UGC NET- Dec2015 - UPSCFEVER

Q.1

How many committees of five people can be chosen from 20 men and 12 women such that each committee contains at least three women?

75240

52492

41800

9900

Ans .

Explanation :

For at least 3 women means there must be 3 or more than 3 women. Therefore we can choose 3 women from 12 women and 2 men from 20 men 4 women from 12 women and 1 men from 20 men 5 women from 12 women and 0 men from 20 men 12C3 x 20C2 + 12C4 x 20C1 + 12C5 x 20C0 = 220 x 190 + 495 x 20 + 79 = 52492

Q.2

Which of the following statement(s) is/are false? (a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree. (b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree. (c) A complete graph (Kn) has a Hamilton Circuit whenever n?3 (d) A cycle over six vertices (C6) is not a bipartite graph but a complete graph over 3 vertices is bipartite. Codes:

(a) only

(b) and (c)

(d) only

Ans .D

Explanation :

G is bipartite if and only if all closed walks in G are of even length. Every minimal cycle in GG has length 4, that is every cycle of length strictly greater than 4 can be divided in cycles of length 4 For solution plz visit to see above theorem proof https://www.cs.sfu.ca/~ggbaker/zju/math/euler-ham.html http://www.sas.upenn.edu/~htowsner/math340/Lecture%207%20(Sep%2019).pdf

Q.3

Which of the following is/are not true? (a) The set of negative integers is countable. (b) The set of integers that are multiples of 7 is countable. (c) The set of even integers is countable. (d) The set of real numbers between 0 and 1/2 is countable.

(a) and (c)

(b) and (d)

(b) only

(d) only

Ans .

Explanation :

An infinite set is countable if and only if it is possible to list the elements of the set in a sequence (indexed by the positive integers). In mathematics, a countable set is a set with the same cardinality (number of elements) as some subset of the set of natural numbers. A countable set is either afinite set or a countably infinite set. Whether finite or infinite, the elements of a countable set can always be counted one at a time and, although the counting may never finish, every element of the set is associated with a natural number. Source https://en.wikipedia.org/wiki/Countable_set So The set of negative integers is countable is true(infinite countable) the integers that are multiples of 7 Answer: Countably infinite; listing: 0, 7, -7, 14, -14, 21, -21, 28, -28, ... ....is True The set of even integers is countable is true(infinite countable) the real numbers between 0 and 1 2 Answer: Uncountable (similar to diagonalization proof for real numbers between 0 and 1)

Q.4

Consider the graph given below:

Check graph from question paper The two distinct sets of vertices, which make the graph bipartite are:

(v1, v4, v6); (v2, v3, v5, v7, v8)

(v1, v7, v8); (v2, v3, v5, v6)

(v1, v4, v6, v7); (v2, v3, v5, v8)

(v1, v4, v6, v7, v8); (v2, v3, v5)

Ans .

Explanation :

A simple graph G=(V,E) is called bipartite if its vertex set can be partitioned into two disjoint subsets V=V1⋃V2, such that every edge has the form e=(a,b) where aϵV1 and bϵV2.
Bipartite graphs are equivalent to two-colorable graphs.

1.            Assign Red color to the source vertex (putting into set V1).
2.            Color all the neighbours with Black color (putting into set V2).
3.            Color all neighbour’s neighbour with Red color (putting into set V1).
4.            This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
5. While assigning colors, if we find a neighbour which is colored with same color as current vertex, then the graph cannot be colored with 2 colors (ie., graph is not Bipartite).
So answer is option (C).

Q.5

A tree with n vertices is called graceful, if its vertices can be labelled with integers 1,2,...,n such that the absolute value of the difference of the labels of adjacent vertices are all different. which of the following trees are graceful ?
(a) UGC NET DEC 15 Solved
(b)
(c)

(a) and (b)

(b) and (c)

(a) and (c)

(a),(b) and (c)

Ans .

Explanation :

Caterpillar tree: In graph theory, a caterpillar is a tree in which all the vertices are within distance 1 of a central path. Theorem: All caterpillars are graceful. So, (a), (b) and (c) are graceful.

Q. 6

Which of the following arguments are not valid ? (a) "if gora gets the job and works hard, then he will be promoted. if gora gets promotion, then he will be happy. he will not be happy, therefore, either he will not get the job or he will not work hard" (b)"Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty." (c) if n is a real number such that n>1, then n

> 1. suppose that n

> 1, then n>1

(a) and (c)

(b) and (c)

(a), (b) and (c)

(a) and (b)

Ans .

Explanation :

(a) P: Gora gets the job Q: Gora works hard R: Gora gets promotion S: Gora will be happy The argument can bet written as (P˄Q)→R R→S ¬S Therefore ¬P˅¬Q (b) Let P: Puneet is not guilty Q: Pankaj is telling the truth The argument can bet written as P˅Q ~Q Therefore, P Thus by disjunctive syllogism, the argument is valid. Disjunctive Syllogism: The disjunctive syllogism rule may be written as: P˅Q, ¬P ⊢Q It may also be written as: ((P˅Q)˄¬P)→Q where P, and Q are propositions expressed in some formal system.

Q.7

Let P(m,n) be the statement “m divides n” where the Universe of discourse for both the variables is the set of positive integers. Determine the truth values of the following propositions. (a) ∃m ∀n P(m,n) (b) ∀n P(1,n) (c) ∀m ∀n P(m,n)

(a)-True; (b)-True; (c)-False

(a)-True; (b)-False; (c)-False

(a)-False; (b)-False; (c)-False

(a)-True; (b)-True; (c)-True

Ans .

NONE OF THE ABOVE

Explanation :

∀m ∀n P(m, n) says that every number divides every other number and result should be a postive integer. Clearly it is a false proposition. Eg: if m=10 n=3 10 divides 3 does not follow the proposition a is false ∀n P(1, n) says that any positive integer is divisible by 1 and result will be a +ve integer. That is correct b is true ∃m∀nP(m,n) says that there are some +ve integers which divides any other +ve integer. The proposition is correct example: 1 c is true Correct option must be a-- false b--true c--true None of the given options follows

Q.8

Match the following terms List - I List -II (a)Vacuous Proof (i) A proof that implication p-->q is true based on the fact that p is false. (b) Trivial proof (ii)A proof that implication p-->q is true based on the fact that q is true. (c)Direct proof (iii)A proof that implication p-->q is true that proceeds by showing that q must be true when p is true (d) Indirect proof (iv)A proof that implication p-->q is true that proceeds by showing that q must be false when p is false Codes : (a) (b) (c) (d)/p>

(i) (ii) (iii) (iv)

(ii) (iii) (i) (iv)

(iii) (ii) (iv) (i)

(iv) (iii) (ii) (i)

Ans .

Explanation :

Types of Proofs. Suppose we wish to prove an implication p ? q. Here are some strategies we have available to try. � Trivial Proof: If we know q is true then p ? q is true regardless of the truth value of p. � Vacuous Proof: If p is a conjunction of other hypotheses and we know one or more of these hypotheses is false, then p is false and so p ? q is vacuously true regardless of the truth value of q. � Direct Proof: Assume p, and then use the rules of inference, axioms, defi- nitions, and logical equivalences to prove q. � Indirect Proof or Proof by Contradiction: Assume p and �q and derive a contradiction r ? �r. � Proof by Contrapositive: (Special case of Proof by Contradiction.) Give a direct proof of �q ? �p. Assume �q and then use the rules of inference, axioms, definitions, and logical equivalences to prove �p.(Can be thought of as a proof by contradiction in which you assume p and �q and arrive at the contradiction p ? �p.) � Proof by Cases: If the hypothesis p can be separated into cases p1 ? p2 ? � � � ? pk, prove each of the propositions, p1 ? q, p2 ? q, . . . , pk ? q, separately. (You may use different methods of proof for different cases.)

Q.9

Consider the compound propositions given below as: (a) p˅~(p˄q) (b) (p˄~q)˅~(p˄q) (c) p˄(q˅r) Which of the above propositions are tautologies?

(a) and (c)

(b) and (c)

(a) and (b)

(a), (b) and (c)

Ans .

NONE OF THE ABOVE

Explanation :

Q.10

Which of the following property/ies a Group G must hold, in order to be an Abelian group? (a) The distributive property (b) The commutative property (c) The symmetric property Codes:

(a) and (b)

(b) and (c)

(a) only

(b) only

Ans .

Explanation :

When a Group G must hold The commutative property then group is Abelian group.

Q.11

Consider the following program:
#include
main()
{
    int i, inp;
    float x, term=1, sum=0;
    scanf(“%d %f”,&inp, &x);
    for(i=1;i<=inp;i++)
    {
                term=term*x/i;
                sum=sum+term;
}
printf(“Result=%f\n”,sum);
}

The program computes the sum of which of the following series?

x+x²/2+x³/3+x⁴/4+...

x+x²/2!+x³/3!+x⁴/4!+...

1+x²/2+x³/3+x⁴/4+...

1+x²/2!+x³/3!+x⁴/4!+...

Ans .

Explanation :

term =1 , sum=0
suppose inp entered = 3 so First iteration
term= 1*x/1 = x
sum = 0 + x =x
Second iteration
term = x*x/2= x²/2 =x²/2! sum= x+x²/2!
Third iteration
term = =x²/2! *x/3 
=x³/3.2.1=x³/3! sum= x+x²/2!+x³/3!
and so on so sum= x+x²/2!+x³/3!+................

Q.12

Consider the following two statements: (a) A publicly derived class is a subtype of its base class. (b) Inheritance provides for code reuse. Which of the following statements is correct?

Both the statements (a) and (b) are correct

Neither of the statements (a) and (b) are correct

Statement (a) is correct and (b) is incorrect

Statement (a) is incorrect and (b) is correct

Ans .

Explanation :

A publicly derived class is a subtype of its base class. Inheritance provides for code reuse. BOTH are correct

Q.13

Consider a “CUSTOMERS” database table having a column “CITY” filled with all the names of Indian cities (in capital letters). The SQL statement that finds all cities that have “GAR” somewhere in its name, is

Select *from customers where city=’%GAR%’;

Select *from customers where city=’$GAR$’;

Select *from customers where city like ‘%GAR%’

Select *from customers where city as ’%GAR’;

Ans .

Explanation :

For Matching Any String in Oracle Like Operator is Used. also % symbol is used as a wild card character to represent multiple characters. so the answer is C.

Q.14

Match the following database terms to their functions:
List-I                                       List-II
(a) Normalization                     (i) Enforces match of primary key to foreign key
(b) Data Dictionary                  (ii) Reduces data redundancy in a database
(c) Referential Integrity           (iii) Define view(s) of the database for particular user(s).
(d) External Schema               (iv) Contains metadata describing database structure.
Codes:
      (a)   (b)   (c)    (d)

(iv) (iii) (i) (ii)

(ii) (iv) (i) (iii)

(ii) (iv) (iii) (i)

(iv) (iii) (ii) (i)

Ans .

Explanation :

Normalization: Reduces data redundancy in a database Data Dictionary: Contains metadata describing database structure Referential Integrity : Enforces match of primary key to foreign key External Schema: Define view(s) of the database for particular user(s).

Q.15

In general, in a recursive and non-recursive implementation of a problem (program):

Both time and space complexities are better in recursive than in non-recursive program

Both time and space complexities are better in non-recursive than in recursive program

Time complexity is better in recursive version but space complexity is better in non-recursive version of the program

Space complexity is better in recursive version but time complexity is better in non-recursive version of the program

Ans .

Explanation :

Both time and space complexities are better in non-recursive than in recursive program

Q.16

A three dimensional array in ‘C’ is declared as int A[x][y][z]. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer):

&A[0][0][0]+w(y*z*q+z*p+r)

&A[0][0][0]+w(y*z*p+z*q+r)

&A[0][0][0]+w(x*y*p+z*q+r)

&A[0][0][0]+w(x*y*q+z*p+r)

Ans .

Explanation :

Base address+ offseet

Q.17

In C++, which system-provided function is called when no handler is provided to deal with an exception?

terminate()

unexpected()v

abort()

kill()

Ans .

Explanation :

This function is provided so that the terminate handler can be explicitly called by a program that needs to abnormally terminate, and works even if set_terminate has not been used to set a custom terminate handler

Q.18

Which of the following provides the best description of an entity type?

A specific concrete object with a defined set of processes (e.g. Jatin with diabetes)

A value given to a particular attribute (e.g. height-230 cm)

A thing that we wish to collect data about zero or more, possibly real world examples of it may exist.

A template for a group of things with the same set of characteristics that may exist in the real world

Ans .

Explanation :

Entity type refers to some thing that may exist in real world with name and attributes

Q.19

Data which improves the performance and accessibility of the database are called:

Indexes

User Data

Application Metadata

Data Dictionary

Ans .

Explanation :

Indexes are used to simplify the access of database

Q.20

A relation R={A,B,C,D,E,F,G} is given with following set of functional dependencies: F={AD?E, BE?F, B?C, AF?G} Which of the following is a candidate key?

ABC

ABD

Ans .

Explanation :

ABD CLOSURE = { ABDE}={ABDEF}={ABCDEFG} SO CANDIDATE KEY IS ABD

Q.21

Which of the following services is not provided by wireless access point in 802.11 WLAN?

Association

Disassociation

Error Correction

Integration

Ans .

Explanation :

IEEE 802.11 Services

Q.22

Which of the following fields in IPv4 datagram is not related to fragmentation?

Type of service

Fragment offset

Flags

Identification

Ans .

Explanation :

Types of services are not related to fragmentation

Q.23

Four channels are multiplexed using TDM. If each channel sends 100bytes/second and we multiplex 1 byte per channel, then bit rate for the link is----------------------------

400 bps

800 bps

1600 bps

3200 bps

Ans .

Explanation :

Each frame carries 1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32 bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte from each channel, the frame rate must be 100 frames per second. The bit rate is 100 × 32 = 3200 bps.

Q.24

In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in adjacent cells. if 840 frequencies are available, how many can be used in a given cell ?

280

210

140

120

Ans .

Explanation :

Each cell has six other adjacent cells, in a hexagonal grid. If the central cell uses frequency group A, its six adjacent cells can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. So the answer is 840/3 = 280 frequencies.

Q.25

using p=3, q=13, d=7 and e=3 in the RSA algorithm, what is the value of ciphertext for a plain text 5?

Ans .

NONE OF THE ABOVE

Explanation :

p=3, q=13, d=7, e=3, M=5, C=? C = M^e mod n n = p*q = 3*13 = 39 C = 5³ mod 39 = 8 Answer is 8.

Q.26

A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries: UGC NET DEC 15 Solved
Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU?

1024, 3072, 4096, 6144

1234, 4012, 5000, 6200

1020, 3012, 6120, 8100

2021, 4050, 5112, 7100

Ans .

Explanation :

The pages which are not in main memory are:

Page 
Address 
Address that will cause page fault
1 
1K 
1024-2047
3 
3K 
3072-4095
4 
4K 
4096-5119
6 
6K 
6144-7167

1020 will not cause page fault (1024-2047)
3012 will not cause page fault (3072-4095)
6120 will not cause page fault (4096-5119)
8100 will not cause page fault (6144-7167)

Page	Address	Address that will cause page fault
1	1K	1024-2047
3	3K	3072-4095
4	4K	4096-5119
6	6K	6144-7167

Q.27

Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available?

60 sec

30 sec

45 sec

10 sec

Ans .

Explanation :

T = Ninstr x 1µs + 15,000 x 2,000 µs = 60s
Ninstr x 1µs = 60,000,000 µs – 30,000,000 µs = 30,000,000 µs
Ninstr = 30,000,000
The number of instruction between two page faults is
Ninstr /NPageFaults = 30,000,000/15,000 = 2,000
If the mean interval between page faults is doubled, the number of instruction between two page faults is also doubled and is 4,000. Now the number of page faults is
30,000,000/4,000 = 7,500
T’            = 30,000,000 µs + 7,500 x 2,000 µs
= 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s
Doubling the memory, doesn’t mean that the program runs twice as fast as on the first system. Here, the performance increase is of 25%

Q.28

Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. what is the time in msec to read a block of 1024 bytes from this disk ?

57 msec

49 msec

48 msec

17 msec

Ans .

Explanation :

Time in msec to read a block of 1024 bytes (Access time or Disk Latency) = seek time + average rotational delay + transfer time
If there are 16384 bytes per track there are 1024/16384 tracks to be read for this block.
Seek time = 40 msec
Rotational delay = 16 msec
Transfer time = (sectors_read/sectors per rev.) x rotational delay = (1024/16384) x 16 = 1
average rotational delay = rotational delay/2 = 16/2 = 8
access time = 40 + 8 + 1 = 49

Q.29

A system has four processes and five allocatable resources. The current allocation and maximum need are as follows

  . Allocated Maximum  Available 
 Process A 1 0 2 1 2 1 1 2 1 3   0 0 x 1 1
Process B  2 0 1 1 0 2 2 2 1 0 . 
Process C  1 1 0 1 0  2 1 3 1 0 . 
Process D    1 1 1 1 0  1 1 2 2 1 .

.	Allocated	Maximum	Available
Process A	1 0 2 1 2	1 1 2 1 3	0 0 x 1 1
Process B	2 0 1 1 0	2 2 2 1 0	.
Process C	1 1 0 1 0	2 1 3 1 0	.
Process D	1 1 1 1 0	1 1 2 2 1	.

The smallest value of x for which the above system in safe state is

Ans .

Explanation :

Let us start with x=0 as smallest value is zero, so Available = 0 0 0 1 1 Need of each process = Maximum - Allocated so Need matrix becomes Need Matrix Process A 0 1 0 0 1 Process B 0 2 1 0 0 Process C 1 0 3 0 0 Process D 0 0 1 1 1 since available = 0 0 0 1 1 is not less than equal to any process need so x is not equal to 0 next smallest we take x = 1 so available = 0 0 1 1 1 since available <= process D need so we allocate now available becomes = allocated + available =1 1 1 1 0 + 0 0 1 1 1 =1 1 2 2 1 now need of process A <= available so process A completed and available becomes = allocated + available = 1 0 2 1 2 + 1 1 2 2 1 = 2 1 4 3 3 now need of process C <= available so process C completed and available becomes = allocated + available = 1 1 0 1 0 + 2 1 4 3 3 = 3 2 4 4 3 now need of process B <= available so process B completed and available becomes = allocated + available = 2 0 1 1 0 + 3 2 4 4 3 = 5 2 5 5 3 So system in safe state. Hence the correct option is (1) 1

Q.30

In Unix, the login prompt can be changed by changing the contents of the file ___-

contrab

init

gettydefs

inittab

Ans .

Explanation :

The file /etc/gettydefs contains information used by getty(1m) to set up the speed and tty settings for a line. It supplies information on what the login-prompt should look like. It also supplies the speed to try next if the user indicates the current speed is not correct by typing a character.

Q.31

A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?

(2n-1)p+1

(p+1)n

Ans .

Explanation :

(a) What is the maximum number of cells possible in the base cuboid? pn. This is the maximum number of distinct tuples that you can form with p distinct values per dimensions. (b) What is the minimum number of cells possible in the base cuboid? p. You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension. (c) What is the minimum number of cells possible in the data cube, C? (2n-1)p+1. The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell. (d) What is the maximum number of cells possible (including both base cells and aggregate cells) in the data cube, C? (p+1)n. The argument is similar to that of part (a), but now we have p+1 because in addition to the p distinct values of each dimension we can also choose ?.

Q.32

Suppose that from given statistics, it is known that meningitis causes stiff neck 50% of the time, that the proportion of persons having meningitis is 1/50000, and that the proportion of people having stiff neck is 1/20. Then the percentage of people who had meningitis and complain about stiff neck is:

0.01%

0.02%

0.04%

0.05%

Ans .

Explanation :

P{M/S} = Probability that a person had meningitis , conditioned by the existence of existence of stiff neck 
P{S/M} = probability that a person complains about stiff neck, conditioned by the existence of meningitis. 
P{S} = proportion of people who complain about stiff neck. = 1/20P{M} = proportion of people who had meningitis. = 1/50,000Then:P{M|S} = (P{S|M}*P{M}) / P{S} 
=(1/2 x 1/50,000) / 1/20 = 0.0002 =( 2/10000)*100=2/100=0.02%

Q.33

.................. system is market oriented and is used for data analysis by knowledge workers including Managers, Executives and Analysts.

OLTP

Data System

Market System

Ans .

Explanation :

OLAP means online Analytical Processing. OLAP performs multidimensional analysis of business data and provides the capability for complex calculations, trend analysis, and sophisticated data modeling.

Q.34

.................. allows selection of the relevant information necessary for the data warehouse.

The Top-Down View

Data Warehouse View

Data source View

Business Query View

Ans .

Explanation :

To design an effective and efficient data warehouse, we need to understand and analyze the business needs and construct a business analysis framework. Each person has different views regarding the design of a data warehouse. These views are as follows: The top-down view - This view allows the selection of relevant information needed for a data warehouse. The data source view - This view presents the information being captured, stored, and managed by the operational system. The data warehouse view - This view includes the fact tables and dimension tables. It represents the information stored inside the data warehouse. The business query view - It is the view of the data from the viewpoint of the end-user.

Q.35

The hash function used in double hashing is of the form:

h(k, i)=(h1(k)+h2(k)+i)mod m

h(k, i)=(h1(k)+h2(k)-i)mod m

h(k, i)=(h1(k)+ih2(k))mod m

h(k, i)=(h1(k)-ih2(k))mod

Ans .

Explanation :

Double hashing uses the idea of applying a second hash function to the key when a collision occurs. The result of the second hash function will be the number of positions form the point of collision to insert. There are a couple of requirements for the second function: it must never evaluate to 0 must make sure that all cells can be probed

Q.36

In the following graph, discovery time stamps and finishing time stamps of Depth First Search (DFS) are shown as x/y where x is discovery time stamp and y is finishing time stamp. UGC NET DEC 15 Solved It shows which of the following depth first forest?

{a,b,e} {c,d,f,g,h}

{a,b,e} {c,d,h} {f,g}

{a,b,e} {f,g} {c,d} {h}

{a,b,c,d} {e,f,g} {h}

Ans .

Explanation :

Check for the Cycles in the graph, you will find that node A,B,E form one cycle. node F,G form another cycle. Node C,D form another cycle. and H is having self loop.

Q.37

The number of disk pages access in B-tree search, where h is height, n is the number of keys, and t is the minimum degree, is:

θ(logn h*t)

θ(logt n*h)

θ(logh n)

θ(logt n)

Ans .

Explanation :

As in the TREE-SEARCH procedure for binary search trees, the nodes encountered during the recursion form a path downward from the root of the tree. The number of disk pages accessed by B-TREE-SEARCH is therefore O(h) = O(logt n), where h is the height of the B-tree and n is the number of keys in the B-tree.

Q.38

The inorder traversal of the following tree is: UGC NET DEC 15 Solved

2 3 4 6 7 13 15 17 18 18 20

20 18 18 17 15 13 7 6 4 3 2

15 13 20 4 7 17 18 2 3 6 18

2 4 3 13 7 6 15 17 20 18 18

Ans .

Explanation :

The In-order Traversal is Left--Root--Right. following this we have first root at 15, then left sub tree root at 13, then left subtree root at 4, then left is 2 and so on which results in 2 4 3 13 7 6 15 17 20 18 18

Q.39

An ideal sort is an in-place-sort whose additional space requirement is ...............

O(log2n)

O(nlog2n)

O(1)

O(n)

Ans .

Explanation :

An ideal sort is an in-place-sort whose additional space requirement is O(1). that is, an in place sort manipulates the elements to be sorted within array or list space that contained the original unsorted input.

Q.40

Which of the following is not a congestion policy at network layer?

Flow Control Policy

Packet Discard Policy

Packet Lifetime Management Policy

Routing Algorithm

Ans .

Explanation :

Various congestion policy at network layer are : Virtual circuit versus datagram Packet queueing and service policy Packet discard policy Routing algorithm Packet lifetime management

Q.41

Loop unrolling is a code optimization technique:

that avoids tests at every iteration of the loop

that improves performance by decreasing the number of instructions in a basic block.

that exchanges inner loops with outer loops

that reorders operations to allow multiple computations to happen in parallel.

Ans .

Explanation :

Loop unrolling is a compiler optimization applied to certain kinds of loops to reduce the frequency of branches and loop maintenance instructions. It is easily applied to sequential array processing loops where the number of iterations is known prior to execution of the loop. The general idea of loop unrolling is to replicate the code inside a loop body a number of times. The number of copies is called the loop unrolling factor. The number of iterations is divided by the loop unrolling factor. Loop unrolling is most effective when computations involving the loop control variable can be simulated by the compiler. For example, when a loop is stepping sequentially through an array, increments to a register that points to array entries can be simulated by the compiler with changes in the displacements in load and store instructions. Loop unrolling can reduce the number of loop maintenance instruction executions by the loop unrolling factor. In effect, the computations are done by the compiler rather than being done during program execution.

Q.42

What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions? Mov al, 15 Mov ah, 15 Xor al, al Mov cl, 3 Shr ax, cl Codes:

0F00 h

0F0F h

01E0 h

FFFF h

Ans .

Explanation :

in assembly AX=AH+BL AL is the lower 8 bits AH is the higher bits 8 HERE AL and Ah both contains 15 or F so AX contains 0F0F or 0000 1111 0000 1111 XOR AL AL will return 0 in AL so AX becomes 0000 1111 0000 0000 MOV CL,3 will store 3 in CL shr ax,cl will shift right ax by 3 so 0000 1111 0000 0000 will become 0000 0001 1110 0000 or 01E0 in hexadecimal.

Q.43

Which of the following statements is false?

Top-down parsers are LL parsers where first L stands for left-to-right scan and second L stands for a leftmost derivation.

(000)* is a regular expression that matches only strings containing an odd number of zeroes, including the empty string.

Bottom-up parsers are in the LR family, where L stands for left-to-right scan and R stands for rightmost derivation

The class of context-free languages is closed under reversal. That is, if L is any context-free language, then the language LR={WR:w?L} is context free.

Ans .

Explanation :

in option B. if (000)^* executed twice then output is 000000 i.e. even number of zeros

Q.44

System calls are usually invoked by using:

A privileged instruction

An indirect jump

A software interrupt

Polling

Ans .

Explanation :

System calls are usually invoked by using a software interrupt

Q.45

The ............... transfers the executable image of a C++ program from hard disk to main memory.

Compiler

Linker

Debugger

Loader

Ans .

Explanation :

The loader transfers the executable image of a C++ program from hard disk to main memory.

Q.46

In software testing, how the error, fault and failure are related to each other?

Error leads to failure but fault is not related to error and failure

Fault leads to failure but error is not related to fault and failure

Error leads to fault and fault leads to failure

Fault leads to error and error leads to failure

Ans .

Explanation :

Error leads to fault and fault leads to failure

Q.47

Which of the following is not a software process model ?

Prototyping

Iterative

Timeboxing

Glassboxing

Ans .

Explanation :

timeboxing allocates a fixed time period, called a time box, to each planned activity. Several project management approaches use timeboxing

Q.48

How many solutions are there for the equation x+y+z+u=29 subject to the constraints that x≥1, y≥2, z≥3 and u≥0?

4960

2600

23751

8855

Ans .

Explanation :

We let y1=x-1, y2=y-2, y3=z-3, y4=u-0 We count the number of solutions for y1+y2+y3+y4=29-6=23 n=4, r=23 The number of solutions is C(n+r-1, r) = C(4+23-1, 23) = C(26,23) = C(26,3) = 26x25x24/1x2x3 = 2600

Q.49

A Unix file system has 1-KB blocks and 4-byte disk addresses. what is the maximum file size if i-nodes contain 10 direct entries and one single,double and triple indirect entry each ?

32 Gb

64 GB

16 Gb

1 GB

Ans .

Explanation :

block = 2^10 block pointer size = 4B entries possible in block = 2^10 / 2^2 = 256 direct pointer gives = 10 * 256 = 10 blocks single indirect gives = 256 * 256 = 2^8 blocks double indirect gives = 256 * 256 * 256 = 2^16 blocks triple indirect gives = 256 * 256 * 256 * 256 = 2^14 blocks total = 16777546 blocks = 16777546 * 1024 = 17180207104 ~ 16 GB

Q.50

................ uses electronic means to transfer funds directly from one account to another rather than by cheque or cash?

M-Banking

E-Banking

O-Banking

C-Banking

Ans .

Explanation :

E-Banking uses electronic means to transfer funds directly from one account to another rather than by cheque or cash.