# UGC NET CS 2014 Dec – II

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?

533 | |

599 | |

467 | |

66 |

**Arithmetic Aptitude**

**UGC NET CS 2014 Dec - II**

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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?

5 | |

n - 3 | |

20 | |

11 |

**Graph Theory**

**UGC NET CS 2014 Dec - II**

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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 __________.

Complete Graph | |

Bipartite Graph | |

Hamiltonian Graph | |

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?

0.5 | |

0.704 | |

0.632121 | |

0.68 |

**Probability**

**UGC NET CS 2014 Dec - II**

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Question 4 Explanation:

Favorable outcome k = 1
Total outcome = 1000000 = 10

^{6}Probability of k = 1; Probability = favorable outcome / total outcome = 1 / 10^{6}= 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 10^{6}printouts: = 0.999*0.999*.......0.9999 (10^{6}time) = 0.999^{106}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) = x

^{4}, g(x) = √(x^{2}+ 1), h(x) = x^{2}+ 72, then the value of the composite functions ho(gof) and (hog)of are given asx ^{8} – 71 and x^{8} – 71 | |

x ^{8}– 73 and x^{8} – 73 | |

x ^{8} + 71 and x^{8} + 71 | |

x ^{8} + 73 and x^{8} + 73 |

**Numerical Methods and Calculus**

**UGC NET CS 2014 Dec - II**

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Question 5 Explanation:

f(x) = x

^{4}, g(x) = √(x^{2}+ 1) and h(x) = x^{2}+ 72 Now ho(gof) = ho√(x^{8}+ 1) = x^{8}+ 72 + 1 = x^{8}+ 73. We have to find (hog)of: (hog)of = (x^{2}+ 1) + 72 = x^{2}+ 73 = x^{8}+ 73. So, option (D) is correct.Question 6 |

The BCD adder to add two decimal digits needs minimum of

6 full adders and 2 half adders | |

5 full adders and 3 half adders | |

4 full adders and 3 half adders | |

5 full adders and 2 half adders |

**Digital Logic & Number representation**

**UGC NET CS 2014 Dec - II**

**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.

(1) | |

(2) and (3) | |

(1) and (3) | |

All are correct. |

**Digital Logic & Number representation**

**UGC NET CS 2014 Dec - II**

**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)?

4 PUSH and 3 POP instructions | |

5 PUSH and 4 POP instructions | |

6 PUSH and 2 POP instructions | |

5 PUSH and 3 POP instructions |

**Computer Organization and Architecture**

**UGC NET CS 2014 Dec - II**

**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 + opeeration 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

2 ^{-128} to (1 – 2^{-23} ) × 2^{127} | |

(1 – 2 ^{–23} ) × 2 ^{–127} to 2^{128} | |

(1 – 2 ^{–23} ) × 2 ^{–127} to 2^{ 23} | |

2 ^{–129} to (1 – 2 ^{–23} ) × 2 ^{127} |

**Digital Logic & Number representation**

**Number Representation**

**UGC NET CS 2014 Dec - II**

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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

2 ^{16} × 8 | |

2 ^{18 }× 10 | |

2 ^{16} × 10 | |

2 ^{18} × 8 |

**OS Memory Management**

**UGC NET CS 2014 Dec - II**

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There are 50 questions to complete.