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Showing posts with label KVS Compuer Science - January 2017 Question Paper. Show all posts
Showing posts with label KVS Compuer Science - January 2017 Question Paper. Show all posts

KVS PGT Computer Science Question Paper - January 2017 (Question No 26-30)

 October 11, 2018     KVS Compuer Science - January 2017 Question Paper     No comments   

KVS PGT Computer Science Question Paper - January 2017 (Question No 26-30)



Q 26. Buying and selling goods over the internet is called :

1)  Hyper-Marketing
2)  E-Commerce
3)  Cyber-Selling
4)  Euro-Conversion

Answer : 2)  E-Commerce


Q 27. Mechanism that is used to convert domain name into ip address is known as _____________.


1)  URL
2)  DNS
3)  FTP
4)  HTTP

Answer : 2)  DNS
Solution : DNS is also a internet protocol and is known as Domain Name Server or Domain Name Service. Its purpose is to translate domain names into IP addresses. Whenever we use a domain name, then a service is used for translating domain names into IP addresses and is known as DNS.


Q 28. An analog signal carries 4 bits in each signal unit. If 1000 signal units are cent per second, then baud rate and bit rate of the signal are _____ and _____ .

1)  1000 bauds/sec & 500 bps
2)  2000 bauds/sec & 1000 bps
3)  4000 bauds/sec & 1000 bps
4)  1000 bauds/sec & 4000 bps

Answer : 4)  1000 bauds/sec & 4000 bps
Solution : baud rate (signal rate) = bit rate/no of bits per signal or bit rate = baud rate x no of bits per signal
so here baud rate is given 1000 bauds(signals)per sec  and no of bits per signal = 4
so bit rate = 4000 bits per sec


Q 29. Which of the following data structure is most suitable for evaluating post-fix expressions?

1)  Stack
2)  Queue
3)  Linked List
4)  Tree

Answer : 1)  Stack
Solution : A postfix expression can be evaluated using the Stack data structure. To evaluate a postfix expression using Stack data structure we can use the following steps...
  1. Read all the symbols one by one from left to right in the given Postfix Expression
  2. If the reading symbol is operand, then push it on to the Stack.
  3. If the reading symbol is operator (+ , - , * , / etc.,), then perform TWO pop operations and store the two popped oparands in two different variables (operand1 and operand2). Then perform reading symbol operation using operand1 and operand2 and push result back on to the Stack.
  4. Finally! perform a pop operation and display the popped value as final result.

Q 30. The complexity of linear search algorithm is

1)  O (n)
2)  O (n*n) 
3)  O (log n)
4)  O (n log n)

Answer : 1)  O (n) 
Solution :  The worst case complexity of linear search is O(n).
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KVS PGT Computer Science Question Paper - January 2017 (Question No 21-25)

 October 11, 2018     KVS Compuer Science - January 2017 Question Paper     No comments   

KVS PGT Computer Science Question Paper - January 2017 (Question No 21-25)

Q 21. Ten signals, each requiring 3000 Hz, are multiplexed onto a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel? Assume that the guard bands are 300 Hz wide. 

   1)  30,000
   2)  33,000
   3)  32,700 
   4)  33,700

  Answer : 3)  32,700
  Solution : Ten signals need 10 x 3000 = 30000 Hz
                   Nine (10-1) Guard bands (or gaps) needs =300 x 9 = 2700 Hz
                   So Minimum Bandwidth need 30000 + 2700 = 32700Hz 



Q 22. In which addressing mode, the operand is given explicitly in the instruction (Example instruction: ADD R4, #3) ?

   1)  Absolute mode 
   2)  Register indirect mode. 
   3)  Immediate mode 
   4)  Based Indexed mode.

Answer : 3)  Immediate mode. 
Solution : An addressing mode is a way in which an operand is specified in an instruction. There are different ways in which an operand may be specified in an instruction. An operand may be specified using the immediate, direct, extended, indexed, and inherent modes.


Q 23. In relational database the natural join of two tables is :

   1)  Combination of Union and filtered Cartesian product.
   2)  Combination of Projection and filtered Cartesian product.
   3)  Combination of Selection and filtered Cartesian product.
   4)  Cartesian product always.

Answer : 2)  Combination of Projection and filtered Cartesian product.
Solution :   Projection to remove duplicate attributes in the resultant relation(common column will be present twice in Cartesian product) and Filtering with match on common column.


Q 24. Which of the following memory improves the speed of execution of a program ?

   1)  Primary Memory
   2)  Cache Memory
   3)  Secondary Memory
   4)  Virtual Memory

Answer : 2)  Cache Memory
Solution : Cache is a small amount of memory which is part of the CPU which is physically closer to the CPU than RAM is. The more cache there is, the more data can be stored closer to the CPU.
Cache memory is beneficial because:
  • Cache memory holds frequently used instructions/data which the processor may require next and it is faster access memory than RAM, since it is on the same chip as the processor.
  • This reduces the need for frequent slower memory retrievals from main memory, which may otherwise keep the CPU waiting.
The more cache the CPU has, the less time the computer spends accessing slower main memory and as a result programs may run faster.


Q 25. Which of the following devices understands the format and contents of the data and translate message from one format to another ?

   1)  Hub
   2)  Router
   3)  Switch
   4)  Gateway

Answer : 4)  Gateway
Solution : A gateway is a device, system, or application that translates data from one format to another.
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KVS PGT Computer Science Question Paper - January 2017 (Question No 16-20)

 October 11, 2018     KVS Compuer Science - January 2017 Question Paper     No comments   

KVS PGT Computer Science Question Paper - January 2017  (Question No 16-20)



Q 16. What is the bit rate for transmitting uncompressed 800 x 600 pixel color frame with 8 bits/pixel at 40 frames/second.

1)   2.4 Mbps
2)   153.6 Mbps
3)  15.36 Mbps  
4)   1536 Mbps

Answer : 2) 153.6 Mbps
Solution :  The number of bit per second is 800 x 600 x 40 x 8 = 153600000 bps
                   (As One megabit per second (Mbps) is equal to 1,000,000 bps or 1,000 Kbps.)
                   So 153600000 bps =     153600000       Mbps = 153.6 Mbps
                                                       (1000 x 1000) 


Q 17. What is the length of an IP address in bytes (Pre-IPv6).

1)  1
2)  4
3)  2
4)  8

 Answer: N.A
 Solution: (The length of an IPv6 address is 128 bits, compared with 32 bits in IPv4.)


Q 18. Program counter contents indicate: 

1)  The time elapsed since execution begins. 
2)  The count of programs being executed after switching the power ON. 
3)  The address where next instruction is stored. 
4)  The time needed to execute a program.

Answer : 3)  The address where next instruction is stored. 
Solution : A program counter (PC) is a CPU register in the computer processor which has the address of the next instruction to be executed from memory. It is a digital counter needed for faster execution of tasks as well as for tracking the current execution point.A program counter is also known as an instruction counter, instruction pointer, instruction address register or sequence control register.


Q 19. Which of the following is the most appropriate format for graphics that are to be embedded within an internet document? 

1)  BMP 
2)  GIF 
3)  TIFF 
4)  HTML


Answer : 2)  GIF (Graphics Interchange Format)


Q 20. The technology that stores only the essential instructions on a microprocessor chip and thus enhances its speed is referred to as: 

1)  CISC 
2)  SIMD 
3)  RISC 
4)  MIMD

Answer : 3)  RISC (Reduced instruction set computing)

Solution : RISC (reduced instruction set computer) is a microprocessor that is designed to perform a smaller number of types of computer instructions so that it can operate at a higher speed (perform more millions of instructions per second, or MIPS).
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KVS PGT Computer Science Question Paper - January 2017 (Question No 11-15)

 August 31, 2018     KVS Compuer Science - January 2017 Question Paper     No comments   

KVS PGT Computer Science Question Paper - January 2017  (Question No 11-15)





Q 11. The problem of indefinite blockage of low-priority jobs in general priority scheduling algorithm can be solved using :
1)   Dirty bit
2)   Compaction
3)   Aging
4)   Swapping

Answer : 3) Aging


Q 12. What is garbage collection in the context of Java ?
1)   The Operation System periodically deletes all of the Java files available on the system.
2)   When all references to an object are gone, then the memory used by the object is automatically reclaimed.
3)   Any java package imported in a program and not being used, is automatically deleted.
4)   The Java Virtual Machine (JVM) checks the output of any Java program and deletes  anything that does not make sense at all.

Answer : 2) When all references to an object are gone, then the memory used by the object is automatically reclaimed.


Q 13. Which of the following represents the life-cycle of software development?
1)   Analysis -->Design -->Coding -->Testing -->Operation and Maintenance
2)   Analysis -->Design -->Coding-->Operation and Maintenance-->Testing
3)   Design -->Analysis-->Coding-->Testing-->Operation and Maintenance
4)   Design -->Analysis -->Coding -->Operation and Maintenance --> Testing

Answer : 1)  Analysis -->Design -->Coding -->Testing -->Operation and Maintenance

Q 14. In a relational database model, cardinality of a relation means :
1)   The number of tuples
2)   The number of tables
3)   The number of attributes
4)   The number of constraints

Answer : 1) The number of tuples

Q 15. The 10Base5 cabling is also known as __________
1)   Thick Ethernet
2)   Gigabit Ethernet
3)   Thin Ethernet
4)   Fast Ethernet

Answer : 1) Thick Ethernet
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KVS PGT Computer Science Question Paper - January 2017 (Question No 6-10)

 August 29, 2018     KVS Compuer Science - January 2017 Question Paper     No comments   

KVS PGT Computer Science Question Paper - January 2017  (Question No 6-10)






Q 6. How many gate(s) would be required to implement the following Boolean expression after simplification?  Expression : AC + ABC
1)       1
2)       3
3)       2
4)       4

 Answer : 1


Q 7. To declare the version of XML, the correct syntax is: (NA)
1)       <? xml version =’1.0’ />
2)       <? xml version =”1.0” />
3)       <* xml version =’1.0’ />
4)       </ xml version =’1.0’ />

 Answer : -NA-
 
XML Declaration Examples
Following are few examples of XML declarations −

XML declaration with no parameters −

<?xml >

XML declaration with version definition −

<?xml version = "1.0">

XML declaration with all parameters defined −

<?xml version = "1.0" encoding = "UTF-8" standalone = "no" ?>

XML declaration with all parameters defined in single quotes −

<?xml version = '1.0' encoding = 'iso-8859-1' standalone = 'no' ?>


 
Q 8. A multimedia project is said to be __________ and user-interactive when users are given navigational control.
1)       Hytertext
2)       Non-liner
3)       Liner
4)       Secure

 Answer : 2

 
Q 9. On which of the following storage media, storage of information is organized as a single continuous spiral groove?
1)       RAM
2)       Floppy disk
3)       Hard disk
4)       CD-ROM

 Answer : 4

 
Q 10. The addressing mode used in an instruction of the form ADD X, Y is :
1)       Absolute
2)       Indirect
3)       Immediate
4)       Index

 Answer : 1
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