The Senior Computer Instructor (SCI) Paper Solution 2026 by Suraku Academy offers complete, reliable, and well-explained answers to every question of the Senior Computer Instructor (SCI) exam 2026 held on 23 August 2026. Our team has carefully prepared this solution to provide clear explanations that make concepts easier to understand and help students build a strong foundation. With this resource, candidates can prepare with confidence and improve their chances of success in the exam.
In addition, this solution guide is not just about answers, it is designed as a learning companion. By studying these explanations, students can strengthen their problem-solving skills and approach future exams with greater clarity and confidence.
Q: Arrange the following components of the CSS Box Model in the correct order from innermost to outermost:
(1) Margin
(2) Content
(3) Border
(4) Padding
Choose the most appropriate answer from the options given below:
Option A
The CSS Box Model describes how every HTML element is represented as a rectangular box. It consists of four main components:
Therefore, from innermost to outermost, the correct order is: Content → Padding → Border → Margin.
CSS Box Model Structure:

Q: In machine language, instructions are executed directly by the:
Option D
Machine Language is the lowest-level programming language that consists of binary instructions (0s and 1s). These instructions are written in a form that the computer's processor can understand and execute directly.
The CPU (Central Processing Unit) is the actual hardware that fetches, decodes, and executes machine-language instructions.

Q: A multiplexer with ‘n’ select lines can select from a maximum of how many input lines?
Option B
A Multiplexer (MUX) is a combinational digital circuit that selects one input from several input lines and sends the selected input to a single output line.
The number of select lines determines how many different input lines can be selected.
If a multiplexer has n select lines, each select line can have two possible values, either 0 or 1. Therefore, n select lines can produce 2n different combinations.
Each combination selects one input line. Hence, a multiplexer with n select lines can select a maximum of 2n input lines.
| Select Lines (n) | Maximum Input Lines (2n) | Multiplexer Name |
|---|---|---|
| 1 | 2 | 2×1 MUX |
| 2 | 4 | 4×1 MUX |
| 3 | 8 | 8×1 MUX |
| 4 | 16 | 16×1 MUX |
| 5 | 32 | 32×1 MUX |
Q: For every element x in Boolean Algebra, there exists a complement x’ such that:
Option B
In Boolean Algebra, every element x has a corresponding complement, represented by x′. The complement has the opposite Boolean value of x.
For a Boolean variable:
The two fundamental complement laws of Boolean Algebra are:
Q: What will be the output of the following PHP code?
<!DOCTYPE html>
<html>
<body>
<?php
$x=”10”;
$y=5;
echo $x + $y;
?>
</body>
</html>
Option A
PHP is a "loosely typed" language. This means you don't have to declare what type of data a variable hold. PHP figures it out automatically and even converts types on its own when needed. This automatic conversion is called "Type Juggling".
In PHP, the + operator is an arithmetic addition operator. It adds the numeric values of its operands.
$x = "10";
$y = 5;
echo $x + $y;
Here, $x contains the value "10" as a string, while $y contains 5 as an integer.
When PHP sees the + operator, it expects both sides to be numbers. So, if one side is a numeric string (a string that looks like a number like "10"), PHP automatically converts it into a real number before doing the math.
Therefore, 10+5 = 15. So, the output is 15.
Note:
Q: Which of the following traversal combinations is insufficient to uniquely construct a general binary tree?
Option D
A binary tree can be "visited" in different orders.
To uniquely construct a general binary tree, the traversal information must be sufficient to determine both:
The Golden Rule:
To uniquely reconstruct a binary tree, you generally need Inorder (provides the left/right subtree boundary) paired with one of Preorder, Postorder, or Level-order.
| Traversal Combination | Possible? | Reason |
|---|---|---|
| Inorder + Preorder | Yes | Preorder identifies the root. Inorder separates the left and right subtrees. |
| Inorder + Postorder | Yes | Postorder identifies the root. Inorder separates the left and right subtrees. |
| Inorder + Level-order | Yes | Level-order identifies the root and, together with Inorder, the nodes can be divided into left and right subtrees recursively. |
| Postorder + Level-order | No | There is no Inorder traversal to uniquely determine whether nodes belong to the left or right subtree. |
Why is Postorder + Level-order insufficient?
Consider two different binary trees:
| Tree 1 | Tree 2 |
|---|---|
| A / B Postorder = B A Level-order = A B | A \ B Postorder = B A Level-order = A B |
Thus, two different binary trees have exactly the same Postorder and Level-order traversals. Therefore, these two traversals cannot uniquely determine a general binary tree.
Q: What will be the output of the following C program?
void fun()
{
static int s=1;
int d=1;
s++;
d++;
printf(“%d %d”,s,d);
}
int main()
{
fun();
fun();
fun();
return 0;
}
Option B
This question tests the difference between a static variable and a local automatic variable in C.
static int s = 1;
int d = 1;
| Function Call | Value of s before s++ | Value of s after s++ | Value of d before d++ | Value of d after d++ | Output |
|---|---|---|---|---|---|
| 1st Call | 1 | 2 | 1 | 2 | 2 2 |
| 2nd Call | 2 | 3 | 1 | 2 | 3 2 |
| 3rd Call | 3 | 4 | 1 | 2 | 4 2 |
Therefore, the complete output is 2 2 3 2 4 2.
Q: In a Binary Search Tree (BST) with n nodes and height h, what is the worst-case time complexity of finding the in-order successor of a given node?
Option B
In a Binary Search Tree (BST), the word successor means the node that comes immediately after a given node in a particular order.
The in-order successor of a node is the node that comes immediately after that node in the In-order Traversal of the BST.

The in-order traversal sequence of above tree is 20 → 30 → 40 → 50 → 60 → 70 → 80.
Therefore:
Successor of 20 = 30
Successor of 30 = 40
Successor of 40 = 50
Successor of 50 = 60
Successor of 60 = 70
Successor of 70 = 80
80 has no successor, because it is the largest node.
So, in other word, “in a BST, the in-order successor of a node is the smallest value that is greater than that node” (किसी node से बड़ी values में सबसे छोटी value = उसका In-order Successor).
Finding the In-order Successor?
Case 1: The node has a Right Subtree
If the given node has a right child or right subtree, go to the right subtree and then keep moving left until the leftmost node is reached.
Case 2: The Node Has No Right Subtree
If a node does not have a right subtree, its successor cannot be found below that node. So, we move upward through its parent and ancestor nodes.
While moving upward, find the first ancestor for which the given node is in its left subtree. That ancestor is the in-order successor.
Time Complexity:
The time required to find the successor depends on the height h of the BST. In the worst case, we may have to move down or up along a path whose length is proportional to the tree height.
Therefore, O(h).
| BST Structure | Height (h) | Time Complexity for Finding In-order Successor |
|---|---|---|
| Balanced BST | h = O(log n) | O(h) = O(log n) |
| Completely Skewed BST | h = O(n) | O(h) = O(n) |
Q: In Karnaugh Map minimization using the Sum of Products (SOP) form, a group of 4 cells produces A’C. An overlapping group of 2 cells produces A’B’C. The final minimized expression is:
Option B
In Karnaugh Map (K-Map) minimization using SOP (Sum of Products), we form groups of 1s. Each group gives a product term, and then all the product terms are combined using the OR (+) operation.
The question gives two groups:
Therefore, the initial SOP expression is: F=A′C+A′B′C
Now we need to check whether the second term is already covered by the first term.
F=A’C+A’B’C
F=A’C(1+B’)
F=A’C.1
F=A’C
Thus, the smaller 2-cell group does not add any new minterms because it is completely covered by the 4-cell group.
The 4-cell group represented by A'C already includes the cells represented by A'B'C. Therefore, adding the smaller group does not change the function. A'C+A'B'C = A'C
Q: Which statement is true about enumeration constants?
Option A
In C programming, an enumeration (enum) is a user-defined data type used to assign meaningful names to a set of integer constants.
enum WeekDay {
MON,
TUE,
WED,
THU
};
By default, the enumeration constants are assigned integer values starting from 0, not 1.
MON = 0
TUE = 1
WED = 2
THU = 3
The values can also be explicitly specified:
enum WeekDay {
MON = 1,
TUE = 5,
WED = 10
};
No matter what values you assign, they are always integers, you cannot assign floating-point (decimal) values like 1.5 to an enum constant in C.
Q: In C++, which access specifier allows class members to be accessed only within the same class and its derived classes?
Option D
In Object-Oriented Programming (OOP), a class contains data (variables) and functions (methods).
Access specifiers control who is allowed to access these members from outside the class. This is part of a core OOP principle called encapsulation. The encapsulation controlling visibility and protecting data from unwanted or unsafe access.
C++ provides three main access specifiers, public, private, and protected.
Members declared as public can be accessed from anywhere, inside the class, in derived (child) classes, and even from outside the class entirely. It is the most unrestricted access level.
Members declared as private can be accessed only within the same class where they are defined. This is the most restrictive access level.
Members declared as protected behave almost like private, except for one important difference that they can be accessed by derived (child) classes through inheritance.
| Access Specifier | Same Class | Derived Class | Outside Class |
|---|---|---|---|
| public | Yes | Yes | Yes |
| protected | Yes | Yes | No |
| private | Yes | No | No |
Q: Why is insertion faster in a linked list?
Option C
In a linked list, each element is stored in a separate node, and nodes are connected using pointers. Unlike an array, the elements of a linked list do not need to be stored in contiguous memory locations.
When a new node is inserted at a suitable position, we generally only need to change the links (pointers) between the existing nodes. The existing elements do not have to be physically moved or shifted.
Note:
Q: Consider the following functions:
f1(n) = n!
f2(n) = 2n
f3(n) = nlog n
f4(n) = n5
Arrange the functions in increasing order of their asymptotic growth rates as n→∞.
Option A
When comparing functions, we care about how fast a function grows as n becomes very large (n → ∞), not how it behaves for small values of n.
There's a well-known ordering of common growth rates, from slowest to fastest:
| Constant | Logarithmic | Linear | Polynomial | Quasi-Polynomial | Exponential | Factorial |
|---|---|---|---|---|---|---|
| O(1) | O(log n) | O(n) | O(nk) | O(nlog n) | O(2n) | O(n!) |
Q: In the context of ADTs, encapsulation means:
Option B
An ADT (Abstract Data Type) defines what operations can be performed on a data type and what those operations do, without requiring the user to know how those operations are implemented internally.
This separation between the interface (what the user can use) and the implementation (how it actually works) is called encapsulation.
For example, consider a Stack ADT. A user may use operations such as:
The user does not need to know whether the stack is implemented using an Array or a Linked List.
Q: In C++, constructors in inheritance are executed in which order?
Option A
In C++ inheritance, when an object of a derived class is created, the base class constructor executes first, followed by the derived class constructor.
This order is important because the derived class may depend on the base class being properly initialized first.
Q: Which search strategy in branch-and-bound is similar to Breadth First Search (BFS)?
Option A
Branch-and-Bound is a problem-solving technique used to explore a State-Space Tree while eliminating branches that cannot lead to a better solution.
The way we explore this tree of possibilities is called the search strategy, and there are a few common strategies used in Branch and Bound.
Why does FIFO Search match BFS?
In FIFO (First-In, First-Out) Branch-and-Bound, the live nodes are maintained in a queue. The node that was generated first is explored first. This is exactly the basic principle of Breadth First Search (BFS).
BFS (Breadth First Search) explores a tree or graph level by level, visiting all nodes at the current depth or level before moving on to the next deeper level. This naturally requires a Queue.
Comparison of Branch-and-Bound Search Strategies:
| Strategy | Data Structure | Similar To | Basic Principle |
|---|---|---|---|
| FIFO Search | Queue | BFS | Explore the oldest live node first. |
| LIFO Search | Stack | DFS | Explore the newest live node first. |
| LC Search | Priority Queue | Best-First Style | Explore the live node with the lowest cost/bound first. |
| DFS Search | Stack | DFS | Explore one branch deeply before backtracking. |
Q: Which of the following statements about file handling in C is incorrect?
Option C
In C file handling, files are accessed using a FILE pointer and functions such as fopen(), fclose(), fscanf(), fprintf(), etc.
The statement in option (C) is incorrect because fclose() does not delete a file. It only closes the file that was previously opened using fopen().
FILE *fp;
fp = fopen("myfile.txt", "r");
/* File operations */
fclose(fp);
Here, fclose(fp) releases the resources associated with the opened file and closes the file stream. The file still exists on the storage device.
| Function | Purpose |
|---|---|
| fopen() | Opens a file and returns a FILE * pointer. If the file cannot be opened, it returns NULL. |
| fclose() | Closes an opened file. It does not delete the file. |
| fscanf() | Reads formatted data from a file. |
| fprintf() | Writes formatted data to a file. |
| fgetc() | Reads one character from a file. |
| fputc() | Writes one character to a file. |
| remove() | Deletes a file from the storage system. |
Q: Which of the following Xpath expressions selects all title elements anywhere in an XML document?
Option C
XPath is a language used to navigate through elements in an XML document, kind of like giving directions to find a specific element inside a tree-structured document. An XPath expression describes the location of a particular node or group of nodes.
Important XPath Symbols:
| Symbol | Meaning |
|---|---|
| / | It is used to specify a direct path from the root node or from a particular context node. |
| // | It is used to select matching elements anywhere in the document, regardless of their level or depth. |
| . | It represents the current node in the XPath expression. |
| .. | It represents the parent node of the current node. |
| @ | It is used to select an attribute of an element. |
| * | It represents any element at the specified location. |
| [ ] | It is used to specify a condition or filter for selecting particular nodes. |
| text() | It is used to select the text content contained inside an element. |
| last() | It is used to select the last node from the nodes currently being selected. |
E.g.:
<?xml version="1.0" encoding="UTF-8"?>
<library>
<book id="101" category="programming">
<title>Learn C++</title>
<author>ABC</author>
<price>500</price>
</book>
<book id="102" category="database">
<title>Learn SQL</title>
<author>DEF</author>
<price>600</price>
</book>
<book id="103" category="web">
<title>Learn HTML</title>
<author>GHI</author>
<price>400</price>
</book>
</library>
| XPath Expression | What It Selects | Result |
|---|---|---|
| /library | The library root element | Entire <library> element |
| /library/book | All book elements directly inside library. | 3 book elements |
| /library/book/title | All title elements directly inside book. | Learn C++, Learn SQL, Learn HTML |
| //book | All book elements anywhere in the document. | 3 book elements |
| //title | All title elements anywhere in the document. | Learn C++, Learn SQL, Learn HTML |
| //book/@id | id attributes of all books. | 101, 102, 103 |
| //book/title/text() | Text contained inside all title elements. | Learn C++, Learn SQL, Learn HTML |
| //book[1] | First book in the selected book node-set. | Book with ID 101 |
| //book[last()] | Last book. | Book with ID 103 |
| //book[@id='102'] | Book whose id is 102. | Book "Learn SQL" |
| //book[@category='web'] | Book whose category is web. | Book "Learn HTML" |
| //book[price>500] | Books whose price is greater than 500. | Book "Learn SQL" |
Q: In C++, operator overloading is an example of:
Option D
Operator overloading in C++ is a form of Compile-Time Polymorphism, also called Static Polymorphism.
Operator overloading allows us to give an existing operator, such as +, -, ==, or *, a special meaning when it is used with objects of a class.
E.g.:
class Complex
{
public:
int real, imag;
Complex operator+(Complex c)
{
Complex temp;
temp.real=real+c.real;
temp.imag=imag+c.imag;
return temp;
}
};
Complex c3 = c1 + c2;
Here, the + operator is overloaded to add two Complex objects.
Q: Which of the following statements regarding fanout is FALSE?
Option D
Fan-out is the maximum number of standard logic-gate inputs that can be connected to the output of a single logic gate while still maintaining valid logic levels and proper operation.
Fan-out is not determined by the driving gate alone. It depends on the electrical characteristics of both the driving gate and the inputs of the gates being driven.
For example, the driving gate must be capable of supplying enough current when its output is HIGH and absorbing enough current when its output is LOW. Therefore, fan-out is checked for both logic states.
Q: Which of the following properties is satisfied by an Abelian group, but need not be satisfied by a group?
Option D
A Group is an algebraic structure in which a set and a binary operation satisfy four basic properties:
An Abelian group satisfies all the properties of a group plus one additional property: Commutativity.
The commutative property means that changing the order of the operands does not change the result : A*B = B*A or A+B = B+A
Therefore, commutativity is guaranteed in an Abelian group but is not required in a general group.
Q: Suppose a BST contains n distinct keys and is completely skewed. The worst-case time complexity for searching an element is:
Option C
A Binary Search Tree (BST) is a binary tree in which, for each node:
The time required for searching in a BST depends mainly on its height.
For a well-balanced BST, the height is approximately O(log n). So, searching can be performed in O(log n) time.
A skewed tree is a BST where every node has only one child (either only a left child, or only a right child), meaning the tree looks less like a proper "tree" and more like a straight line.
In skewed BST, each node has only one child. Therefore, the height (h) of the tree becomes: h=n-1.
In the worst case, the element we are searching for may be at the last node. We may therefore have to examine all n nodes. Hence the complexity becomes O(n).
Balanced vs Completely Skewed BST:
| BST Structure | Height | Worst-Case Search |
|---|---|---|
| Balanced BST | O(log n) | O(log n) |
| Completely Skewed BST | O(n) | O(n) |
Q: In Java, System.out and System.err are object of which class?
Option D
In Java, System.out and System.err are predefined objects of the PrintStream class.
Here, System is a predefined class in Java, provided by the java.lang package. The System class provides several static fields and methods that allow a Java program to interact with the system environment, standard input/output, and other system-level resources.
When we write, System.out.println("Suraku Academy");
Q: In Java, which event class is fired when a button is pressed?
Option B
In Java GUI programming (AWT or Swing), an event is any action performed by the user, like clicking a button, typing on the keyboard, moving the mouse, etc.
Java provides different Event Classes, each specifically designed to represent a particular type of user interaction.
When such an action occurs, Java automatically creates an object of the matching event class and passes it to the appropriate event-handling code.
| Event Class | Used For |
|---|---|
| ActionEvent | Button clicks, menu-item selections, etc. |
| KeyEvent | Keyboard key actions such as pressing or releasing a key |
| MouseEvent | Mouse actions such as clicking, pressing, releasing, or moving |
| TextEvent | Changes in text components |
Q: What will be the output of the following C program?
int main()
{
int i=0;
while(i<5)
{
i++;
if(i==2)
continue;
if(i==4)
break;
printf(“%d”,i);
}
return 0;
}
Option A
This question tests the working of the continue and break statements inside a while loop.
| Iteration | Value after i++ | Condition | Action | Output |
|---|---|---|---|---|
| 1st | 1 | i == 2 : False | printf() executes | 1 |
| 2nd | 2 | i == 2 : True | continue : skip printf() | — |
| 3rd | 3 | i == 2 : False | printf() executes | 3 |
| 4th | 4 | i == 4 : True | break : loop terminates | — |
Therefore, the output is 1 3.
Q: Binary Search is generally not preferred over Linear Search when:
Option B
Linear Search checks elements one by one, from the start to the end, until it finds the target or reaches the end. It works on any list either sorted or unsorted. Its time complexity is O(n).
Binary Search works by repeatedly dividing the search range in half. It checks the middle element, and depending on whether the target is smaller or larger, it eliminates half of the remaining elements each time. Its time complexity is O(log n), which is much faster than linear search for large datasets.
But Binary Search only works on sorted data. If the data is not already sorted, you would need to sort it first before you can apply binary search.
So, in this situation, sorting the data just to use binary search ends up being more expensive than simply doing a linear search directly on the unsorted data.
Q: The DOM hierarchy, the document object is:
Option C
DOM stands for Document Object Model. When a browser loads an HTML page, it converts the entire page into a tree-like structure made of objects, so that JavaScript can access, read, and modify any part of the page. This tree structure is called the DOM.
At the top of the DOM hierarchy is the document object. It represents the entire HTML document and acts as the root from which the elements of the document can be accessed.
For example, JavaScript commonly uses methods such as:
In both cases, we start with document and then access a particular HTML element. This is possible because the document object represents the whole HTML document and provides the starting point for accessing its elements.
E.g.:
<html>
<body>
<h1>Welcome</h1>
<p>Hello World</p>
</body>
</html>
DOM Hierarchy:
Document
|
html
|
body
/ \
h1 p
| |
Text Text
Q: In a 3-bit synchronous binary counter using T flip-flops, the present state is 0 1 1 and the next state is 1 0 0. What are the required flip-flop inputs (TA2, TA1, TA0)?
Option D
A T (Toggle) flip-flop is a memory element used in digital circuits that has a very simple rule:
T Flip-Flop Truth Table:
| Present State (QN) | Next State (QN+1) | T |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
We have a 3-bit counter with bits labeled A2, A1, A0 from left to right.
| State | A2 | A1 | A0 |
|---|---|---|---|
| Present State | 0 | 1 | 1 |
| Next State | 1 | 0 | 0 |
Let's check each bit one by one, comparing Present vs Next:
| Bit | Present | Next | Changed? | T Required |
|---|---|---|---|---|
| A2 | 0 | 1 | Yes | 1 |
| A1 | 1 | 0 | Yes | 1 |
| A0 | 1 | 0 | Yes | 1 |
Therefore, (TA2,TA1,TA0)=(1,1,1)
Q: Which characteristic of CMOS technology makes it the preferred choice for VLSI circuits compared to TTL?
Option B
VLSI stands for Very Large Scale Integration. VLSI refers to the process of creating integrated circuits (ICs) by combining millions or billions of transistors onto a single small chip.
Since so many transistors are packed together in such a tiny space, power consumption and heat generation become extremely important factors when choosing which technology to build these chips with.
CMOS (Complementary Metal-Oxide-Semiconductor) and TTL (Transistor-Transistor Logic) are two important digital logic technologies used to build logic circuits.
VLSI involves placing a very large number of transistors on a single integrated circuit. Therefore, low power consumption is extremely important because millions or billions of devices may operate on the same chip.
The major advantage of CMOS technology is its very low static power dissipation. In a CMOS circuit, ideally, there is almost no direct current path from the power supply to ground when the circuit is in a stable logic state.
Therefore, CMOS is highly suitable for VLSI circuits, where low power consumption, high packing density, and efficient operation are important.
Q: Consider a Binary Search Tree (BST) containing distinct integers from 1 to 1000. During a search for the key 363, which one of the following sequences cannot occur as the sequence of visited nodes?
Option B
In a Binary Search Tree, every node follows this rule:
When you search for a key (say 363), you start at the root and compare:
The Key Idea: Every Visited Node Restricts the Range
As the search proceeds, every node you visit narrows down the possible range for all future nodes in the sequence:
If any node in a given sequence falls outside the range built up so far, that sequence is impossible because it could never actually happen during a real BST search.
Checking Option (A): 924, 220, 911, 244, 898, 258, 362, 363
Starting Range: (-∞, +∞)
| Step | Node | Within Range? | Key(363) vs Node | Action | New Range |
|---|---|---|---|---|---|
| 1 | 924 | Within (-∞, +∞) | 363 < 924 | Go Left | (-∞, 924) |
| 2 | 220 | Within (-∞, 924) | 363 > 220 | Go Right | (220, 924) |
| 3 | 911 | Within (220, 924) | 363 < 911 | Go Left | (220, 911) |
| 4 | 244 | Within (220, 911) | 363 > 244 | Go Right | (244, 911) |
| 5 | 898 | Within (244, 911) | 363 < 898 | Go Left | (244, 898) |
| 6 | 258 | Within (244, 898) | 363 > 258 | Go Right | (258, 898) |
| 7 | 362 | Within (258, 898) | 363 > 362 | Go Right | (362, 898) |
| 8 | 363 | Within (362, 898) | Found. | — | — |
This is a valid, possible search sequence.
Checking Option (B): 925, 202, 911, 240, 912, 245, 363
Starting Range: (-∞, +∞)
| Step | Node | Within Range? | Key(363) vs Node | Action | New Range |
|---|---|---|---|---|---|
| 1 | 925 | Within (-∞, +∞) | 363 < 925 | Go Left | (-∞, 925) |
| 2 | 202 | Within (-∞, 925) | 363 > 202 | Go Right | (202, 925) |
| 3 | 911 | Within (202, 925) | 363 < 911 | Go Left | (202, 911) |
| 4 | 240 | Within (202, 911) | 363 > 240 | Go Right | (240, 911) |
| 5 | 912 | Not Within (240, 911) | — | — | VIOLATION |
This is not a valid, possible search sequence.
Checking Option (C): 2, 399, 387, 219, 266, 382, 381, 278, 363
Starting Range: (-∞, +∞)
| Step | Node | Within Range? | Key(363) vs Node | Action | New Range |
|---|---|---|---|---|---|
| 1 | 2 | Within (-∞, +∞) | 363 > 2 | Go Right | (2, +∞) |
| 2 | 399 | Within (2, +∞) | 363 < 399 | Go Left | (2, 399) |
| 3 | 387 | Within (2, 399) | 363 < 387 | Go Left | (2, 387) |
| 4 | 219 | Within (2, 387) | 363 > 219 | Go Right | (219, 387) |
| 5 | 266 | Within (219, 387) | 363 > 266 | Go Right | (266, 387) |
| 6 | 382 | Within (266, 387) | 363 < 382 | Go Left | (266, 382) |
| 7 | 381 | Within (266, 382) | 363 < 381 | Go Left | (266, 381) |
| 8 | 278 | Within (266, 381) | 363 > 278 | Go Right | (278, 381) |
| 9 | 363 | Within (278, 381) | Found. | — | — |
This is a valid, possible search sequence.
Checking Option (D): 2, 252, 401, 398, 330, 344, 397, 363
Starting Range: (-∞, +∞)
| Step | Node | Within Range? | Key(363) vs Node | Action | New Range |
|---|---|---|---|---|---|
| 1 | 2 | Within (-∞, +∞) | 363 > 2 | Go Right | (2, +∞) |
| 2 | 252 | Within (2, +∞) | 363 > 252 | Go Right | (252, +∞) |
| 3 | 401 | Within (252, +∞) | 363 < 401 | Go Left | (252, 401) |
| 4 | 398 | Within (252, 401) | 363 < 398 | Go Left | (252, 398) |
| 5 | 330 | Within (252, 398) | 363 > 330 | Go Right | (330, 398) |
| 6 | 344 | Within (330, 398) | 363 > 344 | Go Right | (344, 398) |
| 7 | 397 | Within (344, 398) | 363 < 397 | Go Left | (344, 397) |
| 8 | 363 | Within (344, 397) | Found. | — | — |
This is a valid, possible search sequence.
Q: In a circular queue of size n, the queue is considered full when:
Option A
A Circular Queue is a queue in which the last position is connected back to the first position. It follows the FIFO (First In, First Out) principle.
In a normal linear queue, when rear reaches the last position, we may not be able to reuse empty positions at the beginning.
A circular queue solves this problem by allowing rear to wrap around to the beginning using the modulo operator %.
For a circular queue of size n, the next position of rear is calculated as: (rear+1)%n
The queue is considered full when this next position is equal to front: (rear+1)%n==front
Q: In Java, which class is at the top of the exception hierarchy?
Option C
In Java, Throwable is the topmost class in the exception hierarchy. It is the parent class of both Exception and Error.

Q: In HTML, the required attribute of an input field is an example of:
Option D
Validation is the process of checking whether the data entered by a user is present, valid, and in the expected format before it is processed or submitted. There are two main places where validation can happen:
Client-Side Validation:
E.g.:
<form>
<label>Name:</label>
<input type="text" name="username" required>
<button type="submit">Submit</button>
</form>
Here, the required attribute tells the browser that the Name field cannot be left empty. If the user tries to submit the form without entering a name, the browser displays a validation message and prevents the form from being submitted.
Server-Side Validation:
Q: In DHTML, event handling is commonly implemented using:
Option C
DHTML stands for Dynamic HTML. DHTML actually a combination of technologies (HTML + CSS + JavaScript + DOM) used together to make web pages interactive and dynamic.
An event is an action that occurs in a web page, such as:
JavaScript is commonly used to handle these events. JavaScript event handlers contain the code that is executed when a particular event occurs.
E.g.:
<button onclick="showMessage()">Click Me</button>
<script>
function showMessage()
{
alert("Button was Clicked!");
}
</script>
Here, onclick is an event handler. When the user clicks the button, the showMessage() function is executed.
Q: A graph with V vertices and E edges is represented using an adjacency list. The space complexity of the representation is:
Option D
An Adjacency List represents a graph by maintaining a list of adjacent (connected) vertices for each vertex.
Suppose a graph has:
The adjacency-list representation requires space for two main things:
Therefore, the total space is O(V)+O(E) = O(V+E) (Total Entries)
Q: Which operator checks both value and data type equality in JavaScript?
Option C
In JavaScript, comparison operators are used to compare two values. The strict equality operator (===) checks both the value and the data type of the operands.
The == Operator : Loose equality
The === Operator : Strict equality
| Operator | Name | Value Compared? | Data Type Compared? | Example | Result |
|---|---|---|---|---|---|
| == | Loose Equality | Yes | No | 15=="15" | true |
| === | Strict Equality | Yes | Yes | 15==="15" | false |
Thank you so much for taking the time to read my Computer Science MCQs section carefully. Your support and interest mean a lot, and I truly appreciate you being part of this journey. Stay connected for more insights and updates! If you'd like to explore more tutorials and insights, check out my YouTube channel.
Don’t forget to subscribe and stay connected for future updates.