## Nielit Scentist-B [02-12-2018]

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Question 1 |

**For the given recurrence equation**

T(n)=2T(n-1), if n>0.

=1, otherwise

O(nlogn) | |

O(n ^{2}) | |

O(2 ^{n}) | |

O(n) |

Question 1 Explanation:

T(n) = 2T (n - 1) + 1

= 2 [2T(n - 2) + 1] + 1

= 2

⋮

= 2

= 2

= 2

= 2

≅ O( 2

= 2 [2T(n - 2) + 1] + 1

= 2

^{2}T(n - 2) + 3⋮

= 2

^{k}T(n - k) + (2^{k}- 1)= 2

^{n-1}T(1) + (2^{n-1}- 1)= 2

^{n-1}+ 2^{n-1}- 1= 2

^{n}- 1≅ O( 2

^{n})Question 2 |

___ number of leaf nodes in a rooted tree of n nodes, where each node is having 0 or 3 children.

n/2 | |

(2n+1)/3 | |

(n-1)/n | |

(n-1) |

Question 2 Explanation:

→ Any n-ary tree in which every node has either 0 or n children will take L=(n-1)*I +1

[ Where L is the number of leaf nodes and I is the number of internal nodes]

→ Given data n=3.

L=(3-1)I +1

=2I +1 ------------> 1

→ To find total number of nodes is nothing but sum of leaf nodes and internal nodes

n=L+I ------------> 2

With the help of 1 and 2, we get L =(2n+1)/3/

[ Where L is the number of leaf nodes and I is the number of internal nodes]

→ Given data n=3.

L=(3-1)I +1

=2I +1 ------------> 1

→ To find total number of nodes is nothing but sum of leaf nodes and internal nodes

n=L+I ------------> 2

With the help of 1 and 2, we get L =(2n+1)/3/

Question 3 |

___ number of gates are required to implement the boolean function (AB+C) with using only 2 input NOR gates.

2 | |

3 | |

4 | |

5 |

Question 3 Explanation:

NOR is Complement of OR

AB+C

= (A+C)(B+C) ← Distribution of + over

= ((A+C)’+(B+C)’)’

1st NOR- (A+C)’. Let X = (A+C)’

2nd NOR- (B+C)’. Let Y = (B+C)’

3rd NOR- (X+Y)’

AB+C

= (A+C)(B+C) ← Distribution of + over

= ((A+C)’+(B+C)’)’

1st NOR- (A+C)’. Let X = (A+C)’

2nd NOR- (B+C)’. Let Y = (B+C)’

3rd NOR- (X+Y)’

Question 4 |

Time To Live(TTL) field in the IP datagram is used___

To optimize throughput | |

To prevent packet looping | |

To reduce delays | |

To prioritize packets |

Question 4 Explanation:

→ Time to Live (TTL) is a limit on the period of time or transmissions in computer and computer network technology that a unit of data (e.g. a packet) can experience before it should be discarded.

→ If the limit is not defined then the packets can go into an indefinite loop.

→ The packet is discarded when the time to live field reaches 0 to prevent looping.

→ If the limit is not defined then the packets can go into an indefinite loop.

→ The packet is discarded when the time to live field reaches 0 to prevent looping.

Question 5 |

Consider the following finite state automaton. The language accepted by this automaton is given by the regular.

ab*b* | |

a*b* | |

b*b | |

b*ab* |

Question 5 Explanation:

According to the given state transition diagram, the regular expression is

Step-1: Starting state has loop with symbol ‘b’ . It means, we can get n number of times ‘b’.

Step-2: The symbol ‘a’ between the starting state and end state. It means we can get only one

‘a’ after b*.

Step-3: The final state also having self loop with symbol ‘b’.

Step-1: Starting state has loop with symbol ‘b’ . It means, we can get n number of times ‘b’.

Step-2: The symbol ‘a’ between the starting state and end state. It means we can get only one

‘a’ after b*.

Step-3: The final state also having self loop with symbol ‘b’.

Question 6 |

Identify the subnet mask for the given direct broadcast address of subnet of subnet is 201.15.16.31.

255.255.192.192 | |

255.255.255.198 | |

255.255.255.240 | |

255.255.257.240 | |

None of the above |

Question 6 Explanation:

Subnet mask ID: 201.15.16.16/28 and subnet broadcast ID is 201.15.16.31.

The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits, i.e., 27 bit for NID which is 255.255.255.224.

The correct answer would be none of these.

The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits, i.e., 27 bit for NID which is 255.255.255.224.

The correct answer would be none of these.

Question 7 |

The smallest element that can be found in time___ in a binary max heap

O(nlogn) | |

O(logn) | |

O(n) | |

O(n ^{2}) |

Question 7 Explanation:

We need to traverse all nodes in order to identify smallest element because it is not following binary search tree properties. Searching time for ‘n’ elements will take O(n) in worst case.

Question 8 |

The ALU uses ____ to store intermediate result.

Cache | |

Registers | |

Accumulators | |

Stack |

Question 8 Explanation:

→ An accumulator is a register for short-term, intermediate storage of arithmetic and logic data in a computer's CPU (central processing unit).

→ A processor register (CPU register) is one of a small set of data holding places that are part of the computer processor.

→ Cache memory is a small-sized type of volatile computer memory that provides high-speed data access to a processor and stores frequently used computer programs, applications and data.

→ A processor register (CPU register) is one of a small set of data holding places that are part of the computer processor.

→ Cache memory is a small-sized type of volatile computer memory that provides high-speed data access to a processor and stores frequently used computer programs, applications and data.

Question 9 |

___symbol is used to denote derived attribute in ER model

Dashed ellipse | |

Square ellipse | |

Ellipse with attribute name undIrected | |

Rectangular box |

Question 9 Explanation:

Derived attributes are depicted by dashed ellipse.

Example:

Example:

Question 10 |

Among all given option, ___ must reside in the main memory.

Assembler | |

Compiler | |

Linker | |

Loader |

Question 10 Explanation:

→ A loader is the part of an operating system that is responsible for loading programs and libraries.

→ It is one of the essential stages in the process of starting a program, as it places programs into memory and prepares them for execution.

→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.

→ Once loading is complete, the operating system starts the program by passing control to the loaded program code.

Note: The loader is a program that loads the object program from the secondary memory into the main memory for execution of the program

→ It is one of the essential stages in the process of starting a program, as it places programs into memory and prepares them for execution.

→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.

→ Once loading is complete, the operating system starts the program by passing control to the loaded program code.

Note: The loader is a program that loads the object program from the secondary memory into the main memory for execution of the program

Question 11 |

If the period of a signal is 100 ms. Then its frequency in Hertz is___

10 | |

100 | |

1000 | |

10000 |

Question 11 Explanation:

Given data,

Frequency = 1/Period.

Frequency = 1/100 x 10

= 10 Hz

Frequency = 1/Period.

Frequency = 1/100 x 10

^{-3}Hz= 10 Hz

Question 12 |

(A+C’)(B’+C’) simplifies to ______

AC’ +B’ | |

C(A’+B’) | |

BC’+A | |

AB’+C’ |

Question 12 Explanation:

Distributive law:

x + ( y. z) = (x + y) . (x + z)

Proof:

x + ( y. z) = (x + y) . (x + z)

Proof:

Question 13 |

Identify the true statement from the given statements. Program relocation at run time:

- Requires transfer complete block to some memory locations
- Requires both base address and relative address
- Requires only absolute address

(1) | |

(1) and (2) | |

(1),(2) and (3) | |

(1) and (3) |

Question 13 Explanation:

→ Program relocation at run time transfer complete block to some memory locations.

→ This requires as base address and block should be relatively addressed through this base address .This require both base address and relative address

→ This requires as base address and block should be relatively addressed through this base address .This require both base address and relative address

Question 14 |

___ is the worst case time complexity for all operations(i.e. Search,update and delete) in a general binary search tree

O(n) | |

O(nlogn) | |

O(logn) for search and insert, and O(n) for delete | |

None of these |

Question 14 Explanation:

→ Suppose the elements are not in sorted order, it will take O(n) time for worst case time complexity.

→ In question, they are not mentioned about elements are sorted or unsorted. So, worst case we have to consider unsorted elements.

→ All operations(i.e. Search,update and delete) will take O(n).

→ In question, they are not mentioned about elements are sorted or unsorted. So, worst case we have to consider unsorted elements.

→ All operations(i.e. Search,update and delete) will take O(n).

Question 15 |

Maximum number of superkeys for the relation schema R(X,Y,Z,W) with X as the key is:

6 | |

8 | |

9 | |

12 |

Question 15 Explanation:

→ Maximum no. of possible super keys for a relation with n attributes with one candidate key

(only one attribute) = 2n-1

Here, n = 4.

→ So, the possible super keys = 24-1 = 8

→ The super keys are: X, XY, XZ, YZ, XYZ, XYW, YZW, XYZW.

(only one attribute) = 2n-1

Here, n = 4.

→ So, the possible super keys = 24-1 = 8

→ The super keys are: X, XY, XZ, YZ, XYZ, XYW, YZW, XYZW.

Question 16 |

Identify the correct nodes and edges in the given intermediate code:

(1) i=1

(2) t1=5*I

(3) t2=4*t1

(4) t3=t2

(5) a[t3]=0

(6) i=i+1;

(7) if i<15 goto(2)

(1) i=1

(2) t1=5*I

(3) t2=4*t1

(4) t3=t2

(5) a[t3]=0

(6) i=i+1;

(7) if i<15 goto(2)

33 | |

44 | |

43 | |

34 |

Question 16 Explanation:

Step-1: Initialization we are taking one node

Step-2: From 2nd statement to 6th statement we are performing some tasks.

Step-3: It indicate if condition, so it’s another statement. And the back loop indicates goto statement.

Here, total 3 nodes and 3 edges using control flow graph.

In options they have to give like this

3 and 3

4 and 4

4 and 3

3 and 4

But they combines the node value and edge value. It seems different value.

Step-2: From 2nd statement to 6th statement we are performing some tasks.

Step-3: It indicate if condition, so it’s another statement. And the back loop indicates goto statement.

Here, total 3 nodes and 3 edges using control flow graph.

In options they have to give like this

3 and 3

4 and 4

4 and 3

3 and 4

But they combines the node value and edge value. It seems different value.

Question 17 |

__number of queues are needed to implement symbol and is acting as permanent database

Variable Table | |

Terminal Table | |

Keyword Table | |

Identifier Table |

Question 17 Explanation:

Symbol Table is an important data structure created and maintained by the compiler in order to keep track of semantics of variable i.e. it stores information about scope and binding information about names, information about instances of various entities such as variable and function names, classes, objects, etc.

Symbol Table entries: Each entry in symbol table is associated with attributes that support

→ compiler in different phases.

→ Items stored in Symbol table:

→ Variable names and constants

→ Procedure and function names

→ Literal constants and strings

→ Compiler generated temporaries

→ Labels in source languages

Symbol Table entries: Each entry in symbol table is associated with attributes that support

→ compiler in different phases.

→ Items stored in Symbol table:

→ Variable names and constants

→ Procedure and function names

→ Literal constants and strings

→ Compiler generated temporaries

→ Labels in source languages

Question 18 |

Identify the true statement from the given statements.

(1) Lossless, dependency preserving decomposition into 3NF is always possible

(2) Any relation with two attributes is in BCNF

(1) Lossless, dependency preserving decomposition into 3NF is always possible

(2) Any relation with two attributes is in BCNF

(1) | |

(2) | |

(1) and (2) | |

None of these |

Question 18 Explanation:

Both the statements are true.

(1) TRUE: Lossless, dependency preserving decomposition into 3NF is always possible but not in BCNF. In BCNF lossless is mandatory but dependency preserving is not mandatory.

(ii) TRUE: Any relation with two attributes is in BCNF

(1) TRUE: Lossless, dependency preserving decomposition into 3NF is always possible but not in BCNF. In BCNF lossless is mandatory but dependency preserving is not mandatory.

(ii) TRUE: Any relation with two attributes is in BCNF

Question 19 |

The hexadecimal representation of (632)

_{8}is:19A | |

198 | |

29A | |

291 |

Question 19 Explanation:

(632)

Group four bits from right to left to convert into Hexadecimal

So first four bits 1010 means A

1001 means 9

1 means 1

Then final value is (19A)

_{8}in binary form (110011010)Group four bits from right to left to convert into Hexadecimal

So first four bits 1010 means A

1001 means 9

1 means 1

Then final value is (19A)

Question 20 |

Identify the true statements from the given statements.

(1) HTTP may use different TCP connection for different objects of a webpage if non persistent connections are used

(2) FTP uses two TCP connections, one for data and another control

(3) TELNET and FTP can only use TWO connection at a time.

(1) HTTP may use different TCP connection for different objects of a webpage if non persistent connections are used

(2) FTP uses two TCP connections, one for data and another control

(3) TELNET and FTP can only use TWO connection at a time.

(1) | |

(1) and (2) | |

(2) and (3) | |

(1),(2) and (3) |

Question 20 Explanation:

It is very tricky and difficult question to answer. After referred many resources we got this information.

Question 21 |

Which of the following languages over the alphabet {0,1} is described by the given regular expression: (0+1)*1(0+1)*1?

The set of all strings containing the substrings 11 | |

The set of all string containing at most two 1’s | |

The set of all strings containing at least two 1’s | |

The set of all strings that begins and ends with only 0. |

Question 21 Explanation:

(0+1)* → We can take zero or many times

1→ definitely one should be there.

(0+1)* → We can take zero or many times

1→ definitely one should be there.

So, this regular expression containing at least two 1’s.

1→ definitely one should be there.

(0+1)* → We can take zero or many times

1→ definitely one should be there.

So, this regular expression containing at least two 1’s.

Question 22 |

____ number of undirected graphs can be constructed using V=(v

_{1},v_{2},..v_{n}).n ^{3} | |

2 ^{n(n-1)/2} | |

n-½ | |

2 ^{(n-1)/2} |

Question 22 Explanation:

→ With n vertices no. of possible edges =

→ Each subset of these edges will be form a graph.

→ Number of possible undirected graphs is 2(

^{n}C_{2}→ Each subset of these edges will be form a graph.

→ Number of possible undirected graphs is 2(

^{n}C_{2}) 2^{(n(n-1)/2)}Question 23 |

Dijkstra’s algorithm is based on:

Greedy approach | |

Dynamic programming | |

Backtracking paradigm | |

Divide and conquer paradigm |

Question 23 Explanation:

→ Dijkstra's algorithm is following greedy approach. It always selects shortest path among all possibilities.

→ Dijkstra’s algorithm is solving the problem of single source shortest path.

→ Dijkstra’s algorithm is solving the problem of single source shortest path.

Question 24 |

The language {W

^{a}X^{b}Y^{a+b}| a,b >=1} is:Regular | |

Context free but non regular | |

Context sensitive but non context free | |

Type 0 but non context sensitive |

Question 24 Explanation:

It is context free grammar but not regular because it requires to solve one stack of memory.

Question 25 |

The process executes the following code and after execution ____ number of child process get created.
fork();
fork();
fork();
fork();

4 | |

1 | |

15 | |

16 |

Question 25 Explanation:

Step-1: The number of child processes of “n” fork() system calls are 2

Step-2: Total 4 fork() system calls. It means n=4

= 2

= 2

= 15

^{n}-1.Step-2: Total 4 fork() system calls. It means n=4

= 2

^{n}-1= 2

^{4-1}= 15

Question 26 |

_____ is used to convert from recursive to iterative implementation of an algorithm

Array | |

Tree | |

Stack | |

Queue |

Question 26 Explanation:

→ Stack is used to convert recursive to iterative implementation of an algorithm using last in first out method.

→ Stack to hold information about procedure/function calling and nesting in order to switch to the context of the called function and restore to the caller function when the calling finishes.

→ The functions follow a runtime protocol between caller and callee to save arguments and return value on the stack. Stacks are an important way of supporting nested or recursive function calls.

→ This type of stack is used implicitly by the compiler to support CALL and RETURN statements (or their equivalents) and is not manipulated directly by the programmer.

→ Stack to hold information about procedure/function calling and nesting in order to switch to the context of the called function and restore to the caller function when the calling finishes.

→ The functions follow a runtime protocol between caller and callee to save arguments and return value on the stack. Stacks are an important way of supporting nested or recursive function calls.

→ This type of stack is used implicitly by the compiler to support CALL and RETURN statements (or their equivalents) and is not manipulated directly by the programmer.

Question 27 |

In the truth table, f(x,y) represent the boolean function

x ↔ y | |

x ⋀ y | |

x V y | |

x → y |

Question 27 Explanation:

→ The output of a digital logic Exclusive-NOR gate ONLY goes “HIGH” when its two input terminals, A and B are at the “SAME” logic level which can be either at a logic level “1” or at a logic level “0”.

→ In other words, an even number of logic “1’s” on its inputs gives a logic “1” at the output, otherwise is at logic level “0”.

→ This type of gate gives and output “1” when its inputs are “logically equal” or “equivalent” to each other, which is why an Exclusive-NOR gate is sometimes called an Equivalence Gate

→ In other words, an even number of logic “1’s” on its inputs gives a logic “1” at the output, otherwise is at logic level “0”.

→ This type of gate gives and output “1” when its inputs are “logically equal” or “equivalent” to each other, which is why an Exclusive-NOR gate is sometimes called an Equivalence Gate

Question 28 |

Evaluation of the given postfix expression

10 10 +60 6 / * 8 - is:

10 10 +60 6 / * 8 - is:

192 | |

190 | |

110 | |

92 |

Question 28 Explanation:

→ '10' is pushed in the stack

Question 29 |

Identify the total number of tokens in the given statements printf(“A%B=”, &i);

7 | |

8 | |

9 | |

13 |

Question 29 Explanation:

The tokens are

Printf

(

“A%B=”

,

&

I

)

;

Printf

(

“A%B=”

,

&

I

)

;

Question 30 |

_____ transport layer protocol is used to support electronic mail

SNMP | |

IP | |

SMTP | |

TCP |

Question 30 Explanation:

→ TCP is used in transport layer to carry out mail which is initiated by application layer protocol SMTP.

Primary protocols for E-Mail management

Post Office Protocol (POP)

Simple Mail Transfer Protocol (SMTP)

Internet Message Access Protocol (IMAP)

The above protocols are Application Layer Protocols

→ Once a client connects to the E-Mail Server, there may be 0(zero) or more SMTP transactions. If the client has no mail to send, then there are no SMTP transactions. Every e-mail message sent is an SMTP transfer.

→ SMTP is only used to send (push) messages to the server. POP and IMAP are used to receive messages as well as manage the mailbox contents(which includes tasks such as deleting, moving messages etc.).

Primary protocols for E-Mail management

Post Office Protocol (POP)

Simple Mail Transfer Protocol (SMTP)

Internet Message Access Protocol (IMAP)

The above protocols are Application Layer Protocols

→ Once a client connects to the E-Mail Server, there may be 0(zero) or more SMTP transactions. If the client has no mail to send, then there are no SMTP transactions. Every e-mail message sent is an SMTP transfer.

→ SMTP is only used to send (push) messages to the server. POP and IMAP are used to receive messages as well as manage the mailbox contents(which includes tasks such as deleting, moving messages etc.).

Question 31 |

Find the effective access time for the memory for given data.
Page fault service time=8ms
Average memory access time=20ms
One page fault is generated for every memory access=10

^{6}29ns | |

33ns | |

28ns | |

30ns |

Question 31 Explanation:

Given data,

Page fault service time=8ms

Average memory access time=20ms

One page fault is generated for every memory access=10

P = page fault rate

Effective Memory Access = p × page fault service time + (1 – p) × Memory access time

=(1/10

≅27.99998 ns

Page fault service time=8ms

Average memory access time=20ms

One page fault is generated for every memory access=10

^{6}P = page fault rate

Effective Memory Access = p × page fault service time + (1 – p) × Memory access time

=(1/10

^{6})*8*10^{6}+(1-1/10^{6})*20≅27.99998 ns

Question 32 |

Identify the true statement from the given statements

(1) FIFO is non preemptive

(2) Round robin is non preemptive

(3) Multilevel queue scheduling is non preemptive

(1) FIFO is non preemptive

(2) Round robin is non preemptive

(3) Multilevel queue scheduling is non preemptive

(1) | |

(1) and (2) | |

(1),(2) and (3) | |

(2) |

Question 32 Explanation:

Nonpreemptive Scheduling

A scheduling discipline is nonpreemptive if, once a process has been given the CPU, the CPU cannot be taken away from that process.

Eg: FIFO→ Non preemptive

Preemptive Scheduling

A scheduling discipline is preemptive if, once a process has been given the CPU can taken away.

Eg: Round robin is preemptive

Multilevel queue scheduling is preemptive

Priority,SRTF,Multilevel feedback scheduling is preemptive

A scheduling discipline is nonpreemptive if, once a process has been given the CPU, the CPU cannot be taken away from that process.

Eg: FIFO→ Non preemptive

Preemptive Scheduling

A scheduling discipline is preemptive if, once a process has been given the CPU can taken away.

Eg: Round robin is preemptive

Multilevel queue scheduling is preemptive

Priority,SRTF,Multilevel feedback scheduling is preemptive

Question 33 |

Minimum ___ full adders and half adders are required by the BCD adder to add two decimal digits.

3,2 | |

9,4 | |

6,5 | |

5,2 |

Question 33 Explanation:

→ Each digit is represented by a 4-bit BCD code.

→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).

→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.

→ This can be done with 1 Half adder and 2 Full Adder

(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.

→ Here, Half adder is used to add next significant bits).

Total 5 Full Adders and 2 Half Adders are needed

→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).

→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.

→ This can be done with 1 Half adder and 2 Full Adder

(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.

→ Here, Half adder is used to add next significant bits).

Total 5 Full Adders and 2 Half Adders are needed

Question 34 |

Which of the following code replacements is an example of operator strength reduction?

Replace P^2 by P*P | |

Replace P*16 by P<< 4 | |

Replace pow(P,3) by P*P*P | |

Replace (P <<5) -P by P*3 |

Question 34 Explanation:

Strength Reduction :

It is a compiler optimization where expensive operations are replaced with equivalent but less expensive operations. The classic example of strength reduction converts "strong" multiplications inside a loop into "weaker" additions – something that frequently occurs in array addressing.

Examples:

Replacing a multiplication within a loop with an addition

Replacing an exponentiation within a loop with a multiplication

According to options, Option B is most suitable answer.

It is a compiler optimization where expensive operations are replaced with equivalent but less expensive operations. The classic example of strength reduction converts "strong" multiplications inside a loop into "weaker" additions – something that frequently occurs in array addressing.

Examples:

Replacing a multiplication within a loop with an addition

Replacing an exponentiation within a loop with a multiplication

According to options, Option B is most suitable answer.

Question 35 |

Identify the true statement from the given statements

(1) Number of child pointers in a B/ B+ tree node is always equal to number of keys in it plus one.

(2) B/B+ tree is defined by a term minimum depends on hard disk block size, key and address sizes.

(1) Number of child pointers in a B/ B+ tree node is always equal to number of keys in it plus one.

(2) B/B+ tree is defined by a term minimum depends on hard disk block size, key and address sizes.

(1) | |

(1) and (2) | |

(2) | |

None of these |

Question 35 Explanation:

The above two statements are true.

Question 36 |

The following table has two attributes X and Y where X is the primary key and Y is the foreign key referencing X with on delete cascade

The set of all tuples that must be additionally deleted to preserve referential integrity when the tuple (3,4) is deleted is:

The set of all tuples that must be additionally deleted to preserve referential integrity when the tuple (3,4) is deleted is:

(4,3) and (6,4) | |

(2,4) and (7,2) | |

(3,2) and (9,5) | |

(3,4),(4,3) and (6,4) |

Question 36 Explanation:

→ On removal of row (3,4), row (4,3) must also be deleted as they depend on value 3.

→ On removal of row (4,3), row (2,4) and (6,4) must also be deleted as it depends on value 4.

→ As there is no option with row(2,4) and also the question says additional tuples should be deleted. So, option D is eliminated.

→ On removal of row (4,3), row (2,4) and (6,4) must also be deleted as it depends on value 4.

→ As there is no option with row(2,4) and also the question says additional tuples should be deleted. So, option D is eliminated.

Question 37 |

For the given nodes:

89,19,50,17,12,15,2,5,7,11,6,9,100

Minimum _____ number of interchanges are needed to convert it into a max-heap

89,19,50,17,12,15,2,5,7,11,6,9,100

Minimum _____ number of interchanges are needed to convert it into a max-heap

3 | |

4 | |

5 | |

6 |

Question 37 Explanation:

1st swap is: 100 and 15

2nd swap is: 100 and 50

3rd swap is: 100 and 89

2nd swap is: 100 and 50

3rd swap is: 100 and 89

Question 38 |

_____ IP address can be used in WAN

256.0.0.1 | |

172.16.0.10 | |

15.1.5.6 | |

127.256.0.1 |

Question 38 Explanation:

→ Option A, eliminated because the IP address atmost consists of 255.

→ 172.16.x.x is private address which is usen with in local area network.

→ So,option C is correct.

→ Option D, eliminated because the IP address atmost consists of 255.

→ 172.16.x.x is private address which is usen with in local area network.

→ So,option C is correct.

→ Option D, eliminated because the IP address atmost consists of 255.

Question 39 |

___ pairs of traversals is not sufficient to build tree

Preorder and Inorder | |

Postorder and Inorder | |

Postorder and Preorder | |

None of these |

Question 39 Explanation:

We can't build a tree without the in-order traversal.

Consider two different trees,

TREE-1

root=a;

root→ left=b;

root→ left→ right=c;

TREE-2

root=a;

root→ right=b;

root→ right→ left=c;

Both the trees are different, but have same pre-order and post-order sequence.

pre-order - a b c

post-order - c b a

Because we cannot separate the left subtree and right subtree using the pre-order or post-order traversal alone

Consider two different trees,

TREE-1

root=a;

root→ left=b;

root→ left→ right=c;

TREE-2

root=a;

root→ right=b;

root→ right→ left=c;

Both the trees are different, but have same pre-order and post-order sequence.

pre-order - a b c

post-order - c b a

Because we cannot separate the left subtree and right subtree using the pre-order or post-order traversal alone

Question 40 |

Identify the true statement from the given statements:

(1) A recursive formal language is a recursive subset in the set of all possible words over the alphabet of the language

(2) The complement of a recursive language is recursive

(3) The complement of a context free language is context free

(1) A recursive formal language is a recursive subset in the set of all possible words over the alphabet of the language

(2) The complement of a recursive language is recursive

(3) The complement of a context free language is context free

Only (1) | |

(1) and (2) | |

(1),(2) and (3) | |

(2) and (3) |

Question 40 Explanation:

TRUE: A recursive formal language is a recursive subset in the set of all possible words over the alphabet of the language

TRUE: The complement of a recursive language is recursive

FALSE because The complement of a deterministic context free language is deterministic context free.

TRUE: The complement of a recursive language is recursive

FALSE because The complement of a deterministic context free language is deterministic context free.

Question 41 |

In the disk, swap space is used to _____

Save XML files | |

Save process data | |

Save drivers | |

Save HTML files |

Question 41 Explanation:

→ A swap file (or swap space or a pagefile) is a space on a hard disk used as the virtual memory extension of a computer's real memory (RAM).

→ Having a swap file allows your computer's operating system to pretend that you have more RAM than you actually do.

→ Having a swap file allows your computer's operating system to pretend that you have more RAM than you actually do.

Question 42 |

In a system, counting semaphore was initialized to 10. Then 6P(wait) operations and 4V (signal) operations were completed on this semaphore. So ___ is the final value of the semaphore.

7 | |

8 | |

13 | |

12 |

Question 42 Explanation:

Given data,

Counting semaphore(S)=10,

Wait operations(P)=6,

Signal operations(V)=4,

→ P(wait) operations means decrementing the value and V(signal) operations means incrementing the Semaphore values.

→ After completion of 6P operations and 4V operations the final value of S will be

S=10-6+4

= 8

Counting semaphore(S)=10,

Wait operations(P)=6,

Signal operations(V)=4,

→ P(wait) operations means decrementing the value and V(signal) operations means incrementing the Semaphore values.

→ After completion of 6P operations and 4V operations the final value of S will be

S=10-6+4

= 8

Question 43 |

______ to evaluate an expression an execution without any embedded function calls

Two stacks are required | |

One stack is needed | |

Three stacks are required | |

More than three stacks are required |

Question 43 Explanation:

Applications of stack are

Converting infix expression into post/prefix expression

Evaluating post/pre-fix expression

Parenthesis matching

With one stack also we can easily evaluate the expression.

Converting infix expression into post/prefix expression

Evaluating post/pre-fix expression

Parenthesis matching

With one stack also we can easily evaluate the expression.

Question 44 |

A RAM chip has a capacity of 1024 words of 8 bits each (1K x 8). The number of 2 x 4 decoders with enable line needed to construct a 32K x8 RAM from 1K x8 RAM is:

4 | |

5 | |

6 | |

7 | |

11 |

Question 44 Explanation:

Thirty two 1Kx8 RAMs are required to construct 32Kx8 RAM.

To enable one of the thirty two 1Kx8 RAMs, eight 2x4 decoders are needed.

To enable one of the eight decoders, two 2x4 decoders are needed.

To enable one of the two decoders, one 2x4 decoder is needed.

So, total 11 decoders of size 2x4 are needed.

To enable one of the thirty two 1Kx8 RAMs, eight 2x4 decoders are needed.

To enable one of the eight decoders, two 2x4 decoders are needed.

To enable one of the two decoders, one 2x4 decoder is needed.

So, total 11 decoders of size 2x4 are needed.

Question 45 |

____ merges the bodies of two loops.

Loop rolling | |

Loop folding | |

Loop merge | |

Loop jamming |

Question 45 Explanation:

→ Loop fusion (or loop jamming) is a compiler optimization and loop transformation which replaces multiple loops with a single one.

→ Loop fission (or loop distribution) is a compiler optimization in which a loop is broken into multiple loops over the same index range with each taking only a part of the original loop's body.

→ Loop fission (or loop distribution) is a compiler optimization in which a loop is broken into multiple loops over the same index range with each taking only a part of the original loop's body.

Question 46 |

A RAM chip has a capacity of 1024 words of 8 bits each (1K x 8). The number of 2 x 4 decoders with enable line needed to construct a 32K x8 RAM from 1K x8 RAM is:

4 | |

5 | |

6 | |

7 |

Question 46 Explanation:

→ The capacity of the RAM needed = 16K
→ Capacity of the chips available = 1K

→ No. of address lines = 16K/1K = 16

→ Hence we can use 4×16 decoder for this. But we were only given 2×4 decoders.

→ So, 4 decoders are required in inner level as from one 2×4 decoder we have only 4 output lines whereas we need 16 output lines.

→ To point to these 4 decoders, another 2×4 decoder is required in the outer level.

→ Hence no. of 2×4 decoders to realize the above implementation of RAM = 1+4 = 5

→ No. of address lines = 16K/1K = 16

→ Hence we can use 4×16 decoder for this. But we were only given 2×4 decoders.

→ So, 4 decoders are required in inner level as from one 2×4 decoder we have only 4 output lines whereas we need 16 output lines.

→ To point to these 4 decoders, another 2×4 decoder is required in the outer level.

→ Hence no. of 2×4 decoders to realize the above implementation of RAM = 1+4 = 5

Question 47 |

____ merges the bodies of two loops.

Loop rolling | |

Loop folding | |

Loop merge | |

Loop jamming |

Question 47 Explanation:

→ Loop fusion (or loop jamming) is a compiler optimization and loop transformation which replaces multiple loops with a single one.

→ Loop fission (or loop distribution) is a compiler optimization in which a loop is broken into multiple loops over the same index range with each taking only a part of the original loop's body.

→ Loop fission (or loop distribution) is a compiler optimization in which a loop is broken into multiple loops over the same index range with each taking only a part of the original loop's body.

Question 48 |

Given message M=1010001101. The CRC for this given message using the divisor polynomial x

^{5}+ x^{4}+ x^{2}+1 is _____01011 | |

10101 | |

01110 | |

10110 |

Question 48 Explanation:

Degree of generator polynomial is 5. Hence, 5 zeroes is appended before division.

M = 1010001101

Divisor polynomial: 1.x

Divisor polynomial bit= 110101

append 5 zeroes = M = 101000110100000

∴ CRC = 01110

M = 1010001101

Divisor polynomial: 1.x

^{5}+1.x^{4}+0.x^{3}+1.x^{2}+0.x^{2}+1.x^{0}Divisor polynomial bit= 110101

append 5 zeroes = M = 101000110100000

∴ CRC = 01110

Question 49 |

_____sorting algorithms has the lowest worst case complexity

Selection sort | |

Bubble sort | |

Merge sort | |

Quick sort |

Question 49 Explanation:

The Worst case time complexity of Selection,Bubble and Quick sort is O(n

^{2}) where as Merge sort is O(nlog(n)). So Merger sort has lowest worst case time complexityQuestion 50 |

_____ is NOT a part of the ACID properties.

Inconsistency | |

Consistency | |

Atomicity | |

Isolation |

Question 50 Explanation:

ACID means Atomicity, Consistency ,Isolation and Durability

Atomicity: Transactions are often composed of multiple statements. Atomicity guarantees that each transaction is treated as a single "unit", which either succeeds completely, or fails completely: if any of the statements constituting a transaction fails to complete, the entire transaction fails and the database is left unchanged. An atomic system must guarantee atomicity in each and every situation, including power failures, errors and crashes.

Consistency : Consistency ensures that a transaction can only bring the database from one valid state to another, maintaining database invariants: any data written to the database must be valid according to all defined rules, including constraints, cascades, triggers, and any combination thereof. This prevents database corruption by an illegal transaction, but does not guarantee that a transaction is correct.

Isolation : Transactions are often executed concurrently (e.g., reading and writing to multiple tables at the same time). Isolation ensures that concurrent execution of transactions leaves the database in the same state that would have been obtained if the transactions were executed sequentially. Isolation is the main goal of concurrency control; depending on the method used, the effects of an incomplete transaction might not even be visible to other transactions.

Durability :Durability guarantees that once a transaction has been committed, it will remain committed even in the case of a system failure (e.g., power outage or crash). This usually means that completed transactions (or their effects) are recorded in non-volatile memory.

Atomicity: Transactions are often composed of multiple statements. Atomicity guarantees that each transaction is treated as a single "unit", which either succeeds completely, or fails completely: if any of the statements constituting a transaction fails to complete, the entire transaction fails and the database is left unchanged. An atomic system must guarantee atomicity in each and every situation, including power failures, errors and crashes.

Consistency : Consistency ensures that a transaction can only bring the database from one valid state to another, maintaining database invariants: any data written to the database must be valid according to all defined rules, including constraints, cascades, triggers, and any combination thereof. This prevents database corruption by an illegal transaction, but does not guarantee that a transaction is correct.

Isolation : Transactions are often executed concurrently (e.g., reading and writing to multiple tables at the same time). Isolation ensures that concurrent execution of transactions leaves the database in the same state that would have been obtained if the transactions were executed sequentially. Isolation is the main goal of concurrency control; depending on the method used, the effects of an incomplete transaction might not even be visible to other transactions.

Durability :Durability guarantees that once a transaction has been committed, it will remain committed even in the case of a system failure (e.g., power outage or crash). This usually means that completed transactions (or their effects) are recorded in non-volatile memory.

Question 51 |

____ number of queues are needed to implement a stack

1 | |

2 | |

3 | |

4 |

Question 51 Explanation:

→ Implementing stack by using two queues:

push(s,x)

1) Enqueue x to q1 (assuming size of q1 is unlimited). pop(s)

1) One by one dequeue everything except the last element from q1 and enqueue to q2.

2) Dequeue the last item of q1, the dequeued item is result, store it.

3) Swap the names of q1 and q2

4) Return the item stored in step 2.

→ Implementing queue by using two stack:

(1) When calling the enqueue method, simply push the elements into the stack 1.

(2) If the dequeue method is called, push all the elements from stack 1 into stack 2, which reverses the order of the elements. Now pop from stack 2.

push(s,x)

1) Enqueue x to q1 (assuming size of q1 is unlimited). pop(s)

1) One by one dequeue everything except the last element from q1 and enqueue to q2.

2) Dequeue the last item of q1, the dequeued item is result, store it.

3) Swap the names of q1 and q2

4) Return the item stored in step 2.

→ Implementing queue by using two stack:

(1) When calling the enqueue method, simply push the elements into the stack 1.

(2) If the dequeue method is called, push all the elements from stack 1 into stack 2, which reverses the order of the elements. Now pop from stack 2.

Question 52 |

Baud rate measures the number ____ transmitted per second.

Symbols | |

Bits
| |

Byte | |

None of these |

Question 52 Explanation:

Bit rate is nothing but number of bits transmitted per second

Baud rate is nothing but number of signals units transmitted per unit time.

Baud, or baud rate, is used to describe the maximum oscillation rate of an electronic signal. For example, if a signal changes (or could change) 1200 times in one second, it would be measured at 1200 baud.

Baud unit symbol "Bd is synonymous to symbols per second or pulses per second. It is the unit of symbol rate, also known as baud or modulation rate; the number of distinct symbol changes (signalling events) made to the transmission medium per second in a digitally modulated signal or a line code.

Baud rate is nothing but number of signals units transmitted per unit time.

Baud, or baud rate, is used to describe the maximum oscillation rate of an electronic signal. For example, if a signal changes (or could change) 1200 times in one second, it would be measured at 1200 baud.

Baud unit symbol "Bd is synonymous to symbols per second or pulses per second. It is the unit of symbol rate, also known as baud or modulation rate; the number of distinct symbol changes (signalling events) made to the transmission medium per second in a digitally modulated signal or a line code.

Question 53 |

Decreasing the RAM causes ______

Fewer page faults | |

More page faults | |

Virtual memory get increases | |

Virtual memory get decreases |

Question 53 Explanation:

A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process.

When handling a page fault, the operating system generally tries to make the required page accessible at the location in physical memory, or terminates the program in case of an illegal memory access.

So, Decrease the physical RAM on your machine could result in more page faults

When handling a page fault, the operating system generally tries to make the required page accessible at the location in physical memory, or terminates the program in case of an illegal memory access.

So, Decrease the physical RAM on your machine could result in more page faults

Question 54 |

In the given language L={ab,aa,baa}, __ number of strings are in L*

(1) baaaba

(2) aabaaaa

(3) baaabaaaabaa

(4) baaabaaa

(1) baaaba

(2) aabaaaa

(3) baaabaaaabaa

(4) baaabaaa

(1) | |

(2) | |

(3) | |

(4) |

Question 54 Explanation:

Any combination of strings in set {ab, aa, baa} will be in L*.

L = { ab, aa, baa }

Let S1 = ab , S2 = aa and S3 =baa

Option-1: baa ab a , We can’t partition with combinations of S1,S2 and S3.

Option-2: aa baa aa, We can’t partition with combinations of S1,S2 and S3.

Option-3: baa ab aa aa baa, We can partition this into S3S1S2S2S3

Option-4: baa ab aa a, We can’t partition with combinations of S1,S2 and S3.

We can’t partition the above options with combinations of {ab, aa, baa} except Option-3.

So the L* Consists of the string “baaabaaaabaa:

L = { ab, aa, baa }

Let S1 = ab , S2 = aa and S3 =baa

Option-1: baa ab a , We can’t partition with combinations of S1,S2 and S3.

Option-2: aa baa aa, We can’t partition with combinations of S1,S2 and S3.

Option-3: baa ab aa aa baa, We can partition this into S3S1S2S2S3

Option-4: baa ab aa a, We can’t partition with combinations of S1,S2 and S3.

We can’t partition the above options with combinations of {ab, aa, baa} except Option-3.

So the L* Consists of the string “baaabaaaabaa:

Question 55 |

Consider the following functional dependencies in a database:

A → B

B → C

D → E

E → D

F → G

F → H

(E,F) → I

The relation (E,D,A,B) is :

A → B

B → C

D → E

E → D

F → G

F → H

(E,F) → I

The relation (E,D,A,B) is :

2 NF | |

3 NF | |

BCNF | |

None of the above |

Question 55 Explanation:

Functional dependencies of the given subrelation R(E,D,A,B) are:

A → B

D → E

E → D

Since A is not determined by any of the given functional dependencies for R(E,D,A,B) so it must be included in the key now check whether “A” is the Primary key or not.

(A)+ = {A}

(AB)+ = {AB}

(AD)+ = {ABDE}

(AE)+ = {ABDE}

So (AD)+ and (AE)+ are the primary keys of R(E,D,A,B).

A → B, in this given FD there exist a partial dependency because here a prime key attribute(attribute which is a part of primary key but not a primary key itself of a function) which is determining a non-key attribute.

So the relation R(E,D,A,B) is not in 2NF .

A → B

D → E

E → D

Since A is not determined by any of the given functional dependencies for R(E,D,A,B) so it must be included in the key now check whether “A” is the Primary key or not.

(A)+ = {A}

(AB)+ = {AB}

(AD)+ = {ABDE}

(AE)+ = {ABDE}

So (AD)+ and (AE)+ are the primary keys of R(E,D,A,B).

A → B, in this given FD there exist a partial dependency because here a prime key attribute(attribute which is a part of primary key but not a primary key itself of a function) which is determining a non-key attribute.

So the relation R(E,D,A,B) is not in 2NF .

Question 56 |

A Professor passed one sixth of his life in childhood, one twelfth in youth, and one seventh more as a bachelor; five years after his marriage a son was born who died four years before his father at half his final age then what is the age of professor?

84 years | |

74 years | |

64 years | |

54 years |

Question 57 |

Which word does NOT belong with the others?

wing | |

fin | |

beak | |

rudder |

Question 58 |

Look at this series : 5000, 1001, 201, 41,… What number should come next?

9 | |

10 | |

11 | |

42 |

Question 59 |

Candid is to indirect as honest is to :

frank | |

wicked | |

truthful | |

untruthful |

Question 60 |

Let us consider the length of the side of a square represented by 2y + 3. The length of the side of an equilateral triangle is 4y. If the square and the equilateral triangle have equal perimeter, then what is the value of y?

3 | |

4 | |

6 | |

8 |

Question 61 |

If finch related to bird i.e. FINCH: BIRD. Then, find the pair that has a similar relationship.

rog : toad | |

elephant : reptile | |

Dalmatian: dog | |

collie : marsupial |

Question 62 |

Look at this series : 25, 25, 37, 37, __, 51, … What number should fill the blank?

51 | |

39 | |

23 | |

25 |

Question 63 |

Suppose a fraud shopkeeper sells rice to the customer at the cost price, but he uses a false weight of 900 gm for a kg then his percentage gain is _______.

5.75 % | |

5.56 % | |

5.20 % | |

5.00 % | |

None of the above |

Question 64 |

Divide 88 into four parts such as first part known as a, second part b, third part c, and fourth part is d, when 5 is added to the first part, 5 is subtracted from the second part, 3 is multiplied by the third part and the fourth part is divided by 5, then all results are equal. Find the value of a, b, c and d respectively.

7, 17, 4, 60 | |

8, 25, 5, 50 | |

10, 30, 3, 45 | |

17, 7, 4, 60 |

Question 65 |

A software engineer has the capability of thinking 200 lines of code in five minutes and can type 200 lines of code in 10 minutes. He takes a break for five minutes after every ten minutes. How many lines of code will he complete typing after an hour?

200 | |

300 | |

400 | |

500 |

Question 66 |

A construction company ready to finish a construction work in 180 days, hired 80 workers each working 8 hours daily. After 90 days, only 2/7 of the work was completed. How many workers are to be increased to complete the work on time? Note : If additional acquired workers do agree to work for 10 hours daily.

90 workers | |

80 workers | |

65 workers | |

85 workers |

Question 67 |

Look at this sequence SCD TEF UGH _____ WKL and find the missing sequence.

CMN | |

UJI | |

VIJ | |

IJT |

Question 68 |

Here are some words translated from any Language.

Which word could mean “

**diano**means**oak tree****blynot**means**oak leaf****blycrin**means**maple leaf**Which word could mean “

**maple syrup**”?blymuth | |

hupponot | |

patricrin | |

crinweel |

There are 68 questions to complete.