Nielit STA 17-12-2017

Question 1
Correct expression for UDP user datagram length is
A
length of UDP=length of IP - length of IP header
B
length of UDP=length of UDP - length of UDP header
C
length of UDP=length of IP + length of IP header
D
length of UDP=length of UDP + length of UDP header
       Computer-Networks
Question 1 Explanation: 
→ Total length of the datagram in bytes.
→ The size of the data by computing "total length - header length"
→ A user datagram is encapsulated in an IP datagram. There is a field in the IP datagram the defines the total length. There is another field in the IP datagram that defines the length of the header. So if we subtract the length of a UDP datagram that is encapsulated in an IP datagram, we get the length of UDP user datagram.
Question 2
In binary search tree which traversal is used for getting ascending order values?
A
Inorder
B
Preorder
C
Postorder
D
None of the above
       Data-Structures
Question 2 Explanation: 
To traverse a binary search tree in inorder following operations are carried-out
(i) Traverse the left most subtree starting at the left external node,
(ii) Visit the root, and
(iii) Traverse the right subtree starting at the left external node.
The Inorder traversal of the above tree will outputs: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Question 3
The automaton which allows transformation to a new state without consuming any input symbols:
A
NFA
B
DFA
C
NFA-ɛ / NFA-l
D
All of the above
       Theory-of-Computation
Question 3 Explanation: 
→ NFA-l or e-NFA is an extension of Non deterministic Finite Automata which are usually called NFA with epsilon moves or lambda transitions.

We extend the class of NFAs by allowing instantaneous (ε) transitions:
1. The automaton may be allowed to change its state without reading the input symbol.
2. In diagrams, such transitions are depicted by labeling the appropriate arcs with ε.
3. Note that this does not mean that ε has become an input symbol. On the contrary, we assume that the symbol ε does not belong to any alphabet.
Question 4
Complement of a DFA can be obtained by:
A
making starting state as final state
B
make final as a starting state
C
make final states non-final and non final as final
D
None of the options
       Theory-of-Computation
Question 4 Explanation: 
→ If (Q, ∑, δ, q0, F) be a DFA that accepts a language L, then the complement of the DFA can be obtained by swapping its accepting states with its non-accepting states and vice versa. Note: If we want to complement an NFA, we have to first convert it to DFA and then have to swap states as in the previous method.
Question 5
Concatenation operation refers to which of the following set operations:
A
Union
B
Dot
C
Kleene
D
none of the options
       Theory-of-Computation
Question 5 Explanation: 
→ It does not matter in which order we group the expression with the operators as they are associative.
→ If one gets a chance to group the expression, one should group them from left for convenience. For instance, 012 is grouped as (01)2.
Question 6
Which of the following statement is true?
A
mealy and moore machine are language acceptors
B
Finite state automata is language translator
C
NPDA is more powerful than DPDA
D
mealy machine is more powerful than moore machine
       Theory-of-Computation
Question 6 Explanation: 
→ NPDA is more powerful than DPDA because
1.DPDA accept only a proper subset of CFL's ie. LL grammars.
2.NPDA can accept any CFL, which makes them more powerful over DPDA
Question 7
If file size is large and if it is to be accessed randomly then which of the following allocation strategy should be best to use in a system?
A
Linked allocation
B
Indexed allocation
C
Contiguous allocation
D
None of the options
       Data-Structures
Question 7 Explanation: 
Contiguous:All blocks of a file are stored contiguously.
Advantages:
Faster access as all blocks are nearby.
Suitable for small sequentially accessed file
Disadvantages:
Poor performance if file grows or shrinks.

Linked Allocation:Each block stores pointer to next block
Advantages:
No fragmentation.
Suitable for large sequentially accessed file
Disadvantages:
Random access is not possible, If one link is lost, cannot access subsequent blocks
Note: In File Allocation Table (FAT) all links are cached in a table for faster access.

Indexed Allocation:A single bock stores indexes of all blocks of a file.
Advantage:
Suitable for large randomly accessed file
Eg: UNIX inode stores the first 12 or so data block pointers and then singly, doubly, and triply indirect pointers.
Question 8
Microprocessors are used in which generation of computers?
A
Ist generation
B
IInd generation
C
IIIrd generation
D
IV generation
       Micro-Processor
Question 8 Explanation: 
In the history of computing hardware, computers using
1. Vacuum tubes were called the first generation;
2. Transistors and diodes, the second;
3. Integrated circuits, the third;
4. Microprocessors, the fourth.
5. Whereas previous computer generations had focused on increasing the number of logic elements in a single CPU, the fifth generation, it was widely believed at the time, would instead turn to massive numbers of CPUs for added performance.
Question 9
Operating system maintains the page table for:
A
each process
B
each thread
C
each instruction
D
each address
       Operating-Systems
Question 9 Explanation: 
Role of the page table → In operating systems that use virtual memory, every process is given the impression that it is working with large, contiguous sections of memory.
→ Physically, the memory of each process may be dispersed across different areas of physical memory, or may have been moved (paged out) to another storage, typically to a hard disk drive.
→ When a process requests access to data in its memory, it is the responsibility of the operating system to map the virtual address provided by the process to the physical address of the actual memory where that data is stored.
→ The page table is where the operating system stores its mappings of virtual addresses to physical addresses, with each mapping also known as a page table entry (PTE).
Question 10
Close-loop control mechanism try to:
A
Remove congestion after it occurs
B
Remove congestion after sometime
C
Prevent congestion before it occurs
D
Prevent congestion before sending packets
       Computer-Networks
Question 10 Explanation: 
We can divide congestion control mechanisms into two broad categories:
open-loop congestion control (prevention)
closed-loop congestion control (removal).

Question 11
Which multiple access technique is used by IEEE 802.11 standard for wireless LAN?
A
CDMA
B
CSMA/CA
C
ALOHA
D
None of the Options
       Computer-Networks
Question 11 Explanation: 
A wireless LAN system allows a station using infrared, visible light, or radio wave to communicate with similarly configured stations in the vicinity, or to communicate with remote stations via a nearby base station that is connected to a wired network. The most prevalent wireless LANs used today are those based IEEE 802.11. These standards use similar medium access control protocol and frame format as the Ethernet, and therefore IEEE 802.11 - based wireless LANs have been referred to as wireless Ethernets.
A wireless LAN can be characterized by the following attributes
Architecture
Physical layer
Medium access control layer
Architecture
Based on IEEE 802.11 standards, wireless LANs can be classified as follows: infrastructure and ad hoc. With the ad hoc architecture, wireless stations communicate directly with each other on an ad hoc basis.
Physical Layer
The physical layer is characterized by the following three elements: the frequency band, the multiple access method (which are similar to "multiplexing" in the wired environment, and the data rate.
Medium Access Control (MAC) Layer
This layer provide the same function as the MAC layer in the wired LAN environment. IEEE 802.11, instead of using Ethernet's CSMA/CD, specifies a scheme called CSMA/CA.
CA stands for Collision Avoidance. (In the wireless environment, since a station may not be able to hear all other stations, CD, Collision Detection is not feasible)
Question 12
PGP encrypts data using a block cipher called:
A
International data encryption algorithm
B
private data encryption algorithm
C
Internet data encryption algorithm
D
none of the options
       Computer-Networks
Question 12 Explanation: 
→ International data encryption algorithm(IDEA) was used in Pretty Good Privacy (PGP) v2.0 and was incorporated after the original cipher used in v1.0, BassOmatic, was found to be insecure.
→ IDEA is an optional algorithm in the OpenPGP standard. PGP encrypts data by using a block cipher called international data encryption algorithm.
→ IDEA operates on 64-bit blocks using a 128-bit key and consists of a series of 8 identical transformations (a round, see the illustration) and an output transformation (the half-round).
→ The processes for encryption and decryption are similar.
Question 13
If L1 and L2 are regular sets then intersection of these two will be:
A
Regular
B
Non regular
C
Recursive
D
Non Recursive
       Theory-of-Computation
Question 13 Explanation: 
If L1 and If L2 are two regular languages, their intersection L1 ∩ L2 will also be regular.
Example
L1= {am bn | n ≥ 0 and m ≥ 0} and L2= {am bn ∪ bn am | n ≥ 0 and m ≥ 0}
L3 = L1 ∩ L2 = {am bn | n ≥ 0 and m ≥ 0} is also regular.
Question 14
The smallest integer that can be represented by an 8-bit number in 2's complement form is:
A
-256
B
-128
C
-127
D
0
       Digital-Logic-Design
Question 14 Explanation: 
→ For n bit 2’s complement numbers, range of number is -(2(n-1)) to +(2(n-1)-1)
→ The smallest integer that can be represented by an 8-bit number in 2’s complement form is =-(2(n-1))
= -128
Question 15
Non contiguous memory allocation splits program into blocks of memory called ___ that can be loaded in non adjacent holes in main memory
A
Pages
B
Frames
C
Partition
D
Segments
       Operating-Systems
Question 15 Explanation: 
Noncontiguous memory allocation splits programs into blocks of memory called Segments that can be loaded into non adjacent “holes” in main memory.
Question 16
In a full binary tree number of nodes is 63 then the height of the tree is:
A
2
B
4
C
3
D
6
       Data-Structures
Question 16 Explanation: 
⌈log2n⌉
=log263
=6
Question 17
let P,Q,R be a regular expression over Z. If P does not contain null string, then R=Q+RP has a unique solution____
A
Q*P
B
QP*
C
Q*P*
D
(P*Q*)*
       Theory-of-Computation
Question 17 Explanation: 
→ According to arden’s theorem, we can directly find the answer is QP*
→ In order to find out a regular expression of a Finite Automaton, we use Arden’s Theorem along with the properties of regular expressions.
Statement:
Let P and Q be two regular expressions.
If P does not contain null string, then R = Q + RP has a unique solution that is R = QP*
Proof:
R=Q+(Q+RP)P [After putting the value R=Q+RP]
=Q+QP+RPP
When we put the value of R recursively again and again, we get the following equation:
R=Q+QP+QP2+QP3…..
R=Q(ε+P+P2+P3+…. )
R=QP*[As P* represents (ε+P+P2+P3+….)]
Question 18
Total number of simple graphs that can be drawn using six vertices are:
A
215
B
214
C
213
D
212
       Engineering-Mathematics
Question 18 Explanation: 
To get total number of simple graphs, we have a direct formula is 2(n(n-1)/2).
=26(5)/2
=215
Question 19
The 2-3-4 tree is a self balancing data structure, which is also called:
A
2-4 tree
B
B+ tree
C
B-Tree
D
None of the options
       Data-Structures
Question 19 Explanation: 
In computer science, a 2–3–4 tree (also called a 2–4 tree) is a self-balancing data structure that is commonly used to implement dictionaries.
The numbers mean a tree where every node with children (internal node) has either two, three, or four child nodes:
→ a 2-node has one data element, and if internal has two child nodes;
→ a 3-node has two data elements, and if internal has three child nodes;
→ a 4-node has three data elements, and if internal has four child nodes.
Question 20
Which type of algorithm is used to solve the "8 Queens" problem?
A
Greedy
B
Dynamic
C
Divide and conquer
D
Backtracking
       Algorithms       Backtracking
Question 20 Explanation: 
→ 8 queens problem is based backtracking method.
→ Dijkstra’s algorithm is best example for greedy
→ Floyd warshall is best example for dynamic programming.
→ Quicksort is best example for divide and conquer.
Question 21
A 3.5 inch micro floppy high density disk contains he data__
A
720 MB
B
1.44 MB
C
720 KB
D
1.44 KB
       Operating-Systems
Question 21 Explanation: 
→ The 3.5-Inch floppy diskettes have dimensions of 8.9cm in width by 9.3cm in height and are referred to as floppies because of the circular magnetic floppy within the hard shell.
→ 3.5-inch floppy diskettes come in sizes of 720 KB low-density, 1.44 MB high density capacity, and IBM even developed an Extended Density disk capable of holding 2.88 MB.
Question 22
A subnet mask in class C can have ___ 1's with the remaining bits 0's
A
10
B
24
C
12
D
7
       Computer-Networks
Question 22 Explanation: 
→ Subnet mask for class C is 255.255.255.0.
→ Convert 255.255.255.0 into binary format.
11111111.11111111.11111111.00000000.
→ Total number of 1’s are 24. And total number of 0’s are 8.
Question 23
What is the value of acknowledgement field in segment?
A
Number of previous bytes to receive
B
Total number of bytes to receive
C
Number of next bytes to receive
D
Sequence of zero's and one's
       Computer-Networks
Question 23 Explanation: 
Acknowledgement field in a segment defines the number of next bytes to receive.
Question 24
The number (25)6 base 6 is equivalent to __ in binary number system
A
11001
B
10001
C
11000
D
10000
       Digital-Logic-Design
Question 24 Explanation: 
(25)6=(10001)2
Here, First we have to convert (25)6 into decimal number system. Then we have to convert into binary. (25)6=(17)10=(10001)2
Question 25
Which of the following is a class B host address?
A
230.0.0.0
B
130.4.5.6
C
230.7.6.5
D
30.4.5.6
       Computer-Networks
Question 25 Explanation: 
→ Class A addresses only include IP starting from 1.x.x.x to 126.x.x.x only. The IP range 127.x.x.x is reserved for loopback IP addresses.
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.

Note: Class A addresses 127.0.0.0 to 127.255.255.255 cannot be used and is reserved for loopback and diagnostic functions.
Private IP Addresses:

Question 26
There is a need to create a network that has 5 subnets, each with at least 16 hosts. which one is used as classful subnet mask?
A
255.255.255.192
B
255.255.255.248
C
255.255.255.240
D
255.255.255.224
       Computer-Networks
Question 26 Explanation: 
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts.
Question 27
What is compaction?
A
a technique for overcoming internal fragmentation
B
a paging technique
C
a technique for overcoming external fragmentation
D
a technique for overcoming fatal error
       Operating-Systems
Question 27 Explanation: 
→ Compaction refers to combining all the empty spaces together and processes. Compaction helps to solve the problem of fragmentation, but it requires too much of CPU time.
→ It moves all the occupied areas of store to one end and leaves one large free space for incoming jobs, instead of numerous small ones.
→ In compaction, the system also maintains relocation information and it must be performed on each new allocation of job to the memory or completion of job from memory.
Question 28
How can we set a cookie visibility scope to local storage?
A
$
B
%
C
/
D
All of the options
       HTML
Question 28 Explanation: 
Setting the path of a cookie to “/” gives scoping like that of localStorage and also specifies that the browser must transmit the cookie name and value to the server whenever it requests any web page on the site
Question 29
A finite automaton accepts which type of language:
A
Type 0
B
Type 1
C
Type 2
D
Type 3
       Theory-of-Computation
Question 29 Explanation: 

Question 30
Which NetWare protocol provides link state routing?
A
NLSP
B
RIP
C
SAP
D
NCP
       Computer-Networks
Question 30 Explanation: 
→ NetWare Link Services Protocol (NLSP) provides link-state routing. SAP (Service Advertisement Protocol) advertises network services.
→ NCP (NetWare Core Protocol)provides client-to-server connections and applications.
→ RIP is a distance vector routing protocol.
Question 31
Which layer connects the network support layers and user support layers?
A
Transport layer
B
Network layer
C
Data link layer
D
Session layer
       Computer-Networks
Question 31 Explanation: 
→ Physical, data link and network layers are network support layers.
→ Session, presentation and application layers are user support layers.
→ Transport layer layer connects the network support layers and user support layers
Question 32
Process is in a ready state____
A
When process is scheduled to run after some execution
B
When process is unable to run until some task has been completed
C
When process is using the CPU
D
None of the above
       Operating-Systems
Question 32 Explanation: 
When process is unable to run until some task has been completed, the process is in blocked state and if process is using the CPU, it is in running state.
Question 33
(0+ɛ)(1+ɛ) represents:
A
{0,1,01,ɛ}
B
{0,1,ɛ}
C
{0,1,01,11,00,10,ɛ}
D
{0,1}
       Theory-of-Computation
Question 33 Explanation: 
(0+ ε) (1+ε)= (ε+0) (ε+1)
={ ε, 0, 1, 01 }
Question 34
The average search time of hashing, with linear probing will be less if the load factor:
A
is far less than 1
B
equals 1
C
is far greater than 1
D
none of the options
       Data-Structures
Question 34 Explanation: 
→ In other words, insert, remove and search operations can be implemented in O(1), as long as the load factor of the hash table is a constant strictly less than one.
Question 35
How many times the word "PROCESS" will be printed when executing the following program?
main()
{
printf("process");
fflush();
fork();
fork();
}
A
8
B
4
C
6
D
7
       Operating-Systems
Question 35 Explanation: 
→ fflush() is typically used for output stream only. Its purpose is to clear (or flush) the output buffer and move the buffered data to console (in case of stdout) or disk (in case of file output stream).
→ The two fork() calls create 3 child processes, and hence "PROCESS" will be executed 4 times if we don't use fflush.
→ If we put a '\n' at end of printf or use fflush(stdout); only 1 printf will be done.
Question 36
If a planner graph, having 25 vertices divides plane into 17 different regions. Then how many edges are used to connect the vertices in this graph
A
20
B
30
C
40
D
50
       Engineering-Mathematics
Question 36 Explanation: 
Euler’s Formula : For any polyhedron that doesn’t intersect itself (Connected Planar Graph), F(faces) + V(vertices) − E(edges) = 2.
Here as given, F=?,V=25 and E=17
→ F+25-17=2
→ 40
Question 37
Merge sort uses:
A
Divide and conquer
B
Backtracking
C
Heuristic approach
D
Greedy approach
       Algorithms
Question 37 Explanation: 
→ Merge Sort is a divide and conquer algorithm. It works by recursively breaking down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly.
→ The solutions to the sub-problems are then combined to give a solution to the original problem.
→ So Merge Sort first divides the array into equal halves and then combines them in a sorted manner.
Question 38
What is the relation between DFA and NFA on the basis of computational power?
A
DFA>NFA
B
NFA>DFA
C
Equals
D
Can't be said
       Theory-of-Computation
Question 38 Explanation: 
Both NFA and DFA having same computational power, as both defines the same class of regular languages.
Question 39
If a random coin is tossed 11 times, then what is the probability that for 7th toss head appears exactly 4 times?
A
5/32
B
15/128
C
35/128
D
None of the options
       Engineering-Mathematics
Question 39 Explanation: 
→ To find probability that for the 7th toss head appears exactly 4 times. We have find that past 6 run of tosses.
→ Past 6 tosses, What happen we don’t know. But according to 7th toss, we will find that head appeared for 3 times.
→ Any coin probability of a head=½ and probability of a tail=½
→ If we treat tossing of head as success,then this leads to a case of binomial distribution.
→ According to binomial distribution, the probability that in 6 trials we get 3 success is
6C3*(½)3 *(½)3= 5/16 (3 success in 6 trials can happen in 6C3 ways). → 7th toss, The probability of obtaining a head=1/2
→ In given question is not mention that what happens after 7 tosses.
→ Probability for that for 7th toss head appears exactly 4 times is =(5/16)*(1/2)
=5/32.
Question 40
Which of the following is illegal declaration in C language?
A
char*str="Raj is a research scholar";
B
charstr[25]="Raj is a research Scholar";
C
charstr[40]="Raj is a research Scholar";
D
char[] str="Raj is a research Scholar";
       Data-Structures
Question 40 Explanation: 
The initialization of variable in option D is not valid.
Variable initialization syntax : Datatype varible_name[size]="sring"
(or)
Datatype variable_name[ ]=”string”
Question 41
When we use slow start algorithm, the size of the congestion window increase__unit it reaches a threshold
A
Additively
B
Multiplicatively
C
Exponentially
D
None of the options
       Computer-Networks
Question 41 Explanation: 
→ In the slow-start algorithm, the size of the congestion window increases exponentially until it reaches a threshold.
→ ​TCP slow start is an algorithm which balances the speed of a network connection. Slow start gradually increases the amount of data transmitted until it finds the network’s maximum carrying capacity.
​→ TCP slow start is one of the first steps in the congestion control process. It balances the amount of data a sender can transmit (known as the congestion window) with the amount of data the receiver can accept (known as the receiver window).
→ The lower of the two values becomes the maximum amount of data that the sender is allowed to transmit before receiving an acknowledgment from the receiver.
Question 42
In Ipv4 addresses, classful addressing is replaced with:
A
Classless addressing
B
Classful addressing
C
Subnet advertising
D
None of the options
       Computer-Networks
Question 42 Explanation: 
→ Classful addressing is replaced with classless addressing.
→ The problem with this classful addressing method is that millions of class A address are wasted, many of the class B address are wasted, whereas, number of addresses available in class C is so small that it cannot cater the needs of organizations.
→ Class D addresses are used for multicast routing, and are therefore available as a single block only. Class E addresses are reserved.
Note: Because of these problems, Classful networking was replaced by Classless Inter-Domain Routing (CIDR) in 1993.
Question 43
An ethernet destination address 07-01-12-03-04-05 is:
A
Unicast address
B
Multicast address
C
Broadcast address
D
All of the options
       Computer-Networks
Question 43 Explanation: 
Here, Shortcut method to find out using MAC address is
The first octant of MAC address written in binary format.
07=0000 0111.
The least significant bit 0 then unicast, 1 means Multicast and entire 8 bits are 1’s then broadcast.
Question 44
The open file table has a/an __ associated with each file.
A
file content
B
file permission
C
open count
D
close count
       Operating-Systems
Question 44 Explanation: 
Open count indicates the number of processes that have the file open
Question 45
The process of loading the operating system into memory is called:
A
Booting
B
Spooling
C
Thrashing
D
Formatting
       Operating-Systems
Question 45 Explanation: 
The operating system is loaded through a bootstrapping process, more succinctly known as booting. A boot loader is a program whose task is to load a bigger program, such as the operating system.
Question 46
Why is one-time password safe?
A
it is easy to generate
B
it cannot be shared
C
it is different for every access
D
it can be easily decrypted
       Computer-Networks
Question 46 Explanation: 
One time password is safe since it is generated per access and thus cannot be brute forced or deduced.
Question 47
How many DFA's exits with two states over input alphabet {0,1}?
A
16
B
26
C
32
D
64
       Theory-of-Computation
Question 47 Explanation: 
Number of DFA’s = 2n * n(2*n)
=22*24
=4*16
=64
Question 48
The address field of linked list:
A
contain address of next node
B
may contain null character
C
contain address of next pointer
D
both (A) and (B)
       Data-Structures
Question 48 Explanation: 
→ Each record of a linked list is often called an 'element' or 'node'.
→ The field of each node that contains the address of the next node is usually called the 'next link' or 'next pointer'.
→ The remaining fields are known as the 'data', 'information', 'value', 'cargo', or 'payload' fields. Note: Option C is also correct.
Question 49
Given two sorted list of size 'm' and 'n' respectively. he number of comparisons needed in the worst case by the merge sort algorithm will be:
A
m*n
B
minimum of m,n
C
Maximum of m,n
D
M+n-1
       Algorithms
Question 49 Explanation: 
→ To merge two lists of size m and n, we need to do m+n-1 comparisons in worst case.
→ Since we need to merge 2 at a time, the optimal strategy would be to take smallest size lists first.
→ The reason for picking smallest two items is to carry minimum items for repetition in merging.
Question 50
Complement of (a+b)* will be:
A
π
B
C
a
D
b
       Theory-of-Computation
Question 50 Explanation: 
Given expression accept all string so complement will accept nothing
Question 51
In a particular system it is observed that, The cache performance gets improved as a result of increasing the block size of he cache. The primary reason behind this is:
A
Programs exhibits temporal locality
B
Programs have small working set
C
Read operation is frequently required rather than write operation
D
Programs exhibits spatial locality
       Computer-Organization
Question 51 Explanation: 
Helps improve miss rate b/c of principle of locality:
1)Temporal locality says that if something is accessed once, “it will probably be accessed again soon"
2)Spatial locality says that if something is accessed, something nearby it will probably be accessed
Note: Larger block sizes help with spatial locality. Easiest way to reduce miss rate is to increase cache block size
Question 52
Starvation can be avoided by which of the following statement:
i.By using shortest job first resource allocation policy
ii.By using first come first serve resources allocation policy
A
i only
B
i and ii only
C
ii only
D
None of the options
       Operating-Systems
Question 52 Explanation: 
Starvation possible in SJFS allocation policy because it is preemptive and always giving priority to shortest jobs.
→ Starvation never happen into FCFS because it is purely non preemptive allocation strategy.
Question 53
Page fault frequency in an operating system is reduced when the:
A
Process tend to be I/O bound
B
Locality of reference is applicable to the process
C
size of pages is reduced
D
Process tend to be CPU bound
       Operating-Systems
Question 53 Explanation: 
We can reduce page faults using some methods
1.We can increase size of main memory
2.Decreasing the degree of multiprogramming
3.Locality of reference is applicable to the process
4.Possible to increase the page size.
Question 54
The expression 5-2-3*5-2 will evaluate to 19, if:
A
'-' is left associative and '*' has precedence over '-'
B
'-' is right associative and '*' has precedence over '-'
C
'-' is right associative and '-' has precedence over '*'
D
'-' is left associative and '-' has precedence over '*'
       Data-Structures
Question 54 Explanation: 
Evaluation procedure:
Step-1: 5-2-3*5-2 will be 19.
Step-2: if it is treated as (5-(2-3))*(5-2)
Step-3: - has precedence over * and if it associates from the right.
Question 55
Finite automata requires minimum__ number of stacks
A
1
B
0
C
2
D
none of the options
       Theory-of-Computation
Question 55 Explanation: 
Finite automata doesn’t require any stack operation
Question 56
In classless addressing, there are no classes but addresses are still granted in:
A
Codes
B
Blocks
C
IPs
D
Sizes
       Computer-Networks
Question 56 Explanation: 
→ When we have run out of class A and B addresses, and a class C block is too small for most midsize organizations.
→ To overcome the problem of address depletion and give more organizations access to internet, classless addressing was designed and implemented.
→ The size of the block (the number of addresses) varies based on the nature and size of the entry. For example, a household may be given only two addresses; a large organization may be given thousands of addresses. An ISP, may be given thousands or hundreds of thousands based on the number of customer it may serve.
→ To simplify the handling of addresses, the Internet authorities impose three restrictions on classless address blocks:
1. The addresses in a block must be contiguous, one after the other.
2. The number of addresses in a block must be a power of 2 (1, 2, 4,8, .... ).
3. The first address must be evenly divisible by the number of address.
Question 57
In boolean algebra 1+1+1+1...800 times ones=___
A
1
B
0
C
11
D
800
       Aptitude
Question 57 Explanation: 
In Boolean Algebra, the operator + represents OR operation.
I.1+ X = 1. So 1+(1+1…+1)= 1+(X)= 1
II.0+ X = X
Question 58
The function f(x)= (x2 - 1)/(x-1) at x=1 is:
A
Continuous and differentiable
B
Continuous but not differentiable
C
Differentiable but not continuous
D
Neither continuous nor differentiable
       Engineering-Mathematics
Question 58 Explanation: 
For a given f(x) can not be defined at x=1 because the function gives 0/0 .
So we can’t any thing about continuous or differentiable .So suitable options are neither continuous nor differentiable.

Question 59
If X,Y and Z are three exhaustive and mutually exclusive events related with any experiment and the P(X)=0.5P(Y) and P(Z)=0.3P(Y). then P(Y)=___
A
0.54
B
0.66
C
0.33
D
0.44
       Engineering-Mathematics
Question 59 Explanation: 
If X,Y and Z are three exhaustive and mutually exclusive events, the
P(X)+P(Y)+P(Z)=1
Given data is P(X)=0.5P(Y) and P(Z)=0.3P(Y)
Substitute the values in the above equation
0.5P(Y)+P(Y)+0.3P(Y) =1
1.8P(Y)=1
P(Y)=1/1.8=0.5555
Question 60
The demands of the fragmentation are:
A
Complex routers
B
Open to DOS attack
C
No overlapping of fragments
D
(A) and (B) both
       Computer-Networks
Question 60 Explanation: 
→ IP fragmentation is the process of breaking up a single Internet Protocol (IP) packet into multiple packets of smaller size. Every network link has a characteristic size of messages that may be transmitted, called the maximum transmission unit (MTU).
→ The support for fragmentation of larger packets provides a protocol allowing routers to fragment a packet into smaller packets when the original packet is too large for the supporting data link frames.
→ IP fragmentation exploits (attacks) use the fragmentation protocol within IP as an attack vector.
IP fragment over lapped:
The IP fragment overlapped exploit occurs when two fragments contained within the same IP packet have offsets that indicate that they overlap each other in positioning within the packet. This could mean that either fragment A is being completely overwritten by fragment B, or that fragment A is partially being overwritten by fragment B.
Overlapping fragments may also be used in an attempt to bypass Intrusion Detection Systems
There are 60 questions to complete.