NCERT Solutions for Class 12 Computer Science (C++) – Database Concept

NCERT Solutions for Class 12 Computer Science (C++) – Database Concepts

Computer ncert solution of class 12 (C++) – Database Concept, we have covered all the chapter questions. This solution will prove to be beneficial for the board examination as well as for students who are preparing for graduate(B.Tech) in information technology and preparing for Post Graduate(M.C.A).

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NCERT Solutions for Class 12 Computer Science (C++) having 14th Chapter whose Chapter wise Solution given below.

Short Answer Type Questions-I

Question 1:
What is relation? Define the relational data model.
Аnswer:
A relation is a table having atomic values, unique row and unordered rows and columns. The relational mode represent data and relationship among data by a collection of tables known as relation, each of which has a number of columns with unique names.

Question 2:
What are all the domain names possible in gender ?
Аnswer:
Male and Female.

Question 3:
Give a suitable example of a table with sample data and illustrate Primary and Candidate Keys in it.
Аnswer:
A table may have more than one such attribute/ group of attribute that identifies a row/tuple uniquely, all such attribute(s) are known as Candidate Keys. Out of the Candidate keys, one is selected as Primary Key,

Question 4:
Differentiate between cardinality and degree of a table with the help of an example.
Аnswer:
Cardinality is defined as the number of rows in table.
Degree is the number of columns in a table,
eg: Consider the following tables


(1 Mark for writing suitable example/correct definition of a table)
(½ Mark for correct illustration/definition of Candidate Keys)
(½ Mark for correct illustration/definition of Primary Key)
Cardinality of Account table is : 3
Degree of Account table is : 2
[½ mark each for definition of cardinality and degree]
[1 mark for correct demonstration using example]

Question 5:
Observe the following table carefully and write the names of the most appropriate columns, which can be considered as (i) candidate keys and (ii) primary key.

Аnswer:
Candidate keys : Id, Product
Primary keys: Id
(1 Mark for writing correct Candidate keys)
(1 Mark for writing correct Primary key)

Note : No marks to be deducted for mentioning Price and/or Transaction Date as additional candidate keys.

Question 6:
Observe the following table carefully and write the names of the most appropriate columns, which can be considered as (i) candidate keys and (ii) primary key :

Аnswer:
Candidate keys : Code, Item Primary keys : Code
(1 Mark for writing correct Candidate keys)
(1 Mark for writing correct Primary key)
Note : No marks to be deducted for mentioning Price and/or Transaction Date as additional candidate keys.

Question 7:
Define degree and cardinality. Based upon given table write degree and cardinality.

Аnswer:
Degree is the number of attributes or columns present in a table.
Cardinality is the number of tuples or rows present in a table.
Patients Degree = 4
[½ mark for each correct definition]
Cardinality = 5
[ ½ mark for each correct answer]

Question 8:
Differentiate between the Primary key and Alternate key of a table with the help of an example.
Аnswer:
Primary Key – A primary key is a value that can be used to identify a unique row in a table .
Alternate Key – An alternate key is any candidate key which is not selected to be the primary key.
Example:

So, (Bank Account Number, Aadhaar Number) is a candidate key for the table.
Aadhaar Number ” Primary key
Bank Account Number ” Alternate Key

Question 9:
Explain the concept of candidate keys with the help of an appropriate example.
Аnswer:
Candidate key is a column or set of columns that can help in identifying records uniquely.
Example,consider a Table STUDENT

Here, AdminN0  & Roll N0 define Table uniquely.
Hence, they are candidate keys.

Question 10:
A table ‘customer’ has 10 columns but no row. Later, 10 new rows are inserted and 3 rows are deleted in the table. What is the degree and cardinality of the table customer
Аnswer:
Degree = 10[no.of cols]
Cardinality = 10-3=7[no.of rows]

Question 11:
A table ‘student’ has 3 columns and 10 rows and another table ‘student 2’ has the same number of columns and 15 rows. 5 rows are common in both the tables. If we take union, what is the degree and cardinality of the resultant table ?
Аnswer:
Degree-3
Cardinality – 30 = (20 + 15-5)

Question 12:
A table ‘studentl’ has 4 columns and 10 rows and table ‘student 2’ has 5 columns and 5 rows.If we take cartesian product of these two tables,what is the degree and cardinally of the resultant table?
Аnswer:
Degree-4 ×5=20[no.of columns]
Cardinally – 10×5 = 50 [no.of rows]

Question 13:
Observe the following table and answer the parts (i) and (ii) :

1. In the above table, can we have Qty as primary key.[Answer as yes/no]. Justify your answer.
2.  What is the cardinality and degree of the above table ?

Аnswer:

1. We cannot use Qty as primary key because there is dulication of values and primary key value cannot be duplicate.
2. Degree = 4

[1/2 mark for each correct definition]
Cardinity = 5
[1/2 mark for each correct definition]

Question 14:
What do you understand by Union & Cartesian product in the relational algebra ?
Аnswer:

Union of R and S : The Union of two relations is a relation that includes all the tuples that are either in R or in S or in both R and S. Duplicate tuples are eliminated.
Cartesian Product: The Cartesian Product is an operator which works on two sets. It combines the tuples of one relation with all the tuples of the other relation.
Example: Cartesian Product

The Union is an operator which works on two sets. It combines the tuples of one relation with all the tuples of the other relation such that there is no duplication.

Question 15:
Explain the concept of union between two tables, with the help of appropriate example.
Аnswer:
The union operation denoted by ‘U’ combines two or more relations. The resultant of union operation contain tuples that are in either of the table or in both tables.
For example :

Question 16:
In the following 2 tables, find the union value of Student 1 and Student 2.

Аnswer:
The resultant table is

Question 17:
Observe the table ‘Club’ given below :

1. What is the cardinality and degree of the above given table ?
2. If a new column contact_no has been added and three more members have joined the club then how these chages will affect the degree and cardinality of the above given table.

Аnswer:
Cadinality: 4
Degree: 5
(1/2 Mark for each correct answer)
2. Cadinality: 7
Degree: 6
(1/2 Mark for each correct answer)

Question 18:
In which situation can we apply union operation of two table ?
Аnswer:
Each table in the UNION

1. should have the same number of columns
2. similar data types
3. columns must be in the same order [2]

 

Short Answer Type Questions – II

Question 1:
Observe the following STUDENTS and EVENTS tables carefully and write the name of the RDBMS operation which will be used to produce the output as shown in LIST. Also, find the Degree and Cardinality of the LIST.

Аnswer:
Cartesian Product
Degree = 4
Cardinality = 6

Question 2:
Observe the following PARTICIPANTS and EVENTS tables carefully and write the name of the RDBMS operation which will be used to produce the output as shown in RESULT. Also, find the Degree and Cardinality of the

Аnswer:
Degree = no of columns = 4
Cartesian Product
Cardinality = no. of rows = 6

NCERT Solutions for Class 12 Computer Science