1. A relation R(A,B,C,D) has the dependencies A → B, B → C, C → D. The highest normal form of R is:

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

Answer: B
Explanation:

A is a candidate key (A → B → C → D). There is a transitive dependency (A → B and B → C), so 3NF is violated. Since the key is single-attribute, 2NF holds. Hence highest NF = 2NF.

2. If a table contains a composite primary key (A, B) and a functional dependency A → C holds, what can be concluded?

A. Likely violates 1NF
B. Likely violates 2NF
C. Likely satisfies BCNF
D. Must be in 5NF

Answer: B
Explanation:

A partial dependency exists because A (a part of the composite key) determines a non-key attribute C. Partial dependencies violate 2NF.

3. A student table has repeating groups of phone numbers for each student. Which normal form is surely violated?

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

Answer: A
Explanation:

Repeating groups mean attributes are not atomic. That violates 1NF, which requires atomic (non-repeating) attribute values.

What is meant by repeating groups?


Following relation is not in First Normal Form (1NF).

StudentID StudentName Courses
101 Arun Kumar DBMS, AI, Networks
102 Meera Devi DBMS, ML

Why this violates 1NF?

  • The column Courses contains multiple values in a single cell.
  • In 1NF, every attribute must hold atomic (indivisible) values only.
  • A repeating or multi-valued attribute breaks the rule of atomicity.
4. Which among the following ensures removal of partial dependency?

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

Answer: B
Explanation:

Second Normal Form (2NF) is specifically defined to remove partial dependencies that arise when a non-prime attribute depends on part of a composite key.

Refer here to know more about Partial Key Dependency
5. A relation R(A,B,C) has FDs: AB → C, C → A. Which of the following is true?

A. R is in BCNF
B. Only AB is a key
C. C is also a key
D. No candidate keys exist

Answer: B
Explanation:

From AB → C and C → A, AB+ = {A,B,C} so AB is a key. C+ gives {C,A} but not B, so C alone is not a key. Therefore AB is the (only) key.

6. A relation contains transitive dependency X → Y and Y → Z. To remove it, we decompose into:

A. XY + YZ
B. XZ + YZ
C. XY + XZ
D. X + Y + Z

Answer: A
Explanation:

To eliminate the transitive dependency, split R(X,Y,Z) into R1(X,Y) and R2(Y,Z). This preserves the direct dependencies and removes transitivity. This is called dependency preserving decomposition.

7. Which normal form focuses mainly on transitive dependency removal?

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

Answer: C
Explanation:

3NF removes transitive dependencies (non-key attribute depending on another non-key attribute).

What is transitive dependency?

A transitive dependency in a relation exists when a non-prime attribute depends on another non-prime attribute, which in turn depends on a key attribute. In simple words: If A → B and B → C and A is a key attribute, C is not part of any key, then C is transitively dependent on A through B.
8. If a relation is in BCNF, it must also be in:

A. 2NF and 3NF
B. Only in 1NF
C. Only in 2NF
D. 4NF

Answer: A
Explanation:

BCNF is stricter than 3NF and 2NF, so a BCNF relation is necessarily in 2NF and 3NF (and 1NF as well).

9. Consider relation R(P,Q,R,S) with FDs: P → Q, Q → R. Which is a key?

A. P
B. Q
C. PQ
D. PR

Correct (precise) answer: PS — PS is a candidate key (not listed in the original choices).

Reasoning (closures)

  • P+: start with {P}
    P → Q → add Q → now {P,Q}
    Q → R → add R → finally P+ = {P, Q, R}
    → S is not included.
  • Q+: {Q,R} (since Q → R) — not a key.
  • P S (PS)+: start with {P,S}
    using the two FDs we get Q (from P → Q) and then R (from Q → R)
    so (PS)+ = {P, S, Q, R} — all attributes of the relation.

Minimality check:

  • P+ = {P,Q,R} (missing S) → P alone is not a key.
  • S+ = {S} (missing P,Q,R) → S alone is not a key.

Therefore PS is a candidate key (it determines all attributes and is minimal). The original options A–D are all not keys unless the question implicitly stated some FD for S (which it did not).

10. A BCNF decomposition is always:

A. Lossless only
B. Dependency preserving only
C. Both lossless and dependency preserving
D. Lossless but may fail to preserve dependencies

Answer: D
Explanation:

BCNF decomposition algorithms guarantee a lossless join, but they can sometimes break dependency preservation (some FDs may not hold on individual decomposed relations).