clock. 
101%24 = 5 
Hence time after 101 hours will be 4+5=9 o'clock

2.       Let m = (313)₄ and n = (322)₄. Find the base 4 expansion of m + n.

(1) (635)₄

(2) (32312)₄

(3) (21323)₄

(4) (1301)₄

Answer: 4
Explanation:
m = (313)₄ and n = (322)₄
Convert m and n to decimal.
m = 3*4^2 + 1*4^1 + 3*4^0 = 48+4+3 = (55)₁₀
m = 3*4^2 + 2*4^1 + 2*4^0 = 48+8+2 = (58)₁₀
m+n = 55+58 = 113
Now we have to convert 113 to base 4.
113%4 = 1 ---(1)
113/4 = 28
28%4 = 0 ---(2)
28/4 = 7
7%4 = 3 ----(3)
7/4 = 1
1%4 = 1 ----(4)
The answer will be step(4) to step (1) = 1301

3.       Let

Find the boolean product AʘB of the two matrices.

Answer: 1
Explanation:
We can only multiply two matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix.
So the boolean product AB =

So option (1) is the correct answer.

4.       How many distinguishable permutations of the letters in the word BANANA are there?

(1) 720

(2) 120

(3) 60

(4) 360

Answer: 3
Explanation:
Number of permutation of n objects with n1 identical objects of type 1, n2 identical objects of type 2, ..........and nk identical objects of type k is n! / n1!n2!....nk!

Here, first we have to count the total number of letters in it. Here it is 6.
Find out how many letters are repeating in the word. Here A is repeated 3 times. N is repeated 2 times.


So, Permutation of letter BANANA are 6! / 3!2! = 60

5.       Consider the graph given below :

Use Kruskal's algorithm to find a minimal spanning tree for the graph. The List of the edges of the tree in the order in which they are chosen is?

(1) AD, AE, AG, GC, GB, BF

(2) GC, GB, BF, GA, AD, AE

(3) GC, AD, GB, GA, BF, AE

(4) AD, AG, GC, AE, GB, BF

Answer: Marks to all
Explanation:

Kruskal’s Algorithm builds the spanning tree by adding edges one by one into a growing spanning tree. Kruskal's algorithm follows greedy approach as in each iteration it finds an edge which has least weight and add it to the growing spanning tree. 

Algorithm Steps:
* Sort the graph edges with respect to their weights.
* Start adding edges to the MST from the edge with the smallest weight until the edge of the largest weight.

* Only add edges which doesn't form a cycle , edges which connect only disconnected components.
Here all options (1), (2), (3) and (4) are matching with minimum spanning tree.

6.       The Boolean function with the Karnaugh map

is :

(1) (A + C).D + B

(2) (A + B).C + D

(3) (A + D).C + B

(4) (A + C).B + D

Answer: 1
Explanation:

From the above Karnaugh map, we will get B + CD + AD.
It can be reduced as (A+C).D+B

7.       The Octal equivalent of the binary number 1011101011 is:

(1) 7353

(2) 1353

(3) 5651

(4) 5657

Answer: 2
Explanation:
Group all the bits in the binary number in sets of three, starting from the far right. Add zeros to the left of the last digit if you don't have enough digits to make a set of three.
Original Binary: 1011101011
Grouping: 1  011  101  011
Adding Zeros for Groups of Three: 001  011  101  011
001=1, 011=3, 101=5, 011=3
So final answer will be 1353

8.       Let P and Q be two propositions, ⌐(P ↔ Q) is equivalent to:

(1) P ↔ ⌐ Q

(2) ⌐P↔ Q

(3) ⌐P↔ ⌐Q

(4) Q → P

Answer: 1, 2
Explanation:

P	Q	¬P	¬Q	Q → P	⌐(P ↔ Q)	P ↔ ⌐ Q	⌐P↔ Q	⌐P↔ ⌐Q
0	0	1	1	1	0	0	0	1
0	1	1	0	0	1	1	1	0
1	0	0	1	1	1	1	1	0
1	1	0	0	1	0	0	0	1

So both options 1 and 2 are correct.

Important Logical Equivalences.

The logical equivalences below are important equivalences that should be memorized.

Identity Laws:

p ∧ T ⇔ p

p ∨ F ⇔ p

Domination Laws:

p ∨ T ⇔ T

p ∧ F ⇔ F

Idempotent Laws:

p ∨ p ⇔ p

p ∧ p ⇔ p

Double Negation Law:

¬(¬p) ⇔ p

Commutative Laws:

p ∨ q ⇔ q ∨ p

p ∧ q ⇔ q ∧ p

Associative Laws:

(p ∨ q) ∨ r ⇔ p ∨ (q ∨ r)

(p ∧ q) ∧ r ⇔ p ∧ (q ∧ r)

Distributive Laws:

p ∨ (q ∧ r) ⇔ (p ∨ q) ∧ (p ∨ r)

p ∧ (q ∨ r) ⇔ (p ∧ q) ∨ (p ∧ r)

De Morgan’s Laws:

¬(p ∧ q) ⇔ ¬p ∨ ¬q

¬(p ∨ q) ⇔ ¬p ∧ ¬q

Absorption Laws:

p ∧ (p ∨ q) ⇔ p

p ∨ (p ∧ q) ⇔ p

Implication Law:

(p → q) ⇔ (¬p ∨ q)

Contrapositive Law:

(p → q) ⇔ (¬q → ¬p)

Tautology:

p ∨ ¬p ⇔ T

Contradiction:

p ∧ ¬p ⇔ F

Equivalence:

(p → q) ∧ (q → p) ⇔ (p ↔ q)

9.       Negation of the proposition Ǝ x H(x) is:

(1) Ǝ x ⌐H(x)

(2) ∀ x ⌐H(x)

(3) ∀ x H(x)

(4) ⌐ x H(x)

Answer: 2

10.    The output of the following combinational circuit is F.

The value of F is :

(1) P₁+P₂’P₃

(2) P₁+P₂’P₃’

(3) P₁+P₂P₃’

(4) P₁’+P₂P₃

Answer: 2

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