NET JUNE 2015 PAPER-2 SOLVED PART-1


Pages: 1 2 3 4 5


  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

Answer: B

EXPLANATION :

Let n=7 and r=5.

n+r-1Cr-1=330.


  1. 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

Answer: C


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

(A) 1001 (B) 3876

(C) 775 (D) 200

Answer: A

Explanation.

15C1x14C4/15


  1. 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)

Answer: D


  1. 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 v and w not connected by an edge

(A) (a) and (b) (B) (b) and (c)

(C) (a) and (c) (D) (a), (b) and (c)

Answer: D


  1. 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)

Answer: D


  1. 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 C0 (carry output) for the above input values.

(A) S=1 , C0= 0 and S=0 , C0= 1 (B) S=0 , C0= 0 and S=1 , C0= 1

(C) S=1 , C0= 1 and S=0 , C0= 0 (D) S=0 , C0= 1 and S=1 , C0= 0

Answer: A

Explanation.
Assume it, that it is written like (Carry,Sum.) in BINARY that is weighted.
X Y Carry(in) C Sum
0+ 0+0= 0 0 0.
0+ 0+1= 1 0 1.
0+ 1+0= 1 0 1.
0+ 1+1=(2) 1 0.
1+ 0+0=1= 0 1
1+ 0+1=(2) 1 0.
1+ 1+0=(2) 1 0.
1+ 1+1=(3) 1 1.


  1. “lf 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∧b) → r and ¬b → ¬p, then (¬r∧p)→¬c

(B) if (c∨b) → r and ¬b → ¬p, then (r∧p)→c

(C) if (c∧b) → r and ¬p → ¬b, then (¬r∨p)→¬c

(D) if (c∨b) → r and ¬b → ¬p, then (¬r∧p)→¬c

Answer: A


  1. Match the following:

List – I List – II

(a) (p →q) ⇔ (¬q→¬p) (i) Contrapositive

(b) [(p∧q)→r]⇔[p→ (q→r)] (ii) Exportation law

(c) (p→q)⇔[(p∧¬q)→o] (iii) Reductio ad absurdum

(d) (p⇔q)⇔[(p→q)∧(q→p)] (iv) Equivalence

Codes:

(a) (b) (c) (d)

(A) (i) (ii) (iii) (iv)

(B) (ii) (iii) (i) (iv)

(C) (iii) (ii) (iv) (i)

(D) (iv) (ii) (iii) (i)

Answer: A


  1. Consider a proposition given as:

“x≥6, if x2≥25 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

Answer: C

Explanation.

Proof says any number greater than or equal to 6, its square is greater than or equal to 25.

6×6>25. 7×7>25——-.

Proposition 1 Fails because it does not prove that number (6,7,8—) square is not greater than 25.

Proposition 2 fails because there is no requirement of assumption as numbers are already given.

Proposition 3 is correct.


Pages: 1 2 3 4 5


Leave a comment