Ans .

A

1. Explanation :

The postfix expression can be evaluated using a stack. First the operands are pushed into the stack and the moment a operator is encountered, operands are popped out, operation performed on the operands and the result once again pushed into the stack. This series of activities continue till the entire expression is not evaluated.Hence,Option(A)Stack.

Ans .

C

• Explanation :

A binary tree is a data structure which has at most two child nodes. A binary tree can be traversed in three ways. Inorder, preorder and postorder. The three methods of traversal can be explained this way. Left will always be followed by right. But Root would be either in the front or in the middle or at the last depending on whether it is pre, in or postorder traversal. So given a post order traversal of a binary tree, we can know the root first. In the question, the post order traversal is given as DEBFCA. Since Root is the last node to be traversed in a post order traversal we know one thing for sure. A is the root. Next, we are left with only DEBFC. Here some of the nodes belong to the left side of the binary tree and some belong to the right side. How many nodes belong to the left and how many belong to the right. Since left side of the binary tree is considered first, and since every node is expected to have at most two child, DEB will be the left side of the binary tree and FC would be the right. Now, we know that FC is in the right side of the binary tree. Again the last node would be the root of the sub tree and F its left side. Next we come to the left side of the binary tree and it is DEB. Again B would be the root of the sub tree. D and E are its left and right side respectively. So the binary tree would look something like this given below

After constructing the binary tree, writing the preorder traversal is very simple. In preorder traversal root comes first. Since A is the root, A would appear first. Following Root would be the Left and Right sub tree and so the left subtree would be BDE. Again B would be the root of the left sub tree followed by D and E which are the left and right child respectively. SO the preorder traversal till now would be ABDE. Last comes the Right sub tree. Here again C would be the root of the right sub tree followed by C its left child. So the entire preorder traversal of the tree would be ABDECF. So option C is the right answer.

• Ans .

A

1. Explanation :

it should be A.controlled transfer.

Ans .

D

1. Explanation :

According to the Five-colour theorem, Every planar graph is 5-colourable. Any planar graph G with n vertices requires five colors for proper coloring. So the answer must be option D which is 5. But in the UGC answer key it is given as 2, option A. But the correct answer is D.

Ans .

D

1. Explanation :

Packets are the units of exchanging information in network. A bridge connects two or more networks, or segments of the same network. Switches are basically Bridges but usually have multiple ports. Routers forward data packets from one place to another. They forward data depending on the network, not the hardware(MAC)address. So the answer is option D.

Ans .

A

1. Explanation :

Routing is a way to get one packet from one destination to the next. But the Layer – 5 (Application) includes all the higher-level protocols. So the layer for which protocol software is not needed by a router is Layer – 5, the Application Layer which is option A.

Ans .

D

1. Explanation :

The explanation is quite simple. It is a straightforward question with less confusing options. So if we are talking about a multiuser database and if two users wish to update the same record at the same time, they are prevented from doing so by using Record lock.

Ans .

D

1. Explanation :

The statements given in the option A and C are true, is very clear. There could be only a slight confusion whether statement in option B is true or not. The above mentioned properties are mentioned in the book on Data structures:A pseudocode approach with C written by Richard F.Gilberg et al. According to the summary on binary search trees the following properties are mentioned.
All items in the left subtree are less than the root.
All items in the right subtree are greater than or equal to the root.
Each subtree is itself a binary search tree.
So the correct answer for this question would be option D.

Ans .

C

1. Explanation :

ZAP deletes the entire file except the file structure

Ans .

B

1. Explanation :

Answer should be B as we are just copying the structure

Ans .

B

1. Explanation :

B+ trees can store a relatively large number of pointers to child nodes in each node. Fanout means the number of pointers per node. B+ trees minimize the depth of the tree( and so one can reach the leaf node faster)by employing high fanout. That is linking many children from each node. Because of a high fanout the number of I/O operations required to find an element in the tree would be less. So B+ trees are preferred. The option is B.

Ans .

D

1. Explanation :

Transaction manager performs all the operations mentioned in option A,B and C.

Ans .

B

1. Explanation :

In a B-tree leaves are at the same level. A B-tree of order m is an m-way tree which means that for each node there may be upto m children. m should be odd only.

Ans .

C

1. Explanation :

ARP is a protocol in the TCP/IP protocol suite. This protocol is required for basic TCP/IP operations. This protocol is used to find the ethernet(hardware) address from a specific IP number.

Ans .

C

1. Explanation :

Decryption and encryption of data happens in the presentation layer. One more important activity which happens at this layer is the Compression and decompression of data.

Ans .

B

1. Explanation :

Time-division multiplexing (TDM) is a method of transmitting and receiving independent signals over a common signal path by means of synchronized switches at each end of the transmission line so that each signal appears on the line only a fraction of time in an alternating pattern.

Ans .

B

1. Explanation :

There are two major classes of routing methods.One is Distance vector and the other one Link state. In distance vector, each router knows only its neighbors. Decision is taken on two factors one is distance and the other vector, that is the direction to reach. In case of link state all the routers know about the paths reachable by all other routers in the network. All routers possess a synchronized copy of the area’s link-state database. So the answer for this question would be option B.

Ans .

A

1. Explanation :

in PSK 16 level =2^4 so 4 bits needed for signal unit bit rate 9600 bps baud or signal rate
9600/4 =2400 bauds hence ans is A

Ans .

B

1. Explanation :

10BaseT – 100 meters
10Base2 – 185 meters
10Base5 – 500 meters
10BaseFL – 2000 meters
Fast Ethernet is referred by 100BaseX standard. 100BaseX has three specifications. Here 100 refers to the speed 100Mbps. Base refers to Baseband.The specifications are
100BaseT4
100BaseTX
100BaseFX
100BaseFx uses the 2-strand fiber-optic cable and the station to hub distance is 2000 metros.

Ans .

D

1. Explanation :

Main aim of software engineering is to produce software within budget in the given schedule

Ans .

C

1. Explanation :

In CMM model there are five maturity levels identified by the numbers 1 to 5. They are
Initial
Managed
Defined
Quantitatively Managed
Optimizing
If an organization is at level 2, it means it has crossed over level 1 and the same holds true for subsequent levels. At level 4, the process at level 3 is also included. CMM level 5 is not correct because the question only talks about level 4. CMM level 2 is not correct because level 3 includes level 2 as well. So option C is the correct answer.

Ans .

B

1. Explanation :

Validation – Are we building the right productb
Verification – Are we building the product right.

Ans .

C

1. Explanation :

Statistical process control (SPC) is a method of quality control which uses statistical methods. SPC is applied in order to monitor and control a process. Monitoring and controlling the process ensures that it operates at its full potential.
In stochastic analysis, a part of the mathematical theory of probability, a predictable process is a stochastic process whose value is knowable at a prior time.

Ans .

B

1. Explanation :

A function oriented design focusses on creating modules or functions and each module supports its own abstraction. The design purpose is to minimize coupling and maximize cohesion. Cohesion is a way of understanding how close or bound your module is. Coupling is the level of interactivity between modules. For a good design to happen cohesion should be more and coupling should be less.

Ans .

B

1. Explanation :

Line of count and Member of token depend on the programming language. Function count does not depend on programming language. Function count are a unit of measure for software just like a unit of measure for temperature would be degrees.

Ans .

C

1. Explanation :

Actually the answer given in the UGC answer key is option A. But i would beg to differ with that. Using the formula for calculating the degree with the parameters specified above, the answer we are arriving at is 43 which is actually option C. Let me explain how it is done.
Let n be the degree.
Given, Key size(length of the name attribute of STUDENT) = 8 bytes(k)
Index pointer size = 4 bytes (b)
Disk Block size = 512 bytes
Degree of B+ tree can be calculated if we know the maximum number of key a internal node can have. The formula for that is
(n-1)k+n*b= blocksize
(n-1)*8+n*4=512
8n – 8 + 4n = 512
12n=520
n=520/12=43
So, the answer is option C.

Ans .

B

1. Explanation :

In a binary search tree all the elements to the root of the tree will lesser than that of the root. Also elements greater than the root will be to the right of the root. So, in order traversal of such a tree would yield a sorted listing of elements.

Ans .

B

1. Explanation :

Mobile IP is the basic behind how wireless devices offer IP connectivity. Agent discovery and registration are two basic functions involved here. So the options is B.

Ans .

A

1. Explanation :

Every process is allocated a specific time slice in the CPU and it runs for that entire time. In pre-emptive scheduling even before the process’s time slice expires, it is temporarily suspended from its execution. So the option is A.

Ans .

D

1. Explanation :

There are few criterias used for measuring the performance of a particular scheduling algorithm. The turn around time is the interval of time between the submission of a process and its completion.
The wait time is the amount of time a process has been waiting in the ready queue.
The response time is the time taken between the process submission and the first response produced.
In RR algorithm, the value of time quantum or the time slice, plays a crucial role in deciding how effective the algorithm is. If the time quantum is too small, there could be lot of context switching happening which could slow down the performance. If the time quantum is too high, then RR behaves like FCFS. If the time quantum is increased, the average response time varies irregularly. If you take any comprehensive material on operating system, you will come across a graph which depicts this behavior. So the answer is option D.

Ans .

A

1. Explanation :

There are four necessary and sufficient conditions for a deadlock. One of them is Mutual Exclusion which means that the resources involved are non-sharable. So the answer is option A.

Ans .

B

1. Explanation :

Compared to the processor speed, the speed of the primary memory is slow. Cache memory is a small memory which sits in between the processor and primary memory and fetches information to the processor at a much higher speed or it makes it appear so. Caching can be effective based on a property of computer programs called locality of reference. Analysis of program show that the majority of the execution time is spent around a small part of the program may be a simple loop,nested loop or a few functions. The rest of the program is accessed infrequently. There is something called temporal locality and spatial locality also which we need to know when we talk about cache. But cache memory is ideally suited for small loops. Interleaved memory is a technique for increasing the speed of RAM. Here multiple memory chips are grouped together to form what are known as banks.Each of them take turns for supplying data. An interleaved memory with “n” banks is said to be n-way interleaved. Macintosh systems are considered to be one using memory interleaving. So the answer for this question is option B.

Ans .

C

1. Explanation :

The reference string is 4,3,2,1,4,3,5,4,3,2,1,5

The number of frames m = 4
The first 4 references(4,3,2,1) cause page faults and brought into the empty frames.
The next reference (4) is already available and so there is no page fault.
The next reference (3) is also already available and so there is no page fault.
The next reference (5) replaces page 4 which was brought in first.No of page faults=5.
Next reference (4) replaces page 3 which is the next to come in.No of page faults=6.
Next reference (3) replaces 2.No of page faults till now=7.
Next reference (2) replaces 1 which was the last of the pages to come in. No of page faults till now=8.
Next reference (1) replaces 5 which was the first to come in in the second cycle. No of page faults till now=9.
The last reference in the reference string is 5 which will replace 4. No of page faults till now=10
So the answer is option C which is 10.

Ans .

D

1. Explanation :

Check sum is the error detecting mechanism when data is treated as a sequence of character. Parity is the mechanism used when data is treated as a sequence of bits.

Ans .

B

1. Explanation :

In case of signed Magnitude Representation the range is from -(2n-1 - 1) to (2n-1 - 1)
Min no that can be represented in this system is -(2n-1 - 1)
Max no that can be represented in this system is (2n-1 - 1)
In case of 2's complement no system the range is from -2n-1 to 2n-1 - 1
Min no that can be represented in this system is -2n-1
Max no that can be represented in this system is 2n-1 - 1
As they said 2 Bytes = 16 bits
we can use max no that can be represented here which is 2n-1 - 1
216-1 - 1→ 215 - 1

Ans .

B

1. Explanation :

ECL stands for Emitter-Coupled Logic. It is designed for extremely high speed application. It is well suited for large mainframe computer that require high number of operation per second.

Ans .

D

1. Explanation :

There are three types of interrupts. They are
1.External interrupts
2.Internal interrupts
3.Software interrupt
External interrupts come from I/O devices. Internal interrupts are from illegal or wrong use of an instruction or data. A software interrupt is initiated by executing an instruction. The answer key gives the answer as C. But i think it should be D.

Ans .

B

1. Explanation :

The %c format specifier prints the ascii equivalent of the value.

Ans .

C

1. Explanation :

it should be B as we are just copying the structute

ok

Ans .

B

1. Explanation :

Recursive languages are closed under the following operations.
1.Kleene star
2.Concatenation
3.Union
4.Intersection
5.Complement
6.Set difference
Recursively enumerable languages are closed under the following operations.
1.Kleene star
2.Concatenation
3.Union
4.Intersection
Recursively enumerable languages are not closed under complement. So the statements I and II are only true. So the option is B.

Ans .

C

1. Explanation :

RIP(Routing Information Protocol) and IGRP(Interior Gateway Routing Protocol) are examples of Distance Vector routing protocols and Open Shortest Path First(OSPF) is an example of Link State routing protocols. Distance Vector routing protocols are based on Bellma and Ford algorithms. Link state routing protocols are based on Dijkstra algorithms. Here the options OSPF, Link-state are ruled out. Even Dijkestra is ruled out because link state protocols are based on that. The only left out option is C

Ans .

D

1. Explanation :

E-commerce security threats are more than 3 in number and so the incorrect statement here is D.

Ans .

B

1. Explanation :

One of the closest answer to this also could be option A which is quality of the design. But it can also be understood that if a software product has a large number of defects it is unreliable. So if the reliability of the software improves it means that the number of defects has reduced. So reliability of software is dependent on the number of errors present. So it is option B.

Ans .

D

1. Explanation :

E-Commerce security threats can be grouped into three major categories.

Ans .

D

1. Explanation :

Global positioning system is a satellite based tracking system that enables the determination of person's position.

Ans .

D

1. Explanation :

A complete microcomputer system consists of Microprocessor,Memory and Peripheral equipment.

Ans .

C

1. Explanation :

Since option A,B and D are ruled out, the answer is option C.

Ans .

B

1. Explanation :

Pipelining is a technique to build fast processors. It allows the execution of multiple instruction by overlapping them. In an assembly unit every stage has one and only one activity to do. It keeps repeating them again and again. In the same way in a instruction pipeline at every clock cycle one particular step of multiple instruction will be performed. Every instruction has multiple stages. Say at the first clock cycle first step of instruction1 is performed. At the second clock cycle the second step of instruction1 and 1st step of instruction2 would be performed and so on. So pipelining is called instruction prefetch. So the option is B.

Ans .

D

1. Explanation :

Strings,Lists,Queues are linear data structure.

Ans .

C

1. Explanation :

Dequeue and priority are types of queues which is again a linear data structure. If you want to represent hierarchal elements we should go for tree and so answer is option C.