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 |
Discuss it
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 |
Discuss it
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 __________.
This graph is a __________.Complete Graph | |
Bipartite Graph | |
Hamiltonian Graph | |
All of the above |
Discuss it
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 |
Discuss it
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
x8 – 71 and x8 – 71 | |
x8– 73 and x8 – 73 | |
x8 + 71 and x8 + 71 | |
x8 + 73 and x8 + 73 |
Discuss it
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 |
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 |
Discuss it
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. |
Discuss it
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 |
Discuss it
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
2-128 to (1 – 2-23 ) × 2127 | |
(1 – 2 –23 ) × 2 –127 to 2128 | |
(1 – 2 –23 ) × 2 –127 to 2 23 | |
2 –129 to (1 – 2 –23 ) × 2 127 |
Discuss it
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
216 × 8 | |
2 18 × 10 | |
216 × 10 | |
2 18 × 8 |
Discuss it
There are 50 questions to complete.
