GATE 2025 DA Q.57: solution and answer

Official answer: B, DSource: Official answer key, GATE 2025 DA (IIT Roorkee)

1. Closure of $A$: $A\to BCEF$ gives $\{A,B,C,E,F\}$, then $E\to DG$ adds $D,G$. So $A^{+}=ABCDEFG$, which is all of R. $A$ is a superkey, and being a single attribute it is a candidate key.

2. Closure of $BC$: $BC\to A$ gives $A$, and $A^{+}$ is all of R, so $(BC{)}^{+}=ABCDEFG$. Also $B^{+}=\{B\}$ and $C^{+}=\{C\}$ since no FD has $B$ or $C$ alone on its left side. So $BC$ is a minimal superkey, hence a candidate key. Therefore $BC$ is also a candidate key, and (A) is false.

3. Closure of $E$: $E\to DG$ gives $E^{+}=\{E,D,G\}$, which is not all of R. So $E$ is not a key, and (C) is false.

4. Are there any other candidate keys? The only FDs that put $A$ on the right side are $BC\to A$. So a set that does not contain $A$ can reach $A$ only if it contains both $B$ and $C$. Such a set contains $BC$ and is not minimal. Hence the candidate keys are exactly $\{A\}$ and $\{B,C\}$, and (B) is true.

5. BCNF needs the left side of every non-trivial FD to be a superkey. $A\to BCEF$ and $BC\to A$ satisfy this. $E\to DG$ is non-trivial and $E$ is not a superkey (step 3), so it violates BCNF.

6. So R is not in BCNF, and (D) is true. The correct options are (B) and (D).