Plus Two Computer Application Previous Year Question Paper March 2019

Kerala State Board New Syllabus Plus Two Computer Application Previous Year Question Papers and Answers.

Kerala Plus Two Computer Application Previous Year Question Paper March 2019 with Answers

Board SCERT
Class Plus Two
Subject Computer Application
Category Plus Two Previous Year Question Papers

Time: 2 Hours
Cool off time : 15 Minutes

General Instructions to candidates

  • There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
  • Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
  • Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
  • Read questions carefully before you answering.
  • All questions are compulsory and only internal choice is allowed.
  • When you select a question, all the sub-questions must be answered from the same question itself.
  • Calculations, figures and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Part – A

Answer all questions from 1 to 5. Each carries 1 score. (5 × 1 = 5)

Question 1.
The input operator in C++ is ___________
Answer:
>> or cin >>

Question 2.
_________ character is stored at the end of the string.
Answer:
NULL or ‘\0’

Question 3.
The process of breaking large program into smaller sub-programs is called __________
Answer:
Modularization

Question 4.
Name the keyword used to declare variables in JavaScript.
Answer:
var

Question 5.
Expand MIS.
Answer:
Management Information System

Part – B

Answer any 9 questions from 6 to 16. Each carries 2 scores. (9 × 2 = 18)

Question 6.
List the type modifiers in C++.
Answer:
Type modifiers used in C++ are signed, unsigned, short and long.

Question 7.
Rewrite the following code using for loop:

int x = 1;
start:
cout<<x;
x = x + 5;
if (x < = 50)
goto start;

Answer:

for(x=1; x<=50; x+=5)
cout<<x;

Question 8.

  1. Define an Array.
  2. Initialize an integer array with 5 elements.

Answer:

  1. Array: An array is a collection of elements with the same data type store in contiguous memory location.
  2. int mark[] = {40, 42, 44, 46, 48, 50};

Question 9.
Write the port number for the following web services:

  1. Simple Mail Transfer Protocol.
  2. HTTP secure (HTTPS)

Answer:

  1. 25
  2. 443

Question 10.
What is the use of frame tag in HTML? What is its limitation?
Answer:
frame tag helps to view multiple web pages in a single window. The main limitation is that all browsers not supporting the frame tag.

Question 11.
Write the HTML code to display the following using list tag:
i) Biology Science
ii) Commerce
iii) Humanities
Answer:

<html>
<head>
<title>list demo
</title>
</head>
<body bgcolor="red">
<ol type="i">
<li> Biology Science</li>
<li> Commerce</li>
<li> Humanities</li>
</ol>
</body>
</html>

Question 12.
What is the difference between isNaN() and Number() functions in JavaScript?
Answer:
isNaN() function checks the given value is a number or not. If it is not a number it returns a true value otherwise false.
Number() function converts the data into numerical type.

Question 13.
What is CMS? Give two examples.
Answer:
CMS means Content Management System. It is a collection of programs that are used to create, modify, update, and publish website content. CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc. are examples of CMS.

Question 14.
First table containing 4 rows and 3 columns, the second table contains 5 rows and 2 columns, then the Cartesian product table contains ______ rows and ______ columns.
Answer:
The number of rows is the product of rows, i.e. 4 × 5 = 20 rows
The number of columns is the sum of columns, i.e. 3+2 = 5 columns

Question 15.
How Business Process Re-Engineering (BPR) is related to Enterprise Resource Planning (ERP)?
Answer:
ERP and BPR will not make much change if they are in stand-alone. To improve the efficiency of an enterprise integrate both ERP and BPR because they are the two sides of a coin. For better results conducting BPR before implementing ERP, will help an enterprise to avoid unnecessary modules from the software.

Question 16.
Define the terms:
i) Cyber Forensics
ii) Infomania
Answer:
i) Cyber Forensics: Critical evidence of a particular crime is available in electronic format with the help of computer forensics. It helps to identify the criminal with help of blood, skin or hair samples collected from the crime site. DNA, polygraph, finger prints are another effective tool to identify the accused person is a criminal or not.

ii) Infomania: Right information at the right time is considered as the key to success. The information must be gathered, stored, managed and processed well. Infomania is the excessive desire (Infatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application (WhatsApp) and Smart Phones. Due to this, the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired. They give first preference to the Internet than others. They create their own Cyber World and no interaction to the surroundings and the family.

Part – C

Answer any 9 questions from 17 to 27. Each carries 3 scores. (9 × 3 = 27)

Question 17.
Compare the selection statements ‘if’ and ‘switch’.
Answer:
Following are the difference between the switch and if-else if ladder.

  1. Switch can test only for equality but if can evaluate a relational or logical expression.
  2. If else is more versatile.
  3. If else can handle floating values but switch cannot
  4. If the test expression contains more variable if else is used.
  5. Testing a value against a set of constants switch is more efficient than if-else.

Question 18.
Write a program in C++ to accept a string with white space like “good morning” from the keyboard and display the same string.
Answer:

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[80];
cout<<"Enter a string:";
gets(str);
puts(str);
}

Question 19.
Compare static webpage and dynamic webpage.
Answer:

Static web pages Dynamic web pages
Content and layout is fixed Content and layout is changed frequently
Never use database Database is used
Run by the browser It runs on the server and the result gets back to the client(browser)
Easy to develop Not at all easy

Question 20.
i) What is the use of reserved characters for HTML entities? (1)
ii) List any four reserved characters and their use. (2)
Answer:
(i) HTML entities are used to print reserved characters in HTML.
(ii)

Character Entity Description
&nbsp; Nonbreakable space
&quot; Double quotes
&apos; Single quote
& &amp; Ampersand symbol
< &lt; Less than symbol
> &gt; Greater than symbol
© &copy; Copyright symbol
TM &trade; Trademark symbol
&reg; Registered symbol

Question 21.
Write the built-in JavaScript functions used for the following situation:

  1. Display warning message in the screen.
  2. Character at a particular position.
  3. Convert uppercase to lowercase.

Answer:

  1. alert()
  2. charAt()
  3. toLowerCase()

Question 22.
Write the merits and demerits of free Webhosting.
Answer:
The name implies it is free of cost service and the expense is met by the advertisements. Some service providers allow limited facility such as limited storage space, do not allow multimedia (audio and video) files.

Question 23.
What is the key? Explain any two keys in a relational database management system.
Answer:
Key is used to identify or distinguish a tuple in a relation.

  • Candidate key – It is used to uniquely identify the row.
  • Primary key – It is a set of one or more attributes used to uniquely identify a row.
  • Alternate key – A candidate key other than the primary key.
  • Foreign key – A single attribute or a set of attributes, which is a candidate key in another table is called a foreign key.

Question 24.
Define the term Data independence. Explain different levels of data independence.
Answer:
Data Independence – It is the ability to modify the schema definition in one level without affecting the scheme definition at the next higher level.

  • Physical Data Independence – It is the ability to modify the physical scheme without causing application programs to be rewritten.
  • Logical Data Independence – It is the ability to modify the logical scheme without causing application programs to be rewritten.

Question 25.
Explain any three situations to modify the structure of a table with the help of alter command in SQL.
Answer:
We can alter the table in two ways.
We can add a new column to the existing table using the following syntax,
ALTER TABLE <tablename>ADD(<cloumnname> <type> <constraint>);
We can also change or modify the existing column in terms of type or size using the following syntax,
ALTER TABLE<tablename>MODIFY(<column> <newtype>);

Question 26.
Explain the merits of ERP system.
Answer:
Benefits of ERP system
1. Improved resource utilization: Resources such as Men, Money, Material and Machine are utilized maximum hence increase productivity and profit.

2. Better customer satisfaction: Without spending more money and time all the customer’s needs are considered well. Because the customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through the Internet.

3. Provides accurate information: Right information at the right time will help the company to plan and manage the future cunningly. A company can increase or reduce production based upon the right information hence increase productivity and profit.

4. Decision-making capability: Right information at the right time will help the company to take a good decision.

5. Increased flexibility: A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Question 27.
Compare GPRS and EDGE.
Answer:
GPRS (General Packet Radio Services): It is a packet-oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically^fiarged based on the volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

EDGE (Enhanced Data rates for GSM Evolution): It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

Part – D

Answer any 2 questions from 28 to 30. Each carries 5 scores. (2 × 5 = 10)

Question 28.
Identify the built-in C++ function for the following cases:

  1. to convert -25 to 25.
  2. compare ‘computer’ and ‘COMPUTER’ ignoring cases.
  3. to check the given character is a digit or not.
  4. to convert the character from ‘B’ to ‘b’.
  5. to find the square root of 64 or a number.

Answer:

  1. abs()
  2. strcmpi()
  3. isdigit()
  4. tolower()
  5. sqrt()

Question 29.
(i) Write the name of the tag used to group related data in an HTML form. (1)
(ii) Write the HTML code to display the following webpage: (4)
Plus Two Computer Application Previous Year Question Paper March 2019 Q29
Answer:
(i) <fieldset> tag

(ii) <html>
<head>
<title>
login page
</title>
</head>
<BODY BGCOLOR="cyan">
<FORM NAME="frmlogin">
<center>
User Name
<input type="text" name="txtname">
<br><br>
Password
<input type="password" name="txtpass">
<br><br>
<input type="Submit" value="Submit">
<input type="Reset" value="Reset">
</center>
</FORM>
</body>
</html>

Question 30.
Consider the table student with attribute admno, Name, course, percentage. Write the SQL statements to do the following:

  1. Display all the student details. (1)
  2. Modify the course’Commerce1 to’Science1. (1)
  3. Remove the student details with a percentage below 35. (1)
  4. Create a view from the above table with a percentage greater than 90. (2)

Answer:

  1. select * from student;
  2. update student set course=”Science” where course=”Commerce”;
  3. delete from student where percentage<35;
  4. create view stud view as select * from student where percentage > 60;