Normalization is the DBMS topic that decides marks in GATE, UGC NET and PSU papers, and it is where most aspirants lose them: partial versus transitive dependency, the exact line between 3NF and BCNF, and whether a decomposition is lossless. Every question that follows is a previous-year problem, running from GATE 1997 to GATE 2019 and taking in the UGC NET, DSSSB and TPSC papers along the way. Work each one on paper before you read the reasoning under it. If a concept feels shaky, the full theory sits in the DBMS learn module.
Normalization basics: 1NF and 2NF
Q1. For a relation R(a, b, c, d) whose domains hold only atomic values, only the dependencies a → c and b → d (and those inferred from them) hold. This relation is (GATE 1997)
(a) in first normal form but not in second normal form
(b) in second normal form but not in first normal form
(c) in third normal form
(d) none of the above
Answer: (a). Atomic domains make R satisfy 1NF straight away. Since {a, b} together determine every attribute, it is the candidate key, but c depends only on a and d only on b. Those are partial dependencies of non-prime attributes on part of the composite key, which breaks 2NF. So R is in 1NF but not 2NF (see the full solution).
Q2. What is the goal of normalization in database design? (TPSC 2024)
(a) to increase data redundancy
(b) to reduce data anomalies (insertion, deletion, update)
(c) to optimize query performance
(d) all of these
Answer: (b). Normalization organizes a schema to minimize redundancy so that insertion, deletion and update anomalies cannot arise. It does not add redundancy, and it usually adds joins rather than speeding queries, so faster performance is not its purpose. Reducing anomalies is the single correct goal here.
Q3. Relational database schema normalization is NOT for (UGC NET 2016)
(a) reducing the number of joins required to satisfy a query
(b) eliminating uncontrolled redundancy of stored data
(c) eliminating anomalies from inserts and deletes
(d) ensuring that functional dependencies are enforced
Answer: (a). Normalization decomposes tables to remove redundancy, kill anomalies and make dependencies explicit. That decomposition typically increases the joins a query needs, not reduces them. So "reducing joins" is the one aim normalization is not designed for.
Q4. If a relation schema is in 2NF, then it is definitely in (DSSSB 2018)
(a) 1NF
(b) 4NF
(c) 3NF
(d) BCNF
Answer: (a). The normal forms are nested: each higher form assumes every condition of the one below it. A relation cannot reach 2NF without first satisfying 1NF, so 2NF guarantees 1NF. It says nothing about 3NF, BCNF or 4NF, which are stricter and lie above it.
3NF, BCNF and spotting the highest normal form
Q5. Which one of the following statements is FALSE? (GATE 2007)
(a) any relation with two attributes is in BCNF
(b) a relation in which every key has only one attribute is in 2NF
(c) a prime attribute can be transitively dependent on a key in a 3NF relation
(d) a prime attribute can be transitively dependent on a key in a BCNF relation
Answer: (d). In a two-attribute relation any non-trivial dependency has a superkey on its left, so (a) holds. Single-attribute keys leave no partial dependency, so (b) holds, and 3NF does permit a prime attribute to be transitively dependent, so (c) holds. BCNF forbids exactly that, which makes (d) the false statement (see the full solution).
Q6. Which one of the following statements about normal forms is FALSE? (GATE 2005)
(a) BCNF is stricter than 3NF
(b) lossless, dependency-preserving decomposition into 3NF is always possible
(c) lossless, dependency-preserving decomposition into BCNF is always possible
(d) any relation with two attributes is in BCNF
Answer: (c). The 3NF synthesis algorithm always yields a decomposition that is both lossless and dependency-preserving. BCNF decomposition can be made lossless, but it cannot always preserve every functional dependency. That gap is exactly what statement (c) gets wrong (see the full solution).
Q7. For R = (S, T, U, V) with S → T, T → U, U → V and V → S, the relation R is (GATE 1999)
(a) not in 2NF
(b) in 2NF but not in 3NF
(c) in 3NF but not in 2NF
(d) in both 2NF and 3NF
Answer: (d). Taking closures, each of S, T, U and V determines all four attributes, so every single attribute is a candidate key. Every determinant is therefore a superkey, which puts R in BCNF. Since BCNF implies both 3NF and 2NF, the relation sits comfortably in both (see the full solution).
Q8. A table with fields F1, F2, F3, F4, F5 has F1 → F3, F2 → F4 and (F1, F2) → F5. In terms of normalization, this table is in (GATE 2005)
(a) 1NF
(b) 2NF
(c) 3NF
(d) none
Answer: (a). Since {F1, F2} determines F3, F4 and F5, it is the candidate key. But F3 depends only on F1 and F4 only on F2, both partial dependencies on part of the composite key. Those violate 2NF, so the highest form the table reaches is 1NF (see the full solution).
If a chain of these still feels fuzzy, the worked walkthrough in our Normalization in DBMS: 1NF to BCNF guide rebuilds the ladder from scratch with a running example.
Decomposition: lossless join and dependency preservation
Q9. For R(A, B, C, D) with A → B and C → D, decomposing R into R1(A, B) and R2(C, D) is (GATE 2001)
(a) dependency preserving and lossless join
(b) lossless join but not dependency preserving
(c) dependency preserving but not lossless join
(d) neither dependency preserving nor lossless join
Answer: (c). Both dependencies survive: A → B lands in R1 and C → D in R2, so the decomposition is dependency-preserving. But R1 and R2 share no common attribute, so their natural join cannot reconstruct R without spurious tuples. Preserving dependencies yet losing the lossless property gives answer (c) (see the full solution).
Q10. For R(A, B, C, D) with A → B, B → C, C → D and D → B, the decomposition into (A, B), (B, C), (B, D) (GATE 2008)
(a) gives a lossless join and is dependency preserving
(b) gives a lossless join but is not dependency preserving
(c) does not give a lossless join but is dependency preserving
(d) does not give a lossless join and is not dependency preserving
Answer: (a). The common attribute B is a key in the fragments it joins, so each chained natural join reconstructs R with no spurious tuples, making the split lossless. Every given dependency also appears in some fragment once you use the B-cycle, so it is dependency-preserving too (see the full solution).
Q11. Let r be a relation on schema R = (A, B, C, D). Define r1 as the projection on A, B, C and r2 as the projection on A, D. If the decomposition is lossy and S is the natural join of r1 and r2, which is true? (GATE 2005)
(a) S is a proper subset of r
(b) the union of r and S equals r
(c) r is a proper subset of S
(d) the natural join of r and S equals S
Answer: (c). Projecting and re-joining can never drop an original tuple, so every tuple of r survives into S. A lossy decomposition means the join also manufactures spurious tuples, so S holds strictly more. Hence r is a proper subset of S. Drill more of this shape in the lossless join decomposition set.
Q12. F = {QR → S, R → P, S → Q} holds on X = (P, Q, R, S), which is not in BCNF. X is decomposed into Y = (P, R) and Z = (Q, R, S). Consider: (I) both Y and Z are in BCNF; (II) the decomposition is dependency preserving and lossless. Which is correct? (GATE 2019)
(a) I only
(b) neither I nor II
(c) II only
(d) both I and II
Answer: (c). In Y the dependency R → P makes R a key, so Y is in BCNF. In Z the dependency S → Q holds but S does not determine R, so S is not a key and Z violates BCNF, which breaks statement I. The intersection {R} is a key of Y and all three dependencies are recoverable, so the decomposition is lossless and dependency-preserving, making II the only correct claim (see the full solution).
How normalization is examined
Definitions and the 1NF-to-2NF boundary (Q1 to Q4) are the reliable one-mark grabs. Questions 5 to 8 turn on naming the highest form a relation actually reaches: 3NF tolerates a transitively dependent prime attribute while BCNF does not, only 3NF guarantees a lossless dependency-preserving synthesis, and a single partial dependency pins a table at 1NF. The decomposition block (Q9 to Q12) is where two marks are won or lost: master the lossless-join test (does the shared attribute set determine one side?) and dependency preservation, and you can attempt almost any variant unseen.
Work each question a second time a week later, and confirm your reasoning against the theory in the normalization lessons. GATE aspirants get the full DBMS sequence inside GATE Guidance by Sanchit Sir, and you can see where DBMS sits across the paper on the GATE CS Exam page. Solve, review the ones you missed, and the marks get locked in on the second pass.
