Method-1:
Given data,
-- Set A={1, 2, 3, ........, 1000}
-- Set A shall be divisible by 3=?
-- Set A shall be divisible by 5=?
-- Set A shall be divisible by 3 and 5=?
Step-1: To find divisible by 3 numbers are
=⌊1000/3⌋
= 333
Step-2: To find divisible by 5 numbers are
=⌊1000/5⌋
= 200
Step-3: To find divisible by 3 and 5 numbers are
=⌊1000/(3*5)⌋
= 66
These 66 is already part of 333 and 200. So, we have to exclude it from the list.
Total= 333+200-66
= 467
Note: We are using floor because excluding fraction value.
Method-2:
The above problem is in the form of (AUB) = (A)+(B)-(A∩B)
A=1000/3
B=1000/5
(A∩B)=66
(AUB)=467
Note: Getting this idea in exam hall is very difficult. So, better follow method-1 ratherhan method-2

Method-1:
Here, they are clearly mentioned that “certain tree”
The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.
→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.
→ Two vertices of degree 4, we can write into (2*4)
→ One vertex of degree 3, we can write into (1*3)
→ One vertex of degree 2, we can write into (1*2).
→ Other vertices(X) have degree 1, we can write into (X*1)
Step-1: (2*4)+(1*3)+(1*2)+(X*1)
=2(X+4-1) [Note: ‘X’ value is not given ]
X=7
Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.
=X+4
=7+4
=11
Method-2:
Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1
D degree = 2
A degree = 4
K degree = 4
G degree = 3

Option-A: It is not complete graph because it won’t have n(n-1)/2 edges.
Option-B: It is not Bipartite Graph because it takes more than 2 colours. Bipartite Graph have exactly 2 colours.
Option-C: Hamiltonian path is a path in an undirected or directed graph that visits each vertex exactly once.

Hence, Option-C is correct answer.

Probability that the value k=1 appears in the printout at least once is
1-p(x=0)
We will apply the formula of binomial distribution
1-(^{​ 1000000​} C​ ₀​ (1/1000000)​ ⁰​ * (999999/1000000)​ ^{10^6}​ )
=1-(1*1*0.367879)
=0.632121
Hence answer is option C

Given f(x) = x​ ⁴​ , g(x) = √ x ² + 1 , h(x) = x​ 2​ + 72
for, ho(gof)
gof=g(f(x))
=g(x​ ⁴​ )
=√(x​ ⁸​ +1)
ho(gof)=h(gof)
=h(√(x​ ⁸​ +1))
=(√(x​ ⁸​ +1)​ 2​ +72
=x​ ⁸​ +1+72
=x​ ⁸​ +73
√ x ² + 1 , h(x) = x​ ²​ + 72
for, (hog)of,
hog=h(g(x))
=h(√(x​ ²​ +1)
=(√(x​ ²​ +1)​ 2​ +72
=x​ ²​ + 1+72
=x​ ²​ +73
(hog)of=(hog)(f(x))
=(hog)(x​ ⁴​ )
=(x​ ⁴​ )​ ²​ +73
=x​ ⁸​ +73
Hence, option-D is the correct answer

→ 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

The Excess-3 decimal code is a self-complementing code because complement can be generated by inverting each bit pattern.

Given infix notation, we have to change infix notation into postfix notation.
after converting postfix notation the notations are ab* cde /* +

Total 5 PUSH and 4 POP operations performed.

The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is​ ​ 2 ^–129 to (1 – 2​ ^–23​ ) * 2^{​ 127}

The size of the ROM required to build an 8-bit adder/subtractor with mode control, carry input, carry output and two’s complement overflow output is given as 2​ ¹⁸​ * 10.

In this program we are using post increment. Post increment will send the value before updating the value.
printf(“\n%d”, 1 + x++); /* x=128 */
Here, 1+128=129

char *(*(*a[N]) ( )) ( ); /* an array of n pointers to function returning pointers to functions returning pointers to characters */

A friend function of a class is defined outside that class scope but it has the right to access all private and protected members of the class. Even though the prototypes for friend functions appear in the class definition, friends are not member functions.
1. A function can only be declared a friend by a class itself.
2. It can have access to all members of the class, even private ones.

The function can change values in the original array, when an array is passed as parameter to a function because we have to pass it by reference.

→ The DIVISION operation is defined for convenience for dealing with queries that involve universal quantification or the all condition. For a tuple 't' to appear in the result T of the DIVISION, the values in ‘t’ must appear in R in combination with every tuple in S.
→ Note that in the formulation of the DIVISION operation, the tuples in the denominator relation S restrict the numerator relation R by selecting those tuples in the result that match all values present in the denominator. The DIVISION operation can be expressed as a sequence of π, ×, and – operations as follows:
T1 ← πY(R)
T2 ← πY((S × T1) – R)
T ← T1 – T2

NATURAL JOIN requires that the two join attributes (or each pair of join attributes) have the same name in both relations. If this is not the case, a renaming operation is applied first.
OUTER JOIN: A set of operations, called outer joins, were developed for the case where the user wants to keep all the tuples in R, or all those in S, or all those in both relations in the result of the JOIN, regardless of whether or not they have matching tuples in the other relation.
In SQL, the same name can be used for two (or more) attributes as long as the attributes are in different relations. If this is the case, and a multi table query refers to two (or more) attributes with the same name, we must qualify the attribute name with the relation name to prevent ambiguity. The ambiguity of attribute names also arises in the case of queries that refer to the same relation twice. In this case, we are required to declare alternative relation names, called aliases or tuple variables.
An alias can follow the keyword "AS" , as shown below
SELECT E.Fname, E.Lname, S.Fname, S.Lname
FROM EMPLOYEE AS E, EMPLOYEE AS S
WHERE E.Super_ssn=S.Ssn;
In above query relation names E and S, called aliases or tuple variables, for the EMPLOYEE relation.
From above explanation it is clear that only option (D) is correct.

→ Entity types that do not have key attributes of their own are called weak entity types. In contrast,regular entity types that do have a key attribute are called strong entity types. Entities belonging to a weak entity type are identified by being related to specific entities from another entity type in combination with one of their attribute values. We call this other entity type the identifying or owner entity type, and we call the relationship type that relates a weak entity type to its owner the identifying relationship of the weak entity type.
→ A weak entity type always has a total participation constraint (existence dependency) with respect to its identifying relationship because a weak entity can not be identified without an owner entity.

AB → C
AB → D
C → A
D → B
A​ +​ ={A}
B​ +​ ={B}
C​ +​ ={C,A}
D​ +​ ={D,B}
Since single attributes can't determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
BD​ +​ ={B,D}
CD​ +​ ={C,D,B,A}
So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional dependency should be a super key) so the given relation is in 3NF only.

Given functional dependencies are F={A → B, B → C, AC → D}
Step-1: Attribute A and E are not present in the right hand side of any production. So, every key should include AE to be a candidate key.
Step-2: To check whether AE could be a candidate key or not.
(AE)​ +​ = {A,B,C,D,E}
Since AE can uniquely identify each attribute and we can say this is the candidate key for given relation.

→ According to priority (or) infix expression we can write expression like this ((A + B^D) / (E – F)) + G
→ Actual tree structure is

Here, they given sorted list followed by a few ‘random’ elements.
So, we can eliminate option-B and Option-C because it gives worst complexity(O(n​ 2​ )) when array is already sorted.
Bubble sort and insertion sort will give best case complexity(O(n)) but here they given small constraint is “there are few random elements”. So, insertion sort is more appropriate answer.

The directory can be viewed as symbol table that translates filenames into their directory entries.
Symbol table can be organized in many ways using some operations:
1.Search for a file or multiple files
2.Create a file
3.Delete a file
4.List a directory
5.Rename a file
6.Traverse the file system

Here, Row major order generally follows offset formula to find address of the location.
Offset= (Row number * Total number of columns) + Column Number
Step-1: Add the Base address to calculate the effective address. Given matrix is A[20][10], Size of memory word is 4 bytes, address of the location to find A[11][5], Base address is 100.
Step-2: Offset = (11*10) + 5
= 115 (multiply with 4 words)
= 115*4
= 460
Step 3: Initial base address is 100
=Offset+base address
= 460 +100
= 560.

A Binary Tree is full if every node has 0 or 2 children. So, in such case, the binary tree with n leaves contains a total of 2*n-1 nodes.

There are 4 leaf nodes in the above tree.
So, total number of nodes are 2*4-1
=8-1
=7

First we change 10 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10​ ⁻³​ kHz).
10 ms=10*10​ ^-3​ s
=10​ ^-2​ s
Frequency(f)= 1/T
= 1/10​ ^-2 ​ Hz
=100 Hz

The class-A address starts with 0 to 127 but we never used 0 and 127 numbers. The numbers in binary format is 00000001 to 01111110. So, no option is starts with 0.
Class-A is 0
Class-B is 10
Class-C is 110
Class-D is 1110
Class-E is 1111

Flooding:​ Sending packets to all neighbors in the network.
Multi Destination routing:​ It is also sending packets to multi destinations in the network.
Reverse path forwarding (RPF):​ Only sends packet forward if it is received from the next node in the shortest path back to the sender.

Given data,
-- Baud rate = 2000 baud
-- Bit rate = 6000 bps
-- data elements are carried by each signal element=?
Step-1: To find data elements are carried by each signal element using bits per baud.
Bits per baud = bit rate / baud rate
= 6000 / 2000
= 3 bits/baud

Total 19 distinct stages are there in DES algorithm, which is parameterized by a 56-bit key(and +8 parity bits). The block size is 64 bits and uses 16 rounds.

Shift-Reduce parsers perform shift step that advances in the input stream by one symbol and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.

TRUE: Canonical LR parser is LR (1) parser with single look ahead terminal
TRUE: All LR(K) parsers with K > 1 can be transformed into LR(1) parsers.

→ In a two-pass assembler, symbol table is generated in first pass.
→ The first pass of the assembler reads and processes the assembly program one line at a time. In processing a single line of the assembly program the assembler can make addition(s) to the symbol table, add a (possibly partial) SML instruction to the Simpletron's memory.
→ The purpose of the second pass is to complete the partial instructions written in the first pass.

Debugger is a program that allows to set breakpoints, execute a segment of program and display contents of register.

Option A is false: as it generate string "abab" which has even number of a's and b's.
Option B is false: as it generate string "ab" which has odd number of a's and b's.
Option D is false: as it always generate one "a" along with one "b". So it will always generate equal number of a's and b's and will never generate different number of a's and b's.
Hence option C is correct.

Two-level CPU scheduling are using for giving situations:
1. When memory is too small to hold all the ready processes.
2. It facilitates putting some set of processes into memory and a choice is made from that

Given data,
-- Program text=200 K
-- Initial stack=15 K
-- Initialized data=50 K
-- Bootstrap code=70 K
-- Physical memory needed=?
Step-1: Here, given constraint that, all five editors are started simultaneously.
So, all editors to perform all above operations. It need physical memory is
= Program text + Initial stack + Initialized data + Bootstrap code
= 200 K + 15 K + 50 K + 70 K
= 335 K

As per the given options, Option-C is the most appropriate one because all situations is somehow related to critical section problem. But Option-C is not relevant.

Since the average number of pages required per process will be x/y and the amount of space required by the page table will be (x/y)*z. The amount of space lost due to internal fragmentation is y/2. So total space wastage is
Loss(L)=(x/y)*e + y/2
To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero. (dL/dy) =0
y=√(2xz)

Given data,
-- Time to serve empty page fault = 8 m.sec
-- Time to serve modified page fault= 20 m.sec
-- Percentage of page modification= 70%
= 70/100
= 0.7
-- Page to be replaced is not modified= 1-0.7
= 0.3
-- Average access time to service a page fault=?
Step-1: Average access time = (Not modified*Time to serve page fault) + (Modified time * percentage of page modification)
= (0.3*8) + (0.7*20)
= 2.4 + 14
= 16.4 m.sec

→ The umbrella activities occur throughout the software process they are applied evenly across the process, the analysis encompasses a set of work tasks (eg. requirement gathering, elaboration, negotiation specification and validation).
→ These activities include Software project tracking and control, Risk management, Software quality assurance, and formal technical reviews, measurement, Software configuration management, reusability management and work product preparation and production.

Managed maturity level focuses on software metrics. Two metrics are
1. Product Metric: It measures the characteristics of product being developed. Product metrics are size, reliability, understandability etc..,
2. Process Metrics: It reflect the effectiveness of process. Process metrics are average defect correction on time, number of failures detected during testing per LOC etc.,
This level focuses on improving product and process quality. The Key process areas are quantitative process metric and software quality management.

The software Architecture of a program or a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.

Programming language dependent attributes should not be encompassed by effective software metrics.

Cyclomatic complexity uses 3 formulas
1. Number of regions + 1
2. Predicate + 1
3. Edges-Vertices+2

→ ​ Hard handover (or) Hard Handoff:​ Early systems used hard handoff. In a hard handoff, a mobile station only communicates with one base station. When the MS moves from one cell to another, communication must first be broken with the previous base station before communication can be established with the new one.
→ ​ Soft handover (or) Soft Handoff:​ New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.

→ Factless fact table in a data warehouse contains only surrogate keys.
→ The factless fact table is a fact table that does not contain any facts. There are two kinds of factless fact tables:
1. Factless fact table describes event or activity.
2. Factless fact table describes a condition, eligibility or coverage.

Consumer-to-business (C2B)​ is a business model in which consumers (individuals) create value and businesses consume that value.
Business-to-government (B2G)​ (or) "public sector marketing": Which encompasses marketing products and services to various government levels through integrated marketing communications techniques such as strategic public relations, branding, marketing communications (marcom), advertising, and web-based communications.
Customer to customer (C2C)​ markets provide an innovative way to allow customers to interact with each other.
Business-to-business (B2B)​ is a situation where one business makes a commercial transaction with another.

→ Demand-Sensitive Pricing model is designed to bring prices down by increasing the number of customers who buy a particular product at once.
→ Demand sensitivity is referred to the change in demand for a product when its price is changed by a small amount.
Price Elasticity of demand = ((%change in quantity demanded) / (% change in price of a particular product))

Call control protocol→ Connection management
A-bis → Interface between Base Transceiver Station(BTS) and Base Station Controller(BSC)
BSMAP → Works between Mobile Switching Centre(MSC) and Base Station Subsystem (BSS)
CDMA → Spread spectrum