Plus Two

Students can Download Chapter 1 Overview of Computerised Accounting System Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 1 Overview of Computerised Accounting System

Plus Two Accountancy Overview of Computerised Accounting System One Mark Questions and Answers

Question 1.
Accounting Packages are developed on the basis of
(a) Accounting concepts
(b) Accounting conventions
(c) Both Accounting concepts and Conventions
(d) None of the above
Answer:
(c) Both Accounting concepts and Conventions

Question 2.
What type of Software is an Accounting Package?
(a) System Software
(b) Application Software
(c) Utility Software
(d) Basic
Answer:
(b) Application Software

Question 3.
The components of computerised accounting system are
(a) Data, Report, Ledger, software, Hardware
(b) Software, Hardware, People, Procedure, Data
(c) Data, Coding, Procedure, Objective, Output
(d) People, Procedure, Hard ware, software
Answer:
(b) Software, Hardware, People, Procedure, Data

Question 4.
Grouping of Accounts means the classification of data from:
(a) Assets, Capital, and Liabilities
(b) Assets, Capital, Liabilities, Revenues & Expenses
(c) Assets, Owners equity, Revenue & Expenses
(d) Capital, Liabilities, Revenues, & Expenses
Answer:
(b) Assets, Capital, Liabilities, Revenues & Expenses

Question 5.
Codification of Accounts required for the purpose of:
(a) Hierarchical relationship between groups and components
(b) Data processing faster and preparing of final accounts
(c) Keeping data and information secured
(d) None of the above
Answer:
(a) Hierarchical relationship between groups and components

Question 6.
Method of codification should be
(a) Such that it leads to grouping of accounts
(b) An identification mark
(c) Easy to understand and leads to grouping of accounts
(d) None of the above
Answer:
(c) Easy to understand and leads to grouping of accounts

Question 7.
Final account subsystem in Accounting Information System (AIS) deals with
(a) Preparation of budgets
(b) Preparation of Pay Roll
(c) Preparation of Final Accounts
(d) None of the above
Answer:
(c) Preparation of Final Accounts

Question 8.
Pick the odd one out
(a) Password security
(b) Data Audit
(c) Data Bank
(d) Data vault
Answer:
(c) Data Bank

Question 9.
Which among the following is an example of mnemonic codes.
(a) AS03, AS04, AS05
(b) 1925, 1926, 1927, 1928
(c) ACC, ECO, ENG, MAL
(d) 001-100, 101-200, 201-300, 301-400
Answer:
(c) ACC, ECO, ENG, MAL

Question 10.
Which among the following deals with generation of reports that are vital for management decision making?
(a) Costing sub system
(b) Pay Roll Accounting Sub system
(c) Budget Sub System
(d) Management Information System
Answer:
(d) Management Information System

Question 11.
The need of codification is
(a) Easy to process data
(b) Keeping proper records
(c) The generation of block codes
(d) The encryption of data
Answer:
(d) The encryption of data

Question 12.
Choose the correct pair
(a) Cash and Bank sub system – Deals with receipts and payments of cash
(b) Inventory subsystem – deals with recording of sales
(c) Payroll Accounting sub system – deals with the preparation of final accounts
(d) Accounts receivable sub system – deals with expenses.
Answer:
(a) Cash and Bank sub system – Deals with receipts and payments of cash

Plus Two Accountancy Overview of Computerised Accounting System Two Mark Questions and Answers

Question 1.
Name the Five pillers of computerised Accounting System.
Answer:
Procedure, Data, people, Hardware and software

Question 2.
What is computerised accounting?
Answer:
Computerised accounting is the process of recording business transactions and generating financial statements and reports with the help of computer.

Question 3.
List down any Five popular accounting softwares available in India.
Answer:
GNUKhata, Tally, Dac Easy, Tata Ex, Peach Tree

Question 4.
Define Data
Answer:
Data is raw, unorganised facts that need to be processed. Data can be something simple and useless until it is organised.
When data is processed, organised, structured or presented in a given context so as to make it useful, it is called information. A computer is an information processing machine. Computers process data to produce information

Question 5.
Write down any two features of computerised accounting system
Answer:
Features of computerised accounting system:
1. Simple and Integrated:
Computerised accounting system is designed to integrate all the business operations such as sales, finance, purchase, etc.

2. Accuracy and speed:
Computrised Accounting system provides data entry forms for fast and accurate data entry of the transactions.

3. Scalability:
The system can cope easily with the increase in the volume of business transactions. The software can be used for any size and type of the organisation.

Question 6.
What are the account groups of Trading Account under Computerised Accounting System?
Answer:

• Sales account
• Purchase account
• Direct Expenses Account
• Direct Incomes Account

Question 7.
Complete the Table
Name of Account –  Account group in financial Account
Sundry Debtors –  Balance sheet, Asset side

1. Indirect Income __________________________
2. Current Liabilities ________________________
3. Stock in hand ____________________________
4. Sales Account ___________________________

Answer:

1. Profit & Loss Account, Credit side
2. Balances Sheet, Liability side
3. Balance Sheet, Asset side
4. Trading Account, Credit side

Question 8.
What do you mean by sequential code? Give example.
Answer:
Sequential codes:
Here numbers or alphabets are assigned in consecutive order. These codes are applied primarily to source documents such as cheques, invoices, etc.
Example:

Question 9.
What are the basic accounting reports prepared by Computerised Accounting System?
Answer:

• Day Book/ Journals
• Ledges
• Trial balance
• Profit and Loss Account
• Balance Sheet

Plus Two Accountancy Overview of Computerised Accounting System Three Mark Questions and Answers

Question 1.
A computerised accounting software is developed for a company. In Ledger Group, Expenses are divided in to direct expenses and indirect expenses. All the possible expenses of the company are listed under these two sub headings. Appropriate identification numbers are assigned to each such items.

1. Identify the concepts referred to this context
2. Give explanations to each.

Answer:

1. Grouping of Accounts and codification of Accounts
2. Grouping of accounts means classifying the ledger accounts and organizing them under major heads of accounts
   The process of assigning codes to account groups and ledger groups is called codification of accounts.

Question 2.
What are the advantages of computerised Accounting System?
Answer:

• Financial reports can be prepared in time
• Alterations and additions in transactions are easy and gives the changed result in all books of accounts instantly
• It ensures effective control over the system
• Economy in the processing of accounting data
• Confidentiality of data is maintained
• The closing balance of one financial year is automatically carried forward to next financial year

Question 3.
Create an accounting hierarchy for the following ledger account:

• Cash-in-hand
• Provisions for Tax
• Land and Buildings
• Bank Account (Current)
• Trade Investments
• Investments in Govt. Securities
• Deposits with Bank
• Duties and Taxes
• Sundry Debtors

Answer:

• Fixed assets: Land & Building Investment in Govt. Securities
• Current assets: Cash in hand Trade Investments
• Bank: Bank a/c (Current), Deposits with bank
• Sundry debtors: Sundry debtors
• Loans & advances: Provision for Tax
• Duties and Taxes: Duties of Taxes

Question 4.
What are the different types of codes commonly used in CAS?
Answer:
Sequential codes:
Here numbers or alphabets are assigned in consecutive order. These codes are applied primarily to source documents such as cheques, invoices, etc.
Example:

2. Block Codes:
In Block codes, a range of numbers is alloted to a particular account group. Here, numbers within a range follow sequential coding scheme.
Room Numbering System of a Lodge

Code	Account group (group of Legers)
100 – 199	Rooms in First Floor
200 – 299	Rooms in Second Floor
300 – 399	Rooms in third Floor

Coding of Dresses

Code	Account group (group of Legers)
HS400 – HS 499	Half Sleeve Shirts
FS500 – FS 599	Full Sleeve Shirts
MI 600 – MI 699	Mens Inner Wares

3. Mnemonic codes:
A mnemonic code consists of alphabets or abbreviations as symbols to codify an account.
1.

Code	Name of Accounts
1) SLR	Salary Account
2) BOD	Bank Overdraft
3) INV	Inventory

2.

Code	Name of Places
1) TSR	Thrissur
2) TVM	Thiruvananthapuram
3) DLH	Delhi

Question 5.
Distinguish between Data Audit and Data Vault
Answer:
Security Features of Computerised Accounting Software
1. Data Audit:
It enables one to know as to who and what changes have been made in the original data there by helping and fixing the responsibility of the person who has manipulated the data and ensures data integrity.

2. Data Vault:
Software provides additional security through data encryption. Encryption means scrambling the data so as to make its interpretation impossible.

Question 6.
What is the activity sequence of computerised Accounting information processing model?
Answer:

• Collect data
• Organise data
• Communicate Accounting Information

Question 7.
Akhil, your class mate, argues that computerised Accounting System suffers some draw

1. Do you agree with this?
2. Give your justification

Answer:

1. I agree
2. Limitations of CAS.
   • Faster Obsolescence of technology necessitates frequently upgradation in accounting software.
   • Data may be lost or corrupted due to power interruption.
   • Un programmed reports can not be generated
   • Alterations in transactions are easy. This reduces the reliability of accounting work.
   • Work with CAS is expensive.

 

Question 8.
What are the different components of computerised Accounting System?
Answer:
Components of computerised Accounting System:

• Procedure: A logical sequence of actions to perform a task.
• Data: The raw fact for any business operation.
• People: Users of computerised accounting system
• Hardware: The physical components of a computer.
• Software: A set of programmes to do a work.

Plus Two Accountancy Overview of Computerised Accounting System Four Mark Questions and Answers

Question 1.
In the computerised accounting package used by a trading concern, the fixed assets and current assets are combined to form the assets group. The fixed assets are sub divided into tangible and intangible assets.
Show the hierarchy of this ledger group, assuming various assets items.
Answer:

Question 2.
Illustrate with example methodology to develop coding structure.
Answer:
Methodology to Develop Coding Structure:
Let us examine the 15 digit Goods and Services Tax Identification Number (GSTIN). GSTIN is a 15 digit unique code that is assigned to each taxpayer, which will be state-wise and PAN-based.

GSTIN Format or Structure (GSTIN):

From the above details, we can identify that GSTIN is a combination of state code, PAN, number of registration within the state, default digit and check code to detect errors.

Another Example:
Let us examine how to develop a coding structure for students in Thrissur district under DHSE. The first step is to develop a hierarchy of the school system and attributes

Thus we allocate 13 digit code to a student.

Question 3.
What do you mean by AIS? What are the different subsystems of AIS?
Answer:
Accounting Information System (AIS):
Accounting Information System (AIS) and its various subsystems may be implemented through the Computerised Accounting System (CAS). Such system of AIS are described below.

1. Cash and Bank subsystem:
Receipts and payments of cash

2. Sales and Accounts Receivable sub system:
Maintaining of sales and Receivables ledgers.

3. Inventory sub system:
Purchase and sale of goods, Specifying the price, quantity, and date.

4. Purchase and Accounts Payable Sub system:
Maintaining of purchase and payable leadgers.

5. Pav Roll Accounting sub System:
Payment of salaries and wages.

6. Fixed Assets Accounting Sub system:
Purcahses, additions, sale and usage of fixed assets.

7. Expense Accounting sub system:
Various types of expenses.

8. Tax Accounting Sub system:
Deals with GSTIN, Income Tax etc.

9. Final Accounts sub system:
Preparation of final accounts.

10. Costing sub system:
Ascertainment of cost of goods produced.

11. Budget sub system:
Preparation of budgets.

12. Management information sub system (MIS):
Preparation of reports that are vital for management decision making.

Question 4.
What is DATA Encryption? What are the security features of CAS?
Answer:
Security Features of Computerised Accounting Software:
Every accounting software ensures data security, safety, and confidentiality by providing the features like Password Security, Data Audit and Data Vault.

1. Password Security:
Password is the key to allow the access to the system. Computerised accounting system protects the unauthorised persons from accessing to the business data. Only authorised person, who is supplied with the password, can enter to the system.

2. Data Audit:
It enables one to know as to who and what changes have been made in the original data there by helping and fixing the responsibility of the person who has manipulated the data and ensures data integrity.

3. Data Vault:
Software provides additional security through data encryption. Encryption means scrambling the data so as to make its interpretation impossible.

Question 5.
Draw a chart showing different sub systems of Accounting information system.
Answer:
Accounting Information System (AIS):
Accounting Information System (AIS) and its various sub-systems may be implemented through Compterised Accounting System (CAS). Such system of AIS are described below.

1. Cash and Bank subsystem:
Receipts and payments of cash

2. Sales and Accounts Receivable sub system:
Maintaining of sales and Receivables ledgers.

3. Inventory sub system:
Purchase and sale of goods, Specifying the price, quantity, and date.

4. Purchase and Accounts Payable Sub system:
Maintaing of purchase and payable leadgers.

5. Pav Roll Accounting sub System:
Payment of salaries and wages.

6. Fixed Assets Accounting Sub system:
Purcahses, additions, sale and usage of fixed assets.

7. Expense Accounting sub system:
Various types of expenses.

8. Tax Accounting Sub system:
Deals with GSTIN, Income Tax, etc.

9. Final Accounts sub system:
Preparation of final accounts.

10. Costing sub system:
Ascertainment of cost of goods produced.

11. Budget sub system:
Preparation of budgets.

12. Management information sub system (MIS):
Preparation of reports that are vital for management decision making.

Question 6.
Differentiate between Data and Information.
Answer:

Data	Information
1. It is used as input	1. It is the output of processed data
2. Data is the raw material	2. Information is the product
3. It doesn’t carry a meaning	3. It must carry a logical meaning
4. It is an independent value	4. Information depends on data

Students can Download Chapter 9 Structured Query Language Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 9 Structured Query Language

Plus Two Computer Science Structured Query Language One Mark Questions and Answers

Question 1.
………………. form of SQL is designed for use within general purpose programming languages such as COBOL, C, etc.
Answer:
Embedded SQL.

Question 2.
What is TCL?
Answer:
Transaction Control Language- component of SQL includes commands for specifying transactions.

Question 3.
Data dictionary is a special file in DBMS. What is it used for?
Answer:
Table details are stored in this file.

Question 4.
What does the following statement mean?
Name VARCHAR(30)
Answer:
Field Name can store up to 30 characters. It is a column definition.

Question 5.
Is there any data type available in SQL to store your date of birth information?
Answer:
Yes. DATE data type

Question 6.
Pick the odd one out.
(DEC, NUMBER, INT, DATE)
Answer:
DATE.

Question 7.
How do you ensure that the field ‘name’ will have some value always?
Answer:
using the constraint NOT NULL.

Question 8.
How to set the default value of column District in a table to Thrissur?
Answer:
District VARCHAR(30) DEFAULT ‘Thrissur’

Question 9.
…………… symbol is used as substitution operator in SQL.
Answer:
&

Question 10.
Is there any method to find the strings starting with letter ‘a’ from a field in SQL?
Answer:
Use LIKE operator, LIKE ‘a%’

Question 11.
How to check whether a particular field contains null values or not?
Answer:
Use operator IS NULL

Question 12.
‘ORDER BY’ clause is used for …………
Answer:
Sorting the results of a query by ascending(Asc) or Descending (Desc)

Question 13.
The built-in functions in SQL that return just a single value for a group of rows in a table are called ………..
Answer:
Summary functions or aggregate functions

Question 14.
How to find the number of values in a column in the table?
Answer:
Using the function COUNT()

Question 15.
………………. is the clause in SQL used for categorization.
Answer:
GROUP BY

Question 16.
Thomas wants to remove a table that he were using. How could you help him doing this?
Answer:
Using DROP TABLE command.

Question 17.
Pick the odd one out.
(SELECT, UPDATE, DELETE, DROP TABLE)
Answer:
DROP TABLE

Question 18.
Name the aggregate function that can be used to find the total number of records.
Answer:
COUNT()

Question 19.
Which of the following is an essential clause used with SELECT command?
(GROUP BY, ORDER BY, WHERE, FROM)
Answer:
FROM

Question 20.
Which of the following is not a column constraint?
(CHECK, DISTINCT, UNIQUE, DEFAULT)
Answer:
DISTINCT

Question 21.
Which of the following is a DDL command?
(SELECT, UPDATE, CREATE TABLE, INSERT INTO)
Answer:
CREATE TABLE

Question 22.
What are the logical operators used in SQL?
Answer:
And, Or, Not.

Question 23.
The …………. operator pf SQL is used to match a pattern with the help of %. February 2009
(a) BETWEEN
(b) AND
(c) LIKE
(d) OR
Answer:
(c) LIKE

Question 24.
The structure of a table Book is given below.

BookNo	Integer
Title	Varchar (200)
Author	Varchar(100)
Price	Dec{5, 2)

Write SQL query to Insert an additional column Purchase date of date type to the Books table.
Answer:
Alter table Book add(PurchaseDate Date);

Question 25.
Suppose we want to include a column in a table in which serial numbers are to be stored automatically on adding new records. Which constraint is to be used for that column during table creation?
Answer:
The constraint Autojncrement is used.

Question 26.
Which of the following cannot be used to name a table in SQL? Give the reason.
Answer:
(a) Studnt50
(b) Table
(c) $Employee
(d) Stock_123
Answer:
(b) Table. This is a keyword hence it cannot be used.

Question 27.
Which of the following commands is used to view the structure of a table?
(a) SHOW TABLES
(b) DESC
(C) SELECT
(d) DISPLAY
Answer:
(b) DESC

Question 28.
The command to eliminate the table CUSTOMER from a database is:
(a) REMOVE TABLE CUSTOMER
(b) DROP TABLE CUSTOMER
(c) DELETE TABLE CUSTOMER
(d) UPDATE TABLE CUSTOMER
Answer:
(b) DROP TABLE CUSTOMER

Question 29.
Which SQL command is used to open a database?
(a) OPEN
(b) SHOW
(c) USE
(d) CREATE
Answer:
(c) USE

Question 30.
Which is the keyword used with SELECT command to avoid duplication of rows in the selection?
Answer:
DISTINCT

Question 31.
Pick odd one out and write reason:
(a) WHERE
(b) ORDER BY
(c) UPDATE
(d) GROUP BY
Answer:
UPDATE. It is a command and others are clauses.

Question 32.
Which of the following clause is not used with SELECT command in SQL?
(a) GROUP BY
(b) WHERE
(c) SET
(d) ORDER BY
Answer:
(c) SET. This clause is used with UPDATE.

Question 33.
……………… operator in SQL is used with wild card characters for selection of records.
(a) LIKE
(b) IN
(c) NOT IN
(d) IN and NOT IN
Answer:
(a) LIKE

Plus Two Computer Science Structured Query Language Two Mark Questions and Answers

Question 1.
How is SQL different from other computer high-level languages?
Answer:
SQL means Structured Query Language. It is a relational database language, not a programming language like other high-level languages. It provides facilities to create a table, insert data into a table, retrieve information from a table, modify data in the table, etc.

Question 2.
Distinguish the SQL keywords UNIQUE and DISTINCT.
Answer:
1. Unique – It ensures that no two rows have the same value in a column while storing data. It is used with create command.

2. Distinct – This keyword is used to avoid duplicate values in a column of a table while retrieving data. It is used with select command.

Question 3.
Identify errors in the following SQL statement and rewrite it correctly. Underline the corrections.
CREATE student TABLE (admno PRIMARY KEY, roll no INT, name CHAR);
Answer:
The correct SQL statement is as follows.
CREATE TABLE student( admno INT PRIMARY KEY, roll_no INT, name CHAR(3));

Question 4.
Suppose a column named Fee does not contain any value for some records in the table named STUDENT. Write SQL statements to fill these blanks with 1000.
Answer:
UPDATE STUDENT SET Fee = 1000 WHERE Fee IS NULL;

Question 5.
Identify the errors in the following SQL statement and give reason for the error.
SELECT FROM STUDENT ORDER BY Group WHERE Marks above 50;
Answer:
In this query Group is the keyword hence it cannot be used. The correct query is as follows.
SELECT * FROM STUDENT WHERE Marks > 50 ORDER BY Marks;

Question 6.
Differentiate CHAR and VARCHAR data types of SQL.
Answer:

1. Char – It is used to store fixed number of characters. It is declared as char(size).
2. Varchar – It is used to store characters but it uses only enough memory.

Question 7.
Assume that CUSTOMER is a table with columns Cust_code, Cust_name, Mob_No and Email. Write an SQL statement to add the details of a customer who has no e-mail id.
Answer:
INSERT INTO CUSTOMER VALUES(1001, ‘ALVIS’, 9447024365, NULL);

Question 8.
Find the correct clause from the 2nd column for each SQL command in the 1_st column.

Answer:

Question 9.
Is ALL a keyword in SQL? Explain
Answer:
Yes, ALL retains all the duplicate values of a field

Question 10.
Differentiate LIKE ‘a%’ and LiKE ‘a_’
Answer:
%, it replaces a string but _ replaces only one character at a time.LIKE ‘a%’ retrieve all strings of any length that start with letter ‘a’ while LIKE ‘a_’ retrieve all two letter strings that start with letter ‘a’.

Question 11.
Is multiple sorting possible using ORDER BY clause? Explain
Answer:
Yes, by giving multiple field names separated by comma in the ORDER BY clause.

Question 12.
While inserting records into a table, Raju finds it is not possible to give more than 20 characters in the ‘name’ field. How can you help Raju solve this problem?
Answer:
Using ALTER TABLE command he can modify the width of field so that it can accommodate more than 20 characters.

Question 13.
After executing a query Mohan gets a message like ‘Table Altered’. What would have he done? Give the syntax.
Answer:
He would have used the ALTER TABLE command.
Syntax: ALTER TABLE ADD/MODIFY ();

Question 14.
During discussion one student said that DELETE and DROP TABLE commands are same. Do you agree with that? Justify.
Answer:
No, DROP TABLE removes a table entirely from the database whereas DELETE command deletes only records from an existing table.

Question 15.
1. Pick odd one out.
DROP TABLE, DELETE, ALTER TABLE, CREATE VIEW
2. Justify your answer
Answer:

1. DELETE
2. All others are DDL commands

Question 16.

1. Pick odd one out from the following.
2. Justify your answer.

Answer:

1. Group By
2. Group By is not a Constraint.

Question 17.
Consider the following SQL statements.
DELETE Name.
From Student
Where name = ‘Raju”
Find the error in the SQL if any and correct it.
Answer:
Delete from student where name = ‘Raju’

Question 18.
Consider the following Query in SQL.
From Employee Where branch = ‘Kannur’ Group By Department Having avg (salary) > 7000, Compare the ‘where’ clause and ‘Having’ Clause using the above query.
Answer:
‘where’ clause applies on single rows but Having clause applies on a group.

Question 19.
Some constraints in SQL are called column constraints. Some constraints are called table constraints. How do they differ?
Answer:
Column constraints are specified while defining each column, table constraints are specified once for the entire table at the end of table definition

Question 20.
1. Choose the odd one from the following.
Primary key, Unique, Distinct, Default, Check
2. Justify your choice
Answer:

1. Distinct
2. Distinct is used with select command while others are column constraints of create table command.

Question 21.
Name the most appropriate SQL data type required to store the following data.

1. Name of a student
2. Date of Birth of a student.
3. RollNo. of a student (in the range 1 to 50)
4. Percentage of marks obtained

Answer:

1. Varchar(70)
2. Date
3. Smallint OR Integer Or decimal
4. Dec(5, 2)

Question 22.
1. From the list given below select the names that cannot be used as a table name.
Adm_No, Date, Salary2006, Table, Column_Name, Address.
2. Justify your selection.
Answer:

1. Date, Table
2. There is a data type Date and Table is a keyword used to create a table. So these two are not used.

Question 23.
Classify the following SQL elements into two and give proper title for each category.
NOT NULL, AVG, COUNT, CHECK, SUM, DEFAULT
Answer:

Plus Two Computer Science Structured Query Language Three Mark Questions and Answers

Question 1.
As a part of your school project, you are asked to create a table Student with the fields RollNo, Name, Date of Birth and Score in IT. The constraints required are RollNo. is the primary key. Name cannot be empty.
Answer:
Create table student(RollNo decimal(2) not null primary key, Name varchar(20) not null, DOB date, Score number(2));

Question 2.
A table Employee consists of fields EmployeeNo, Name, Designation and Salary.
Consider that you are forced to give access to this table for an engineer from another company. But for security reasons you need to hide Salary from him. Employee No, Name, Designation, Salary.

1. Name the concept that provides this engineer, a facility to work on this table without viewing salary. Salary view
2. Write SQL query for implementing this SQL query

Answer:

1. Views
2. Create view Empview as select employeeno, Name, Designation from Employee.

Question 3.
Explain the keywords in the following query.
Answer:
SELECT allows to retrieve a subset of rows from the table, DISTINCT avoids duplication of courses from the table STUDENT, FROM is a keyword used to specify the name of the table

Question 4.
A table named student is given below.

Write answers for the questions based on the above table.

1. SQL statement to display the different courses available without duplication.
2. SQL statement to display the Name and Batch of the students whose percentage has a null value.
3. Output of the query select count (percentage) from Student.

Answer:

1. Select distinct batch from Student;
2. Select name, batch from Student where percent is null
3. 5

Question 5.
Once the creation of a table is over, one can perform two changes in the schema of the table. What are they? Give syntax.
Answer:
We can alter the table in two ways.
1. We can add a new column to the existing table using the following
syntax: ALTER TABLE ADD( );

2. We can also change or modify the existing column in terms of type or size using the following
syntax: ALTER TABLEMODIFY( );

Question 6.
Explain how pattern matching can be done in SQL with an example.
Answer:
pattern matching can be done using the operator LIKE while setting the condition with pattern matching. for eg., to display the names of all students whose name begins with letters ‘ma’, we can write the following query, SELECT name FROM STUDENT WHERE name LIKE ‘ma%’; here the character ‘%’ – substitutes any number of characters in the value of the specified column. Another character substitutes only one character of the specified column.

Question 7.
During the discussion of study your friend say that table and view are the same. How can you correct him?
Answer:
Tables and views are different. A view is a single virtual table that is derived from other physically existing tables. When we access a view we actually access the base tables.

We can use all the DML commands with the views but care should be taken as operation actually reflects in the base tables. The advantage of view is that without sparing extra storage space, we can use same table as different virtual tables. It also implements sharing along with privacy.

Question 8.
Find the errors if any of the following code.
Create table emp (name char, RolINo int(20));
Answer:
Here the argument size of data type char of field name is missing. So we can store a character only. The second error is the data type int has no argument. The correct statement is as follows Create table emp (name char(20), RolINo int);

Question 9.
Consider the following variable Declaration in SQL.
a. name char (25)
b. name Varchar (25)

1. Considering the utilisation of memory, which variable declaration is more suitable.
2. Justify your answer

Answer:
1. name varchar (25)
2. Because char data type is fixed length. It allocates maximum memory i.e, here it allocates memory for 25 characters maybe there is a chance of memory wastage. But Varchar allocates only enough memory to store the actual size.

Question 10.
Mr. Dilip wants to construct a table and implement some restrictions on the table.

1. Column can never have empty values.
2. One of the columns must be a key to identify the rows etc.

Can you help him to create that table satisfying the above restrictions with an example
Answer:

1. Use of create table, Not NULL.
2. Primary key Eg: Create table employee(RollNo decimal(3) not null primary key,Name Varchar(70) not null, Design Varchar(30), DOB Date,Salary Dec(7,2));

Question 11.
Write SQL query to construct a table student with fields Reg No, Stud Name, Sex, Course, grade, etc. As per the following conditions

1. Reg No should not be empty and using this column any rows in that table can be identified. Reg. No.
2. Student name should have some value.
3. Default value of sex be Female.
4. Grade should be any of the values: A+, A, B+, B, C+, C, D+, D, E

Answer:
Create table Student(RollNo decimal(3) not null primary key, Name Varchar(70) not null, sex char default ‘Female’,Grade char(2));

Question 12.
Ramu create Table employee with fields empId, empname, Designation, salary using SQL statements. Later he found that data type of emPId is typed as Integer instead of character and missed a field ‘Department’. Can you help him to solve this problem without recreating ‘employee’ table?
Answer:
Alter Table Employee Modify(Empld char(4)); AlterTable Employee Add(Dept char(15));

Question 13.
While creating a table Alvis give “Emp Details” for table name. Is it Possible? Write down the rules for naming a Table.
Answer:
It is not possible because there is a space between Emp and Details. The rules are given below:

1. It must not be a keyword(Keywords are reserved words and have predefined meaning)
2. It must begin with alphabets
3. Digits can be used followed by alphabets
4. Special characters cannot be used except underscore
5. We cannot give the name of another table

Plus Two Computer Science Structured Query Language Five Mark Questions and Answers

Question 1.
Consider the table ITEMS.

a) SELECT ITEMCODE, NAME FROM ITEMS WHERE CATEGORY = ‘Stationery’;
b) SELECT * FROM ITEMS WHERE SALES_ PRICE < UNIT_PRICE;
c) SELECT CATEGORY, COUNT(*) FROM ITEMS GROUP BY CATEGORY;
Answer:

Question 2.
Write SQL queries to SQL Query to
1. Create a table Employee with the fields given below.

2. List the name of all employees whose name’s second letter is ‘a’.
3. List the Name and Designation of employees whose Designation is not ‘Manager’
4. Increase the salary of ail employees by 10 percent.
5. Remove all managers whose salary is less than Rs. 10,000 from the table.
Answer:

1. Create table employee(Name Varchar(70), Desgn Varchar(30), DOB Date, Salary Dec(7,2));
2. Select Name from Employee where Name like
3. Select Name, Desgn from Employee where Desg no’Manager’;
4. Update Employee set Salary = Salary + Salary * .01;
5. Delete from Employee where Desgn = ‘Manager’ and Salary < 10000;

Question 3.
A table Student consists of fields Roll No, Name, Batch and Percent. Write SQL statements to

1. Display RolINo and Name of students whose percentage is Iessthan90 and greater than 70.
2. Display RolINo and Name of all students in science batch whose percentage is more than 90.
3. Display Names of all students in commerce and science batches.
4. Display the average Percent of students in each batch.
5. Display RolINo and Name in the ascending order of Batch and descending order of Percent.

Answer:

1. Select RollNo, Name from Student where percent < 90 and percent > 70;
2. Select RollNo, Name from Student where Batch = ’Science’ and percent > 90;
3. Select Name from Student where batch = ’Science’ Or ’batchCommerce’;
4. Select batch, Avg(percent) from Student group by batch;
5. Select RollNo, Name from Student order by batch, percent desc;

Question 4.
1. Classify the following SQL commands.
Create table, Insert Into, AlterTable, Delete, Up-date, Drop Table, Select.
2. List the features of each category.
Answer:
1. DDL – Create Table, Alter Table, Drop Table.
DML- Insert Into, Delete, Update, Select.

2. DDL – DDL means Data Definition Language. It is used to create the structure of a table, modify the structure of a table and delete the structure of a table. DML -DML means Data Manipulation Language. It is used to insert records into a table, modify the records of a table, delete the records of a table and retrieve the records from a table.

Question 5.
Explain the available database integrity constraints.
Answer:
1. NOT NULL: it specifies that a column can never have null values, i.e., not empty

2. UNIQUE: it ensures that no two rows have same value in the specified column.

3. PRIMARY KEY: it declares a column or a set of columns as the primary key of the table. This constraint makes a column NOT NULL and UNIQUE.

4. DEFAULT: it sets a default value for a column when the user does not enter a value for that column.

5. Auto_increment: This constraint is used to perform auto increment the values in a column. That automatically generates serial numbers. Only one auto-increment column per table is allowed.

Question 6.
Consider a table student with fields Reg No, Stud Name, Sex, Course, total score.
Write SQL queries for the following:
RegNo, Studname, Sex, Course, Total score.

1. Enter a Record
2. List the Detail’s of all students
3. Display Details of the student whose name ends with ‘Kumar’.
4. List all Female students who got more than 50 marks.
5. List all students who are studying either science or in Commerce group.
6. List details of those students who are studying Humanities in the Descending order of their names.

Answer:

1. Insert into Student values (52, ‘JOSE’, ‘Male’, ‘Science’, 500);
2. Select*from Student;
3. Select* from Student where name like ’%Kumar’;
4. Select * from Student where sex=’Female’ and Total >50;
5. Select * from Student where course = ‘science’ or course =‘commerce’;
6. Select* from Student where course-‘Humanities’ order by name desc;

Question 7.
Construct a Table Product with the following fields using SQL.
Product code – Consist of Alphanumeric code.
Product name – Consist of maximum of 30 Alphabets
Unit price – Numeric values
Quantity – Numeric values
Product price – Numeric Values
Write Query for the following:

1. Enter 5 records in the table (not give values for product prices)
2. Calculate product price (unit price × quantity)
3. List all product whose unit price ranging from 10 to 20
4. Calculate total price of all products.

Answer:
Create table Product(ProdCode Varchar(20) not null primary key, ProdName char(30), UPrice decimal(7,2), Qty decimal(6), ProdPrice decimal(7,2));

1. Insert into Product(Product Code,Product Name, Unit Price, Quantity) values (‘101 ’, ‘LUX’, 29.50,500); Similarly insert four more records
2. Update Product set ProdPrice=Qty * UPrice;
3. Select ProdName from Product where UPrice be-tween 10 and 20;
4. Select sum(ProdPrice) from Product;

Question 8.
Mr. Wilson wants to store the details of students. The details consist of different types of data. Explain different data types used in SQL to store data.
Answer:
The different data types are
1. Char (Fixed):
It is declared as char (size). This data type is used to store alphanumeric characters. We have to specify the size. If no size is specified, by default we can store only one character. Eg: name char(20)

2. Variable Character:
It is declared as varchar (size). This data type is also used to store alphanumeric characters. But there is a slight difference. lt allocates only enough memory to store the actual size.
Eg: name varchar(20)

3. Decimal:
It is declared as Dec(size, scale), where size is the number of digits and scale is the maximum number of digits to the right of the decimal point.
Eg: weight dec(3, 2)

4. Integer:
It is declared as int. It does not have any arguments. It takes more memory. Eg: ROllNO int.

5. Small Integer:
It is declared as smallint. It takes less memory ROllNO int.

6. Date:
It is used to store date. Eg: DOB date.

7. Time:
It is used to store time. Eg: Joining_Time Time.

Question 9.
A company wants to create a table to store its employees details. Write SQL Commands for the following :

1. Create a table with EMP table having fields EMPNO primary key varchar(10), Name varchar(20), Salary number(6), Department varchar(3)
2. Insert values to table
3. List all employees whose salary > 10,000 Salary 10,000
4. Display name and salary in the order of name.

Answer:

1. Create table EMP (EMPNO varchar(10) not null primary key, Name varchar(20), Salary decimal(6), Deptvarchar(3));
2. Insert into EMP values (‘1001 ’, ‘ALVIS’, 50000, ‘Sales’);
3. Select * from EMP where salary > 10000;
4. Select name, salary from EMP order by name;

Question 10.
Create a table ‘Savings’ with the following fields. Acc No (Integer), Name (char), age (Integer) and balance (Number) where Acc No is the primary key. Write SQL commands for the following.

1. Insert data in all the fields for 3 records inserted
2. Display the list of account holders in the ascending order Of their names.
3. Display the list of all account holders having age between 20 and 30
4. Display the name and Acc No of customers having a balance > 10 lakhs

Answer:
Create table Savings(Accno decimal(4) not null primary key, name char(2), age decimal(3), Balance decimal(8, 2));

1. Insert into savings values (501, ‘Andrea’, 18,45000)
2. Select * from Savings order by name
3. Select * from Savings where age between 20 and 30
4. Select name.accno from Savings where balance >100000

Question 11.
Which are the components of SQL? How do they help to manage database?
Answer:
The components of SQL are given below.
a. DDL commands ( 3 commands):

1. Create table – Used to create a table.
2. Alter table – Used to modify existing column or add new column to an existing table. There are 2 keywords used ADD and MODIFY.
3. Drop table – Used to remove a table from the memory.

b. DML commands ( 4 commands):

1. Select – Used to select rows from a table. The keyword From is used with this. Where clause is used to specify the condition.
2. Insert – Used to insert new records into a table. So the keyword used is INTO.
3. Delete – Used to delete records in a table.
4. Update – Used to modify the records in a table the keyword used is set.

c. DCL (Data Control Language) commands:

1. Grant – It grants permission to the users to the database
2. Revoke – It withdraws user’s rights given by using Grant command

Students can Download Chapter 10 Vector Algebra Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 10 Vector Algebra

Plus Two Maths Vector Algebra Three Mark Questions and Answers

Question 1.
Find \(\bar{a}+\bar{b}, \bar{a}-\bar{b}\) and \(\bar{b}+\bar{c}\) using the vectors.
\(\bar{a}\) = 3i + 4j + k, \(\bar{b}\) = 2i – 7 j – 3k and \(\bar{c}\) = 2i + 3j – 9k.
Answer:
\(\bar{a}+\bar{b}\) = 3i + 4j + k + 2i – 7j – 3k = 5i – 3j – 2k
\(\bar{a}-\bar{b}\) = 3i + 4j + k – (2i – 7j -3k) = i + 11j + 4k
\(\bar{b}+\bar{c}\) = 2i – 7j -3k + 2i +3j – 9k
= 4i – 4j – 12k.

Question 2.

1. Find the vector passing through the point A( 1, 2, -3) and B(-1, -2, 1).
2. Find the direction cosines along AB.

Answer:
1. \(\overline{A B}\) = \(\overline{O B}\) – \(\overline{O A}\) = -i – 2j + k – (i + 2j – 3k) = -2i – 4j + 4k.

2. Unit Vector

Direction cosines are \(\frac{-2}{6}\), \(\frac{-4}{6}\), \(\frac{4}{6}\).

Question 3.
Show that the points A, B and C with position vectors \(\bar{a}\) = 3i – 4j – 4k, \(\bar{b}\) = 2i – j + k and \(\bar{c}\) = i – 3j – 5k respectively from the vertices of a right angled triangle.
Answer:

41 = 35 + 6 ⇒ BC² = AB² + CA²
Hence right angled triangle.

Question 4.
Prove that \([\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}+\bar{a}]=2[\bar{a} \bar{b} \bar{c}]\).
Answer:
LHS

Note: If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar then so is \([\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}+\bar{a}]\).

Question 5.
Consider the vector \(\bar{p}\) = 2i – j + k. Find two vectors \(\bar{q}\) and \(\bar{r}\) such that \(\bar{p}\), \(\bar{q}\) and \(\bar{r}\) are mutually perpendicular.
Answer:
Find a vector \(\bar{q}\) such that \(\bar{p} \cdot \bar{q}\) = 0, for this use any \(\bar{q}\) whose two components are randomly selected. Let \(\bar{q}\) = 2i + 2j + xk
\(\bar{p} \cdot \bar{q}\) = (2i – j + k) . (2i + 2 j + xk) = 0
⇒ 4 – 2 + x = 0 ⇒ x = -2

= 6j + 6k.

Question 6.

Answer:

= i(-12 + 7) – j(-9 – 2) + k(-21 – 8)
= -5i + 11j – 29k

= i(63 + 9) – j(-18 + 6) + k(6 – 14)
= 72i + 12 j – 8k.

Question 7.
If \(\bar{a}\) = 3i + j + 2k,
(i) Find the magnitude of \(\bar{a}\). (1)
(ii) If the projection of \(\bar{a}\) on another vector \(\bar{b}\) is \(\sqrt{14}\), which among the following could be \(\bar{b}\) ? (1)
(a) i + j + k
(b) 6i + 2j + 4k
(c) 3i – j + 2k
(d) 2i + 3j + k
(iii) If \(\bar{a}\) makes an angle 60° with a vector \(\bar{c}\), find the projection of \(\bar{a}\) on \(\bar{c}\) (1)
Answer:

(ii) Since projection of \(\bar{a}\) on another vector \(\bar{b}\) and magnitude of \(\bar{a}\) is \(\sqrt{14}\), then \(\bar{a}\) and \(\bar{b}\) are parallel, (b) 6i + 2j + 4k.

(iii) Projection of \(\bar{a}\) on \(\bar{c}\)
= |\(\bar{a}\)|cos60° = \(\sqrt{14}\) × \(\frac{1}{2}\) = \(\frac{\sqrt{14}}{2}\).

Question 8.
(i) The projection of the vector 2i + 3j + 2k on the vector i + j + k is (1)

(ii) Find the area of a parallelogram whose adjacent sides are the vectors 2i + j + k and 6i – j (2)
Answer:

(ii) Let \(\bar{a}\) = 2i + j + k, \(\bar{b}\) = i – j

= i(0 + 1) – j(0 – 1) + k(-2 – 1 ) = i + j -3k
Area =

Question 9.
(i) The angle between the vectors i + j and j + k is (1)
(a) 60°
(b) 30°
(c) 45°
(d) 90°

Answer:
(i) (a) 60°

Question 10.

Answer:

(ii) Given, \(\bar{a}\) + \(\bar{b}\) + \(\bar{a}\) = \(\bar{0}\), squaring both sides we get

Plus Two Maths Vector Algebra Four Mark Questions and Answers

Question 1.
Let A (2, 3), B (1, 4), C (0, -2) and D (x, y) are vertices of a parallelogram ABCD.

1. Write the position vectors A, B, C and D. (2)
2. Find the value of x and y. (2)

Answer:
1. Position vector of A = 2i + 3 j
Position vector of B = i + 4j
Position vector of C = 0i – 2j
Position vector of D = xi + yj.

2. Since ABCD is a parallelogram, then

(1) ⇒ -i + j = -xi – (y + 2 )j
x = 1, -2 – y = 1 ⇒ y = -3
∴ D is (1, -3).

Question 2.
Find the position vector of a point R which divides the line joining the two points P and Q whose vectors i + 2j – k and -i + j + k in the ratio 2:1

1. internally and
2. externally.

Answer:
\(\overline{O P}\) = i + 2j – k, \(\overline{O Q}\) = -i + j + k
Let R be the position vector of the dividing point,
1.

2.

Question 3.
(i) Choose the correct answer from the bracket. If a unit vector \(\widehat{a}\) makes angles \(\frac{\pi}{4}\) with i and \(\frac{\pi}{3}\) with j and acute angle θ with k. then θ is
(a) \(\frac{\pi}{6}\),
(b) \(\frac{\pi}{4}\),
(c) \(\frac{\pi}{3}\),
(d) \(\frac{\pi}{2}\) (1)
(ii) Find a unit vector \(\widehat{a}\) (1)
(iii) Write down a unit vector in XY plane, making an angle 60°of with the positive direction of x – axis. (2)
Answer:
(i) (c), Since I = cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\), m = cos\(\frac{\pi}{3}\) = 1/2;
n = cos θ
l² + m² + n² = 1
n² = 1 – (\(\frac{1}{2}\))² – (\(\frac{1}{\sqrt{2}}\))² = 1/4
n = \(\frac{1}{2}\), cosθ = 1/2 , θ = \(\frac{\pi}{3}\).

(ii)

Question 4.
Let the vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) denoted the sides of a triangle ABC.
(i) Prove that (2)

(ii) Find the projection of the vector i + 3j + 7k on the vector 7i – j + 8k (2)
Answer:

(ii) Projection of the vector i + 3j + 7k on the vector 7i – j + 8k

Question 5.
(i) If \(\bar{a}\) and \(\bar{b}\) are any two vectors, then axb is (1)
(a) a vector on the same plane where \(\bar{a}\) and \(\bar{b}\) lie.
(b) ab cosθ, if θ is the angle between them.
(c) a vector parallel to both \(\bar{a}\) and \(\bar{b}\).
(d) a vector perpendicular to both \(\bar{a}\) and \(\bar{b}\).
(ii) Let \(\bar{a}\) = 2i + 4j – 5k, \(\bar{b}\) = i + 2j + 3k. Then find a unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\). (2)
(iii) Find a vector of magnitude 5 in the direction perpendicular to both \(\bar{a}\) and \(\bar{b}\) (1)
Answer:
(i) (d) a vector perpendicular to both \(\bar{a}\) and \(\bar{b}\).

(ii) \(\bar{a}\) = 2i + 4j-5k, \(\bar{b}\) = i + 2j+3k

= i(12 + 10) – j(6 + 5) + k(4 – 4) = 22i – 11j

Therefore unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is

(iii) 5 × unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\)

Question 6.
Consider a vector that is inclined at an angle 45° to x-axis and 60° to y-axis

1. Find the dc’s of the vector. (2)
2. Find a unit vector in the direction of the above vector. (1)
3. Find a vector which is of magnitude 10 units in the direction of the above vector. (1)

Answer:
1. Let l, m, n are the direction ratios.
Given that, l = cos 45° = \(\frac{1}{\sqrt{2}}\), m = cos 60° = \(\frac{1}{2}\)
⇒ l² + m² + n² = 1

∴ the dc’s of the vector are \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), \(\frac{1}{2}\)

2. A unit vector in the direction of the above vector is given by li + mj + nk ⇒ \(\frac{1}{\sqrt{2}}\)i + \(\frac{1}{2}\)j + \(\frac{1}{2}\)k.

3. A vector, which is of magnitude 10 units in the direction of the above vector is given by

Question 7.
Consider the point A(2, 1, 1) and B(4, 2, 3)

1. Find the vector \(\overline{A B}\) (1)
2. Find the direction cosines of \(\overline{A B}\) (2)
3. Find the angle made by \(\overline{A B}\) with the positive direction of x-axis. (1)

Answer:
1. \(\overline{A B}\) = 2i + j + 2k

2. |\(\overline{A B}\)| = \(\sqrt{4+1+4}\) = 3
The direction cosines are \(\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{2}{3}\).

3. cos α = \(\frac{2}{3}\) ⇒ α = cos^-1(\(\frac{2}{3}\)).

Question 8.
If i + j + k, 2i + 5j, 3i + 2 j – 3k, i – 6j – k respectively are the position vector of points A, B,C and D. Then

1. Find \(\overline{A B}\) and \(\overline{C D}\). (1)
2. Find the angle between the vectors \(\overline{A B}\) and \(\overline{C D}\). (2)
3. Deduce that \(\overline{A B}\) parallel to \(\overline{C D}\). (1)

Answer:
1.

2.

3. Since the angle between \(\overline{A B}\) and \(\overline{C D}\) is π they are parallel.

Question 9.
Let ABCD be a parallelogram with sides as given in the figure.

1. Find area of the parallelogram. (2)
2. Find the distance between the sides AB and DC. (2)

Answer:
1. Given;
\(\overline{A B}\) = i – 3j + k and \(\overline{A D}\) = i + j + k

2. Let h be the distance between the parallelsides AB and DC. Then ; Area = Base × h _____(2)
Here, Base = |\(\overline{A B}\)|
|i – 3j + k| = \(\sqrt{1+9+1}=\sqrt{11}\)
From (1) and (2)

Question 10.
Consider \(\bar{a}\) = i + 2j – 3k, \(\bar{b}\) = 3i – j + 2k, \(\bar{c}\) = 11i + 2j

1. Find \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\).\(\bar{b}\) (2)
2. Find the unit vector in the direction of \(\bar{a}\) + \(\bar{b}\). (1)
3. Show that \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\) – \(\bar{b}\) are orthogonal. (1)

Answer:

(ii) Unit vector in the direction of

(iii) We have,

Therefore, they are orthogonal.

Question 11.
Let A (1, -1, 4), B ( 2, 1, 2 ) and C (1, -2, -3 )

1. Find \(\overline{A B}\). (1)
2. Find the angle between \(\overline{A B}\) and \(\overline{A C}\).(2)
3. Find the area of the parallelogram formed by \(\overline{A B}\) and \(\overline{A C}\) as adjacent sides. (1)

Answer:
1. \(\overline{A B}\) = P.v of B – P.v of A
= 2 i + j + 2 k – (i – j + 4k) = i + 2 j – 2k

2. \(\overline{A C}\) = P.v of C – P.v of A
= i – 2 j – 3 k -(i – j + 4k) = – j – 7k
Let A be the angle between \(\overline{A B}\) and \(\overline{A C}\)

3.

Area of the parallelogram

Plus Two Maths Vector Algebra Six Mark Questions and Answers

Question 1.
Using this figure answer the following questions.

1. Find \(\overline{O A}\), \(\overline{O B}\), \(\overline{O C}\) (2)
2. Find \(\overline{O D}\) (2)
3. Find the coordinate of the vertex D. (2)

Answer:
1. \(\overline{O A}\) = (3 – 1)i + (-1 – 2)j + (7 – 3)k = 2i – 3j + 4k
\(\overline{O B}\) = (2 – 1)i + (4 – 2)j +(2 – 3)k = i + 2j – k
\(\overline{O C}\) = (4 – 1)i + (1 – 2 )j + (5 – 3 )k = 3i – j + 2 k.

2. From the figure,

3. Let the vertex of D be (x , y , z),
Then, \(\overline{O D}\) = (x – 1)i + (y – 2)j + (z – 3)k.
But we have,
\(\overline{O D}\) = 6i – 2j + 5k = (x – 1)i + (y – 2)j +(z – 3)k
x – 1 = 6 ⇒ x = 7, y – 2 = -2 ⇒ y = 0, z – 3 = 5 ⇒ z = 8.

Question 2.
OABCDEFG is a cube with edges of length 8 units and axes as shown. L, M, N are midpoints of the edges FG, GD, GB respectively.

1. Find p.v’s of F, B,D and G. (1)
2. Show that the angle between the main diagonals is θ = cos^-1\(\left(\frac{1}{3}\right)\). (2)
3. Find the p.v’s of L, M, N. (1)
4. Show that \(\overline{L M}+\overline{M N}+\overline{N L}=0\). (1)

Answer:
1. \(\overline{O F}\) = 8 j + 8k, \(\overline{O B}\) = 8i + 8k, \(\overline{O D}\) = 8i + 8k, \(\overline{O G}\) = 8i + 8j + 8k.

2. Consider the main diagonals \(\overline{O G}\) and \(\overline{E B}\)

3. P.V of L = \(\overline{O L}\) = 4i + 8j + 8k
P.V of M = \(\overline{O M}\) = 8i + 8j + 4k
P.V of N = \(\overline{O N}\) = 8i + 4j + 8k

4.

Question 3.
Using the figure answer the following questions

1. Evaluate \(\overline{A B}\).\(\overline{A C}\) (2)
2. Find \(\overline{A D}\) . (2)
3. Find the coordinates of D.

Answer:
1. \(\overline{A B}\) = p.v of B – p.v of A= -4i + 0j + 3k
\(\overline{A C}\) = p.v of C – p.v of A = 0i – 4 j + 4k
\(\overline{A B}\).\(\overline{A C}\) = -4 × 0 + 0 × -4 + 3 × 4 = 12

2.

3. Let the coordinate of D be (x, y ,z)
⇒ \(\overline{A D}\) = (x – 3)i + (y – 2)j + (z – 1)k,

Question 4.
Consider the Parallelogram ABCD

1. Find \(\overline{A B}\) and \(\overline{A D}\) (1)
2. Find the area of the parallelogram ABCD. (1)
3. Find \(\overline{A C}\). (2)
4. Find co-ordinate of C. (2)

Answer:
1. \(\overline{A B}\) = p.v of B – p. v of A
= 3i + 5j + 8k – (i + 2j + k) = 2i + 3j + 7k
\(\overline{A D}\) = p.v of D – p. v of A
= i + 3j + 2k – (i + 2j + k)= 0i + j + k.

2.

3. By triangle inequality;

4. Let the co-ordinate of C be (x, y, z)
Then, \(\overline{A C}\) = (x – 1)i + (y – 2)j + (z – 1)k = 2i + 4j + 8k
x – 1 = 2 ⇒ x = 3, y – 2 = 4 ⇒ y = 6,
z – 1 = 8 ⇒ z = 9
Co-ordinate of C is (3, 6, 9).

Question 5.
Consider the following quadrilateral ABCD in which P, Q, R, S are the midpoints of the sides.

1. Find \(\overline{P Q}\) and \(\overline{S R}\) in terms of \(\overline{A C}\) (2)
2. Show that PQRS is a parallelogram. (2)
3. If \(\bar{a}\) is any vector, prove that (2)

Answer:
1. Using triangle law of addition, we get

2. From the above explanation we have,

and parallel. Similarly, |\(\overline{S P}\)| = |\(\overline{R Q}\)|
Therefore, PQRS is a parallelogram.

3. Let \(\bar{a}\) = a₁ i + a₂ j + a₃ k

Students can Download Chapter 12 Linear Programming Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming

Plus Two Maths Linear Programming Four Mark Questions and Answers

Question 1.
Solve the following LPP Graphically;
Maximise; Z = 60x + 15y
Subject to constraints;
x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0.
Answer:
1. In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(30, 0), B(20, 30), C(0, 50).

Given; Z = 60x + 15y

Corner points	Value of Z
O	Z = 0
A	Z = 60(30) + 15(0) = 1800
B	Z = 60(20) + 15(30) = 1650
C	Z = 60(0) + 15(50) = 750

Since maximum value of Z occurs at A, the soluion is Z = 1800, (30, 0).

Question 2.
Solve the following LPP Graphically;
Minimise; Z = -3x + 4y
Subject to constraints;
x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Answer:
In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(4, 0) B(2, 3), C(0, 4).

Given; Z = -3x + 4y

Corner points	Value of Z
O	Z = 0
A	Z = -3(4) + 4(0) = -12
B	Z = -3(2) + 4(3) = 6
C	Z = -3(0) + 4(4) = 16

Since minimum value of Z occurs at A, the soluion is Z = -12, (4, 0).

Question 3.
Solve the following LPP Graphically;
Maximise; Z = 3x + 5y
Subject to constraints;
x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0
Answer:
In the figure the shaded region ABC is the fesible region. Here the region is unbouded.

The corner points are A(3, 0), B\(\left(\frac{3}{2}, \frac{1}{2}\right)\), C(0, 2)
Given; Z = 3x + 5y

Corner points	Value of Z
A	Z = 3(3) + 5(0) = 9
B	Z = 3\(\left(\frac{3}{2}\right)\) + 5\(\left(\frac{1}{2}\right)\) = 7
C	Z = 3(0) + 5(2) = 10

Form the table, minumum value of Z is 7 at B\(\left(\frac{3}{2}, \frac{1}{2}\right)\). The feasible region is unbounded, so consider the inequality 3x + 5y < 7. Clearly the feasible region has no common points with 3x + 5y < 7, Thus minimum value of Z occurs at B, the soluion is Z = 7.

Plus Two Maths Linear Programming Six Mark Questions and Answers

Question 1.
One kind of a cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1kg of fat assuming that there is no shortage of the other ingredients, used in making the cake.
Answer:
Let the number of cakes made of type I are x and that of type II are y. Then the total number of cakes will be Z = x + y
Flour constraint 200x + 100y ≤ 5000
Fat constraint 25x + 50y ≤ 1000
Therefore;
Maximise; Z = x + y
2x + y ≤ 50; x + 2y ≤ 40; x ≥ 0, y ≥ 0

In the figure the shaded region OABC is the feasible region. Here the region is bounded. The corner points are O(0, 0), A(25, 0), B(20, 10), C(0, 20)
Given; Z = x + y

Corner points	Value of Z
O	Z = 0
A	Z = 25 + 0 = 25
B	Z = 20 + 10 = 30
C	Z = 0 + 20 = 20

Since maximum value of Z occurs at B, the soluion is Z = 30, (20, 10).

Question 2.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine and 3hours of craftman’s time in its making, while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has availability of not more than 42 hours of machine time and 24 hours of craftman’s time.

1. What no. of rackets and bats must be produced if the factory is to work at full capacity?
2. If the profit on a racket and a bat is 10 find maximum profit.

Answer:
Let the number of rackets made = x and that of bats = y.
Maximise; Z = x + y
Machine constraints 1.5x + 3y ≤ 42
Craftsman’s constraint 3x + y ≤ 24
Therefore; Maximise; Z = x + y
x + 2y ≤ 14, 3x + y ≤ 24, x ≥ 0, y ≥ 0
In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(8, 0), B(4, 10), C(0, 14).

Given; Z = x + y

Corner points	Value of Z
O	Z = 0
A	Z = 8 + 0 = 8
B	Z = 4 + 12 = 16
C	Z = 0 + 14 = 14

Since maximum value of Z occurs at B, the soluion is Z = 16, (4, 12).

Question 3.
Two godowns A and B have grains capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E, and F whose requirement are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops is given in the following table; Transportation cost per quintal(in Rs.)

Hence should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Answer:

Express the problem diagrammatically as shown above. The total transportation cost is given by
Z = 6x + 3y + 2.5{100 – (x + y)} + 4(60 – x) + 2(50 – y) + 3(-60 + (x + y))
⇒ Z = 2.5x + 1.5y + 410
100 – (x + y) ≥ 0 ⇒ x + y ≤ 100
60 – x ≥ 0 ⇒ x ≤ 60
50 – y ≥ 0 ⇒ y ≤ 50 – 60 + x + y ≥ 0 ⇒ x + y ≥ 60
Then the given LPP is
Minimise; Z = 2.5x + 1.5y + 410
x + y ≤ 100, x + y ≥ 60
0 ≤ x ≤ 60, 0 ≤ y ≤ 50

In the figure the shaded region ABCD is the feasible region. Here the region is bounded. The corner points are
A(60, 0), B(60, 40), C(50, 50), D(10, 50).
Given; Z = 2.5x + 1.5y + 410

Corner points	Value of Z
A	Z = 2.5(60) + 1.5(0) + 410= 560
B	Z = 2.5(60) + 1.5(40) + 410 = 620
C	Z = 2.5(50) + 1.5(50) + 410 = 610
D	Z = 2.5(10) + 1.5(50) + 410 = 510

Since minimum value of Z occurs at D, the soluion is Z = 510.

Question 4.
(i) Choose the correct answer from the bracket. If an LPP is consistent, then its feasible region is always
(a) Bounded
(b) Unbounded
(c) Convex region
(d) Concave region
(ii) Maximize Z = 2x + 3y subject to the constraints x + y ≤ 4, x ≥ 0, y ≥ 0.
Answer:
(i) (c) Convex region.

(ii)

Corner points of the feasible region are as follows

Corner points	Z = 2x + 3y
0(0, 0)	0
A(0, 4)	12 → Maximum
B(4, 0)	8

∴ the maximum value of Z is 12 attained at (0, 4).

Question 5.
The graph of a linear programming problem is given below. The shaded region is the feasible region. The objective function is Z = px + qy

1. What are the co-ordinates of the comers of the feasible region.
2. Write the constraints
3. If the Max. Z occurs at A and B, what is the relation between p and q?
4. If q = 1, write the objective function
5. Find the Max Z

Answer:
1. Corner points are O(0, 0), A(5, 0), B(3, 4), C(0, 5).

2. Constraints are 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0.

3. At (3, 4), Z = 3p + 4q
At (5, 0), Z = 5p
⇒ 3p + 4q = 5p ⇒ p = 2q.

4. If q = 1, p=2
Then the objective function is,
Maximize Z = 2x + y.

5. At (3, 4) Z = 2 × 3 + 4 = 10 is the maximum value.

Question 6.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs Rs 4 per unit food and F₂ costs Rs 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum costs for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Answer:
Let x units of food F₁ and y units of food F₂ be in the diet
Total cost Z = 4x + 6y
Then the LPP is
Minimize Z = 4x + 6
Subject to the constraints
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0

The feasible region is unbounded

As the feasible region is unbounded, 104 may or may not be the minimum value of Z. For this we draw a graph of the inequality 4x + 6y < 104 or 2x + 3y < 52 and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common points with 2x + 3y < 52 Therefore minimum cost of the mixture will be 104.

Students can Download Chapter 6 Client-Side Scripting Using Java Script Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 6 Client-Side Scripting Using Java Script

Plus Two Computer Application Client-Side Scripting Using Java Script One Mark Questions and Answers

Question 1.
Among the following which one is the most correct. JavaScript is used mostly at the
(a) client side
(b) server side
(c) client side and server side
Answer:
(a) client side

Question 2.
Name the tag that is used to embed scripts in a web page.
Answer:
<SCRIPT>

Question 3.
In JavaScript, a variable is declared using the keyword______.
Answer:
var

Question 4.
_____are small programs embedded in the HTML pages.
Answer:
Scripts

Question 5.
Who developed JavaScript?
Answer:
Brendan Eich

Question 6.
_____are the scripts that are executed by the browser (client).
Answer:
Client Scripts

Question 7.
_____are the scripts that are executed by the web server.
Answer:
Server Scripts

Question 8.
_____Script is a platform-independent script.
Answer:
Java Script

Question 9.
______Script is a platform dependent script.
Answer:
VB Script

Question 10.
______makes the tags meaningful.
Answer:
Attribute

Question 11.
_____attribute specifies the name of the scripting language used.
Answer:
Language

Question 12.
State True or False. The identifiers are case sensitive Identifiers case.
Answer:
True

Question 13.
Which part of the browser executes the JavaScript.
Answer:
JavaScript engine

Question 14.
Odd one out
(a) Google Chrome
(b) Internet Explorer
(c) Mozila FireFox
(d) C++
Answer:
(d) C++, It is a programming language others are browsers.

Question 15.
A group of codes with a name is called_____.
Answer:
function

Question 16.
To declare a function the keyword_____is used.
Answer:
function

Question 17.
A function contains a function____and function_____.
Answer:
header, body

Question 18.
State true or false
Even though a function is defined within the body section, it will not be executed, if it is not called.
Answer:
True

Question 19.
Write down the purpose of the following code snippet
function print()
{
document .write (“ Welcome to JS”);
}
Answer:
This code snippet is used to display the string, “Welcome to JS” on the screen(monitor).

Question 20.
From the following select which one is Not the data type in JavaScript
(a) Number
(b) String
(c) Boolean
(d) Time
Answer:
(d) Time

Question 21.
_____keyword is used to declare a variable in JavaScript.
Answer:
var

Question 22.
______function is used to return the data type.
Answer:
typeof()

Question 23.
_____is a special data type to represent variables that are not defined using var.
Answer:
undefined

Question 24.
Odd one out
(a) +
(b) –
(c) %
(d) ==
Answer:
(d) ==, it is a relational operator the others are arithmetic operator.

Question 25.
Odd one out
(a) &&
(b) ||
(c) !
(d) %
Answer:
(d) %, it is an arithmetic operator, others are logical operator.

Question 26.
Odd one out
(a) <
(b) >
(c) ==
(d) !
Answer:
(d) ! , it is a logical operator, the others are relational operator.

Question 27.
Consider the following declaration var b;
From the following which value can be used for the variable b as boolean value.
(a) true
(b) TRUE
(c) True
(d) FALSE
Answer:
(a) true, The value is case sensitive.

Question 28.
Predict the output of the following.
var a, b;
a = ”0480″;
b = 2828159;
document.write(a+b);
Answer:
The out Put is “04802828159”. That is the string “0480” concatenates(joins) the number 2828159. The output is a string not a number.

Question 29.
Predict the output
var x, y;
x = ”8″;
y = 3;
document.write(x+y);
Answer:
The output is a string “83”; String addition means concatenation.

Question 30.
From the following which method is used to display a message (dialog box) on the screen.
(a) alert()
(b) isNaN()
(c) toUpperCase()
(d) toLowerCase()
Answer:
(a) alert()

Question 31.
Raju wants to convert a lower case text to Upper case text, which function is to be used.
Answer:
toUpperCase()

Question 32.
Christy wants to convert an upper case text to lower case text, which function is to be used.
Answer:
toLowerCase()

Question 33.
Andrea wants to check a value is a number or not. From the following which function is used for that.
(a) isNumb()
(b) isNaN()
(c) isNotNumb()
(d) isNotNumber()
Answer:
(b) isNaN().

Question 34.
Predict the output of the following code snippet
var x=” HIGHER SECONDARY”;
alert(x.charAt(4));
Answer:
It displays a message box with character ‘E’ from the fifth (4 + 1) place

Question 35.
Read the following three statements regarding JavaScript.

1. JavaScript can be used at the client side for data validation.
2. JavaScript statements are case sensitive.
3. JavaScript can be used only for creating web pages.

Answer:
All the three statements are correct.

Question 36.
Write the output of the following web page.

Answer:
welcome welcome.

Question 37.
Following is the web page that accepts a string from a text box, converts in to uppercase and display it on the screen. Complete the missing portion in the page.


Answer:
document.forml.text1.value

Question 38.
Name the attribute of <SCRIPT> tag that is used to include an external JavaScript file into a web page.
Answer:
src

Plus Two Computer Application Client-Side Scripting Using Java Script Two Mark Questions and Answers

Question 1.
A javaScript code has the following three variables and values.
x = “Script”;
y = “3”;
z = “2”;
Then match the following table.

A	B
x.length()	false
isNaN(x)	5
isNaN(y)	6
y + z	true
	32

Answer:

A	B
x.length()	6
isNaN(x)	true
isNaN(y)	false
y + z	32

Question 2.
“Placing JavaScript as an external file has some advantages”. Do you agree with this statement?
Why?
Answer:
External (another) JavaScript file:
We can write scripts in a file and save it as a separate file with the extension .js. The advantage is that this file can be used across multiple HTML files and can be enhance the speed of page loading.

Question 3.
Explain the difference between the statements.
documerit.write (“welcome”);
and
alert (“welcome)”;
Answer:

1. document.write () is a JavaScript command used to print anything on the browser window.
2. document write (“welcome”) prints “welcome” on the browser window.
3. alert (“welcome”). This is a built in function used to display a message here the message “welcome” in a separate window.

Question 4.
Is it necessary to use Language – ‘JavaScript” in the <SCRIPT>tag to specify the JavaScript code? Why?
Answer:
No, it is not nece^ary. If the language attribute is not specified, it will take the default value as Javascript.

Question 5.
Write the output of the following web page:

Answer:
It will not display anything on the screen. This code snippet contains a function that will not do anything unless it is invoked(called).

Question 6.
Write the output of the following web page:

Answer:

• The output is as follows
• welcome to JavaScript
• welcome to JavaScript
• The message repeats 2 times.

Question 7.
Among the following, identify the data types used in JavaScript
int, float, number, char, boolean, long
Answer:
From the list there is only two, number and boolean are the types used in JavaScript.

Question 8.
Write the output of the following web page and jus¬tify your answer.


Answer:
x = “10” means x is a string variable
y = “20” means y is a string variable
x + y means the string x and y will be concatenated Hence it displays 1020.

Question 9.
What do you mean by Scripts? Explain?
Answer:
Scripts are small programs embedded in the HTML pages.
<SCRIPT> tag is used to write scripts The attributes used are

• Type-To specify the scripting language
• Src-Specify the source file
• Two types of scripts
  1. Client scripts: These are scripts executed by the browser.
     Eg: VB Script, Javascript, etc.
  2. Server scripts: These are scripts executed by the server.
     Eg: ASP, JSP, PHP, Perl, etc.
• The languages that are used to write scripts are known as scripting languages.

Question 10.
Suppose you have written a JavaScript function named checkData(). You want to execute the function when the mouse pointer is just moved over the button. How will you complete the following to do the same?
<INPUT Type=”button”_____= “checkData()”>
Answer:
<INPUT Type=”button” onMouseEnter = “checkData()”>

Plus Two Computer Application Client-Side Scripting Using JavaScript Three Mark Questions and Answers

Question 1.
Categorize the following as UpperCamelCase or lowerCamelCase and explain.
(a) DateOfBirth
(b) dateOfJoining
(c) timeOfJoining
(d) PlaceOfBirth
Answer:
a. CamelCase: An identifier does not use special characters such as space hence a single word is formed using multiple words. Such a naming method is called CamelCase (without space between words and all the words first character is in upper case letter).
These are two types

1. Upper CamelCase: when the first character of each word is capitalised,
2. lowerCamelCase: when the first character of each word except the first word is capitalised.

b. UpperCamelCase: DateOfBirth, PlaceOfBirth

c. lowerCamelCase: dateOfJoining, timeOfJoining

Question 2.
Explain the method of working of a JavaScript.
Answer:
Every browser has a JavaScript engine. If the code snippet contains JavaScript code, it is passed to the JavaScript engine for processing, the engine executes the code.

If there is no script then it processes without the help of script engine. Hence an HTML file without JavaScript is faster than with JavaScript code.

Question 3.
Write down the various data types used in JavaScript.
Answer:

1. Number: Any number(whole or fractional) with or without sign.
   Eg: +1977,-38.0003,-100, 3.14157,etc
2. String: It is a combination of characters enclosed within double quotes.
   Eg: “BVM”, “jobi_cg@rediffmail.com”, etc
3. Boolean: We can use either true or false.lt is • case sensitive. That means can’t use TRUE OR
   FALSE

Question 4.
Explain how a variable is declaring in JavaScript.
Answer:
For storing values you have to declare a variable, for that the keyword var is used. There is no need to specify the data type.
Syntax:
var<variable name1 > [, <variable name2>, <variable name3>,etc…]
Here square bracket indicates optional.
Eg: var x, y, z;
x= 11;
y = ”BVM”;
z = false;
Here x is of number type, y is of string and z is of Boolean type.

Question 5.
What are the different ways to add Scripts to a web page?
Answer:
The three different ways to add Scripts as follows

1. Inside <BODY> section: Scripts can be placed inside the <BODY> section.
2. Inside <HEAD> section: Scripts can be placed inside the <HEAD> section. This method is widely accepted method
3. External (another) JavaScript file

We can write scripts in a file and save it as a separate file with the extension .js. The advantage is that this file can be used across multiple HTML files and can be enhance the speed of page loading.

Question 6.
Create a web page that checks whether a student has passed or not?
Answer:

Question 7.
Create a web page to display the squares of first 10 numbers Sp <HTML>
Answer:

Question 8.
Create a web page to display even numbers upto 10.
Answer:

Question 9.
Following web page is used to show “Passed” or “Failed” based on a mark. Mark less than 30 is cosidered as failed. There are some errors in the code. Correct them.

Answer:

Plus Two Computer Application Client-Side Scripting Using Java Script Five Mark Questions and Answers

Question 1.
Consider the following declarations
var a, b, c, d;
a= “BVM”;
b=100;
c=true;
d=3.14157;
Predict the output of the following

1. document.write(typeof(a)); (1)
2. document.write(typeof(b)); (1)
3. document.write(typeof(c)); (1)
4. document.whte(typeof(d)); (1)
5. document.write(typeof(e)); (1)

Answer:

1. string
2. number
3. boolean
4. number
5. undefined

Question 2.
Create a web page to print the day of a week
Answer:

Question 3.
What is an operator? Explain in detail.
Answer:
Operators are the symbols used to perform an operation.
1. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction(-), division (/), multiplication(*), modulus(%-gives the remainder), increment(++) and decrement(—) operations.
Eg. If x=10 and y=3 then

lf x=10 then
document.write(++x); → It prints 10+1=11
If x=10 then
document.write(x++); → It prints 10 itself.
If x=10 then
document.write(—x); It prints 10-1=9
lf x=10 then
document.write(x—); → It prints 10 itself.

2. Assignment operators:
If a=10 and b=3 then a=b. This statement sets the value of a and b are same, i.e. it sets a to 3.
It is also called short hands If X=10 and Y= 3 then

3. Relational(Comparison) operators:
It is used to perform comparison or relational operation
between two values and returns either true or false.
Eg: If X=10 and Y=3 then

4. Logical operators:
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true or false
If X=true and Y=false then

Both operands must be true to get a true value in the case of AND(&&) operation
If X=true and Y=false then

Either one of the operands must be true to get a true value in the case of OR(||) operation
If X= true and Y=false then

!X	!Y
false	true

5. String addition operator(+):
This is also called concatenation operator. lt joins(concatenates)two strings and forms a string.
Eg: var x, y, z;
x= “BVM HSS;
y= “Kalparamba”;
z=x+y;
Here the variable z becomes “BVM HSS Kalparamba”.
Note: If both the operands are numbers then addition operator(+) produces number as a result otherwise it produces string as a result. Consider the following

Question 4.
Write down the control structures used in JavaScript.
Answer:
Control structures in JavaScript. In general the execution of the program is sequential, we can change the normal execution by using the control structures.
1. simple if
Syntax:
if(test expression)
{
statements;
}
First the test expression is evaluated, if it is true then the statement block will be executed otherwise not.

if-else
Syntax:

First the test expression is evaluated, if it is true then the statement block1 will be executed otherwise statement block? will be evaluated.

2. switch
It is a multiple branch statement. Its syntax is
given below.
switch(expression)

First expression evaluated and selects the statements with matched case value.
Eg.
switch (n)

3. for loop
If a statement wants to execute more than once.
Loop is used.
for loop is an entry controlled loop.
The syntax of for loop is given below
For(initialisation; testing; updation)
{
Body of the for loop;
}

4. while loop
If a statement wants to execute more than once.
Loop is used.
It is also an entry controlled loop The syntax is given below
Loop variable initialised
while(expression)
{
Body of the loop;
Update loop variable;
}
Here the loop variable must be initialised out side the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.

Question 5.
Write down the different mouse events used in JavaScript.
Answer:
Different mouse events and their description is given below

Event	Description
onClick	It occurs when the user clicks on an object by using mouse
onMouseEnter	It occurs when the mouse pointer is moved onto an object
onMouseLeave	It occurs when the mouse pointer is moved out of an object
onKeyDown	It occurs when the user presses a key on the keyboard
onKeyUp	It occurs when the user releases a key on the keyboard

Question 6.
Create a web page that displays the capital of a state.
Answer:

Plus Two Computer Application Client-Side Scripting Using Java Script Let Us Practice Questions and Answers

Question 1.
Develop a web page to display the following screen. (3 Mark)

User can enter a number in the first text box. On clicking the show button, product of all numbers from 1 to the entered limit should be displayed in the second text box.
Answer:

Question 2.
Develop a web to display the following screen. (5 Mark)

User can enter a number in the first text box. On clicking the show button, Even or Odd should be displayed in the second text box depending on whetherthe number is even or odd.
Answer:

Question 3.
Develop a web page to display the following screen. (5 Mark)

The user can enter an age in the text box. If the user enters an alphabet, instead of a number in the text box, on clicking the show button, it should display a message “Invalid Age” to the user. Otherwise it should display a message “Correct Data”.
Answer:

Question 4.
Develop a login page as shown in the following figure. (5 Mark)

The page must contain one text box for entering the username and one password box for entering the password. The user name must contain at least 4 characters and the password must contain at least 6 characters. The first two characters in the password must be numbers. On clicking the show button, if the valid data are given in boxes, a message “Correct Data” should be displayed. Otherwise, “Wrong Data” message should be displayed.
Answer:

Question 5.
Develop a web page to implement a simple calculator. The page should have two text boxes to enter two numbers. It should also have 4 buttons to add, subtract, multiply and divide the two numbers. The answer should be displayed in a third text box on clicking the button. The web page should be as shown in the following figure. (5 Mark)

Answer:

Plus Two Computer Application Client-Side Scripting Using Java Script Let Us Assess Questions and Answers

Question 1.
Write the value of the variable z in each of the following. (5 Mark)
(a) var x, y, z;
x = 5;
y = 3;
z = ++x – y—;
Answer:
z = 6 – 3 = 3 Hence z=3.

(b) var x, y, z;
x = ”12″;
y = 13;
z = x+y;
Answer:
z = “12” + 13 = 1213. If one of the operand is a string the two strings concatenates the strings. That is it concatenates two strings.

(c) var x, y, z;
x = 20;
y = 8;
x % = y;
z = x++;
Answer:
x % = y means x = x%y.
that is x= 20 % 8.
i. e. x= 4.
Then z= 4. (x++ first use the value then change).

(d) var x, y, z;
x = 1;
y = 4;
z = !(x<y);
Answer:
The answer is false. z = !(1 <4) z = !(true) z = false.

(e) var x, y, z;
x = 5;
y = 6;
Z = (x>y) || (y% 2 == 0);
Answer:
The answer is true.
Steps
z = (5>6) || (6%2==0)
= (false) || (0 == 0)
= (false) || (true)
= true.

Question 2.
Predict the output of the following. (5 Mark)
(a)

Answer:

(b)

Answer:
The output is 2500.

(c)

Answer:
The output is 275.

(d)

Answer:
The output is 120(5 factorial).

Question 3.
Following is an html code segment in a web page (1 Mark)
<FORM Name=”frmStud”>
<INPUT Type=”text” Name=”studentName”>
</FORM>
Fill in the blanks to store the value of the text box to the variable n.
var n;
n =______;
Answer:
n = document.frmstud.studentName.value;

Question 4.
Suppose you have written a JavaScript function named checkData(). You want to execute the function when the mouse pointer is just moved over the button. How will you complete the following to do the same? (1 Mark)
<INPUT Type=”button”_____= “checkData()”>
Answer:
<INPUT Type=”button” onMouseEnter= “checkData()”>

Question 5.
Explain <SCRIPT> tag and its attributes. (2 Mark)
Answer:
Scripts are small programs embedded in the HTML pages, to write scripts <SCRIPT> tag is used.
Language attribute specifies the name of the scripting language used.
Example:
<SCRIPT Language=”JavaScript”>
_______
</SCRIPT>

Question 6 .
Write the syntax of a built in function in JavaScript. (5 Mark)
Answer:
Built in functions(methods)
1. alert(): This is used to display a message(dialogue box) on the screen.
eg: alert(“Welcome to JS”);

2. isNaN(): To check whether the given value is a number or not. It returns a Boolean value. If the value is not a number(NaN) then this function returns a true value otherwise it returns a false value.
Eg:

• isNaN(“BVM”); returns true
• isNaN(8172); returns false
• isNaN(“680121”); returns false
• alert(isNaN(8172); displays a message box as false.

3. toUpperCase(): This is used to convert the text to uppercase.
Eg:
var x=“bvm”;
alert(x.toUpperCase());
Output is as follows

4. toLowerCase(): This is used to convert the text to lower case.
Eg:
var x=“BVM”;
alert(x.toLowerCase());
Output is as follows

5. charAt(): It returns the character at a particular position.
Syntax: variable.charAt(index);
The index of first character is 0 and the second is 1 and so on.
Eg:
var x=”HIGHER SECONDARY”;
alert(x.charAt(4));
Output is as follows

Eg 2.
var x=”HIGHER SECONDARY”;
alert(“The characters @ first position is “+ x.charAt(0));

6. length property: It returns the number of characters in a string.
Syntax: variable.length;
Eg.
var x=”HIGHER SECONDARY”;
alert(“The number of characters is “+ x.length);
Output is as follows(note that space is a character)

Question 7.
Classify the following values in JavaScript into suitable data types. (3 Mark)
“Welcome”, “123”, “true”, 67.4, .98, false, “hello”
Answer:

• Number: 67.4 and .98.
• String: “Welcome”, “123″, “true” and “hello”.
• Boolean: false.

Question 8.
What is meant by undefined data type in JavaScript mean? (2 Mark)
Answer:
undefined: It is a special data type to represent variables that are not defined using var.

Question 9.
Explain operators in JavaScript. (5 Mark)
Answer:
Operators in JavaScript. Operators are the symbols used to perform an operation
1. Arithmetic operators:
It is a binary operator. It is used to perform add i-tion (+), subtraction(-), division(/), multiplication(*), modulus(%-givesthe remainder), increment(++) and decrement(—) operations. Eg. If x = 10 and y = 3 then

If x = 10 then
document.write(++x); → It prints 10+1=11
If x = 10 then
document.write(x++); → It prints 10 itself.
If x = 10 then
document.write(—x); It prints 10-1=9
If x = 10 then
document.write(x—);→ It prints 10 itself.

2. Assignment operators:
If a = 10 and b = 3 then a = b. This statement sets the value of a and b are same,i.e. it sets a to 3. It is also called short hands
lf X = 10 and Y = 3 then

3. Relational(Comparison) operators:
It is used to perform comparison or relational operation between two values and returns either true or false.
Eg:
lf X = 10 and Y= 3 then

4. Logical operators:
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true or false
If X = true and Y= false then

Both operands must be true to get a true value in the case of AND(&&) operation If X = true and Y = false then

Either one of the operands must be true to get a true value in the case of OR(||) operation If X = true and Y = false then

!X	!Y
false	true

5. String addition operator(+):
This is also called concatenation operator. It joins (concatenates) two strings and forms a string.
Eg:
var x, y, z;
x= “BVM HSS”;
y= “Kalparamba”;
z = x + y;
Here the variable z becomes “BVM HSS Kalparamba”.
Note: If both the operands are numbers then addition operator(+) produces number as a result otherwise it produces string as a result. Consider the following.

Eg:

• 8(number) + 3(number) = 11 (Result is a number)
• 8 (number)+ “3”( string) = “83″ (Result is a string)
• “8” (string) + 3 (number) = “83”(Result is a string)
• “8” (string) + “3” (string) = “83” (Result is a string).

Question 10.
Write JavaScript functions to perform the following (5 Mark)

1. To check whether a variable N contains a number
2. To convert the string “scert” to all capitals.
3. To convert the string “HTML” to all small letters.
4. To display a message “Welcome to functions”.
5. To display the third character in the string “Computer”.

Answer:
1.

2. <HTML>
<head>
<title>
check
</title>
<SCRIPT Language=”JavaScript”>
function convert()

3. <HTML>
<head>
<title>
check
</title>
<SCRIPT Language=”JavaScript”>
function convert()

4. <HTML>
<head>
<title>
check
</title>
<SCRIPT Language=”JavaScript”>
function display()
{

5. <HTML>
<head>
<title>
check
</title>
<SCRIPT Language=”JavaScript”>
function display()

Question 11.
Write JavaScript code to display the length of the string “Computer”. (3 Mark)
Answer:
<HTML>
<head>
<title>
length of a string
</title>
</head>
<BODY>
<SCRIPT Language=”JavaScript”>
var str.len;
str=”Computer”;
len=str.length;
alert(“The length of the string Computer is “+len);
</SCRIPT>
</BODY>
</HTML>

Question 12.
A web page contains a putton. Write HTML code for the button which executes a function Message() on the occurance of the following events. (2 Mark

1. When a user clicks the mouse on the button.
2. When user moves the mouse over the button.

Answer:

1. <input type=”button”value=”Message” onClick=”Message()”>
2. <inputtype-button”value=”Message” onMouseEnter=”Message()”>

Question 13.
What are the advantages of writing JavaScript code in the head section of an HTML page? (2 Mark)
Answer:
Scripts can be placed inside the <HEAD> section. This method is widely accepted method. The main reason for this is that body section contains a large volume of text contents to be displayed on the web page. More over the head section is loaded before the body section.

Question 14.
Design an HTML page that contains a text box to enter the marks in a given subject. (5 Mark)
(a) Write HTML code for this web page
(b) Provide validations for this text box in a separate JavaScript file and link it with the HTML file. The validations are

• it should not be empty
• it should be a number
• it should be between 0 and 60.

(c) List the advantages of writing the script in a separate file.
Answer:
(a)

(b) Step 1.
Take a Notepad and type the following and save it as check.js.
function checkData()

Step 2.
Take another Notepad and type the following and save it as valid.html and execute valid.html file

(c) We can write scripts in a file and save it as a separate file with the extension .js. The advantage is that this file can be used across multiple HTML files and can be enhance the speed of page loading

Students can Download Chapter 11 Trends and Issues in ICT Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 11 Trends and Issues in ICT

Plus Two Computer Application Trends and Issues in ICT One Mark Questions and Answers

Question 1.
SIM is_____.
(a) Subscriber Identity Module
(b) Subscriber Identity Mobile
(c) Subscription Identity Module
(d) Subscription Identity Mobile
Answer:
(a) Subscriber Identity Module

Question 2.
The protocol used to send SMS message is______.
Answer:
SS7 (Signalling System No.7)

Question 3.
_____is a standard way to send and receive short text message using mobile phone.
Answer:
Short Message Service(SMS)

Question 4.
_____is a standard way to send and receive message with multimedia content using mobile phone.
Answer:
Multimedia Messaging Service (MMS)

Question 5.
From the following which generation network has more speed?
(a) 1G
(b) 3G
(c) 2G
(d) 4G
Answer:
(d) 4G

Question 6.
GSM stands for______.
Answer:
Global System for Mobiles.

Question 7.
CDMA stands for_____.
Answer:
Code Division Multiple Access

Question 8.
GPRS stands for______.
Answer:
Global Packet Radio Sysetm.

Question 9.
GPS stands for______.
Answer:
Global Positioning System.

Question 10.
EDGE stands for____.
Answer:
Enhanced Data rates for GSM Evolution

Question 11.
_____technology provides speed, voice quality, wider coverage, better security to the mobile network than GSM network.
Answer:
CDMA

Question 12.
Write down the popular standards introduced by 2G network.
Answer:
GSM and CDMA

Question 13.
From the following which one is used analog signals instead of digital signals.
(a) 2G
(b) 3G
(c) 1G
(d) 4G
Answer:
(c) 1G

Question 14.
Consider a person while using internet through mobile phone, it shows G on the network coverage icon. Which type of network is he using and name the communication feature.
Answer:
The network is 2G and the communication feature is GPRS.

Question 15.
Consider a person while using internet through mobile phone, it shows E on the network coverage icon. Which type of network is he using and name the communication feature.
Answer:
The network is 2G and the communication feature is EDGE.

Question 16.
Consider a person while using internet through mobile phone, it shows H on the network coverage icon. Which type of network is he using and name the technology.
Answer:
The network is 3G and the technology is WCDMA

Question 17.
_____is the technology used in 3G.
Answer:
WCDMA (Wideband Code Division Multiple Access)

Question 18.
______is the technology used in 4G.
Answer:
OFDMA(Orthogonal Frequency Division Multiple Access)

Question 19.
From the following which generation network provides good quality images and videos than TV.
(a) 1G
(b) 2G
(c) 3G
(d) 4G
Answer:
(d) 4G

Question 20.
SMS stands for_______.
Answer:
Short Message Service

Question 21.
MMS stands for_______.
Answer:
Multimedia Messaging Service

Question 22.
______is a plastic card with small chip.
Answer:
Smart card

Question 23.
MOS stands for_______.
Answer:
Mobile Operating systems

Question 24.
Name a Mobile Operating System.
Answer:
Android.

Question 25.
KitKat, Jelly Bean, Donut, Cupcake are the different versions of______OperatingSystem.
KitKat, Jelly Bean, Donut, Cupcake 4rmloi
Answer:
Android.

Question 26.
RFID stands for______.
Answer:
Radio Frequency Identification

Question 27.
_______technology helped in Business Logistics to identify, track, sort or detect objects.
Answer:
RFID technology

Question 28.
IPR stands for_______.
Answer:
Intellectual Property Right

Question 29.
WIPO stands for_______.
Answer:
World Intellectual Property Organisation

Question 30.
_______is the exclusive rights to prevent unauthorized copying of inventions by a Creator from the Unauthorised person or company.
Answer:
Patent

Question 31.
______is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company.
Answer:
Trademark

Question 32.
A product or article is designed so beautifully to attract the customers. This type of design is called______.
Answer:
Industrial Design.

Question 33.
Aranmuia Kannadi, Palakkadan Matta, Marayoor Sarkkara, etc are example of______.
Answer:
Geographical indications.

Question 34.
_______is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.
Answer:
Copyright

Question 35.
From the following which is the symbol for copyright.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(b) ©

Question 36.
From the following which is the symbol for Unregistered trademark.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(d) ™

Question 37.
From the following which is the symbol for Registered trademark.
(a)  $
(b) ©
(c) ®
(d) ™
Answer:
(c) ®

Question 38.
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called_______.
Answer:
Intellectual Property Infringement.

Question 39.
______prevents others from the unauthorized or intentional copying or use of Patent without the permission of the creator.
Answer:
Patent Infringement

Question 40.
______is the illegal copying, distribution, or use of software.
Answer:
Piracy

Question 41.
_______prevents others from the unauthorized or intentional copying or use of Trademark without the permission of the creator.
Answer:
Trademark Infringement

Question 42.
_______prevents others from the unauthorized or intentional copying or use of Copy right without the permission of the creator.
Answer:
Copy right Infringement

Question 43.
______is a virtual environment created by computer systems connected to the internet.
Answer:
Cyber space

Question 44.
A person committing crimes and illegal activities with the use of computer over Internet. This crime is included as______crime.
Answer:
Cyber crime

Question 45.
State True or False.
Cyber crimes can be classified into three categories such as against individual, property, and Government.
Answer:
True

Question 46.
Phishing, hacking, denial of service attacks, etc are______crimes.
Answer:
Cyber

Question 47.
Odd one out
(a) Identity theft
(b) Harrassment
(c) vilation of privacy
(d) credit card fraud
Answer:
(d) credit card fraud, it is a cyber crime against individual others are cyber crimes against property

Question 48.
Odd one out
(a) Credit card theft
(b) Intellectual property theft
(c) Internet time theft
(d) Dissemination of obscene material
Answer:
(d) Dissemination of obscene material, It is cyber crime against individual, the others are cyber against property

Question 49.
Odd one out
(a) cyber terrorism
(b) Attacks against e-Governance websites
(c) Impersonation and cheating
(d) Website defacement
Answer:
(c) Impersonation and cheating, it is cyber crime against individual others are cyber crimes against Government

Question 50.
IT Act amended in_______
(a) 2015
(b) 2008
(c) 1900
(d) 1998
Answer:
(b) 2008

Question 51.
IT Act passed in Indian parliament is_______.
Answer:
2000

Question 52.
The laws to prevent cyber crimes is termed as______.
Answer:
Cyber law

Question 53.
_______ is the excessive enthusiasm for acquiring knowledge.
Answer:
Infomania

Question 54.
Phishing is an example of_______.
Answer:
Cyber crime

Question 55.
Expand the term GPRS.
Answer:
General Packet Radio Services.

Question 56.
SMS messages are exchanged using the protocol called_______.
Answer:
SS7(Signaling System No.7)

Question 57.
Pick the odd one out and justify your answer.
(A) SMS
(B) MMS
(C) GPS
(D) Android
Answer:
(D) Android. It is a mobile operating System. Others are mobile communication services.

Question 58.
Which among the following statements is wrong in relation with 2G network?
(A) It support MMS.
(B) It can provide only voice Service.
(C) GSM and CDMA are two popular standards introduced in 2G.
(D) 2G Network were later expanded to include GPRS and EDGE.
Answer:
(B) It can provide only voice service.

Question 59.
Pick the odd one out.
(A) Windows
(B) Android
(C) IOS
(D) Linux
Answer:
(D) Linux. Others are mobile Operating Systems.

Question 60.
_____refers to the exclusive right given to a person over the creation of his/her mind fora period of time.
Answer:
Patent / Intellectual Property Right

Question 61.
Which among the following are considered as violation to privacy?

1. Keeping hidden cameras in private places
2. Publishing private photos of individual in social media without their permission
3. Use of unauthorized software
4. Using simple password

(A) All the above are correct
(B) 1, 2 and 3 only
(C) 1 and 4 only
(D) 1 and 2 only
Answer:
(D) 1 and 2 only

Plus Two Computer Application Trends and Issues in ICT Two Mark Questions and Answers

Question 1.
What is mobile computing?
Answer:
The drawbacks of Desk computers are, it is heavy and power consumption rate is high and it is not portable (not mobile).

The advancements in computing technology, light weight, and low power consumption have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc.

Nowadays instead of desktops, light weight and low power consumption devices are used because they are cheap and common. Moreover, people are able to connect to others through internet even when they are in move.

Question 2.
Write short notes on SMS.
Answer:
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems.

It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

Question 3.
Expand GPS? Explain?
Answer:
It is a space-based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites.

The system provides critical capabilities to military, civil and commercial users around the world. It is maintained by the United States government and is freely accessible to anyone with a GPS receiver. GPS was created and realized by the U.S.

Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Question 4.
What is a smart card? How is it useful?
Answer:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data. The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Question 5.
How do trademark and industrial design differ?
Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 6.
Explain the exclusive right given to the owner by IPR?
Answer:
The exclusive right given to the owner by IPR is owner can disclose their creations for money.

Question 7.
What is piracy?
Answer:
It is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copy right act and hence the person committed deserve the punishment.

Question 8.
What do you meant by infringement?
Answer:
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called intellectual property lnfringement(violation). It is a punishable offence.

Question 9.
Match the following.

(a) 1 G	(i)  LTE (Long Term Evolution)
(b) 2 G	(ii) GSM & CDMA
(c) 3 G	(iii) Analog system which support only voice facility
(d) 4 G	(iv) WCDMA

Answer:

• (a) – (iii)
• (b) – (ii)
• (c) – (iv)
• (d) – (i)

Plus Two Computer Application Trends and Issues in ICT Three Mark Questions and Answers

Question 1.
Compare GSM and CDMA standards.
Answer:
1. Global System for Mobile (GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

2. 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 (ETSlJ GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

3. 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.

4. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel.

This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

Question 2.
Differentiate GPRS and EDGE?
Answer:
1. 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 charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

2. 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.

Question 3.
Explain the features of Android OS.
Answer:
It is a Linux based OS for Touch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends. In 2005, Google acquired this.

Ateam led by Rubin developed a mobile device platform powered by the Linux Kernel. The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects.

In 2007 onwards this OS is used in many mobile phones and tablets. Android SDK(Software Development Kit) is available to create applications(apps) like Google Maps, FB, What’s App.etc. It is of open source nature and many Apps are available for free download from the Android Play Store hence increase the popularity.

Question 4.
What is cyberspace?
Answer:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too. Latertelephoneswere used for voice communication.

Nowadays telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

Question 5.
Why is cyberspace called a virtual world?
Answer:
The telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

This is an imaginary world. We can see persons with different behaviour. Because of good and bad people we can’t believe blindly. If we search a solution fora problem thousands of answers will get instantly and may confused us.

Question 6.
What is copyright? How does it differ from patent?
Answer:
1. Copyright:
The trade mark is ©, copy right is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

2. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

Question 7.
Why is Cyber law important?
Answer:
Just like normal crimes (theft, trespassing private . area, destroy, etc.) Cyber crimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc.) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc. and some of them commit suicide.

Cyber law ensures the use of computers and Internet by the people safely and legally. It consists of rules and regulations like Indian Penal Code (IPC) to stop crimes and for the smooth functions of Cyber world. Two Acts are IT Act 2000 and IT Act Amended in 2008.

Question 8.
“Infomania has became a psychological problem”. Write your opinion.
Answer:
Infomania is the excessive desire(lnfatuation) 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 Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Question 9.
What do you mean by big data in business? Explain big data analytics.
Answer:
Earlier before buying a product people may consult two or three shop keepers or local friends and take decisions. But nowadays before taking decisions people search shopping sites, social network groups(Facebook, WhatsApp, Instagram, twitter, etc), web portals etc.forthe best prices. Almost all online sites have product comparison menus.

By this we can compare the price, features, etc. Earlier a product is created and customers are forced to buy. But today customer is the King of the market, so products are created for the choices of the customers.

So companies gathering information about the customers from various sources such as social medias like Internet forums, social blogs, Micro blogs, etc. The volume of such data is very large and considered as big data in business.

With the help of a s/w analysis this big data and generate a report that contain all the information such as choices, taste, needs, status, etc of a customer.

Question 10.
What do you mean by business logistics?
Answer:
It is the management of the flow(transportation) of resources such as food, consumer goods, services, animals, etc in a business between the point of origin (source) and the point of consumption (destination) in order to meet the needs of companies and customers.

Business logistics consists of many more complexities. The effective use of hardware and software reduces the complexities faced in Business logistics. For this the hardware used is RFID(Radio Frequency Identification) tag and the reader.

It is like the barcode. The RFID tag contains all the details of a product and it consists of a combination of transmitter and a receiver. The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

Question 11.
How does RFID improve the way business is done?
Answer:
The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

In business the tag(contains details about Product code, Price, Batch no., Manufacturing Date, Expiry date, etc.) is stick on the cartons and by using RFID reader process it speedily.

Question 12.
Define Mobile computing.
Answer:
1. Mobile computing:
The drawbacks of Desk computers are, it is heavy and power consumption rate is high and it is not portable(not mobile).

The advancements in computing technology, light weight, and low power consumption have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc.

Nowadays instead of desktops, light weight and low power consumption devices are used because they are cheap and common. Moreover, people are able to connect to others through internet even when they are in move.

Question 13.
“2G networks introduced data services for the mobile. Two popular standards of 2G systems are GSM and CDMA.”

1. Expand CDMA.
2. Discuss the different technologies that are used to enhance data communication features of GSM.

Answer:
1. Code Division Multiple Access.

2. different technologies:

a. 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 charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

b. 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.

Question 14.
“Mobile communication offers many services apart from the basic voice calling facility”

1. Name the service which helps to locate a geographical position anywhere on the earth.
2. Differentiate SMS and MMS.

Answer:
1. GPS

2. Differentiate SMS and MMS:

a. Short Message Service(SMS):
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems. It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/ 12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

b. Multimedia Messaging Service (MMS):
It allows sending Multi-Media(text, picture, audio and video file) content using mobile phones. It is an extension of SMS.

Question 15.
Write a short note on GPS.
Answer:
Global Positioning System(GPS):
It is a space-based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites.

The system provides critical capabilities to military, civil and commercial users around the world. It is maintained by the United States government and is freely accessible to anyone with a GPS receiver.

GPS was created and realized by the U.S. Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Question 16.
Write a short note on Android OS.
Answer:
1. Android OS:
It is a Linux based OS for touch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends.

In 2005, Google acquired this. A team led by Rubin developed a mobile device platform powered by the Linux Kernel. The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects. In 2007 onwards this OS is used in many mobile phones and tablets.

Question 17.
“In some of the states ration cards have been replaced by smart cards”. What is a smart card? List any two advantages of replacing ration card using smart cards?
Answer:
Smart Cards:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data.

The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Question 18.
Business firms have started watching the conversation and opinions posted in social media. Why?
Answer:
Business firms gathering information about the customers from various sources such as social medias like Internet forums, social blogs, Micro blogs, etc. The volume of such data is very large and considered as big data in business.

With the help of a s/w, analyses this big data and generate a report that contain all the information such as choices, taste, needs, status, opinions, suggestions etc of a customer.

Question 19.
“The following is a figure of a tag which helps in business logistics”

1. Name the tag shown in the figure.
2. How it helps business logistics?

Answer:
1. RFID tag

2. The data stored in the RFID tag can be accessed by a special reader and to read the data no need of RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

In business the tag(contains details about Product code, Price, Batch no., Manufacturing Date, Expiry date, etc.) is stick on the cartons and by using RFID reader process it speedily.

Question 20.
“IPR (Intellectual Property Right) encourages innovation” Justify.
Answer:
Some people spend lots of money, time body and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc.

These type of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR). It enables to earn recognition, financial benefit, can sell the innovation, etc. It motivates further innovation.

Question 21.
Write a short note on

1. Trademark
2. Industrial design

Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 22.
Compare patent and Trade mark.
Answer:
1. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

2. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Question 23.
Write a short note on intellectual property theft.
Answer:
Intellectual property theft: The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres are less significantly and undergoes huge loss. (Eg: Premam, Bahubali, etc).

Question 24.
What is cyberspace?
Answer:
a. CyberSpace:
Earlier Traditional communication services such as postal service (Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used.

Its speed was high but it has lot of limitations and expensive too. Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual (unreal) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

Question 25.
Write a short note on the importance of IT Act 2000.
Answer:
Information Technology Act 2000 (amended in 2008):
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data and Information in Electronic format and provide legal infrastructure for E-commerce, in India.

This is developed to promote IT industry, control e-commerce, also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person those who violate this will be prosecuted. In India IT bill introduced in the May 2000 Parliament Session and it is known as Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008.

Question 26.
“Informania affects people’s lives and their loved ones” Justify.
Answer:
Infomania is the excessive desire(lnfatuation) 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 Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Plus Two Computer Application Trends and Issues in ICT Five Mark Questions and Answers

Question 1.
Explain generations in mobile communication?
Answer:
The mobile phone was introduced in the year 1946. Early stage it was expensive and limited services hence its growth was very slow. To solve this problem. Cellular communication concept was developed in 1960’ satBell Lab. 1990’s onwards cellular technology became a common standard in our country.
The various generations in mobile communication are:

a. First Generation networks (1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

b. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.

1. Global System for Mobile (GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

(i) GPRS (General Packet Radio Services):
It is a packet oriented mobile data seivice on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

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

2. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

c. Third Generation networks(3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

d. Fourth Generation networks(4G):
It is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

e. Fifth Generation networks:
This is the next generation network and expected to come in practice in 2020. It is more faster and cost effective than other four generations. More connections can be provided and more energy efficient.

Question 2.
Explain different categories of cyber crimes in detail.
Answer:
Just like normal crimes(theft, trespassing private area, destroy, etc,) Cyber crimes(Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc and some of them commit suicide.

A. Cyber crimes against individuals:

1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number, etc.), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friendship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake sms, fake emails, etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

B. Cyber crimes against property:
Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading latest movies, etc, are considered as cyber crimes against property.

1. Credit card fraud:
Stealing the details such as credit card number, company name, expiry date, cvv number, password, etc. and use these details to make payment for purchasing goods or transfer funds also.

2. Intellectual property theft:
The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself.

Hence the revenue from the theatres are less significantly and undergoes huge loss. (Eg: Premam, Bahubali, etc). Copying a person’s creation and present as a new creation is called plagiarism. This can be identified some tools (programs) available in the Internet.

3. Internet time theft:
This is deals with the misuse of WiFi Internet facility. If it is not protected by good password there is a chance of misuse our devices (Modem/Router) to access Internet without our consent by unauthorized persons. Hence our money and volume of data (Package) will lose and we may face the consequences if others make any crimes.

C. Cyber crimes against government:
The cyber crimes against Govt, websites is increased significantly. For example in 2015 the
website of Registration Department of Kerala is hacked and destroys data from 2012 onwards.

1. Cyber terrorism:
It is deals with the attacks against very sensitive computer networks like computer-controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc. This is a very serious matter and may lead to huge loss (money and life of citizens). So Govt is very conscious and give tight security mechanism for their services.

2. Website defacement:
It means spoil or hacking websites and posting bad comments about the Govt.

3. Attacks against e-governance websites:
Its main target is a Web server. Due to this attack the Web server/ computer forced to restart and this results refusal of service to the genuine users.

If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server. Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Question 3.
Discuss various generations of mobile communication.
Answer:
The various generations in mobile communication are:

a. First Generation networks(1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

b. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.

1. Global System for Mobile(GSM):
It is the most successful standard. It uses narrow band TDMA (Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

(i) 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 charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

(ii) 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.

2. Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

c. Third Generation networks (3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

d. Fourth Generation networks (4G):
It is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

e. Fifth Generation networks (5G):
This is the next generation network and expected to come in practice in 2020. It is more faster and cost effective than other four generations. More connections can be provided and more energy efficient.

Question 4.
“Due to anonymous nature of Internet it is possible for the people to engage in variety of criminal activities”. Justify the statement with special reference to cyber crimes taking place against Individual.
Answer:
Cyber crimes against individuals:
1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number, etc), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friendship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias(Facebook,Twitter,etc), fake sms, fake emails etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

Students can Download Chapter 5 Accounting Software Package – GNUKhata Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 5 Accounting Software Package – GNUKhata

Plus Two Accountancy Accounting Software Package – GNUKhata One Mark Questions and Answers

Question 1.
_________ is a free and open source software for accounting, developed by Digital Freedom Foundation.
Answer:
GNUKhata

Question 2.
The first step in GNUKhata is ________
(a) To create an organisation
(b) To create Ledger account
(c) Voucher entry
(d) None of these
Answer:
(a) To create an organisation

Question 3.
Write the path for starting GNUKhata
Answer:
Applications → Office → GNUKhata

Question 4.
Which among the following is not a level of users in GNUKhata
(a) Managing Director
(b) Admin
(c) Manager
(d) Operator
Answer:
(a) Managing Director

Question 5.
GNUKhata has ………… (a) …………… predetermined groups and ……… (b) ……….. subgroups
Answer:
(a) 13
(b) 16

Question 6.
Amount received by way of cash or cheque is recorded in GNUKhata by using ________ voucher type
Answer:
Receipt Voucher (F4)

Question 7.
Short cut keys for activating Credit Note is _______
(a) Ctrl + 1
(b) Ctrl + 2
(c) Ctrl + 3
(d) Ctrl + 3
Answer:
(d) Ctrl + 3

Question 8.
To clone a voucher means
(a) To delete a voucher
(b) To copy a voucher
(c) To edit a voucher
(d) To print a voucher
Answer:
(b) To copy a voucher

Question 9.
From ________ menu, we will enable to view Ledger,
Trial Balance, Balance sheet, Profit and Loss AC etc.
Answer:
Report menu

Question 10.
Which among the following is NOT a feature of GNUKhata?
(a) Free and open source accounting software
(b) Based on double entry book keeping
(c) Coding of Ledger account
(d) Linking of sales and purchase transactions to invoices
Answer:
(c) Coding of Ledger account

Question 11.
What is FOSS?
(a) Free and Open Security Software
(b) Financial Accounting Open Source Software
(c) Free and Open System Software
(d) Free and Open Source Software
Answer:
(c) Free and Open Source Software

Question 12.
Left side of the GNUKhata welcome screen shows __________
(a) Three buttons and dialog box
(b) Description about double entry principles
(c) Printer settings and network
(d) Description about the advantages and developer of GNUKhata
Answer:
(d) Description about the advantages and developer of GNUKhata

Question 13.
Short Cut keys Shift + Ctrl + R is used for ________
Answer:
Create Organisation

Question 14.
Choose the statement which is NOT True
(a) We can create a new group but a new sub group can not be created
(b) We can not create a new group but a new sub group can be created
(c) We cannot delete a group or a sub group
(d) We cannot create a sub group of a sub group
Answer:
(a) We can create a new group but a new sub group can not be created

Question 15.
______________ is prepared by an account holder to check whether all cheques deposited by him are cleared and all cheques
issued by him are presented on not.
(a) Bank Reconciliation statement
(b) Trading, Profit and Loss Account & Balance Sheet
(c) Cash Book and Pass Book
(d) Trial Balance
Answer:
(a) Bank Reconciliation statement

Question 16.
Sub Group “Sundry Debtors” coming under ________ Group
Answer:
Current Asset

Question 17.
Choose the right statement
(a) Once organisation is created, the financial year can be changed whenever necessary
(b) Once organisation is created, the financial year can not be changed.
Answer:
(b) Once organisation is created, the financial year can not be changed.

Question 18.
Create is mandatory in GNUkhata
(a) Internal Auditor
(b) Operator
(c) Manager
(d) Admin
Answer:
(d) Admin

Question 19.
Which one of the following is free and open source software for Accounting
(a) Tally
(b) PeechTree
(c) GNUKhata
(d) None of these
Answer:
(c) GNUKhata

Question 20.
Which accounting software can be easily transformed into Indian languages?
Answer:
GNUKhata

Question 21.
When we open GNUKhata for the first time, we will see ______
(a) Create Organisation window
(b) Voucher Entry window
(c) Welcome screen
(d) User name and Password window
Answer:
(c) Welcome screen

Question 22.
Which type of organisation can be created in GNUKhata?
(a) Profit Making Organisations
(b) Not for Profit Organisations
(c) (a) and (b)
(d) None of these.
Answer:
(c) (a) and (b)

Question 23.
Only ______ user can log in as ‘Admin’
(а) One
(b) Two
(c) Many
(d) None
Answer:
(a) One

Question 24.
Write path for deleting an organisation.
Answer:
Log in as ‘Admin user’ → Administration menu → Select ‘Delete Organisation’.

Question 25.
The sub Group ‘Plant and Machinery’ is coming under _______________ group.
Answer:
Fixed Assets

Question 26.
Which among the following is not a Profit and loss A/c Group in GnuKhata.
(a) Direct Income
(b) Direct Expenses
(c) Indirect Income
(d) None of these
Answer:
(d) None of these

Question 27.
There are system generated ledger accounts in GNUkhata
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(d) Four

Question 28.
In GNUKhata, while creating a Ledger account, we have to enter opening balance only if _________
(a) The account belongs to Trading, P/La/c group
(b) The account belongs to a Balance sheet group
(c) It is a cash account
(d) It is a Liability account
Answer:
(b) The account belongs to a Balance sheet group

Question 29.
Contra Voucher is used to record.
(a) Deposits or withdrawals of cash from the bank
(b) Transfer of fund from one bank to another
(c) Transfer of cash to petty cash
(d) All the above
Answer:
(d) All the above

Plus Two Accountancy Accounting Software Package – GNUKhata Two Mark Questions and Answers

Question 1.
Write the Path to display Trial Balance.
Answer:
Report → Trial Balance → Enter Date → Select Trial Balance Type → Click on view button

Question 2.
What is BRS?
Answer:
A Bank Reconciliation Statement (BRS) is a Statement prepared by the depositor for the purpose of reconciling the cash book balance with the pass book balance on a certain date.

Question 3.
There are four levels of users in GNUKhata. Who are they?
Answer:

1. Admin
2. Manager
3. Operator.
4. IntemalAuditor.

Question 4.
Match the following
Group Name Sub-Group Name

Group Name	Sub-Group Name
(a) Fixed Assets	(i) Sundry Creditors
(b) Current Liability	(ii) Loans and Advances
(c) Investments	(iii) Furniture
(d) Current Assets	(iv) Fixed Bank Deposits

Answer:
(a)- (iii); (b) – (i); (c) – (iv); (d) – (ii)

Question 5.
List down the Income and Expenditure account groups in GNUKhata
Answer:

• Direct Income
• Direct Expenses
• Indirect Income
• Indirect Expenses

Question 6.
Identify the group and sub group under which the following ledger accounts are to be created.
(a) Depreciation
(b) Carriage inward
(c) Bills Receivable
(d) Drawings
Answer:

Question 7.
List down any two features of GNUKhata.
Answer:

1. It is free and open source accounting software.
2. It is based on double entry book keeping.
3. All financial reports can be prepared.
4. Display of dual ledger facility.
5. Attachment of source document to vouchers is possible.
6. Linking of sales ad purchases tranactions to invoice.
7. Export or import of data from spread sheet is possible.
8. It ensures password security and data audit facility.

Question 8.
Define Groups and Sub Groups in GNUKhata.
Answer:
Grouping of account is a method of organising the large number of ledger accounts into sequential arrangement. GNUKhata has 29 predefined Groups and Sub Groups. Out of these, 13 are Groups and 16are Sub Groups.

Question 9.
What are the three buttons in Right side of the GNUKhata welcome screen?
Answer:

1. Select Existing Organisation
2. Create Organisation
3. Language

Question 10.
What is the use of ‘Theme’ tab in GNUKhata Menu Bar?
Answer:
‘Theme’ tab is used to change the theme Background of the screen.

Plus Two Accountancy Accounting Software Package – GNUKhata Three Mark Questions and Answers

Question 1.
What are the specialities of system generated Ledger Accounts
Answer:

1. GNUKhata has four system-generated ledger account.
2. We cannot change the name of these accounts.
3. We can not delete these accounts.

Question 2.
What are the limitations of GNUKhata?
Answer:

1. Only trained employees can use GNUKhata.
2. Updations and additions in GNUKhata Software may seriously affect the accounting system.

Question 3.
Mr. Roby prepared a chart for grouping the ledger account. Some errors are identified by his teacher. Help him to rectify the errors.

Ledger Account	Group
(a) Office expenses	Direct Expenses
(b) Carriage Inward	Direct Expenses
(c) Salary	Indirect Expenses
(d) Sales Tax	Direct Expenses
(e) Purchases	Indirect Expenses

Answer:

Ledger Account	Group
(a) Office expenses	Indirect Expenses
(b) Carriage Inward	Direct Expenses
(c) Salary	Indirect Expenses
(d) Sales Tax	Indirect Expenses
(e) Purchases	Direct Expenses

Question 4.
Three options are available to view Trail Balance. What are they?
Answer:

1. Net Trial Balance
2. Gross Trial Balance
3. Extended Trial Balance

Question 5.
Define the following
(a) Credit note voucher
(b) Journal voucher
(c) Sales voucher
Answer:

(a) Credit note	For recording a reduction in the price charged to a customer. Function key: Ctrl + 3
(b) Journal	For recording rectifying entries, transfer entries, adjustment entries, purchase of fixed assets on credit and sale of fixed assets on credit. Function key: F9
(c) Sales	For recording the return of goods by a customer Function key: Ctrl + 1

Question 6.
Write the Debit and Credit for the following accounting Transactions
(a) Returned goods to Vinu Rs. 5000
(b) Price of goods sold to Mohan reduced by Rs. 500
(c) Vinu returned goods worth Rs. 3000
Answer:

Question 7.
What is the importance of ‘Admin’ user in GNUKhata?
Answer:

• Only one user can login as Admin
• The creator of organisation is always considered as Admin User
• The ‘Admin’ user can create the other users ‘Admin’ user

Question 8.
List down the major Sub Groups in the Group ‘Fixed Assets”?
Answer:

• Building
• Furniture
• Land
• Plant and Machinery

Question 9.
Match the following

Answer:
(a) → ii → x
(b) → iii → y
(c) → i → z

Question 10.
Rearrange the diagram in the right way

Answer:

Plus Two Accountancy Accounting Software Package – GNUKhata Four Mark Questions and Answers

Question 1.
How to record the opening balance amount in opening Stock Account?
Answer:
stage 1:

• Step 1 → Click on Master tab.
• Step 2 → Select Edit account option from the drop down list.
• Step 3 → Select Stock at the beginning from the edit account window.
• Step 4 → Click on Edit Button.
• Step 5 → Enter opening balance amount.
• Step 6 → Click on Save button to save the amount.

Stage 2:

• Step 1 → Click on Transaction tab.
• Step 2 → Select Journal option to activate Journal Voucher.
• Step 3 → Enter voucher number.
• Step 4 → Enter opening date.
• Step 5 → Select Opening stock a/c in Dr. Account and enter opening stock amount.
• Step 6 → Select stock at the beginning a/c in the Cr. account and enter the opening stock amount.
• Step 7 → Give narration.
• Step 8 → Click on Save button or Press Enter key.

Question 2.
What are the reasons behind the difference in pass book balance and cash book balance.
(a) Cheques issued to suppliers, but not cleared
(b) Cheques deposited, but not honoured
(c) Accounts directly deposited by customers in bank account through electronic transfer
(d) Charges levied by the bank
(e) Interest on deposits credited by the bank
Answer:
S

Question 3.
Give a brief explanation about Bank Reconciliation Statement in GNUKhata
Answer:

1. Bank Reconciliation Statement is done to identify why there is difference between the two balances (Cash Book balance and pass book balance)
2. The date of transaction is the Transaction Date
3. The date on which a particular transaction appears in a Pass Book is called the Clearance Date.
4. The period for which Bank Reconciliation is done is called Reconciliation Period.
5. The reconciliation is done by comparing the Trans-action Date and Clearance Date.

Question 4.
GNUKhata provides the facility to view Single Ledger Account and Dual Ledger Account. Explain.
Answer:
GNUKhata provides the facility to view two ledger accounts simultaneously for comparing entries. We can view side by side:

• Two different Ledger accounts for the same period.
• Two different ledger accounts, each for different period.
• Same ledger accounts for different periods.

Plus Two Accountancy Accounting Software Package – GNUKhata Five Mark Questions and Answers

Question 1.
What do you mean by ‘Year End Activities’?
Answer:
Year End Activities consists of the following Year End Activities

• Closing Ledger Accounts.
• Opening Ledger Accounts for the next year.

(A) Closing of Accounts:
GNUKhata has a module to Close books at the click of a mouse, when this module is activated, balances in all expenses and incomes accounts are transferred to Profit and Loss/ Income & Expenditure account and the accounts are closed. After Close
books is done, transactions cannot be recorded but ledger accounts can be viewed.

(B) Opening Ledger Accounts for the next year:
GNUKhata has a module to Roll over ledger accounts at the click of a mouse. This module opens accounts for the next year automatically and since it is automatic, it is error free. When Roll over is done.

• New organisation with the same name and type is created by GNUKhata for the next accounting year.
• Assets and liabilities accounts automatically open under its respective Groups and Sub Groups with opening balances.
• Expenses and income accounts are opened under their respective Groups and Sub Groups without opening balances
• The closing stock gets transferred to the next year as opening stock.

This module can only be activated after close Books is done.

Question 2.
Write the steps in the preparation of final accounts by using GNUKhata.
Answer:

• Step 1 – Creation of new organisation
• Step 2 – Creation of Admin and Login
• Step 3 – Creation of Sub Groups
• Step 4 – Creation of Ledger Accounts
• Step5 – Voucher Entry
• Step 6 – Editing opening stock and Closing stock account
• Step 7 – Display Ledger Accounts
• Step 8 – Display Trial Balance
• Step 9 – Display P & L Account
• Step 10 – Display Balance Sheet

Question 3.
What are the different vouchers in GNUKhata?
Answer:

Question 4.
Write the Journal entry, voucher Type and Short cut keys of the following transactions.

1. 1. Cash purchases worth Rs. 10,000
   2. Sold goods to Akhil Rs. 6, 000
   3. Akhil returned goods worth Rs. 2,000
   4. Price of goods sold to Akhil reduced by Rs. 500
   5. Credit Purchases from Linjo Rs, 8,000
   6. Returned goods to Linjo Rs. 3,000

1. Linjo reduced the price of goods sold to us by Rs. 1,000
2. Cash received from Akhil Rs.3,000
3. Cash paid to Linjo Rs. 4,000
4. Cash sales ₹ 800

Answer:

Question 5.
Menu Bar has several Tabs. Can you list down a few of them? Also, mention the short cut keys and activities of such Tabs.
Answer:
Menubar has the following Tabs, The shortcuts for and the activities included in these Tabs are:

Question 6.
Create an organisation with the following particulars.

• Name: JOSHWIN ZIAN Enterprises
• Accounting Year: 01-01-2018 to 31-12-2018
• Type of organisation: Profit Making
• Maintain Inventory: No
• Records Admin Name and Password: Appu, Appose
• Security Question & answer: Best friends name? Chikku
• Address: 1234, M. C. Road, Velappaya, Tcr.

Answer:

• Step1 – Start GNUKhata (Application → Office → GNUKhata)
• Step 2 – Click on “Create Organisation”.
• Step 3 – Enter Organisation Name (JOSHWIN ZIAN Enterprises), Press Enter/Tab
• Step 4 – Select the case, Use down arrow key to select any one of the case and press Enter
• Step 5 – Select organisation type – ‘Profit making’ and press Enter/Tab
• Step 6 – Enter From date – 01-01-2018, Press Enter
• Step 7- Skip the box of Inventory
• Step 8 – Click on Proceed or Press Enter to open Create Admin window
• Step 9 – Enter user Name – Appu, Password – Appose and Press Enter/ Tab
• Step 10 – Enter security Question – “Best friends’s Name’, Press Enter/Tab
• Step 11 – Enter Answer to security Question Chikku, Press Enter to Login as Admin
• Step 12 – Select Edit organisation particulars from Master Menu
• Step13 – Enter Organisation – 1234, MC Road, Velappaya, TCr, and Click on Save

Plus Two Accountancy Accounting Software Package – GNUKhata Practical Work Questions and Answers

Question 1.
Create an organisation with the following particulars

• Name: TALLMEN WELFARE ASSOCIATION
• Accounting year: 1-1-2018 to 31-12-2018
• Type of organisation: Not for profit
• Maintain Inventory Records: No
• Admin Name & password: Visal, TCR2018
• Security Question and Answer Mothers’maiden name? Meenakshy

Procedure:
Step 1 – Start GNUKhata
Applications → Office → GNUKhata

Step 2 – Create New Organisation
Click on ‘Create Organisation’ Tab Enter the following informations.

• Organisation Name: TALLMEN WELFARE ASSOCATION
• Case: As-Is
• Organisation type: Not for Profit
• Financial year: 1-1 -2018 to 31-12-2018
• Inventory: (Leave it blank)

Click on the Proceed button. Now the organisation is created

Step 3 – Create Admin and Login
Press Enter or Alt+Shift+P to Open Create Admin module.

• User Name: Visal
• Password: TCR2018
• Confirm Password: Retype the password
• Security Question: Mothers’ Maiden name
• Answer to the security Question: Meenakshy

Press enter to Login as Admin

Step 4 – Select Edit organisation Particulars from Master Menu. Enter the address of the organisation. Press Enter key to save the details.
Output:

Question 2.
Create an organisation with the following particulars:

• Name: Mahindra Enterprises
• Accounting year: 1-04-2018 to 31-3-2019
• Type of organisation: Profit-making
• Maintain inventory Records: No
• Admin Name & Password: Vinesh, Vinumon
• Security Question & answer: Best friend’s name? Rohit
• Address: 268, Poonkunnam, Thrissur

Procedure:
Step 1 – Start GNUKhata
Applications → Office → GNUKhata

Step 2 – Create New Organisation Click on ‘Create Organisation’ Tab
Enter the following informations.

• Organisation Name: Mahindra Enterprises
• Case: As-Is
• Organisation Type: Profit Making
• Financial year: 1-04-2018 to 31-3-2019
• Inventory : (Leave it blank)

Click on the Proceed button, is created. Now the organisation

Step 3 – Create Admin and Login
Press Enter or Alt+Shift+P to Open Create Admin module.

• User Name: Vinesh
• Password: Vinumon
• Confirm Password: Retype the password
• Security Question: Best friend’s name?
• Answer to the security Question: Rohit

Press enter to log in as Admin

Step 4 – Select Edit organisation Particulars from Master Menu. Enter the address of the organisation
Press Enter key to save the details.
Output:

Question 3.
Create the following ledger accounts of Najim Traders

procedure:
Step 1 – Create a New Organisation ‘Najim Traders’- Profit Making with other imaginary data.

Step 2 – Creation of Ledger Accounts
Master menu → Create Account
Analysis Table

Step 3 – Create all the ledger accounts and save the details

Step 4 – Display ledger Accounts Report → Ledger view
OR
Display Trial Balance ReportTrial balance → (Select the type of Trial balance) → view
Output:

Question 4.
Create the following Ledger accounts of Anusree agencies.

procedure:
Step 1 – Create New Organisation ‘Anusree Agencies’ – Profit Making with other imaginary data
Enter the following information.

Step 2 – Creation of Ledger Accounts
Master Menu → Create Account

Question 5.
Record the following transactions in appropriate vouchers

• 1/6/2018 – Riyas started a business with cash – Rs. 1,00,000
• 2/6/2018 – Opened a bank account with Canara Bank – Rs. 30,000
• 3/6/2018 – Bought furniture for cash – Rs.10,000
• 4/6/2018 – Purchase goods from Niyastores – Rs,25,000
• 5/6/2018 – Cash sales – Rs.12,500
• 6/6/2018 – Paid for stationery – Rs. 500
• 7/6/2018 – Cash paid to Niya stores – Rs. 20,000
• 8/6/2018 – Cash withdraw from Canara bank – Rs.10,000

Procedure:
Step 1 – Create a New Organisation – Profit Making – with imaginary data

Step 2 – Creation of Ledger Accounts
Master menu → Create Account
Analysis Table

Step 3 – Create all the ledger accounts and save the details

Step 4 – Enter all the transactions in appropriate vouchers

1. Select the appropriate voucher form Voucher menu
2. Enter the Voucher Number and date
3. Select the Debit account name and enter the amount, then press Enter
4. Select the credit account name and enter the amount
5. Enter narration and click on Save.

Output:

Question 6.
Enter the following transactions of Royal Enterprises and display Trial Balance

Procedure:
Step 1 – Open GNUKhata
Application Office GNUkhata

Step 2 – Create Organisation “Royal Enterprises – Profit making – with Imaginary data.

Step 3 – Create Admin
Enter the details (Imaginary)

Step 4 – Create appropriate Ledger Accounts
Master → Create account → Select Group → Sub Group → Type Account name → Save
Analysis Table

Step 5 – Enter all transactions in appropriate vouchers
Voucher menu → (appropriate voucher) → Enter voucher number → Date → Select Debit account → enter debit amount → Press Enter → Select Credit Account → Enter credit amount → Press Enters → Enter narration → Click on Save

Step 6 – Display Trial balance Report → Trial Balance
Output:

Question 7.
From the following Trial Balance, Prepare Trading Profit and loss account and Balance sheet of Athulya Ltd. as on 31/3/2018
Trial Balance as on 1/4/2017

Transaction during the year 2017 – 18 is given below

• Sales (Cash) – Rs. 51000
• Sales (Credit) – Rs. 34000
• Purchases (Credit) – Rs. 68000
• Office expenses – Rs. 2400
• Wages. – Rs. 1.600
• Cash received from Debtors – Rs. 16,000
• Cash paid to creditors – Rs. 35,000
• Discount received – Rs. 1000
• Salary – Rs. 5000
• Rent – Rs. 11,000
• Carriage Outward – Rs. 700
• Commission Paid – Rs. 500
• Insurance Premium – Rs. 4000

Other Adjustments:

• Salalry Prepaid – Rs. 1500
• Rent outstanding – Rs. 1000
• Wages Outstnading – Rs. 800
• Insurance Premium prepaid – Rs. 2000
• Closing stock – Rs. 3700
• Depreciation of Machinery – Rs. 10%

Procedure:
Step 1 – Create an organisation Athulya Ltd- Profit making for the year 1 -4-2017 to 31/3/2018 with imaginary data.

Step 2 – Creation of Ledger Accounts
Master → Create account
Analysis Table (Trail Balance)

Create all the ledger a/c under appropriate group and sub Group.

Step 3 – Voucher Entry
Enter all the transactions using appropriate voucher type
Voucher menu /Transaction menu → Select appropriate voucher type → Enter voucher number → Enter voucher Date → Enter Debit account and amount → Enter Credit account and amount → Save
Analysis Table (Transactions)

Analysis Table (Transactions)

Step 4 – Enter the value of opening stock and closing stock

1. Master menu → Edit Account → Select Stock at the beginning a/c → Click Edit Button
2. Enter 14000 as Opening Balance. Click on Save button
3. Enter the values of Opening Stock and closing stock through appropriate voucher

Voucher menu/Transaction menu → Journal
Analysis Table (Stock)

Step 5 – Display Trial Balance
Report → Trial Balance → Enter Date → Type → Click on the View button

Step 6 – Display Profit and loss Account
Report → Profit and loss → Enter Date → Click on View Button

Step 7 – Display Balance Sheet
Report → Balance sheet → Enter Date → Type → Click on View Button
Output:

Question 8.
Prepare a Bank Reconciliation Statement of Cili Traders, a Profit making organisation, for the period 1-1-2018 to 31-3-2018

While comparing with Cash book with Pass book, the following details were noted.

1. Clearance date of cheque No.7125 was on 7/4/18
2. Cheque No. 9614 cleared only on 2/4/2018
3. Cheque No. 7127 is cleared on 1/4/18
4. Clearance date of cheque No.4528 was on 12/2/18
5. Cheque No. 7129 was cleared on 19/3/2018
6. All other transaction’s clearance date is same as transaction date

procedure:
Step 1 – Start GNUKhata
Application → Office → GNUKhata

Step 2 – Create Organisation
Create an organisation → Cili Traders → Profit making → 1-1-18 to 31-12-2018 → Create admin with imaginary data.

Step 3 – Create Ledger accounts
Master Menu → Create Account → Select Group → SubGroup → Account Name → Save
Analysis Table

Step 4 – Voucher Entry
Transaction/Voucher → Select appropriate voucher → Number → Date → DebitCredit → Save

Step 5 – Preparation of Bank Reconciliation Statement Press Alt +R

OR

1. Master → Bank rEconciliation statement → Select Bank’s Name (Canara Bank) → Set Reconciliation period → Click on view button to view Bank reconciliation screen
2. Enter clearance date → Click on View statement button
3. To show the cleared Transactions, click on cleared items
4. To show the uncleared transactions, click on uncleared items

Output:

Question 9.
Enter the following transactions of Sibi Traders and Display Final Accounts as on 31/3/2018

procedure:
Step 1 – Open GNUKhata
Application Office → GNUKhata

Step 2 – Create Organisation
Click on Create organisation → Enter Details → Sibi Traders → Proft Making → 1-1-2018 to 31-12-2018

Step 3 – Create Admin
Enter imaginary data

Step 4 – Creation of Ledger accounts
Master Menu → Create account → Select Group → SubGroup → Type Account name → Save
Analysis Table

Step 5 – Voucher Entry (based on above Analysis Table)
Transaction menu/Voucher menu → Select appropriate voucher → Enter voucher No → Date → Debit account and amount → Credit account and amount → Narration → Save

Step 6 – Display Trial balance
Report → Trial balance
Display Profits and loss Account
Report → Profits Loss account

Step 8 – Display Balance Sheet
Report → Balance sheet
Output:

Question 10.
From the following transactions of Fijo and Joshy associates, prepare Trial Balance, Trading Profit, and Loss account and Balance Sheet.

Adjustments

1. Depreciation of Furniture Rs. 640
2. Closing stock valued at Rs.2900

Procedure:
Step 1 – Create Organisation
Click on create Organisation Tab Enter the details
Fijo and Joshy Associates – Profit Making- 1/1/2018-31/12/2018. Enter all other details with imaginary data.

Step 2 – Create Ledger Account
Master → Create Account
Analysis Table


Create all the above ledger accounts under appropriate Groups and Subgroups.

Step 3 – Voucher Entry
Enter the adjustment entries in appropriate voucher type
Voucher menu/Transaction menu → Select Voucher type → Number → Date → Debit & Credit → Save
Analysis Table (Adjustments)

Step 4 – Display Trial Balance
Report → Trial Balance

Step 5 – Display Profit & Loss Account
Reports → Profit & Loss Account

Step 6 – Display Balance Sheet
Report → Balance Sheet
Output:

Question 11.
Given below the Trial Balance of Akhil &Athulya associates as a 31/3/2018. Prepare Final Account for the year 2018-19

Other Information:

1. Depreciation of Machinery-4500
2. Rent outstanding-3000
3. Building depreciated at 1.0%
4. Salary Outstanding – 3500
5. Value of closing stock -18000
6. Value of Motor vehicle depreciated by 5%

Procedure:
Step 1 – Open GNUKhata
Applications → Office → GNUKhata

Step 2 – Create Organisation
Click on Create Organisation tab. Enter the details
Akhil and Athulya Associates – Profit making 1-4-2018 to 31-3-2019.

Step 3 – Create Admin & Login Use imaginary data.

Step 4 – Creation of new subgroup
Here we need to create a new Sub Group Vehicles under the Group Fixed Assets. Master → Create Account → Select Group Fixed Assets → Select New Sub Group in Sub Group Name → Enter the new Sub Group Name ‘Vehicles’ → Enter the Account Name Motor Vehicles in that Sub Groups Enter the Opening Balance Rs. 1,50, 000 → Save.

Step 5 – Create Ledger Account
Master → Create Account
Analysis Table (Trial Balance)

Create all the ledger accounts under ap¬propriate Groups and Sub Groups

Step 6 – Voucher Entry
Enter all adjustments in appropriate voucher type
Voucher menu/Transaction Menu → Select Voucher → Type → Number → Date Debit account and amount → Credit account and amount → Save.
Analysis Table (Adjustments)

Step 7 – Enter the value of opening stock and closing stock

1. Master Menu → Edit Account → Select stock at the beginning a/c → Click Edit button
2. Enter 15000 as opening balances click on. Save button
3. Enter the values of opening stock and closing stock through appropriate voucher

Voucher Menu/Transaction menu → Journal
Analysis Table (Stock)

Step 8 – Display Trial Balance
Report → Trial Balance

Step 9 – Display profit and loss Account
Report → Profit & Loss

Step 10 – Display Balance Sheet
Report → Balance Sheet
Output:

Question 12.
Enter the following transactions in appropriate vouchers of Jos & Rejina Enterprises and prepare Bank Reconciliation Statement
Step 1 – Open GNUKhata

Verification of the Cash Book with Bank Pass Book revealed that the Cheque Nos. 8117, 8238 & 5314 were cahsed only on 3^rd, 5^th and 8^th July respectively and the clearance date of all other transactions are same as transaction date.
Procedure:
Step 1 – Open GNUKhata
Application → Office → GNUKhata

Step 2 – Create Organisation
Click on Create Organisation → Enter Name → Jos & Rejina Enterprises – Profit Making → (other imaginary data)

Step 3 – Create Admin – Enter the details (imaginary Data)

Step 4 – Create Ledger Account
Master → Create account → Select Group → Sub Group → Type Account name → Save
Analysis Table


Step 5 – VoucherEntry
Transaction / Voucher → Select Appropriate → Enter details.

Step 6 – Preparation of Bank reconciliation statement Press Alt + R
OR

1. 1. Master → Bank Reconciliation statement → Select Bank’s Name (Union Bank) → Set Reconciliation period → Click on the View button to view Bank Reconciliation screen
   2. Enter clearance data Click on View Statement button
   3. To show the cleared Reconciliation click on cleared items
   4. To know the uncleared Transactions, Click on Uncleared items.

Students can Download Chapter 4 Web Technology Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 4 Web Technology

Plus Two Computer Application Web Technology One Mark Questions and Answers

Question 1.
Specify an attribute of HTML tag.
Answer:
DIR: Indicates documents direction. It can take values rtl or Itr.
Eg. <HTML DIR= “rtl”> This specify that document is to read from right to left.

Question 2.
Mention the default value of size attribute of <BASEFONT>tag.
Answer:
3.

Question 3.
Name the tag which has ‘Noshade’ attribute.
Answer:
<HR> Tag

Question 4.
Maximum possible value of the size attribute of the <BASEFONT>tag?
Answer:
7.

Question 5.
Salim developed a personal website. In which he has to create an e-mail link. Can you suggest the protocol used to achieve this task?
Answer:
Mailto:
Eg. <A href= “Mailto:cgj@rediffmail.Com” >Mail to me</A>.

Question 6.
You want to connect your webpage to the web portal www.yahoo.com. Mention the tag that can be used for this.
Answer:
<A> Anchor Tag
Eg. <A href=”www.yahoo.com”>Yahoo</A>

Question 7.
Specify the main attribute of <IMG> tag used to include an image file in web page.
Answer:
SRC. It specify the name of the image file to be included in the page.
Eg. <IMG SRC=“C:/home,Jpg”>

Question 8.
Select the attribute associated with <IMG> tag from the following: (Name, size, Align, value)
Answer:
Align.

Question 9.
Write HTML code forgiving hyper link in webpage.
Answer:
<A HREF= “page2.htm”>Page2</A>

Question 10.
The default colour of Vlink is______.
(a) Blue
(b) Green
(c) Red
(d) Yellow
Answer:
(c) Red

Question 11.
The default colour of A link is______.
(a) Blue
(b) Green
(c) Red
(d) Yellow
Answer:
(b) Green

Question 12.
Saritha is assigned with a task of writing explana¬tory notes in an HTML code. Which tag she can utilize.
Answer:
Explanatory notes can be given using <comment> tag. Using two mark up elements. as<!—and →

Question 13.
Pick the odd man out.
(a) BODY
(b) HTML
(c) CENTER
(d) ALIGN
Answer:
(d) ALIGN, all others are tag.

Question 14.
Your school has a web site, www.myschool.org. You want to create a link to this site from your website. Write the code for implementing this.
Answer:
<A href=”www. myschool.org”>My School</A>

Question 15.
HTML was developed by_______.
Answer:
Tim Berners-Lee.

Question 16.
What are the two major sections of an HTML document?
Answer:
Head section and Body section.

Question 17.
An HTML file is saved with______extension.
(a) .vbp
(b) .mdb
(c) .htm
(d) .fnm
Answer:
(c) .htm or .html

Question 18.
The software used to view web page is______.
(a) Notepad
(b) Web Browser
(c) Web server
(d) Web Editor
Answer:
(b) Web Browser

Question 19.
The default alignment of image in HTML is______.
(a) Left
(b) Right
(c) Center
(d) Inline with content
Answer:
(a) Left

Question 20.
______is an alternative for centralizing a paragraph other than using <P> tag
(a) <ALIGN>
(b) <C>
(c) <CENTER>
(d) <CENTRE
Answer:
(d) <CENTER>

Question 21.
_______attribute of <A> tag specifies the URL of the hyper linked document.
(a) Name
(b) Target
(c) HREF
(d) SRC
Answer:
(c) HREF

Question 22.
What does HTML stands for?
Answer:
HyperText Markup Language.

Question 23.
Communication on web can be classified into_____and______.
Answer:
Client to Server and Server to Server

Question 24.
The protocol which is responsible for splitting the data into smaller packets is______.
Answer:
TCP

Question 25.
The protocol which is responsible for the routing of data packets through the correct destination is____.
Answer:
IP (Internet Protocol)

Question 26.
TCP / IP stands for_____.
Answer:
Transmission Control Protocol/lntemet Protocal.

Question 27.
Identify the protocol responsible fore-mail commu-nication.
(a) DNS
(b) HTTP
(c) TCP/IP
(d) SMTP
Answer:
(d) SMTP

Question 28.
In server to server communication, authentication is done with help of______.
(a) HTTP
(b) Digital certificate
(c) Client
(d) DNS
Answer:
(b) Digital certificate

Question 29.
Which server acts between merchant server and bank server for transferring data in encrypted format?
Answer:
Payment Gateway.

Question 30.
Identify the name of a place where servers and networking systems are placed with high security,
(a) Head office
(b) DNS
(c) Data centre
(d) IIS
Answer:
(c) Data centre

Question 31.
Identify the port number which request for the service of sending e-mail communication.
(a) 22
(b) 25
(c) 53
(d) 80
Answer:
(b) 25

Question 32.
The IP address corresponding to a domain name is present in____server.
Answer:
DNS

Question 33.
Programs embedded in HTML documents are termed as____.
Answer:
Scripts

Question 34.
Running of ______scripts can be blocked by the user.
(a) Client side
(b) Server side
(c) Both client side and server side
(d) None of these
Answer:
(a) Client side

Question 35.
A platform-independent server-side scripting language is_____.
Answer:
PHP

Question 36.
Which among the following tools is used for easy formatting and defining style of a document written in HTML?
(a) Ajax
(b) CSS
(c) JSP
(d) JavaScript
Answer:
(b) CSS

Question 37.
Pick the Odd one from the following list and give reason. (IMG, FONT, BR, ALIGN, PRE)
Answer:
ALIGN which is an attribute, all others are tags.

Question 38.
Choose the correct HTML statement to display an image with file name “kerala.jpg” as the background of the web page.
(a) <IMG src=”kerala.jpg”>
(b) <BODY src=”kerala.jpg”>
(c) <BODY bgcolor=”kerala.jpg”>
(d) <BODY background=”kerala.jpg”>
Answer:
(d) <BODYbackground=”kerala.jpg”>

Question 39.
Two ofthe following HTMLtags have same attribute Align’. Identify them. (<IMG>, <MARQUEE>, <B>, <P>, <BODY>)
Answer:
<IMG> AND <P>

Question 40.
Identify the correct HTML statement to draw a hori-zontal line with half the width of the screen.
(a) <HR width=“50%” size= “3″>
(b) <HR length=“50%” size= “3″>
(c) <HR size= “50%” width= “3″>
(d) <HR width= “50%” length= “3″>
Answer:
(a) <HR width=“50%” size= “3″>

Question 41.
A student wants to display a poem in a web page just like as he entered in the text editor. Which tag in HTML will help him?
Answer:
<PRE>tag

Question 42.
A student created a webpage about his school. The school name is displayed in the page. He wanted to change the style, colour, and size of the school name. Identify the most appropriate tag in HTML needed for that.
Answer:
<FONT> tag

Plus Two Computer Application Web Technology Two Mark Questions and Answers

Question 1.
A student developed a web page about India. He wanted to display a scrolling text moving from right side to left side with a background colour blue. The text is “I Love My Country”.

1. Identify the tag needed for it.
2. Write the HTML statement to do the task.

Answer:

1. <MARQUEE>
2. <MARQUEE direction-’left” bgcolor=”blue” > I Love My country </MARQUEE>

Question 2.
Write HTML statement for displaying the following text items:

1. A₂B³
2. A>B

Answer:

1. A<SUB> 2 </SUB> B <SUP>3</SUP>
2. A&gt;

Question 3.
Two HTML tags are given. They are <BODY> and <FONT>. Identify and write the attribute of each from the following list. (Size, Text, Link, Bgcolor, Color).
Answer:
The attributes of <BODY> tag are Text, Link and Bgcolor The attributes of <FONT> tag are size and color.

Question 4.
Write the use of Border and Aft attribute of <IMG> tag.
Answer:
Border: This attribute is used forgiving border to an image.

Alt: This attribute issued forgiving an alternate text. When there is no image in the specified location or the browser doesn’t support the image them this text will be displayed.

Question 5.
When a client send request to a server, the server must know which service is demanded by the client.

1. How does the server identify the type of service requested?
2. Write the name of any one of the services in the web server.

Answer:

1. Port number
2. Any service like FTP, SMTP, HTTP, etc.

Question 6.
Following are steps for searching the IP address of a domain name by a browser. Rearrange them in proper order.
(a) Look in the local memory of ISP
(b) Look in the DNS servers starting from the root server
(c) Look in the local memory of brower
(d) Look in the local memory of Operating System
Answer:
Correct order is c, d, a, b.

Question 7.
Categorise the following tags into Containertags and empty tags.
<B>, <BR>,<A> ,<FRAME>, <FRAMESET>,<LI>, <HR>
Answer:
1. Container tags:
<B>,<A> <FRAMESET>

2. Empty tags:
<BR>,<HR>,<LI>, <FRAME>.

Question 8.
Write the HTML statement to get the following output.

1. Commerce
2. Humanities

Answer:

1. <B>Commerce</B>
2. <l>Humanities</l>

Question 9.
Write the HTML statement to get the following output.

1. H₂S0₄
2. a² + b²
3. Computer

Answer:

1. H₂SO₄
2. a² + b ²
3. <b>Computer</b>

Question 10.
Write HTML code to display as follows. The <IMG> tag is used for placing images.
Answer:
The & lt IMG &gt tag is used for placing images.

Question 11.
Name some browsers.
Answer:

1. Internet Explorer
2. Netscape Navigator
3. Opera
4. Eudora

Question 12.
What is a website?
Answer:
A website is a collection of webpages. A webpage is created by using HTML tags.

Question 13.
What is the role of attributes in an HTML tag?
Answer:
Attributes are parameters for providing additional information within a tag. Attribute values specify certain characteristics of the tag.
Eg: <P align=”right”>

Question 14.
What is an HTML?
Answer:
HTML stands for Hyper Text Markup Language. It is used to create webpages. It has two types of tags empty and container. The important thing we have to remember in container tag is first opened tag must be closed last.

Question 15.
Name the main attributes of <HR> tag.
Answer:

1. Size: It specifies the line thickness.
2. Width: It specifies the length
3. Noshade: It specifies no shade is given for the line.
4. Color: It specifies the color.

Question 16.
Mary wants to display her name in various headings. Name the heading tags available in HTML.
Answer:
Heading can be given in six levels from <H1> to <H6>.The tag <H1 > produces big heading. The tag <H6> produces small heading. The heading size reduce from <H1> to <H6> sequentially.

Question 17.
Differentiate between <FONT>and<BASEFONT>tags.
Answer:
<BASEFONT>tag sets the normal font for the entire document text. The font specified will be taken as the default font for the entire document. The main attributes are size, face, and color.

The <FONT> tag defines the font characteristics of the text enclosed. <Font> tag change the font property of the text enclosed within <Font> and</Font> whereas <BaseForrt> tag specifies the default font characteristics.

Question 18.
Explain the different types of hyperlinks.
Answer:
The two types of Hyper Links are

1. External Hyper Link: This is used two connect the locations of two different web pages
2. Internal Hyper Link: This is used to connect the different locations of the same web page.

Question 19.
Write the HTML code to display the following list:

1. Form
2. TextBox
3. Label
4. Command Button

Answer:

Question 20.
The body section forms the content displayed in the browser window. Briefly explain any four attributes in the BODY tag.
Answer:

1. Bgcolor – It is used to set background colour.
2. Background – It is used to set a background picture.
3. Text – it is used to set the foreground colour.
4. Left margin – It is used to set the left margin.

Question 21.
HTML has the facility to provide External links as well as Internal links.

1. Which tag is used to include an External link?
2. How will you construct an Internal link?

Answer:
1. Anchor Tag i.e. <a href=”filename”>

2. Internal link is used to link two places of the same web page

Question 22.
Categorise the following tags in HTML and write the criterian for the categorisation.
<BR>, <P>, <BODY>, <B>, <HR>, <IMG>
Answer:

Empty_tag	Container tag
<BR>	<P>
<HR>	<BODY>
<IMG>	<B>

Question 23.
Differentiate empty tags and container tag with example.
Answer:
1. Empty tags: It has opening tag only, no closing tag
Eg: <hr>, <br> etc.

2. Container tag: It as both opening and closing tag.
Eg: <html> </html>
<body></body>etc.

Question 24.
Write True or False

1. Text is an attribute of <BODY> tag to insert a text matter in the web page.
2. <EM>tagfunctionssimilarto<l>tag.

Answer:

1. False
2. True.

Plus Two Computer Application Web Technology Three Mark Questions and Answers

Question 1.
Write and explain any four text formatting tags in HTML.
Answer:
Text formatting tags are given below.

1. <B> – This tag is based to make the text Bold Eg: <B> Computer application </B>
2. <l>: This tag is used to make the text in italics eg: <l> computer aplication</l>
3. <U>:This tag is used to underline the text eg: <U> computer aplication</U>
4. <S>: This tag is used for striking out the text eg: <S> computer aplication</S>

Question 2.
Match the following.

Answer:
Correct match as given below
<H2> -Heading-Align
<MARQUEE> – Scrolling text – Bgcolor
<IMG> – Inserting picture – Src.

Question 3.
Briefly explain the use of tags <Q>, <PRE> and <ADDRESS> tags.
Answer:

• <Q>: It is used to give text within double-quotes.
• <PRE>: This tag is used to display the content as we entered in the text editor.
• <ADDRESS>: This tag is used to provide information of the author or owner.

Question 4.
Write a HTML code to develop a web page about Kerala state as shown below:

KERALA STATE

God’s Own Country

Capital: Thiruvananthapuram

The specifications for the page are:

1. The main heading must be of bigger in size, centralised and bold.
2. Sub headings must be lesser size than main heading and in italics.
3. There should be a picture at the center of the page with file name“tree.jpg”.
4. The background colour of the page must me blue.

Answer:

Question 5.
PHP is a popular scripting language.

1. Write whether it is client side or server side.
2. Write a brief note on PHP.

Answer:

1. Server side
2. PHP (PHP Hypertext Preprocessor)
   • It is an open-source, general-purpose scripting language.
   • It is a server side scripting language
   • Introduced by Rasmus Lendorf
   • A PHP file with extension .php
   • It support data base programming the default DBMS is MySQL
   • It is platform-independent
   • PHP interpreter in Linux is LAMP(Linux, Apache, MySQL, PHP)

Question 6.
How client side scripting differs from server side scripting?
Answer:
Following are the differences

Client Side Scripting	Server Side Scripting
Script is copied to client browser	to the web server
Executed by the client	Executed by the server and result is get back to the browser window
Used for Client level validation	Connect to the database in the server
It is possible to block by the user	Cannot possible
Client side scripts depends the type and version of the browser	It does not depend the type and version of the browser

Question 7.
Briefly explain the two types of communication on the web.
Answer:
The two types of communication on the web are given below.
1. Client to Web server communication:
This communication is carried out between client to the web server (shopping site). The technology used to protect data that are transferred from client to web server is https.

2. Web sever to web server communication:
This communication is usually carried out between web .sever (seller) to another web server (normally bank). For the safe transaction. Digital certificate issued by third party web sites are used.

Question 8.
Compare static and dynamic web pages.
Answer:

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

Question 9.
Differentiate Empty tag and Containertag.
Answer:
There are two types of tags, opening tag, and closing tag

• Empty tag: It has only opening tag and no closing tag
  Eg: <br>, <hr>,..
• Container tag: It has both opening and closing tags. These tag contains some text data Eg:<html>,<head>,<body>,….

Question 10.
Create a webpage using HTML to display the following message. ‘The symbol H₂O represents water’.
Answer:

Question 11.
While designing a webpage Neena wants to explain the meaning of each step. Can you help her?
Answer:
Neena can use Comments while writing the code. It is a good programming practice. Comments improve readability. It is not the part of a program. Comments
<!—and — >

Question 12.
Complete the following table.

RGB code	Colour Name
#0000FF	_________
________	Yellow
#FF0000	_________
#00FF00	_________
________	White
________	Black

Answer:

RGB code	Colour Name
#0000FF	Blue
#FFFF00	Yellow
#FF0000	Red
#00FF00	Green
#FFFFFF	White
#000000	Black

Question 13.
Give HTML tag to display the sentence “WELCOME TO HTML” as centralized heading, having red colour.
Answer:

Question 14.
Explain the Main attributes of font tag.
Answer:
<FONT>tag defines the font properties of text enclosed
The main attributes are

1. Face-This specifies the type of font.
2. Color-This specifies the colour of the text enclosed
3. Size- This specifies the font size

Eg. <FONT Face= “Arial” size = “3” color= “magenta”>

Question 15.
John visited a website, it is found that when clicking on a particular text the browser open a new web page. Name the feature and Identify the tag used for this purpose. Write the HTML code to link to a file name “main.html”.
Answer:

1. Hyper Linking. By clicking on hypertext we can see or go to other webpages or to other section of same document.
2. <A> tag is used
3. <Ahref=”main.html”>Main</A>

Question 16.
Antony visited his school website, he could not see the picture of his school instead of that there is a text message “Your browser could not support images”. Why it is happened and write the html code for this.
Answer:
Because the browser he used cannot load the image. He can use Alt attribute of IMG Tag.
<IMG SRC =” D:\school.jpg” Alt=“Your browser could not support images”>

Question 17.
Suppose you want to display a picture named school.jpg located in the ‘Photos’ sub directory of the directory ‘My documents’ of C drive in your web page.

1. Name the tag used for this purpose.
2. Write HTML code for the web page.

Answer:
1. <IMG>tag. It is used to display images in webpage

2. <HEAD><TITLE>
</HEAD>
<Body>
<IMG SRC= ‘C;/My documents/Photos/School.jpg’>
</Body>
<html>

Plus Two Computer Application Web Technology Five Mark Questions and Answers

Question 1.
Explain the main list tags in HTML?
Answer:
HTML provides three basic types of lists-unordered, ordered and definition list.
1. Unordered list:
Unordered list arranges the list items with bullet symbols in front. <UL> and </UL> tag encloses an Unondered list. List items are specified by <LI> tag. The tag <UL> can take values square, circle or disc. The default type is disc.
Eg : <UL>
<LI>COMPUTER
<LI>BIOLOGY
</UL>

2. Ordered list:
In Ordered list, the list items are numbered in sequence. <OL> and </OL> tag encloses an Or-dered list. List items are specified by <LI> tag. The tag <OL> can take values as follows

• type = 1 for 1, 2, 3,….
• type = i for i, ii, iii,……
• type = I for I, II, III,……
• type = a for a, b, c,…
• type = A for A, B, C,…..

Eg :<OL>
<LI>COMPUTER
<LI>BIOLOGY
</OL>

3. Definition List:
It is formed by a group of definitions and their descriptions. No bullet symbol or number is provided for the list items. The <DL> and </DL> tags enclose the definition list. The <DT> tag contains the definition term and <DD> tag specifies the description.
Eg: <DL>
<DT>Eeheque
<DD>Electronic cheque
</DL>

Question 2.
Explain the use of <BODY> tag and list any four of its attributes
Answer:
Web page contents are given int the body section. Attributes of body tag are:

1. BGCOLOR-Specifies background color for the document Body
   Eg. <BODY BGCOLOR= “RED”>
2. BACKGROUND -Sets the image as background for the document body
   Eg. <BODY BACKGROUNG= “C:\result.jpg”>
3. TEXT-Specifies the color of the text content of the page
   Eg. <BODYTEXT= “Red”>
4. LINK- Specifies colour of the hyperlinks that are not visited by the user
5. ALINK-Specifies the colour of hyperlinks
6. VLINK-Specifies the color of hyperlinks which are already visited by the viewer.
   Eg. < BODY ALINK= “Cyan” LINK=” Magenta” VLINK= “Orange’’> *
7. Left margin and Right margin-Sets margin from left and top of the document window.

Students can Download Chapter 8 Database Management System Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 8 Database Management System

Plus Two Computer Application Database Management System One Mark Questions and Answers

Question 1.
Select the property which is desirable for a database.
(a) Redundancy
(b) Inconsistency
(c) Integrity
(d) Complexity
Answer:
(c) Integrity

Question 2.
Pick the odd man out.
(a) Create
(b) Select
(c) Update
(d) Insert
Answer:
(a) Create

Question 3.
______is the ability to modify a schema definition in one level without affecting the schema definition in the next higher level.
Answer:
Data Independance.

Question 4.
For accessing data from a database, provides an interface with programming languages.
Answer:
SQL (or DML)

Question 5.
Give an example for RDBMS package.
Answer:
Packages such as Oracle, My SQL, etc.

Question 6.
Name the key that acts as a candidate key but not a primary key.
Answer:
Alternate key

Question 7.
With the help of_______the process of storing, retrieving and modifying date are greatly simplified.
Answer:
DBMS

Question 8.
If______is controlled , DBMS can guarantee that database is never inconsistent.
Answer:
Redundancy.

Question 9.
The property of a DBMS that guarantees that the database is never inconsistent.
Answer:
Redundancy.

Question 10.
______and_____are the two types of integrity checks.
Answer:
Range checks, Value checks.

Question 11.
Which component of DBMS provides interfaces with programming Languages.
Answer:
DML

Question 12.
The level of database abstraction that describes how the data is actually stored in the storage medium.
Answer:
Physical level.

Question 13.
The level of database abstraction that describes what data are stored in the database.
Answer:
Logical level.

Question 14.
Data Base Administrators (DBA) are more concerned with level______of Abstraction.
Answer:
Logical level.

Question 15.
The programmers are connected with______level of abstraction.
Answer:
Logical level.

Question 16.
The teller at a bank sees only that part of the database that has information on customer accounts. Which level of Database abstraction he is at?
Answer:
View level.

Question 17.
As part of project work, Ashish defines the type of data and the relationship among them. He is at_______level of database abstraction.
Answer:
Logical.

Question 18.
_______is the other name for logical level.
Answer:
Conceptual level

Question 19.
If the modifications made on storage format does not affect the structure of data, then we achieve______data independence.
Answer:
Physical data independence.

Question 20.
Pick the odd one out.
(a) net work model
(b) hybrid model
(c) relational model
(d) hierarchical model
Answer:
(b) Hybrid model.

Question 21.
“I am a data model. My records can have more than one parent record” Who am I?
Answer:
Network model.

Question 22.
Name the language that enables user to access or manipulate data as organized by RDBMS.
Answer:
DBML

Question 23.
ADatabaseAdministratorisableto modify the structure a programmer changes data types and length of a database without affecting certain fields in a database of a bank and the program. Identify the data independence associated in it.
Answer:
Logical data Independence.

Question 24.
Name the language that used to define a database scheme.
Answer:
DDL

Question 25.
Name the person who has central control over the database and programs in DBMS
(a) Naive user
(b) Programmer
(c) Database Administrator
(d) System Analyst
Answer:
(c) Database Administrator.

Question 26.
Oracle DBMS package is based on______model.
Answer:
Relational

Question 27.
Match the following.

A	B
(a) Relation	(i)  Field
(b) Tuple	(ii)  Table
(c) Atribute	(iii)  Record

Answer:
(a) – (ii), (b) – (iii), (c) – (i).

Question 28.
Name the Relational operation which selects certain columns from the table while discarding others.
Answer:
Project.

Question 29.
How to define the Domain of a column ‘subject 1’ of MARKS relation.
Answer:
Range of values from 0 to 100.

Question 30.
State whetherTrue or False. A view is a kind of table whose contents are taken from other tables.
Answer:
True

Question 31.
State True or False. A view can be queried, inserted into, updated and deleted from.
Answer:
True.

Question 32.
Name an efficient way to provide only required data to users hiding other data from the database.
Answer:
View

Question 33.
Pick the key which can not be used to uniquely identify a tuple on a relation:
(Candidate Key, Primary Key, Alternate Key, Super Key, None of these)
Answer:
None of these.

Question 34.
π(pi) Greek letter is used to denote________operation in relational algebra.
Answer:
Project.

Question 35.
Which relational Algebra operation returns all possible combinations of tuples from two relations.
Answer:
Cartesian product.

Question 36.
Pick the odd one out.
(Select, Cartesian product, Union, intersection)
Answer:
Select – unary operator.

Question 37.
What will be the cardinality of the resultant table if after the following operation if the cardinality of STUDENT is 5 and INSTRUCTOR is 3?
Answer:
15

Question 38.
Consider two relations FOOTBALL AND CRICKET, How to get the names of players play only cricket not also FOOTBALL.
Answer:
CRICKET – FOOTBALL

Question 39.
Why we call a Foreign Key so?
Answer:
It is a candidate Key in another table, A foreigner.

Question 40.
_______is range of values from which actual values are appearing in a given columns are drawn.
Answer:
Domain

Question 41.
Name the table that does not contain data of its own, but is derived from a base table.
Answer:
view

Question 42.
Name a way to uniquely identify a tuple in a relation.
Answer:
By using primary key.

Question 43.
Give two Unary operations performed on a relation in Relational Algebra.
Answer:
Select, Project

Question 44.
Data redundancy is not a desirable property. But All redundancy can not or should not be eliminated. Do you agree with this statement. Justify.
Answer:
Yes. Because sometimes there can be technical or business reasons for maintaining several distinct copies of same data.

Question 45.
Anju is able to do all the internal operations in a DBMS. What type of user is she? What are the other type of users?
Answer:
DBA, Other type users are Appl Prograammer and Naive users.

Question 46.
Which of the following statements are true?

1. DBMS facilitates storage, retrieval, and management of databases.
2. We must keep more copies of the same data in databases.
3. Data inconsistency is eliminated in DBMS.
4. DBMS allows sharing of data, but does not ensure security.

Choose the correct option from the following:
(a) Both 1 and 3 are true
(b) Statements 1, 3 and 4 are true
(c) Statements 1, 2 and 4 are true
(d) All statements are true
Answer:
(a) Both 1 and 3 are true

Question 47.
Which of the following refers to duplication of data in files?
(a) Data redundancy
(b) Data inconsistency
(c) Data integrity
(d) Data security
Answer:
(a) Data redundancy

Question 48.
The following are some responsibilities of database users. Which of them belong to Database Administrator?

1. Design the conceptual schema of the database.
2. Develops programs to interact with the database.
3. Interacts with the database through queries.
4. Ensures authorised and secured access of data

(a) Both 1 and 3
(b) Except 2 and 3
(c) 1, 2 and 4
(d) All the four
Answer:
(b) Except 2 and 3

Question 49.
Which of the keys in a relation do not allow null values? Choose the most appropriate option from the following.
(a) Primary key
(b) Candidate key
(c) Both primary key and candidate key
(d) Either primary key or candidate key
Answer:
(c) Both primary key and candidate key

Question 50.
Choose the level of database abstraction that describes what data is stored in the database and what relationships exist among them.
(a) External
(b) Logical
(c) Physical
(d) View
Answer:
(b) Logical

Question 51.
Which of the following operations can extract the specified columns of a table?
(a) Selection
(b) Projection
(c) Intersection
(d) Set Difference
Answer:
(b) Projection

Plus Two Computer Application Database Management System Two Mark Questions and Answers

Question 1.
The schema of a table is EMPLOYEE(emp_code, emp_name, designation, salary). Write down the relational expressions for the following:

1. To get the name and designation of all employees.
2. To get the details of employees whose salary is above 25000.
3. To get the names of employees who designation is Manager.
4. To get the details of Managers with salary less than 25000.

Answer:

1. π_{name, designation} (EMPLOYEE)
2. σ_salary>25000 (EMLOYEE)
3. π_name( σ_{designation=”Manager”}(EMPLOYEE))
4. σ_{designation = “Manager” and salary < 25000} (EMLOYEE).

Question 2.
Data sharing is an essential feature of DBMS. How data sharing reduces the data inconsistency in a database? Data sharing is an essential feature of DBMS. How data sharing reduces the data inconsistency in a database?
Answer:
Instead of storing more than one copy of the same data, it stores only one copy. This can be shared by several users. If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

Question 3.
Pick the odd one out and justify your answer:
(a) Column
(b) Attribute
(c) Field
(d) Tuple
Answer:
(d) Tuple. The other three terminologies indicate the same characteristic of a table.

Question 4.
Suppose a table (relation) contains the details of customers in a bank. Which attribute of the customer will be set as primary key for the table? Give reason for your opinion.
Answer:
Account number can be set as primary key since account number is different for different customers. That is it is unique hence it can be set as primary key.

Question 5.
How many distinct tuples and attributes are there in a relation with cardinality 22 and degree 7. Answer:

• Cardinality is the number of rows (tuples)
• Hence number of tuples is 22
• Degree is the number of coloums (attributes)
• Hence number of attributes 7

Question 6.
Distinguish primary key and alternate key.
Answer:

1. Primary key: It is a set of one or more attributes used to uniquely identify a row Alternate
2. key: A candidate key other than the primary key.

Question 7.
Write an example for relational data model.
Answer:

Question 8.
Observe the following table and choose the correct match from the following options:

Column A	Column B
(1) Cardinality	(A) Row of a table
(2) Degree	(B) Table
(3) Relation	(C) Number of rows
(4) Tuple	(D) Number of columns
	(E) Attribute

(a) 1 → B, 2 → D, 3 → E, 4 → C
(b) 1 → C, 2 → D, 3 → E, 4 → A
(c) 1 → C, 2 → D, 3 → B, 4 → A
(d) 1 → D, 2 → C, 3 → B, 4 → E
Answer:
(c) 1 → C, 2 → D, 3 → B, 4 → A

Question 9.
Consider the table with the following fields Name, RollNumberand Mark for a set of students. Suggest a field among them, which is suitable for primary key. Justify your answer.
Answer:
field RollNumberis suitable for the primary key. The name and mark can have same values so they are not suitable for the primary key.

Question 10.
Raju is confused with the statement ‘logical data independence is more difficult to achieve than physical data independence’. How can you help Raju to understand the statement.
Answer:
Because Appl. Programs heavily dependent on the logical structure of data. So any change in structure means chance of rewriting Appl. Programs.

Question 11.
Match the following.

A	B
(a) DBA	(i) querying and updation
(b) Application programmer	(ii) ensures consistency
(c) Naive users	(iii) defines conceptual view

Answer:
(a) – (ii), (b) – (iii), (c) – (i)

Question 12.
Match the following.

A	B
(a) Hierarchical model	(i) data as tables
(b) Network model	(ii) Network as storage medium
(c) Relational model	(iii) Child record can have more than one parent
	(iv) Tree structure

Answer:
(a) – (iv), (b) – (iii), (c) – (i)

Question 13.
Your friend tells you that only relational model is used nowadays as DBMS. Will you agree with that? Justify.
Answer:
Yes. Other two models are complex. In RDMS, no redundancy and relationships can be formed easily.

Question 14.
The telephone number of Gokul is entered in Library file as 802111 and in admission register file as 802171.

1. Can you correlate this problem with a concept in DBMS
2. Can you propose a solution to avoid this.

Answer:

1. Consistency problem
2. Remove data redundancy.

Question 15.
What is a DBMS?
Answer:
A DBMS is used to store large volume of data and it is used to retrieve data whenever needed, edit the existing data, update the data and it is possible to delete also.

Question 16.
“View provides an excellent way to access data from data.” Do you agree with this statement? Justify your answer.
Answer:
Yes. Views can have data from more than one table, view can be queried, inserted into, deleted from and updated like a normal table.

Question 17.
A relation is given below.

Mark the following:
Tuple, Attributes, Cardinality, Degree
Answer:

• Tuple- It is the Rows
• Attributes – It is the columns
• Cardinality – 3(Number of Rows)
• Degree – 4 (Number of Columns)

Question 18.
State whetherTrue or False.

1. Primary key cannot be composite key.
2. Only a candidate key can become a Primary Key.
3. Foreign key of a table is a candidate key in another table.
4. Super key uniquely identifies a row in a relation.

Answer:

1. False
2. True
3. True
4. True

Question 19.
Explain the meaning of following operations.
Answer:
select the tuples whose department is sales and who have salary >5000.

Question 20.
Two relations are given below.
FOOTBALL

Name	Age
Jose	23
Raju	22

CRICKET

Name	Age
Jomon	22
Raju	22

Is it possible to find the players those who play both FOOTBALL AND CRICKET by applying any of the Relational Algebra Operations? Explain.
Answer:
Intersection operation.
FOOTBALL ∩ CRICKET.

Question 21.
How will you differentiate Primary key and Super Key? Answer:
Answer:

• primary key- one of the candidate keys chosen to uniquely identifies the rows of a table.
• Super key – Combination of a Primary key with any other attribute or group of attributes.

Question 22.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible. If the cardinality of result T1 ∪ T2 is 13, then what is the cardinality of T1 ∩ T2? Justify your answer.
Answer:
Cardinally of table T1 is 10 means it has 10 rows Cardinally of table T2 is 8 means it has 8 rows Normally T1 ∪ T2 is 10 + 8 = 18 But Here T1 ∪ T2 is 13 means after eliminating duplication of 5 rows this happened. This means 5 rows are common. That is T1 ∩ T2 is 5.

Question 23.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of T1 ∪ T2?
2. What will be the minimum possible cardinality of T1 ∩ T2?

Answer:
1. Degree(CD) -the number of Columns is the Degree
Cardinality (RC)-: the number of Rows is the Cardinality
T1 ∪ T2 = Sum of cardinalities of Table 1 and Table 2
i.e. T1 ∪ T2 = 10 + 8 = 18.

2. T1 ∩ T2 is the common rows(tuples) in T1 and T2 If there is no common tuples then T1 n T2 is 0 hence the cardinality is 0.

Plus Two Computer Application Database Management System Three Mark Questions and Answers

Question 1.
For catering to the needs of users, a database is implemented through three general levels. Name the three levels and discuss them.
Answer:

1. Physical Level is the lowest level.lt describes how the data is actually stored in the storage medium. At physical level complex low-level data structures are described in detail.
2. Logical level describes what data are stored in the database and what relationships exist among data. Here database is described in terms of simple structures. Records are defined in this level. Programmers work at this level.
3. View level is the highest level of data abstraction. It is concerned with the way in which the users view the database. It describes only part of the database.

Question 2.
Consider the following table and write relational algebra operations for the following DEPOSIT.

1. To display those tuples from DEPOSIT relation where amount is greater than 25,000.
2. To display only AccNo. and Amoun to fall depositors.

Answer:

1. σ_amount >25000 (Deposit)
2. π_{AccNo, amount} (deposit)

Question 3.

Show the output of the following relational operations.

1. R₁ – R₂
2. R₁ ∩ R₂
3. R₁ ∪ R₂

Answer:
1.

2.

3.

Question 4.
Developers hide the complexity of Database system three several levels of abstraction. What are they?
Answer:

1. Physical – (how data)
2. logical – (what data)
3. view level – (view data)

1. Physical Level is the lowest level. It describes how the data is actually stored in the storage medium. At physical level complex low-level data structures are described in detail.

2. Logical level describes what data are stored in the database and what relationships exist among data. Here database is described in terms of simple structure. Records are defined in this level. Programmers work at this level.

3. View level is the highest level of data abstraction. It is concerned with the way in which the users view the database. It describes only part of the database.

Question 5.
How data are organized in a database.
Answer:

1. Field: Smallest unit of data. Eg: RolINo, Name.
2. Record: Collection of related fields. Eg: The information of a particular student.
3. File: Collection of related records. Eg: The informations of 10 students.

Question 6.
Salih check his account details using an ATM machine.

1. Identify the levels of abstraction associated with this?
2. Specify other levels.

Answer:

1. View level
2. Logical level, Physical level

Question 7.
Match the following.

A	B
(1) Database Administrator	(a) Not concerned with or even aware of details of the DBMS
(2) Application Programer	(b) Person who has a central control over definition and DBMS
(3) Users	(c) Computer professionals who interact with the DBMS through Application programs

Answer:
(1) – (b), (2) – (c), (3) – (a)

Question 8.
Match the following.

A	B
Domain	Table
Tuple	No. of rows in a relation
Attribute	No. of columns in a relation
Cardinality	Rows in a relation
Degree	A pool of values
Relation	Column in a Table

Answer:

A	B
Domain	A pool of values
Tuple	Rows in a relation
Attribute	Column in a Table
Cardinality	No. of rows in a relation
Degree	No. of columns in a relation
Relation	Table

Question 9.
Explain the major components of DBMS.
Answer:
Components of DBMS:

1. Databases – It is the main component.
2. Data Definition Language (DDL) – It is used to define the structure of a table.
3. Data Manipulation Language (DML) – It is used to add, retrieve, modify and delete records in a database.
4. Users – With the help of programs users interact with the DBMS.

Question 10.
Categorise the users of DBMS and write their functions.
Answer:
Users of Database:

1. Database Administrator: It is a person who has a central control over the DBMS.
2. Application Programmer: These are computer professionals who interact with the DBMS through programs.
3. Naive users: He is an end user. He does not know the details of DBMS.

Question 11.
A table with three columns is given below. For each relational operation given in the 1^st column find the best matches from 2^nd and 3^rd columns.

Answer:

1. Select → c) σ → (ii)
2. Union → d) ∪ → (iv)
3. Set difference → b) → (i)

Question 12.
Observe the given table BOOK and write down the outputs of the following relational expressions:

Answer:
1. This query returns all the tuples(rows) that contain BPB in column Publisher.

2. This query returns the column Book_Title with price<200.
Book Title

• Computer Fundamentals
• C++ Programming
• Mystery of Chemistry

Plus Two Computer Application Database Management System Five Mark Questions and Answers

Question 1.
Cardinality of a table A is 10 and of table, B is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of (A ∪ B) and (A ∩ B)?
2. What will be the minimum possible cardinality of (A ∪ B) and (A ∩ B)?

Give justifications for your answers.
Answer:
There are two possibilities:

1. Both relations contain different tuples (rows).
2. The 8 tuples (rows) of table B are same as that of table A.

Case 1.
If both relations contain different tuples then the maximum possible cardinality of A ∪ B is 10 + 8 = 18.

Case 2.
If 8 tuples of table B are same as that of table A then the maximum possible cardinality of A ⊂ B is 8.

Case 3
If 8 tuples of table B are same as that of table then the minimum possible cardinality of A ∪ B is 10.

Case 4.
If both relations contain different tuples then the minimum possible cardinality of A ∩ B is 0.

Question 2.
There are different data models of which one is currently used in all business transactions. Specify it and discuss in detail.
Answer:
Relational data model is currently used in all business transactions. It is based on the concept introduced by E F Codd. It is composed of one or more tables. Tables are made up of rows and columns.

Here tables are called relations, rows are called tuples and the columns are called attributes. The advantages of this model is neither data redundancy nor complexity. Eg:

Customer	Address
Gita	Add1
Lata	Add 2
Ram	Add3

Question 3.
You have to present a seminar on the topic “Keys in RDBMS”. Prepare the seminar report.
Answer:
1. Candidate Key:
It is a set of attributes that uniquely identifies a row. There may be more than candidate key and may be a combination of more than one attribute.

2. Primary Key:
A primary key is one of the Candidate Keys. It is a set of one or more attributes that can uniquely identify tuples in a relation.

3. Alternate Key:
The Candidate key that is not the primary key is called the alternate key.

4. Super Key:
A combination of a primary key with any other attribute or group of attributes is called a super key.

5. Foreign Key:
A single attribute or a set of attributes, which is a candidate key in another table, is called foreign key.

Question 4.

Consider the relations STUDENT and GRADE given above and predict the output of the following relational operations in table format.

1. σ (STUDENT)
   Score > 80
2. π (STUDENT)
   RegNo, Score
3. σ (GRADE)
   RegNo <103
4. σ (GRADE)
   Grade = ‘B+’
5. π_{RegNo, Name}( σ_{Scores >70}(STUDENT))

Answer:
1.

2.

RegNo	Scores
101	70
102	68
103	94
104	87
105	77

3.

4.

5.

RegNo	Name
103	Ruksana
104	Nancy
105	Anu

Question 5.
Explain any 4 Relational Algebra Operations.
Answer:
A. SELECT operation:
SELECT operation is used to select tuples in a relation that satisfy a selection condition. Greek letter σ (sigma) is used to denote the operation. Syntax,
σ_condition (relation)
eg. σ_{salary < 10000} (EMPLOYEE) – selects tuple whose salary is less than 10000 from EMPLOYEE relation.

B. PROJECT operation:
PROJECT operation selects certain columns from the table and discards the other columns. Greek letter π(pi) is used to denote PROJECT operation. Syntax,
π_condition (relation)
eg. π_{name, salary} (EMPLOYEE) displays only the name and salary of all employees.

C. UNION operation:
This operation returns a relation consisting of all tuples appearing in either or both of the two specified relations. It is denoted by U. duplicate tuples are eliminated.

Union operation can take place between compatible relations only, i.e., the number and type of attributes in both the relations should be the same and also their order.

e.g. SCIENCE ∪ COMMERCE gives all the tuples in both COMMERCE and SCIENCE.

D. INTERSECTION operation:
This operation returns a relation consisting of all the tuples appearing in both of the specified relations. It is denoted by n. It can takes place only on compatible relations, e.g. FOOTBALL ∩ CRICKET returns the players who are in both football and cricket teams.

Question 6.
Why should you choose a database system instead of simply storing data in conventional files?
Answer:
Advantages of DBMS over conventional files:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access:
It stored huge amount of data. efficiently and can be retrieved whenever a need arise.

4. Data can be shared:
The data stored in the database can be shared by the users or programs.

5. Standards can be enforced:
The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained:
It ensures that the data is to be entered in the database is correct.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 7.
We have admission register, attendance register, marks register, etc. in our school to keep various details of students. Briefly describe how DBMS can replace these registers by stating any five merits.
Answer:
Advantages of DBMS:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Data can be shared:
The data stored in the database can be shared by the users or programs.

4. Standards can be enforced:
The data in the database follows some standards. Eg : a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

5. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

6. Integrity can be maintained:
It ensures that the data is to be entered in the databse is correct.

7. Efficient data access:
It stored huge amount of data efficiently and can be retrieved whenever a need arise.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Plus Two Computer Application Database Management System Let Us Assess Questions and Answers

Question 1.
Who is responsible for managing and controlling the activities associated with the database? (1 Mark)
(a) Database administrator
(b) Programmer
(c) Native user
(d) End user
Answer:
(a) Database administrator

Question 2.
In the relational model, cardinality is the (1 Mark)
(a) number of tuples
(b) number of attributes
(c) number of tables
Answer:
(a) number of tuples

Question 3.
Cartesian product in relational algebra is (1 Mark)
(a) a Unary operator
(b) a Binary operator
(c) a Ternary operator
(d) not defined
Answer:
(b) a Binary operator

Question 4.
Abstraction of the database can be viewed as (1 Mark)
(a) two levels
(b) four levels
(c) three levels
(d) one level
Answer:
(c) three level

Question 5.
In a relational model, relations are termed as (1 Mark)
(a) tuples
(b) attributes
(c) tables
(d) rows
Answer:
(c) tables

Question 6.
In the abstraction of a database system the external level is the (1 Mark)
(a) physical level
(b) logical level
(c) conceptual level
(d) view level
Answer:
(d) view level

Question 7.
Related fields in a database are grouped to form a (1 Mark)
(a) datafile
(b) data record
(c) menu
(d) bank
Answer:
(b) data record

Question 8.
A relational database developer refers to a record as (1 Mark)
(a) criteria
(b) relation
(c) tuple
(d) attribute
Answer:
(c) tuple

Question 9.
An advantage of the database management approach is (1 Mark)
(a) data is dependent on programs
(b) data redundancy increases
(c) data is integrated and can be accessed by multiple programs
(d) none of the above
Answer:
(c) data is integrated and can be accessed by multiple programs

Question 10.
Data independence means (1 Mark)
(a) data is defined separately and not included in programs
(b) programs are not dependent on the physical attributes of data
(c) programs are not dependent on the logical attributes of data
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 11.
Key to represent relationship between tables is called (1 Mark)
(a) primary key
(b) candidate Key
(c) foreign Key
(d) alternate Key
Answer:
(c) foreign key

Question 12.
Which of the folowing operations is used if we are interested only in certain columns of a table? (1 Mark)
(a) PROJECTION
(b) SELECTION
(c) UNION
(d) SELECT
Answer:
(a) PROJECTION

Question 13.
Which of the following operations need the participating relations to be union compatible? (1 Mark)
(a) UNION
(b) INTERSECTION
(c) SET DIFFERENCE
(d) All of the above
Answer:
(d) All of the above

Question 14.
Which database level is closest to the users? (1 Mark)
(a) External
(b) Irttemal
(c) Physical
(d) Conceptual
Answer:
(a) View level (External)

Question 15.
The result of the UNION operation between R1 and R2 is a relation that includes (1 Mark)
(a) all the tuples of R1
(b) all the tuples of R2’
(c) all the tuples of R1 and R2
(d) all the tuples of R1 and R2 which have common columns
Answer:
(c) All the tuples of R1 and R2 (eliminating the duplication)

Question 16.
A file manipulation command that extracts some of the records from a file is called (1 Mark)
(a) SELECT
(b) PROJECT
(c) JOIN
(d) PRODUCT
Answer:
(a) SELECT

Question 17.
An instance of relational schema R (A, B, C) has distinct values of A including NULL values. Which one of the following is true? (1 Mark)
(a) A is a candidate key
(b) A is not a candidate key
(c) A is a primary key
(d) Both (a) and (c)
Answer:
(a) A is a candidate key

Question 18.
How many distinct tuples are there in relation instance with cardinality 22? (1 Mark)
(a) 22
(b) 11
(c) 1
(d) none
Answer:
(a) 22

Question 19.
A set of possible data values is called (1 Mark)
(a) Attribute
(b) Degree
(c) Tuple
(d) Domain
Answer:
(d) Domain

Question 20.
Why should you choose a database system instead of simply storing data in conventional files? (5 Mark)
Answer:
Advantages of DBMS over conventional files:
1. Data Redundancy:
It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided:
If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access:
It stored huge amount of data efficiently and can be retrieved whenever a need arise.

4. Data can be shared:
The data stored in the database can be shared by the users or programs.

5. Standards can be enforced:
The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied:
The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained:
It ensures that the data is to be entered in the database is correct.

8. Crash recovery:
Some times all or a portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 21.
Explain the different levels of data abstraction in DBMS? (3 Mark)
Answer:
Levels of Database Abstraction:

1. Physical Level (Lowest Level): It describes how the data is actually stored in the storage medium.
2. Logical Level (Next Higher Level): It describes what data are stored in the database.
3. View Level (Highest level): It is closest to the users. It is concerned with the way in which the individual users view the data.

Question 22.
How are schema layers related to the concepts of logical and physical data independence? (3 Mark)
Answer:
Data Independence:
It is the ability to modify the scheme definition in one level without affecting the scheme definition at the next higher level.

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

Question 23.
Consider the instance of the EMPLOYEE relation shown in the following table. Identify the attributes, degree, cardinality, and domain of Name and Emp_code. (3 Mark)

Answer:

1. Attributes: These are column names, i.e, Emp_Code, Name, Department, Designation, and Salary.
2. Degree(CD): The number of Columns is the Degree i.e Degree is 5(Here 5 columns).
3. Cardinality (RC): the number of Rows is the Cardinality.

i.e. Cardinality is 4(Here 4 rows)
Domain is the pool of possible values. Domain of Name is a String(Sudheesh, Dhanya,
Fathima, Shajan.etc). Domain of Emp_Code is a number (1000,1001,1002, 1003, etc).

Question 24.
Identify primary key, candidate keys and alternate keys in the instance of EMPLOYEE relation in Question 23. (3 Mark)
Answer:
Candidate key: It is used to uniquely identify the row.

1. Emp_code and Emp_Code + Department (Composite) are the candidate keys:
   Primary key: It is a set of one or more attributes used to uniquely identify a row.
2. Emp_code is the primary key:
   Alternate key: A candidate key other than the primary key.

We set Emp_code as the primary key then Emp_code+ Department is the alternate key.

Question 25.
Consider the instance of the STUDENT relation shown in the following table Assume Reg_no as the primary key. (3 Mark)

1. Identify the candidate keys and alternate keys in the STUDENT relation
2. How are the primary key and the candidate key related?

Answer:

1. Reg_no and Reg_no+Batch are the candidate keys. We set Reg_no as the primary key hence Reg_no+Batch is the alternate key
2. Candidate Key: It is a set of attributes that uniquely identifies a row. There may be more than candidate key and may be a combination of more than one attribute.
3. Primary Key: A primary key is one of the Candidate Keys. It is a set of one or more attributes that can uniquely identify tuples in a relation.

Question 26.
What is a database? Describe the advantages and disadvantages of using DBMS. (5 Mark)
Answer:
A Database is a collection of large volume of data.
Advantages of DBMS:
1. Data Redundancy: It means duplication of data. DBMS eliminates redundancy. DBMS does not store more than one copy of the same data.

2. Inconsistency can be avoided: If redundancy occurs there is a chance to inconsistency. If redundancy is removed then inconsistency cannot occur.

3. Efficient data access: It stored huge amount of data efficiently and can be retrieved whenever a need arise.

4. Data can be shared: The data stored in the database can be shared by the users or programs.

5. Standards can be enforced: The data in the database follows some standards. Eg: a field ‘Name’ should have 40 characters long. Some standards are ANSI, ISO, etc.

6. Security restrictions can be applied: The data is of great value so it must be kept secure and private. Data security means the protection of data against accidental or intentional disclosure or unauthorized destruction or modification by unauthorized person.

7. Integrity can be maintained: It ensures that the data is to be entered in the database is correct.

8. Crash recovery: Some times all ora portion of the data is lost when a system crashes. A good DBMS helps to recover data after the system crashed.

Question 27.
What is data independence? Explain the difference between physical and logical data independence. (3 Mark)
Answer:
Data Independence: It is the ability to modify the scheme definition in one level without affecting the scheme definition at the next higher level.

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

Question 28.
Enforcement of standard is an essential feature of DBMS. How are these standards applicable in a database? (3 Mark)
Answer:
There is a standard BIS (Bureau of Indian Standards) in the field of Gold and ISBN (International Standard Book Number) in the field of publication. Similarly here is also some standards like ANSI(American National Standards Institute), ISO (International Organization for standardization), etc.. For example a filed “Name” should have 40 characters is a standard.

Question 29.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible. If the cardinality of result T1 ∪ T2 is 13, then what is the cardinality of T1 ∩ T2? Justify your answer. (3 Mark)
Answer:
Cardinalty of table T1 is 10 means it has 10 rows Cardinalty of table T2 is 8 means it has 8 rows Normally T1 ∪ T2 is 10 + 8 = 18 But Here T1 ∪ T2 is 13 means after eliminating duplication of 5 rows this happened. This means 5 rows are common. That is T1 ∩ T2 is 5.

Question 30.
Cardinality of a table T1 is 10 and of table, T2 is 8 and the two relations are union compatible.

1. What will be the maximum possible cardinality of T1 ∪ T2?
2. What will be the minimum possible cardinality of T1 ∩ T2? (3 Mark)

Answer:
1. Degree(CD): the number of Columns is the Degree.
Cardinality (RC)-: the number of Rows is the Cardinality
T1 ∪ T2 = Sum of cardinalities of Table1 and Table2
i.e. T1 ∪ T2 = 10 + 8 = 18.

2. T1 ∩ T2 is the common rows(tuples) in T1 and T2 If there is no common tuples then T1 ∩ T2 is 0 hence the cardinality is 0.

Question 31.
Consider the relations, City (city_name, state) and Hotel (name, address, city_name). Answer the following queries in relational algebra (5 Mark)

1. Find the names and address of hotels in Kochi.
2. List the details of cities in Kerala state.
3. List the names of the hotels in Thrissur.
4. Find the names of different hotels.
5. Find the names of hotels in Kozhikode or Munnar.

Answer:

1. π_{name, address} (σCity_name = “Kochi” (Hotel))
2. π_{city_name} (σstate = “Kerala” (City))
3. π_name (σcity_name = “Thrissur” (Hotel))
4. π_name (Hotel)
5. π_name(σcity_name = “Kozhikode” V city_name = “Munnar” (Hotel)).

Question 32.
Using the instance of the EMPLOYEE relation shown in question 23, write the result of the following relational algebra expressions. (5 Mark)

1. σ_{Departments=“Sales”}(EMPLOYEE).
2. σ_{salary> 20000 ∧ Department = “Sales”} (EMPLOYEE).
3. σ_{Salary>20000 ∨ Department = “Sales”} (EMPLOYEE).
4. π_{name, salary} (EMPLOYEE).
5. π_{name, salary} (σ_{Designations=“Manager”} (EMPLOYEE)).
6. π_{name, Department} (σ_{Designation = “Clerk” salary > 20000}(EMPLOYEE)).

Answer:
1.

2.

3.

4.

Name	Salary
Sudheesh	25000
Dhanya	25000
Fathima	12000
Shajan	13000

5.

Name	Salary
Sudheesh	25000
Dhanya	25000

6. No rows selected.

Question 33.
Consider the instance of the BORROWER and DEPOSITOR relations shown in following figure which stores the details of customers in a Bank. Answer the following queries in relational algebra. (5 Mark)

1. Display the details of the customers who are either a depositor or a borrower.
2. Display the name of customers who are both a depositor and a borrower.
3. Display the details of the customers who are d positors but not borrowers.
4. Display the name and amount of customer who is a borrower but not depositor.

Answer:
1.

Acc_No	Name
AC 123	Albin
AC105	Shabana
AC116	Vishnu
AC 108	Aiswarya
AC 103	Rasheeda
AC 106	Vishnu

2.

Acc_No	Name
AC123	Albin
AC108	Aiswarya

3.

Acc_No	Name
AC 105	Shabana
AC116	Vishnu

4.

Name	Amount
Rasheeda	25000
Vishnu	25000

Question 34.
Consider the instance of the CUSTOMER and BRANCH relations shown in the following table. Write the Cartesian Product of the two relations. (3 Mark)


Answer:

Students can Download Chapter 5 Continuity and Differentiability Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 5 Continuity and Differentiability

Plus Two Maths Continuity and Differentiability Three Mark Questions and Answers

Question 1.
Consider the function g(x) = \(\frac{|x-2|}{x-2}\)

1. Find the domain and range of the function g(x). (2)
2. Check whether the g(x) is continuous at x = 2. (1)

Answer:
1. The function is not defined at points where denominator is zero.
i.e., x – 2 = 0 ⇒ x = 2.
∴ domain is R – {2}.
g(x) = \(\frac{|x-2|}{x-2}\) = \(\left\{\begin{array}{c}{\frac{x-2}{x-2}, \quad x-2>0} \\{\frac{-(x-2)}{x-2}, \quad x-2<0}\end{array}=\left\{\begin{array}{ll}{1,} & {x>2} \\{-1,} & {x<2}\end{array}\right.\right.\)
∴ range is {-1, 1}

2. The function g(x) is not defined at x = 2. Therefore discontinuous.

Question 2.
(i) If f(x) = x+|x| + 1, then which of the follow represents f (x) (1)


(ii) Test whether f (x) is continuous at x=0. Explain. (2)
Answer:
(i) (b) Since, f(x) = \(\left\{\begin{array}{c}{x+x+1, x \geq 0} \\{x-x+1, \quad x<0}
\end{array}=\left\{\begin{array}{c}{2 x+1, x \geq 0} \\{1, x<0}\end{array}\right.\right.\)

(ii) We have ,f (0) = 1,

Continuous at x = 0.

Question 3.
Consider the function f(x) = \(\left\{\begin{array}{ll}{\frac{\sin x}{x}} & {, x<0} \\{x+1} & {, x \geq 0}\end{array}\right.\)

1. Find \(\lim _{x \rightarrow 0} f(x)\) (2)
2. Is f (x) continuous at x= 0? (1)

Answer:
1. To find \(\lim _{x \rightarrow 0} f(x)\) we have to find f(0^– )and f(0⁺)
f(0^–) = \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\),
f(0⁺) = \(\lim _{x \rightarrow 0}\) + 1 = 0 + 1 = 1
f(0^–) = f(0⁺) = 1 .Therefore \(\lim _{x \rightarrow 0} f(x)\) = 1

2. Here, f (0) = 0 + 1 = 1 = f(0^–) = f(0⁺) = 1
Therefore continuous at x = 0.

Question 4.
Consider the figure and answer the following questions.

(i) Identify the graphed function. (1)


(ii) Discuss the continuity of the above function at x = 2. (2)
Answer:
(i) (b) f(x) = \(\left\{\begin{array}{ll}{\frac{|x-2|}{x-2},} & {x \neq 2} \\{0,} & {x=2}\end{array}\right.\)

(ii) From the figure we can see that
f(2^–) = 1, f(2⁺) = -1 and f(2) = 0
Therefore, f(2^–) = 1 ≠ f(2⁺) = -1 ≠ f(2) = 0. Discontinuous.

Question 5.
Consider f(x) = \(\left\{\begin{array}{ll}{2 x} & {\text { if } x<2} \\{2} & {\text { if } x=2} \\ {x^{2}} & {\text { if } x>2}\end{array}\right.\)
(a) Find \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) (2)
(b) f(x) is continuous. If not so, how can you make it continuous. (1)
Answer:


Therefore f(x) is not continuous at x = 2.
If f(2) = 4, then f(x) becomes continuous.

Question 6.
If y = log₁₀x + log_x10 + log_xx + log₁₀10. Find \(.
Answer:
y = log_e x log₁₀e + log_e10 log_xe + 1 + 1

Question 7.
Examine the continuity of the function

Answer:
In both the intervals x = 1 and x < 1 the function f(x) is a polynomial so continuous. So we have to check the continuity at x = 1.

f(1) = 2
Since

f(x) is continuous at x = 1.

Question 8.
Find [latex]\frac{d y}{d x}\) of the following (3 score each)

1. 2x + 3y = sinx
2. xy + y² = tanx + y
3. x³ + x²y + xy² + y³ = 81
4. sin²x + cos²y = 1
5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1
6. x² + xy + y² = 7
7. x²(x – y) = y²(x + y)
8. xy² + x²y = 2
9. sin y = xcos (a + y)

Answer:
1. Given; 2x + 3y = sinx
Differentiating with respect to x;

2. Given; xy + y² = tanx + y
Differentiating with respect to x;

3. Given; x³ + x²y + xy² + y³ = 81
Differentiating with respect to x;

4. Given; sin²x + cos²y = 1
Differentiating with respect to x;

5. \(\sqrt{x}\) + \(\sqrt{y}\) = 1
Differentiating with respect to x;

6. x² + xy + y² = 7
Differentiating with respect to x;

7. x²(x – y) = y²(x + y)
Differentiating with respect to x;

8. xy² + x²y = 2
Differentiating with respect to x;

9. sin y = xcos (a + y)
⇒ x = \(\frac{\sin y}{\cos (a+y)}\)
Diff. with respect to y.

Question 9.
Find \(\frac{d y}{d x}\) of the following (3 score each)

(viii) x = a(t – sint), y = a(1 – cost)
(ix) y = e^t cost, x = e^t sint.
Answer:
(i) We know; y = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\)
⇒ y = = 2 tan^-1(x)
Differentiating with respect to x;

(ii) We know; y = tan^-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan^-1x

⇒ y = tan^-1(tan3θ)
⇒ y = 3 tan^-1(x)
Differentiating with respect to x;

(iii) We know; y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Put x = tanθ ⇒ θ = tan^-1x

(iv) We know; y = sec^-1 = \(\left(\frac{1}{2 x^{2}-1}\right)\)
Put x = cosθ ⇒ θ = cos^-1x

⇒ y = sec^-1 sec2θ = 2θ
⇒ y = 2cos^-1(x)
Differentiating with respect to x;

(v) ∴ y = tan^-1\(\sqrt{\tan ^{2} x / 2}\) = tan^-1 tanx/2 = x/2
\(\frac{d y}{d x}=\frac{1}{2}\).

= π – 2tan^-1x
\(\frac{d y}{d x}=\frac{-2}{1+x^{2}}\).

(vii) Let x = sinθ and \(\sqrt{x}\) = sinφ

= sin^-1 (sinθcosφ + cosφsinφ)
= sin^-1 (sin(θ + φ)) = θ + φ
= sin^-1x + sin^-1\(\sqrt{x}\)
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}}\).

(ix) y = e^tcost ⇒ \(\frac{d y}{d t}\) = – e^t sin t + e^t
x = e^t sint ⇒ \(\frac{d x}{d t}\) = e^t cos t + e^t sin t

Question 10.
If y = log\(\left(\frac{1}{x}\right)\), Show that \(\frac{d y}{d x}\) + e^y = 0.
Answer:
Given, y = log\(\left(\frac{1}{x}\right)\) ? \(\left(\frac{1}{x}\right)\) = e^y …….(1)

Question 11.
If e^y (x+1) = 1. Show that

1. \(\frac{d y}{d x}\) = -e^y (2)
2. \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\) (1)

Answer:
1. e^y (x+1) = 1
Differentiating e^y +e^y(x +1) \(\frac{d y}{d x}\) = 0

2.

Question 12.
(i) Evaluate \(\lim _{x \rightarrow 0} \frac{k \cos x}{\pi-2 x}\) (2)
(Hint: Put π – 2x = y , where Iris a constant)
(ii) Find the value of k if f (x) is a continuous function given by (1)

Answer:
(i) \(\lim _{x \rightarrow \pi / 2} \frac{k \cos x}{\pi-2 x}=k \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\pi-2 x}\),
Put π – 2x = y when Put x → π/2, y → 0

(ii) Since f (x) is continuous

Question 13.
If f(x) \(f(x)=\left\{\begin{array}{cc}{x-[x]} & {, x<.2} \\{0} & {; x=2} \\{3 x-5} & {, x>2}
\end{array}\right.\)

1. Find \(\lim _{x \rightarrow 2} f(x)\) (2)
2. Is f(x) continuous at x = 2? (1)

Answer:
1. To find \(\lim _{x \rightarrow 2} f(x)\)
we have to find f(2^–) and f(2⁺)
f(2^–) = \(\lim _{x \rightarrow 2} x-[x]\) = 2 -1 = 1,
f(2⁺) = \(\lim _{x \rightarrow 2}\) 3x – 5 = 6 -5 = 1
f(2^–) = f(2⁺) = 1.
Therefore \(\lim _{x \rightarrow 2}\) f(x) = 1

2. Here, f(2) = 0 ≠ f(2^–) = f(2⁺) = 1.
Therefore discontinuity at x = 2.

Plus Two Maths Continuity and Differentiability Four Mark Questions and Answers

Question 1.
If x = 2cosθ; y = 3sinθ

1. Find \(\frac{d y}{d x}\).
2. Find \(\frac{d^{2} y}{d x^{2}}\)

Answer:
1. x = 2cosθ ⇒ \(\frac{d x}{d θ}\) = -2sinθ
y = 3sinθ ⇒ \(\frac{d x}{d θ}\) = 3cosθ

2.

Question 2.
If y = (tan^-1 x)², show that (x² +1)² y₂ + 2x(x² +1) y₁ = 2.
Answer:
y = (tan^-1 x)²
⇒ y₁ = 2(tan^-1 x) \(\frac{1}{1+x^{2}}\)
⇒ (1 + x²)y₁ = 2(tan^-1 x)
⇒ (1 + x²)y₂ + y₁2x = 2 \(\frac{1}{1+x^{2}}\)
⇒ (1 + x²)² y₂ + x(1 + x²)y₁ = 2.

Question 3.
Find \(\frac{d y}{d x}\) if y = sin^-1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1.
Answer:
Put x = tanθ

Question 4.
Let f(x) = \(\left\{\begin{array}{ll}{\cos x,} & {0 \leq x \leq c} \\{\sin x,} & {c<x \leq \pi}\end{array}\right.\)

1. Find the value of c if / is continuous on [0, π].
2. Show that is f not differentiable at the point c.

Answer:
1. Since f is continuous on [0, π], we have;
\(\lim _{x \rightarrow c^{+}}\) f(x) = \(\lim _{x \rightarrow c^{-}}\) f(x) = f(c)
⇒ \(\lim _{x \rightarrow c^{+}}\) sinx = \(\lim _{x \rightarrow c^{-}}\) cosx = cosc
⇒ sinc = cosc ⇒ c = \(\frac{π}{4}\).

2. \(f^{\prime}(x)=\left\{\begin{array}{ll}{-\sin x,} & {0 \leq x \leq c} \\{\cos x,} & {c<x \leq \pi}\end{array}\right. \)
Left derivative at \(\frac{\pi}{4}\) = – sin \(\frac{\pi}{4}\) = –\(\frac{1}{\sqrt{2}}\)
Right derivative at \(\frac{\pi}{4}\) = cos \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)
Left derivative at \(\frac{\pi}{4}\) ≠ Right derivative at \(\frac{\pi}{4}\)
Therefore is not differentiable at the point c.

Question 5.

1. Find \(\frac{d y}{d x}\) if x = 2sinθ; y = 3cosθ
2. Which among the following functions is differentiable on R?

(a) |sinx|
(b) |cosx|
(c) cos|x|
(d) sin|x|
Answer:
1. \(\frac{d x}{d θ}\) = 2cosθ; \(\frac{d y}{d θ}\) = -3 sinθ ⇒ \(\frac{d y}{d x}\) = \(-\frac{3}{2}\)tanθ

2. (c) cos|x|
(Since cos x is an even function, it treats x and -x in the same way).

Question 6.
(i) Examine whether the function defined by \(f(x)=\left\{\begin{array}{ll}{x+5,} & {x \leq 1} \\{x-5,} & {x>1}\end{array}\right.\) is continuous or not. (2)
(ii) If x = ^sin-1t, y = ^cos-1t, a > 0, show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Answer:

f(x) is not continuous.

Question 7.
(i) If \(f(x)=\left\{\begin{array}{ll}{1-x,} & {0 \leq x \leq 1} \\{1+x,} & {1<x \leq 2}\end{array}\right.\) then which of the following is not true (1)
(a) f is continuous in ( 0, 1 )
(b) f is continuous in (1, 2 )
(c) f is continuous in [ 0, 2 ]
(d) f is continuous in [ 0,1 ]
if \(\left\{\begin{array}{cc}{1,} & {x \leq 3} \\{a x+b} & {, \quad 3<x<5} \\{7,} & {5 \leq x}\end{array}\right.\)
(ii) Find f(3⁺) and f(5^–) (1)
(iii) Hence find the value of ‘a’ and ‘b’ so that f(x) is continuous. (2)
Answer:
(i) (c) Since f is not continuous at x = 1.

(iii) Since f (x) is continuous, it is continuous at x = 3 and x = 5
∴ f(3⁺) = f(3) ⇒ 3a + b = 1 ____(1)
and f(5^–) = f(5) ⇒ 5a + b = 7 ____(2)
(2) – (1) ⇒ 2a = 6, a = 3
(1) ⇒ b = 1 – 3 a ⇒ b = -8
∴ a = 3, b = – 8.

Question 8.
Consider f(x) = \(\left\{\begin{array}{ll}{2 x+3,} & {x \leq 2} \\{x+2 k,} & {x>2}\end{array}\right.\)

1. Find f(2) (1)
2. Evaluate \(\lim _{x \rightarrow 2^{+}}\)f(x) (1)
3. Find the value of k, if is continuous at x = 2. (2)

Answer:
1. f(2) = 2(2) + 3 = 7

2. Here, f(x) = x + 2k for x > 2.
\(\lim _{x \rightarrow 2^{+}}\)f(x) = \(\lim _{x \rightarrow 2}\)(x + 2k) = 2 + 2k.

3. Since f (x) is continuous at x = 2
We have, f(2) = \(\lim _{x \rightarrow 2^{+}}\)f(x)
⇒ 7 = 2 + 2k ⇒ k = \(\frac{5}{2}\)

Question 9.
Find \(\frac{d y}{d x}\) of the following (4 score each)

1. y = (logx)^cosx
2. x = 2at², y = at⁴
3. x = a(cosθ + θsinθ), y = a(sinθ – θcosθ)
4. y=x^x
5. y =(x log x)^log(logx)
6. y = \(\sqrt{\sin x \sqrt{\sin x+\sqrt{\sin x+\ldots .}}}\)
7. y^x = x^siny
8. y =(log x)^x + x^logx
9. y = (sinx)^x + sin^-1 \(\sqrt{x}\)

Answer:
1. Given; y = (logx)^cosx, taking log on both sides;
log y = cosxlog(logx),
Differentiating with respect to x;

2. Given; x = 2at² ⇒ \(\frac{d x}{d t}\) = 4at

3. Given; x = a(cosθ + θsinθ)
\(\frac{d x}{d \theta}\) = a(-sinθ + θcosθ + sinθ) = aθcosθ
y = a(sinθ – θcosθ)
\(\frac{d x}{d \theta}\) = a(cosθ – θ(-sinθ) – cosθ) = aθ sinθ
\(\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta\)

4. y= x^x; Taking log on both sides;
log y = x log x

5. y = (x log x)^{log logx}
Taking log on both sides;
log y = (log log x) [log (xlogx)]

6. y = \(\sqrt{\sin x+y}\) ⇒ y² = sinx + y
⇒ y² – y = sinx

7. y^x = x^{sin y};
Taking log on both sides;
xlogy = siny log x

8. y = (log x)^x + x^logx = u + v

9. y = (sinx)^x + sin^-1\(\sqrt{x}\)
Let u = (sinx)^x ⇒ log u = x log sinx

Question 10.
Find \(\frac{d^{2} y}{d x^{2}}\) of the following

1. y = x² + 3x + 2 (2)
2. y = tan^-1x (2)

Answer:
1. Given; y = x² + 3x + 2
Differentiating with respect to x;
\(\frac{d y}{d x}\) = 2x + 3
Differentiating again with respect to x;
\(\frac{d^{2} y}{d x^{2}}\) = 2.

2. Given; y = tan^-1x
Differentiating with respect to x; \(\frac{d y}{d x}\) = \(\frac{1}{1+x^{2}}\)
Differentiating again with respect to x;
\(\frac{d^{2} y}{d x^{2}}\) = \(-\frac{1}{\left(1+x^{2}\right)^{2}} \cdot 2 x\).

Question 11.
Match the following. (4)

Answer:

Question 12.
If x – sint and y = sinmt show that
(i) y = sin(m sin^-1x) (1)
(ii) \(\frac{d y}{d x}\) (1)
(iii) (1 – x²) y₂ – xy₁ + m²y = 0 (2)
Answer:
(i) x = sint, y = sinmt
t = sin^-1x ⇒ y = sin(msin^-1x).


multiplying with \(\sqrt{1-x^{2}}\)
(1 – x²) y₂ – xy₁ = -m²y
(1 – x²) y₂ – xy₁ + m²y = 0.

Question 13.
Consider the function f(x) = x(x – 2), x ∈ [1, 3]. Verify mean value theorem for the function in[1, 3].
Answer:
f(x) = x(x – 2) = x² – 2x ⇒ f'(x) = 2x – 2.
As f is a polynomial, it is continuous in the interval [1, 3] and differentiable in the interval (1, 3).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (1,3)such that.

Hence MVT is verified.

Question 14.
Verify Lagranges’ Mean value theorem for the function f(x) = 2x² – 10x + 29 in [2, 9]
Answer:
f(x) = 2x² – 10x + 29; f'(x) = 4x – 10.
As f is a polynomial, it is continuous in the interval [2, 9] and differentiable in the interval (2, 9).
Therefore two conditions of MVT are satisfied and so there exists c ∈ (2, 9) such that.

Hence MVT is verified.

Question 15.
Let f(x) = x(x – 1)(x – 2), x ∈ [0, 2]

1. Find f(0) and f(2) (1)
2. Find f'(x) (1)
3. Find the values of x when f'(x) = 0 verify Rolle’s theorem. (2)

Answer:
1. f(0) = 0, f(2) = 2(2 – 1)(2 – 2) = 0

2. We have, f(x) = x³ – 3x² + 2x
⇒ f'(x) = 3x² – 6x + 2.

3. f'(x) = 3x² – 6x + 2 = 0

Clearly all the three conditions of Rolle’s theorem are satisfied and 1 ± \(\frac{1}{\sqrt{3}}\) ∈ (0, 2).

Question 16.
Verify Rolle’s Theorem for the function
f(x) = x² + 2x – 8, x ∈ [-4, 2]
Answer:
f(x) = x² + 2x – 8, f'(x) = 2x + 2.
As f is a polynomial, it is continuous in the interval [-4, 2] and differentiable in the interval (-4, 2).
f(-4) = 16 – 8 – 8 = 0
f(2) = 4 + 4 – 8 = 0
f'(c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1 ∈ (-4, 2) Hence Rolle’s Theorem is verified.

Question 17.
Examine that Rolle’s Theorem is applicable to the following function in the given intervals, justify your answer.

1. f(x) = [x], x ∈ [5, 9]
2. f(x) = x² – 1, x ∈ [1, 2]

Answer:
1. The function f(x) = [x] is not differentiable and continuous at integral values. So in the given interval [5, 9] the function is neither differentiable nor continuous at x = 6, 7 ,8. Therefore Rolle’s Theorem is not applicable.

2. The function f(x) = x² – 1 is a polynomial function so differentiable and continuous.
f(1) = 1 – 1 = 0, f(2) = 4 – 1 = 3
f(1) ≠ f(2) . Therefore Rolle’s Theorem is not applicable.

Question 18.
Examine the continuity of the function
\(f(x)=\left\{\begin{array}{cc}{|x|+3,} & {x \leq-3} \\{-2 x,} & {-3<x<3} \\{6 x+2,} & {x \geq 3}\end{array}\right.\)
Answer:
In the intervals x ≤ -3, f(x) is the sum of a constant function and modulus function so continuous. In the intervals -3 < x < 3 and x ≥ 3the function f(x) is a polynomial so continuous. Hence we have to check the continuity at x = -3, x = 3.
At x = -3
f(-3) = 6

f(x) is continuous at x = -3.
At x = 3
f(3) = 6(3) + 2 = 20

Since \(\lim _{x \rightarrow 3^{-}}\)f(x) = f(3), f(x) is not continuous at x = 3.

Question 19.
Test continuity for the following functions.

Answer:

= 0 × a finite quantity between -1 and 1 = 0 Also f(0) = 0
Therefore f(x) is continuous at x = 0.

Therefore f(x) is discontinuous at x = 0

But f(1) = 2
∴ f(x) is discontinuous at x = 1.

Question 20.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) prove that
(i) (1 – x²) y² = (sin^-1x)² (1)
(ii) (1 – x²)y₁ – xy = 1 (1)
(iii) (1 – x²) y₂ – 3xy₁ – y = 0 (2)
Answer:

(ii) Differentiating
(1 – x²) 2y.y₁ + y²x – 2x = \(\frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}}\)
(1 – x²)2yy₁ – 2xy² = 2y
(1 – x²) y₁ – xy = 1.

(iii) Again differentiating
(1 – x²) y₂ + y₁ x – 2x – xy₁ – y = 0
(1 – x²) y₂ – 3xy₁ – y = 0.

Question 21.
At what point on the curve y = x², x ∈ [-2, 2] at which the tangent is parallel to x-axis?
Answer:
Y = x², a continuous function on [-2, 2] and differentiable on [-2, 2] f(2) = 4 = f(-2). All conditions of Rolles theorem is satisfied. Given the tangent is parallel to x-axis.
f¹ (x) = 2x
f¹(c) = 2c
f¹(c) = 0 ⇒ 2c = 0 ⇒ c = 0 ∈ [-2, 2]
where c = 0, y = 0
Therefore (0, 0) is the required point.

Question 22.

is continuous in the interval [-1 1].
(a) Find \(\lim _{x \rightarrow 0}\)f(x) (2)
(b) Find f(0). (1)
(c) Find P. (1)
Answer:


(c) Since f is continuous in [-1 1] it is continuous at 0.
Therefore P = \(-\frac{1}{2}\).

Question 23.
If ax² + 2hxy + by² = 1

1. Find \(\frac{d y}{d x}\) (1)
2. Find \(\frac{d^{2} y}{d x^{2}}\) (3)

Answer:
1. We have, ax² + 2hxy + by² = 1 ___(1)
Differentiating w.r.t.x, we get,

2. Differentiating (2) w.r.tx, we get,

Question 24.
Consider the function y = x^x \(\sqrt{x}\)

1. Express the above function as logy = \(\left(x+\frac{1}{2}\right)\) logx (2)
2. Find \(\frac{d y}{d x}\) (2)

Answer:
1. Given, y = x^x \(\sqrt{x}\). Take log on both sides,

2. We have, logy = \(\left(x+\frac{1}{2}\right)\) logx
Differentiating w.r.t x, we get,

Plus Two Maths Continuity and Differentiability Six Mark Questions and Answers

Question 1.

1. Verify mean value theorem for the function f(x) = (x – 2)² in [1, 4].
2. Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (1, 1) and (4, 2)
3. Find a point on the above curve at which the tangent is parallel to the x-axis.

Answer:
1. f(x) = (x – 1)², x ∈ [1, 4]
f(x) is continuous in [1, 4]
f'(x) = 2(x – 2) is differentiable in [1, 4]
Then there exists c ∈ [1, 4] so that

Hence Mean Value Theorem is verified.

2. c = \(\frac{5}{2}\) will be the x-coordinate to the point of contact of tangent and the curve, then y = (x – 2)² ⇒ y = (\(\frac{5}{2}\) – 2)² = \(\frac{1}{4}\)
Therefore the point is (\(\frac{5}{2}\), \(\frac{1}{4}\)).

3. The tangent parallel to x- axis will have
f'(c) = 0 ⇒ 2(c – 2) = 0 ⇒ c = 2
Then; x = 2 ⇒ y = (2 – 2)² = 0
Therefore the point is (2, 0).

Question 2.

1. Differentiate x^sinx w.r.t.x (2)
2. If x = at², y = 2at, then find \(\frac{d y}{d x}\) (2)
3. If y = sin^-1(cosx) + cos^-1(sinx), then find \(\frac{d y}{d x}\). (2)

Answer:
1. Let y = x^sinx, take log on both sides,
log y = sinx logx differentiate w. r.t.x, we get

2.

3. Given, y = sin^-1(cosx) + cos^-1(sinx)

Differentiate w.r.t. x, we get \(\frac{d y}{d x}\) = -2.

Question 3.

1. Differentiate \(\frac{x-1}{x-3}\) with respect to x.(2)
2. Differentiate \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\) with respect to x. (4)

Answer:
1. Let y = \(\frac{x-1}{x-3}\) Differentiate w.r.t. x, we get;

2. Given, y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Take log on both sides;

Differentiate w.r.t. x, we get;

Question 4.
(i) Define |x|
(a) |x| = \(\sqrt{x^{2}}\)
(b) |x| = x
(c) |x| = -x
(d) |x| = x²
(ii) At which point \(\frac{d}{d x}\)|x| does not exist?
Find \(\frac{d}{d x}\) |x|. (2)
(iii) Find \(\frac{d}{d x}\)|x³ – 7x| . Also, find the point at which the derivative exists. (3)
Answer:
(i) (a) |x| = \(\sqrt{x^{2}}\).

(ii) At x = 0, \(\frac{d}{d x}\) |x| does not exist.


Does not exist at
x³ – 7x = 0 ⇒ x(x² – 7) = 0
⇒ x = 0, x² – 7 = 0 ⇒ x = ±\(\sqrt{7}\).

Question 5.
(i) Match the following (4)

(ii) If log (x² + y²) = 2 tan^-1\(\left(\frac{y}{x}\right)\), then, show that \(\frac{d y}{d x}=\frac{x+y}{x-y}\) (2)
Answer:
(i)

(ii) Given, log (x² + y²) = 2 tan^-1\(\left(\frac{y}{x}\right)\).
Differentiate w.r.to x, we get;

Question 6.
If x = a sec³θ and y = a tan³θ

Answer:
(i) Given, x = a sec³θ
Differentiate w.r.to θ, we get;
\(\frac{d x}{d \theta}\) = 3a sec²θ. secθ. tanθ = 3a sec³ θ. tan θ
Given, y = a tan³θ .
Differentiating w.r.to θ, we get
\(\frac{d x}{d \theta}\) = 3a tan² θ. sec²θ.

(iii) We have, \(\frac{d y}{d x}\) = sinθ
Differentiating w.r.to x, we get

(iv) We have,

Question 7.
Consider the function \(f(x)=\left\{\begin{array}{cc}{1-x} & {, \quad x<0} \\{1} & {x=0} \\ {1+x} & {, \quad x>0}\end{array}\right.\)
(i) Compete the following table. (2)

(ii) Draw a rough sketch of f (x). (2)
(iii) What is your inference from the graph about Its continuity. Verify your answer using limits. (2)
Answer:

Since, f (- 2) = 1 – (- 2) = 3, f (-1) = 1 – (-1) = 2,
f(1) = 1 + (1) = 2, f (2) = 1 + (2) = 3.

(ii)

(iii) From the graph we can see that there is no break or jump at x = 0. Therefore continuous.
From the figure we can see that
f(0^–) = 1 f(0⁺) = 1 and f(0) = 1
Hence, f(0^–) = f(0⁺) = f(0) = 1.
Therefore continuous.

Question 8.
Consider the equation \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\)
(i) Simplify the above equation to sin^-1x – sin^-1y = 2cot^-1 a by giving suitable substitution.
(ii) Prove that \(\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Answer:

(ii) We have; sin^-1 x – cos^-1y = 2cot^-1a.
Differentiating w.r.t x, we get,

Question 9.
(i) Match the following. (4)

(ii) If y = e^{a cos-1x}, then show that (1 – x²)y₂ – xy₁ – a²y = 0 (2)
Answer:
(i)

(ii) Given, y = e^{a cos-1x} ____(1)
Differentiating w.r.to x,

Again differentiating w. r.to x

⇒ (1 – x²)y₂ – xy₁ = a². e^{a cos-1x}
⇒ (1 – x²)y₂ – xy₁ = a²y
⇒ (1 – x²)y₂ – xy₁ – a²y = 0.

Question 10.
(i) Match the following (3)

Differentiate the following
(ii) y =\(\frac{1}{5 x^{2}+3 x+7}\) (1)
(iii) y = 3cosec⁴(7x) (1)
(iv) y = e^{2log tan 5x} (1)
Answer:
(i)

(iii) Given,

(iv) Given,