# UGC NET CS 2014 Dec – II

• Last Updated : 22 Oct, 2021

 Question 1
Consider a set A = {1, 2, 3, ........, 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5?
 A 533 B 599 C 467 D 66
UGC NET CS 2014 Dec - II    Arithmetic Aptitude
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Question 1 Explanation:
From set A numbers {3,6,9,........999} which are divisible by 3 are 999 / 3 (A)= 333 . From set A numbers {5,10,......995,1000} which are divisible by 5 are 1000 / 5 (B)= 200. From set A numbers {15, 30, ...990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66. So, numbers divisible by 3 or by 5 or by both 3 and 5: (A ∨ B) = A + B - (A ∧ B) (A ∨ B) = 333 + 200 - 67. (A ∨ B) = 467. So, option (C) is correct.
 Question 2
A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph?
 A 5 B n - 3 C 20 D 11
UGC NET CS 2014 Dec - II    Top MCQs on Graph Theory in Mathematics
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Question 2 Explanation:
There are 2 vertices of degree 4, 1 vcertex of degree 3, 1 vcertex of degree 2 and vertex of degree one is unknown. Let's assume k be the nof vertex of degree one. Total vertex = 2 + 1 + 1 + k = k + 4. Number of edges = vertex - 1 i.e. k + 4 - 1 = k + 3. Now apply handshaking lemma(For more information on handshaking lemma Refer:Handshaking Lemma and Interesting Tree Properties) 2 * 4 + 1 * 3 + 1 * 2 + 1 * K = 2 * (No of edges) i.e. 13 + k = 2 * (k + 3) k = 7. Total vertex = 7 + 4 = 11. So, option (D) is correct.
 Question 3
Consider the Graph shown below: This graph is a __________.
 A Complete Graph B Bipartite Graph C Hamiltonian Graph D All of the above
UGC NET CS 2014 Dec - II
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Question 3 Explanation:
There is no edge between D and B , A and C and one more thing for complete graph edge must be n * (n - 1) / 2 which is not valid in given graph. If a graph is two colorable then it is bipartite but this graph is not bipartite. According to Dirac's Theorem If each vertex have degree greater then n / 2 then it is Hamiltonian For more information on Hamiltonian Refer:Mathematics | Euler and Hamiltonian Paths Option (C) is correct.
 Question 4
A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout atleast once?
 A 0.5 B 0.704 C 0.632121 D 0.68
UGC NET CS 2014 Dec - II    Probability
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Question 4 Explanation:
Favorable outcome k = 1 Total outcome = 1000000 = 106 Probability of k = 1; Probability = favorable outcome / total outcome = 1 / 106 = 10-6 Probability of k != 1: = 1 - Probability of k = 1; 1 - 10-6 = 0.999 Probability that k = 1 is never printed in all 106 printouts: = 0.999*0.999*.......0.9999 (106 time) = 0.999106 Probability that 1 is printed at least once: = 1- probability that 1 is never printed = 1-0.99910^6 =0.6321 So, option (C) is correct.
 Question 5
If we define the functions f, g and h that map R into R by : f(x) = x4 , g(x) = √(x2 + 1), h(x) = x2 + 72, then the value of the composite functions ho(gof) and (hog)of are given as
 A x8 – 71 and x8 – 71 B x8– 73 and x8 – 73 C x8 + 71 and x8 + 71 D x8 + 73 and x8 + 73
UGC NET CS 2014 Dec - II    Numerical Methods and Calculus
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Question 5 Explanation:
f(x) = x4, g(x) = √(x2 + 1) and h(x) = x2 + 72 Now ho(gof) = ho√(x8 + 1) = x8 + 72 + 1 = x8 + 73. We have to find (hog)of: (hog)of = (x2 + 1) + 72 = x2 + 73 = x8 + 73. So, option (D) is correct.
 Question 6
UGC NET CS 2014 Dec - II    Digital Logic & Number representation    Combinational Circuits
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 Question 7
The Excess-3 decimal code is a self-complementing code because (1)The binary sum of a code and its 9’s complement is equal to 9. (2)It is a weighted code. (3)Complement can be generated by inverting each bit pattern. (4)The binary sum of a code and its 10’s complement is equal to 9.
 A (1) B (2) and (3) C (1) and (3) D All are correct.
UGC NET CS 2014 Dec - II    Digital Logic & Number representation    Combinational Circuits
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Question 7 Explanation:
The Excess-3 decimal code is a self-complementing code because the binary sum of a code and its 9’s complement is equal to 9 and complement can be generated by inverting each bit pattern. So, option (C) is correct.
 Question 8
How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A ∗ B) + (C ∗ D / E)?
 A 4 PUSH and 3 POP instructions B 5 PUSH and 4 POP instructions C 6 PUSH and 2 POP instructions D 5 PUSH and 3 POP instructions
UGC NET CS 2014 Dec - II    Computer Organization and Architecture    CPU control design and Interfaces
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Question 8 Explanation:
Reverse polish notation is a system of formula notation without brackets or special punctuation. To evaluate (A ∗ B) + (C ∗ D / E): First avoid brackets and punctuation and convert it into postfix form i.e. AB+CDE/*+ Now push AB On * pop AB and perform A * B. Now push back the result(say it X). Push CDE. On / pop DE and push back the result(say it Y). On * Pop CY and perform * operation and push the result(say it z). On + pop XZ and perform + operation and and push back the final answer. Above computation include 5 PUSH and 4 POP instructions. So, option (B) is correct.
 Question 9
The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is
 A 2-128 to (1 – 2-23 ) × 2127 B (1 – 2 –23 ) × 2 –127 to 2128 C (1 – 2 –23 ) × 2 –127 to 2 23 D 2 –129 to (1 – 2 –23 ) × 2 127
UGC NET CS 2014 Dec - II    Digital Logic & Number representation    Number Representation
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 Question 10
The size of the ROM required to build an 8-bit adder / subtractor with mode control, carry input, carry output and two’s complement overflow output is given as
 A 216 × 8 B 2 18 × 10 C 216 × 10 D 2 18 × 8
UGC NET CS 2014 Dec - II    OS Memory Management
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