The variable is static variable, the value is retained during multiple function calls. Initial value is 3
“A--” is post decrement so it will print “3”
if(2) condition is true and main() function will call again , Here the “a” value is 2.
“A--” is post decrement so it will print “2”
if(1) condition is true and main() function will call again Here the “a” value is 1.
“A--” is post decrement so it will print “1”
if(0) condition is false it won’t call main() function

● The number of nodes n in a full binary tree, is at least n = 2 h + 1 and at most n = 2 ^h+1 − 1 , where h is the height of the tree. A tree consisting of only a root node has a height of 0.
● The number of leaf nodes l in a perfect binary tree, is l = ( n + 1 )/2 because the number of non-leaf (a.k.a. internal) nodes

● This means that a perfect binary tree with l leaves has n = 2 l − 1 nodes. ● In a balanced full binary tree, h = ⎡ log ₂ (l)⎤ + 1 = ⎡ log ₂ ((n + 1 )/2)⎤ + 1 = ⎡ log ₂ (n + 1 )⎤
● In a perfect full binary tree, l = 2 ^h thus n = 2 ^h+1 − 1
● The maximum possible number of null links (i.e., absent children of the nodes) in a complete binary tree of n nodes is ( n + 1 ) , where only 1 node exists in bottom-most level to the far left.
●The number of internal nodes in a complete binary tree of n nodes is ⎣ n/2⎦ .

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called hypertext or hypermedia.

→ The purpose of any analysis activity in the software life cycle is to create a model of the system's functional requirements that is independent of implementation constraints.
→ The main difference between object-oriented analysis and other forms of analysis is that by the object-oriented approach we organize requirements around objects, which integrate both behaviors (processes) and states (data) modeled after real world objects that the system interacts with. In other or traditional analysis methodologies, the two aspects: processes and data are considered separately.
→ For example, data may be modeled by ER diagrams, and behaviors by flow charts or structure charts.

In computing, an opcode (abbreviated from operation code, also known as instruction syllable, instruction parcel or opstring) is the portion of a machine language instruction that specifies the operation to be performed. Beside the opcode itself, most instructions also specify the data they will process, in the form of operands. In addition to opcodes used in the instruction set architectures of various CPUs, which are hardware devices, they can also be used in abstract computing machines as part of their byte code specifications.
Note: Instruction opcode not in a form of main memory.

Queue can be generally thought as horizontal in structure i.e, breadth/width can be attributed to it - BFS, whereas Stack is visualized as a vertical structure and hence has depth - DFS.

M has n rows. If all the elements of a row are multiplied by , the determinant value becomes 2*5. Multiplying all the n rows by 2, will make the determinant value 2​ ⁿ ​ *5=40. Solving n= 3.

If null were an Object, it would support the methods of java.lang.Object such as equals().
However, this is not the case - any method invocation on a null results in a NullPointerException.
There is also a special null type, the type of the expression null, which has no name. Because the null type has no name, it is impossible to declare a variable of the null type or to cast to the null type. The null reference is the only possible value of an expression of null type. The null reference can always be cast to any reference type. In practice, the programmer can ignore the null type and just pretend that null is merely a special literal that can be of any reference type

Priority scheduling algorithm could result starvation. Starvation is nothing but indefinite blocking.

The goal of software engineering is high cohesion and low coupling. The desirable property of module is independency.

A low level event that indicates that a window has changed its status. This low-level event is generated by a Window object when it is opened, closed, activated, deactivated, iconified, or deiconified, or when focus is transferred into or out of the Window.

Everyone thought that answer in worst case time is O(n) for both the cases but using circular queue, we can implement in constant amount of time.

Here, i=5 and putchar(i+100) it means actually putchar(5+100) ⇒ putchar(105);
It will print the ASCII equivalent of 105 which is lower case ' i '. The printf statement prints the current value of i. i.e. 5 and then decrements it. So, h4 will be printed in the next pass. This continues until ' i ' becomes 0, at which point the loop gets terminated.

Apply master theorem.
a=8,b=2,k=0,p=0;
So, according to masters theorem, a>b​ ^k
=O(n^log​ ₂^{​ 8}​ )
=O(n^{​ 3}​ )

A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.

Let principal be ‘x’
S.I = P T R / 100
S.I = X * 3 * 10 / 100
C.I = [ P (1 + (R / 100)^T) - P]
C.I = [ X (1+ (10 / 100)³) - X]
C.I = 331X / 1000
Difference between Compound interest and Simple interest is 77.5
( 331X / 1000 ) - ( 3X / 10) = 77.5
31x / 1000 = 77.5
x = 2500.

Here in this question 5000g means 5kg only
So from that question Aamir (A) Birju (B) Charles (C)
Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then
1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30
A + B + C = 13/240
1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240
So C alone can complete the work in 240 min.

Amount of copper in first alloy is 5x/8 and zinc is 3x/8.
Amount of copper in second alloy is 8x/13 and tin is 5x/13.
So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x
Tin present in 3rd alloy is 5/26.

A number that is divisible only by itself and 1 is called prime number.

Given data,
AB=24cm
CB=12cm
θC=5cm(Assume)
θB=5​ ²​ +12​ ²​ =169
θB=13cm
Note: They forgot to mention the value 5. If we are considering value=5 then answer should be A.

This option is both near the center of town and in a location where children and their parents are sure to be around. This is the only option that meets both of Monisha’s requirements.

Option-A FALSE: The author never says that reading is dull.
Option-B,C FALSE: Not related to given paragraph.
Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

In between r and T there is letter is s
Likewise m and O letter is n
Same h is the next letter to fit in that sequence

Qmelaqali→ fruitcake
We can divide artificial language into parts.
Qmela→ fruit
qali→ cake
Qalitiimmeo→ cakewalk
We can divide artificial language into parts.
Qali→ cake
tiimmeo→ walk
Useguamao → buttercup
We can divide artificial language into parts.
Usegu→ butter
amao→ cup
cupcake→ ?
amao→ cup
qali→ cake
So, option-D is correct answer.

In the word DROVE we have two vowels
Those are O and E
After consonants are arranged in alphabetically we got the word like DRV
But we have two place vowels in between two pairs of consonants we got finally DEROV
So right from fourth letter is E

From that question we can decide Pali Quereshi sant and timber but Rohan is not mentioned there. So Data inadequate is answer.

Q $ R means R’s father is Q
R @ T means T’s brother is R
T * M means M’s Daughter T
Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.