[SOLUTION] SET A PABSON SLC PRE-BOARD EXAM 2081(2025)
PABSON
SLC PRE-BOARD EXAM
2081
Grade: XII
Subject: Computer Full Marks: 50 Time: 2:15 hrs
Candidates
are required to give their answer according to the given instructions.
Attempt all questions
Group A (Multiple choice question)
Select the best answer from the following
alternatives. 1×9 = 9
1. In which of the
following switching methods, the message is divided into small packets?
a. Message switching b. Packet switching
c. Virtual switching d. None of the these
Note: In packet switching,
messages are divided into small packets before transmission.
2. A _________ set of
rules that governs data communication
a. Protocols b. Standards
c. RFCs d. Servers
Note: Protocols are a set
of rules that govern data communication.
3. Which of the following
is not the phase of classical waterfall model?
a. Feasibility Study b. Requirement Analysis
c. Building Protype d. Maintenance
Note: The classical
waterfall model does not include prototyping; it follows a linear and
sequential approach.
4. Which of the following
is generally used for performing tasks like creating the structure of the
relations, deleting relation?
a. DML(Data Manipulation
Language)
b. Query
c. Relational Schema
d. DDL(Data Definition
Language)
Note: DDL is used for
defining and modifying database structures.
5. The term
"DFD" stands for?
a. Data file diagram
b. Data flow document
c. Data flow diagram
d. None of the above
Note: DFD stands for Data
Flow Diagram, which represents the flow of data in a system.
6. What will be the output
of the following code snippet?
#include <stdio.h>
int main() {
int a = 3, b = 5;
int t = a;
a = b;
b = t;
printf("%d
%d", a, b);
return 0;
}
Note: The code swaps the values
of a and b using a temporary variable t. So, Answer is 5,3
7. What is the size of a C
structure.?
a. C structure is always
128 bytes.
b. Size of C structure is
the total bytes of all elements of structure.
c. Size of C structure is
the size of largest element.
d. None of the above
Note: The size of a
structure depends on the sum of its elements, considering padding and
alignment.
8. String values in PHP
must be enclosed within -
a. Double Quotes
"" b. Single
Quotes ' '
c. Both (a) and (b) d. None
of the above
Note: String values in PHP
can be enclosed in either single (' ') or double (" ") quotes.
9. Which of the following
variable name is invalid?
a. $newVar b. $new_Var
c. $new-var d. All of the above
Note: In PHP, variable
names cannot contain hyphens (-).
Group B:
Short Answer Questions (5 × 5=25)
1. What is data security?
How can it be implemented?
Ans:
Data security is the practice of protecting digital
information from unauthorized access, corruption, or theft throughout its
entire lifecycle.
It involves a set of strategies, processes, and
technologies designed to ensure the confidentiality, integrity, and
availability of data.
Data security can be implemented through multiple techniques as
given below:
a) Authentication & Access Control: Use strong
passwords, multi-factor authentication (MFA), and role-based access control
(RBAC) to restrict unauthorized access.
b)
Data
Encryption: Secure sensitive data using encryption algorithms (AES, RSA) and
protect data transmission with SSL/TLS protocols.
c)
Firewalls
& Network Security: Monitor and block cyber threats using firewalls,
intrusion detection, and prevention systems (IDPS).
d)
Backup &
Disaster Recovery: Regularly back up data and implement a disaster recovery
plan to restore lost information.
e) Regular Security Updates: Keep software, databases,
and systems updated with the latest security patches to prevent
vulnerabilities.
OR
What is RDBMS? List
Out the functions of RDBMS.
Ans:
RDBMS (Relational
Database Management System) is a type of database management system that stores
data in a structured format using tables (relations) with rows and columns. It
follows the relational model, which was introduced by E.F. Codd in 1970.
RDBMS provides various
functionalities for efficient database management as given below:
a) Data Storage & Retrieval: Efficiently stores large
amounts of structured data and retrieves it using SQL queries.
b)
Data Integrity
& Consistency: Ensures accuracy through constraints like Primary Key,
Foreign Key, Unique, and Check.
c)
Transaction
Management: Maintains ACID (Atomicity, Consistency, Isolation, Durability)
properties for reliable transactions.
d)
Data Security:
Provides authentication, access control, and encryption to protect data from
unauthorized access.
e) Data Manipulation: Allows CRUD operations (Create,
Read, Update, Delete) using SQL commands.
2. Write a JavaScript
program to calculate simple interest using Principal, rate and time.
Ans:
<!DOCTYPE html>
<html>
<head>
<title>Simple
Interest Calculator</title>
</head>
<body>
<script>
function
calculateSimpleInterest()
{
var p, r, t, si;
p =
parseFloat(prompt("Enter Principal Amount:"));
r =
parseFloat(prompt("Enter Annual Interest Rate (%):"));
t =
parseFloat(prompt("Enter Time (years):"));
si = (p * r
* t) / 100;
document.write("Simple Interest: " + si);
}
calculateSimpleInterest();
</script>
</body>
</html>
OR
Write a PHP program to
insert data in a database.
Ans:
A
server-side scripting code to insert data into the table student having fields
(first name, last name, mark and email) is as follows:
<?php
// Step 1: Database configuration
$servername = "localhost";
$username = "root";
$password = "";
$database = "studentDB";
//Step2: Database connection
$con = new mysqli($servername, $username, $password,
$database);
// Step 3: Check connection
if ($con->connect_error)
{
die("Connection failed: "
. $con->connect_error);
} else {
echo "Connected
successfully!";
}
// Step 4: Prepare the SQL statement for insertion
$sql = "INSERT INTO student (first_name,
last_name, mark, email)
VALUES
('Madhav', 'Sapkota', 30, 'madhav@gmail.com')";
// Step 5: Execute the query and check if data is
inserted
if ($con->query($sql) === TRUE)
{
echo "Data inserted
successfully";
} else {
echo "Error: " . $sql .
"<br>" . $con->error;
}
// Step 5: Close connection
$con->close();
?>
3. Explain the concept of
classes, objects, polymorphism, and inheritance in OOP.
Ans:
Class:
A class is a blueprint or template for creating
objects. It defines a set of properties (attributes) and methods (functions)
that the objects created from the class will have. The class acts as a
prototype from which individual objects are created, encapsulating data for the
object and methods to manipulate that data.
Object:
An object is an instance of a class. It is a specific
realization of a class with actual values for the properties defined by the
class. Objects can interact with one another through methods, making them the
building blocks of an application. The object represents a real-world entity
that combines both attributes(data) and behaviours
(methods/functions).
Inheritance:
In Object-Oriented Programming (OOP), inheritance is a
fundamental concept that allows a class (called a subclass or derived class) to
inherit properties and behaviours (methods and attributes) from another class
(called a superclass or base class).
Polymorphism:
Polymorphism refers the ability of an object to take on different
forms depending upon situations. The same method(function) can behave
differently on different classes.
4. What is a feasibility
study? Explain different types of feasibility studies.
Ans:
A
feasibility study is a detailed analysis that assesses the practicality and
viability of a proposed project or plan. It aims to determine whether the
project is likely to succeed by evaluating various factors
(level/types/components) as follows:
1.
Technical Feasibility: Assesses the availability and suitability of the
required technology and resources. Example: A
company wants to develop a mobile app but checks if their developers have the
right skills and tools to create it.
2.
Economic Feasibility: Evaluates the financial viability of the project,
including cost-benefit analysis, return on investment, and funding options. Example: A startup plans to build an e-commerce website and
calculates if they have enough budget for development and maintenance.
3. Legal
Feasibility: Examines potential legal and regulatory implications, such as
intellectual property rights, data privacy, and licensing requirements. Example: A company wants to launch a social media app and checks
if it complies with data privacy laws.
4.
Operational Feasibility: Assesses the impact of the project on existing
operations, including organizational changes, user acceptance, and training
needs. Example: A school wants to introduce
an attendance tracking system and checks if teachers and students can easily
use it.
5.
Scheduling Feasibility: Determines the project timeline, resource
allocation, and potential risks and develops contingency plans. Example: A software company plans to release a new game and
ensures it can be completed before the holiday season.
5. What is e-governance?
Explain its importance in country like Nepal.
Ans:
E-Government refers to the use of technology, especially the
internet, to provide government services, enhance public administration, and
engage with citizens. Examples: Online tax filing, digital voter registration,
and e-governance portals.
E-Government plays a
crucial role in a developing country like Nepal, where geographical challenges,
bureaucratic inefficiencies, and limited access to public services have long
been obstacles.
Some key Importances of
E-Government in Nepal are as follows:
a) Better Accessibility – Citizens in remote areas can
access services online without traveling.
b) Increased Efficiency – Reduces paperwork, speeds up
processes, and minimizes bureaucracy.
c) Transparency & Accountability – Online platforms
reduce corruption and ensure open governance.
d) Economic Growth – Encourages digital literacy,
attracts investment, and supports IT sector development.
e) Disaster Management – Helps in quick response during
emergencies like earthquakes and pandemics.
f)
Cost-Effective
Governance – Reduces administrative costs and improves service delivery.
Conclusion: E-Government is key to making governance
in Nepal more efficient, transparent, and citizen-friendly.
Group C:
Long Answer Questions (2 × 8=16)
6. What do you mean by
transmission media? Explain Guided Media and Unguided Media.
Ans:
A
transmission medium, also known as communication media is a pathway that
carries the information from the sender to the receiver. It can be either
guided (wired) or unguided (wireless) as explained below:
1. Guided Media: Guided media refers to transmission that uses a physical pathway to
direct signals from the sender to the receiver.
Key Characteristics:
a) The signal is confined to a path/medium/channel.
b) Usually more secure and less affected to
interference and environmental factors like weather compared to unguided media.
c) Suitable for short to medium distances.
Examples:
a) Twisted Pair Cable: Used in telephone lines and
Ethernet cables.
b) Coaxial Cable: Used for cable TV and broadband
connections.
c) Optical Fiber: Used for high-speed internet and
long-distance communication.
Applications: Local area networks (LANs), wired telephony, and high-speed internet.
2. Unguided Media: Unguided media refers to wireless transmission where the signal travels
through free space (air, vacuum, or water) without a physical conductor.
Key Characteristics:
a) The signal is broadcast in all directions and can
be intercepted easily.
b) Usually less secure and more affected to
interference and environmental factors like weather.
c) Suitable for long distances or areas where laying
physical cables is impractical.
Examples:
a) Radio Waves: Used in AM/FM radio, television
broadcasts, and mobile phones.
b) Microwaves: Used in satellite communication and
point-to-point links.
c) Infrared and Laser Signals: Used in remote controls
and point-to-point communication in limited ranges.
Applications: Mobile networks, Wi-Fi, satellite communication, and broadcasting.
7. What is structure?
Create a structure to store the name and salary of 100 employees and display
the names of the employee getting salary between 10000 and 25000.
Ans:
A structure in C is a user-defined data type that allows
grouping variables of different types under a single name. This helps to
logically organize and manage related data. It is defined and declared using
the ‘struct’ keyword followed by the structure name and a set of curly braces.
For example:
struct Student
{
int id;
float marks;
} s1;
A structure to store the name and salary of 100 employees and
display the names of the employee getting salary between 10000 and 25000 is as
follows:
#include <stdio.h>
#include <conio.h>
struct staff
{
char
name[20];
int sal;
};
void main()
{
struct staff
s[100];
int i ;
clrscr();
for (i = 0;
i < 100; i++)
{
printf("Enter Name: ");
scanf("%19s", s[i].name);
printf("Enter Salary: ");
scanf("%d", &s[i].sal);
}
printf("\nEmployees
with salary between 10,000 and 25,000:\n");
printf("\nNAME\tSALARY");
for (i = 0;
i < 100; i++)
{
if
(s[i].sal > 10000 && s[i].sal < 25000)
{
printf("\n%s\t%d", s[i].name, s[i].sal);
}
}
getch();
}
OR
Write a C program to
get the name, address and grade of 15 students. Create a data file name
“std.txt” and store the provided details.
Ans:
#include <stdio.h>
#include <conio.h>
void main()
{
char name[50],
address[100], grade[10];
int i;
FILE *fptr;
clrscr();
fptr =
fopen("std.txt", "w");
if (fptr == NULL)
{
printf("Error
opening file!");
return;
}
for (i = 0; i < 15;
i++)
{
printf("Enter
name: ");
scanf("%s", name);
printf("Enter
address: ");
scanf("%s", address);
printf("Enter
grade: ");
scanf("%s", grade);
fprintf(fptr,
"%s\t%s\t%s\n", name, address, grade);
}
fclose(fptr);
fptr =
fopen("std.txt", "r");
if (fptr == NULL)
{
printf("Error
opening file!");
return;
}
printf("NAME\tADDRESS\tGRADE\n");
while (fscanf(fptr,
"%s %s %s", name, address, grade) != EOF)
{
printf("%s\t%s\t%s\n", name, address, grade);
}
fclose(fptr);
getch();
}
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