NIC-NIELIT Scientist-B 2020
October 8, 2023Cache
October 8, 2023Database-Management-System
|
Question 2
|
Consider the following entity relationship diagram (ERD), where two entities E1 and E2 have a relation R of cardinality 1 : m.
The attributes of E1 are A11, A12 and A13 where A11 is the key attribute. The attributes of E2 are A21, A22 and A23 where A21 is the key attribute and A23 is a multi-valued attribute. Relation R does not have any attribute. A relational database containing minimum number of tables with each table satisfying the requirements of the third normal form (3NF) is designed from the above ERD. The number of tables in the database is
|
2
|
|
|
3
|
|
|
5
|
|
|
4
|
Question 2 Explanation:
One table for E1, two tables for E2 (A21, A22 and A21, A23) because we need to make a separate table for multivalued attribute to satisfy minimum 1NF condition that requires atomic attributes.
Then, we get
T1: {A11, A12, A13} – key is A11
T2: {A21, A22, A11} – key is A21
T3: {A21, A23} – key is {A21, A23}
Then, we get
T1: {A11, A12, A13} – key is A11
T2: {A21, A22, A11} – key is A21
T3: {A21, A23} – key is {A21, A23}
Correct Answer: B
Question 2 Explanation:
One table for E1, two tables for E2 (A21, A22 and A21, A23) because we need to make a separate table for multivalued attribute to satisfy minimum 1NF condition that requires atomic attributes.
Then, we get
T1: {A11, A12, A13} – key is A11
T2: {A21, A22, A11} – key is A21
T3: {A21, A23} – key is {A21, A23}
Then, we get
T1: {A11, A12, A13} – key is A11
T2: {A21, A22, A11} – key is A21
T3: {A21, A23} – key is {A21, A23}
Subscribe
Login
0 Comments