UGC NET Dec 2013 paper 3

Q.1

If the primal Linear Programming problem has unbounded solution,then it's dual problem will have

feasible solution

alternative solution

no feasible solution at all

no bounded solution at all

Ans .

Explanation :

If the Primal problem is feasible, but unbounded in the direction of optimisation, then the dual has no feasible solution. Otherwise, if the Primal problem has an optimal solution, then the dual has also an optimal solution

Q.2

All of these

Ans .

Explanation :

All the three conditions are necessary for local maxima

Q.3

feasible solution

No feasible Solution

Unbounded Solution

Single point as Solution

Ans .

Explanation :

draw lines for two inequalities nothing is common in shaded area so infeasible solution

Q.4

Given a flow graph with 10 nodes. 13 edges and one connected components, the number of regions and the number of predicate(decision) nodes in the flow graph will be

4,5

5,4

3,1

13,8

Ans .

Explanation :

Predicate (decision) nodes are those nodes which has out degree atleast 2. Region =13-10+2 = 5 [i.e. edges - vertex + (p+1)] since p= 1] predicate nodes(P) region = P+1 = 5 P = 4

Q.5

Function points can be calculated by

UFP*CAF

UFP*FAC

UFP*Cost

UFP*Productivity

Ans .

Explanation :
UFP is Unadjusted Function Point CAF is Complixity Adjustment Factor Function Point FP = UFP*CAF

Q.6

Match the following:
List-I                     List-II
a. Data coupling        i. Module A and Module B have shared data
b. Stamp coupling      ii. Dependency between modules is based on the fact they communicate by only passing of data. 
c. Common coupling    iii. When complete data structure is passed from one module to another.
d. Content coupling     iv.When the control is passed from one module to the middle of another.

Codes: a b c d

iii ii i iv

ii iii i iv

ii iii iv i

iii ii iv i

Ans .

Explanation :a. Data coupling ii. Dependency between modules is based on the fact they communicate by only passing of data
b. Stamp coupling iii. When complete data structure is passed from one module to another
c. Common coupling i. Module A and Module B have shared data
d.Content coupling iv. When the control is passed from one module to the middle of another

Q.7

A process which defines a series of tasks that have the following four primary objectives is known as 1. to identify all items that collectively defines the software configuration. 2. to manage changes to one or more of these items. 3. to facilitate the construction of different versions of an application. 4. to ensure that software quality is maintained as the configuration evolves over time.

Software Quality Management Process

Software Configuration Management Process

Software Version Management Process

Software Change Management Process

Ans .

Explanation :

Software Configuration Management Process has following primary objective: 1. to identify all items that collectively define the software configuration 2. to manage changes to one or more of these items 3. to facilitate the construction of different versions of an application 4. to ensure that software quality is maintained as the configuration evolves over time

Q.8

One weakness of boundary value analysis and equivalent partitioning is

they are not effective.

they do not explore combinations of input circumstances.

they explore combinations of input circumstances

none of the above

Ans .

Explanation :

both are black box testing techniques as we only test at boundaries and at certain points they do not explore combinations of input circumstances for more detail refer http://reqtest.se/testing-sv/what-is-boundary-value-analysis-and-equivalence-partitioning/

Q.9

Which one of the following is not a software myth?

Once we write the program and get it to work, our job is done.

If we get behind schedule, we can add more programmers and catch up.

If an organization does not understand how to control software project internally, it will invariably struggle when it outsources software projects.

Ans .

Explanation :

Rest are myths (programmer myths and management's myths)

Q.10

Match the following with respect to relationship between objects and classes:
List-I                 List-II
a. State diagram       i. Useful for both abstract modeling and for designing actual program.
b. Object diagram    ii. Describes object classes.
c. Class diagram     iii. Useful for documenting test cases.
d. Instance diagram iv. Describing the behavior of a single class of object.

Codes: a b c d

iv i ii iii

ii iii iv i

iii iv ii i

ii iv i iii

Ans .

Explanation :

1.state diagram is a type of diagram used in computer science and related fields to describe the behavior of systems. State diagrams require that the system described is composed of a finite number of states;
2.In software engineering, a class diagram in the Unified Modeling Language (UML) is a type of static structure diagram that describes the structure of a system by showing the system's classes, their attributes, operations (or methods), and the relationships among objects.
3.An InstanceDiagram is a part of the UnifiedModelingLanguage that one does not see mentioned too often. The basic idea is to make a static snap shot of instances (not classes) in your system or subsystem. Make it show exactly who points to whom.
4.Object diagrams are derived from class diagrams so object diagrams are dependent upon class diagrams.Object diagrams represent an instance of a class diagram. The basic concepts are similar for class diagrams and object diagrams. Object diagrams also represent the static view of a system but this static view is a snapshot of the system at a particular moment.Object diagrams are used to render a set of objects and their relationships as an instance.

Q.11

Match the following style rules for reusability:
List-I               			List-II
a. Keep methods coherent   		i. Write a method to get the last element of a list.
b. Keep methods small   		ii. Maintain parallel structure when possible.
c. Keep methods consistent	  	iii. Breaking a method into smaller parts.
d. Provide uniform coverage	    iv. Perform a single function or a group of closely related function.

Codes: a b c d

iv iii ii i

ii i iv iii

iii iv ii i

ii iii iv i

Ans .

Explanation :

Keep methods coherent  --->    Performs a single function or a group of a closely related functions
Keep methods small----> Breaking a method into smaller parts
Keep methods consistent--->Maintain parallel structure when possible
Provide uniform coverage--->Write a method to get the last element of a list

Q.12

Which is the protocol for performing RPCs between applications in a language and system independent way?

Hyper Text Transmission Protocol (HTTP)

Simple Network Management Protocol (SNMP)

Simple Object Access Protocol (SOAP)

Simple Mail Transfer Protocol (SMTP)

Ans .

Explanation :

SOAP is an acronym for Simple Object Access Protocol. It is an XML-based messaging protocol for exchanging information among computers. SOAP is an application of the XML specification.
Although SOAP can be used in a variety of messaging systems and can be delivered via a variety of transport protocols, the initial focus of SOAP is remote procedure calls(RPC) transported via HTTP.

Q.13

The document that is used by XSLT to indicate, how to transform the element of the XML document to another format is

HTML page

DOC type procedure

Style sheet

Store procedure

Ans .

Explanation :

Any prefix can be used, but the prefix "xsl" is used most frequently with XSLT. The prefix and the element name are separated by a colon (:). In other words, "xsl:…" is notated to separate transformation commands and output data. Since the XSLT stylesheet is an XML document, a root element is required.

Q.14

Which of the following concepts means adding new concept to a program as it runs?

Data hiding

Dynamic loading

Dynamic typing

Dynamic binding

Ans .

Explanation :

Dynamic loading is a mechanism by which a computer program can, at run time, load a library (or other binary) into memory, retrieve the addresses of functions and variables contained in the library, execute those functions or access those variables, and unload the library from memory. In OOPs Dynamic Binding refers to linking a procedure call to the code that will be executed only at run time. The code associated with the procedure in not known until the program is executed, which is also known as late binding.

Q.15

Which of the following correctly describes overloading of functions?

Virtual polymorphism

Transient polymorphism

Ad-hoc polymorphism

Pseudo polymorphism

Ans .

Explanation :

Ad-hoc polymorphism allows functions with the same name act differently for each type In programming languages, ad hoc polymorphism is a kind of polymorphism in which polymorphic functions can be applied to arguments of different types, because a polymorphic function can denote a number of distinct and potentially heterogeneous implementations depending on the type of argument(s) to which it is applied.

Q.16

Match the following with respect to programming languages:
List-I            				List-II
a. Structured Language	   			i. JAVA
b. Non-structured Language	    		ii. BASIC
c. Object oriented Programming Language   	iii. PASCAL
d. Interpreted Programming Language   		iv. FORTRAN

Codes: a b c d

iii iv i ii

iv iii ii i

ii iv i iii

ii iii iv i

Ans .

Explanation :Since JAVA is Object Oriented language

Q.17

The compiler converts all operands upto the type of the largest operand is called

Type Promotion

Type Evaluation

Type Conversion

Type Declaration

Ans .

Explanation :

Q.18

C++ actually support the following two complete dynamic system

one defined by C++ and the other not defined by C.

one defined by C and one specific to C++.

Both are specific to C++

Both of them are improvements of C.

Ans .

Correct Answer is none

Explanation :

Q.19

Important advantage of using new and delete operation in C++ is

Allocation of memory

Frees the memory previously allocated

Initialization of memory easily

Allocation of memory and frees the memory previously allocated.

Ans .

Explanation :

Important advantage of using new and delete operators in C++ is we can allocate and deallocate memory at any time.

Q.20

Match the following control strategies of prolog:
List-I            	List-II 
a. Forward movement	i. Variable can be done with a constant, another variable or a function
b. Unification   	ii. The entire conjunctive goal is executed.
c. Deep backtracking   	iii. Previous sub goal to find alternative solutions.
d. Shallow backtracking	iv. Choose sub goal with possible unifier.

Codes: a b c d

iv i ii iii

ii iv i iii

iii i iv ii

ii iii iv i

Ans .

Explanation :

a.Forward movement ---->iv. Choose sub goal with possible unifier.
b. Unification ---->i. Variable can be done with a constant, another variable or a function
c. Deep backtracking ---->ii. The entire conjunctive goal is executed.
d. Shallow backtracking ---->iii. Previous sub goal to find alternative solutions.

Q.21

Given the following statements S1 : The grammars S-->asb | bsa | ss | a and S-->asb|bsa|a are not equivalent. S2: The grammars S-->ss|sss|asb|bsa|e and S-->ss|asb|bsa|e are equivalent. Which one of the following is true ?

S1 is correct and S2 is not correct

Both S1 and S2 are correct

S1 is not correct and s2 is correct

Both S1 and S2 are not Correct

Ans .

Explanation :
By Expanding both the Grammars you can verify the answers.

Q.22

What are the final value of Q1 and Q0 after 4 clock cycles, if initial values are 00 in the sequential circuit shown below

Ans .

Explanation :

 T flip flop toggles the state if input T is 1, and does not change state if input T is 0.
T flip flop is a clocked (synchronous or edge-triggered) flip flop.
Edge triggered flip flops can only change its values at the edge of a clock, in absence of edge does not change its state, remains same.
In given circuit, clock symbol is showing that it is positive edge triggering (low to high change). Note that there are mainly four different types of clock-triggering methods. 
Now, we need to find out Q0Q1 after 4 clock cycles. 
Initial: 00 States Q0 and Q1 are zero 
Clock 1: 11 State Q0 changed to 1 because input T is 1 and previous state was 0 and State Q1 changed to 1 because input T is 1 and previous state was 0. 
Clock 2: 01 State Q0 changed to 0 because input T is 1 and previous state was 1 and State Q1 remains same because clk value is 0. 
Clock 3: 10 State Q0 changed to 1 because input T is 1 and previous state was 0 and State Q1 changed to 0 because input T is 1 and previous state was 1. 
Clock 4: 00 State Q0 changed to 0 because input T is 1 and previous state was 1 and State Q1 remains same because clk value is 0.

Q.23

High level knowledge which relates to the use of sentences in different contexts and how the context affect the meaning of the sentences?

Morphological

Syntactic

Semantic

Pragmatic

Ans .

Explanation :

A process in which the speaker implies and a listener infers. Simply Pragmatics studies language that is not directly spoken. The speaker hints at or suggests a meaning , and the listener assume the correct intention. Pragmatics study how words can be interpreted in different ways based on the situation.

Q.24

The objective of ___________ procedure is to discover at least one ________ that causes two literals to match

unification, validation

unification, substitution

substitution, unification

minimax, maximax

Ans .

Explanation :

The objective of unification procedure is to discover at least one substitution that causes two literals to match

Q.25

If h* represents an estimate of the cost of getting from the current node N to the goal and h represents actual cost of getting from current node to the goal node, then A* algorithm gives an optimal solution if

h* us equal to h

h* overestimates h

h* underestimates h

none of these

Ans .

Explanation :

h* underestimates h . Rule is an algorithm gives an optimal solution if h* never overestimates h (admissible heuristics so b is out underestimating h means algorithm is looking for better solution which may be found if it is there(optimal solution)

Q.26

The mean-end analysis process centers around the detection of difference between the current state and the goal state. Once such a difference is isolated, an operator that can reduce the difference must be found. But perhaps that operator can not be applied to the current state. So a sub-problem of getting to a state in which it can be applied is set up. The kind of backward chaining in which operators are selected and then sub goals are set up to established the precondition of operators is called

backward planning

goal stack planning

operator subgoaling

operator overloading

Ans .

Explanation :

Most of the search strategies either reason forward of backward however, often a mixture o the two directions is appropriate. 
Such mixed strategy would make it possible to solve the major parts of problem first and solve the smaller problems the arise when combining them together. 
Such a technique is called "Means - Ends Analysis".

Operator subgoaling is a computer problem solving method in which the inability of the computer to take the desire  next step at any point in the problem solving process leads to a subtotal of making step feasible
Steps:
Computing the current state s1 to a goal state s2 and computing their difference D12.
Satisfy the preconditions for some recommended operator op is selected, then to reduce the difference D12.
The operator OP is applied if possible. If not the current state is solved a goal is created and means- ends analysis is applied recursively to reduce the sub goal.

Q.27

In alpha-beta pruning, ______ is used to cut off the search at maximizing level only and _______ is used to cut off the search at minimizing level only.

alpha,beta

beta,alpha

alpha,alpha

beta,beta

Ans .

Explanation :

Alpha-Beta pruning is not actually a new algorithm, rather an optimization technique for minimax algorithm. It reduces the computation time by a huge factor. This allows us to search much faster and even go into deeper levels in the game tree. It cuts off branches in the game tree which need not be searched because there already exists a better move available. It is called Alpha-Beta pruning because it passes 2 extra parameters in the minimax function, namely alpha and beta.

Q.28

If A and B are two Fuzzy Sets with membership Functions

Then the value of μ_{A ∩ B} will be

{0.2,0.5,0.6,0.7,0.9}

{0.2,0.5,0.2,0.1,0.8}

{0.1,0.5,0.6,0.1,0.8}

{0.1,0.5,0.2,0.1,0.8}

Ans .

Explanation :

The fuzzy intersection of two fuzzy sets A and B on universe of discourse X: μ_A∩B(x) = min [μ_A(x), μ_B(x)] , where x belongs to X The minimum of 0.2 and 0.1 will be 0.1, The second value is min(0.5,0.5)=0.5 The third value is min(0.6,0.2)=0.2 The fourth value is min(0.1,0.7)=0.1 The last value is min(0.9,0.8)=0.8 So the answer is option D.

Q.29

The height h(a) of a Fuzzy set is defined as the the fuzzy set A is called normal when

h(a)=0

h(a)<0

h(a)=1

h(a)<1

Ans .

Explanation :

when h(A)=1 that is height or membership function is 1 fuzzy set become normal or crisp set
choice A means element does not belong to set so neither a fuzzy nor normal
choice B is not defined (noon-meaningful) and so is choice D both can not be same though so some misprint if it is >0 then it will be only fuzzy set and not normal

Q.30

Ans .

Explanation :

ans is D (some slight change as sigma should come only one time) weighted sum is calculated by multiplying inputs with their respective weights and then adding them all which can be shown as Σ wi . xi

Q.31

Data Compression

Data Redundancy

Data Representation

Ans .

Explanation :

Data compression ratio is defined as the ratio between the uncompressed size and compressed size Thus a representation that compresses a 10 MB file to 2 MB has a compression ratio of 10/2 = 5, often notated as an explicit ratio, 5:1 (read "five" to "one"), or as an implicit ratio, 5/1.
Note that this formulation applies equally for compression, where the uncompressed size is that of the original; and for decompression, where the uncompressed size is that of the reproduction.
Sometimes the space savings is given instead, which is defined as the reduction in size relative to the uncompressed size: Thus a representation that compresses a 10MB file to 2MB would yield a space savings of 1 - 2/10 = 0.8, often notated as a percentage, 80%.

Q.32

Find the false statement:

In Modern Cryptography symmetric key algorithms use same key both for Encryption and Decryption.

The Symmetric cipher DES(Data Encryption Standard) was widely used in the industry for security product.

The AES(Advanced Encryption Standard) cryptosystem allows variable Key lengths of size 56 bits and 124 bits.

Public key algorithms use two different keys for Encryption and Decryption.

Ans .

Explanation :

The AES (Advanced Encryption Standard) cryptosystem allows variable key lengths of size 128 bits , 192 bits and 256 bits . 56 is used by DES.

Q.33

The message 11001001 is to be transmitted using the CRC Polynomial x³+1 to protect it from errors. The message that should be Transmitted is

110010011001

11001001

110010011001001

11001001011

Ans .

Explanation :

As given polynomial is x³+1 =>1001
with this directly we can say option D because we add 3 digits more for the data and this satisfied only by option D

But general solution could be

1001) 11001001000(1100011 
       1001 
     ---------------- 
       01011 
        1001 
     ---------------- 
        00100 
         0000 
     ---------------- 
         01000 
          1001 
     ---------------- 
          0001100 
             1001 
     ---------------- 
             01010 
              1001 
     ---------------- 
              0011 

so the message transmitted is 11001001011 

Option D

Q.34

_____________ comparisons are necessary in the worst case to find both the maximum and minimum of n numbers

2n-2

n + floor(lg n) - 2

floor(3n/2) - 2

2 lg n - 2

Ans .

Explanation :

minimum number of comparison require to find minimum and maximum is: Approach is divide and conquer ....
  T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2 
  T(2) = 1 // if two element then compare both and return max and min
  T(1) = 0 // if one element then return both max and min same
  If n is a power of 2, then we can write T(n) as:

   T(n) = 2T(n/2) + 2 
  After solving above recursion, we get

  T(n)  = 3/2n -2 
 Thus, the approach does 3/2n−2 comparisons if n is a power of 2. And it does more than 3/2n−2 comparisons if n is not a power of 2.

Q.35

Let A and B be two n x n matrices. The efficient algorithm to Multiply the two matrices has the time complexit

O(n³)

O(n^2.81)

O(n^2.67)

O(n²)

Ans .

Explanation :

The efficient algorithm to multiply the two matrices has the time complexity O(n^2.81) using Strassen’s algorithm.

Q.36

The recurrence relation T(n)=mT(n/2)tan2 is satisfied by

O(n²)

O(n^{lg m})

O(n² lg n)

O(n lg n)

Ans .

Explanation :


Now according to Master Theorem a=m , b=2 and f(n)=an2 
n^{log m} > f(n) for all m>4    [Using case-1:  Master Theorem]
so solution is O(n^{lg m}) 
Ans : B
Note :  inplace of

Q.37

The longest common subsequence of the sequence X= and Y= has length

Ans .

Explanation :

4 is ans i.e. BDAB , BCAB , BCBA .

Q.38

Assuming there are n keys and each keys is in the range [0,m-1] The run time of bucket sort is

O(n)

O(n lgn)

O(n lgm)

O(n+m)

Ans .

Explanation :

Best and Avg. case= O(n+m) worst case= O(n^2)

Q.39

A __________ complete subgraph and a________ subset of vertices of a graph G=(V,E) are a clique and a vertex cover respectively.

minimal, maximal

minimal, minimal

maximal, maximal

maximal, minimal

Ans .

Explanation :

Clique- A clique is a subset of vertices of a graph such that every two distinct vertices are adjacent( Every two vertices adjacent means a complete subgraph). Minimal Vertex Cover- A minimal vertex cover of a graph is set of minimum vertices such that each edge of the graph is incident to at least one vertex of set. Ans - Maximal , Minimal

Q.40

<=m,<=1

<=m,>=1

>=m,<=1

>=m,>=1

Ans .

Explanation :

Let L be an infinite context-free language. Then there is some positive integer m such that, if S is a string of L of length at least m, then
S = uvwxy (for some u, v, w, x, y)
|vwx| <= m
|vx|  1>=
uvⁱwxⁱy is a member of L
for all nonnegative values of i.

B is ans.

Q.41

The Greibach normal form grammar for the language L = { aⁿ bⁿ⁺¹| n>0} is

S-->a SB, B-->bB|λ

S-->a SB, B-->bB|b

S-->a SB|b, B-->b

S-->aSb |b

Ans .

Explanation :

(A)S→aSB , B→bB ∣ λ
From this grammar we can't able to generate any string because there is no terminal string present for S.hence we can eliminate this option.
(B)S→aSB , B→bB ∣ b
From this grammar we can't able to generate any string because there is no terminal string present for S.hence we can eliminate this option.
(C)S→aSB ∣ b , B→b
From this we can generate all required string (b,abb,aabbb.....).
(D)S→aSb ∣ b
From this we can generate all required string (b,abb,aabbb.....) but question asking for GNF and this grammar is not GNF.

Hence,Option(C)S→aSB ∣ b , B→b is the correct choice.

Q.42

Given the following statements: S1: Every context-sensitive language L is recursive. S2: There exists a recursive language that is not context sensitive. Which statement is correct?

S1 is not correct and S2 is not correct.

S1 is not correct and S2 is correct.

S1 is correct and S2 is not correct.

S1 is correct and S2 is correct

Ans .

Explanation :

Every context sensitive language is recursive

There is a recursive language that is not context sensitive

Q.43

What is the bit rate for transmitting uncompressed 800 x 600 pixel color frames with 8 bits/pixel at 40 frames/second

2.4 Mbps

15.36 Mbps

153.6 Mbps

1536 Mbps

Ans .

Explanation :

Uncompressed pixel = 800*600
each of 8 bit = 800*600 *8
Bit rate for transmitting = 800*600 *8 *40 =153600000 = 153.6Mbps
C is ans

Q.44

In IPv 4, the IP address 200.200.200.200 belongs to

Class A

Class B

Class C

Class D

Ans .

Explanation :

200.200.200.200 belongs to class C [192-223].

Q.45

Which layer of OSI reference model is responsible for decomposition of message and generation of sequence numbers to ensure correct re-composition from end to end the network?

Physical

Data-link

Transport

Application

Ans .

Explanation :

Transport layer is the fourth layer in OSI Reference Model. It responds to service requests from Session layer and issues service requests to Network layer. Transport Layer is responsible for delivering messages between hosts. In Transport layer, data travels in the form of segments. Transport layer is responsible for creating an end to end connection between source IP and the destination IP. For establishing this end to end connection, Transport layer is using two major protocols TCP and UDP. TCP is connection oriented protocol where UDP is connection less protocol. So by using TCP we can create an end to end reliable connection between source and destination hosts. Transport layer is called host-to-host transport layer in TCP/IP model.

Q.46

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in responses to a request? (Note that the speed of light in air is 3,00,000 km/second).

133.33 m sec

266.67 m sec

400.00 m sec

533.33 m sec

Ans .

Explanation :

Q.47

The start and stop bits are used in serial communication for

error detection

error correction

synchronization

slowing down the communication

Ans .

Explanation :

The start and stop bits are used in serial communication for synchronization i.e. to tell the receiver data is come after me ready to accept it , and data is finish after me.

Q.48

______________ is a type of transmission impairment in which the signal looses strength due to the resistance of the transmission medium

Attenuation

Distortion

Noise

Decibel

Ans .

Explanation :

Attenuation is a general term that refers to any reduction in the strength of a signal. Attenuation occurs with any type of signal, whether digital or analog. Sometimes called loss, attenuation is a natural consequence of signal transmission over long distances.
Distortion, in acoustics and electronics, any change in a signal that alters the basic waveform or the relationship between various frequency components; it is usually a degradation of the signal. Straight amplification or attenuation without alteration of the waveform is not usually considered to be distortion.
Noise is unwanted electrical or electromagnetic energy that degrades the quality of signals and data. Noise occurs in digital and analog systems, and can affect files and communications of all types, including text, programs, images, audio, and telemetry.
The decibel (dB) is a logarithmic unit used to express the ratio of two values of a physical quantity, often power or intensity. One of these values is often a standard reference value, in which case the decibel is used to express the level of the other value relative to this reference

Q.49

Match the following:
List-I              		List-II
a. Indexed Addressing   	i. is not used when an operand is moved from memory into a register or from a register to memory.
b. Direct Addressing   		ii. Memory address is computed by adding up two registers plus an (optional) offset.
c. Register Addressing	   	iii. Addressing memory by giving a register plus a content offset.
d. Base-Indexed Addressing	iv. can only be used to access global variables whose address is known at compile time.

Codes: a b c D

ii i iv iii

ii iv i iii

iii iv i ii

iii i iv ii

Ans .

Explanation :

a. Indexed Addressing   	iii. Addressing memory by giving a register plus a content offset.
b. Direct Addressing   		iv. can only be used to access global variables whose address is known at compile time.
c. Register Addressing	   	i. is not used when an operand is moved from memory into a register or from a register to memory.
d. Base-Indexed Addressing	ii. Memory address is computed by adding up two registers plus an (optional) offset.

Q.50

Which of the following is a design criteria for instruction formats?

The size of instructions

The numbers of bits in the address fields

The sufficient space in the instruction format to express all the operations desired.

All of these

Ans .

Explanation :

Design Criteria for Instruction Formats (How to outlast the current processor.)
1. Size of the Instructions balance instruction size vs. required capabilities, addressing, and inst. fetch times typically word length or multiple word lengths 1a. Consider the instruction fetch and instruction building required multiple word fetches require more time … leading to fewer instructions per second (average memory access time and I-cache make a difference)
2. Include enough bits to define “ALL” operations allow for future growth provide some “unassigned” opcodes
3. Determine any address range or size limitations a 32-bit address by byte accesses only 4GB of space the number of RISC registers defines the address field size (32 reg - 5 bits each)

Q.51

Synchronization is achieved by a timing device called a ________ which generates a periodic train of _________.

clock generator, clock pulse

master generator, clock pulse

generator, clock

master clock generator, clock pulse

Ans .

Explanation :

Q.52

Serial access memories are useful in applications where

Data consists of numbers

Short access time is required.

Each stored word is processed differently.

None of these

Ans .

Explanation :

serial access memories are used where data naturally needs to flow in and out in serial form

Q.53

What will be the output of the following logic diagram

x OR y

x AND y

x NOR y

x XNOR y

Ans .

Explanation :

Q.54

The essential difference between traps and interrupt is

traps are asynchronous and interrupt are synchronous with the program

traps are synchronous and interrupt are asynchronous with the program

traps are synchronous and interrupt are asynchronous with the I/O devices.

None of these

Ans .

Explanation :

A trap is an exception in a user process. It's caused by division by zero or invalid memory access. It's also the usual way to invoke a kernel routine (a system call) because those run with a higher priority than user code. Handling is synchronous (so the user code is suspended and continues afterwards). In a sense they are "active" - most of the time, the code expects the trap to happen and relies on this fact.
An interrupt is something generated by the hardware (devices like the hard disk, graphics card, I/O ports, etc). These are asynchronous (i.e. they don't happen at predictable places in the user code) or "passive" since the interrupt handler has to wait for them to happen eventually.

Q.55

Consider the following ER diagram

The minimum number of tables required to represents M,N,P,R1,R2 is

Ans .

Explanation :

First strong entity types are made to tables. So, we get two tables M and P.
I assume R1 is 1:1 or 1:n as that would minimize the number of tables as asked in question. 
Now participation of M in R1 is total (indicated by double arrow) meaning every entity of M participate in R1. Since R1 is not having an attribute, we can simple add the primary key of P to the table M and add a foreign key reference to M. This handles R1 and we don't need an extra table. So, M becomes {M1, M2, M3, P1}. 
N here is a weak entity weakly related to P. So, we form a new table N, and includes the primary key of P (P1) as foreign key reference. Now (P1, N1) becomes the primary key of N. 
Thus we get 3 tables.
M: {M1, M2, M3, P1} - M1 primary key, P1 references P
P: {P1, P2} - P1 primary key
N: {P1, N1, N2} - (P1, N1) primary key, P1 references P. 
So, answers is A.

Q.56

Consider the following schemas: Branch = (Branch_name, Assets, Branch_city) Customer = (Customer-name, Bank-name,Customer-city) Borrow = (Branch-name, loan-number, customer-name, Balance) Deposit = (Branch-name, Account-number, Customer-name, Balance) Using relational Algebra, the query that finds customers who have balance More than 10,000 is ________

πcustomer-name(σbalance > 10000(Deposit))

σcustomer-name(σbalance > 10000(Deposit))

πcustomer-name(σbalance > 10000(Borrow))

σcustomer-name(πbalance > 10000(Deposit))

Ans .

Explanation :

πcustomer-name(σbalance >10000(Deposit)) we have to apply projection and use table Deposite. A is ans

Q.57

Find the false statement:

The relationship construct known as the weak relationship type was defined by Dey, Storey & Barron (1999).

A weak relationship occurs when two relationship types are linked by either Event-Precedent sequence or Condition-Precedent sequence

Conceptual model is not accurate representation of "Universe of interest".

Ternary, Quaternary and Quintary relationships are shown through a series of application scenario's and vignette's

Ans .

Explanation :

option c
reference
all others are true

Q.58

Consider the table Student(stuid, name, course, marks).

Which one of the following two queries is correct to find the highest marks student in course 5?
Q.1. Select S.stuid From student S Where not exists (select * from student e where e course='5' and e marks >= s marks)
Q.2. Select S.stu.id From student S Where s.marks > any (select distinct marks from student S where s.couse = 5)

Q.1

Q.2

Both Q.1 and Q.2

Neither Q.1 nor Q.2

Ans .

Explanation :

1. Select S.stuid from student S where not exist (select * from student e where e course ='5' and e marks ≥ s marks) = false
Select S.stuid from student S where not exist (select * from student e where e course ='5' and e marks > s marks) = correct
Q.2. select S.stuid From student S where s.marks > any (select distinct marks from student S where s.course =5) = false
select S.stuid From student S where s.marks >All (select distinct marks from student S where s.course =5) = correct

Q.59

Armstrong (1974) proposed systematic approach to derive functional dependencies. Match the following w.r.t. functional dependencies :

List I 				List II
a. Decomposition rule		i. If X→Y and Z→W then {X,Z}→{Y,W}
b. Union rule			ii. If X→Y and {Y,W}→Z then {X,W}→Z
c. Composition rule    		iii. If X→Y and X→Z then X→{Y,Z}
d. Pseudo transitivity rule	iv. If X→{Y,Z} then X→Y and X→Z

Codes: (a) (b) (c) (d)

iii ii iv i

i iii iv ii

ii i iii iv

iv iii i ii

Ans .

Explanation :

Decomposition rule : If x->yz then x->y , x->z
Union rule : If x->y , x->z then x->yz
Composition rule: if x->y and z->w then xz-> yw
Pseudo transitivity rule : If x-> y and yw->z then xw-> z [ y replace by x].
X→Z

Q.60

Match the following

List I 				List II
(a) Secondary Index     	(i) Functional dependency
(b) Non-procedural Query	(ii) B-tree
(c) Closure set attribute	(iii) Relational Algebraic Operation
(d) Natural Join     		(iv) Domain Calculus

Codes: (a) (b) (c) (d)

i ii iv iii

ii i iv iii

i iii iv ii

ii iv i iii

Ans .

Explanation :

 It is easy to observe that B-tree has secondary index 
Natural join is Relational Algebraic Operation
Closure of set of attributes is related to Functional Dependency
Non-procedural Query language is related to  Domain Calculus
Hence ans is D

Primary index:
A primary index is an index on a set of fields that includes the unique primary key for the field and is guaranteed not to contain duplicates. 
eg. Employee ID can be Example of it.

Secondary index:
A Secondary index is an index that is not a primary index and may have duplicates. eg. Employee name can be example of it. 
Because Employee name can have similar values.

Q.61

Which of the following is not true with respect to a trackball and/or spaceball I. A trackball is a two dimensional positioning device while as a spaceball provides six degree of freedom II. Unlike the trackball a spaceball does not actually move. III. A trackball is a three dimensional positioning device while as a spaceball provides six degree of freedom.

I & II

II & III

II only

III only

Ans .

Explanation :

option D is the ans as statement III is incorrect
First two statements are correct
A spaceball is a "six degrees of freedom" device. It allows full rotation in multiple directions, such as with a CAD model. a trackball, however is merely a standard mouse. rather than having the ball moved by the mouse platform, you manually actuate the ball. it just moves the cursor around. the libsball page has a nice image showing the spaceball's range and type of motion. a trackball just does XY axis motion.

Q.62

Which of the following statement(s) is(are) true? I. Two successive translations are additive. II. Two successive rotations are additive. III. Two successive scaling operations multiplicative.

I and II

I and III

II and III

All of above.

Ans .

Explanation :

Q.63

Given below are three basic rules: I. Squash and Stretch II. Slow-in and Slow out III. To stage the action properly These rules are applied in case of

Rendering

Morphing

Animation

All of above

Ans .

Explanation :

ans is C animation refer

Q.64

Which of the following points lies on the same side as the origin, with reference to the line 3x+7y=2 ?

(3,0)

(1,0)

(0.5,0.5)

(0.5,0)

Ans .

Explanation :
Put the value of x and y in the equation 3x+7y if 3x+7y >2 then it lies on the same side of origin else if 3x+7y > 2 it does not lies on the same side of origin

Q.65

The transformation matrix required for conversion of CMY color model to RGB color model is given as

Ans .

Explanation :

Q.66

What step shall be required to rotate an object about the point P1 and its placement such that what was at P1 is now reduced and is at P2. I. Translate P1 to origin II. Scale as required III. Rotate IV. Translate to the final position P2.

I,II and III

II,III and IV

I,III and IV

All of the above

Ans .

Explanation :

Q.67

In Unix, how do you check that two given strings a and b are equal

test $a -eq $b

test $a -equal $b

test $a=$b

sh -C test $a==$b

Ans .

Explanation :

Q.68

In windows 2000 operating system all the processor-dependent code is isolated in a dynamic link library called

NTFS file system

Hardware Abstraction layer

Microkernel

Process Manager

Ans .

Explanation :

Q.69

To place a sound into a word document, following feature of windows is used

Clip board

Task switching

C Win App

OLE

Ans .

Explanation :

OLE is used to place a sound into a word document

Q.70

Translation Look-aside Buffer(TLB) is

a cache-memory in which item to be searched is compared one-by-one with the keys.

a cache-memory in which item to be searched is compared with all the keys simultaneously.

an associative memory in which item to be searched is compared one-by-one with the keys.

an associative memory in which item to be searched is compared with all the keys simultaneously.

Ans .

Explanation :

Translation Look-aside Buffer (TLB) is an associative memory in which item to be searched is compared with all the keys simultaneously.When the associative memory is presented with an item, the item is compared with all keys simultaneously. If the item is found, the corresponding value field is returned.

Q.71

Simplest way of deadlock recovery is

Roll back

Preempt resource

Lock one of these processes

Kill one of the processes

Ans .

Explanation :

Recovery through killing processes is the Crudest but simplest way to break a deadlock.
– Kill one of the processes in the deadlock cycle: the other processes get its resources.
– Choose process that can be rerun from the beginning

Q.72

The directory structure used in Unix file system is called

Hierarchical directory

Tree structure directory

Directed acyclic graph

Graph structured directory

Ans .

Explanation :

Q.73

Which statement is not true about process 0 in the Unix operating system?

Process 0 is called init process.

Process 0 is not created by fork system call.

After forking process 1, process 0 becomes swapper process.

Process 0 is a special process created when system boots.

Ans .

Explanation :

Q.74

Which one of the following commands would return process_id of sleep command?

Sleep 1 and echo $?

Sleep 1 and echo $#

Sleep 1 and echo $x

Sleep 1 and echo $!

Ans .

Explanation :

Q.75

Possible thread states in Windows 2000 operating system include:

Ready,running and waiting

Ready, standby, running, waiting, transition and terminated.

Ready, running, waiting, transition and terminated

Standby, running, transition and terminated.

Ans .

Explanation :

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