Explanation: cloud,river and mountain are three different things

Atmosphere consists of various gases. Oxygen and nitrogen are some of the gases present in the atmosphere.

II number is 3*1+2 =5 III number is 5*2+2=12 IV number is 12*3+2=38 V number is 38*4+2=154 VI number is 154*5+2-772 So wrong number is 39

The wrong number is 348. The correct sequence is 701,349,173,85,41,19,8 second number is 701- 352 = 349 third number is 349 - 176 = 173 fourth number is 173 - 88 = 85 fifth number is 85 - 44 = 41 sixth number is 41 - 22 = 19 and the seventh number is 19 - 11 = 8. to get the next number in the sequence we are subtracting the sequence is 352,176,88,44,22,11. (which is every time we are dividing the number with 2)

the correct sequence is 1,9,25,49,81,121= 12, 32 , 52, 72, 92, 112 ( which is squares of odd numbers)

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P and S are unmarried ladies and do not participate in any game, so S is not a hockey player. Q is the brother of R and is neither a chess player nor a hockey player. so Q is not a hockey player. There is a married couple in which T is husband so R is wife. None of the ladies plays chess and football so R is a hockey player.

P and S are unmarried ladies and R is the wife of T. So the correct group of ladies is P, R, and S.

R is a hockey player, S wont play any game, T is chess player, so Q is football player.

N is good at Volleyball, Cricket whereas O is only good at baseball according to the mentioned games in the question.

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we know that, ∟DAB+ ∟ ABC=180 dividing by 2, ½ ∟ DAB+½ ∟ ABC= 180/2 ∟ PAB + ∟PBA=90. ………(1 say) hence, in triangle APB, ∟PAB +∟PBA + ∟APB=180. (Angle sum prop) from 1, angAPB+90= 180 ang APB=90

Volume of the cuboid : L*B*H = 24*9*8 = 1728 cm^3 Volume of the cube = (side)^3 = 3^3 = 27cm^3 Number of cubes possible = 1728 cm^3 / 27cm^3 = 64

Let, Length = l say and breadth = b say. According to given question we can write, (l+b) - l2 + b2 = l/2 (l+b) -l/2 = l2 + b2 l + 2b = 2 l2 + b2 => l2+ 4b2 + 41lb = 4(l2 + b2) b/l = 3/4

Let fraction is x/y numerator increased by 25% then x will become 1.25x denominator diminished by 10% then y will become 0.9x From the given question we can write 1.25x0.9y= 59 xy= 25

Let us assume the common ratio of workers x and common ratio of wages is y. So, 5x=20, or x= 4 So, skilled worker=8x4=32, unskilled worker= 20, clerks=4 Hence total wages=32x5y+20x2y+4x3y=212y So, 212 y=318 or, y=1.5 So, daily wages of skilled workers = 5*1.5 = 7.5 Daily wages of unskilled worker=2*1.5 = 3 And daily wages of clerks =3*1.5= 4.5 Therefore, Wages paid to skilled workers = 7.5 * 32 = 240 wages paid to unskilled worker = 3*20 = 60 wages paid to clerks = 4.5*4 = 18

Given that, ½ log10 25 – 2 log10 3 + log10 18 = ½ log10 5^2 – 2 log10 3 + log10 (9*2) = log10 5 – log10 9 + log10 9 + log10 2 = log10 5 + log10 2 = log10 10 = 1

In a mixture of 60L, the ratio of milk and water is 2: 1, then milk is 40L and water is 20L. if the ratio becomes 1:2, we need to add 60L of water then water becomes 80L.

Explanation: (-1,1) satisfies both the given equations.

We are given that four students got equal marks. On one bit of paper, one arrangement of rank is to be written. Let the students be named as A,B,C and D. Now, A can be treated as having rank I in 4 ways. B can be treated as having rank II in 3 ways. C can be treated as having rank III in 2 ways. D can be treated as having rank IV in 1 way. Therefore, total number of bits of paper required for all arrangement =4×3×2×1=24

Volume of the sphere is 43r3 = 43(302)3= 4500 Then, the volume of the water displaced = 4500 cm3 Let, h be the increased level, then (602)2h = 4500 h = 5cm.

In 2015, percentage = (2513/7894) *100 = 31.83 In 2016, percentage = (2933/8562) *100 = 34.25 In 2017, percentage = (2468/8139) *100 = 30.32 In 2018, percentage = (3528/9432) *100 = 37.40 In 2017, the percentage was least in semi-urban area.

Percentage drop = (8562-8139)/8562 = 423/8562 = 4.9% = 5%

Talk shop means discussing one's work even when not working.

Principle: a fundamental truth or proposition that serves as the foundation for a system of belief or behavior or for a chain of reasoning. Principal: Head of the school.

DELETERIOUS: Harmful often in a subtle or unexpected way.

If all transactions obey the Two phase Locking protocol, then all possible interleaved schedules (non – serial schedules) are serializable.

Monitor is the collection of condition variables and procedures combined together in a special kind of module or a package. A protected variable which can be accessed and changed by a particular set of operations.

A wireless access point (wireless AP) is a network device that transmits and receives data over a wireless local area network (WLAN). Access point in a wireless network is any device that will allow the wireless devices to a wired network. A router is the best example of an Access Point.

The channel capacity of a noise free channel having M symbols is given by log M base 2.

Type 0 grammar generates recursively enumerable languages which are recognized by Turing Machine.

A bottom up parser generates rightmost derivation in reverse.

Let m is the number of parity bits and n is the number of message bits. So, m=5 and n=26. Then, 2^m ≥ n+m+1 should be satisfied. 25 ≥ 26+5+1 32 ≥ 32

Merge sort → Worst case time complexity is O(nlogn) Bubble, Quick and insertion sort → Worst case time complexity is O(n^2)

WPA → Wi-fi protected access

In Software Modeling ‘IS A’ represents Inheritance relationship.

Every production |LHS| <= |RHS| and LHS contains more than one symbol in second and fourth production, thus it is a Type 1 grammar.

Delivery of the software product on time is considered as Planning

(a + b)²=(a+b)(a+b) which is concatenation of (a+b) with (a+b) (a+b)(a+b)= {aa, ab, ba, bb}

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Both presentation and session layers are not presented in the TCP/IP model.

A variable is live at some point if it holds a value that may be needed in the future, or equivalently if its value may be read before the next time the variable is written to. Thus, Value of p at m would be used along some path in the flow graph starting at point m then variable p is said to be live at point m.

Each layer of the OSI model receives services or data from a above and below layer

If a developer wants to transform a model into source code is also known as Reverse Engineering

If the total number of edges in all adjacency lists is E, then there will be a total of E number of iterations, hence there will be a total of at most E decrease key operations.

One root of x<3>TEXT – x – 4 – 4 = 0 lies on (1, 2). In the bisection method, after the first iteration the root lies in the interval (1.5, 2)

A wireless network interface controller (WNIC) is a network interface controller which connects to a wireless network, such as Wi-Fi or Bluetooth, rather than a wired network, such as a Token Ring or Ethernet. An 802.11 WNIC can operate in two modes known as infrastructure mode and ad hoc mode Infrastructure mode: In an infrastructure mode network the WNIC needs a wireless access point: all data is transferred using the access point as the central hub. All wireless nodes in an infrastructure mode network connect to an access point. All nodes connecting to the access point must have the same service set identifier (SSID) as the access point, and if a kind of wireless security is enabled on the access point (such as WEP or WPA), they must share the same keys or other authentication parameters. Ad hoc mode: In an ad hoc mode network the WNIC does not require an access point, but rather can interface with all other wireless nodes directly. All the nodes in an ad hoc network must have the same channel and SSID.

Dijkstra’s Algorithm cannot be applied on Graphs having negative weights and negative weight cycle.

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Bubble sort worst case time complexity is O(n^2) Worst case comparisons are (n(n-1))/2 For 100 elements = (100(100-1))/2 = 19,800 comparisons. 19,800*c= 100 sec (n(n-1))/2 = 400 sec (n(n-1))/2 = 79,200 sec n=200

The Data flow diagram enables the software engineer to develop models of the information domain and functional domain at the same time.

Cardinality and Modality are the indicators of the business rules around a relationship. Cardinality refers to the maximum number of times an instance in one entity can be associated with instances in the related entity. Modality refers to the minimum number of times an instance in one entity can be associated with an instance in the related entity. Cardinality can be 1 or Many and the symbol is placed on the outside ends of the relationship line, closest to the entity, Modality can be 1 or 0 and the symbol is placed on the inside, next to the cardinality symbol

An Eulerian trail is a trail in a finite graph that visits every edge exactly once. An Eulerian circuit or Eulerian cycle is an Eulerian trail that starts and ends on the same vertex.

AB’+AB => A(B’+B) =A (B’+B is 1)

By examining the possibilities, we find 1 graph with 0 edges, 1 graph with 1 edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. Altogether, we have 11 non-isomorphic graphs on 4 vertice

NP is a set of problems that can be solved by a Non-deterministic Turing Machine in Polynomial time.

Whitespace and comments are ignored by lexical analysers, so these are not any categories of lexical analysis phase.

A port is a communication endpoint. At the software level, within an operating system, a port is a logical construct that identifies a specific process or a type of network service. A port is identified for each transport protocol and address combination by a 16-bit unsigned number, known as the port number. The most common transport protocols that use port numbers are the Transmission Control Protocol (TCP) and the User Datagram Protocol (UDP).

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The number of different entity sets participating in a relationship set is called the degree of a relationship set.

Three-address code(TAC) is an intermediate code used by optimizing compilers to aid in the implementation of code-improving transformations. Each TAC instruction has at most three operands and is typically a combination of assignment and a binary operator. For example, t1 := t2 + t3. The name derives from the use of three operands in these statements even though instructions with fewer operands may occur.

A+B*C ===> A+[BC*] => ABC*+

Worst Case Analysis: T(N) = N + T(N-1) Recurrence Relation: T(0) = T(1) = 0 (base case) T(N) = N + T(N-1) T(N-1) = (N-1) + T(N-2) T(N-2) = (N-2) + T(N-3) ;; T(3) = 3 + T(2) T(2) = 2 + T(1) T(1) = 0 Hence, T(N) = N + (N-1) + (N-2) ... + 3 + 2 ≈ (N^2)/2 =O(N^2)

Condition coverage → White-box testing Equivalence class partitioning → Black-box testing Volume testing → Performance testing Alpha testing → System testing

Bn full binary trees with b vertices Recurrence relation is Bn = 2Bn – 1 + O(1) Time complexity is O(2n)

The equation for the output of the diagram is (A+B)(A’+B’). (A+B)(A’+B’)= AA’+AB’+BA’+BB’ = 0+AB’+A’B+0 = AB’+A’B = A ⊕ B

If A and B are sets, then a binary relation R from A to B is a subset of the Cartesian product of A and B (A x B)

BigOmega→ Lower bound [ Generally we call it as Best case) BigTheta → Average bound [ Generally we call it as Average case) BigOh → Upper bound [ Generally we call it as Worst case)

One bit is used for signs . Remaining bits are 15 Maximum value stored in signed integer is 2₁₅ – 1

Class diagram is a static diagram. It represents the static view of an application. Class diagram is not only used for visualizing, describing, and documenting different aspects of a system but also for constructing executable code of the software application

A processor in a computer running Windows has two different modes: user mode and kernel mode. The processor switches between the two modes depending on what type of code is running on the processor. Applications run in user mode, and core operating system components run in kernel mode

Turnaround time (TAT) means the amount of time taken to complete a process or fulfill a request

An Access Control Matrix or Access Matrix is an abstract, formal security model of protection state in computer systems, that characterizes the rights of each subject with respect to every object in the system. An access matrix can be envisioned as a rectangular array of cells, with one row per subject and one column per object. The entry in a cell – that is, the entry for a particular subject-object pair – indicates the access mode that the subject is permitted to exercise on the object. Each column is equivalent to an access control list for the object; and each row is equivalent to an access profile for the subject

If the size of the logical address space is 2m and the page size is 2n addressing units, then the high order “m-n” bits of a logical address designate the page number and the “m” low order bits designate the page offset

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

Network layer converts Packets to Frames and Frames to Packets in the OSI model

The given code swaps(exchanges ) two variable values without using a third variable.