Solved Paper II UGC NET Computer Science November 2017
1.
If the time is now 4 O'clock, what will be the time
after 101 hours from now?
(1) 9 O'clock
(2) 8 O'clock
(3) 5 O'clock
(4) 4 O'clock
Answer: 1
Explanation:
After 24 hours , time
will again be 4 O'clock.
101%24 = 5
Hence time after 101
hours will be 4+5=9 o'clock
2.
Let m = (313)4 and n = (322)4.
Find the base 4 expansion of m + n.
(1) (635)4
(2) (32312)4
(3) (21323)4
(4) (1301)4
Answer: 4
Explanation:
m = (313)4 and
n = (322)4
Convert m and n to
decimal.
m = 3*4^2 + 1*4^1 + 3*4^0 =
48+4+3 = (55)10
m = 3*4^2 + 2*4^1 + 2*4^0 =
48+8+2 = (58)10
m+n = 55+58 = 113
Now we have to convert 113 to base
4.
113%4 = 1 ---(1)
113/4 = 28
28%4 = 0 ---(2)
28/4 = 7
7%4 = 3 ----(3)
7/4 = 1
1%4 = 1 ----(4)
The answer will be step(4) to step
(1) = 1301
3.
Let
Find the boolean product AʘB of the two matrices.
Answer: 1
Explanation:
We can only multiply
two matrices if the number of columns in the first matrix is the same as the
number of rows in the second matrix.
So the boolean product
AB =
So option (1) is the
correct answer.
4.
How many distinguishable permutations of the letters
in the word BANANA are there?
(1) 720
(2) 120
(3) 60
(4) 360
Answer: 3
Explanation:
Number of permutation
of n objects with n1 identical objects of type 1, n2 identical objects of type
2, ..........and nk identical objects of type k is n! / n1!n2!....nk!
Here, first we have to count the total number of letters in it. Here it is 6.
Find out how many
letters are repeating in the word. Here A is repeated 3 times. N is repeated 2
times.
So, Permutation of
letter BANANA are 6! / 3!2! = 60
5.
Consider the graph given below :
Use Kruskal's algorithm to find a minimal spanning
tree for the graph. The List of the edges of the tree in the order in which
they are chosen is?
(1) AD, AE, AG, GC, GB, BF
(2) GC, GB, BF, GA, AD, AE
(3) GC, AD, GB, GA, BF, AE
(4) AD, AG, GC, AE, GB, BF
Answer: Marks to all
Explanation:
Kruskal’s Algorithm
builds the spanning tree by adding edges one by one into a growing spanning
tree. Kruskal's algorithm follows greedy approach as in each iteration it finds
an edge which has least weight and add it to the growing spanning tree.
Algorithm Steps:
* Sort the graph edges
with respect to their weights.
* Start adding edges
to the MST from the edge with the smallest weight until the edge of the largest
weight.
* Only add edges which
doesn't form a cycle , edges which connect only disconnected components.
Here all options (1),
(2), (3) and (4) are matching with minimum spanning tree.
6. The
Boolean function with the Karnaugh map
is :
(1) (A + C).D
+ B
(2) (A + B).C
+ D
(3) (A + D).C
+ B
(4) (A +
C).B + D
Answer: 1
Explanation:
From the above Karnaugh map, we will get B + CD + AD.
It can be reduced as (A+C).D+B
7. The
Octal equivalent of the binary number 1011101011 is:
(1) 7353
(2) 1353
(3) 5651
(4) 5657
Answer: 2
Explanation:
Group
all the bits in the binary number in sets of three, starting from the
far right. Add zeros to the left of the last digit if you don't have
enough digits to make a set of three.
Original Binary: 1011101011
Grouping: 1 011 101 011
Adding Zeros for Groups of Three: 001 011 101 011
001=1, 011=3, 101=5, 011=3
So final answer will be 1353
8. Let
P and Q be two propositions, ⌐(P ↔ Q) is equivalent to:
(1) P ↔ ⌐ Q
(2) ⌐P↔ Q
(3) ⌐P↔ ⌐Q
(4) Q → P
Answer: 1, 2
Explanation:
P
|
Q
|
¬P
|
¬Q
|
Q → P
|
⌐(P ↔ Q)
|
P ↔ ⌐ Q
|
⌐P↔ Q
|
⌐P↔ ⌐Q
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
So both options 1 and 2 are correct.
Important Logical Equivalences.
The logical equivalences below are important equivalences that should be memorized.
Identity Laws:
p ∧ T ⇔ p
p ∨ F ⇔ p
Domination Laws:
p ∨ T ⇔ T
p ∧ F ⇔ F
Idempotent Laws:
p ∨ p ⇔ p
p ∧ p ⇔ p
Double Negation Law:
¬(¬p) ⇔ p
Commutative Laws:
p ∨ q ⇔ q ∨ p
p ∧ q ⇔ q ∧ p
Associative Laws:
(p ∨ q) ∨ r ⇔ p ∨ (q ∨ r)
(p ∧ q) ∧ r ⇔ p ∧ (q ∧ r)
Distributive Laws:
p ∨ (q ∧ r) ⇔ (p ∨ q) ∧ (p ∨ r)
p ∧ (q ∨ r) ⇔ (p ∧ q) ∨ (p ∧ r)
De Morgan’s Laws:
¬(p ∧ q) ⇔ ¬p ∨ ¬q
¬(p ∨ q) ⇔ ¬p ∧ ¬q
Absorption Laws:
p ∧ (p ∨ q) ⇔ p
p ∨ (p ∧ q) ⇔ p
Implication Law:
(p → q) ⇔ (¬p ∨ q)
Contrapositive Law:
(p → q) ⇔ (¬q → ¬p)
Tautology:
p ∨ ¬p ⇔ T
Contradiction:
p ∧ ¬p ⇔ F
Equivalence:
(p → q) ∧ (q → p) ⇔ (p ↔ q)
9. Negation
of the proposition Ǝ x H(x) is:
(1) Ǝ x ⌐H(x)
(2) ∀ x ⌐H(x)
(3) ∀ x H(x)
(4) ⌐ x H(x)
Answer: 2
10. The
output of the following combinational circuit is F.
The value of
F is :
(1) P1+P2’P3
(2) P1+P2’P3’
(3) P1+P2P3’
(4) P1’+P2P3
Answer: 2
The value of
F is :
(1) P1+P2’P3
(2) P1+P2’P3’
(3) P1+P2P3’
(4) P1’+P2P3
Answer: 2
11.
'ptrdata' is
a pointer to a data type. The expression *ptrdata++ is evaluated as (in C++):
(1) *(ptrdata++)
(2) (*ptrdata++)
(3) *(ptrdata)++
(4) Depends
on compiler
Answer: 1
12. The
associativity of which of the following operators is Left to Right, in C++?
(1) Unary
Operator
(2) Logical
not
(3) Array
element access
(4)
addressof
Answer: 3
| Category | Operator | Associativity |
|---|
| Postfix | () [] -> . ++ - - | Left to right |
| Unary | + - ! ~ ++ - - (type)* & sizeof | Right to left |
| Multiplicative | * / % | Left to right |
| Additive | + - | Left to right |
| Shift | << >> | Left to right |
| Relational | < <= > >= | Left to right |
| Equality | == != | Left to right |
| Bitwise AND | & | Left to right |
| Bitwise XOR | ^ | Left to right |
| Bitwise OR | | | Left to right |
| Logical AND | && | Left to right |
| Logical OR | || | Left to right |
| Conditional | ?: | Right to left |
| Assignment | = += -= *= /= %= >>= <<= &= ^= |= | Right to left |
| Comma | , | Left to right |
13. A
member function can always access the data in ................ , (in C++).
(1) the
class of which it is member
(2) the object
of which it is a member
(3) the
public part of its class
(4) the
private part of its class
Answer: 1
14. Which
of the following is not correct for virtual function in C++?
(1) Must be
declared in public section of class.
(2) Virtual
function can be static.
(3) Virtual
function should be accessed using pointers.
(4) Virtual
function is defined in base class.
Answer: 2
15. Which
of the following is not correct (in C++)?
(1) Class
templates and function templates are instantiated in the same way.
(2) Class
templates differ from function templates in the way they are initiated.
(3) Class
template is initiated by defining an object using the template argument.
(4) Class
templates are generally used for storage classes.
Answer: 2,3,4
16. Which
of the following is/are true with reference to 'view' in DBMS?
(a) A 'view'
is a special stored procedure executed when certain event occurs.
(b) A 'view'
is a virtual table, which occurs after executing a pre-compiled query.
Code:
(1) Only (a)
is true
(2) Only (b)
is true
(3) Both (a)
and (b) are true
(4) Neither
(a) nor (b) are true
Answer: 2
17. In
SQL, .................. is an Aggregate function.
(1) SELECT
(2) CREATE
(3) AVG
(4) MODIFY
Answer: 3
18. Match
the following with respect to RDBMS:
List - I
(a) Entity
integrity
(b) Domain
integrity
(c) Referential
integrity
(d)
Userdefined integrity
List - II
(i) enforces
some specific business rule that do not fall into entity or domain
(ii) Rows
can't be deleted which are used by other records
(iii)
enforces valid entries for a column
(iv) No
duplicate rows in a table
Code:
(a)
(b) (c) (d)
(1) (iii) (iv)
(i) (ii)
(2) (iv) (iii)
(ii) (i)
(3)
(iv) (ii) (iii) (i)
(4)
(ii) (iii) (iv)
(i)
Answer: 2
19. In
RDBMS, different classes of relations are created using ................... technique
to prevent modification anomalies.
(1)
Functional Dependencies
(2) Data
integrity
(3)
Referential integrity
(4) Normal
Forms
Answer: 4
20. ..................
SQL command changes one or more fields in a record.
(1) LOOK-UP
(2) INSERT
(3) MODIFY
(4) CHANGE
Answer: 3
21.
Consider an array representation of an n
element binary heap where the elements are stored from index 1 to index n of
the array. For the element stored at index i of the array (i< = n), the
index of the parent is:
(1) floor
((i +1)/2)
(2) ceiling
((i + 1)/2)
(3) floor
(i/2)
(4) ceiling
(i/2)
Answer: 3
Explaination:
A binary heap is a heap data structure that takes the form of a binary tree. Binary heaps are a common way of implementing priority queues.
A heap is a Tree which satisfies the following two properties:
1. All the child nodes must be less (or greater) than its parent node for a min (or max) heap.
2. The leaves must either fill the h or h-1 node where h is the height of the tree; it means it must be a complete binary tree.
22. The
following numbers are inserted into an empty binary search tree in the given
order:
10, 1, 3, 5,
15, 12, 16. What is the height of the binary search tree?
(1) 3
(2) 4
(3) 5
(4) 6
Answer: 1
The height is the maximum distance of a leaf node from the root
Constructed binary search tree will be..
10
/ \
1 15
\ / \
3 12 16
\
5
23. Let
G be an undirected connected graph with distinct edge weight. Let Emax
be the edge with maximum weight and Emin the edge with minimum
weight. Which of the following statements is false?
(1) Every
minimum spanning tree of G must contain Emin.
(2) If Emax
is in minimum spanning tree, then its removal must disconnect G.
(3) No
minimum spanning tree contains Emax.
(4) G has a
unique minimum spanning tree.
Answer: 3
24. A
list of n strings, each of length n, is sorted into lexicographic order using
merge - sort algorithm. The worst case running time of this computation is:
(1) O(n log
n)
(2) O(n2
log n)
(3) O(n2
+ log n)
(4) O(n3)
Answer: 2
25. Postorder
traversal of a given binary search tree T produces following sequence of keys:
3, 5, 7, 9,
4, 17, 16, 20, 18, 15, 14
Which one of
the following sequences of keys can be the result of an in-order traversal of
the tree T?
(1) 3, 4, 5,
7, 9, 14, 20, 18, 17, 16, 15
(2) 20, 18,
17, 16, 15, 14, 3, 4, 5, 7, 9
(3) 20, 18,
17, 16, 15, 14, 9, 7, 5, 4, 3
(4) 3, 4, 5,
7, 9, 14, 15, 16, 17, 18, 20
Answer: 4
Explanation: Inorder traversal of a
BST always gives elements in increasing order. Among all four options, 4 ) is the only increasing order sequence.
26. Which
of the following devices takes data sent from one network device and forwards
it to the destination node based on MAC address?
(1) Hub
(2) Modem
(3) Switch
(4) Gateway
Answer: 3
27. ..................
do not take their decisions on measurements or estimates of the current traffic
and topology.
(1) Static
algorithms
(2) Adaptive
algorithms
(3) Non -
adaptive algorithms
(4)
Recursive algorithms
Answer: 3
28. The
number of bits used for addressing in Gigabit Ethernet is ...................
(1) 32 bits
(2) 48 bits
(3) 64 bits
(4) 128 bits
Answer: 2
29. Which
of the following layer of OSI Reference model is also called end-to-end layer?
(1) Network
layer
(2) Datalink
layer
(3) Session
layer
(4)
Transport layer
Answer: 4
30. The
IP address ................... is used by hosts when they are being booted.
(1) 0.0.0.0
(2) 1.0.0.0
(3) 1.1.1.1
(4)
255.255.255.255
Answer: 1
31.
Consider the following program fragment in
assembly language:
mov ax, 0h
mov cx, 0A h
doloop:
dec ax
loop doloop
What is the
value of ax and cx registers after the completion of the doloop?
(1) ax =
FFF5 h and cx = 0 h
(2) ax =
FFF6 h and cx = 0 h
(3) ax =
FFF7 h and cx = 0A h
(4) ax =
FFF5 h and cx = 0A h
Answer: 2
32. Consider
the following assembly program segment:
stc
mov
al, 11010110b
mov
cl, 2
rcl
al, 3
rol
al, 4
shr
al, cl
mul
cl
The contents
of the destination register ax (in hexadecimal) and the status of Carry Flag
(CF) after the execution of above instructions are:
(1) ax =
003CH; CF = 0
(2) ax =
001EH; CF = 0
(3) ax =
007BH; CF = 1
(4) ax =
00B7H; CF = 1
Answer: 1
33. Which
of the following regular expressions, each describing a language of binary
numbers (MSB to LSB) that represents non-negative decimal values, does not
include even values?
(1) 0*1+0*1*
(2) 0*1*0+1*
(3) 0*1*0*1+
(4) 0+1*0*1*
Where {+, *}
are quantification characters.
Answer: 3
34. Which
of the following statements is/ are TRUE?
(a) The
grammar S → SS a is ambiguous. (Where S is the start symbol)
(b) The
grammar S → 0S1 | 01S | ϵ is ambiguous. (The special symbol ϵ represents the
empty string) (Where S is the start symbol)
(c) The
grammar (Where S is the start symbol)
S → T/U
T → x S y |
xy | ϵ
U → yT
generates a
language consisting of the string yxxyy.
(1) Only (a)
and (b) are TRUE.
(2) Only (a)
and (c) are TRUE.
(3) Only (b)
and (c) are TRUE.
(4) All of
(a), (b) and (c) are TRUE.
Answer: 4
35. Match
the description of several parts of a classic optimizing compiler in List - I,
with the names of those parts in List - II:
List - I
(a) A part
of a compiler that is responsible for recognizing syntax.
(b) A part
of a compiler that takes as input a stream of characters and produces as output
a stream of words along with their associated syntactic categories.
(c) A part
of a compiler that understand the meanings of variable names and other symbols
and checks that they are used in ways consistent with their definitions.
(d) An
IR-to-IR transformer that tries to improve the IR program in some way
(Intermediate representation).
List - II
(i)
Optimizer
(ii)
Semantic Analysis
(iii) Parser
(iv) Scanner
Code:
(a) (b) (c) (d)
(1) (iii)
(iv) (i) (ii)
(2) (iv)
(iii) (ii) (i)
(3) (ii)
(iv) (i) (iii)
(4) (ii)
(iv) (iii) (i)
Answer: 1
36. In
Distributed system, the capacity of a system to adapt the increased service
load is called .................
(1)
Tolerance
(2)
Scalability
(3)
Capability
(4) Loading
Answer: 2
37. In
.................... disk scheduling algorithm, the disk head moves from one
end to other end of the disk, serving the requests along the way. When the head
reaches the other end, it immediately returns to the beginning of the disk
without serving any requests on the return trip.
(1) LOOK
(2) SCAN
(3) C - LOOK
(4) C - SCAN
Answer: 4
Explaination:
In operating systems, seek time is very important. Since all device
requests are linked in queues, the seek time is increased causing the system
to slow down. Disk Scheduling Algorithms are used to reduce the total seek
time of any request.
38. Suppose
there are six files F1, F2, F3, F4, F5, F6 with corresponding sizes 150 KB,
225KB, 75 KB, 60 KB, 275 KB and 65 KB respectively. The files are to be stored
on a sequential device in such a way that optimizes access time. In what order
should the files be stored?
(1) F5, F2,
F1, F3, F6, F4
(2) F4, F6,
F3, F1, F2, F5
(3) F1, F2,
F3, F4, F5, F6
(4) F6, F5,
F4, F3, F2, F1
Answer: 2
F4 F6 F3 F1 F2 F4
60 65 75 150 225 275
Explanation: For access time optimization sort the files in increasing order of file size.
So, option (2) is correct.
39. Which
module gives control of the CPU to the process selected by the short - term scheduler?
(1)
Dispatcher
(2)
Interrupt
(3) Scheduler
(4)
Threading
Answer: 1
Explaination:-
The dispatcher is the module that gives control of the CPU to the
process selected by the short-time scheduler(selects from among the
processes that are ready to execute).
The function involves :
Swithching context
Switching to user mode
Jumping to the proper location in the user program to restart that program.
40. Two
atomic operations permissible on Semaphores are .................. and
...............
(1) wait,
stop
(2) wait,
hold
(3) hold,
signal
(4) wait,
signal
Answer: 4
41.
Software does not wear-out in the traditional
sense of the term, but software does tend to deteriorate as it evolves,
because:
(1) Software
suffers from exposure to hostile environments.
(2) Defects
are more likely to arise after software has been used often.
(3) Multiple
change requests introduce errors in component interactions.
(4) Software
spare parts become harder to order.
Answer: 3
42. Software
re-engineering is concerned with:
(1)
Re-constructing the original source code from the existing machine (low -
level) code program and modifying it to make it more user - friendly.
(2) Scrapping
the source code of a software and re-writing it entirely from scratch.
(3)
Re-organising and modifying existing software systems to make them more
maintainable.
(4)
Translating source code of an existing software to a new machine (low - level)
language.
Answer: 3
43. Which
of the following is not a key issue stressed by an agile philosophy of software
engineering?
(1) The
importance of self-organizing teams as well as communication and collaboration
between team members and customers.
(2)
Recognition that change represents opportunity.
(3) Emphasis
on rapid delivery of software that satisfies the customer.
(4) Having a
separate testing phase after a build phase.
Answer: 4
44. What
is the normal order of activities in which traditional software testing is
organized?
(a)
Integration Testing
(b) System
Testing
(c) Unit
Testing
(d)
Validation Testing
Code:
(1) (c),
(a), (b), (d)
(2) (c),
(a), (d), (b)
(3) (d),
(c), (b), (a)
(4) (b),
(d), (a), (c)
Answer: 2
45. Which
of the following testing techniques ensures that the software product runs
correctly after the changes during maintenance?
(1) Path
Testing
(2)
Integration Testing
(3) Unit
Testing
(4)
Regression Testing
Answer: 4
46. Which
of the following Super Computers is the fastest Super Computer?
(1) Sun-way
TaihuLight
(2) Titan
(3) Piz
Daint
(4) Sequoia
Answer: 1
47. Which
of the following statements about ERP system is true?
(1) Most ERP
software implementations fully achieve seamless integration.
(2) ERP
software packages are themselves combinations of seperate applications for
manufacturing, materials, resource planning, general ledger, human resources,
procurement and order entry.
(3)
Integration of ERP systems can be achieved in only one way.
(4) An ERP
package implemented uniformly throughout an enterprise is likely to contain
very flexible connections to allow charges and software variations.
Answer: 2
48. Which
of the following is not a Clustering method?
(1) K - Mean
method
(2) Self
Organizing feature map method
(3) K -
nearest neighbour method
(4)
Agglomerative method
Answer: 3
49. Which
of the given wireless technologies used in IoT, consumes the least amount of
power?
(1) Zigbee
(2)
Bluetooth
(3) Wi-Fi
(4) GSM/CDMA
Answer: 2
50. Which
speed up could be achieved according to Amdahl's Law for infinite number of
processes if 5% of a program is sequential and the remaining part is ideally
parallel?
(1) Infinite
(2) 5
(3) 20
(4) 50
Answer: 3
Explanation: According to Amdahl’s law speed up for infinite number of process:
S = 1 / (1-P)
where p is parallel part of program
Given, sequential part of program is 5%. So parallel part of the program (P)
= 1 - sequential part
= 1 - 0.05 (or 5%)
= 0.95 (or 95%)
Now S = 1 / (1-P)
ie S = 1 / (1-0.95)
S = 1 / 0.05
S = 20
So, option (3) is correct.
From the above Karnaugh map, we will get B + CD + AD.
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