1. The postfix expression AB + CD – * can be evaluated using a
(A) stack
(B) tree
(C) queue
(D) linked list
(A) stack
(B) tree
(C) queue
(D) linked list
Ans:-A
Explanation:-This is a simple question. 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. So the answer is A.
2. The post order traversal of a binary tree is DEBFCA. Find out the pre- order traversal.
(A) ABFCDE
(B) ADBFEC
(C) ABDECF
(D) None of the above
(B) ADBFEC
(C) ABDECF
(D) None of the above
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.
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.
| TYPE | FIRST | SECOND | THIRD |
|---|---|---|---|
| Inorder | L – Left | Ro – Root | R – Right |
| Preorder | Ro – Root | L – Left | R – Right |
| Postorder | L – Left | R – Right | Ro – Root |
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.
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.
3.The branch logic that provides making capabilities in the control unit is known as
(A) Controlled transfer
(B) Conditional transfer
(C) Unconditional transfer
(D) None of the above
(B) Conditional transfer
(C) Unconditional transfer
(D) None of the above
Ans:-A
Explanation:- This question is available in lot of websites and other materials and the answer given there is unconditional transfer which is option C. But UGC answer key is given as option A. I am just going to go with A, but will find out more about this and post it. Meanwhile, if there are any explanation for this question,please post it, so that everyone can benefit.
4. The number of colours required to properly colour the vertices of every planer graph is
(A) 2
(B) 3
(C) 4
(D) 5
(B) 3
(C) 4
(D) 5
Ans:-D
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.
5. Networks that use different technologies can be connected by using
(A) Packets
(B) Switches
(C) Bridges
(D) Routers
(B) Switches
(C) Bridges
(D) Routers
Ans:-D
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.
6.Both hosts and routers are TCP/IP protocol software. However, routers do not use protocol from all layers. The layer for which protocol software is not needed by a router is
(A) Layer – 5 (Application)
(B) Layer – 1 (Physical)
(C) Layer – 3 (Internet)
(D) Layer – 2 (Network Interface)
(B) Layer – 1 (Physical)
(C) Layer – 3 (Internet)
(D) Layer – 2 (Network Interface)
Ans:-A
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.
7. In multiuser database if two users wish to update the same record at the same time, they are prevented from doing so by
(A) Jamming
(B) Password
(C) Documentation
(D) Record lock
(B) Password
(C) Documentation
(D) Record lock
Ans:-D
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.
8. A binary search tree is a binary tree :
(A) All items in the left subtree are less than root
(B) All items in the right subtree are greater than or equal to the root
(C) Each subtree is itself a binary search tree
(D) All of the above
(B) All items in the right subtree are greater than or equal to the root
(C) Each subtree is itself a binary search tree
(D) All of the above
Ans:-D
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.
9. What deletes the entire file except the file structure ?
(A) ERASE
(B) DELETE
(C) ZAP
(D) PACK
(B) DELETE
(C) ZAP
(D) PACK
Ans:-C
10. Which command is the fastest among the following ?
(A) COPY TO
(B) COPY STRUCTURE TO
(C) COPY FILE
(D) COPY TO MFILE-DA T DELIMITED
(B) COPY STRUCTURE TO
(C) COPY FILE
(D) COPY TO MFILE-DA T DELIMITED
Ans:-B
Explanation:- I am considering the same answer as what is given in the UGC answer key. I will find out why that command is supposedly faster than the rest. Give me some time. Meanwhile, i do not want to get stuck in this question and i want to proceed. So we are off to the next one but will visit the same sometime later.
11. B+ tree are preferred to binary tree in Database because
(A) Disk capacity are greater than memory capacities
(B) Disk access is much slower than memory access
(C) Disk data transfer rates are much less than memory data transfer rate
(D) Disks are more reliable than memory
(B) Disk access is much slower than memory access
(C) Disk data transfer rates are much less than memory data transfer rate
(D) Disks are more reliable than memory
Ans:-B
Explanation:- Actually the same question was given in GATE CS(2000) paper and the solution is available in a webpage called geeksforgeeks. A good page with solutions to some gate questions. Everyone knows that disk access is slower compared to memory access. When we compare a binary tree to a B+ tree, B+ tree is supposed to have a higher fanout. What is a fanout actually?. 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.
12. A Transaction Manager is which of the following ?
(A) Maintains a log of transactions
(B) Maintains before and after database images
(C) Maintains appropriate concurrency control
(D) All of the above
(B) Maintains before and after database images
(C) Maintains appropriate concurrency control
(D) All of the above
Ans:-D
Explanation:- Transaction manager performs all the operations mentioned in option A,B and C.
13. Leaves of which of the following trees are at the same level ?
(A) Binary tree
(B) B-tree
(C) A VL-tree
(D) Expression tree
(B) B-tree
(C) A VL-tree
(D) Expression tree
Ans:-B
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.
14. Which of the following TCP/IP Internet protocol is diskless machine uses to obtain its IP address from a server ?
(A) RAP
(B) RIP
(C) ARP
(D) X.25
(B) RIP
(C) ARP
(D) X.25
Ans:-C
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.
15. Decryption and encryption of data are the responsibility of which of the following layer ?
(A) Physical layer
(B) Data Link layer
(C) Presentation layer
(D) Session layer
(B) Data Link layer
(C) Presentation layer
(D) Session layer
Ans:-C
Explanation:- Not a very complicated question. 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.
16. In which circuit switching, delivery of data is delayed because data must be stored and retrieved from RAM ?
(A) Space division
(B) Time division
(C) Virtual
(D) Packet
(B) Time division
(C) Virtual
(D) Packet
Ans:-B
17. In which Routing Method do all the routers have a common database ?
(A) Distance vector
(B) Link state
(C) Link vector
(D) Dijkestra method
(B) Link state
(C) Link vector
(D) Dijkestra method
Ans:-B
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.
18. Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e. modulation rate).
(A) 2400 bauds
(B) 1200 bauds
( C) 4800 bauds
(D) 9600 bauds
(B) 1200 bauds
( C) 4800 bauds
(D) 9600 bauds
Ans:-A
Explanation:- Inorder to know why option A is the answer, let us understand the following table.
| MODULATION | BAUD RATE | BIT RATE |
|---|---|---|
| 4-PSK | N | 2N |
| 8-PSK | N | 3N |
| 16-PSK | N | 4N |
In the question it is mentioned as page shift keying, but it is actually Phase Shift Keying. Two values are provided. One is bps which is nothing but Bits per second which is 9600 and also level which is 16. According to the table given above, for a 16-PSK, given the baud rate N, the bit rate would be 4N. Here bps is provides which is 9600. In order to calculate the baud rate we have to divide bps/4, because the signal level is 16. So the answer would be 9600/4=2400 baud. That is how the answer is arrived at.
19. The station to hub distance in which it is 2000 metres.
(A) 100 Base-Tx
(B) 100 Base-Fx
(C) 100 Base-T4
(D) 100 Base-T1
(B) 100 Base-Fx
(C) 100 Base-T4
(D) 100 Base-T1
Ans:-B
Explanation:- Traditional Ethernet technologies had four standards for 10Mbps. They are,
- 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.
20.Main aim of software engineering is to produce
(A) program
(B) software
(C) within budget
(D) software within budget in the given schedule
(B) software
(C) within budget
(D) software within budget in the given schedule
Ans:-D
Explanation:- Not much explanation needed here.
21. Key process areas of CMM level 4 are also classified by a process which is
(A) CMM level 2
(B) CMM level 3
(C) CMM level 5
(D) All of the above
(B) CMM level 3
(C) CMM level 5
(D) All of the above
Ans:-C
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.
22. Validation means
(A) are we building the product right
(B) are we building the right product
(C) verification of fields
(D) None of the above
(B) are we building the right product
(C) verification of fields
(D) None of the above
Ans:-B
Explanation:-
Validation – Are we building the right product
Verification – Are we building the product right.
Validation – Are we building the right product
Verification – Are we building the product right.
23. If a process is under statistical control, then it is
(A) Maintainable
(B) Measurable
(C) Predictable
(D) Verifiable
(B) Measurable
(C) Predictable
(D) Verifiable
Ans:-C
Explanation:-
The answer is available in the book “Integrated approach to software engineering” by Pankaj Jalote. A predictable process is also said to be under statistical control.
The answer is available in the book “Integrated approach to software engineering” by Pankaj Jalote. A predictable process is also said to be under statistical control.
24. In a function oriented design, we
(A) minimize cohesion and maximize coupling
(B) maximize cohesion and minimize coupling
(C) maximize cohesion and maximize coupling
(D) minimize cohesion and minimize coupling
(B) maximize cohesion and minimize coupling
(C) maximize cohesion and maximize coupling
(D) minimize cohesion and minimize coupling
Ans:-B
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.
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.
25. Which of the following metric does not depend on the programming language used ?
(A) Line of code
(B) Function count
(C) Member of token
(D) All of the above
(B) Function count
(C) Member of token
(D) All of the above
Ans:-B
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.
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.
26. A/B+ tree index is to be built on the name attribute of the relation STUDENT. Assume that all students names are of length 8 bytes, disk block are of size 512 bytes and index pointers are of size 4 bytes. Given this scenario what would be the best choice of the degree (i.e. the number of pointers per node) of the B+ tree ?
(A) 16
(B) 42
(C) 43
(D) 44
(B) 42
(C) 43
(D) 44
Ans:-C
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
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.
27. The Inorder traversal of the tree will yield a sorted listing of elements of tree in
(A) Binary tree
(B) Binary search tree
(C) Heaps
(D) None of the above
(B) Binary search tree
(C) Heaps
(D) None of the above
Ans:-B
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.
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.
28. Mobile IP provides two basic functions.
(A) Route discovery and registration
(B) Agent discovery and registration
(C) IP binding and registration
(D) None of the above
(B) Agent discovery and registration
(C) IP binding and registration
(D) None of the above
Ans:-B
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.
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.
29. Pre-emptive scheduling is the strategy of temporarily suspending a gunning process
(A) before the CPU time slice expires
(B) to allow starving processes to run
(C) when it requests I/O
(D) to avoid collision
(B) to allow starving processes to run
(C) when it requests I/O
(D) to avoid collision
Ans:-A
Explanation:-
Again one of those rare non ambiguous question. 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.
Again one of those rare non ambiguous question. 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.
30. In round robin CPU scheduling as time quantum is increased the average turn around time
(A) increases
(B) decreases
(C) remains constant
(D) varies irregularly
(B) decreases
(C) remains constant
(D) varies irregularly
Ans:-D
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.
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.
31. Resources are allocated to the process on non-sharable basis is
(A) mutual exclusion
(B) hold and wait
(C) no pre-emption
(D) circular wait
(B) hold and wait
(C) no pre-emption
(D) circular wait
Ans:-A
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.
32. Cached and interleaved memories are ways of speeding up memory access between CPU’s and slower RAM. Which memory models are best suited (i.e. improves the performance most) for which programs ?
(i) Cached memory is best suited for small loops.
(ii) Interleaved memory is best suited for small loops
(iii) Interleaved memory is best suited for large sequential code.
(iv) Cached memory is best suited for large sequential code.
(A) (i) and (ii) are true.
(B) (i) and (iii) are true.
(C) (iv) and (ii) are true.
(D) (iv) and (iii) are true.
(ii) Interleaved memory is best suited for small loops
(iii) Interleaved memory is best suited for large sequential code.
(iv) Cached memory is best suited for large sequential code.
(A) (i) and (ii) are true.
(B) (i) and (iii) are true.
(C) (iv) and (ii) are true.
(D) (iv) and (iii) are true.
Ans:-B
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.
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.
33. Consider the following page trace : 4,3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5 Percentage of page fault that would occur if FIFO page replacement algorithm is used with number of frames for the JOB m = 4 will be
(A) 8
(B) 9
(C) 10
(D) 12
(B) 9
(C) 10
(D) 12
Ans:-C
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.
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.
34. Check sum used along with each packet computes the sum of the data, where data is treated as a sequence of
(A) Integer
(B) Character
(C) Real numbers
(D) Bits
(B) Character
(C) Real numbers
(D) Bits
Ans:-B
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.
35. If an integer needs two bytes of storage, then the maximum value of a signed integer is (A) 216– 1
(B) 215 – 1
(C) 216
(D) 215
(B) 215 – 1
(C) 216
(D) 215
Ans:-B
36.Which of the following logic families is well suited for high-speed operations ?
(A) TTL
(B) ECL
(C) MOS
(D) CMOS
(B) ECL
(C) MOS
(D) CMOS
Ans:-B
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.
37. Interrupts which are initiated by an instruction are
(A) Internal
(B) External
(C) Hardware
(D) Software
(B) External
(C) Hardware
(D) Software
Ans:-D
Explanation:- There are three types of interrupts. They are
- External interrupts
- Internal interrupts
- 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.
38. printf(“%c”, 100);
(A) prints 100
(B) prints ASCII equivalent of 100
(C) prints garbage
(D) none of the above
(B) prints ASCII equivalent of 100
(C) prints garbage
(D) none of the above
Ans:-B
Explanation:- The %c format specifier prints the ascii equivalent of the value.
39. For the transmission of the signal, Bluetooth wireless technology uses
(A) time division multiplexing
(B) frequency division multiplexing
(C) time division duplex
(D) frequency division duplex
(B) frequency division multiplexing
(C) time division duplex
(D) frequency division duplex
Ans:-C
40. Consider the following statements :
I. Recursive languages are closed under complementation.
II. Recursively enumerable languages are closed under union.
III. Recursively enumerable languages are closed under complementation.
Which of the above statements are true ?
I. Recursive languages are closed under complementation.
II. Recursively enumerable languages are closed under union.
III. Recursively enumerable languages are closed under complementation.
Which of the above statements are true ?
(A) I only
(B) I and II
(C) I and III
(D) II and III
(B) I and II
(C) I and III
(D) II and III
Ans:-B
Explanation:-
Recursive languages are closed under the following operations.
Recursive languages are closed under the following operations.
- Kleene star
- Concatenation
- Union
- Intersection
- Complement
- Set difference
Recursively enumerable languages are closed under the following operations.
- Kleene star
- Concatenation
- Union
- Intersection
Recursively enumerable languages are not closed under complement. So the statements I and II are only true. So the option is B.
41. What is the routing algorithm used by RIP and IGRP ?
(A) OSPF
(B) Link-state
(C) Dynamic
(D) Dijkestra vector
(B) Link-state
(C) Dynamic
(D) Dijkestra vector
Ans:-
Explanation:- Actually in the answer key it is given as D which is Dijkestra vector. 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 which is Dynamic and which does not comply with the answer key. So i am just not entering any particular option for it.
42. Identify the incorrect statement :
(A) The overall strategy drives the E-Commerce data warehousing strategy.
(B) Data warehousing in an E-Commerce environment should be done in a classical manner.
(C) E-Commerce opens up an entirely new world of web server.
(D) E-Commerce security threats can be grouped into three major categories.
(B) Data warehousing in an E-Commerce environment should be done in a classical manner.
(C) E-Commerce opens up an entirely new world of web server.
(D) E-Commerce security threats can be grouped into three major categories.
Ans:-D
Explanation:- E-commerce security threats are more than 3 in number and so the incorrect statement here is D.
43. Reliability of software is directly dependent on
(A) quality of the design
(B) number of errors present
(C) software engineers experience
(D) user requirement
(B) number of errors present
(C) software engineers experience
(D) user requirement
Ans:-B
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.
44. ______ is not an E-Commerce application.
(A) House banking
(B) Buying stocks
(C) Conducting an auction
(D) Evaluating an employee
(B) Buying stocks
(C) Conducting an auction
(D) Evaluating an employee
Ans:-D
Explanation:- This is a simple one and in fact does not need any explanation.
45. ______ is a satellite based tracking system that enables the determination of person’s position.
(A) Bluetooth
(B) WAP
(C) Short Message Service
(D) Global Positioning System
(B) WAP
(C) Short Message Service
(D) Global Positioning System
Ans:-D
Explanation:- Again quite a easy one and does not need much explanation.
46. A complete microcomputer system consists of
(A) Microprocessor
(B) Memory
(C) Peripheral equipment
(D) All of the above
(B) Memory
(C) Peripheral equipment
(D) All of the above
Ans:-D
Explanation:- Again no explanation needed!!!.
47. Where does a computer add and compare data ?
(A) Hard disk
(B) Floppy disk
(C) CPU chip
(D) Memory chip
(B) Floppy disk
(C) CPU chip
(D) Memory chip
Ans:-C
Explanation:- Since option A,B and D are ruled out, the answer is option C.
48. Pipelining strategy is called implement
(A) instruction execution
(B) instruction prefetch
(C) instruction decoding
(D) instruction manipulation
(B) instruction prefetch
(C) instruction decoding
(D) instruction manipulation
Ans:-B
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.
49. Which of the following data structure is linear type ?
(A) Strings
(B) Lists
(C) Queues
(D) All of the above
(B) Lists
(C) Queues
(D) All of the above
Ans:-D
No explanation needed
50. To represent hierarchical relationship between elements, which data structure is suitable ?
(A) Dequeue
(B) Priority
(C) Tree
(D) All of the above
(B) Priority
(C) Tree
(D) All of the above
Ans:-C
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.
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