Ans .

A

1. Explanation :

The resulting network is a local area network that has been extended.

Ans .

C

1. Explanation :

The extended LAN transfers frames, and so it still belongs in the data link layer

Ans .

C

1. Explanation :

In an infrastructured network, a cell is known as a BSA. It is the area containing a number of mobile hosts supported by a base station.

Ans .

B

1. Explanation :

Ans .

A

1. Explanation :

The classifications defined by Flynn are based upon the number of concurrent instruction (or control) streams and data streams available in the architecture

Ans .

C

1. Explanation :

Solved by master's theorem case 1.

Ans .

B

1. Explanation :

1-3,3-2,2-5,3-6,4-3

Ans .

C

1. Explanation :

both are the worst cases of quick sort. (assuming pivot is either first or last element) (1) is sorted in ascending order. (2) is sorted in descending order.

Ans .

C

1. Explanation :

Either using DFT or BFT :: O(V+E)

Ans .

C

1. Explanation :

In the first pass we can create [105/5]=21 sorted sub-files,each 5 pages long. In the second pass by 4-way merging (4 because one of the 5 available buffers has to be reserved for holding the output), We can create [21/4]=6 sorted sub-files,each 20 pages long (except the last). Again applying 4-way merge sort,we get create [6/4]=2 sorted sub-files. These two can be merged to get the final sorted file. We need a total of 4 passes. Total cost will be 2 x 105 x 4=840 units. Hence option C

Ans .

A

1. Explanation :

DFD : Data flow diagram software FSM : Field Service Management Software ERP : Enterprise Resource Planning Software Petri net : A Petri net, also known as a place/transition (PT) net, is one of several mathematical modeling languages for the description of distributed systems. A Petri net is a directed bipartite graph, in which the nodes represent transitions (i.e. events that may occur, represented by bars) and places (i.e. conditions, represented by circles). Data Dictionary : The terms data dictionary and data repository indicate a more general software utility.

Ans .

D

1. Explanation :

Variant of waterfall model in Software Engineering is incremental method.

Ans .

B

1. Explanation :

In software engineering, the fundamental part of agile approach is basically used by incremental model.

Ans .

A

1. Explanation :

In Software Engineering, spiral model have specialty of having combination in change avoidance with change tolerance.

Ans .

D

1. Explanation :

Ans .

C

1. Explanation :

Exciting, expected and normal are requiremnet classification used in QFD.

Ans .

B

1. Explanation :

Ans .

C

1. Explanation :

SRS document should be unambiguous

Ans .

A

1. Explanation :

Except option “A” all other tasks/activities are present in Spiral Model as well.

Ans .

B

1. Explanation :

Dataflow diagram comes under Logical View of the system. A data flow diagram (DFD) is a graphical representation of the "flow" of data through an information system, modelling its process aspects. A DFD is often used as a preliminary step to create an overview of the system without going into great detail, which can later be elaborated.

Ans .

A

1. Explanation :

Maintenance and Evaluation is the phase where all upgrade and bug fixed reported is implemented on the system.

Ans .

B

1. Explanation :

In Review phase the quality and performance of a system is measured and commented on.

Ans .

C

1. Explanation :

Requirements determination is the beginning sub phase of analysis. In this sub phase, analysts should gather information on what the system should do from as many sources as possible.

Ans .

D

1. Explanation :

Basically all the stake holders who are involved in the development

Ans .

C

1. Explanation :

Self Exploratory

Ans .

A

1. Explanation :

Self Exploratory

Ans .

A

1. Explanation :

DSS - Analysis, decision Support MIS - Information processing EIS - Status Access

Ans .

C

1. Explanation :

Common mistake: using ∧ as the main connective with ∀

Ans .

D

1. Explanation :

Ans .

A

1. Explanation :

This is minimal finite automata. Since its endings with therefore n+1 states.

Ans .

B

1. Explanation :

We need to find the number of different configurations possible for memory and each of these will be a state in FSM. (At any time memory will be in one configuration and in next instance it either remains same or goes to a different configuration) A word is of n bits. And we have m such words. So, total number of bits = m*n. We need a separate state for each bit combination. So, no. of states = 2^(mn).

Ans .

A

1. Explanation :

X is undecidable but partially decidable. We have the TM M. Just make the state q the final state and make all other final states non-final and get a new TM M'. Give input w to M'. If w would have taken M to state q (yes case of the problem), our new TM M' would accept it. So, the given problem is partially decidable.

Ans .

C

1. Explanation :

L1 - CSL L2 - Palindrome so CFL L3 - Linear Power and regular expression can be stated as (00)∗ L4 - non linear power So CSL

Ans .

B

1. Explanation :

In DFA any subset of the N states (for N element set 2^N subsets possible) can become a new state and they can remain even when the DFA is minimized. So, maximum we can get 2^N states for the minimized DFA. (at least in question must be a typo for at most).

Ans .

A

1. Explanation :

Babel Fish was a free Web-based multilingual translation application.

Ans .

A

1. Explanation :

The minimum criterion takes the least value in the set given. In an AND operation all the values have to be true in order for the whole expression to be true. Hence taking the least of the values ensures that the range of all values are met.

Ans .

A

1. Explanation :

A knowledge base (KB) is a technology used to store complex structured and unstructured information used by a computer system. So it contains Rules, facts and relationships.

Ans .

B

1. Explanation :

Self Explanatory

Ans .

C

1. Explanation :

Let the amount of B1 be x (in kg) Let the amount of B2 be y (in kg) Minimize 5x + 8y Subject To: x + y = 5 (since the weight of the brick must be 5 kg) x ≤ 4 (since there must be no more than 4 kg of B1) y ≥ 2 (since there must be a min. of 2 kg of B2) x ≥ 0

Ans .

C

1. Explanation :

Self Explanatory

Information for question 42 and 43 : A modern IT park has 7 buildings viz. B1,.....,B7 and 9 roads containing them. B1 is centrally located and is connected to all the 6 buildings B2,.....,B7 surrounding it. The pair (B2,B3),(B4,B5) and (B6,B7) are connected. cleaning of the roads in the IT park is done by cleaning truck and the depot of the vehicle is located on the road joining B4 and B5. The truck operator needs some help in planning the cleaning tour.Help him by identifying the correct statements about the tour

Ans .

C

1. Explanation :

Solve the problem using Hamaltonion Path.

Ans .

B

1. Explanation :

Since it does not contain any hamaltonion cycle and B1 is least costly path therefore its obvious it would be visited more.

Ans .

A

1. Explanation :

n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)

Information for question 45 and 46 : A multi-layer feed-forward artificial neural network has 16 neurons in the input layer. The number of neurons in every following layer is half of that in previous layer . There is only one 1 neuron in the output layer.

Ans .

B

1. Explanation :

Non-linear functions are those which have degree more than one and they have a curvature when we plot a Non-Linear function. Now we need a Neural Network Model to learn and represent almost anything and any arbitrary complex function which maps inputs to outputs. Neural-Networks are considered Universal Function Approximators.

Ans .

A

1. Explanation :

Ans .

A

1. Explanation :

Ans .

D

1. Explanation :

When a training data set is not available, a classifier can be designed from prior knowledge and expertise. Once trained, the classifier is ready for operation on unseen objects.

Ans .

B

1. Explanation :

P(B|A)/P(B)

Ans .

B

1. Explanation :

There is no such file as user32.exe user16.exe and user.dll in 32 but windows

Ans .

C

1. Explanation :

\$\$ contains current shell process id.

Ans .

B

1. Explanation :

.o is a simple program in passive state, .h is c++ header file format while .a is unix static library extension

Ans .

B

1. Explanation :

Loader's job is to bring a program from passive state to active by loading it and running it from memory

Ans .

A

1. Explanation :

Ans .

B/C/D

1. Explanation :

The function of an IC (integrated circuit) chip is to replace many separate electronic components which could possibly have been used to build a particular electronic circuit. Most of those separate components are replaced by just one tiny IC chip that has been manufactured ("fabricated" is the correct technical word) to include extremely miniature circuits which imitate the behavior of all those separate components.

Ans .

A

1. Explanation :

Stack is used to store critical data duriing subroutines and interrupt.

Ans .

C

1. Explanation :

CS is Code Segment.

Ans .

B

1. Explanation :

The code segment register is used for addressing a memory location in the code segment of memory, where executable program is stored.

Ans .

D

1. Explanation :

An integrated circuit or monolithic integrated circuit (also referred to as an IC, a chip, or a microchip) is a set of electronic circuits on one small flat piece (or "chip") of semiconductor material, normally silicon.

Ans .

B

1. Explanation :

By atomic , either all the effects of the transaction are reflected in the database, or none are (after rollback).

Ans .

C

1. Explanation :

It is an abstraction through which relationships are treated as higher level entities Aggregation. (In ER diagram, aggregation is used to represent a relationship as an entity set.

Ans .

B

1. Explanation :

IS – Intent to get S lock(s) at finer granularity.

Ans .

B

1. Explanation :

Explanation : Strong entities E1 and E2 are represented as separate tables. In addition to that many-to-many relationships(R2) must be converted as seperate table by having primary keys of E1 and E2 as foreign keys. One-to-many relaionship (R1) must be transferred to ‘many’ side table(i.e. E2) by having primary key of one side(E1) as foreign key( this way we need not to make a seperate table for R1). Let relation schema be E1(a1,a2) and E2( b1,b2). Relation E1( a1 is the key) Relation E2( b1 is the key, a1 is the foreign key, hence R1(one-many) relationship set satisfy here ) Relation R2 ( {a1, b1} combined is the key here , representing many-many relationship R2 )

Ans .

B

1. Explanation :

In domain relational calculus we create variable for every column.

Ans .

C

1. Explanation :

To insert an element, we need to search for its place first. The search operation may take O(n)

Ans .

A

1. Explanation :

HTTP is an application layer protocol. Since firewal is at layer 4, it cannot block HTTP data.

Ans .

A

1. Explanation :

Lexical analysis is the first step in compilation. In lexical analysis, program is divided into tokens. Lexical analyzers are typically based on finite state automata. Tokens can typically be expressed as different regular expressions: An identifier is given by [a-zA-Z][a-zA-Z0-9]* The keyword if is given by if. Integers are given by [+-]?[0-9]+.

Ans .

B

1. Explanation :

Minimum spanning tree for 2 nodes would be (v1) _ (v2) Total weight 3 Minimum spanning tree for 3 nodes would be (v1) _ (v2) | (v3) Total weight= 3 + 4 = 7 Minimum spanning tree for 4 nodes would be (v1) _ (v2) _ (v4) | (v3) Total weight= 3 + 4 + 6 = 13 Minimum spanning tree for 5 nodes would be (v1) _ (v2) _ (v4) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 = 21 Minimum spanning tree for 6 nodes would be (v1) _ (v2) _ (v4) _ (v6) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 + 10 = 31 We can observe from above examples that when we add kth node, the weight of spanning tree increases by 2k-2. Let T(n) be the weight of minimum spanning tree. T(n) can be written as T(n) = T(n-1) + (2n-2) for n > 2 T(1) = 0, T(2) = 0 and T(2) = 3 The recurrence can be written as sum of series (2n – 2) + (2n-4) + (2n-6) + (2n-8) + …. 3 and solution of this recurrence is n^2 – n + 1.

Ans .

D

1. Explanation :

If the graph is planar, then it must follow below Euler’s Formula for planar graphs v - e + f = 2 v is number of vertices e is number of edges f is number of faces including bounded and unbounded 10 - 15 + f = 2 f = 7 There is always one unbounded face, so the number of bounded faces = 6

Ans .

B

1. Explanation :

There can be total 6C4 ways to pick 4 vertices from 6. The value of 6C4 is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45.

Ans .

C

1. Explanation :

Lossy compression

Ans .

B

1. Explanation :

The main idea for using Spanning Trees is to avoid loops.

Ans .

B

1. Explanation :