Consider the following two tables along with their structures and sample data.…
2026
Consider the following two tables along with their structures and sample data.
Table: STUDENT
AdmissionNo | Name | DOB | AadhaarNo | ClassSec |
|---|---|---|---|---|
2026103 | Sameer Jha | 2013-09-12 | 112233445566 | 6B |
2026112 | Sonika Jain | 2014-01-08 | 123456123456 | 6A |
2026125 | Zoya Javed | 2013-11-08 | 654321123456 | 6A |
Table: CLASSTEACHER
ClassSec | CTName | RoomNo |
|---|---|---|
6A | Gurpreet Kaur | 112 |
6B | KJ Jacob | 106 |
Note: Both tables include more relevant rows.
Answer the following questions based on the given tables and concepts of relational databases:
(A) Identify all possible candidate keys that the STUDENT table may have.
(B) If the STUDENT table contains records of 100 students, determine the degree and cardinality of the STUDENT table.
(C) If ClassSec is the common attribute between the two tables, identify in which table it will act as a foreign key.
(D) Determine the total number of attributes and tuples in the resulting table obtained after performing a natural join between the STUDENT and CLASSTEACHER tables. Assume that each class section has exactly one class teacher and every student belongs to a valid class section.
Show answer & explanation
(A) Candidate Keys: AdmissionNo, AadhaarNo (both uniquely identify a student).
(B) Degree = Number of attributes = 5
Cardinality = Number of records = 100
(C) ClassSec acts as a Foreign Key in STUDENT table (refers to CLASSTEACHER).
(D) Natural Join (on ClassSec):
Attributes = 5 (STUDENT) + 3 (CLASSTEACHER) – 1 (common) = 7
Tuples = 3 (each student matches one teacher)
Conclusion: Keys ensure uniqueness, while joins combine related data efficiently.