Plus Two Computer Application Chapter Wise Questions and Answers Chapter 8 Database Management System

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 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) network 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 users to access or manipulate data as organized by RDBMS.
Answer:
DBML

Question 23.
ADatabaseAdministratorisableto modifies 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 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.

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 whether True 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:

(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.

Answer:
(a) – (ii), (b) – (iii), (c) – (i)

Question 12.
Match the following.

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 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 whether True 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

CRICKET

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 Acc No. 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 the 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.

Answer:
(1) – (b), (2) – (c), (3) – (a)

Question 8.
Match the following.

Answer:

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 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:

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.

3.

4.

5.

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 follow 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 follow 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:
Sometimes 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 maybe 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: Sometimes 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 schema 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.

5.

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.

2.

3.

4.

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:

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