Ans .
B
answer B We can download 56 Kb in one sec. Data downloaded in 2 minutes = 56 K x 2 x 60 =6720 Kb = 6.72 Mbits
Ans .
D
p-Persistent CSMA Protocol: 1. If the medium is idle, transmit with probability p, and delay for one time unit with probability (1 - p) 2. If the medium is busy, continue to listen until medium becomes idle, then go to Step 1 3. If transmission is delayed by one time unit, continue with Step 1.
Ans .
B
Polyalphabetic rest two are one to one.
Ans .
A
Answer : 100 Meter The maximum allowed length of a Cat 6 cable is 100 meters or 328 feet.
Ans .
D
Answer :Common Gateway Interface Common Gateway Interface (CGI) is a standard way for web servers to interface with executable programs installed on a server that generate web pages dynamically. Such programs are known as CGI scripts or simply CGIs; they are usually written in a scripting language. Reference : CGI Short for HyperText Markup Language, the authoring language used to create documents on the World Wide Web. HTML defines the structure and layout of a Web document by using a variety of tags and attributes. Reference : HTML The File Transfer Protocol (FTP) is a standard network protocol used to transfer computer files between a client and server on a computer network. HTTP is short for HyperText Transfer Protocol. HTTP is the underlying protocol used by the World Wide Web and this protocol defines how messages are formatted and transmitted.
Ans .
D
Answer C Simple explanation of count to infinity problem The core of the count-to-infinity problem is that if A tells B that it has a path somewhere, there is no way for B to know if the path has B as a part of it. To see the problem clearly, imagine a subnet connected like A–B–C–D–E–F, . Now suppose that A is dead. In the vector-update-process B notices that the A is down. The problem is, B also gets an update from C, and C says B that A is only two hops from C (C to B to A), which is false. Since B doesn't know that the path from C to A is through itself (B), it updates its table with the new value "B to A = 2 + 1". Later on, B forwards the update to C and due to the fact that A is reachable through B (From C point of view), C decides to update its table to "C to A = 3 + 1". This slowly propagates through the network until it reaches infinity
Ans .
D
The IEEE single-precision and double-precision format to represent floating-point numbers, has a length of 32 and 64 respectively.
Ans .
D
The n vertex graph with the maximal number of edges that is still disconnected is a Kn-1 a complete graph Kn-1 with n-1 vertices has (n-12)(n-12)edges, so (n-1)(n-2)2(n-1)(n-2)2 edges. Adding any possible edge must connect the graph, so the minimum number of edges needed to guarantee connectivity for an n vertex graph is (n-1)(n-2)2(n-1)(n-2)2+1 The maximum number of edges to be included in G so that graph is not connected =(n-1)(n-2)2(n-1)(n-2)2=(99)(98)2(99)(98)2=4851 Hence,Option(C)4851.
Ans .
D
Answer : Search, Insert and Delete.
Amortized analysis is a method for analyzing a given algorithm's time complexity, or how much of a resource,
especially time or memory in the context of computer programs, it takes to execute.
A splay tree is a self-adjusting binary search tree with the additional property that recently accessed elements
are quick to access again. It performs basic operations such as insertion, look-up and removal in O(log n) amortized time.
Ans .
D
A maximum depth path always takes the larger part of partition
In question given that always β <=0.5
So maximum partition will be (1-β ) n
We have to find maximum depth d, means how many times quick sort loop will be executed!!
In case of maximum depth path ( let maximum depth be d)
(1-β)d n = 1
(1-β)d = 1/n
Take logarithm on both sides,
d log ( 1-β) = -log n
d = -log n/log ( 1-β)
Similarly we can find minimum depth d' = -log n/log β
Ans .
C
d + 1. Notice that its just a binary tree. A binary tree can have maximum of two children, but not more. Here its asking the minimum, hence we will consider only one node at each level. Hence for the depth of d, we need d + 1 nodes. For example consider this tree. Each and every node 1 child.
Ans .
C
Doubly linked list
Reason:
only the node to be inserted/deleted needs to be manipulated and no traversals in forward or backward direction required for the operation so it is more efficient.
Ans .
C
ans is C 5 states
Ans .
D
The Language here is L={?,0,1,01,011,101,0101..................}
But
for A) 010010 possible
B)1001 possible
C)0100 possible
These all violates the condition for no pair of consecutive 0
Ans .
C
A is the answer.
L1 is regular but not L2.
In L2 we have to remember the value of l (no of b's) so that it can be compared with k.
regular grammar cannot remember a value (It has no stack)
Thus it is not regular
Ans .
C
Answer is C
in option A) n > 0 and m >0 in every condition so m=0 is not satisfied
in option B) There is no way to end the cycle S1-> bS1c
in option D) same condition as A
Ans .
A
Answer A: S1 is correct but S2 is wrong!
For a Context free language: Membership, Emptiness, Finiteness are decidable. ie we can make an algorithm to check whether a given CGLis infinite or not.
But we cannot design an algorithm to check equivalence of two CFLS.
Following are decidable for various types of languages
Regular language: {Membership, Emptiness, Finiteness, Equivalence, Everything (S*), Ambiguity, Regularity, Disjointedness}
Context free language: {Membership, Emptiness, Finiteness}
Context sensitive : {Membership }
Recursive: {Membership }
Recursive Enumerable : None.
Hope now you can easily solve this type of questions
Ans .
A
Number of 8-bit strings beginning with 111 are 32. First 3 bits are fixed and the remaining 5 bits can be 0 or 1. So the total combinations are 25=3225=32. Same is the case with the strings starting with 101. So total number of strings are 32+32=6432+32=64
Ans .
B
Answer B
Total number of ways 12 offices can be painted = 12!
But 3 of them will be green, 2 of them pink, 2 of them yellow and 5 of them white.
Answer = 12! /(3!* 2 ! * 2! * 5!) = 166320
Ans .
C
Lets go one by one,
(i) A graph in which there is a unique path between every pair of vertices is a tree.
==> This statement is true, Because graph can have unique path only when it does not have cycle. And according to the definition of tree, its a graph without cycel. Hence this is a valid statement.
(ii) A connected graph with e = v − 1 is a tree.
==> This statement is true. Not every graph with e = v − 1, will be a tree. But if the graph is connected hence its true.
(iii) A graph with e = v − 1 that has no circuit is a tree.
==> This statement is also true, Here he has not mentioned the connected thing, but mentioned that it has no circuit. It means that its connected.
Ans .
C
If we consider given In-order and Post Order and construct the tree then no option is going to match .
This is the tree with given in-order and post-order
If you traverse it from top to bottom left to right and print each element second time then we are going to get the exact in-order given in question .
if we print the element last time then we are going to get the post-order. But for pre-order it does not give us any option from given options, So correct pre-order should be
Preorder : abejknpofdglcmhi
Ans .
B
i=1 C(ni, 2) edges
A simple graph with n vertices is connected if if has more than (n−1)(n−2)2(n−1)(n−2)2 edges.
The n vertex graph with the maximal number of edges that is still disconnected is a Kn-1
a complete graph Kn−1 with n−1 vertices has (n−12)(n−12)edges, so (n−1)(n−2)2(n−1)(n−2)2 edges.
Adding any possible edge must connect the graph, so the minimum number of edges needed to guarantee connectivity for an n vertex graph is (n−1)(n−2)2(n−1)(n−2)2+1
Hence,Option(B)More than (n−1)(n−2)2(n−1)(n−2)2.
Ans .
C
Ex-OR and Ex-NOR are best suited for parity checking and Parity generating. Hence,Option(C)Ex-OR and Ex-NOR.
Ans .
C
m=n≥ 2 If G is connected simple graph with n vertices where n ≥ 3, then G has a Hamilton circuit if the degree of each vertex is at least n/2.
Ans .
B
let the F & G be two boolean function of degree 1:
with degree n =2^2^n total boolean function
with degree n= 2^2^1=4 boolean function
F^1-->F & G^1-->G
F=( 4 boolean function)
G= (4 boolean function)
F+G=( 8 boolean function)
F*G= (16 boolean function)
so answer is B
Ans .
A
Since 48 cards of a 52-card deck are not aces, you are guaranteed to have at least two aces only if you have at least 50 cards.
Ans .
B
B has perfect pairs.
Ans .
D
The Object Request Broker (ORB) manages interaction between clients and servers. This includes the distributed computing responsibilities of location, referencing and 'marshaling' of parameters and results. The Object Request Broker or ORB takes care of all of the details involved in routing a request from client to object, and routing the response to its destination.
Ans .
A
answer Waterfall model
Other flaws of waterfall model are:
Adjusting scope during the life cycle can kill a project
High risk and uncertainty.
Poor model for complex and object-oriented projects.
Poor model for long and ongoing projects.
Poor model where requirements are at a moderate to high risk of changing.
Ans .
C
C has perfect pairs.
Ans .
B
Answer B Black box. Equivalence partitioning. Divides the input domain into classes of data, 1.If an input condition specifies a range, define one valid and two invalid equivalence classes. 2.If an input condition requires a specific value, define one valid and two invalid equivalence classes. 3.If an input condition specifies a member of a set, define one valid and one invalid equivalence class. 4.If an input condition is Boolean, define one valid and one invalid class.
Ans .
D
Let total page =100. Effective Access Time = (30+100)* .7 +( 30+100+100)*(1- .7)=91+69=160
Ans .
D
D has perfect pairs.
Ans .
B
example of banker's algorithm.
Ans .
B
B has perfect pairs.
Ans .
C
option C will generate required output.
Ans .
B
Answer B Deadlock situation. It breaks one of the necessary four conditions of deadlock. Exclusive access. Processes request exclusive access to resources. Wait while hold. Processes hold previously acquired resources while waiting for additional resources. No preemption. A resource cannot be preempted from a process without aborting the process. Circular Wait. There exists a set of blocked processes involved in a circular wait. It is not a desirable condition for a system.
Ans .
A
XSTL :eXtensible Style Sheet Language Transformations is a language for transforming XML documents into other XML documents, or other formats such as HTML for web pages, plain text or into XSL Formatting Objects, which may subsequently be converted to other formats, such as PDF, PostScript and PNG.
Ans .
A
Answer A: hiding.
What is a view?
Views are subset of table.
View also has set of records in the form of rows and columns.
But it is created based on the records in one or more tables.
A table will have large number of data and table will be fired with specific frequently. In such case, instead of rewriting the query again and again, a name is given to the query and it will be called whenever it is required.
Hence view is also called as named query or stored query.
Ans .
A
Decision Tree Classifier is a classification technique. Classification technique is a systematic approach to build classification models from an input data set. For example, decision tree classifiers, rule-based classifiers, neural networks, support vector machines, and naive Bayes classifiers are different technique to solve a classification problem.
Ans .
B
A Crosstab query is a type of select query which allows data to view as both horizontally and vertically so that the data can be more compact and easier to read.
Ans .
D
PGP stands for Pretty Good Privacy is a data encryption and decryption computer program that provides cryptographic privacy and authentication for data communication. PGP is a popular program used to encrypt and decrypt email over the Internet, as well as authenticate messages with digital signatures and encrypted stored files.
Ans .
B
Explanation of Thomas write rule It is timestamp-based concurrency control. It states that, if a more recent transaction has already written the value of an object, then a less recent transaction does not need perform its own write since it will eventually be overwritten by the more recent one. Label each value with a write timestamp (WTS) WTS indicates the timestamp of the last transaction to modify the value. If the write timestamp of the object is greater than the time stamp of the transaction performing a write,, the write is discarded.
Ans .
C
friend function allows one class to access other unrelated class.
Ans .
B
When TLB hot occur, TLB+ memory = 20+100 =120 ns
When TLB miss occur TLB+ page table+ memory =220
Average access time = 80%*120+20%*220 =140 ns
Ans .
C
example of External level data hiding.
Ans .
A
maximum file size is get when we consider double indirect block address. Each indirect block stores 256/2 =128 pointers. double indirect block can point 128 x 128 blocks Maximum file size is 128x128x256 = 4194304 B= 4096 KB =4MB
Ans .
C
by default any variable is available only in the shell in which it is defined. if we want variables to be available to all sub shells we mst export them from the current shell by command
$ export variable_name
the proper syntax of using export variable to obtain a list of ALL exported variable is simply say export at shell prompt $ export
so here first $ title=WELCOME Then $ export title and finally sh −c ″echo $$title″
Ans .
B
THIS WAS THE IMAGE AND ANS IS B XNOR
A'B'+AB=(A'B+AB')'
Ans .
A
Without pipeline one task needs tn time. n tasks need ntn time. With pipeline, First task needs k cycles to finish. So time will be k tp Other n−1 tasks needs tp time only to finish. Total time = (k+ n −1 ) tp Speed up = T pipeline / T without pipeline = n t n / (k+n−1) t p.