# UGC NET CS 2015 Jun – II

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Question 1 |

How many strings of 5 digits have the property that the sum of their digits is 7?

66 | |

330 | |

495 | |

99 |

**Arithmetic Aptitude**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 1 Explanation:

We have to make string which have sum = 7.There are:

2,2,1,1,1 = 5! / 2! * 3! = 10 2,2,2,1,0 = 5! / 3! = 20 3,1,1,1,1 = 5! / 4! = 5 3,2,1,1,0 = 5! / 2! = 60 3,2,2,0,0 = 5! / 2! * 2! = 30 3,3,1,0,0 = 5! / 2! * 2! = 30 4,2,1,0,0 = 5! / 2! = 60 4,3,0,0,0 = 5! / 3! = 20 4,1,1,1,0 = 5! / 3! = 20 5,1,1,0,0 = 5! / 2! * 2! = 30 5,2,0,0,0 = 5! / 3! = 20 6,1,0,0,0 = 5! / 3! = 20 7,0,0,0,0 = 5! / 4! = 5total = 10 + 20 + 5 + 60 + 30 + 30 + 60 + 20 + 20 + 30 + 20 + 20 + 5 = 330. So, option (B) is correct.

**Alternative method -**Lets digits are a, b, c, d, and e. Therefore,a + b + c + d + e = 7Total number of combinations are (n-1+r)C(r) = (5-1+7)C7 = 11C7 = 11C4 = 330.

Question 2 |

Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die?

22 / 36 | |

12 / 36 | |

14 / 36 | |

6 / 36 |

**General Aptitude**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 2 Explanation:

We have two dice one is black other one is red:
There will be total 36 outcome.
We have to find the probability of outcome of first dice divides outcome of second one.
1 divides 1, 2, 3, 4, 5, 6.
2 divides 2, 4, 6.
3 divides 3, 6.
4 divides 4.
5 divides 5.
6 divides 6.
There are 14 number which will be counted in favourable outcome;
probability = favorable outcome / total outcome.
i.e. = 14 / 36.
So, option (C) is correct.

Question 3 |

In how many ways can 15 indistinguishable fish be placed into 5 different ponds, so that each pond contains atleast one fish ?

1001 | |

3876 | |

775 | |

200 |

**General Aptitude**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 3 Explanation:

According to given question we have 15(n) fish and 5(r) pond:
We have to place fish in each pond so that there will be atleast one fish in each pond.
i.e. x

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}= 15^{n - 1}C_{r - 1}=^{15 - 1}C_{5 - 1}=^{14}C_{4}= 14! / (10!* 4!) = 14 * 13 * 12 * 11 * 10! / (10! * 4!) = 14 * 13 * 12 * 11 / 4! = 1001 So, option (A) is correct.Question 4 |

Consider the following statements:
(a) Depth - first search is used to traverse a rooted tree.
(b) Pre - order, Post-order and Inorder are used to list the vertices of an ordered rooted tree.
(c) Huffman’s algorithm is used to find an optimal binary tree with given weights.
(d) Topological sorting provides a labelling such that the parents have larger labels than their children.
Which of the above statements are true?

(a) and (b) | |

(c) and (d) | |

(a), (b) and (c) | |

(a), (b), (c) and (d) |

**Tree Traversals**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 4 Explanation:

- Depth - first search is used to traverse a rooted tree.
**Correct** - Pre - order, Post-order and Inorder are used to list the vertices of an ordered rooted tree.
**Correct** - Huffman’s algorithm is used to find an optimal binary tree with given weights.
**Correct**- Topological sorting provides a labelling such that the parents have larger labels than their children.
**Correct**

Question 5 |

Consider a Hamiltonian Graph (G) with no loops and parallel edges. Which of the following is true with respect to this Graph (G) ?
(a) deg (v) ≥ n / 2 for each vertex of G
(b) |E(G)| ≥ 1 / 2 (n - 1) (n - 2) + 2 edges
(c) deg (v) + deg (w) ≥ n for every n and v not connected by an edge.

(a) and (b) | |

(b) and (c) | |

(a) and (c) | |

(a), (b) and (c) |

**Graph Theory**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 5 Explanation:

In an Hamiltonian Graph (G) with no loops and parallel edges:
According to Dirac's theorem in a n vertex graph, deg (v) ≥ n / 2 for each vertex of G.
According to Ore's theorem deg (v) + deg (w) ≥ n for every n and v not connected by an edge is sufficient condition for a graph to be hamiltonian.
If |E(G)| ≥ 1 / 2 * [(n - 1) (n - 2)] then graph is connected but it doesn't guaranteed to be Hamiltonian Graph.
(a) and (c) is correct regarding to Hamiltonian Graph.
So, option (C) is correct.

Question 6 |

Consider the following statements :
(a)Boolean expressions and logic networks correspond to labelled acyclic digraphs.
(b)Optimal boolean expressions may not correspond to simplest networks.
(c)Choosing essential blocks first in a Karnaugh map and then greedily choosing the largest remaining blocks to cover may not give an optimal expression.
Which of these statement(s) is/are correct ?

(a) only | |

(b) only | |

(a) and (b) | |

(a), (b) and (c) |

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 6 Explanation:

- Boolean expressions and logic networks correspond to labelled acyclic digraphs.
**Correct** - Optimal boolean expressions may not correspond to simplest networks.
**Correct** - Choosing essential blocks first in a Karnaugh map and then greedily choosing the largest remaining blocks to cover may not give an optimal expression.
**Correct**

Question 7 |

Consider a full - adder with the following input values:
(a)x = 1, y = 0 and C

_{i}(carry input) = 0 (b)x = 0, y = 1 and C_{i}= 1 Compute the values of S(sum) and C_{o}(carry output) for the above input values.S = 1, C _{o} = 0 and S = 0, C_{o}= 1 | |

S = 0, C _{o} =0 and S =1, C_{o} = 1 | |

S = 1, C _{o} = 1 and S = 0, C_{o} = 0 | |

S _{o} = 0, C_{o} = 1 and S = 1, C_{o} = 0 |

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 7 Explanation:

In full adder:
For input x, y and C

_{i}S_{o}= x ⊕ y ⊕ C_{i}C_{o}= xy + C_{i}(x + y) For x = 1, y = 0 and C_{i}(carry input) = 0 i.e. S_{o}= 1 ⊕ 0 ⊕ 0 = 1. C_{o}= 1*0 + 0(1 + 0) = 0. For x = 0, y = 1 and C_{i}= 1 i.e. S_{o}= 0 ⊕ 1 ⊕ 1 = 0 C_{o}= 0*1 + 1(0 + 1) = 1. So, option (A) is correct.Question 8 |

“If my computations are correct and I pay the electric bill, then I will run out of money. If I don’t pay the electric bill, the power will be turned off. Therefore, if I don’t run out of money and the power is still on, then my computations are incorrect.”
Convert this argument into logical notations using the variables c, b, r, p for propositions of computations, electric bills, out of money and the power respectively. (Where ¬ means NOT)

if (c Λ b)→r and ¬b→p, then (¬r Λ p)→¬c | |

if (c ∨ b)→r and ¬b→¬p, then (r Λ p)→c | |

if (c Λ b)→r and ¬p→b, then (¬r ∨ p)→¬c | |

if (c ∨ b)→r and ¬b→¬p, then (¬r Λ p)→¬c |

**Propositional and First Order Logic.**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 8 Explanation:

“If my computations are correct and I pay the electric bill, then I will run out of money." : if (c Λ b)→r
"If I don’t pay the electric bill, the power will be turned off. Therefore, if I don’t run out of money and the power is still on, then my computations are incorrect.” : (¬r Λ p)→¬c .
So, option (A) is correct.

Question 9 |

Match the following:

(1) | |

(2) | |

(3) | |

(4) |

**Propositional and First Order Logic.**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 9 Explanation:

- (p → q) ⇔ (¬q → ¬p) is
**Contrapositive** - [(p ∧ q) → r] ⇔ [p → (q → r)] is exportation law.
- (p → q) ⇔ [(p ∧ ¬q) → o] is
**Reductio ad absurdum** - (p ↔ q)⇔[(p → q)∧(q → p)] is
**Equivalence.**

Question 10 |

Consider a proposition given as : “ x ≥ 6, if x

^{2}≥ 5 and its proof as: If x ≥ 6, then x^{2}= x.x ≥ 6.6 = 36 ≥ 25 Which of the following is correct w.r.to the given proposition and its proof? (a)The proof shows the converse of what is to be proved. (b)The proof starts by assuming what is to be shown. (c)The proof is correct and there is nothing wrong.(a) only | |

(c) only | |

(a) and (b) | |

(b) only |

**Propositional and First Order Logic.**

**UGC NET CS 2015 Jun - II**

**Discuss it**

Question 10 Explanation:

“ x ≥ 6, if x

^{2}≥ 5 and its proof as: If x ≥ 6, then x^{2}= x.x ≥ 6.6 = 36 ≥ 25 it is clear from the above statement that proof starts by assuming what is to be shown and it is clear that the proof shows the converse of what is to be proved hence the proof is not correct. Statement (a) and (b) are correct but (3) statement is not correct. So, option (C) is correct.
There are 49 questions to complete.