UGCNET-Dec2014 Paper 2

Q.1

A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph?

n-3

Ans .

Explanation :
So count of all vertices is 11
OR
A tree of n vertices have n-1 edges.
the sum of all the degree in a graph is equal to twice the no of edge
Hence
2*4+1*3+1*2+(n-4)*1=2(n-1)
13+n-4=2n-2
hence n=11.

Q.2

The size of the ROM required to build an 8-bit adder/subtractor with mode control, carry input, carry output and two's complement overflow output is given as

2¹⁶ x 8

2¹⁸ x 10

2¹⁶ x 10

2¹⁸ x 8

2¹⁷ x 10

Ans .

Explanation :
Total input to the rom decoder will be (8+8 ( two 8 bit number ) +1( mode ) +1( carry in))
so total number of words out of decoder will be 2^18 . result will be 8 bit so 8 vertical lines +( 1 for carry ) +1 ( for saying underflow).
answer will be B.

Q.3

The best normal form of relation scheme R(A. B, C, D) along with the set of functional dependencies F = {AB → C, AB → D, C → A, D → B} is

Boyce-Codd Normal form

Third Normal form

Second Normal form

First Normal form

None of these

Ans .

Explanation :
C -> A and D-> B should not be partial dependency,
because here in C -> A, A is a Key element and in D -> B, B is key element.
Partial dependency- { subset of CKs } -> { Non prime attribute key element }.
If all the attributes of a relation are key attributes then relation is automatically in 3NF.

Q.4

In a two-pass assembler, symbol table is

Generated in first pass

Generated in second pass

Not generated at all

Generated and used only in second pass

Generated third pass

Ans .

Explanation :
In first pass it does following things
It allocates space for the literals.
It computes the total length of the program.
It builds the symbol table for the symbols and their values

Q.5

Match the following :
List - I List - II
a. Call control protocol i. Interface between Base Transceiver Station (BTS) and Base Station Controller (BSC)
b. A-bis ii. Spread spectrum
c. BSMAP iii. Connection management
d. CDMA iv. Works between Mobile Switching Centre (MSC) and Base Station Subsystem (BSS)

Code:
a b c d
iii iv i ii

Code:
a b c d
iii i iv ii

Code:
a b c d
i ii iii iv

Code:
a b c d
iv iii ii i

Code:
a b c d
iv ii iii i

Ans .

Explanation :
Call control protocol=Call control protocol
A-bis=Interface between Base Transceiver Station (BTS) and Base Station Controller (BSC)
BSMAP =Works between Mobile Switching Centre (MSC) and Base Station Subsystem (BSS)
CDMA=Spread spectrum

Q.6

Consider a set A {1, 2, 3,……… 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5?

533

599

467

Ans .

Explanation :
467 is the correct answer
1000/3 = 333.33 = 333
1000/5 = 200
1000/(3*5) = 1000/15=66
333+200-66=467

Q.7

You have to sort a list L, consisting of a sorted list followed by a few 'random' elements. Which of the following sorting method would be most suitable for such a task?

Bubble sort

Selection sort

Quick sort

Insertion sort

Merge sort

Ans .

Explanation :
Insertion sort is efficient for small elements which are sorted already.So in above problem insertion sort is most efficient.

Q.8

An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element?

0.336 bits baud

3 bits baud

120,00,000 bits baud

4 bits baud

None of the above

Ans .

Explanation :
Simply speaking, the analog signnal sends 2000 baud or elements in one second.
To send this 6000 bits are transmitted in one second.
Thus 6000/2000 = 3 bits/ baud is transmitted

Q.9

A full binary tree with n leaves contains

n nodes

log₂ n nodes

2n-1 nodes

2ⁿ nodes

Ans .

Explanation :
Binary tree with n leaves have 2n-1 nodes.

Q.10

Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A → B, B → C, AC → D}

Ans .

Explanation :
A is the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A → B, B → C, AC → D}

Q.11

Shift-Reduce parsers perform the following :

Shift step that advances in the input stream by K(K > 1) symbols and Reduce step that applies a
completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.

Shift step that advances in the input stream by one symbol and Reduce step that applies a
completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.

Shift step that advances in the input stream by K(R = 2) symbols and Reduce step that applies a completed grammar rule to form a single tree.

Shift step that does not advance in the input stream and Reduce step that applies a completed grammar rule to form a single tree.

None of these

Ans .

Explanation :
A Shift step advances in the input stream by one symbol. That shifted symbol becomes a new single-node parse tree.
A Reduce step applies a completed grammar rule to some of the recent parse trees, joining them together as one tree with a new root symbol.

Q.12

A computer program selects an integer in the set {k : 1 < k < 10,00,000} at random and prints out the result.
This process is repeated 1 million times. What is the probability that 'he value k = 1 appears in the printout atleast once?

0.5

0.704

0.632121

0.68

0.70

Ans .

Explanation :
Probability of k=1 is 1/1000000= 10^-6 Probability of k!= 1 is 1- 10^-6 = 0.999 Probability that k=1 is never printed in all 10⁶ print outs = 0.999*0.999*.......0.9999 (10⁶ times) = 0.999^{10^6} Probability that 1 is printed at least once = 1- probability that 1 is never printed = 1-0.999^{10^6} =0.6321

Q.13

When an array is passed as parameter to a function, which of the following statements is correct?

The function can change values in the original array.

In C, parameters are passed by value, the function cannot change the original value in the array.

It results in compilation error when the function tries to access the elements in the array.

While developed countries are on the decline emerging ones are rising.Results in a run time error when the function tries to access the elements in the array.

None of these

Ans .

Explanation :
When array name is passed as an argument, it pass only as a reference (ie address of first element) so the ans is A

Q.14

In a classful addressing, first four bits in Class A IP address is

1010

1100

1011

1110

0000

Ans .

Explanation :
In class A, the first bit of the first octet is always set to 0 (zero).
Class A 0xxx : x can be 0 or 1
Class B 10xx
Class C 110x
Class D 1110
No option is correct

Q.15

Which of the following conditions does not hold good for a solution to a critical section problem?

No assumptions may be made about speeds or the number of CPUs.

No two processes may be simultaneously inside their critical sections.

Processes running outside its critical section may block other processes.

Processes do not wait forever to enter its critical section.

None of these

Ans .

Explanation :
If a process is running outside the critical section we don't care whatever it is doing out there either
it is blocking some processes or not it is not going to effect the processes which are running inside the critical section anyway.

Q.16

The software _______ of a program or a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.

Design

Architecture

Process

Requirement

None of these

Ans .

Explanation :
Study operating system Theory

Q.17

Which e-business model allows consumers to name their own price for products and services?

B2 B

B2 G

C2 C

C2 B

None of these

Ans .

Explanation :
its D) C2B (consumer to business) which allows consumers to name their own price for products and services
C2B model, also called a reverse auction or demand collection model, enables buyers to name or demand their own price, which is often binding,
for a specific good or service. The website collects the demand bids then offers the bids to participating sellers
C2C, or customer-to-customer, or consumer-to-consumer, is a business model that facilitates the transaction of products or services between customers.
It is one of four categories of e-commerce, along with B2B (business to business), C2B (customer to business) and B2C (business to customer) B2B and B2G has nothing to concern about consumer price setting

Q.18

odd numbers of a's and odd numbers of b's

even numbers of a's and even numbers of b's

equal numbers of a's and b's

different numbers of a's and b's

None of theory

Ans .

Explanation :
Option A would be the correct answer.
Because we can generate the string aabb (Which is a contradiction to Option B and C).
Given CFL also can generate ab (Which is contradiction to the option D).

Q.19

How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A * B) + (C * D/E)?

4 PUSH and 3 POP instructions

5 PUSH and 4 POP instructions

6 PUSH and 2 POP instructions

5 PUSH and 3 POP instructions

None of these

Ans .

Explanation :
The algorithm for evaluating any postfix expression is fairly straightforward:
While there are input tokens left
Read the next token from input.
If the token is a value
Push it onto the stack.
Otherwise, the token is an operator (operator here includes both operators and functions).
It is already known that the operator takes n arguments.
If there are fewer than n values on the stack
(Error) The user has not input sufficient values in the expression.
Else, Pop the top n values from the stack.
Evaluate the operator, with the values as arguments.
Push the returned results, if any, back onto the stack.
If there is only one value in the stack
That value is the result of the calculation.
Otherwise, there are more values in the stack
(Error) The user input has too many values Here
5 push and 4 pop (B)

Q.20

Division operation is ideally suited to handle queries of the type :

customers who have no account in any of the branches in Delhi.

customers who have an account at all branches in Delhi.

customers who have an account in atleast one branch in Delhi.

customers who have only joint account in any one branch in Delhi.

None of these

Ans .

Explanation :
Study Theory

Q.21

In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. If an empty frame is available or if the replaced page is not modified, and it takes 20 m.sec., if the replaced page is modified.
What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time?

11.6 m.sec.

16.4 m.sec.

28 m.sec.

14 m.sec.

14.6 m.sec.

Ans .

Explanation :
Service a page fault=8msec.
Page is not modified=20m.sec.
Page replace=70% of time
the average access time to service a page fault=0.7⨉20+0.3⨉8=14+2.4=16.4ms

Q.22

Which one of the following is used to compute cyclomatic complexity?

The number of regions - 1

E - N + 1, where E is the number of flow graph edges and N is the number of flow graph nodes.

P – 1, where P is the number of predicate nodes in the flow graph G.

P + 1, where P is the number of predicate nodes in the flow graph G.

None of these

Ans .

Explanation :
Cyclomatic complexity is a metric to measure logical complexity of a program.
Cyclomatic complexity defines the number of independent paths in a program
It gives an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once
An independent path is any path through the program that introduces at least one new set of processing statements or a new condition.
Cyclomatic complexity is computed in one of three ways:
1. The number of regions of the flow graph
2. Cyclomatic complexity, V(G), for a flow graph, G, is defined as V(G) = E - N+ 2k where E is the number of flow graph edges,
N is the number of flow graph nodes and k is the number of connected components or number of exit points.
3. V(G) = P+ 1 where P is the number of predicate (condition) nodes contained in the flow graph G.
Answer (D)

Q.23

The directory can be viewed as _______ that translates filenames into their directory entries.

Symbol table

Partition

Swap space

Cached

None of these

Ans .

Explanation :
Answer : Symbol Table
The symbol table is created by the compiler front-end as a stand-alone file. The purpose of the table is to provide information that the linker and the debugger need to perform their respective functions.
Connection of Directory with symbol Table
The symbol table is comprised of several subtables. The symbolic header acts as a directory for the subtables. it provides the locations of the subtables and gives their sizes.

Q.24

What does the following expression means?
char *(*(*a[N]) ()) ();

a pointer to a function returning array of n pointers to function returning character pointers.

a function return array of N pointers to functions returning pointers to characters.

an array of n pointers to function returning pointers to characters.

an array of n pointers to function returning pointers to functions returning pointers to characters.

None of these

Ans .

ABCD

Explanation :>C language theory

Q.25

Consider the following S1 and S2 :
S1 : A hard handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell.
S2 : A soft handover is one in which the channel in the source cell is retained and only then the channel in the target cell is engaged.

S1 is true and S2 is not true.

S1 is not true and S2 is true.

Both S1 and S2 are true.

Both S1 and S2 are not true.

none of these

Ans .

Explanation :
A hard handover is one in which the channel in the source cell is released and only then the channel in the target cell is engaged.
Thus the connection to the source is broken before or 'as' the connection to the target is madeâ€”for this reason such handovers are also known as break-before-make
A soft handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell. In this case the connection to the target
is established before the connection to the source is broken, hence this handover is called make-before-break. The interval, during which the two connections
are used in parallel, may be brief or substantial.

Q.26

The BCD adder to add two decimal digits needs minimum of

6 full adders and 2 half adders

5 full adders and 3 half adders

4 full adders and 3 half adders

5 full adders and 2 half adders

None of these

Ans .

Explanation :
Study DELD theory

Q.27

What will be the output of the following 'C' code?
main ( )
{ int x = 128;
printf("\n%d”, 1 + x ++);
}

128

129

130

131

132

Ans .

Explanation :
x++ is post increment operator so it first prints the value of x and then increments it. so 128+1=129.

Q.28

What kind of mechanism is to be taken into account for converting a weak entity set into strong entity set in entity-relationship diagram?

Generalization

Aggregation

Specialization

Adding suitable attributes

None of these

Ans .

Explanation :
We can convert weak entity set to strong entity set by adding primary keys of strong entity to the weak entity.This approach causes redundancy storing the primary keys more than once.
Existence of weak entity set depends on Existence of strong entity set,Hence if strong entity set is deleted,weak entity set is also get deleted.Thus if database contains a entity set that depends on another entity set, then it is necessary to mark such as weak entity set.
Hence,Option(D)

Q.29

A specific editor has 200 K of program text, 15 K of initial stack, 50 K of initialized data, and 70 K of bootstrap code.
If five editors are started simultaneously, how much physical memory is needed if shared text is used?

1135 K

335 K

1065 K

320 K

350 K

Ans .

Explanation :
5 editor works simultaneously
Each editor must have separate code in each.
But they could share initial stack, initialized data and bootstrap code
So, total memory required 200*5+15+50+70=1135 K

Q.30

Fact-less fact table in a data warehouse contains

only measures

only dimensions

keys and measures

only surrogate keys

None of these

Ans .

Explanation :Study database theory

Q.31

How many distinct stages are there in DES algorithm, which is parameterized by a 56-bit key?

Ans .

Explanation :
DES(Data Encryption Standard) algorithm which is parameterized by a 56 bit key has 19 distinct stages including 16 rounds or repetition so ans is D

Q.32

The Excess-3 decimal code is a self-complementing code because

The binary sum of a code and its 9's complement is equal to 9.

It is a weighted code.

Complement can be generated by inverting each bit pattern.

The binary sum of a code and its 10's complement is equal to 9.

None of these

Ans .

Explanation :
The key feature of the Excess-3 code is .that it is self complementing. In other words, the l's complement of an Excess- 3 number is the Excess- 3 code for the 9's complement of the corresponding decimal number.
For example, the Excess- 3 code for decimal 6 is 1001. The l's complement of 1001 is 0110, which is the Excess-3 code for decimal 3, and 3 is the 9's complement of 6.
Thus,
The decimal sum (not binary sum) of a code and its 9′s complement is equal to 9
Complement can be generated by inverting each bit pattern.
Hence, ans: C

Q.33

_______ model is designed to bring prices down by increasing the number of customers who buy a particular product at once.

Economic Order Quantity

Inventory

Data Mining

Demand-Sensitive Pricing

None of these

Ans .

Explanation :
Demand-Sensitive Pricing model is designed to bring prices down by increasing the number of customers who buy a particular product at once.

Q.34

The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is

2^-128 to (1 - 2^-23) x 2¹²⁷

(1 - 2^-23) x 2^-127 to 2¹²⁸

(1 - 2^-23) x 2^-127 to 2²³

2^-129 to (1 - 2^-23) x 2¹²⁷

None of these

Ans .

Explanation :
The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is 2^-129 to (1 - 2^-23) x 2¹²⁷

Q.35

Debugger is a program that

allows to examine and modify the contents of registers

does not allow execution of a segment of program

allows to set breakpoints, execute a segment of program and display contents of register

All of the above

None of these

Ans .

Explanation :
Debugger is a program that allows to set breakpoints, execute a segment of program and display contents of register

Q.36

Which of the follow mg algorithms is not a broadcast routing algorithm?

Flooding

Multidestination routing

Reverse path forwarding

All of the above

None of these

Ans .

Explanation :
Study routing protocol theory

Q.37

Which of the following differentiates between overloaded functions and overridden functions?

Overloading is a dynamic or runtime binding and overridden is a static or compile time binding.

Overloading is a static or compile time binding and overriding is dynamic or runtime binding.

Redefining a function in a friend class is called overloading, while redefining a function in a derived class is called as overridden function.

Redefining a function in a derived class is called function overloading, while redefining a function in a friend class is called function overriding.

None of these

Ans .

Explanation :
Overloading is a static or compile time binding and overriding is dynamic or runtime binding is the difference.

Q.38

Consider an array A[20. 10], assume 4 words per memory cell and the base address of array A is 100. What is the address of A[11,5]? Assume row major storage.

560

565

570

575

580

Ans .

Explanation :
A[20, 10]
W = 4 words
Base (A) =100
A [11, 5] in row major m x n
loc [ a [i] [j] = base (a) + w [n (Row location of item)] + column location of item]
= 100 + 4 [10 (11) + 5]
= 100 + 4[110+5]
=100 + 115 *4
= 560

Q.39

Consider the following justifications for commonly using the two-level CPU scheduling :
I. It is used when memory is too small to hold all the ready processes.
II. Because its performance is same as that of the FIFO.
III. Because it facilitates putting some set of processes into memory and a choice is made from that.
IV. Because it does not allow to adjust the set of in-core processes.
Which of the following is true?

I, III and IV

I and II

III and IV

I and III

None of these

Ans .

Explanation :
Study operating system theory

Q.40

Which of the following is not a member of class?

Static function

Friend function

Const function

Virtual function

None of these

Ans .

Explanation :
Friend function is never a member of class

Q.41

For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each page entry requires z bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by

x/2

xz/2

√2xz/2

√xz/2

z/2

Ans .

Explanation :
Average amount of internal fragmentation per segment is y/2.
Average no of pages per process segment is x/y.
Each page requires 'z' bytes of page table.
so each process segment requires page table size of xz/y bytes.
Total overhead per segment, therefore, due to internal fragmentation and page table entries, is
xz/y + y/2
To minimise the overhead, differentiate with respect to page size, p, and equate to 0:
d(xz/y + y/2)/dy = -xz/y2 + ½ = 0
=> y = √(2xz)
Answer (C)

Q.42

Convert the following infix expression into its equivalent post fix expression (A + B^D)/(E - F) + G

ABD^ + EF-/G+

ABD + ^EF-/G+

ABD + ^EF/-G+

ABD^ + EF/-G+

None of these

Ans .

Explanation :
Answer : A
Given Expression is (A + B^D ) / (E - F) + G
first we look at precedence and associativity and a/c to that "()" has higher precedence among all operators so we are going to evaluate them first .Lets take this first (A + B^D )
inside this again we have 2 operators one is "+" and other is "^" in which Exponentiation operator has higher precedence .so
it will evaluate it like this
A+ BD^ then ABD^+ now let move to the second one which is (E - F) it will be EF- till now we have
ABD^+ / EF- + G
Now among both operators which has to be evaluated "/" has higher precedence so we'll evaluate it first
ABD^+EF- /+ G
now finally we are going to evaluate "+"
Final Postfix expression will be ABD^+EF-/G+

Q.43

Requirement Development, Organizational Process Focus, Organizational Training, Risk Management and Integrated Supplier Management are process areas required to achieve maturity level

Performed

Managed

Defined

Optimized

None of these

Ans .

Explanation :
Maturity Level 3 -
Defined DAR - Decision Analysis and Resolution
IPM - Integrated Project
OPD - Organizational Process Definition
OPF - Organizational Process Focus
OT - Organizational Training
PI - Product Integration
RD - Requirements Development
RSKM - Risk Management
TS - Technical Solution
VAL - Validation
VER - Verification

Q.44

Which one of the following set of attributes should not be encompassed by effective software metrics?

Simple and computable

Consistent and objective

Consistent in the use of units and dimensions

Programming language dependent

None of these

Ans .

Explanation :
Effective software metrics should be simple, computable, consistent , objective and consistent in the use of units and dimensions
but it should not be programming language dependent as then it will not be consistent in measuring the software hence ans is D

Q.45

Which one of the following set of attributes should not be encompassed by effective software metrics?

Simple and computable

Consistent and objective

Consistent in the use of units and dimensions

Programming language dependent

None of these

Ans .

Explanation :
Effective software metrics should be simple, computable, consistent , objective and consistent in the use of units and dimensions
but it should not be programming language dependent as then it will not be consistent in measuring the software hence ans is D

Q.46

Which of the following is true?
I. Implementation of self-join is possible in SQL with table alias.
II. Outer-join operation is basic operation in relational algebra.
III. Natural join and outer join operations are equivalent.

I and II are correct.

II and III are correct.

Only III is correct.

Only I is correct.

None of these

Ans .

Explanation :
Study databse theory

Q.47

_______ are applied throughout the software process.

Framework activities

Umbrella activities

Planning activities

Construction activities

None of these

Ans .

Explanation :
Overall Umbrella Activities cover following points
Umbrella activities span all the stages of the SDLC
The concept of umbrella activities focuses on Requirement Traceability Matrix
Requirement traceability matrix is needed to be maintained by projects to ensure that the requirements are adequately addressed
Not maintaining a requirement traceability matrix results in problems including unsatisfied requirements, problems during delivery and maintenance
Software Peer Review needs to be planned, performed, and logged.

Q.48

The period of a signal is 10 ms. What is its frequency in Hertz?

100

1000

10000

The backlash against globalization is serious and must be handled carefully.

Ans .

Explanation :
1m sec= 1000 hertz
period = 10ms
frequency=1/period=1/10 * 1000
=100
correct answer B

Q.49

If we define the functions f, g and h that map R into R by :
f(x) = x4, g(x) = √x2 + 1, h(x) = x2 + 72, then the value of the composite functions ho(gof) and (hog)of are given as

x⁸ - 71 and x⁸ - 71

x⁸- 73 and x⁸ - 73

x⁸ + 71 and x⁸ + 71

x⁸ + 73 and x⁸ + 73

None of these

Ans .

Explanation :
Ho(go(f()) is Ho(G(x⁴))=Ho(sqrt(X⁴+1))
=Ho(x⁴+1)=x⁸+1+72
= x⁸+73 hence answer would be D

Q.50

Which of the following is true?

Canonical LR parser is LR (1) parser with single look ahead terminal

All LR(K) parsers with K > 1 can be transformed into LR( 1) parsers

Both (A) and (B)

Either (A) OR (B)

None of the above

Ans .

Explanation :
A canonical LR parser or LR(1) parser is an LR(k) parser for k=1, i.e. with a single look-ahead terminal.
The special attribute of this parser is that all LR(k) parser's with k>1 can be transformed into a LR(1) parser.

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