## UGC NET CS 2018-DEC Paper-2

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Question 1 |

Data warehouse contains___________ data that is never found in operational environment.

Scripted | |

Encoded | |

Encrypted | |

Summary |

Question 1 Explanation:

Scripted data : A data that you access using java script is termed as scripted data.

Summary Data : Summary data means a new record is created to store values in a row of a relation.

Summary Data : Summary data means a new record is created to store values in a row of a relation.

Question 2 |

Suppose a system has 12 instances of some resources with n processes competing for that resource. Each process may require 4 instances of the resource. The maximum value of n for which the system never enters into deadlock is

3 | |

4 | |

5 | |

6 |

Question 2 Explanation:

→ Here, every process requirement is 4 instances of the resource.

→ If we allocates 3 instance( one instance less than the requirement of each process) of the resource and to one process we allocate its minimum requirement then in that way with limited available instance of resource, without entering into deadlock we can fulfill requirement of maximum number of processes.

→ Now in question it is given that we have 12 instance then using above strategy we can allocate resources to 3 process without entering into deadlock.

→ If we allocates 3 instance( one instance less than the requirement of each process) of the resource and to one process we allocate its minimum requirement then in that way with limited available instance of resource, without entering into deadlock we can fulfill requirement of maximum number of processes.

→ Now in question it is given that we have 12 instance then using above strategy we can allocate resources to 3 process without entering into deadlock.

Question 3 |

To overcome difficulties in Readers-Writers problem, which of the following statement/s is/are true?

1) Writers are given exclusive access to shared objects

2) Readers are given exclusive access to shared objects

3) Both readers and writers are given exclusive access to shared objects.

Choose the correct answer from the code given below:

1) Writers are given exclusive access to shared objects

2) Readers are given exclusive access to shared objects

3) Both readers and writers are given exclusive access to shared objects.

Choose the correct answer from the code given below:

1 only | |

Both 2 and 3 | |

2 only | |

3 only |

Question 3 Explanation:

In Readers-Writers problem, more than one Reader is allowed to read simultaneously but if a Writer is writing then no other writer or any reader can have simultaneous access to that shared object. So Writers are given exclusive access to shared objects.

Question 4 |

Consider the following terminology and match List 1 and List 2 and choose the correct answer from the code given below

b= branch factor

d= depth of shallowest solution

M= Maximum depth of the search tree

I= depth limit

b= branch factor

d= depth of shallowest solution

M= Maximum depth of the search tree

I= depth limit

(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) | |

(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) | |

(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) | |

(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii) |

Question 4 Explanation:

BFS → O( b

DFS → O(bm) worst case space complexity

Depth- Limited Search → O(bl)

Iterative deepening Search → O(bd)

Note: Based upon BFS and DFS we can find the solution.

^{d}) worst case space complexityDFS → O(bm) worst case space complexity

Depth- Limited Search → O(bl)

Iterative deepening Search → O(bd)

Note: Based upon BFS and DFS we can find the solution.

Question 5 |

The third generation mobile phone are digital and based on

AMPS | |

Broadband CDMA | |

CDMA | |

D-AMPS |

Question 5 Explanation:

→ AMPS, D-AMPS, CDMA, Broadband CDMA are the standard methods used for cellular communication.

→ AMPS(Advanced Mobile Phone Service): It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.

→ D-AMPS(Digital- Advanced Mobile Phone Service) : It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.

→ CDMA(Code Division Multiple Access) : In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.

→ Broadband CDMA : Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.

→ AMPS(Advanced Mobile Phone Service): It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.

→ D-AMPS(Digital- Advanced Mobile Phone Service) : It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.

→ CDMA(Code Division Multiple Access) : In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.

→ Broadband CDMA : Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.

Question 6 |

Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.

Synchronization | |

Asynchronization | |

Errors | |

Flow control |

Question 6 Explanation:

→ The session layer is the network dialog controller. It establishes, maintains, and synchronizes the interaction among communicating systems.

→ Dialog control: The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.

→ Synchronization : The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.

→ Dialog control: The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.

→ Synchronization : The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.

Question 7 |

Consider the following two languages:

L

L

Which one of the following is correct?

L

_{1} = {x | for some y with | y| = 2^{|x|},xy∈ L and L is regular language}L

_{2} = { x | for some y such that |x| = |y|, xy∈ L and L is regular language}Which one of the following is correct?

Both L _{1} and L_{ 2} are regular languages | |

Both L _{1} and L _{ 2} are not regular languages | |

Only L _{1} is regular language | |

Only L _{ 2} is regular language |

Question 7 Explanation:

if L is a regular language then we always have a string “w” which can be broken into “xy” such that |y|=2|x|

Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, .......}

then L

DFA for L

DFA for L

By same logic L

Consider a language L= {a 3n | n ≥ 0} , the strings in L are {ε, aaa, aaaaaa, .......}

then L

_{1} = {a^{ n} | n ≥ 0} as every string in L can be broken into three equal parts .DFA for L

DFA for L

_{1}By same logic L

_{2} is also regular, here we have to break each string of L into two equal parts and also every even length string from L will belongs to L_{ 1} , since we can break only even length string into two equal parts such that |x| = |y| if “xy ε L”Question 8 |

An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

245.248.132.0/22 and 245.248.132.0/21 | |

245.248.136.0/21 and 245.248.128.0/22 | |

245.248.136.0/24 and 245.248.132.0/21 | |

245.248.128.0/21 and 245.248.128.0/22 |

Question 8 Explanation:

Question 9 |

The Software Requirement Specification(SRS) is said to be _________ if and only if no subset of individual requirements described in it conflict with each other.

Correct | |

Consistent | |

Unambiguous | |

Verifiable |

Question 9 Explanation:

The Software Requirement Specification(SRS) is said to be Consistent if and only if no subset of individual requirements described in it conflict with each other.

Unambiguous : SRS said to be unambiguous if there exist no word that have more than one meaning and this will not confuse testers to get the exact reference.

Correct : SRS is said to be correct if it is error free and accurate.

Unambiguous : SRS said to be unambiguous if there exist no word that have more than one meaning and this will not confuse testers to get the exact reference.

Correct : SRS is said to be correct if it is error free and accurate.

Question 10 |

Consider the vocabulary with only four propositions A,B,C and D. How many models are there for the following sentence?

( ⌐ A ∨ ⌐ B ∨ ⌐ C ∨ ⌐ D)

( ⌐ A ∨ ⌐ B ∨ ⌐ C ∨ ⌐ D)

8 | |

7 | |

15 | |

16 |

Question 10 Explanation:

Here, number of models is nothing but number of TRUEs in final statement. In this proposition logic we got total 15 number of models.

Question 11 |

Which of the following statements is/are false?

P: The clean-room strategy to software engineering is based on the incremental software process model.

Q: The clean-room strategy to software engineering is one of the ways to overcome “unconscious” copying of copyrighted code.

Choose the correct answer from the code given below:

P: The clean-room strategy to software engineering is based on the incremental software process model.

Q: The clean-room strategy to software engineering is one of the ways to overcome “unconscious” copying of copyrighted code.

Choose the correct answer from the code given below:

Neither P and Q | |

P only | |

Both P and Q | |

Q only |

Question 11 Explanation:

Clean-room Strategy:

Cleanroom software engineering (CSE) is a process model that removes defects before they can precipitate serious hazards. It is a team-oriented, theory based software, which is developed using the formal methods, correctness verification and Statistical Quality Assurance (SQA).

→ Clean room management is based on the incremental model of software development, which accumulates into the final product. The approach combines mathematical-based methods of software specification, design and correctness verification with statistical, usage-based testing to certify software fitness for use.

→ The main goal of clean room engineering is to produce zero error-based software by allowing correct designs, which avoid rework.

Cleanroom software engineering (CSE) is a process model that removes defects before they can precipitate serious hazards. It is a team-oriented, theory based software, which is developed using the formal methods, correctness verification and Statistical Quality Assurance (SQA).

→ Clean room management is based on the incremental model of software development, which accumulates into the final product. The approach combines mathematical-based methods of software specification, design and correctness verification with statistical, usage-based testing to certify software fitness for use.

→ The main goal of clean room engineering is to produce zero error-based software by allowing correct designs, which avoid rework.

Question 12 |

If the frame buffer has 10-bits per pixel and 8-bits are allocated for each of the R,G and B components then what would be the size of the color lookup table(LUT)

(2 ^{10} +2^{ 11} ) bytes | |

(2 ^{10} +2^{ 8} ) bytes | |

(2 ^{10} + 2^{ 24} ) bytes | |

(2 ^{8} + 2^{ 9} ) bytes |

Question 12 Explanation:

10-bits per pixel it means we will have 2

8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).

24-bits = 3 Bytes

So the size of lookup table is = Number of entries * size of each entry

So the size of lookup table is = (2

the size of lookup table is = 3072 Bytes = (2

^{ 10} entries in color lookup table.8-bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24-bits( 8-bits for each of the R, G and B component).

24-bits = 3 Bytes

So the size of lookup table is = Number of entries * size of each entry

So the size of lookup table is = (2

^{ 10} * 3) Bytesthe size of lookup table is = 3072 Bytes = (2

^{10} + 2^{ 11} ) BytesQuestion 13 |

Suppose P,Q and R are co-operating processes satisfying Mutual Exclusion condition. Then if the process Q is executing in its critical section then

‘P’ executes in critical section | |

‘R’ executes in critical section | |

Neither ‘P’ nor ‘Q’ executes in their critical section | |

Both ‘P’ and ‘R’ executes in critical section |

Question 13 Explanation:

● A mutual exclusion (mutex) is a program object that prevents simultaneous access to a shared resource.

● This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource.

● Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts.

● When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource

● In the question, Three process are cooperating processes and satisfying mutual exclusion condition. If process Q is executing in its critical section means remaining two processes in wait stating and they won’t enter into critical section.

● This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource.

● Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts.

● When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource

● In the question, Three process are cooperating processes and satisfying mutual exclusion condition. If process Q is executing in its critical section means remaining two processes in wait stating and they won’t enter into critical section.

Question 14 |

A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?

7,7,18 | |

18,7,7 | |

7,6,18 | |

6,7,18 |

Question 14 Explanation:

An instruction size is given as 32-bits.

Now, the instruction is divided into four parts :

An indirect bit

Register code part : Since number of registers given as 64(2 6 ) so to identify each register uniquely 6-bits are needed.

Address part : 256K(2

Operation code:

Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)

Size of Operation code= 7-bits

Now, the instruction is divided into four parts :

An indirect bit

Register code part : Since number of registers given as 64(2 6 ) so to identify each register uniquely 6-bits are needed.

Address part : 256K(2

^{18} ) word memory is mentioned so to identify each word uniquely 18-bits are needed.Operation code:

Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)

Size of Operation code= 7-bits

Question 15 |

Consider the following set of processes and the length of CPU burst time given in milliseconds:

Assume that processes being scheduled with Round-Robin Scheduling Algorithm with Time Quantum 4ms. Then The waiting time for P4 is _________ ms.

Assume that processes being scheduled with Round-Robin Scheduling Algorithm with Time Quantum 4ms. Then The waiting time for P4 is _________ ms.

0 | |

4 | |

12 | |

6 |

Question 15 Explanation:

Waiting time = Turnaround time- Burst time

Turnaround time = Completion time - Arrival time.

Since in question Arrival time is not given so consider it as “0” for all the processes .

Turnaround time = Completion time - Arrival time.

Since in question Arrival time is not given so consider it as “0” for all the processes .

Question 16 |

Which of the following HTML5 codes will affect the horizontal as well as vertical alignment of the table content?

< t d halign = " middle" valign = " center" > BASH < / td > | |

< t d style = " text − a lign : center; vertical − a lign : middle; " > BASH < / td > | |

< t d align = "middle" valign = " center" > B ASH < / td > | |

< t d style = " horizontal − a lign : c enter; vertical − a lign : m iddle; " > B ASH < / td > |

Question 16 Explanation:

● The text-align property sets the horizontal alignment (like left, right, or center) of the content in or .

● The vertical-align property sets the vertical alignment (like top, bottom, or middle) of the content in or .

● The vertical-align property sets the vertical alignment (like top, bottom, or middle) of the content in or .

Question 17 |

The number of substrings that can be formed from string given by “a d e f b g h n m p” is

10 | |

45 | |

56 | |

55 |

Question 17 Explanation:

If we have no repetition in a string then the number of substrings can be found using the formula :

n*(n+1)/2 + 1

We have added 1 because it may include a NULL string also.

The number of substrings = 10*(11)/2 +1

The number of substrings = 56

n*(n+1)/2 + 1

We have added 1 because it may include a NULL string also.

The number of substrings = 10*(11)/2 +1

The number of substrings = 56

Question 18 |

Which of the following statement/s is/are true?

(i) Firewall can screen traffic going into or out of an organization.

(ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Choose the correct answer from the code given below:

(i) Firewall can screen traffic going into or out of an organization.

(ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Choose the correct answer from the code given below:

(i) only | |

Neither (i) nor(ii) | |

Both (i) and (ii) | |

(ii) only |

Question 18 Explanation:

Statement 1 is correct because firewall works on the Application layer, so it can screen the traffic going into and out of the traffic.

Statement 2 is correct.Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Statement 2 is correct.Virtual private networks cam simulate an old leased network to provide certain desirable properties.

Question 19 |

Match the List 1 and List 2 and choose the correct answer from the code given below

(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) | |

(a)-(ii), (b)-(i),(c)-(iii), (d)-(iv) | |

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) | |

(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) |

Question 19 Explanation:

Propositions r and s are logically equivalent if the statement r ↔ s is a tautology. ∼q⇒ ∼p

According to above table,

equivalence means p ⇔ q

Contrapositive means p⇒ q : ∼q⇒ ∼p

Converse means p⇒ q : q⇒ p

Implication means p ⇔ q

According to above table,

equivalence means p ⇔ q

Contrapositive means p⇒ q : ∼q⇒ ∼p

Converse means p⇒ q : q⇒ p

Implication means p ⇔ q

Question 20 |

Consider the following language:

L

L

Which one of the following is correct?

L

_{1} = { a^{ n+m}b^{ n}a^{ m}| n, m ≥ 0 }L

_{2} = { a^{ n+m}b^{ n+m}a^{ n+m}|n, m ≥ 0 }Which one of the following is correct?

Only L _{ 1} is Context Free Language | |

Both L _{1} and L_{ 2} are not Context Free Language | |

Only L _{1} is Context Free Language | |

Both L _{ 1} and L_{ 2} are Context Free Language |

Question 20 Explanation:

→ L

Since we can have a pushdown automata for L

→ L

So for every “b” we can POP a’s came before “b” but for “a” which came after “b” we have nothing to POP on the top of stack. Since we can’t have a pushdown automata that can accept L

L

_{1} is Context Free language because we can push all the a’s that come as input and when b’s come as input pop “n” number of a’s for “n” b’s from top of the stack and when again a’s come as input pop-up all the remaining a’s(i.e. “M” number of a’s) from top of the stack for “m” number of a’s coming as input.Since we can have a pushdown automata for L

_{1} so we can say L_{1} is a CFL.→ L

_{2} is not a context free language because L is equivalent to language= {a^{p} b^{ p} a^{ p} | p ≥ 0 }.So for every “b” we can POP a’s came before “b” but for “a” which came after “b” we have nothing to POP on the top of stack. Since we can’t have a pushdown automata that can accept L

_{ 2} .L

_{2} is not a CFL.Question 21 |

Consider the following statements related to AND-OR Search algorithm.

S 1 : A solution is a subtree that has a goal node at every leaf.

S 2 : OR nodes are analogous to the branching in a deterministic environment

S 3 : AND nodes are analogous to the branching in a non-deterministic environment.

Which one of the following is true referencing the above statements?

Choose the correct answer from the code given below:

S 1 : A solution is a subtree that has a goal node at every leaf.

S 2 : OR nodes are analogous to the branching in a deterministic environment

S 3 : AND nodes are analogous to the branching in a non-deterministic environment.

Which one of the following is true referencing the above statements?

Choose the correct answer from the code given below:

S1- False, S2- True, S3- True | |

S1- True, S2- True, S3- True | |

S1- False, S2- True, S3- False | |

S1- True, S2- True, S3- False |

Question 21 Explanation:

● An and–or tree is a graphical representation of the reduction of problems (or goals) to conjunctions and disjunctions of subproblems (or subgoals).

● A solution in an AND-OR tree is a sub tree whose leaves are included in the goal set

● A solution in an AND-OR tree is a sub tree whose leaves are included in the goal set

Question 22 |

Which homogeneous 2D matrix transforms the figure (a) on the left side to the figure (b) on the right ?

Question 22 Explanation:

The homogeneous coordinates of the cartesian point (x,y) is (x,y,1) and the transformation Matrix is a 3x3 Matrix.

The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.

The image in figure a is translated(by 1 unit in x axis), scaled( 2 units in y axis) and rotated(90 degrees counterclockwise) to figure in b.

Question 23 |

Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted.

HTTP GET request, DNS query, TCP SYN | |

DNS query, TCP SYN, HTTP GET request | |

TCP SYN, DNS query, HTTP GET request | |

DNS query, HTTP Get request, TCP SYN |

Question 23 Explanation:

Sequence in which the given packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted :

Step 1: DNS query is sent to Domain name server to convert the Domain name into it’s IP address.

Step 2: TCP SYN request packet is sent by sender to establish a connection for data transmission.

Step 3: HTTP GET request is used to request data from a specified IP address.

Step 1: DNS query is sent to Domain name server to convert the Domain name into it’s IP address.

Step 2: TCP SYN request packet is sent by sender to establish a connection for data transmission.

Step 3: HTTP GET request is used to request data from a specified IP address.

Question 24 |

Consider a system with 2 level cache. Access times of Level 1, Level 2 cache and main memory are 0.5 ns, 5 ns and 100 ns respectively. The hit rates of Level1 and Level2 caches are 0.7 and 0.8 respectively. What is the average access time of the system ignoring the search time within cache?

20.75 ns | |

7.55 ns | |

24.35 ns | |

35.20 ns |

Question 24 Explanation:

Average access time = level 1 hit rate( level 1 access time)+ (level1 miss rate)(level 2 hit rate(level 2 access time)+ (level 1 miss rate)( level 2 miss rate) (main memory access time)

Average access time = 0.7(0.5)+ 0.3(0.8)(5)+ 0.3(0.2)(100)

Average access time = 7.55 ns

Average access time = 0.7(0.5)+ 0.3(0.8)(5)+ 0.3(0.2)(100)

Average access time = 7.55 ns

Question 25 |

In computers, subtraction is generally carried out by

1’s complement | |

10’s complement | |

2’s complement | |

9’s complement |

Question 25 Explanation:

● In computers, subtraction is generally carried out by 2’s complement.

● In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value;

● In the subtraction there may possibility of negative number as a result.

● In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value;

● In the subtraction there may possibility of negative number as a result.

Question 26 |

Consider the following tables (relations):

Primary keys in the tables are shown using Underline. Now, Consider the following query:

SELECT S.Name, Sum (P.Marks)

FROM Students S, Performance P

WHERE S.Roll-No = P.Roll-No

GROUP BY S.Name

The number of rows returned by the above query is

Primary keys in the tables are shown using Underline. Now, Consider the following query:

SELECT S.Name, Sum (P.Marks)

FROM Students S, Performance P

WHERE S.Roll-No = P.Roll-No

GROUP BY S.Name

The number of rows returned by the above query is

3 | |

2 | |

0 | |

1 |

Question 26 Explanation:

The following table is returned as the result of executing FROM and WHERE commands in given query.

Since in query “GROUP BY S.Name” is given so, firstly Group names having same“Name” value and and then perform SUM() operation on those values. The below table is returned as the result of given query :

So the number of rows returned by given query are 2.

Since in query “GROUP BY S.Name” is given so, firstly Group names having same“Name” value and and then perform SUM() operation on those values. The below table is returned as the result of given query :

So the number of rows returned by given query are 2.

Question 27 |

Which of the following is true for semi-dynamic environment ?

The environment itself does not change with the passage of time but the agent’s performance score does. | |

The environment change while the agent is deliberating. | |

Even if the environment changes with the passage of time while deliberating, the performance score does not change. | |

Environment and performance score, both change simultaneously. |

Question 27 Explanation:

Semi- Dynamic environment : The environment itself does not change with the passage of time but the agent’s performance score does.

Question 28 |

In 3D Graphics, which of the following statements about perspective and parallel projection is/are true?

P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.

Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away

R: Perspective projections do not preserve straight lines.

P: In a perspective projection, the farthest an object is from the center of projection, the smaller it appears.

Q: Parallel projection is equivalent to a perspective projection where the viewer is standing infinitely far away

R: Perspective projections do not preserve straight lines.

P and R only | |

P,Q and R | |

Q and R only | |

P and Q only |

Question 28 Explanation:

Perspective Projection :

Perspective projection is representing or drawing objects which resemble the real thing.

Perspective projection preserves the straight line.

In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.

Parallel lines do not remain parallel

Distance and angles are not preserved.

Parallel Projection :

In this projection drawing objects looks less realistic.

In this projection parallel lines remains parallel.

Angles are not preserved in this projection.

It is good for exact measurements.

Perspective projection is representing or drawing objects which resemble the real thing.

Perspective projection preserves the straight line.

In perspective projection, objects that are far away appear smaller, and objects that are near appear bigger.

Parallel lines do not remain parallel

Distance and angles are not preserved.

Parallel Projection :

In this projection drawing objects looks less realistic.

In this projection parallel lines remains parallel.

Angles are not preserved in this projection.

It is good for exact measurements.

Question 29 |

Which of the following statements is/are true?

P: Software Reengineering is preferable for software products having high failure rates, having poor design and/or having poor code structure

Q: Software Reverse Engineering is the process of analyzing software with the objective of recovering its design and requirement specification.

P: Software Reengineering is preferable for software products having high failure rates, having poor design and/or having poor code structure

Q: Software Reverse Engineering is the process of analyzing software with the objective of recovering its design and requirement specification.

P only | |

Neither P nor Q | |

Q only | |

Both P and Q |

Question 29 Explanation:

Re-Engineering: Re-engineering refers to the rewriting the the part or whole system without changing its functionality. Re-Engineering is done for maintaining a system, to make the system compatible with the latest technologies.

Reverse engineering: It is also called as back engineering which is done to examine the working of a product.

Reverse engineering: It is also called as back engineering which is done to examine the working of a product.

Question 30 |

Software coupling involves dependencies among pieces of software called modules. Which of the following are correct statements with respect to module coupling?

P: Common Coupling occurs when two modules share the same global data

Q: control Coupling occur when modules share a composite data structure and use only part of it.

R: Content coupling occurs when one module modifies or relies on the internal working of another module.

P: Common Coupling occurs when two modules share the same global data

Q: control Coupling occur when modules share a composite data structure and use only part of it.

R: Content coupling occurs when one module modifies or relies on the internal working of another module.

P and Q only | |

P and R only | |

Q and R only | |

All of P,Q and R |

Question 30 Explanation:

Control Coupling : It exists between two modules if data from one module to other is used to direct the order of instruction execution in other module.

Common Coupling : Common coupling take place when two modules share global data.

Content Coupling : Content coupling exists between two modules if their code is shared.

Common Coupling : Common coupling take place when two modules share global data.

Content Coupling : Content coupling exists between two modules if their code is shared.

Question 31 |

Consider the following pseudo-code fragment, where m is a non-negative integer that has been initialized:

p=0;

k=0;

while(k < m)

p = p + 2

k=k+1;

end while

Which of the following is a loop invariant for the while statement?

(Note: a loop variant for a while statement is an assertion that is true each time guard is evaluated during the execution of the while statement).

p=0;

k=0;

while(k < m)

p = p + 2

^{k};k=k+1;

end while

Which of the following is a loop invariant for the while statement?

(Note: a loop variant for a while statement is an assertion that is true each time guard is evaluated during the execution of the while statement).

p = 2 ^{k} − 1 and 0≤k < m | |

p = 2 ^{k+1} − 1 and 0≤k < m | |

p = 2 ^{k − 1} and 0≤k≤m | |

p = 2 ^{k+1} − 1 and 0≤k≤m |

Question 31 Explanation:

For k=0, P=0+2

k=1, P=1+2

k=2,P=3+2

k=3,P=7+2

Only the option-3 satisfied to the loop variant.

Ex: m=2, The loop will execute for 3 times for k value is 0,1 and 2.

Then options-3 gives P=1,P=3 and P=7 for the k values 0,1 and respectively

^{0} =1k=1, P=1+2

^{1} =3k=2,P=3+2

^{2} =7k=3,P=7+2

^{3} =15Only the option-3 satisfied to the loop variant.

Ex: m=2, The loop will execute for 3 times for k value is 0,1 and 2.

Then options-3 gives P=1,P=3 and P=7 for the k values 0,1 and respectively

Question 32 |

The K-coloring of an undirected graph G=(V,E) is a function C: V➝{0,1,......,K-1} such that c(u)≠c(v) for every edge (u,v) ∈ E
Which of the following is not correct?

G has no cycles of odd length | |

G has cycle of odd length | |

G is 2-colorable | |

G is bipartite |

Question 32 Explanation:

● A k-colouring of a graph G consists of k different colours and G is then called k-colourable.

● A cycle of length n ≥ 3 is 2-chromatic if n is even and 3-chromatic if n is odd.

● A graph is bi-colourable (2-chromatic) if and only if it has no odd cycles.

● A nonempty graph G is bicolourable if and only if G is bipartite

● A cycle of length n ≥ 3 is 2-chromatic if n is even and 3-chromatic if n is odd.

● A graph is bi-colourable (2-chromatic) if and only if it has no odd cycles.

● A nonempty graph G is bicolourable if and only if G is bipartite

Question 33 |

Consider the following problems:

(i) Whether a finite automaton halts on all inputs?

(ii) Whether a given Context Free Language is Regular?

(iii) Whether a Turing Machine computes the product of two numbers?

Which one of the following is correct?

(i) Whether a finite automaton halts on all inputs?

(ii) Whether a given Context Free Language is Regular?

(iii) Whether a Turing Machine computes the product of two numbers?

Which one of the following is correct?

Only (ii) and (iii) are undecidable problems | |

(i), (ii) and (iii) are undecidable problems | |

Only (i) and (ii) are undecidable problems | |

Only (i) and (iii) are undecidable problems |

Question 33 Explanation:

● A decision problem is decidable if there exists a decision algorithm for it. Otherwise it is undecidable.

● Regular languages are useful for many practical applications due to the fact that “all natural” questions concerning regular languages are decidable

● Regular languages are useful for many practical applications due to the fact that “all natural” questions concerning regular languages are decidable

Question 34 |

Consider the following statements

S1: A heuristic is admissible if it never overestimates the cost to reach the goal

S2: A heuristic is monotonous if it follows triangle inequality property.

Which one of the following is true referencing the above statements?

S1: A heuristic is admissible if it never overestimates the cost to reach the goal

S2: A heuristic is monotonous if it follows triangle inequality property.

Which one of the following is true referencing the above statements?

Statement S1 is true but statement S2 is false. | |

Statement S1 is false but statement S2 is true. | |

Neither of the statements S1 and S2 are true | |

Both the statements S1 and S2 are true. |

Question 34 Explanation:

A heuristic function is said to be admissible if it never overestimates the cost of reaching the goal, i.e. the cost it estimates to reach the goal is not higher than the lowest possible cost from the current point in the path

Question 35 |

Consider the following two statements:

S1: TCP handles both congestion and flow control

S2: UDP handles congestion but not flow control

Which of the following option is correct with respect to the above statements (S1) and (S2)?

S1: TCP handles both congestion and flow control

S2: UDP handles congestion but not flow control

Which of the following option is correct with respect to the above statements (S1) and (S2)?

Both S1 and S2 are correct | |

Neither S1 nor S2 is correct | |

S1 is not correct but S2 is correct | |

S1 is correct but S2 is not correct |

Question 35 Explanation:

TCP is a reliable protocol for data transmission.

TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.

UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.

TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.

UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.

Question 36 |

Match the following secret key algorithm (List 1) with the corresponding key lengths (List 2) and choose the correct answer from the code given below,

(a)-(ii),(b)-(iii), (c)- (iv), (d)-(i) | |

(a)-(iv),(b)-(iii), (c)- (ii), (d)-(i) | |

(a)-(iii),(b)-(iv), (c)- (ii), (d)-(i) | |

(a)-(iii),(b)-(iv), (c)- (i), (d)-(ii) |

Question 36 Explanation:

● Blowfish is an encryption algorithm that can be used as a replacement for the DES or IDEA algorithms. It is a symmetric (that is, a secret or private key) block cipher that uses a variable-length key, from 32 bits to 448 bits, making it useful for both domestic and exportable use.

● DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits. ● IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms

● RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.

● DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits. ● IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms

● RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.

Question 37 |

The boolean expression A’⋅B+A.B’+A.B is equivalent to

A+B | |

A.B | |

(A+B)’ | |

A’.B |

Question 37 Explanation:

Question 38 |

If a graph (G) has no loops or parallel edges and if the number of vertices(n) in the graph is n≥3, then the graph G is Hamiltonian if

(i) deg(v) ≥n/3 for each vertex v

(ii) deg(v) + deg(w) ≥ n whenever v and w are not connected by an edge.

(iii) E (G) ≥ 1/3 (n − 1 )(n − 2 ) + 2

(i) deg(v) ≥n/3 for each vertex v

(ii) deg(v) + deg(w) ≥ n whenever v and w are not connected by an edge.

(iii) E (G) ≥ 1/3 (n − 1 )(n − 2 ) + 2

(i) and (iii) only | |

(ii) and (iii) only | |

(iii) only | |

(ii) only |

Question 38 Explanation:

→ Dirac's theorem on Hamiltonian cycles, the statement that an n-vertex graph in which each vertex has degree at least n/2 must have a Hamiltonian cycle

→ Dirac's theorem on chordal graphs, the characterization of chordal graphs as graphs in which all minimal separators are cliques

→ Dirac's theorem on cycles in k-connected graphs, the result that for every set of k vertices in a k-vertex-connected graph there exists a cycle that passes through all the vertices in the set

→ Dirac's theorem on chordal graphs, the characterization of chordal graphs as graphs in which all minimal separators are cliques

→ Dirac's theorem on cycles in k-connected graphs, the result that for every set of k vertices in a k-vertex-connected graph there exists a cycle that passes through all the vertices in the set

Question 39 |

Consider a disk pack with 32 surfaces, 64 tracks and 512 sectors per pack. 256 bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is

19 | |

20 | |

18 | |

22 |

Question 39 Explanation:

There are 32(2

5+6+9 = 20-bits are needed.

^{5} ) surfaces, each surface have 64(2^{6} ) tracks and each surface have 512(2^{9} ) sectors. So to identify each sector uniquely,5+6+9 = 20-bits are needed.

Question 40 |

Match each UML diagram in List 1 with Appropriate description in List 2

(a)-(i), (b)-(iv), (c) -(iii), (d)-(ii) | |

(a)-(iv), (b)-(ii), (c) -(i), (d)-(iii) | |

(a)-(iv), (b)-(i), (c) -(ii), (d)-(iii) | |

(a)-(i), (b)-(iv), (c) -(ii), (d)-(iii) |

Question 40 Explanation:

State diagram: Used to describe the dynamic behaviour of objects and could also be used to describe the entire system behaviour.
Statechart diagram describes the flow of control from one
state to another state. States are defined as a condition in which an object exists and it changes when some event is triggered. The most important purpose of Statechart diagram is to model
lifetime of an object from creation to termination.

Use case diagram: Describe how the external entities (people, devices) can interact with the system. Use case diagrams are used to gather the requirements of a system including internal and external influences.

Class diagram: The purpose of class diagram is to model the static view of an application. Class diagrams are the only diagrams which can be directly mapped with object-oriented languages and thus widely used at the time of construction.

Activity diagram: Usd to show the flow of a business process, the steps of a use case or the logic of an object behaviour.

Use case diagram: Describe how the external entities (people, devices) can interact with the system. Use case diagrams are used to gather the requirements of a system including internal and external influences.

Class diagram: The purpose of class diagram is to model the static view of an application. Class diagrams are the only diagrams which can be directly mapped with object-oriented languages and thus widely used at the time of construction.

Activity diagram: Usd to show the flow of a business process, the steps of a use case or the logic of an object behaviour.

Question 41 |

Consider the C/C++ function f() given below:

void f(char w [ ] )

{

int x = strlen(w); //length of a string

char c;

For (int i = 0; i < x; i++)

{

c = w[i];

w[i] = w[x - i - 1];

w[x - i - 1] = c;

}

}

Which of the following is the purpose of f() ?

void f(char w [ ] )

{

int x = strlen(w); //length of a string

char c;

For (int i = 0; i < x; i++)

{

c = w[i];

w[i] = w[x - i - 1];

w[x - i - 1] = c;

}

}

Which of the following is the purpose of f() ?

It output the content of the array with the characters rearranged so they are no longer recognized a the words in the original phrase. | |

It output the contents of the array with the characters shifted over by one position. | |

It outputs the contents of the array in the original order. | |

It outputs the contents of the array in the reverse order. |

Question 41 Explanation:

It outputs the contents of the array in the original order because it swaps pairs of elements at locations (i and x - i - 1) twice .

Question 42 |

Data Scrubbing is

A process to upgrade the quality of data after it is moved into a data warehouse | |

A process to upgrade the quality of data before it is moved into a data warehouse | |

A process to lead the data in the warehouse and to create the necessary indexes. | |

A process to reject data from the data warehouse and to create necessary indexes. |

Question 42 Explanation:

Data scrubbing which is also called data cleaning, is the process of amending or removing data in a database that is incorrect, incomplete, improperly formatted, or duplicated.

It is the process carried out before the data is moved into a data warehouse.

It is the process carried out before the data is moved into a data warehouse.

Question 43 |

Dirty bit is used to show the

Page with low frequency occurrence | |

Wrong page | |

Page with corrupted data | |

Page that is modified after being loaded into cache memory |

Question 43 Explanation:

The Dirty bit in a PCB(process control block) is set if the page is modified after being loading into cache memory. This bit helps in maintaining updated data in the hard disk.

Question 44 |

A legacy software system has 940 modules. The latest release require that 90 of these modules be changed. In addition, 40 new modules were added and 12 old modules were removed. Compute the software maturity index for the system.

0.725 | |

0.923 | |

0.849 | |

0.524 |

Question 44 Explanation:

Software Maturity Index(SMI) = [M

Where, M

F

F

SMI = [940-(90+40+12)]/940

SMI = 0.849

_{ T} - (F_{ c} + F_{ a} + F_{ d} )] / M_{ T}Where, M

_{ T} = The number of modules in the current releaseF

_{ c}= The number of modules in the current release that have been changed F_{ a} =The number of modules in the current release that have been addedF

_{d}=The number of modules from the preceding release that were deleted in the current releaseSMI = [940-(90+40+12)]/940

SMI = 0.849

Question 45 |

Match List 1 with List 2 and choose the correct answer from the code given below

(a)-(iv), (b)-(iii), (c) -(ii), (d)-(i) | |

(a)-(i), (b)-(iv), (c) -(iii), (d)-(ii) | |

(a)-(iv), (b)-(i), (c) -(ii), (d)-(iii) | |

(a)-(i), (b)-(ii), (c) -(iii), (d)-(iv) |

Question 45 Explanation:

→ Greedy Best-first Search: Selects a node for expansion if optimal path to that node has been found. Best-first search is a search algorithm which explores a graph by expanding the most
promising node chosen according to a specified rule. Best-first search as estimating the promise of node n by a "heuristic evaluation function f(n) which, in general, may depend on the
description of n, the description of the goal, the information gathered by the search up to that point, and most important, on any extra knowledge about the problem domain."

→ A* Search: Time complexity depends upon the quality of heuristic.

→ Recursive Best-First Search: Suffers from excessive node generation. RBFS depends on how widely the promising nodes are separated in the search tree, and is harder to anticipate.

→ Iterative-deepening A* Search: Avoids substantial overhead associated with keeping the sorted queue of nodes

→ A* Search: Time complexity depends upon the quality of heuristic.

→ Recursive Best-First Search: Suffers from excessive node generation. RBFS depends on how widely the promising nodes are separated in the search tree, and is harder to anticipate.

→ Iterative-deepening A* Search: Avoids substantial overhead associated with keeping the sorted queue of nodes

Question 46 |

The relation ≤ and < on a boolean algebra are defined as :

x ≤ y and only if x ∨ y = y

x < y means x ≤ y but x ≠ y

x ≥ y means y ≤ x and

x > y means y <x

Consider the above definitions, which of the following is not true in the boolean algebra ?

(i)If x ≤ y and y ≤ z, then x ≤ z

(ii)If x ≤ y and y ≤ x, then x=y

(iii)If x < y and y < z, then x ≤ y

(iv)If x < y and y < z, then x < y

x ≤ y and only if x ∨ y = y

x < y means x ≤ y but x ≠ y

x ≥ y means y ≤ x and

x > y means y <x

Consider the above definitions, which of the following is not true in the boolean algebra ?

(i)If x ≤ y and y ≤ z, then x ≤ z

(ii)If x ≤ y and y ≤ x, then x=y

(iii)If x < y and y < z, then x ≤ y

(iv)If x < y and y < z, then x < y

(iv) only | |

(iii) only | |

(i) and (ii) only | |

(ii) and (iii) only |

Question 46 Explanation:

iii) “If x < y and y < z, then x ≤ y” is not true.

Because x < y means x ≤ y but x ≠ y.

ii)If x ≤ y and y ≤ x, then x=y is true

Because

x ≤ y implies x v y =x

y ≤ x implies x v y = y

X v y = x = y

Note: From the given definitions, x
In option 3, the condition x≠y is missing.

Key point: y < z has nothing to do with x ≤ y. So, we are ignoring.

Because x < y means x ≤ y but x ≠ y.

ii)If x ≤ y and y ≤ x, then x=y is true

Because

x ≤ y implies x v y =x

y ≤ x implies x v y = y

X v y = x = y

Note: From the given definitions, x

Key point: y < z has nothing to do with x ≤ y. So, we are ignoring.

Question 47 |

In a ternary tree, the number of internal nodes of degree 1,2 and 3 is 4,3 and 3 respectively. The number of leaf nodes in the ternary tree is

11 | |

12 | |

10 | |

9 |

Question 47 Explanation:

Number of internal nodes= 4+3+3= 10

Number of leaf nodes = Sum of degrees of all internal nodes - number of internal nodes +1

= (4*1 + 3*2 + 3*3 )-10 +1

= 19- 10 +1=10

Formula : nd-n +1 where n is number of nodes and d is degree of the tree.

Number of leaf nodes = Sum of degrees of all internal nodes - number of internal nodes +1

= (4*1 + 3*2 + 3*3 )-10 +1

= 19- 10 +1=10

Formula : nd-n +1 where n is number of nodes and d is degree of the tree.

Question 48 |

The four byte IP Address consists of

Neither network nor Host Address | |

Network Address | |

Both Network and Host Address | |

Host Address |

Question 48 Explanation:

The IP address of 32 bit (4 byte) consists of both network and host address.

Question 49 |

Consider the schema R=(A, B, C, D, E, F) on which the following functional dependencies hold :

A➝B

B,C➝D

E➝C

D➝A

What are the candidate keys of R ?

A➝B

B,C➝D

E➝C

D➝A

What are the candidate keys of R ?

AEF, BEF and DEF | |

AEF, BEF and BCF | |

AE and BE | |

AE, BE and DE |

Question 49 Explanation:

EFA

EFB

EFC

EFD

So EFA, EFB, EFD are the keys for the given relation R=(A, B, C, D, E, F).

^{+} = {EFABCD}EFB

^{+}= {EFABCD}EFC

^{+}= {EFC}EFD

^{+} = {EFDCAB}So EFA, EFB, EFD are the keys for the given relation R=(A, B, C, D, E, F).

Question 50 |

A survey has been conducted on methods of commuter travel. Each respondent was asked to check Bus, Train or Automobile as a major methods of travelling to work. More than one answer was permitted. The results reported were as follows : Bus 30 people; Train 35 people; Automobile 100 people; Bus and Train 15 people; Bus and Automobile 15 people; Train and Automobile 20 people; and all the three methods 5 people. How many people complete the survey form ?

160 | |

120 | |

165 | |

115 |

Question 50 Explanation:

Let “A” denotes people travelled by Bus.

Let “B” denotes people travelled by Train.

Let “C” denotes people travelled by Automobile.

Let “(A⋂B)” denotes people travelled by Bus and Train.

Let “(A⋂C)” denotes people travelled by Bus and Automobile.

Let “(B⋂C)” denotes people travelled by Train and Automobile.

Let “(A⋂B⋂C)” denotes people travelled by Bus, Train and Automobile.

Number of people completed the survey = A+B+C-(A⋂B)-(A⋂C)-(B⋂C)+(A⋂B⋂C)

= 30+35+100-15-15-20+5

= 120

Let “B” denotes people travelled by Train.

Let “C” denotes people travelled by Automobile.

Let “(A⋂B)” denotes people travelled by Bus and Train.

Let “(A⋂C)” denotes people travelled by Bus and Automobile.

Let “(B⋂C)” denotes people travelled by Train and Automobile.

Let “(A⋂B⋂C)” denotes people travelled by Bus, Train and Automobile.

Number of people completed the survey = A+B+C-(A⋂B)-(A⋂C)-(B⋂C)+(A⋂B⋂C)

= 30+35+100-15-15-20+5

= 120

Question 51 |

Consider the graph shown below :

Use Kruskal’s algorithm to find the minimum spanning tree of the graph. The weight of this minimum spanning tree is

Use Kruskal’s algorithm to find the minimum spanning tree of the graph. The weight of this minimum spanning tree is

13 | |

16 | |

17 | |

14 |

Question 51 Explanation:

The weight of this minimum spanning tree is 16.

Question 52 |

Consider the following two C++ programs P

S

S

What can you say about the statement S

_{1} and P_{ 2} and two statements S_{ 1} andS_{2} about the programs :S

_{1} : P_{1} prints out 3S

_{2} : P_{ 2} prints out 4:2What can you say about the statement S

_{1} and S_{ 2} ?Neither S _{1} nor S_{ 2} is true | |

Only S _{1} is true | |

Only S _{2} is true | |

Both S _{1} and S_{ 2} are true |

Question 52 Explanation:

The program p

The definition of the function first modifies “i” value by 2 and later by “3” and that modification of the value is automatically reflects in the main program as “c” is reference variable. The program code p

_{1}code consists of function with three parameters integer , pointer and reference variable.The definition of the function first modifies “i” value by 2 and later by “3” and that modification of the value is automatically reflects in the main program as “c” is reference variable. The program code p

_{2}will gives the output of 4:5 as the function will take reference variable as parameter and return type also reference object.Question 53 |

The grammar S ⟶ (S) | SS | ∈ is not suitable for predictive parsing because the grammar is

An Operator Grammar | |

Right Recursive | |

Left Recursive | |

Ambiguous |

Question 53 Explanation:

The grammar is ambiguous, as to derive string ()()() more than one parse tree exists.

Question 54 |

An attribute A of data type varchar(20) has the value ‘xyz’ and attribute B of data type char(20) has the value “Imnop” , then the attribute A has ________ spaces and attribute B has ________ spaces.

20 , 20 | |

3 ,20 | |

3, 5 | |

20, 5 |

Question 54 Explanation:

→ char is a fixed-length data type, the storage size of the char value is equal to the maximum size for this column.

→ varchar is a variable-length data type, the storage size of the varchar value is the actual length of the data entered, not the maximum size for this column.

→ So, A varchar(20) has only 3 spaces because its initialized to 'xyz' and B char(20) has 20 spaces as char data type occupies the storage space equivalent to the maximum size.

→ varchar is a variable-length data type, the storage size of the varchar value is the actual length of the data entered, not the maximum size for this column.

→ So, A varchar(20) has only 3 spaces because its initialized to 'xyz' and B char(20) has 20 spaces as char data type occupies the storage space equivalent to the maximum size.

Question 55 |

A binary search tree is constructed by inserting the following numbers in order:

60, 25, 72, 15, 30, 68, 101, 13, 18, 47, 70, 34

The number of nodes is the left subtree is

60, 25, 72, 15, 30, 68, 101, 13, 18, 47, 70, 34

The number of nodes is the left subtree is

7 | |

6 | |

3 | |

5 |

Question 55 Explanation:

Question 56 |

A full joint distribution for the Toothache, Cavity and Catch is given in the table below:

Which is the probability of Cavity, given evidence of Toothache ?

Which is the probability of Cavity, given evidence of Toothache ?

< 0.2, 0.8 > | |

< 0.6, 0.8 > | |

< 0.4, 0.8 > | |

< 0.6, 0.4 > |

Question 56 Explanation:

Probability cavity given that toothache

P(Cavity/Toothache)=P(Cavity ⋀ Toothache) / P(toothache)

=(0.108 + 0.012) / (0.108 + 0.012 + 0.016 + 0.064)

=(0.12)/(0.2)

=0.6

P(~Cavity / Toothache) = (0.016 + 0.064) / 0.2

= 0.4

P(Cavity/Toothache)=P(Cavity ⋀ Toothache) / P(toothache)

=(0.108 + 0.012) / (0.108 + 0.012 + 0.016 + 0.064)

=(0.12)/(0.2)

=0.6

P(~Cavity / Toothache) = (0.016 + 0.064) / 0.2

= 0.4

Question 57 |

Consider the singly linked list. What is the worst case time complexity of the best-known algorithm to delete the node a, pointer to this node is q, from the list ?

O(1) | |

O(lg n) | |

O(n) | |

O(n lg n) |

Question 57 Explanation:

If we have pointer to the tail of the list in order to delete it, which can be obtained by traversing the list which takes O(n) time.

Question 58 |

Which of the following statement/s is/are true ?

(i) Facebook has the world’s largest Hadoop cluster.

(ii) Hadoop 2.0 allows live stream processing of real time data

(i) Facebook has the world’s largest Hadoop cluster.

(ii) Hadoop 2.0 allows live stream processing of real time data

Neither (i) nor (ii) | |

Both (i) and (ii) | |

(i) only | |

(ii) only |

Question 58 Explanation:

→ The Data warehouse Hadoop cluster at Facebook has become the largest known Hadoop storage cluster in the world.

Here are some of the details about this single HDFS cluster:

1. 21 PB of storage in a single HDFS cluster

2. 2000 machines

3. 12 TB per machine (a few machines have 24 TB each)

4. 1200 machines with 8 cores each + 800 machines with 16 cores each

5. 32 GB of RAM per machine

6. 15 map-reduce tasks per machine

That's a total of more than 21 PB of configured storage capacity! This is larger than the previously known Yahoo!'s cluster of 14 PB.

→ Hadoop 2.0 allows live stream processing of real time data

Here are some of the details about this single HDFS cluster:

1. 21 PB of storage in a single HDFS cluster

2. 2000 machines

3. 12 TB per machine (a few machines have 24 TB each)

4. 1200 machines with 8 cores each + 800 machines with 16 cores each

5. 32 GB of RAM per machine

6. 15 map-reduce tasks per machine

That's a total of more than 21 PB of configured storage capacity! This is larger than the previously known Yahoo!'s cluster of 14 PB.

→ Hadoop 2.0 allows live stream processing of real time data

Question 59 |

Consider the following sequence of two transactions on a bank account(A) with initial balance 20,000 that transfers 5,000 to another account (B) and then apply 10% interest.

(i) T1 start

(ii) T1 A old=20000 new 15,000

(iii) T1 B old=12000 new=17000

(iv) T1 commit

(v) T2 start

(vi) T2 A old=15000 new=16500

(vii) T2 commit

Suppose the database system crashes out just before log record (vii) is written. When the system is restricted, which one statement is true of the recovery process ?

(i) T1 start

(ii) T1 A old=20000 new 15,000

(iii) T1 B old=12000 new=17000

(iv) T1 commit

(v) T2 start

(vi) T2 A old=15000 new=16500

(vii) T2 commit

Suppose the database system crashes out just before log record (vii) is written. When the system is restricted, which one statement is true of the recovery process ?

We can apply redo and undo operation in arbitrary order because they are idempotent. | |

We must redo log record (vi) to set A to 16,500. | |

We must undo log record (vi) to set A to 16,500 and then redo log record (ii) and (iii). | |

We need not redo records (ii) and (iii) because transaction T1 has committed. |

Question 59 Explanation:

In log based recovery we must perform 'Redo' operation for those transactions that contains both start and commit log record.

We perform 'undo' operation for those transaction that contains only start but not commit log record.

Therefore we perform 'Redo' of T1 and 'Undo' of T2.

Note: Actually they given option-3 is “ We must redo log record (vi) to set A to 16,500 and then redo log record (ii) and (iii)”. But we found it is wrong. Given correct option instead of wrong one.

We perform 'undo' operation for those transaction that contains only start but not commit log record.

Therefore we perform 'Redo' of T1 and 'Undo' of T2.

Note: Actually they given option-3 is “ We must redo log record (vi) to set A to 16,500 and then redo log record (ii) and (iii)”. But we found it is wrong. Given correct option instead of wrong one.

Question 60 |

Which of the following statement/s is/are true ?

(i) windows XP supports both peer-peer and client-server networks.

(ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.

(i) windows XP supports both peer-peer and client-server networks.

(ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.

Both (i) and (ii) | |

Neither (i) nor (ii) | |

(ii) only | |

(i) only |

Question 60 Explanation:

→ Windows XP's Peer-to-Peer Networking Wizard allows you to set up a firewall-protected network.

→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine

→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine

Question 61 |

consider the following postfix expression with single digit operands :

6 2 3 * / 4 2 * + 6 8 * -

The top two elements of the stack after second * is evaluated, are :

6 2 3 * / 4 2 * + 6 8 * -

The top two elements of the stack after second * is evaluated, are :

6, 3 | |

8, 1 | |

8, 2 | |

6, 2 |

Question 61 Explanation:

For evaluating a postfix expression we start traversing from first element of the expression

While traversing the expression if you find an element value then push it on the top of stack this way the top element of the stack will represent the second operand of an operation and the element presents after top element will represent the first operand of the operation. While traversing the postfix expression if you find an operation symbol then pop 2 elements from the top of the stack and then after performing operation store it’s result on the top of stack.

While traversing the expression if you find an element value then push it on the top of stack this way the top element of the stack will represent the second operand of an operation and the element presents after top element will represent the first operand of the operation. While traversing the postfix expression if you find an operation symbol then pop 2 elements from the top of the stack and then after performing operation store it’s result on the top of stack.

Question 62 |

The elements 42, 25, 30, 40, 22, 35, 26 are inserted one by one in the given order into a max-heap. The resultant max-heap is stored in an array implementation as

<42, 35, 40, 22, 25, 26, 30> | |

<42, 35, 40, 22, 25, 30, 26> | |

<42, 40, 35, 25, 22, 26, 30> | |

<42, 40, 35, 25, 22, 30, 26> |

Question 62 Explanation:

After inserting each element we will apply MAX-Heapify operation to get the MAX-Heap.

The resultant MAX_Heap will look like

< 4 2, 40, 35, 25, 22, 30, 26 >

The resultant MAX_Heap will look like

< 4 2, 40, 35, 25, 22, 30, 26 >

Question 63 |

The host is connected to a department network which is a part of a university network. The university network, in turn, is part of the internet. The largest network, in which the Ethernet address of the host is unique, is

The university network | |

The internet | |

The subnet to which the host belongs | |

The department network |

Question 63 Explanation:

The Ethernet address, known as MAC (Media Access Control) address, is a 48-bit hardware address which is burned on NIC. Ethernet address is supposed to be unique. If two devices with same Ethernet address are on two different networks then those devices can be uniquely identified. This is because of different Network Ids.

But within the network any two devices should have different MAC addresses.

But within the network any two devices should have different MAC addresses.

Question 64 |

let r= a(a + b)*, s=aa*b and t=a*b be three regular expressions. Consider the following:

(i) L(s) ⊆ L(r) and L(s) ⊆ L(t)

(ii). L(r) ⊆ L(s) and L(s) ⊆ L(t)

(i) L(s) ⊆ L(r) and L(s) ⊆ L(t)

(ii). L(r) ⊆ L(s) and L(s) ⊆ L(t)

Only (i) is correct | |

Both (i) and (ii) are correct | |

Only (ii) is correct | |

Neither (i) nor (ii) is correct |

Question 64 Explanation:

L(s) = {∊, a, b, ab, bb, aa, ba, aab, baa,aaab, aaab,......}

L(r) = { ab, aab, aaab, aaaab, ............}

L(t) = { b, ab, aab, aaab, aaaab, .........}

L(s) can generate every possible string over a, b but L(r) and L(t) can’t generate every possible

string over a, b. So L(s)⊆L(r) and L(t)⊆L(r)

L(r) can’t generate “b” but L(t) can’t. So L(s)⊆L(t).

L(r) = { ab, aab, aaab, aaaab, ............}

L(t) = { b, ab, aab, aaab, aaaab, .........}

L(s) can generate every possible string over a, b but L(r) and L(t) can’t generate every possible

string over a, b. So L(s)⊆L(r) and L(t)⊆L(r)

L(r) can’t generate “b” but L(t) can’t. So L(s)⊆L(t).

Question 65 |

Match List1 with List 2 and choose the correct answer from the code given below :

Where V and E are the number of vertices and edges in graph respectively.

Where V and E are the number of vertices and edges in graph respectively.

(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii) | |

(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv) | |

(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) | |

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) |

Question 65 Explanation:

→ Dijkstra's algorithm (or Dijkstra's Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the shortest paths between nodes in a graph and time complexity is O(V

→ Kruskal's algorithm is a minimum-spanning-tree algorithm which finds an edge of the least possible weight that connects any two trees in the forest and time complexity is O(E log E)

→ A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v,u comes before v in the ordering Θ(V + E)

→ Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights. A single execution of the algorithm will find the lengths of shortest paths between all pairs of vertices

^{2} )→ Kruskal's algorithm is a minimum-spanning-tree algorithm which finds an edge of the least possible weight that connects any two trees in the forest and time complexity is O(E log E)

→ A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v,u comes before v in the ordering Θ(V + E)

→ Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights. A single execution of the algorithm will find the lengths of shortest paths between all pairs of vertices

Question 66 |

An agent can improve its performance by

Learning | |

Responding | |

Observing | |

Perceiving |

Question 66 Explanation:

→ An intelligent agent (IA) is an autonomous entity which observes through sensors and acts upon an environment using actuators (i.e. it is an agent) and directs its activity towards achieving goals (i.e. it is "rational", as defined in economics).

→ Intelligent agents may also learn or use knowledge to achieve their goals. They may be very simple or very complex. A reflex machine, such as a thermostat, is considered an example of an intelligent agent.

→ Intelligent agents may also learn or use knowledge to achieve their goals. They may be very simple or very complex. A reflex machine, such as a thermostat, is considered an example of an intelligent agent.

Question 67 |

The clustering index is defined on the fields which are of type

key and ordering | |

key and non-ordering | |

non-key and ordering | |

non-key and non-ordering |

Question 67 Explanation:

Primary Index − Primary index is defined on an ordered data file. The data file is ordered on a key field. The key field is generally the primary key of the relation.

Secondary Index − Secondary index may be generated from a field which is a candidate key and has a unique value in every record, or a non-key with duplicate values.

Clustering Index − Clustering index is defined on an ordered data file. The data file is ordered on a non-key field.

Ordered Indexing is of two types −Dense Index and Sparse Index

Secondary Index − Secondary index may be generated from a field which is a candidate key and has a unique value in every record, or a non-key with duplicate values.

Clustering Index − Clustering index is defined on an ordered data file. The data file is ordered on a non-key field.

Ordered Indexing is of two types −Dense Index and Sparse Index

Question 68 |

The second smallest of n elements can be found with _______ comparisons in worst case.

n + ceil(lg n) -2 | |

n-1 | |

lg n | |

3n/1 |

Question 68 Explanation:

This takes n-1 comparisons and each element involved is compared at most log n times. Finally, we get the largest element A1’’’ . This element which is largest, has won comparison with the second largest element at some point. So, keeping track of items compared with the largest element will give us a list of (log n) elements. In this list we find out largest, which will be the second largest of all.

So, no. of comparisons =(n-1)+log

_{2} (n)-1

Example: 512 elements

In this case no. of comparisons will be, (512-1)+log

_{2} (512) -1

=511+(9-1)

=511+8

=519

Question 69 |

Suppose that everyone in a group of N people wants to communicate secretly with (N-1) other people using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is

2N | |

N(N-1) | |

N(N-1)/2 | |

(N-1) _{2} |

Question 69 Explanation:

→ If one person in a group of N people want to communicate with remaining (N-1) people using symmetric key cryptographic system then he needs (N-1) keys.

→ We have N people in group so number of keys needed are N(N-1)

→ But two people in a group can use same keys then no need to use 2(N-1) keys they can communicate using (N-1) keys.

So total number of keys needed [N(N-1)]/2

→ We have N people in group so number of keys needed are N(N-1)

→ But two people in a group can use same keys then no need to use 2(N-1) keys they can communicate using (N-1) keys.

So total number of keys needed [N(N-1)]/2

Question 70 |

What does the following java function perform ? (Assume int occupies four bytes of storage)

Public static int f(int a)

{

//Pre-conditions : a>0 and no overflow/underflow occurs

int b=0;

for(int i=0; i<32; i++)

{

b=b <<1;

b= b | (a & 1);

a = a >>> 1; // This is a logical shift

}

Return b;

}

Public static int f(int a)

{

//Pre-conditions : a>0 and no overflow/underflow occurs

int b=0;

for(int i=0; i<32; i++)

{

b=b <<1;

b= b | (a & 1);

a = a >>> 1; // This is a logical shift

}

Return b;

}

Return the int that represents the number of 0’s in the binary representation of integer a. | |

Return the int that represents the number of 1’s in the binary representation of integer a. | |

Return the int that has the reversed binary representation of integer a. | |

Return the int that has the binary representation of integer a. |

Question 70 Explanation:

→ Consider the a value is 5 .

The initial value of b is 0 and b<<1 means b will get value of 2

b=b|(a&1) which means 2|(5&1) then b will get value “2”

a=a<<1 means value a will reduce to 3

→ Repeat the above procedure for 31 iterations .

→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1

→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1)= 4|(3&1) then b is 11 and a becomes 0

→ And this procedure will repeat until “i” value is 32.

→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.

The initial value of b is 0 and b<<1 means b will get value of 2

b=b|(a&1) which means 2|(5&1) then b will get value “2”

a=a<<1 means value a will reduce to 3

→ Repeat the above procedure for 31 iterations .

→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1

→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1)= 4|(3&1) then b is 11 and a becomes 0

→ And this procedure will repeat until “i” value is 32.

→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.

Question 71 |

Consider the following x86 - assembly language instructions :

MOV AL, 153

NEG AL

The contents of the destination register AL (in 8-bit binary notation), the status of Carry Flag(CF) and Sign Flag(SF) after the execution of above instructions, are

MOV AL, 153

NEG AL

The contents of the destination register AL (in 8-bit binary notation), the status of Carry Flag(CF) and Sign Flag(SF) after the execution of above instructions, are

AL = 0110 0110; CF = 0; SF =0 | |

AL = 0110 0111; CF = 0; SF =1 | |

AL = 0110 0110; CF = 1; SF =1 | |

AL = 0110 0111; CF = 1; SF =0 |

Question 71 Explanation:

153 = 1001 1001

NEG(1001 1001) = 1's complement of(1001 1001)= 0110 0110

MOV AL 153

AL= 1001 1001

NEG AL

AL = 0110 0101

MSB is 0. Hence SF=0

By default CF=0 and it will not change with above operations( MOV, NEG)

So, option 1 is the correct answer.

Note: Original UGC Key given correct answer is option-(4).

NEG(1001 1001) = 1's complement of(1001 1001)= 0110 0110

MOV AL 153

AL= 1001 1001

NEG AL

AL = 0110 0101

MSB is 0. Hence SF=0

By default CF=0 and it will not change with above operations( MOV, NEG)

So, option 1 is the correct answer.

Note: Original UGC Key given correct answer is option-(4).

Question 72 |

In mathematical logic, which of the following are statements ?

(i) There will be snow in January

(ii) What is the time now ?

(iii) Today is Sunday

(iv) You must study Discrete Mathematics.

(i) There will be snow in January

(ii) What is the time now ?

(iii) Today is Sunday

(iv) You must study Discrete Mathematics.

(iii) and (iv) | |

(i) and (ii) | |

(i) and (iii) | |

(ii) and (iv) |

Question 72 Explanation:

In mathematical logic, the term statement is variously understood to mean either:

(a) a meaningful declarative sentence that is true or false, or

(b) the assertion that is made by a true or false declarative sentence.

From the above four statements , statement 2 and 4 wont give meaning like true or false answers and statements 1 and 3 will give either true or false answers

(a) a meaningful declarative sentence that is true or false, or

(b) the assertion that is made by a true or false declarative sentence.

From the above four statements , statement 2 and 4 wont give meaning like true or false answers and statements 1 and 3 will give either true or false answers

Question 73 |

In Linux operating system environment ________ command is used to print a file.

print | |

lpr | |

ptr | |

pr |

Question 73 Explanation:

lpr command is used to print a file in LINUX operating system.

→ lpr submits files for printing. Files named on the command line are sent to the named printer (or the system default destination if no destination is specified). If no files are listed on the command-line, lpr reads the print file from the standard input.

→ lpr submits files for printing. Files named on the command line are sent to the named printer (or the system default destination if no destination is specified). If no files are listed on the command-line, lpr reads the print file from the standard input.

Question 74 |

Consider the language L given by

L = { 2

The minimum number of states of finite automaton which accept the language L is

L = { 2

_{nk}| k > 0 , and n is non − n egative integer number }The minimum number of states of finite automaton which accept the language L is

n | |

n+1 | |

2 ^{n} | |

n (n + 1 )/2 |

Question 74 Explanation:

n is a positive integer constant (means it is fixed)

assume n= 2 (for instance)

then we have : L= a

so k's value will be {1,2,3,4,........} and accordingly in L we have strings {a

→ It is clearly even number of a's (except zero a's) and for this we have 3 state DFA,

→ If n=3 then we will have a

→ So, we require 4 state DFA

hence answer will be "n+1" states are required.

assume n= 2 (for instance)

then we have : L= a

^{2k} | k>0so k's value will be {1,2,3,4,........} and accordingly in L we have strings {a

^{2} ,a^{ 4} ,a^{ 6} , a^{ 8} .......}→ It is clearly even number of a's (except zero a's) and for this we have 3 state DFA,

→ If n=3 then we will have a

^{3} , a^{ 6} , a^{ 9} ..........→ So, we require 4 state DFA

hence answer will be "n+1" states are required.

Question 75 |

Which of the following problem is decidable for recursive language (L) ?

Is L = ф ? | |

Is w∈L, where w is a string ? | |

Is L=R, where R is given regular set ? | |

Is L = Σ* ? |

Question 75 Explanation:

→ A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w.

→ Every decidable language is Turing-Acceptable.

→ For recursive language membership problem is decidable, i.e., a string “w” belongs to a recursive language “L” is decidable. Rest all are undecidable.

→ If language is recursive then it means , we have a Turing machine for the language which always halt. So when a string “w” is given to this turing machine it always halt either by accepting “w” or by rejecting “w”. Hence “Is w∈L” is decidable.

→ Every decidable language is Turing-Acceptable.

→ For recursive language membership problem is decidable, i.e., a string “w” belongs to a recursive language “L” is decidable. Rest all are undecidable.

→ If language is recursive then it means , we have a Turing machine for the language which always halt. So when a string “w” is given to this turing machine it always halt either by accepting “w” or by rejecting “w”. Hence “Is w∈L” is decidable.

Question 76 |

Consider the following Grammar G :

S ➝ A | B

A➝ a | c

B➝ b | c

Where {S,A,B} is the set of non-terminals, {a,b,c} is the set of terminals.

Which of the following statement(s) is/are correct ?

S 1 : LR(1) can parse all strings that are generated using grammar G.

S 2 : LL(1) can parse all strings that are generated using grammar G.

S ➝ A | B

A➝ a | c

B➝ b | c

Where {S,A,B} is the set of non-terminals, {a,b,c} is the set of terminals.

Which of the following statement(s) is/are correct ?

S 1 : LR(1) can parse all strings that are generated using grammar G.

S 2 : LL(1) can parse all strings that are generated using grammar G.

Both S 1 and S 2 | |

Only S 2 | |

Neither S 1 nor S 2 | |

Only S 1 |

Question 76 Explanation:

For generating string “c” we have two different parse trees.

Since the grammar is Ambiguous so the strings generated by the grammar G can’t be parsed by LR(1) or LL(1) parser.

Since the grammar is Ambiguous so the strings generated by the grammar G can’t be parsed by LR(1) or LL(1) parser.

Question 77 |

Consider the statements below :

“ There is a country that borders both India and Nepal. “

Which of the following represents the above sentence correctly ?

“ There is a country that borders both India and Nepal. “

Which of the following represents the above sentence correctly ?

∃c Border(Country(c), India ∧ Nepal) | |

∃c Country(c) ∧ Border(c, India) ∧ Border(c, Nepal) | |

[∃c Country(c)] ⇒ [Border(c,India) ∧ Border(c, Nepal)] | |

∃c Country(c) ⇒ [ Border(c, India) ∧ Border(c, Nepal)] |

Question 77 Explanation:

→ ∃c Country(c) which represents there is one country “c”

→ Border(c, India) which represents border between c and india

→ Border(c, Nepal) which represents border between c and Nepal

→ “∧” represents both

Option-2 represents the sentence “ There is a country that borders both India and Nepal.

→ Border(c, India) which represents border between c and india

→ Border(c, Nepal) which represents border between c and Nepal

→ “∧” represents both

Option-2 represents the sentence “ There is a country that borders both India and Nepal.

Question 78 |

Suppose for a process P, reference to pages in order are 1, 2, 4, 5,2,1,2,4. Assume that main memory can accommodate 3 pages and the main memory has already pages 1 and 2 in the order 1 - first, 2- second. At this moment, assume FIFO Page Replacement Algorithm is used then the number of page faults that occur to complete the execution of process P is

6 | |

4 | |

3 | |

5 |

Question 78 Explanation:

In this problem, they are mentioned that “main memory has already pages 1 and 2 in the order 1 - first, 2- second”. So total page faults are 7-already 2 available.

=7-2

=5

Question 79 |

Consider the following recursive Java function f that takes two long arguments and returns a float value :

public static float f(long m, long n)

{

float result = (float) m / (float) n;

if (m < 0 || n < 0)

return 0⋅0f;

else

result += f(m*2, n*3);

result result;

}

Which of the following integers best approximates the value of f(2,3) ?

public static float f(long m, long n)

{

float result = (float) m / (float) n;

if (m < 0 || n < 0)

return 0⋅0f;

else

result += f(m*2, n*3);

result result;

}

Which of the following integers best approximates the value of f(2,3) ?

0 | |

3 | |

1 | |

2 |

Question 79 Explanation:

public static float f(long m, long n)

{

float result = (float) m / (float) n;

if (m < 0 || n < 0)

return 0⋅0f;

else

result += f(m*2, n*3);

result result;

}

function call→ f(2,3)

→ result=2.0/3.0

=0.666

if condition is false, else will be executed result = 0.666+f(4,9) function call→ f(4,9)

→ result=4.0/9.0

=0.444

result=0.4444+f(8,27)

function call→ f(8,27)

→ result=8.0/27.0

=0.2962

result=0.2962+f(16,54)

function call→ f(16,81)

→ result=16.0/81.0

=0.1975

result=0.1975+f(32,243)

function call→ f(32,243)

→ result= 32.0/243.0

= 0.13168

result=0.13168+f(64,729)

and so on

f(128,2187),f(256,6561),f(512,19683),f(1024,59049),f(2048,177147)

result=0.6666+0.4444+0.2962+0.1975+0.13168+0.08779+0.0390+0.026+0.0173+0.0115

The best approximate value is sum of all the function call result values which is equal 2.

{

float result = (float) m / (float) n;

if (m < 0 || n < 0)

return 0⋅0f;

else

result += f(m*2, n*3);

result result;

}

function call→ f(2,3)

→ result=2.0/3.0

=0.666

if condition is false, else will be executed result = 0.666+f(4,9) function call→ f(4,9)

→ result=4.0/9.0

=0.444

result=0.4444+f(8,27)

function call→ f(8,27)

→ result=8.0/27.0

=0.2962

result=0.2962+f(16,54)

function call→ f(16,81)

→ result=16.0/81.0

=0.1975

result=0.1975+f(32,243)

function call→ f(32,243)

→ result= 32.0/243.0

= 0.13168

result=0.13168+f(64,729)

and so on

f(128,2187),f(256,6561),f(512,19683),f(1024,59049),f(2048,177147)

result=0.6666+0.4444+0.2962+0.1975+0.13168+0.08779+0.0390+0.026+0.0173+0.0115

The best approximate value is sum of all the function call result values which is equal 2.

Question 80 |

Consider R to be any regular language and L

_{1} .L_{2} be any two context-free languages Which one of the following is correct ?(L _{1} ⋃ L_{ 2} ) − R is context free | |

L _{1} − R is context free | |

L _{1} is context free | |

L _{1} ⋂ L_{ 2} is context free |

Question 80 Explanation:

Option 1: L

Option 2: L

L

R = (a+b)* which can generate language {∈, a, b, aa, bb, ab, ba, .......}

So L

And since every regular language is also a CFL so option 2 is correct.

Option 3: context free languages are not closed under complementation. So option 3 is incorrect.

Option 4: context free languages are not closed under Intersection. So option 4 is incorrect.

_{1} and L_{ 2} are context free languages and context free languages are closed under union but not closed under complementation. It means intersection result may or may not br CFL. So when we will do subtraction the result may or may not be CFL.Option 2: L

_{1} is a CFL and R is a Regular language and if we do L_{1} - R the result will Regular language.L

_{1} = {a n b n | n>0} is a context free language which can generate strings {ab, aabb, aaabbb, .......}R = (a+b)* which can generate language {∈, a, b, aa, bb, ab, ba, .......}

So L

_{1} -R = {ф} which is a regular language.And since every regular language is also a CFL so option 2 is correct.

Option 3: context free languages are not closed under complementation. So option 3 is incorrect.

Option 4: context free languages are not closed under Intersection. So option 4 is incorrect.

Question 81 |

Consider the following boolean equations :

(i). wx + w(x + y) + x(x + y)=x+wy

(ii). (wx’(y+xz’)+w’x’)y=x’y

What can you say about the above equations ?

(i). wx + w(x + y) + x(x + y)=x+wy

(ii). (wx’(y+xz’)+w’x’)y=x’y

What can you say about the above equations ?

Both (i) and (ii) are true | |

(i) is true and (ii) is false | |

Both (i) and (ii) are false | |

(i) is false and (ii) is true |

Question 81 Explanation:

(i) wx + w(x + y) + x(x + y)

= (wx + wx) + wy + (x + xy)

= wx + wy + x(1 + y)

= wx + wy + x

= (w + 1)x + wy

= x + wy

= (wx + wx) + wy + (x + xy)

= wx + wy + x(1 + y)

= wx + wy + x

= (w + 1)x + wy

= x + wy

Question 82 |

Consider the following statements :

(i) Auto increment addressing mode is useful in creating self-relocating code.

(ii) If auto addressing mode is included in an instruction set architecture, then an additional ALU is required for effective address calculation.

(iii) In auto increment addressing mode, the amount of increment depends on the size of the data item accessed.

Which of the above statements is/are true ?

(i) Auto increment addressing mode is useful in creating self-relocating code.

(ii) If auto addressing mode is included in an instruction set architecture, then an additional ALU is required for effective address calculation.

(iii) In auto increment addressing mode, the amount of increment depends on the size of the data item accessed.

Which of the above statements is/are true ?

(iii) only | |

(ii) and (iii) only | |

(i) and (ii) only | |

(ii) only |

Question 82 Explanation:

→ Auto increment addressing mode is useful in implementing arrays.

→ After determining the effective address, the value in the base register is incremented by the size of the data item that is to be accessed. For example, (A7)+ would access the content of the address register A7, then increase the address pointer of A7 by 1 (usually 1 word).

→ Within a loop, this addressing mode can be used to step through all the elements of an array or vector.

→ After determining the effective address, the value in the base register is incremented by the size of the data item that is to be accessed. For example, (A7)+ would access the content of the address register A7, then increase the address pointer of A7 by 1 (usually 1 word).

→ Within a loop, this addressing mode can be used to step through all the elements of an array or vector.

Question 83 |

Find the boolean expression for the logic circuit shown below :

(1-NAND gate, 2-NOR gate, 3-NOR gate)

(1-NAND gate, 2-NOR gate, 3-NOR gate)

AB | |

AB’ | |

A’B’ | |

A’B |

Question 83 Explanation:

Question 84 |

The decimal floating point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is

0xC2206000 | |

0xC2006666 | |

0xC2006000 | |

0xC2206666 |

Question 84 Explanation:

1. Fraction part can be converted into binary form by multiplying it with 2 and taking non fractional part of the product. Take fractional part and multiply again as explained above.

0.1 x 2= 0.2 → 0

0.2 x 2= 0.4 → 0

0.4 x 2= 0.8 → 0

0.8 x 2= 1.6 → 1

0.6 x 2= 1.2 → 1

(0.1)

(40)

101000.00011

Normalize the number

1.0100000011 x 2

Biased exponent= 5+127= 132=(1000 0100)

Mantissa= 01000000110000000000000

Sign= 1

0.1 x 2= 0.2 → 0

0.2 x 2= 0.4 → 0

0.4 x 2= 0.8 → 0

0.8 x 2= 1.6 → 1

0.6 x 2= 1.2 → 1

(0.1)

_{10} = (0.00011)_{ 2}(40)

_{10} = (101000)_{ 2}101000.00011

Normalize the number

1.0100000011 x 2

^{5}Biased exponent= 5+127= 132=(1000 0100)

_{ 2}Mantissa= 01000000110000000000000

Sign= 1

Question 85 |

Consider two sequences X and Y :

X = <0,1,2,1,3,0,1 >

Y = <1,3,2,0,1,0 >

The length of longest common subsequence between X and Y is

X = <0,1,2,1,3,0,1 >

Y = <1,3,2,0,1,0 >

The length of longest common subsequence between X and Y is

4 | |

2 | |

3 | |

5 |

Question 85 Explanation:

Possible subsequences are :130, 1210, 1301 and length of longest common subsequence is 4.

Question 86 |

Consider the following method :

int f(int m, int n, boolean x, boolean y)

{

int res=0;

if(m<0) {res=n-m;}

else if(x || y) {

res= -1;

if( n==m) { res =1; }

}

else {res=n; }

return res;

} /*end of f */

If P is the minimum number of tests to achieve full statement coverage for f() and Q is the minimum number of tests to achieve full branch coverage for f(), then (P,Q) =

int f(int m, int n, boolean x, boolean y)

{

int res=0;

if(m<0) {res=n-m;}

else if(x || y) {

res= -1;

if( n==m) { res =1; }

}

else {res=n; }

return res;

} /*end of f */

If P is the minimum number of tests to achieve full statement coverage for f() and Q is the minimum number of tests to achieve full branch coverage for f(), then (P,Q) =

(3,4) | |

(3,2) | |

(2,3) | |

(4,3) |

Question 86 Explanation:

→ Code coverage is a measure which describes the degree of which the source code of the program has been tested

→ Statement coverage is a white box test design technique which involves execution of all the executable statements in the source code at least once. It is used to calculate and measure the number of statements in the source code which can be executed given the requirements.

→ Branch coverage is a testing method, which aims to ensure that each one of the possible branch from each decision point is executed at least once and thereby ensuring that all reachable code is executed.

→ In the branch coverage, every outcome from a code module is tested. For example, if the outcomes are binary, you need to test both True and False outcomes.

→ Statement coverage is a white box test design technique which involves execution of all the executable statements in the source code at least once. It is used to calculate and measure the number of statements in the source code which can be executed given the requirements.

→ Branch coverage is a testing method, which aims to ensure that each one of the possible branch from each decision point is executed at least once and thereby ensuring that all reachable code is executed.

→ In the branch coverage, every outcome from a code module is tested. For example, if the outcomes are binary, you need to test both True and False outcomes.

Question 87 |

______________ system call creates new process in Unix.

fork | |

fork new | |

create | |

create new |

Question 87 Explanation:

fork system call is used to creates a new process in Unix.

Question 88 |

Use Dual Simplex Method to solve the following problem :

Maximize z= -2x

subject to:

x

2x

x

x

Maximize z= -2x

_{1} -3x_{ 2}subject to:

x

_{1} + x_{ 2} ≥ 22x

_{1} + x_{ 2} ≤ 10x

_{1} + x_{ 2} ≤ 8x

_{1} , x_{ 2} ≥ 0x _{1} =6, x_{ 2} =2 and z = -18 | |

x _{1} =2, x_{ 2} =6 and z = -22 | |

x _{1} =2, x_{ 2} =0 and z= -4 | |

x _{1} =0, x_{ 2} =2 and z= -6 |

Question 88 Explanation:

→ One simple method is substitute the option values in the maximize equation Z to know the maximum value of the problem and remaining equations to satisfies the conditions which are
given in the problem statement.

→ Option 3 satisfies all conditions and maximize property in the given problem

→ Option 3 satisfies all conditions and maximize property in the given problem

Question 89 |

Consider the midpoint (or Bresenham) algorithm for rasterizing lines given below :

(1) Input (x

(2) y=y

(3) d=f(x

(4) for x=x

(5) do

(6) plot(x,y)

(7) if(d<0)

(8) then

(9) y=y+1

(10) d=d+(y

(11) else

(12) d=d+(y

(13) end

(14) end

Which statements are true ?

P: For a line with slope m>1, we should change the outer loop in line (4) to be over y.

Q: Lines (10) and (12) update the decision variable d through an incremental evaluation of the line equation f.

R: The algorithm fails if d is ever 0.

(1) Input (x

_{1} ,y_{ 1} ) and (x_{ 2} ,y_{ 2} )(2) y=y

_{ 1}(3) d=f(x

_{1} +1, y_{1} +1⁄2) // f is the implicit form of a line(4) for x=x

_{ 1} to x_{ 2}(5) do

(6) plot(x,y)

(7) if(d<0)

(8) then

(9) y=y+1

(10) d=d+(y

_{ 1} - y_{2} ) + (x_{ 2} - x_{ 1} )(11) else

(12) d=d+(y

_{ 1} - y_{ 2} )(13) end

(14) end

Which statements are true ?

P: For a line with slope m>1, we should change the outer loop in line (4) to be over y.

Q: Lines (10) and (12) update the decision variable d through an incremental evaluation of the line equation f.

R: The algorithm fails if d is ever 0.

Q and R only | |

P only | |

P and Q only | |

P,Q and R |

Question 89 Explanation:

From the code given in question gives information that the algorithm will work if d is ever 0, So the statement R is false.

Line 10 and 12 will update the value of d , So the statement Q is true.

Line 10 and 12 will update the value of d , So the statement Q is true.

Question 90 |

In PERT/CPM, the merge event represents___________ of two or more events.

splitting | |

completion | |

beginning | |

joining |

Question 90 Explanation:

→ The program (or project) evaluation and review technique (PERT) is a statistical tool used in project management, which was designed to analyze and represent the tasks involved in completing a given project.

→ The critical path method (CPM), or critical path analysis (CPA), is an algorithm for scheduling a set of project activities. It is commonly used in conjunction with the program evaluation and review technique (PERT).

→ A critical path is determined by identifying the longest stretch of dependent activities and measuring the time required to complete them from start to finish.

→ Event: An event represents a point in time signifying the completion of some activities and the beginning of new ones. This is usually represented by a circle in a network which is also called a node or connector. The events are classified into three categories

1. Merge event – When more than one activity comes and joins an event such an event is known as merge event.

2. Burst event – When more than one activity leaves an event such an event is known as burst event.

3. Merge and Burst event – An activity may be merge and burst event at the same time as with respect to some activities it can be a merge event and with respect to some other activities it may be a burst event.

→ The critical path method (CPM), or critical path analysis (CPA), is an algorithm for scheduling a set of project activities. It is commonly used in conjunction with the program evaluation and review technique (PERT).

→ A critical path is determined by identifying the longest stretch of dependent activities and measuring the time required to complete them from start to finish.

→ Event: An event represents a point in time signifying the completion of some activities and the beginning of new ones. This is usually represented by a circle in a network which is also called a node or connector. The events are classified into three categories

1. Merge event – When more than one activity comes and joins an event such an event is known as merge event.

2. Burst event – When more than one activity leaves an event such an event is known as burst event.

3. Merge and Burst event – An activity may be merge and burst event at the same time as with respect to some activities it can be a merge event and with respect to some other activities it may be a burst event.

Question 91 |

A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be green ?

1/35 | |

1/14 | |

1/9 | |

3/7 |

Question 91 Explanation:

→ Total red balls are 6

→ Green balls are 4

→ Total 10 balls in the box. We need to select 4 balls from 10 balls in which two red balls out of 6 red balls and two green balls from 4 green balls.

→ Total of number of ways for selecting 4 balls out of 10 balls is C(10,4)

→ Number of ways for selecting two red balls from 6 and two green balls from 4 is C(6,2)*C(4,2)

→ Probability of selecting two balls is number of ways for selecting four balls / total number of ways P= C(6,2)*C(4,2) / C(10,4) = C(6,2) is 6x5/2=15

= C(4,2) is 4x3/2=6

= C(10,4) is (10x9x8x7) / (4x3x2x1)=210

P=C(6,2)*C(4,2) / C(10,4)

=15x6/210=

= 3/7, So option 4 is correct

→ Green balls are 4

→ Total 10 balls in the box. We need to select 4 balls from 10 balls in which two red balls out of 6 red balls and two green balls from 4 green balls.

→ Total of number of ways for selecting 4 balls out of 10 balls is C(10,4)

→ Number of ways for selecting two red balls from 6 and two green balls from 4 is C(6,2)*C(4,2)

→ Probability of selecting two balls is number of ways for selecting four balls / total number of ways P= C(6,2)*C(4,2) / C(10,4) = C(6,2) is 6x5/2=15

= C(4,2) is 4x3/2=6

= C(10,4) is (10x9x8x7) / (4x3x2x1)=210

P=C(6,2)*C(4,2) / C(10,4)

=15x6/210=

= 3/7, So option 4 is correct

Question 92 |

A process residing in main memory and ready and waiting for execution, is kept on

Job Queue | |

Execution Queue | |

Wait Queue | |

Ready Queue |

Question 92 Explanation:

A process residing in main memory and ready and waiting for execution, is kept on Ready queue.

Question 93 |

Consider the following minimax game tree search

What will be the value propagated at the root ?

What will be the value propagated at the root ?

6 | |

3 | |

5 | |

4 |

Question 93 Explanation:

Question 94 |

The solution of recurrence relation : T(n)=2T(sqrt(n))+lg(n) is

O(lg (n) lg(n)) | |

O(lg (n)) | |

O(n lg (n)) | |

O(lg (n) lg(lg(n))) |

Question 94 Explanation:

T (n) = l og log (n) · l og (n)

T (n) = l og(n) · l og log (n)

Question 95 |

Which of the following is not one of the principles of agile software development method ?

Following the plan | |

Embrace change | |

Customer involvement | |

Incremental delivery |

Question 95 Explanation:

Agile Software Development is based on twelve principles:

1. Customer satisfaction by early and continuous delivery of valuable software.

2. Welcome changing requirements, even in late development.

3. Deliver working software frequently (weeks rather than months)

4. Close, daily cooperation between business people and developers

5. Projects are built around motivated individuals, who should be trusted

6. Face-to-face conversation is the best form of communication (co-location)

7. Working software is the primary measure of progress

8. Sustainable development, able to maintain a constant pace

9. Continuous attention to technical excellence and good design

10. Simplicity—the art of maximizing the amount of work no done—is essential

11. Best architectures, requirements, and designs emerge from self-organizing teams

12. Regularly, the team reflects on how to become more effective, and adjusts accordingly

1. Customer satisfaction by early and continuous delivery of valuable software.

2. Welcome changing requirements, even in late development.

3. Deliver working software frequently (weeks rather than months)

4. Close, daily cooperation between business people and developers

5. Projects are built around motivated individuals, who should be trusted

6. Face-to-face conversation is the best form of communication (co-location)

7. Working software is the primary measure of progress

8. Sustainable development, able to maintain a constant pace

9. Continuous attention to technical excellence and good design

10. Simplicity—the art of maximizing the amount of work no done—is essential

11. Best architectures, requirements, and designs emerge from self-organizing teams

12. Regularly, the team reflects on how to become more effective, and adjusts accordingly

Question 96 |

In 3D Graphics, which of the following statement/s is/are true ?

P: Back-face culling is an example of an image-precision visible-surface determination.

Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

P: Back-face culling is an example of an image-precision visible-surface determination.

Q: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

P only | |

Q only | |

Neither P nor Q | |

P and Q |

Question 96 Explanation:

Back Face Culling:

→ Back-face culling (an object space algorithm) works on 'solid' objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.

→ Back-face culling is not an example of an image-precision visible-surface determination.

→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box - the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.

1. Back faces: faces of opaque object which are “pointing away” from viewer

2. Back face culling – remove back faces (supported by OpenGL)

→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

→ Back-face culling (an object space algorithm) works on 'solid' objects which you are looking at from the outside. That is, the polygons of the surface of the object completely enclose the object.

→ Back-face culling is not an example of an image-precision visible-surface determination.

→ Back-face culling can very quickly remove unnecessary polygons. Unfortunately there are often times when back-face culling can not be used. For example if you wish to make an open-topped box - the inside and the outside of the box both need to be visible, so either two sets of polygons must be generated, one set facing out and another facing in, or back-face culling must be turned off to draw that object.

1. Back faces: faces of opaque object which are “pointing away” from viewer

2. Back face culling – remove back faces (supported by OpenGL)

→ TRUE: Z-Buffer is a 16-bit, 32-bit, or 64-bit field associated with each pixel in a frame buffer that can be used to determine the visible surface at each pixel.

Question 97 |

________ command is used to remove a relation from an SQL database.

Update table | |

Remove table | |

Delete table | |

Drop table |

Question 97 Explanation:

Update table : Update table command is used to update the content of the table.

Delete table : Delete table command is used to delete the data

stored in the table. When we use Delete command table is not deleted only data stored in the table is deleted.

Drop table : Using table command we can delete a table/relation.

Delete table : Delete table command is used to delete the data

stored in the table. When we use Delete command table is not deleted only data stored in the table is deleted.

Drop table : Using table command we can delete a table/relation.

Question 98 |

Suppose a cloud contains software stack such as Operating system, Application software, etc. This model is referred as _________ model.

SaaS | |

IaaS | |

MaaS | |

PaaS |

Question 98 Explanation:

The following explains the 3 services offered by cloud computing for businesses:

1) Platform as a Service (PaaS): PaaS clouds are created, often times inside IaaS clouds by specialists to deliver the scalability and distribution of any application and to aid make a company’s expenses predictable.

2) Infrastructure as a Service (IaaS): This service provides business access to essential web infrastructure such as servers, connections, storage space, without the need to buy or manage internet infrastructure themselves.

3) Software as a Service (SaaS): This service offered by cloud computing is relatively mature and its phrases use those included in cloud computing. Cloud applications permit the cloud to be leveraged for software infrastructure. This reduces the burden of support, maintenance and operations because the application is run on computers that are owned by the vendor.

1) Platform as a Service (PaaS): PaaS clouds are created, often times inside IaaS clouds by specialists to deliver the scalability and distribution of any application and to aid make a company’s expenses predictable.

2) Infrastructure as a Service (IaaS): This service provides business access to essential web infrastructure such as servers, connections, storage space, without the need to buy or manage internet infrastructure themselves.

3) Software as a Service (SaaS): This service offered by cloud computing is relatively mature and its phrases use those included in cloud computing. Cloud applications permit the cloud to be leveraged for software infrastructure. This reduces the burden of support, maintenance and operations because the application is run on computers that are owned by the vendor.

Question 99 |

Which of the following statements are true ?

(i) Every logic network is equivalent to one using just NAND gates or just NOR gates.

(ii) Boolean expressions and logic networks correspond to labelled acyclic digraphs.

(iii) No two Boolean algebras with n atoms are isomorphic.

(iv) Non-zero elements of finite Boolean algebras are not uniquely expressible as joins of atoms.

(i) Every logic network is equivalent to one using just NAND gates or just NOR gates.

(ii) Boolean expressions and logic networks correspond to labelled acyclic digraphs.

(iii) No two Boolean algebras with n atoms are isomorphic.

(iv) Non-zero elements of finite Boolean algebras are not uniquely expressible as joins of atoms.

(i) and (iv) only | |

(i) and (ii) only | |

(i), (ii) and (iii) only | |

(ii), (iii) and (iv) only |

Question 99 Explanation:

→ A universal logic gate is a logic gate that can be used to construct all other logic gates.

→ NAND and NOR are universal gates, by using these gates we can construct all gates.

→ An atom of a Boolean algebra is an element x such that there exist exactly two elements y satisfying y ≤ x, namely x and 0. A Boolean algebra is said to be atomic when every element is a sup of some set of atoms (the bottom element is always the empty sup).

→ So the options (iii) and (iv) are false

→ NAND and NOR are universal gates, by using these gates we can construct all gates.

→ An atom of a Boolean algebra is an element x such that there exist exactly two elements y satisfying y ≤ x, namely x and 0. A Boolean algebra is said to be atomic when every element is a sup of some set of atoms (the bottom element is always the empty sup).

→ So the options (iii) and (iv) are false

Question 100 |

Software products need perfective maintenance for which of the following reasons ?

To rectify bugs observed while the system is in use. | |

When the customers need the product to run on new platforms | |

To support the new features that users want it to support | |

To overcome wear and tear caused by the repeated use of the software. |

Question 100 Explanation:

Maintenance: Maintenance can be referred as a process of enhancement in the software product according to the changing requirements of user.

4 types of maintenance

1. Adaptive – modifying the system to cope with changes in the software environment (DBMS, OS).

2. Perfective – implementing new or changed user requirements which concern functional enhancements to the software

3. Corrective – diagnosing and fixing errors, possibly ones found by users.

4. Preventive – increasing software maintainability or reliability to prevent problems in the future.

4 types of maintenance

1. Adaptive – modifying the system to cope with changes in the software environment (DBMS, OS).

2. Perfective – implementing new or changed user requirements which concern functional enhancements to the software

3. Corrective – diagnosing and fixing errors, possibly ones found by users.

4. Preventive – increasing software maintainability or reliability to prevent problems in the future.

There are 100 questions to complete.