# UGC NET CS 2015 Jun – II

 Question 1
How many strings of 5 digits have the property that the sum of their digits is 7?
 A 66 B 330 C 495 D 99
Arithmetic Aptitude    UGC NET CS 2015 Jun - II
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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!      = 5 ```
total = 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 = 7 `
Total 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?
 A 22 / 36 B 12 / 36 C 14 / 36 D 6 / 36
General Aptitude    UGC NET CS 2015 Jun - II
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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 ?
 A 1001 B 3876 C 775 D 200
General Aptitude    UGC NET CS 2015 Jun - II
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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. x1 + x2 + x3 + x4 + x5 = 15 n - 1Cr - 1 = 15 - 1C5 - 1 = 14C4 = 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 (a) and (b) B (c) and (d) C (a), (b) and (c) D (a), (b), (c) and (d)
Tree Traversals    UGC NET CS 2015 Jun - II
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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
So, option (D) is 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 (a) and (b) B (b) and (c) C (a) and (c) D (a), (b) and (c)
Graph Theory    UGC NET CS 2015 Jun - II
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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 (a) only B (b) only C (a) and (b) D (a), (b) and (c)
UGC NET CS 2015 Jun - II
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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
So, option (D) is correct.
 Question 7
Consider a full - adder with the following input values: (a)x = 1, y = 0 and Ci(carry input) = 0 (b)x = 0, y = 1 and Ci = 1 Compute the values of S(sum) and Co (carry output) for the above input values.
 A S = 1, Co = 0 and S = 0, Co= 1 B S = 0, Co =0 and S =1, Co = 1 C S = 1, Co = 1 and S = 0, Co = 0 D So = 0, Co = 1 and S = 1, Co = 0
UGC NET CS 2015 Jun - II
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Question 7 Explanation:
In full adder: For input x, y and Ci So = x ⊕ y ⊕ Ci Co = xy + Ci(x + y) For x = 1, y = 0 and Ci(carry input) = 0 i.e. So = 1 ⊕ 0 ⊕ 0 = 1. Co = 1*0 + 0(1 + 0) = 0. For x = 0, y = 1 and Ci = 1 i.e. So = 0 ⊕ 1 ⊕ 1 = 0 Co = 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)
 A if (c Λ b)→r and ¬b→p, then (¬r Λ p)→¬c B if (c ∨ b)→r and ¬b→¬p, then (r Λ p)→c C if (c Λ b)→r and ¬p→b, then (¬r ∨ p)→¬c D if (c ∨ b)→r and ¬b→¬p, then (¬r Λ p)→¬c
Propositional and First Order Logic.    UGC NET CS 2015 Jun - II
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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: A (1) B (2) C (3) D (4)
Propositional and First Order Logic.    UGC NET CS 2015 Jun - II
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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.
So, option (A) is correct.
 Question 10
Consider a proposition given as : “ x ≥ 6, if x2 ≥ 5 and its proof as: If x ≥ 6, then x2 = 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 (a) only B (c) only C (a) and (b) D (b) only
Propositional and First Order Logic.    UGC NET CS 2015 Jun - II
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Question 10 Explanation:
“ x ≥ 6, if x2 ≥ 5 and its proof as: If x ≥ 6, then x2 = 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.
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