UGC NET CS 2004 Dec-Paper-2

Question 1
AVA=A is called :
A
Identity law
B
De Morgan’​ s law
C
Idempotent law
D
Complement law
       Engineering-Mathematics       Propositional-Logic
Question 1 Explanation: 
→ ​ De Morgan’s Laws:
(i). (A ​ V ​ B)’ = A' ​ ∧ ​ B'
(ii). (A ​ ∧ ​ B)’ = A' ​ V ​ B'
→ ​ Identity Law​ :
(i). 1 AND A = A
(ii). 0 OR A = A
→ ​ Complement law:
(i). A AND A'=1
(ii). A OR A'=0
→ ​ Idempotent law:
The idempotence in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means "(the quality of having) the same power", from idem + potence (same + power).
(i). A V A=A
(ii). A ∧ A=A
According to boolean algebra
Question 2
If f(x) = x+1 and g(x)=x+3 then f0 f0 f0 f is :
A
g
B
g+1
C
g​ 4
D
None of the above
       Engineering-Mathematics       Relations-and-Functions
Question 2 Explanation: 
Given data,
f(x)=x+1
g(x)=x+3
Constraint is f0 f0 f0 f
Step-1: We can write into fo fo fo f is f(f(f(x+1)))
We can write into f(f(x+2)) and f(x+3).
Step-2: Above constraint is equal to "x+4" because f(x+3)+1
Step-3: We can also write into fog(x)=x+4 and gof(x)=x+4.
So, g+1 is appropriate answer.
Question 3
The context-free languages are closed for :
(i) Intersection
(ii) Union
(iii) Complementation
(iv) Kleene Star
then
A
(i) and (iv)
B
(i) and (iii)
C
(ii) and (iv)
D
(ii) and (iii)
       Theory-of-Computation       Context-Free-Language
Question 3 Explanation: 
The context free languages are closed under union and kleene star but it is not closed under intersection and complementation.
Note: Except intersection and complementation will closed under all operations in CFL.
Question 4
The following lists are the degrees of all the vertices of a graph :
(i) 1, 2, 3, 4, 5
(ii) 3, 4, 5, 6, 7
(iii) 1, 4, 5, 8, 6
(iv) 3, 4, 5, 6
then
A
(i) and (ii)
B
(iii) and (iv)
C
(iii) and (ii)
D
(ii) and (iv)
       Engineering-Mathematics       Graph-Theory
Question 4 Explanation: 
Every graph is following basic 2 properties:
1. Sum of degrees of the vertices of a graph should be even.
2. Sum of degrees of the vertices of a graph is equal to twice the number of edges.
Statement-(i) is violating property-1.
= 1+2+3+4+5
= 15 is odd number.
Statement-(ii) is violating property-1.
= 3+4+5+6+7
= 25 is odd number.
Statement-(iii) is violating property-1.
= 1+4+5+8+6
= 24 is even number
Statement-(iv) is violating property-1
= 3+4+5+6
= 18 is even number
Question 5
If I​ m denotes the set of integers modulo m, then the following are fields with respect to the operations of addition modulo m and multiplication modulo m :
(i) Z​ 23
(ii) Z​ 29
(iii) Z​ 31
(iv) Z​ 33
Then
A
(i) only
B
(i) and (ii) only
C
(i), (ii) and (iii) only
D
(i), (ii), (iii) and (iv)
       Engineering-Mathematics       Set-Theory
Question 6
An example of a binary number which is equal to its 2​ ’s complement is :
A
1100
B
1001
C
1000
D
1111
       Digital-Logic-Design       Number-Systems
Question 6 Explanation: 
Option-A: 1100 convert into 2’s complement is
1100
1’s complement: 0011
2’s complement: 1
--------
0100
Option-B: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement:
1
--------
0111
---------
Option-C: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement:
1
--------
1000
---------
Option-D: 1111 convert into 2’s complement is
1111
1’s complement: 0000
2’s complement:
1
--------
0001
---------
So, Option-C is correct answer.
Question 7
When a tri - state logic device is in the third state, then :
A
it draws low current
B
it does not draw any current
C
it draws very high current
D
it presents a low impedance
Question 7 Explanation: 
Three Stage logic name itself implies there are 3 states of the buffer.
1. High (1)
2. Low (0)
3. High Impedance (Z)
→ High impedance means that a point in a circuit (a node) allows a relatively small amount of current through, per unit of applied voltage at that point.
→ High impedance circuits are low current, high voltage, whereas low impedance circuits are the opposite.
Truth table of Tri-state logic Inverter
Question 8
An example of a connective which is not associative is :
A
AND
B
OR
C
EX-OR
D
NAND
       Digital-Logic-Design       Logic-Gates
Question 8 Explanation: 
→ ​ OR is associative:
(i) (0 OR 1)OR 1
= 1 OR 1
= 1
(ii). 0 OR (1 OR 1)
= 0 OR 1
= 1
→ NAND is not associative
(i). (0 NAND 1)NAND 1
= 1 NAND 1
= 0
(ii). 0 NAND (1 NAND 1)
= 0 NAND 0
= 1
→ ​ Ex-OR is associative
(i). (0 XOR 1) XOR 1
= 1 XOR 1
= 0
(ii). 0 XOR(1 XOR 1)
= 0 XOR 0
= 0
→ ​ AND is associative:
(i) (0 AND 1) AND 1
= 0 AND 1
= 0
(ii). 0 AND (1 AND 1)
= 0 AND 1
= 0
Question 9
Essential hazards may occur in :
A
Combinational logic circuits
B
Synchronous sequential logic circuits
C
Asynchronous sequential logic circuits working in the fundamental mode
D
Asynchronous sequential logic circuits working in the pulse mode
       Digital-Logic-Design       Sequential-Circuits
Question 9 Explanation: 
→ Essential hazards may occur in asynchronous sequential logic circuits working in the fundamental mode.
→ Asynchronous circuits is called essential hazard ​ is caused by unequal delays along two or more paths that originate from the same same input.
→ It​ cannot be corrected by adding redundant gates and it can only be corrected by adjusting the amount of delay in the affected path.
Question 10
The characteristic equation of a T flip-flop is:__[Note: The symbols used have the usual meaning]
A
Q​ n+1​ =T Q n​ + T Q​ n
B
Q​ n+1​ =T+Q​ n
C
Q​ n+1​ =TQ​ n
D
Q​ n+1​ = T Q n
       Digital-Logic-Design       Sequential-Circuits
Question 10 Explanation: 
T-Flip flop Truth Table:

T-Flip-Flop Characteristic Table:

T-Flip-Flop Characteristic equation:
Q​ next​ = TQ' + T'Q
Question 11
Suppose x and y are two Integer Variables having values 0x5AB6 and 0x61CD respectively. The result (in hex) of applying bitwise operator and to x and y will be :
A
0 x 5089
B
0 x 4084
C
0 x 78A4
D
0 x 3AD1
       Programming       Operator
Question 11 Explanation: 
Given two integer numbers are 0x5AB6 and 0x61CD
Step-1: Convert hexadecimal numbers into binary number because we want to perform AND operation.
0x5AB6 equivalent into binary number is 0101 1010 1011 0110
0x61CD equivalent into binary number is 0110 0001 1100 1101
Step-2: Perform Bitwise AND operation
0101 1010 1011 0110
0110 0001 1100 1101
-----------------------------
0100 0000 1000 0100(Bitwise AND operation)
------------------------------
Step-3: Convert result into hexadecimal number.
0100 0000 1000 0100 equivalent into hexadecimal number is 0x4084
Question 12
Consider the following statements,
int i=4, j=3, k=0;
k= ++i- - -j + i++ - - - j +j++;
What will be the values of i, j and k after the statement.
A
7, 2, 8
B
5, 2, 10
C
6, 2, 8
D
4, 2, 8
       Programming       Operator
Question 12 Explanation: 
Given values are i=4, j=3 and k=0
Step-1: k= ++i- --j + i++ - --j +j++;
k= 5 - 2 + 5 - 1 + 1
k= (5-2)+(5-1)+1
= 3 + 4 + 1
= 8
Step-2: The value of i=6,j=2 and k=8
Question 13
What is the value of the arithmetic expression (Written in C) 2*3/4-3/4*2
A
0
B
1
C
1.5
D
None of the above
       Programming       Operator
Question 13 Explanation: 
Given expression is 2*3/4-3/4* 2
Step-1: In C language multiplication and division having higher priority than addition and subtraction.
Step-2: Evaluating procedure is ((2*3)/4)-((3⁄4)* 2)
= ((2*3)/4)-((3⁄4)* 2)
= (6/4)-((3⁄4)* 2)
= 1 - ((3⁄4)* 2)
= 1 - 0*2
= 1 - 0
= 1
Question 14
A function object :
A
is an instance of a class for which operator ( ) is a member function.
B
is an instance of a class for which operator → is a member function.
C
is a pointer to any function
D
is a member function of a class
       Programming-in-c++       Functions
Question 14 Explanation: 
A function object is an object to which the function call operator can be applied. Typically, it is a class that defines the function call operator (operator()()) as a member function. When a function object is used as a function, the function call operator is invoked whenever the function is called.
Question 15
Polymorphism means :
A
A template function
B
Runtime type identification within a class hierarchy
C
Another name for operator overloading
D
Virtual inheritance
       Programming-in-c++       Properties
Question 15 Explanation: 
→ Polymorphism means "many forms", and it occurs when we have many classes that are related to each other by inheritance.
→ Polymorphism means runtime type identification within a class hierarchy.
Question 16
The E-R model is expressed in terms of :
(i) Entities
(ii) The relationship among entities
(iii) The attributes of the entities
Then
A
(i) and (iii)
B
(i) and (ii)
C
(ii) and (iii)
D
None of the above
       Database-Management-System       ER-Model
Question 16 Explanation: 
The E-R model is expressed in terms of:
Entities: An entity is a object in Database. An entity is expressed using a rectangular box in an E-R diagram. Example: person, car etc.
Attributes: Nouns which describe nouns. Attributes helps to describe an entity in a better way.
Example: If person is an entity then person name, address, phone number can be its attributes.
Relationship: Any verb which you are going to describe. Example: A person owns a car. Here "owns" is a verb describing relationship between two entities(person and car).

In above diagram Person and Car are two entities
"owns" is defining the relationship between Person and Car.
name, address are the attributes of "Person" entity and number and model are the attributes of "Car" entity.
Note: Excluded for evaluation, Given wrong options.
Question 17
Specialization is a __________ process.
A
Top - down
B
Bottom - up
C
Both (A) and (B)
D
None of the above
Question 17 Explanation: 
→ Generalization - the process of defining a general entity type from a collection of specialized entity types.
→ Specialization - the opposite of generalization, since it defines subtypes of the supertype and determines the relationship between the two.
→ Specialization is a top down approach. The following example "Employee" could be a developer or a tester.relationship between the two.
→ Specialization is a top down approach. The following example "Employee" could be a developer or a tester.
Question 18
The completeness constraint has rules :
A
Supertype, Subtype
B
Total specialization, Partial specialization
C
Specialization, Generalization
D
All of the above
Question 18 Explanation: 
→ Completeness constraints are both total specialization and partial specialization.
→ The completeness constraints decide whether a supertype instance must also be a membe of at least one subtype.
→ The total specialization rule demands that every entity in the superclass belong to some subclass. Just as with a regular ERD, total specialization is symbolized with a double line connection between entities.
→ The partial specialization rule allows an entity to not belong to any of the subclasses. It is represented with a single line connection.
Question 19
The entity type on which the __________ type depends is called the identifying owner.
A
Strong entity
B
Relationship
C
Weak entity
D
E - R
       Database-Management-System       ER-Model
Question 19 Explanation: 
A weak entity of a relation do not have any primary key to uniquely identify it's each tuple. To avoid this problem, the weak entity is associated with the keys of it's identifying owner which is a strong entity.
Question 20
Match the following :
A
i-a, ii-c, iii-b, iv-d
B
i-d, ii-c, iii-a, iv-b
C
i-d, ii-c, iii-b, iv-a
D
i-a, ii-b, iii-c, iv-d
E
None of the above
       Database-Management-System       Normalization
Question 20 Explanation: 
→ 2NF ensures no partial dependency in the relation.
→ A relation is in 3NF if it is in eliminates transitive dependency from the relation.
→ 4NF eliminates multivalued dependency from a relation.
→ 5NF eliminates join dependency from a table.
Question 21
What item is at the root after the following sequence of insertions into an empty splay tree :
1, 11, 3, 10, 8, 4, 6, 5, 7, 9, 2, ?
A
1
B
2
C
4
D
8
       Data-Structures       Binary-Trees
Question 21 Explanation: 
A splay tree is a self-balancing binary search tree with the additional property that recently accessed elements are quick to access again. It performs basic operations such as insertion, look-up and removal in O(log n) amortized time. For many sequences of non-random operations, splay trees perform better than other search trees, even when the specific pattern of the sequence is unknown.



Question 22
Suppose we are implementing quadratic probing with a Hash function, Hash(y)=X mod 100. If an element with key 4594 is inserted and the first three locations attempted are already occupied, then the next cell that will be tried is :
A
2
B
3
C
9
D
97
       Data-Structures       Hashing
Question 22 Explanation: 
Given quadratic probing with a Hash function,
-- Hash(y)=X mod 100
-- Key=4594
-- First 3 locations are already occupied.
-- Next cell=?
Step-1: Quadratic Probing function is​ ​ h(k,i) = (h'(k) +c​ 1​ i+c​ 2​ i​ 2​ )mod m 0≤ i ≤ m-1
where c1 and c2 constants ≠0
Step-2: First pass: 4594 % 100
= 94
Sep-3: Second pass: (4594 + 1​ 2​ ) % 100
= (94 + 1) % 100
= 95
Step-3: Third pass: (4594 + 2​ 2​ ) % 100
= (94 + 4) %100
= 98
Step-4: Fourth pass: (4594 + 1​ 2​ ) % 100
= (94 + 9) %100
= 103 % 100
= 3
Question 23
Weighted graph :
A
Is a bi-directional graph
B
Is directed graph
C
Is graph in which number associated with arc
D
Eliminates table method
       Data-Structures       Graphs-and-Tree
Question 23 Explanation: 
→ Weighted graph is a graph in which number associated with arc. A weighted graph is a graph in which each branch is given a numerical weight.
→ A weighted graph is therefore a special type of labeled graph in which the labels are numbers.
Question 24
What operation is supported in constant time by the doubly linked list, but not by the singly linked list ?
A
Advance
B
Backup
C
First
D
Retrieve
       Data-Structures       Linked-List
Question 24 Explanation: 
→ Backup operation is supported in constant time by the doubly linked list, but not by the singly linked list.
Question 25
How much extra space is used by heapsort ?
A
O(1)
B
O(Log n)
C
O(n)
D
O(n​ 2​ )
       Algorithms       Sorting
Question 25 Explanation: 
→ Heap sort uses Max-Heapify function which calls itself but it can be made using a simple while loop and thus making it an iterative function which intern takes no space. So, the space complexity of Heap Sort can be reduced to O(1)
→ The heapify function does not take O(logn) space, it will take O(logn) time in worst case. Here we don't do heapify through any recursion call. We will do through iterative method and we don't require any auxiliary space in this sorting. We use only input array.
So, we do heap sort in O(1) space complexity.
Question 26
Error control is needed at the transport layer because of potential error occuring___.
A
from transmission line noise
B
in router
C
from out of sequence delivery
D
from packet losses
       Computer-Networks       Transport-layer-Protocol
Question 26 Explanation: 
→ Error control is needed at the transport layer because of potential error occuring in router.
→ The error control in the transport layer usually refers to the guaranteed delivery mechanism with TCP, which attempts to safeguard against frames/packets getting lost entirely.
Question 27
Making sure that all the data packets of a message are delivered to the destination is _________ control.
A
Error
B
Loss
C
Sequence
D
Duplication
       Computer-Networks       Error-Control-Methods
Question 27 Explanation: 
Making sure that all the data packets of a message are delivered to the destination is error control. Error Control:
Error Control is performed end to end in this layer to ensure that the complete message arrives at the receiving transport layer without any error. Error Correction is done through retransmission.
Question 28
Which transport class should be used with a perfect network layer ?
A
TP0 and TP2
B
TP1 and TP3
C
TP0, TP1, TP3
D
TP0, TP1, TP2, TP3, TP4
       Computer-Networks       Network-Layer
Question 28 Explanation: 
Five transport layer protocols exist in the OSI suite, ranging from Transport Protocol Class 0 through Transport Protocol Class 4 (TP0, TP1, TP2, TP3 & TP4).
→ TP0 to TP3 work only with connection-oriented communications, in which a session connection must be established before any data is sent.
→ TP4 also works with both connection-oriented and connectionless communications. Transport Protocol Class 0 (TP0)​ performs segmentation (fragmentation) and reassembly functions. TP0 discerns the size of the smallest maximum protocol data unit (PDU) supported by any of the underlying networks, and segments the packets accordingly. The packet segments are reassembled at the receiver.
Transport Protocol Class 2 (TP2)​ performs segmentation and reassembly, as well as multiplexing and demultiplexing of data streams over a single virtual circuit.
Transport Protocol Class 4 (TP4)​ offers error recovery, performs segmentation and reassembly, and supplies multiplexing and demultiplexing of data streams over a single virtual circuit. TP4 sequences PDUs and retransmits them or reinitiates the connection if an excessive number are unacknowledged. TP4 provides reliable transport service and functions with either connection-oriented or connectionless network service. TP4 is the most commonly used of all the OSI transport protocols.
Question 29
Which transport class should be used with residual-error network layer ?
A
TP0, TP2
B
TP1, TP3
C
TP1, TP3, TP4
D
TP0, TP1, TP2, TP3, TP4
       Computer-Networks       Network-Layer
Question 29 Explanation: 
Transport Protocol Class 1 (TP1)​ performs segmentation (fragmentation) and reassembly, plus error recovery. TP1 sequences protocol data units (PDUs) and will retransmit PDUs or re-initiate the connection if an excessive number of PDUs are unacknowledged.
Transport Protocol Class 3 (TP3)​ offers error recovery, segmentation and reassembly, and multiplexing and demultiplexing of data streams over a single virtual circuit. TP3 also sequences PDUs and retransmits them or re-initiates the connection if an excessive number are unacknowledged.
Question 30
Virtual circuit is associated with a __________ service.
A
Connectionless
B
Error-free
C
Segmentation
D
Connection-oriented
       Computer-Networks       Switching
Question 30 Explanation: 
Virtual circuit communication resembles circuit switching, since both are connection oriented, meaning that in both cases data is delivered in correct order, and signalling overhead is required during a connection establishment phase.
Question 31
Which activity is not included in the first pass of two pass assemblers ?
A
Build the symbol table
B
Construct the intermediate code
C
Separate mnemonic opcode and operand fields
D
None of the above
       Compiler-Design       Assembler
Question 31 Explanation: 
Two Pass Assemblers:
Pass-1:
1. Assign addresses to all statements in the program
2. Save the values assigned to all labels for use in Pass2
3. Perform some processing of assembler directives
Pass-2:
1. Assemble instructions Generate data values defined by BYTE,WORD
2. Perform processing of assembler directives not done in Pass 1
3. Write the object program and the assembly listing
Question 32
Which of the following is not collision resolution technique ?
A
Hash addressing
B
Chaining
C
Both (A) and (B)
D
Indexing
       Data-Structures       Hashing
Question 32 Explanation: 
→ Hash collisions are practically unavoidable when hashing a random subset of a large set of possible keys.
→ Two keys mapping to the same location in the hash table is called “Collision”.
Collision Resolving techniques
1. Separate chaining
2. Open addressing
i. Linear Probing
ii. Quadratic Probing
iii. Double hashing
Question 33
Code optimization is responsibility of :
A
Application programmer
B
System programmer
C
Operating system
D
All of the above
       Compiler-Design       Code-Optimization
Question 33 Explanation: 
Code optimization is responsibility of system programmer.
Question 34
Which activity is included in the first pass of two pass assemblers ?
A
Build the symbol table
B
Construct the intermediate code
C
Separate mnemonic opcode and operand fields
D
None of these
E
A,B and C
       Compiler-Design       Assembler
Question 34 Explanation: 
Two Pass Assemblers:
Pass-1:
1. Assign addresses to all statements in the program
2. Save the values assigned to all labels for use in Pass2
3. Perform some processing of assembler directives
Pass-2:
1. Assemble instructions Generate data values defined by BYTE,WORD
2. Perform processing of assembler directives not done in Pass-1
3. Write the object program and the assembly listing.
Activities:
1. Build the symbol table
2. Construct the intermediate code
3. Separate mnemonic opcode and operand fields
Question 35
In two pass assembler the symbol table is used to store :
A
Label and value
B
Only value
C
Mnemonic
D
Memory Location
       Compiler-Design       Assembler
Question 35 Explanation: 
In two pass assembler the symbol table is used to store memory location.
Question 36
Semaphores are used to :
A
Synchronise critical resources to prevent deadlock
B
Synchronise critical resources to prevent contention
C
Do I/o
D
Facilitate memory management
       Operating-Systems       Process-Synchronization
Question 36 Explanation: 
→ Semaphores are implemented to prevent deadlocks using wait and signal operation.
→ A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system.
→ A semaphore is simply a variable. This variable is used to solve critical section problems and to achieve process synchronization in the multi processing environment.
Question 37
In which of the following storage replacement strategies, is a program placed in the largest available hole in the memory ?
A
Best fit
B
First fit
C
Worst fit
D
Buddy
       Operating-Systems       Memory-Management
Question 37 Explanation: 
First fit:​ Allocate the first hole that is big enough. Searching can start either at the beginning of the set of holes or at the location where the previous first-fit search ended. We can stop searching as soon as we find a free hole that is large enough.
Best fit: ​ Allocate the smallest hole that is big enough. We must search the entire list, unless the list is ordered by size. This strategy produces the smallest leftover hole.
Worst fit: ​ Allocate the largest hole. Again, we must search the entire list, unless it is sorted by size. This strategy produces the largest leftover hole, which may be more useful than the smaller leftover hole from a best-fit approach.
Question 38
Remote computing system involves the use of time sharing systems and :
A
Real time processing
B
Batch processing
C
Multiprocessing
D
All of the above
       Operating-Systems       Types-of-Operating-System
Question 38 Explanation: 
→ Remote computing system involves the use of time sharing systems and Batch processing.
→ Batch processing in which an input device is located at a distance from the main installation and has access to a computer through a communication link.
Question 39
Non modifiable procedures are called
A
Serially useable procedures
B
Concurrent procedures
C
Reentrant procedures
D
Top down procedures
Question 39 Explanation: 
→ A computer program or subroutine is called reentrant if it can be interrupted in the middle of its execution and then safely be called again ("re-entered") before it’s previous invocation complete execution.
→ The interruption could be caused by an internal action such as a jump or call, or by an external action such as an interrupt or signal. The previous invocations may resume correct execution before the reentered invocation completes, unlike recursion, where the previous invocations may only resume correct execution once the reentered invocation completes.
Rules:
1. Reentrant code may not hold any static or global non-constant data.
2. Reentrant code may not modify itself.
3. Reentrant code may not call non-reentrant computer programs or routines.
Question 40
Match the following
A
a-3, b-4, c-2, d-1
B
a-4, b-3, c-2, d-1
C
a-2, b-4, c-1, d-3
D
a-3, b-4, c-1, d-2
       Operating-Systems       Process-Scheduling
Question 40 Explanation: 
Disk scheduling→ SCAN,C-SCAN,LOOK,C-LOOK
Batch processing→ FIFO
Time sharing→ Round robin with the help of time quantum.
Interrupt processing→ LIFO
Question 41
The main objective of designing various modules of a software system is :
A
To decrease the cohesion and to increase the coupling
B
To increase the cohesion and to decrease the coupling
C
To increase the coupling only
D
To increase the cohesion only
       Software-Engineering       Software-design
Question 41 Explanation: 
→ Cohesion is a measure of internal strength within a module, whereas coupling is a measure of inter dependency among the modules.
→ So in the context of modular software design there should be high cohesion and low coupling.
Question 42
Three essential components of a software project plan are :
A
Team structure, Quality assurance plans, Cost estimation
B
Cost estimation, Time estimation, Quality assurance plan
C
Cost estimation, Time estimation, Personnel estimation
D
Cost estimation, Personnel estimation, Team structure
       Software-Engineering       Software-requirements
Question 42 Explanation: 
→ A project is a success if it meets the objectives of time, cost, technical and business.
Essential components:
1.Software size estimation
2.Effort estimation
3.Time estimation
4.Cost estimation
Question 43
Reliability of software is dependent on :
A
Number of errors present in software
B
Documentation
C
Testing suites
D
Development Processes
       Software-Engineering       Software-Reliabiliy
Question 43 Explanation: 
→ Reliability of software is dependent on number of errors present in software.
→ Software Reliability is the probability of failure-free software operation for a specified period of time in a specified environment.
Question 44
In transform analysis, input portion is called :
A
Afferent branch
B
Efferent branch
C
Central Transform
D
None of the above
Question 44 Explanation: 
Transform Analysis:
It identifies the primary functional components (modules) and the high level inputs and outputs for these components.
1. Input
2. Logical processing
3. Output
The input portion of the DFD includes processes that transform input data from physical (e.g. character from terminal) to logical forms (e.g. internal tables, lists, etc.). Each input portion is called an afferent branch.
The output portion of a DFD transforms output data from logical to physical form. Each output portion is called an efferent branch. The remaining portion of a DFD is called the central transform.
Question 45
The Function Point (FP) metric is :
A
Calculated from user requirements
B
Calculated from Lines of code
C
Calculated from software​ s complexity assessment
D
None of the above
       Software-Engineering       Cyclomatic-metric
Question 45 Explanation: 
→ The Function Point (FP) metric is calculated from software​ s complexity assessment.
→ A Function Point (FP) is a unit of measurement to express the amount of business functionality, an information system (as a product) provides to a user.
→ FPs measure software size. They are widely accepted as an industry standard for functional sizing.
Question 46
Data Mining can be used as _________ Tool.
A
Software
B
Hardware
C
Research
D
Process
Question 46 Explanation: 
Data Mining can be used as research tool. Data mining is the process of discovering patterns in large data sets involving methods at the intersection of machine learning, statistics, and database systems
Question 47
The processing speeds of pipeline segments are usually :
A
Equal
B
Unequal
C
Greater
D
None of these
Question 47 Explanation: 
→ The processing speeds of pipeline segments are usually unequal.
→ Pipelining attempts to keep every part of the processor busy with some instruction by dividing incoming instructions into a series of sequential steps (the eponymous "pipeline") performed by different processor units with different parts of instructions processed in parallel.
Question 48
The cost of a parallel processing is primarily determined by :
A
Time complexity
B
Switching complexity
C
Circuit complexity
D
None of the above
Question 48 Explanation: 
The cost of a parallel processing is primarily determined by switching complexity.
Question 49
A data warehouse is always _________.
A
Subject oriented
B
Object oriented
C
Program oriented
D
Compiler oriented
Question 49 Explanation: 
→ A data warehouse is a subject-oriented, integrated, time-variant and non-volatile collection of data in support of management's decision making process.
Subject oriented:​ A data warehouse can be used to analyze a particular subject area. For example, "sales" can be a particular subject.
Question 50
The term ​ ’hacker’​ was originally associated with :
A
A computer program
B
Virus
C
Computer professionals who solved complex computer problems.
D
All of the above
Question 50 Explanation: 
→ A computer hacker is any skilled computer expert that uses their technical knowledge to overcome a problem. While "hacker" can refer to any skilled computer programmer, the term has become associated in popular culture with a "security hacker", someone who, with their technical knowledge, uses bugs or exploits to break into computer systems.
There are 50 questions to complete.
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