# 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 |

**Set Theory & Algebra**

**UGC NET CS 2016 July – II**

**Discuss it**

Question 2 |

The number of different spanning trees in complete graph, K

_{4}and bipartite graph, K_{2,2}have ______ and _______ respectively.14, 14 | |

16, 14 | |

16, 4 | |

14, 4 |

**Graph Theory**

**UGC NET CS 2016 July – II**

**Discuss it**

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 K_{4}will be 4^{(4 - 2)}. i.e. 4^{2}= 16. Spanning trees in a bipartite graph K_{m,n}is equal to m^{(n-1)}* n^{(m-1)}. So, spanning trees in K_{2,2}will be 2^{(2-1)}* 2^{(2-1)}. i.e. 2^{1}* 2^{1}.= 4. So, option (C) is correct.Question 3 |

Suppose that R

_{1}and R_{2}are reflexive relations on a set A. Which of the following statements is correct ?R _{1} ∩ R_{2} is reflexive and R_{1} ∪ R_{2} is irreflexive. | |

R _{1} ∩ R_{2} is irreflexive and R_{1} ∪ R_{2} is reflexive. | |

Both R _{1} ∩ R_{2} and R_{1} ∪ R_{2} are reflexive. | |

Both R _{1} ∩ R_{2} and R_{1} ∪ R_{2} are irreflexive. |

**Set Theory & Algebra**

**UGC NET CS 2016 July – II**

**Discuss it**

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 |

**Probability**

**UGC NET CS 2016 July – II**

**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 |

**UGC NET CS 2016 July – II**

**Discuss it**

Question 5 Explanation:

Total 5 clique will be there.
So, option (C) is correct.

Question 6 |

Which of the following logic expressions is incorrect?

1 ⊕ 0 = 1 | |

1 ⊕ 1 ⊕ 1 = 1 | |

1 ⊕ 1 ⊕ 0 = 1 | |

1 ⊕ 1 = 0 |

**Digital Logic & Number representation**

**UGC NET CS 2016 July – II**

**Logic functions and Minimization**

**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 |

**Digital Logic & Number representation**

**Number Representation**

**UGC NET CS 2016 July – II**

**Discuss it**

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 |

**Set Theory & Algebra**

**UGC NET CS 2016 July – II**

**Discuss it**

Question 8 Explanation:

So, option (C) is correct.

Question 9 |

The simplified form of a Boolean equation (AB` + AB`C + AC) (A`C` + B`) is :

AB` | |

AB`C | |

A`B | |

ABC |

**UGC NET CS 2016 July – II**

**Discuss it**

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 |

**Digital Logic & Number representation**

**UGC NET CS 2016 July – II**

**Sequential circuits**

**Discuss it**

There are 50 questions to complete.