UGCNET-June2014 - UPSCFEVER

Q.1

Infrared signals can be used for short range communication in a closed area using ................... propagation.

ground

sky

line of sight

space

Ans .

Explanation :
Infrared signals can be used for short range communication in a closed area using line of sight propagation.

Q.2

A bridge has access to ............... address in the same network.

Physical

Network

Data link layer

Application

Ans .

Explanation :
A bridge has access to Physical address in the same network.

Q.3

The minimum frame length for 10 Mbps Ethernet is ............. bytes and maximum is ................ bytes.

64 & 128

128 & 1518

1518 & 3036

64 & 1518

Ans .

Explanation :
The minimum size of an Ethernet frame is 64 bytes. The breakup of this size between the fields is: Destination Address (6 bytes) + Source Address (6 bytes) + Frame Type (2 bytes) + Data (46 bytes) + CRC Checksum (4 bytes). The minimum number of bytes passed as data in a frame must be 46 bytes. If the size of the data to be passed is less than this, then padding bytes are added. The maximum size of an Ethernet frame is 1518 bytes. The breakup of this size between the fields is: Destination Address (6 bytes) + Source Address (6 bytes) + Frame Type (2 bytes) + Data (1500 bytes) + CRC Checksum (4 bytes). The maximum number of bytes of data that can be passed in a single frame is 1500 bytes.

Q.4

The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is ...............

500 baud/sec

1000 baud/sec

3000 baud/sec

18000 baud/sec

Ans .

Explanation :
Baud rate = 3000 / 6 = 500 baud/s

Q.6

The minimum frame length for 10 Mbps Ethernet is ............. bytes and maximum is ................ bytes.

64 & 128

128 & 1518

1518 & 3036

64 & 1518

Ans .

Explanation :
The minimum frame length for 10 Mbps Ethernet is 64 bytes and maximum is 1518 bytes.

Q.7

A grammar G is LL(1) if and only if the following conditions hold for two distinct productions A -> α | β
I. First (α) n First (β) ≠ {a} where a is some terminal symbol of the grammar.

II. First (α) n First (β) ≠ λ

III. First (α) n Follow(A) = Ψ if λ ε First (β)

I and II

I and III

II and III

I, II and III

Ans .

Explanation :
Grammar G is called LL(1) if and only if whenever, If A->α|β are two distinct productions of G, the following conditions hold :- 1. (a) FIRST(α) , FIRST(β) must be disjoint. This is to be able to deterministically guess the production. (b) At most one of the strings α or β can derive ? (Since FIRST(α), FIRST(β) are disjoint. 2. If α-> ε then FIRST(β) and FOLLOW(A) must be disjoint

Q.8

Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar ?

Removing left recursion alone.

Removing the grammar alone

Removing left recursion and factoring the grammar

None of the above

Ans .

Explanation :
Removing left recursion and factoring the grammar do not suffice to convert an arbitrary CFG to LL(1) grammar.

Q.9

A shift reduce parser suffers from..

shift reduce conflict only

reduce reduce conflict only

both shift reduce conflict and reduce reduce conflict

shift handle and reduce handle conflicts

Ans .

Explanation :
A shift-reduce parser scans and parses the input text in one forward pass over the text, without backing up. A shift/reduce parser has two different actions it can take SHIFTING and REDUCING. Shift: the shift action is when the parser takes the first unanalysed word from the sentence, looks up its dictionary rule, and puts a token of category of the word onto the stack. Reduce: the reduce action is when the parser tries to match the topmost category tokens on the stack to the right hand side of the rule. If this succeeds, the tokens are removed and a token for the category on the left hand side of the rule is put on the stack instead. the shift-reduce parser faces four types of conflicts:
1. shift vs. shift;

2. shift vs. reduce left;

3. shift vs. reduce right;

4. reduce left vs. reduce right

Q.10

The context free grammar for language L = {aⁿb^mc^k | k = |n - m|, n=0,m=0,k=0} is

S->S1S3, S1->aS1c |S2|λ, S2->aS2b|λ, S3->aS3b|S4| λ, S4->bS4c|λ

S->S1|S2, S1->aS1S2c|λ, S2->aS2b|λ, S3->aS3b|S4| λ, S4->bS4c|λ

S->S1|S3, S1->aS1c |S2|λ, S2->aS2b|λ, S3->aS3b|S4| λ, S4->bS4c|λ

Ans .

Explanation :
S->S1|S3, S1->aS1c |S2|λ, S2->aS2b|λ, S3->aS3b|S4| λ, S4->bS4c|λ this grammer gives appropriate solution.

Q.11

The regular grammar for the language L = {w |n_a(w) and n_b(w) are both even, w∊{a, b}*} is given by :
(Assume, p, q, r and s are states)

p->aq|br, q->bs|ap r->as|bp, s->ar|bq, p and s are initial and final states

p->aq|br, q->bs|ap r->as|bp, s->ar|bq, p is both initial and final state

Ans .

Explanation :
p->aq|br|λ, q->bs|ap r->as|bp, s->ar|bq,p is both initial and final state gives above solution.

Q.12

The regular grammar for the language L = {w |n_a(w) and n_b(w) are both even, w∊{a, b}*} is given by :
(Assume, p, q, r and s are states)

p->aq|br, q->bs|ap r->as|bp, s->ar|bq, p and s are initial and final states

p->aq|br, q->bs|ap r->as|bp, s->ar|bq, p is both initial and final state

Ans .

Explanation :
p->aq|br|λ, q->bs|ap r->as|bp, s->ar|bq,p is both initial and final state gives above solution.

Q.13

KPA in CMM stands for

Key Process Area

Key Product Area

Key Principal Area

Key Performance Area

Ans .

Explanation :
Key Process Areas : It identifies a cluster of related activities that, when performed collectively, achieve a set of goals considered important.

Q.14

Which one of the following is not a risk management technique for managing the risk due to unrealistic schedules and budgets?

Detailed multi source cost and schedule estimation

Design Cost

Incremental development

Information hiding

Ans .

Explanation :
Risk management is the area that tries to ensure that the impact of risks on cost,multi source cost, quality, and schedule is minimal. Like configuration management which minimizes the impact of change, risk management minimizes the impact of risks. risk in incremental development is considered. Risk management can be considered as dealing with the possibility and actual occurrence of those events that are not ?regular? or commonly expected. Information Hiding is not a Risk Management Technique because Information hiding is the principle of segregation of the design decisions in a computer program that are most likely to change, thus protecting other parts of the program from extensive modification if the design decision is changed.Information hiding is a powerful programming technique because it reduces complexity.

Q.15

................ of a system is the structure or structures of the system which comprise software elements, the externally visible properties of these elements and the relationship amongst them.

Software construction

Software evolution

Software architecture

Software reuse

Ans .

Explanation :
The software architecture of a program or computing system is the structure or structures of the system, which comprise software elements, the externally visible properties of those elements, and the relationships among them. Architecture is concerned with the public side of interfaces; private details of elements?details having to do solely with internal implementation?are not architectural.

Q.16

In function point analysis, the number of complexity adjustment factors is

Ans .

Explanation :
There are 14 Relative Complexity Adjustment Factors. They are 1. Requirement for reliable backup and recovery 2. Requirement for data communication 3. Extent of distributed processing 4. Performance requirements 5. Expected operational environment 6. Extent of online data entries 7 Extent of multi-screen or multi-operation online data input. 8. Extent of online updating of master files 9. Extent of complex inputs, outputs, online queries and files 10. Extent of complex data processing 11. Extent that currently developed code can be designed for reuse 12. Extent of conversion and installation included in the design 13. Extent of multiple installations in an organization and variety of customer organizations 14. Extent of change and focus on ease of use

Q.17

Regression testing is primarily related to

Functional testing

Development testing

Data flow testing

Maintenance testing

Ans .

Explanation :
Regression testing is primarily a maintenance activity. The purpose of regression testing is to validate the modified software and detect whether the unmodified code is added.

Q.18

How many different truth tables of the compound propositions are there that involve the propositions p & q ?

Ans .

Explanation :
2^4 = 16

Q.19

A Boolean function F is called self-dual if and only if F(x1,x2,x3,...,xn) = F(x1',x2',x3',...,xn') How many Boolean functions of degree n are self-dual ?

2ⁿ

(2)^{2^n}

(2)^{n^2}

(2)^{2^n-1}

Ans .

Explanation :
A function F is called self dual if it has equal number of minterms and maxterms, also mutually exclusive terms should not be included. the number of mutually exclusive terms is (pair wise) is 2ⁿ/2 = 2^n-1. Number of functions possible by taking any of the one term from above mentioned mutually exclusive pair is = 2^2^n-1

Q.20

Which of the following statement(s) is(are) not correct ?
i. The 2's complement of 0 is 0.

ii. In 2's complement, the left most bit cannot be used to express a quantity.

iii. For an n-bit word (2's complement) which includes the sign bit, there are 2n-1 positive integers, 2n+1 negative integers and one 0 for a total of 2n unique states.

iv. In 2's complement the significant information is contained in the 1's of positive numbers and 0's of the negative numbers.

i and iv

i and ii

iii

Ans .

Explanation :
For an n-bit word (2's complement) which includes the sign bit, there are 2^n-1 - 1 positive integers, 2^{n -1} negative integers and one 0 for a total of 2n unique states. so option iii is wrong hence answer C

Q.21

The notarion Ε!xp(x) denotes the proposition "there exists a unique x such that P(x) is true". Give the truth values of the following statements :
I. Ε!xP(x) -> ΕxP(x)

II. Ε!x Γ P(x) -> Γ ∀xp(x)

Both I and II are true

Both I and II are false

I-false, II-true

I-true, II-false

Ans .

Explanation :
If we put Ε!xp(x) denoting "there exists a unique x such that P(x) is true" we can get both options correct.

Q.22

Give a compound proposition involving propositions p, q and r that is true when exactly two of p, q and r are true and is false otherwise.

(p∨q∧ ¬r) ∧ (p∧ ¬q∧ r) ∧ (¬p∧ q∧ r)

(p∧ q∧ ¬r) ∧ (pp∨q∧ ¬r) ∧ (¬p∧q∧r)

(p∧q∧¬r) p∨ (p∧¬q∧r) ∧ (¬p∧q∧r)

(p∨q∨¬r) ∧ (p∨¬q∨r) ∧ (¬p∨q∨r)

Ans .

Explanation :

Q.23

The upper bound and lower bound for the number of leaves in a B-Tree of degree K with height h is given by:

K^h and 2[K/2]^h-1

K*h and 2[K/2]^h-1

K^h and 2[K/2]^h-1

K*h and 2[K/2]^h-1

Ans .

Explanation :
The upper bound and lower bound for the number of leaves in a B-Tree of degree K with height h is K^h and 2[K/2]^h-1

Q.24

Consider a complete bipartite graph km,n. For which values of m and n does this, complete graph have a Hamilton circuit

m=3, n=2

m=2, n=3

m=n ≥ 2

m=n≥ 3

Ans .

Explanation :
m=n≥ 2 If G is connected simple graph with n vertices where n ≥ 3, then G has a Hamilton circuit if the degree of each vertex is at least n/2.

Q.25

Big-O estimates for the factorial function and the logarithm of the factorial function i.e. n! and log n! is given by

O(n!) and O(n log n)

O(nⁿ) and O(n log n)

O(n!) and O(log n!)

O(nⁿ) and O(log n!)

Ans .

Explanation :
B is appropriate solution.

Q.26

How many cards must be chosen from a deck to guarantee that atleast i. two aces of two kinds are chosen. ii. two aces are chosen. iii. two cards of the same kind are chosen. iv. two cards of two different kinds are chosen

50, 50, 14, 5

51, 51, 15, 7

52, 52, 14, 5

51, 51, 14, 5

Ans .

Explanation :
Since 48 cards of a 52-card deck are not aces, you are guaranteed to have at least two aces only if you have at least 50 cards.

Q.27

Match the following with respect to the mobile computing technologies:

List-I	List-II
a. GPRS	i. An integrated digital radio standard
b. GSM	ii.3G wireless/Mobile technology
c. UMTS	iii. Nine different schemes for modulation and error correction
d. EDGE	iv. An emerging wireless service that offers a mobile data

iii iv ii i

iv i ii iii

ii iii iv i

ii i iv iii

Ans .

Explanation :
B has perfect pairs.

Q.28

Object Request Broker (ORB) is I. A software program that runs on the client as well as on the application server. II. A software program that runs on the client side only. III. A software program that runs on the application server, where most of the components reside

I,II and III

I & II

II & III

I only

Ans .

Explanation :
The Object Request Broker (ORB) manages interaction between clients and servers. This includes the distributed computing responsibilities of location, referencing and 'marshaling' of parameters and results. The Object Request Broker or ORB takes care of all of the details involved in routing a request from client to object, and routing the response to its destination.

Q.29

A software agent is defined as I. A software developed for accomplishing a given task. II. A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task. III. An open source software for accomplishing a given task.

III

All of the above

Ans .

Explanation :
A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task.

Q.30

Match the following

List-I	List-II
a. Classification	i. Principal Component Analysis
b. Clustering	ii. Branch and Bound
c. Feature Extraction	iii. K-nearest neighbour
d. Feature Selection	iv. K-means

iii iv ii i

iv iii i ii

iii iv i ii

iv iii ii i

Ans .

Explanation :
C has perfect pairs.

Q.31

SET, an open encryption and security specification model that is designed for protecting credit card transactions on the internet, stands for

Secure Electronic Transaction

Secular Enterprise for Transaction

Security Electronic Transmission

Secured Electronic Termination

Ans .

Explanation :
SET :Secure Electronic Transaction

Q.32

In a paged memory management algorithm the hit ratio is 70%. If it takes 30 nanoseconds to search Translation look aside buffer(TLB) and 100 nanoseconds to access memory, the effective memory access time is

91 ns

69 ns

200 ns

160 ns

Ans .

Explanation :
Let total page =100. Effective Access Time = (30+100)* .7 +( 30+100+100)*(1- .7)=91+69=160

Q.33

Match the following

List-I	List-II
a. Multilevel Feedback queue	i.Time-Slicing
b. FCFS	ii. Criteria to move processes between queues
c. Shortest Process next	iii. Batch processing
d. Round Robin scheduling	iv. Exponential smoothening

i iii ii iv

iv iii ii i

iii i iv ii

ii iii iv i

Ans .

Explanation :
D has perfect pairs.

Q.34

Consider a system with five processes P0 through P4 and three resource types A,B,C. Resource type A has 10 instances ,resource type B has 5 instances and resource type C has 7 instances . Suppose that, at time T0 the following snapshot of the system has been taken Assume that now the process P1 requests one additional instance of type R1 and Two instances of Resource type R3. the state resulting after this allocation will be. UGC NET DEC 15 Solved

Ready State

Safe State

Blocked State

Unsafe State

Ans .

Explanation :
example of banker's algorithm.

Q.35

Match the following

List-I	List-II
a. Contiguous Allocation	i.This scheme supports very large file sizes
b.Linked Allocation	ii. This allocation technique supports only sequential files
c. Indexed Allocation	iii. Number of disks required to access file is minimal.
d. Multi-Level Indexed	iv.This technique suffers from maximum wastage of space in storing pointers.

iii iv ii i

iii ii iv i

i ii iv iii

i iv ii iii

Ans .

Explanation :
B has perfect pairs.

Q.36

Which of the following commands will output "OneTwoThree" ?

for val; do echo-n $val; done

for one two three; do echo-n; done

for n in one two three; do echo-n $n; done

for n in one two three {echo-n $n}

Ans .

Explanation :
option C will generate required output.

Q.37

Mergesort makes two recursive calls. Which statement is true after these recursive calls finish, but before the merge step?

The array elements form a heap.

Elements in each half of the array are sorted amongst themselves.

Elements in the first half of the array are less than or equal to elements in the second half of the array.

All of the above.

Ans .

Explanation :
Elements in each half of the array are sorted amongst themselves.

Q.38

A text is made up of the characters a, b, c, d, e each occurring with the probability .12, .4, .15, .08 and .25 respectively. The optimal coding technique will have the average length of

1.7

2.15

3.4

3.8

Ans .

Explanation :
Using Hoffman's algorithm, code for a is 1111; b is 0; c is 110; d is 1110; e is 10. Average code length is 4 x.12 + 1 x .4 + 3 x.15 + 4 x.08 + 2 x.25 = 2.15

Q.39

Searching for an element in the hash table requires O(1) time for the ____ time, whereas for direct addressing it holds for the _____ time

worst-case,average

worst-case,worst-case

average,worst-case

best-case,average

Ans .

Explanation :
Searching for an element in the hash table requires O(1) time for the average time, whereas for direct addressing it holds for the worst case time.

Q.40

An algorithm is made up of 2 modules Ml and M2. If time complexity of M1 is h(n) and M2 is g(n) then the time complexity of the algorithm is

min (h(n) ,g(n) )

max (h(n) ,g(n))

h(n) + g(n)

h(n) * g(n)

Ans .

Explanation :
An algorithm is made up of 2 modules Ml and M2. If time complexity of M1 is h(n) and M2 is g(n) then the time complexity of the algorithm is max (h(n) ,g(n)).

Q.41

What is the maximum number of parenthesis that will appear on the stack at any one time for parenthesis expression given by ( ( ) ( ( ) ) ( ( ) ) )

Ans .

Explanation :
3 is correct number.

Q.41

Match the following

List-I	List-II
a. Automatic storage class	i.scope of the variable is global
b.Register storage class	ii. Value of the variable persists between different function calls.
c. Static Storage class	iii. Value stored in memory and local to the block in which the variable is defined.
d. External storage class	iv.Value stored in CPU registers

iii iv i ii

iii iv ii i

iv iii ii i

iv iii i ii

Ans .

Explanation :
B has perfect pairs.

Q.42

when we pass an array as an argument to a function, what actually gets passed?

Address of the array

Value of elements in array

Base address of the array

Number of elements of the array

Ans .

Explanation :
we pass base address of the array when we pass array.

Q.43

While (87) printf("computer"); The above C statement will

print "computer" 87 times

print "computer" 0 times

print "computer" 1 times

print "computer" infinite times

Ans .

Explanation :
"computer" will get printed infinitely.

Q.44

A friend function can be used to

avoid arguments between classes

allow access to classes whose source code is unavailable

allow one class to access an unrelated class

None of the above

Ans .

Explanation :
friend function allows one class to access other unrelated class.

Q.45

Which of the following is the correct value returned to the operating system upon the successful completion of a program?

-1

Program do not return value

Ans .

Explanation :
successful completion of a program is indicated by 0.

Q.46

Manager's salary details are hidden from the employee . This is called as

Conceptual level data hiding

Physical level data hiding

External level data hiding

Local level data hiding

Ans .

Explanation :
example of External level data hiding.

Q.47

Which one of the following statements is FALSE?

Any relation with two attributes is in BCNF

A relation in which every key has only one attribute is in 2NF

A prime attribute can be transitively dependent on a key in a 3 NF relation.

A prime attribute can be transitively dependent on a key in a BCNF relation.

Ans .

Explanation :
A prime attribute cannot be transitively dependent on a key in a BCNF relation.

Q.48

A clustering index is created when ______.

primary key is declared and ordered

no key ordered

foreign key ordered

there is no key and no order

Ans .

Explanation :
A clustering index is created when primary key is declared and ordered.

Q.49

Let R= (A, B, C, D, E, F) be a relation scheme with the following dependencies: C->F, E->A, EC->D, A->B. Which of the following is a key for R?

Ans .

Explanation :
Find the closure set of all the options give. If any closure covers all the attributes of the relation R then that is the key. Algorithm to find Closure Set Step1: Equate an attribute or attributes to X for which closure needs to be identified. Step2: Take each FD (functional dependency) one by one and check whether the left side of FD is available in X, if yes then add the right side attributes to X if it is not available. Step3: Repeat step 2 as many times as possible to cover all FD's. Step4: After no more attributes can be added to X declare it as the closure set. FDs: C->F, E->A, EC->D, A->B Find closure set for CD. X = CD = CDF {C->F} No more attributes can be added to X. Hence closure set of CD = CDF Find closure set for EC. X = EC = ECF {C->F} = ECFA {E->A} = ECFAD {EC->D} = ECFADB {A->B} Closure set of EC covers all the attributes of the relation R..

Q.50

Match the following

List-I	List-II
a. DDL	i.LOCK TABLE
b. DML	ii. COMMIT
c. TCL	iii. Natural Difference
d. BINARY OPERATION	iv. REVOKE

ii i iii iv

i ii iv iii

iii ii i iv

iv i ii iii

Ans .

Explanation :
D has perfect pairs.

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