UGC NET CS 2011 June-Paper-2
Question 1 |
Any integer composed of 3n identical digits divisible by
2n | |
3n | |
5n | |
7n |
Question 1 Explanation:
Identical digits means similar(same) digits.
Any integer composed of 3n identical digits.
For example n=1, 3n= 31=3 which means three identical digits (111,222,333,444 and so on)
n=2, 32 =9 which means nine identical digits (111111111, 222222222 and so on)
The above numbers are is divisible by 3
111 is divisible by 3.
222, 333 and 444 are multiple of 3
Any integer composed of 3n identical digits.
For example n=1, 3n= 31=3 which means three identical digits (111,222,333,444 and so on)
n=2, 32 =9 which means nine identical digits (111111111, 222222222 and so on)
The above numbers are is divisible by 3
111 is divisible by 3.
222, 333 and 444 are multiple of 3
Question 2 |
The circumference of the two concentric disks are divided into 100 sections each. For the outer disk, 100 of the sections are painted red and 100 of the sections are painted blue.For the inner disk, the sections are painted red and blue in an arbitrary manner. It is possible to align the two disks so that____of the sections on the inner disks have their colours matched with the corresponding section on outer disk.
100 or more | |
125 or more | |
150 or more | |
175 or more
|
Question 2 Explanation:
→ We fix the larger disk first, then place the smaller disk on the top of the larger disk so that the centers and sectors coincide. There are 200 ways to place the smaller disk in such a manner.
→ For each such alignment, some sectors of the two disks may have the same color. Since each sector of the smaller disk will match the same color sector of the larger disk 100 times among all the 200 ways and there are 200 sectors in the smaller disk, the total number of matched color sectors among the 200 ways is 100 × 200 = 20,000.
→ Note that there are only 200 ways. Then there is at least one way that the number of matched color sectors is 20,000 / 200 = 100 or more.
→ For each such alignment, some sectors of the two disks may have the same color. Since each sector of the smaller disk will match the same color sector of the larger disk 100 times among all the 200 ways and there are 200 sectors in the smaller disk, the total number of matched color sectors among the 200 ways is 100 × 200 = 20,000.
→ Note that there are only 200 ways. Then there is at least one way that the number of matched color sectors is 20,000 / 200 = 100 or more.
Question 3 |
The proposition ~p ∨ q is equivalent to
p → q | |
q → p | |
p ↔ q | |
p ∨ q |
Question 3 Explanation:

Question 4 |
The absorption law in Boolean algebra say that
X + X = X | |
X . X = X | |
x + x . y = x | |
None of the above |
Question 4 Explanation:
Absorption laws in Boolean algebra:
X • (X+Y) = X
X + X•Y = X
X • (X+Y) = X
X + X•Y = X
Question 5 |
The number of 1’s present in the binary representation of 10 × 256 + 5 × 16 + 5 is
5 | |
6 | |
7 | |
8 |
Question 5 Explanation:
Step-1: Evaluate given numbers 10 * 256 + 5 * 16 + 5
= 2645
Step-2: Convert decimal number into binary number
(2645)10 =(?)2
=(101001010101)2
Hence, total 6 number of 1’s present in binary number.
= 2645
Step-2: Convert decimal number into binary number
(2645)10 =(?)2
=(101001010101)2
Hence, total 6 number of 1’s present in binary number.
Question 6 |
The hexadecimal number equivalent to (1762.46)8 is
3F2.89 | |
3F2.98 | |
2F3.89 | |
2F3.98 |
Question 6 Explanation:
(1762.46)8
→ For making the conversion easy, first convert number into binary
(001 111 110 010. 100 110)2
→ Now convert above binary no. into hexadecimal, we need 4-bits to represent a no. into hexadecimal

(3F2.98)16
→ For making the conversion easy, first convert number into binary
(001 111 110 010. 100 110)2
→ Now convert above binary no. into hexadecimal, we need 4-bits to represent a no. into hexadecimal

(3F2.98)16
Question 7 |
(A + B) A’B’ is equivalent to
A ⊕ B | |
A ʘ B | |
(A ⊕ B) ʘ A | |
(A ʘ B) ⊕ A |
Question 7 Explanation:
We can solve this boolean expression into 2 ways.
1. Boolean simplification
2. Based on truth table
Second method is good when they asked small expression

1. Boolean simplification
2. Based on truth table
Second method is good when they asked small expression

Question 8 |
A latch is constructed using two cross-coupled
AND and OR gates | |
AND gates | |
NAND and NOR gates | |
NAND gates |
Question 8 Explanation:
Latches can be constructed either with NAND gates or NOR gates.
Ex: SR latch can be constructed by using 2 cross coupled NAND gates( or NOR gates)
Question 9 |
A multiplexer is a logic circuit that
accepts one input and gives several output | |
accepts many inputs and gives many output | |
accepts many inputs and gives one output | |
accepts one input and gives one output |
Question 9 Explanation:
A multiplexer is a logic circuit that accepts many inputs and gives one output.
Demultiplex is just reverse of it.
Question 10 |
8-bit 1’s complement form of –77.25 is
01001101.0100 | |
01001101.0010 | |
10110010.1011 | |
10110010.1101 |
Question 10 Explanation:
(-77.25)10
Step 1: Convert (77)10into binary form.




Question 11 |
From the point of view of the programmer, what are the major advantages of using a high-level language rather than internal machine code or assembly language ?
Program portability | |
Easy development
| |
Efficiency | |
None of the above |
Question 11 Explanation:
High-level language refers to the higher level of abstraction from machine language. Rather than dealing with registers, memory addresses and call stacks, high-level languages deal with variables, arrays, objects, complex arithmetic or boolean expressions, subroutines and functions, loops, threads, locks, and other abstract computer science concepts, with a focus on usability over optimal program efficiency.
Question 12 |
What features make C++ so powerful ?
Easy implementation | |
Reusing old code | |
Easy memory management
| |
All of the above |
Question 12 Explanation:
C++ Features:
1. Simple (or) Easy implementation
2. Portability
3. Reusing old code
4. Platform dependent
5. Object oriented oriented
6. Case sensitive
7. Compiler based
8. Syntax based language
9. Easy memory management
1. Simple (or) Easy implementation
2. Portability
3. Reusing old code
4. Platform dependent
5. Object oriented oriented
6. Case sensitive
7. Compiler based
8. Syntax based language
9. Easy memory management
Question 13 |
The goal of operator overloading is
to help the user of a class | |
to help the developer of a class | |
to help define friend function | |
None of the above |
Question 14 |
The scheme of which interpreter translates the source program is known as
Paragraph by paragraph | |
Instruction by instruction | |
Line by line | |
None of the above |
Question 14 Explanation:
→ Compiler translates the entire source program into once. So, it is faster than interpreter.
→ Interpreter translates the source program into line by line. So, it is slower than compiler.
→ Interpreter translates the source program into line by line. So, it is slower than compiler.
Question 15 |
Portable program means
Program with wheels | |
Independent from its authors | |
Independent of platform | |
None of the above |
Question 15 Explanation:
Portable program means independent of platform. We can run same program in any operating system like windows,linux,unix,etc..,
Question 16 |
Which of the following is the recovery management technique in DDBMS ?
2PC (Two Phase Commit) | |
Backup | |
Immediate update | |
All of the above |
Question 16 Explanation:
2PC (Two Phase Commit): The protocol achieves its goal even in many cases of temporary system failure (involving either process, network node, communication, etc. failures), and is thus widely used. It has commit-request phase (or voting phase) and commit phase.
→ Sometimes, power failures are happens backup and immediate update techniques help to recover.
→ Sometimes, power failures are happens backup and immediate update techniques help to recover.
Question 17 |
Which of the following is the process by which a user’s privileges ascertained ?
Authorization | |
Authentication | |
Access Control | |
None of these |
Question 17 Explanation:
→ Authorization is the process by which a user’s privileges ascertained.
→ A smart card, chip card, or integrated circuit card (ICC) is a physical electronic authorization device, used to control access to a resource.
→ A smart card, chip card, or integrated circuit card (ICC) is a physical electronic authorization device, used to control access to a resource.
Question 18 |
The basic variants of time-stamp based method of concurrency control are
Total timestamp-ordering | |
Partial timestamp ordering | |
Multiversion Timestamp ordering | |
All of the above |
Question 18 Explanation:
The basic variants of time-stamp based method of concurrency control are
1. Total timestamp-ordering
2. Partial timestamp ordering
3. Multiversion Timestamp ordering
1. Total timestamp-ordering
2. Partial timestamp ordering
3. Multiversion Timestamp ordering
Question 19 |
A transaction can include following basic database access operations :
Read_item(X) | |
Write_item(X) | |
Both (A) and (B) | |
None of these |
Question 19 Explanation:
Basic operations are read and write
1. read_item(X): Reads a database item named X into a program_variable. To simplify our notation, we assume that the program variable is also named X, i.e X=r(x)
steps:
1. Find the address of the disk block that contains item X.
2. Copy that disk block into a buffer in main memory (if that disk block is not already in some main memory buffer).
3. Copy item X from the buffer to the program variable named X.
2. write_item(X): Writes the value of program variable X into the database item named X.
steps:
1. Find the address of the disk block that contains item X.
2. Copy that disk block into a buffer in main memory (if that disk block is not already in some main memory buffer).
3. Copy item X from the program variable named X into its correct location in the buffer.
4. Store the updated block from the buffer back to disk (either immediately or at some later point in time).
1. read_item(X): Reads a database item named X into a program_variable. To simplify our notation, we assume that the program variable is also named X, i.e X=r(x)
steps:
1. Find the address of the disk block that contains item X.
2. Copy that disk block into a buffer in main memory (if that disk block is not already in some main memory buffer).
3. Copy item X from the buffer to the program variable named X.
2. write_item(X): Writes the value of program variable X into the database item named X.
steps:
1. Find the address of the disk block that contains item X.
2. Copy that disk block into a buffer in main memory (if that disk block is not already in some main memory buffer).
3. Copy item X from the program variable named X into its correct location in the buffer.
4. Store the updated block from the buffer back to disk (either immediately or at some later point in time).
Question 20 |
Decomposition help in eliminating some of the problems of bad design
Redundancy | |
Inconsistencies | |
Anomalies | |
All of the above |
Question 20 Explanation:
Decomposition help in eliminating redundancy, inconsistencies and anomalies.
Question 21 |
The number of different trees with 8 nodes is
256 | |
255 | |
248 | |
None of these |
Question 21 Explanation:
Given data
-- nodes=8
-- Total number of trees=?
Possibility-1: To find total number of trees using nn-2.
= 88-2
= 262144 trees are possible.
Possibility-2: According to given problem, they are using binary tree based on below formula
= 2n -n
= 28 -8
= 248
Note: They are not given proper input.
-- nodes=8
-- Total number of trees=?
Possibility-1: To find total number of trees using nn-2.
= 88-2
= 262144 trees are possible.
Possibility-2: According to given problem, they are using binary tree based on below formula
= 2n -n
= 28 -8
= 248
Note: They are not given proper input.
Question 22 |
Given a binary tree whose inorder and preorder traversal are given by
In order : E I C F B G D J H K
Preorder : B C E I F D G H J K
The post order traversal of the above binary tree is
In order : E I C F B G D J H K
Preorder : B C E I F D G H J K
The post order traversal of the above binary tree is
I E F C G J K H D B | |
I E F C J G K H D B | |
I E F C G K J H D B | |
I E F C G J K D B H |
Question 22 Explanation:

Question 23 |
The number of disk accesses performed by insertion operation in B-tree of height ‘h’ is
O(1) | |
O(lgh) | |
O(h) | |
None of these |
Question 23 Explanation:
The number of disk accesses performed by insertion operation in B-tree of height ‘h’ is O(h).
Question 24 |
Consider a hash table of size m = 10000 and the hash function h(k) = ⌊m(kA mod 1)'⌋ for A=(√5 – 1)/2 . The location to the key k=123456 is
46 | |
47 | |
41 | |
43 |
Question 24 Explanation:
Given data,
-- size m=10000
-- hash function h(k) = ⌊m(kA mod 1)'⌋
-- A=(√5 – 1)/2
-- key k=123456 location =?
Step-1: Find out h(123456) = ⌊(10000 * (123456 * (√5 − 1) / 2) mod 1)⌋
= ⌊(10000 * (76300.004115 mod 1)⌋
= ⌊(10000 * (.004115))⌋
= ⌊41.15⌋
= 41
-- size m=10000
-- hash function h(k) = ⌊m(kA mod 1)'⌋
-- A=(√5 – 1)/2
-- key k=123456 location =?
Step-1: Find out h(123456) = ⌊(10000 * (123456 * (√5 − 1) / 2) mod 1)⌋
= ⌊(10000 * (76300.004115 mod 1)⌋
= ⌊(10000 * (.004115))⌋
= ⌊41.15⌋
= 41
Question 25 |
When the priority queue is represented by max heap, the insertion and deletion of an element can be performed in (queue containing n elements)
θ(n) and θ(1) respectively | |
θ(n) and θ(n) respectively | |
θ(1) and θ(1) respectively | |
None of the above |
Question 25 Explanation:

Question 26 |
A comparison of frequency division and time division multiplexing system shows that
FDM requires a lower bandwidth, but TDM has greater noise immunity. | |
FDM has greater noise immunity and requires lower bandwidth than TDM. | |
FDM requires channel synchronization, while TDM has greater noise immunity | |
FDM requires more multiplex while TDM requires band pass filter.
|
Question 26 Explanation:
Comparison:
TDM:
1. TDM stands time division multiplexing.
2. Cross talk like problem not occurs in TDM.
3. TDM requires channel synchronization
4. TDM has greater noise immunity.
FDM:
1. FDM stands for Frequency division multiplexing.
2. FDM suffers from cross-talk like problems due to imperfect band pass filter.
3. FDM not requires channel synchronization
4. FDM requires a lower bandwidth
TDM:
1. TDM stands time division multiplexing.
2. Cross talk like problem not occurs in TDM.
3. TDM requires channel synchronization
4. TDM has greater noise immunity.
FDM:
1. FDM stands for Frequency division multiplexing.
2. FDM suffers from cross-talk like problems due to imperfect band pass filter.
3. FDM not requires channel synchronization
4. FDM requires a lower bandwidth
Question 27 |
If carrier modulated by a digital bit stream, has one one of the possible phase of 0, 90, 180 and 270 degrees, then modulation called
BPSK | |
QPSK | |
QAM | |
MSK |
Question 27 Explanation:
We can easily differentiate by using name itself
→ Binary phase-shift keying (BPSK) possible phases are 00 and 1800.
→ Quadrature phase-shift keying (QPSK) possible phases are 00, 900, 1800 and 2700.
→ Minimum Shift keying(QSM), the two carrier waves of the same frequency are out of phase with each other by 90°, a condition known as orthogonality and as quadrature.
→ Quadrature Amplitude Modulation(QAM) mapping changes in continuous phase. Each bit time, the carrier phase changes by ±90°
→ Binary phase-shift keying (BPSK) possible phases are 00 and 1800.
→ Quadrature phase-shift keying (QPSK) possible phases are 00, 900, 1800 and 2700.
→ Minimum Shift keying(QSM), the two carrier waves of the same frequency are out of phase with each other by 90°, a condition known as orthogonality and as quadrature.
→ Quadrature Amplitude Modulation(QAM) mapping changes in continuous phase. Each bit time, the carrier phase changes by ±90°
Question 28 |
Consider the following statement :
(i) The bandwidth of the A.M.,wave depends on the band width of the modulating signal.
(ii) The bandwidth of the A.M.,wave depends on the modulation index.
(iii) The bandwidth of the F.M, wave for all practical purpose depends on the amplitude of the carrier.
Which Of these statements the correct statements are
(i) The bandwidth of the A.M.,wave depends on the band width of the modulating signal.
(ii) The bandwidth of the A.M.,wave depends on the modulation index.
(iii) The bandwidth of the F.M, wave for all practical purpose depends on the amplitude of the carrier.
Which Of these statements the correct statements are
(i, ii) | |
(i, iii) | |
(ii, iii) | |
All of the above |
Question 28 Explanation:
TRUE: The bandwidth of the A.M.,wave depends on the band width of the modulating signal.
False: The bandwidth of the A.M.,wave depends on the modulation index.
TRUE: The bandwidth of the F.M, wave for all practical purpose depends on the amplitude of the carrier.
False: The bandwidth of the A.M.,wave depends on the modulation index.
TRUE: The bandwidth of the F.M, wave for all practical purpose depends on the amplitude of the carrier.
Question 29 |
A smart modem can dial, hangup and answer incoming calls automatically. Can you tell who provides the appropriate instructions to the modem for this purpose ?
Communication software | |
Error detection protocols | |
Link access procedure (LAP) | |
Telecommunications |
Question 29 Explanation:
A smart modem can dial, hangup and answer incoming calls automatically. Communication software provides the appropriate instructions to the modem for this purpose.
Question 30 |
Which of the following switching techniques is most suitable for interactive traffic ?
Circuit switching | |
Message switching | |
Packet switching | |
All of the above |
Question 30 Explanation:
→ Circuit switching is suitable for light and/or intermediate load under interactive traffic efficiently.
→ Packet switching is suitable for heavy amount of data and efficient link utilization. For fast rate, packet switching may be applicable to interactive real time data transfer.
→ Store and forwarding switching is not applicable to interactive real time traffic.
→ Packet switching is suitable for heavy amount of data and efficient link utilization. For fast rate, packet switching may be applicable to interactive real time data transfer.
→ Store and forwarding switching is not applicable to interactive real time traffic.
Question 31 |
Object modules generated by assembler that contains unresolved external references are resolved for two or more object module by a/an
Operating system | |
Loader | |
Linker | |
Compiler |
Question 31 Explanation:
Object modules generated by assembler that contains unresolved external references are resolved for two or more object module by a Linker.
Question 32 |
A special software that is used to create a job queue is called
Drive | |
Spooler | |
Interpreter | |
Linkage editor |
Question 32 Explanation:
A special software that is used to create a job queue is called spooler.
Question 33 |
Which of the following permanent database that has an entry for each terminal symbol ?
Literal table | |
Identifier table
| |
Terminal table | |
Source table |
Question 33 Explanation:
Terminal table is permanent database that has an entry for each terminal symbol.
Question 34 |
In which way(s) a macro processor for assembly language can be implemented ?
Independent two-pass processor | |
Independent one-pass processor | |
Expand macro calls and substitute arguments | |
All of the above |
Question 34 Explanation:
Different ways a macro processor for assembly language can be implemented.
1. Independent two-pass processor
2. Independent one-pass processor
3. Expand macro calls and substitute arguments
1. Independent two-pass processor
2. Independent one-pass processor
3. Expand macro calls and substitute arguments
Question 35 |
Which of the following can be accessed by transfer vector approach of linking ?
External data segments | |
External subroutine | |
Data located in other procedure | |
All of the above |
Question 35 Explanation:
→ External subroutines are routines that are created and maintained separately from the program that will be calling them.
→ External subroutine can be accessed by transfer vector approach of linking.
→ External subroutine can be accessed by transfer vector approach of linking.
Question 36 |
Block or Buffer caches are used to
improve disk performance | |
handle interrupts | |
increase the capacity of main memory | |
speed up main memory Read operations |
Question 36 Explanation:
Block or Buffer caches are used to speed up main memory Read operations. By reading the information from disk only once and then keeping it in memory until no longer needed, one can speed up all but the first read. This is called disk buffering, and the memory used for the purpose is called the buffer cache.
Question 37 |
A file organization component VSAM file is
Relative records data set | |
Keyed sequential data set | |
Entry sequential data set | |
All of the above |
Question 37 Explanation:
A file organization component VSAM file is relative records data set(RRDS), keyed sequential data set(KSDS) and entry sequential data set(ESDS).
Question 38 |
A relationship between processes such that each has some part (critical section) which must not be executed while the critical section of another is being executed, is known as
Semaphore | |
Mutual exclusion | |
Multiprogramming | |
Message passing |
Question 38 Explanation:
A relationship between processes such that each has some part (critical section) which must not be executed while the critical section of another is being executed, is known as Mutual exclusion.
Question 39 |
How many states can a process be in ?
3 | |
4 | |
2 | |
5 |
Question 39 Explanation:
Process states
1. New→ The process is being created.
2. Running→ Instructions are being executed.
3. Waiting→ The process is waiting for some event to occur
4. Ready→ The process is waiting to be assigned to a processor
5. Terminated→ The process has finished execution
1. New→ The process is being created.
2. Running→ Instructions are being executed.
3. Waiting→ The process is waiting for some event to occur
4. Ready→ The process is waiting to be assigned to a processor
5. Terminated→ The process has finished execution
Question 40 |
Which command allows you to view your file 24 lines at a time ?
More | |
Cat | |
Pg | |
None of the above |
Question 40 Explanation:
"cat command" is used to display the file content and also used for creating files.
"more command" is used to Displays text of the file, one screen at a time.
The pg command is a filter that allows the examination of filenames one screenful at a time on a CRT. If the user types a RETURN, another page is displayed;
Syntax : pg -number file-name
"more command" is used to Displays text of the file, one screen at a time.
The pg command is a filter that allows the examination of filenames one screenful at a time on a CRT. If the user types a RETURN, another page is displayed;
Syntax : pg -number file-name
Question 41 |
Which one of the items listed below is not one of the software engineering layers ?
Process | |
Manufacturing | |
Method | |
Tools |
Question 41 Explanation:
- Software Engineering is divided into four layers:
1. A quality Process
- Any engineering approach must rest on an quality.
- The "Bedrock" that supports software Engineering is Quality Focus.
2. Process
- Foundation for software engineering is the Process Layer.
- Software engineering process holds all the technology layers together and enables the timely development of computer software.
- It forms the base for management control of software project.
3. Methods
- Software engineering methods provide the "Technical Questions" for building Software.
- Methods contain a broad array of tasks that include communication requirement analysis, design modelling, program construction testing and support.
4. Tools
- Software engineering tools provide automated or semi-automated support for the "Process" and "Methods".
- Tools are integrated so that information created by one tool can be used by another.
1. A quality Process
- Any engineering approach must rest on an quality.
- The "Bedrock" that supports software Engineering is Quality Focus.
2. Process
- Foundation for software engineering is the Process Layer.
- Software engineering process holds all the technology layers together and enables the timely development of computer software.
- It forms the base for management control of software project.
3. Methods
- Software engineering methods provide the "Technical Questions" for building Software.
- Methods contain a broad array of tasks that include communication requirement analysis, design modelling, program construction testing and support.
4. Tools
- Software engineering tools provide automated or semi-automated support for the "Process" and "Methods".
- Tools are integrated so that information created by one tool can be used by another.
Question 42 |
What is the first stage in program development ?
Specification and design | |
System Analysis | |
Testing | |
None of the above |
Question 42 Explanation:
Stages in program development:
1. System Analysis
2. Specification and design
3. Coding and unit testing
4. Maintenance
1. System Analysis
2. Specification and design
3. Coding and unit testing
4. Maintenance
Question 43 |
By means of a data flow diagram, the analyst can detect
Task duplication | |
Unnecessary delays | |
Task overlapping | |
Task overlapping |
Question 43 Explanation:
Data flow diagram, the analyst can detect task duplication, unnecessary delays and task overlapping.
Question 44 |
Which of these are the 5 generic software engineering framework activities ?
Communication, planning, modelling, construction, deployment | |
Communication, risk management, measurement, production, reviewing | |
Analysis, designing, programming, Debugging, maintenance | |
Analysis, planning, designing, programming, testing |
Question 44 Explanation:
The following generic process framework is applicable to the vast majority of S/W projects.
1,Communication: involves heavy communication with the customer (and other stakeholders) and encompasses requirements gathering.
2. Planning: Describes the technical tasks to be conducted, the risks that are likely, resources that will be required, the work products to be produced and a work schedule.
3. Modeling: encompasses the creation of models that allow the developer and customer to better understand S/W req. and the design that will achieve those req.
4. Construction: combines code generation and the testing required uncovering errors in the code.
5. Deployment: deliver the product to the customer who evaluates the delivered product and provides feedback.
1,Communication: involves heavy communication with the customer (and other stakeholders) and encompasses requirements gathering.
2. Planning: Describes the technical tasks to be conducted, the risks that are likely, resources that will be required, the work products to be produced and a work schedule.
3. Modeling: encompasses the creation of models that allow the developer and customer to better understand S/W req. and the design that will achieve those req.
4. Construction: combines code generation and the testing required uncovering errors in the code.
5. Deployment: deliver the product to the customer who evaluates the delivered product and provides feedback.
Question 45 |
Many causes of the software crisis can be traced to mythology based on
Management Myths | |
Customer Myths | |
Practitioner Myths | |
All of the above |
Question 45 Explanation:
Many causes of the software crisis can be traced to mythology based on management Myths,Customer Myths and Practitioner Myths.
Question 46 |
Data warehousing refers to
storing data offline at a separate site | |
backing up data regularly | |
is related to data mining | |
uses tape as opposed to disk |
Question 46 Explanation:
Data warehousing refers to is related to data mining
Question 47 |
Virtual memory is
related to virtual reality | |
a form of ROM | |
a form of RAM | |
None of the above |
Question 47 Explanation:
Virtual memory is a form of RAM.
Question 48 |
“M-Commerce” refers to
a myth which does not exist in reality | |
the ability of business to reach potential customers wherever they are | |
the ability to have large capacity of memory storage dealing trade and commerce | |
None of the above |
Question 48 Explanation:
M-Commerce is "the delivery of electronic commerce capabilities directly into the consumer’s hand, anywhere, via wireless technology."
Question 49 |
The principal electronic payment systems for electronic commerce is
Credit Card | |
Digital Wallet | |
Electronic Cheque | |
All of the above |
Question 49 Explanation:
The principal electronic payment systems for electronic commerce is Credit Card, Digital Wallet and Electronic Cheque.
Question 50 |
Which is the most valuable electronic commerce to the individual customer in long run ?
Business to Customer | |
Business to Business | |
Customer to Customer | |
None of the above |
Question 50 Explanation:
Business to Customer is the most valuable electronic commerce to the individual customer in long run.
There are 50 questions to complete.