UGC NET CS 2016 July – II
Question 1 |
How many different equivalence relations with exactly three different equivalence classes are there on a set with five elements?
10 | |
15 | |
25 | |
30 |
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Question 2 |
The number of different spanning trees in complete graph, K4 and bipartite graph, K2,2 have ______ and _______ respectively.
14, 14 | |
16, 14 | |
16, 4 | |
14, 4 |
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Question 2 Explanation:
Spanning trees in complete graph is equal to n(n-2)(where n is no of sides or regularity in complete graph).
So, spanning trees in complete graph K4 will be 4(4 - 2).
i.e. 42 = 16.
Spanning trees in a bipartite graph Km,n is equal to m(n-1) * n(m-1).
So, spanning trees in K2,2 will be 2(2-1) * 2(2-1).
i.e. 21 * 21.= 4.
So, option (C) is correct.
Question 3 |
Suppose that R1 and R2 are reflexive relations on a set A. Which of the following statements is correct ?
R1 ∩ R2 is reflexive and R1 ∪ R2 is irreflexive. | |
R1 ∩ R2 is irreflexive and R1 ∪ R2 is reflexive. | |
Both R1 ∩ R2 and R1 ∪ R2 are reflexive. | |
Both R1 ∩ R2 and R1 ∪ R2 are irreflexive. |
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Question 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 colour as the one side we observed?
3/4 | |
2/3 | |
1/2 | |
1/3 |
Discuss it
Question 4 Explanation:
There are BB RR and BR, total outcome is 3, probability that the opposite side is the same color as the one side we observed:
It is clear that BB and RR will show the same color on opposite side, so favorable outcome will be 2.
Probability will be favorable outcome / total outcome
i.e. 2 / 3.
So, option (B) is correct.
Question 5 |
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?


2 | |
4 | |
5 | |
6 |
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Question 5 Explanation:

Question 6 |
Which of the following logic expressions is incorrect?
1 ⊕ 0 = 1 | |
1 ⊕ 1 ⊕ 1 = 1 | |
1 ⊕ 1 ⊕ 0 = 1 | |
1 ⊕ 1 = 0 |
Discuss it
Question 6 Explanation:
We know that x ⊕ y = x`y + xy`
So,
- 1 ⊕ 0 = 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 = 1
- 1 ⊕ 1 ⊕ 1 = (1 ⊕ 1) ⊕ 1 = (1`1 + 11`) ⊕ 1 = (0) ⊕ 1 = 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 = 1
- 1 ⊕ 1 ⊕ 0 = 1 ⊕ (1 ⊕ 0 ) = 1 ⊕ ( 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 ) = 1 ⊕ 1 = 1`1 + 11` = 0
- 1 ⊕ 1 = 1`1 + 11` = 0 So, option (C) is correct.
Question 7 |
The IEEE-754 double-precision format to represent floating point numbers, has a length of _____ bits.
16 | |
32 | |
48 | |
64 |
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Question 7 Explanation:
The IEEE-754 double-precision format to represent floating point numbers, has a length of 64 bits.
1 bit for sign, 11 bit for exponent and rest 52 is for fraction.
So, option (D) is correct.
Question 8 |
Simplified Boolean equation for the following truth table is:


F = yz` + y`z | |
F = xy` + x`y | |
F = x`z + xz` | |
F = x`z + xz` + xyz |
Discuss it
Question 8 Explanation:

Question 9 |
The simplified form of a Boolean equation (AB` + AB`C + AC) (A`C` + B`) is :
AB` | |
AB`C | |
A`B | |
ABC |
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Question 9 Explanation:
(AB` + AB`C + AC) (A`C` + B`)
= AB` +AB`C +AB`C
= AB` +AB`C
= AB`(1 + C)
= AB`
So, option (A) is correct.
Question 10 |
In a positive-edge-triggered JK flip-flop, if J and K both are high then the output will be _____ on the rising edge of the clock.
No change | |
Set | |
Reset | |
Toggle |
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There are 50 questions to complete.