U.G.C. NET Exam. Dec. 2020-June 2021 Paper II (COMPUTER SCIENCE & APPLICATIONS)

Total Questions: 100

21. Which of the following are logically equivalent?

A. ¬p→(q – r) and q →(pvr)
B. (p →q) →r and p → (q→r)
C. (p → q) → (r→ s) and (p → r) → (q →s)
Choose the correct answer from the options given below : 

Correct Answer: (a) A only
Solution:A only

22. Which of these statements about the floor and ceiling functions are correct?

Correct Answer: (c) Statement I is true but Statement II is false
Solution:

23. How many ways are there to assign 5 different jobs to 4 different employees if every employee is assigned at least 1 job?

Correct Answer: (c) 240
Solution:240

24. A company stores products in a warehouse. Storage bins in this warehouse are specified by their aisle, location in the aisle, and self. There are 50 aisles, 85 horizontal locations in each aisle, and 5 shelves throughout the warehouse. What is the least number of products the company can have so that at least two products must be stored in the same bin?

Correct Answer: (d) 21251
Solution:

21251
total product
= 50×85×5
= 21250
= 1 product in 1 bin
= 21250 + 1
= 21251

25. Let us assume a person climbing the stairs can take one stair or two stairs at a time. How many ways can this person climb a flight of eight stairs

Correct Answer: (d) 34

26. For which value of n is Wheel graph Wn regular?

Correct Answer: (e) *
Solution:According to UGC two option (b & c) are correct.

27. Which of the following Graphs is (are) planer?


Choose the correct answer from the option given below :

Correct Answer: (a) A and B only
Solution:

A planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their end points. In other words, it can be drawn in such a way that no edges cross each other.
For a simple, connected, planar graph with v vertices and e edges and f faces, the following simple conditions hold for v ≥ 3 :
• Theorem 1. e ≤ 3v – 6,
• Theorem 2. If there are no cycles of length 3, then e ≤ 2v – 4
• Theorem 3. f ≤ 2v – 4.

28. Match List-I with List-II

Correct Answer: (c) A ‐ III, B ‐I , C ‐ IV, D ‐ II
Solution:

Identify Law :- The law of identity states that each thing is identical with itself.
x + 0 = x
Absorption Law :- To affirm that either some proposition is true or else that proposition and some other proposition are both true is equivalent to affirming the first proposition.
x + xy = x
Idempotent Law :- It is the property of certain operations in mathematics and computer science that they can be applied multiple time without changing the result beyond the initial application.
x + x = x
1 + 1 = 1 and 0 × 0 = 0
Domination Law :- F∧X = F; if the first thing in an AND expression is false, then it doesn't matter whether the second thing is true or false. The AND expression is going to be false either way.
T∨X = T: If the first thing in an OR expression is true, then, it doesn't matter whether the second thing is true or false the OR expression is going to be true either way.

29. Let (X, *) be a semigroup. Furthermore, for every a and b in X, if a ≠ b, then a*b ≠ b*a. Based on the defined semigroup, choose the correct equalities from the options given below:

A. For every a in X, a*a = a
B. For every a, b in X, a*b *a= a
C. For every a, b, c in X, a*b *c= a*c

Correct Answer: (d) A, B and C
Solution:

Let a*a = b.(a*a)*a = b*a
Since (X, *) is a semi group, * is closed and associative. So,
(a*a)*a = a*(a*a) ⇒ a*b = b*a
Which is possible only if a = b . Thus we proved a*a = a
b. Let (a*b) * a = c
⇒ (a * b) *a *a = c * a
⇒ a*b*a = c*a
⇒ c * a = a
Similarly,
a * (a*b*a) = a*c
⇒ a * a * ( b * a) = a * c
⇒ a * (b * a) = a * c
⇒ a * c = a = c * a
So, c = a
c. Let (a * b)*c = d
⇒ (a * b)* c * c = d * c
⇒ a * b * c = d * c
⇒ d * c = d
Similarly
a*(a*b*c) = a * d
⇒ a * a *(b * c) = a* d
⇒ a * (b * c) = a * d
⇒ a * d = d
Thus d * c = a * d = d
Now,
c * d * c = c * a * d = c * d
⇒ c = c * a * d = c * d
and
d* c * a = a * d * a = d * a
⇒ d * c * a = a = d * a
So,
= a * c = (d * a) * (c * d)
= d * (a * c) * d= d
Thus a * b * c = a * c

30. Consider the following linear optimization problem:

Maximize Z = 6x+5y
Subject to 2x ‐ 3y <= 5
x+3y <= 11
4x + y <=15
and x>=0, y >= 0.
The optimal solution of the problem is:

Correct Answer: (c) 31.72
Solution:31.72