21. When does a table reach 5NF?

A. No transitive dependencies remain
B. No join dependency remains
C. No multivalued dependency remains
D. All FDs have prime attributes

Answer: B
Explanation:

5NF (Projection-Join NF) addresses join dependencies — a relation is in 5NF when it cannot be non-trivially decomposed into smaller relations without loss.

22. Normalization to BCNF may cause:

A. Lossless join always
B. Loss of dependency preservation sometimes
C. Removal of multi-valued dependencies
D. Need for denormalization

Answer: B
Explanation:

BCNF ensures lossless decomposition, but it may break dependency preservation — some FDs might not be enforceable without joins.

23. If all attributes depend on the whole key and no non-key attributes determine others, table is at least:

A. 2NF
B. 3NF
C. BCNF
D. 4NF

Answer: B
Explanation:

If there are no partial dependencies (attributes depend on whole key) and no transitive dependencies (non-keys don't determine others), the relation is at least in 3NF.

24. Transitive dependency leads to:

A. Insertion anomaly
B. Deletion anomaly
C. Update anomaly
D. All of these

Answer: D
Explanation:

Transitive dependencies cause redundancy and can result in insertion, deletion and update anomalies — so all of these.

25. A table storing OrderID → CustomerID → CustomerPhone violates:

A. 1NF only
B. 2NF only
C. 3NF
D. BCNF

Answer: C
Explanation:

OrderID → CustomerID and CustomerID → CustomerPhone creates a transitive dependency OrderID → CustomerPhone via CustomerID. This violates 3NF.

26. A relation R(A,B,C,D) has FD: A → B, C → D. There is no relationship between {A,B} and {C,D}. What is the likely problem?

A. Partial dependency
B. Multivalued dependency
C. Transitive dependency
D. Join dependency

Answer: B
Explanation:

Two independent sets of attributes cause multivalued dependency. A non-trivial MVD suggests need for 4NF decomposition.

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27. A relation is in BCNF but still shows redundancy. The reason may be:

A. Presence of MVDs
B. Presence of transitive dependencies
C. FD LHS not superkey
D. Repeating groups exist

Answer: A
Explanation:

BCNF handles functional dependencies. MVDs are a different redundancy source, handled only in 4NF.

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28. For a table to be in 2NF, it must be:

A. Free from partial dependency
B. Free from transitive dependency
C. Free from MVDs
D. Free from join dependency

Answer: A
Explanation:

A table in 2NF must first be in 1NF and then remove partial dependencies on composite keys.

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29. Which normal form ensures no non-key attribute determines another non-key attribute?

A. 1NF
B. 2NF
C. 3NF
D. 4NF

Answer: C
Explanation:

3NF eliminates transitive dependencies where non-key determines non-key.

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30. A → BC, D → A. Which can be a key of R(A,B,C,D)?

A. A
B. D
C. AD
D. B

Answer: B
Explanation:

D → A and A → BC implies D → ABC. With D itself present, it determines all — so D is a key.