UGCNET-Dec2014 Paper 2 - UPSCFEVER






Ans .

D


  1. 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.





Ans .

B


  1. 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.





Ans .

B


  1. 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.



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Ans .

A


  1. 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





Ans .

B


  1. 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





Ans .

C


  1. Explanation :

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





Ans .

D


  1. Explanation :

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





Ans .

B


  1. 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





Ans .

C


  1. Explanation :

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





Ans .

A


  1. 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}





Ans .

B


  1. 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.





Ans .

C


  1. 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 (10 6 times) = 0.99910^6 Probability that 1 is printed at least once = 1- probability that 1 is never printed = 1-0.99910^6 =0.6321





Ans .

A


  1. 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





Ans .

A


  1. 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





Ans .

C


  1. 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.





Ans .

B


  1. Explanation :

    Study operating system Theory





Ans .

D


  1. 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





Ans .

A


  1. 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).





Ans .

B


  1. 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)





Ans .

B


  1. Explanation :

    Study Theory





Ans .

B


  1. 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





Ans .

D


  1. 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)





Ans .

A


  1. 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.





Ans .

ABCD


  1. Explanation :>C language theory





Ans .

D


  1. 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 made—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.





Ans .

D


  1. Explanation :

    Study DELD theory





Ans .

B


  1. Explanation :

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





Ans .

D


  1. 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)





Ans .

A


  1. 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





Ans .

D


  1. Explanation :Study database theory





Ans .

D


  1. 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





Ans .

C


  1. 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





Ans .

D


  1. Explanation :

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





Ans .

D


  1. 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





Ans .

C


  1. Explanation :

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





Ans .

D


  1. Explanation :

    Study routing protocol theory





Ans .

B


  1. Explanation :

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





Ans .

A


  1. 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





Ans .

D


  1. Explanation :

    Study operating system theory





Ans .

B


  1. Explanation :

    Friend function is never a member of class





Ans .

C


  1. 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)





Ans .

A


  1. 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+





Ans .

C


  1. 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





Ans .

D


  1. 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





Ans .

D


  1. 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





Ans .

D


  1. Explanation :

    Study databse theory





Ans .

B


  1. 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.





Ans .

B


  1. Explanation :

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





Ans .

D


  1. 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





Ans .

C


  1. 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.