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.
| S.N. | Name of Subjects | No. of Questions Asked |
|---|---|---|
| 01. | Pedagogy | 00 |
| 02. | Mental Ability | 00 |
| 03. | Computer Fundamentals | 00 |
| 04. | Operating System (OS), Computer Organization & Architecture (COA) | 00 |
| 05. | Data & Computer Networks (DCN) | 00 |
| 06. | Basics of Communication (Channel Capacity, Attenuation, PCM, WDM, GSM, CDMA) | 00 |
| 07. | Network Security (Symmetric and Asymmetric Encryption, Viruses, Firewalls, Backup & Restoring, Ethical Hacking) | 00 |
| 08. | Programming Languages &OOPS Concepts (C, C++, Java, Python, .NET, AI & ML) | 00 |
| 09. | Data Structure and Algorithms (DSA) | 00 |
| 10. | Database Management System (DBMS) | 00 |
| 11. | Web Development (HTML, DOM, DHTML, CSS, XML, PHP, JavaScript) | 00 |
| 12. | Software Engineering (SE) | 00 |
| 13. | Digital Logic System (Boolean Algebra, TTL and CMOS Logic Families, Combinational & Sequential Circuit) | 00 |
Q: 1 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: 2 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: 3 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: 4 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: 5 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: 6 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: 7 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: 8 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: 9 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: 10 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.
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