GATE 2026 DA Q.17: solution and answer

GATE 2026 · DAQ.17 · 1 mark · MCQ

Let R(A,B,C,D,E) be a relational schema with functional dependency set F={A→BC, CD→E, E→A}.

Which of the following statements is correct?

(A)
AD, ED and CD are the only candidate keys of R.
(B)
AD and ED are the only candidate keys of R.
(C)
A, E and CD are the only candidate keys of R.
(D)
A and CD are the only candidate keys of R.

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Answer and solution

ANSWER

Option (A): AD, ED and CD are the only candidate keys of R.

Official answer: ASource: Official answer key, GATE 2026 DA (IIT Guwahati)

Attribute $D$ appears on no right-hand side of $F$ , so no other attribute can derive it and $D$ belongs to every candidate key. Attribute $B$ appears on no left-hand side, so it is never needed to derive anything and belongs to no candidate key.
Every candidate key is therefore $D$ together with a subset of ${A, C, E}$ . Since $D^{+} = {D}$ , $D$ alone is not a key.
${A, D}^{+}$ : $A \to B C$ gives ${A, B, C, D}$ . Then $C D \to E$ gives ${A, B, C, D, E}$ . So $A D$ is a key.
${E, D}^{+}$ : $E \to A$ gives ${A, D, E}$ . Then $A \to B C$ gives ${A, B, C, D, E}$ . So $E D$ is a key.
${C, D}^{+}$ : $C D \to E$ gives ${C, D, E}$ . Then $E \to A$ gives ${A, C, D, E}$ . Then $A \to B C$ gives ${A, B, C, D, E}$ . So $C D$ is a key.
Each of $A D$ , $E D$ , $C D$ is minimal: removing an attribute leaves a single attribute, and $A^{+}$ , $E^{+}$ , $C^{+}$ do not contain $D$ (so none is a key), while $D^{+} = {D}$ . Any other candidate key must contain $D$ plus at least one of $A, C, E$ (and not $B$ ), so it contains one of $A D$ , $C D$ , $E D$ and would not be minimal.
Hence the candidate keys are exactly $A D$ , $E D$ and $C D$ , which is option (A). Options (B), (C) and (D) are false: (B) omits $C D$ , and (C) and (D) name $A$ , $E$ or $C$ alone as keys, but $A^{+} = {A, B, C}$ and $E^{+} = {E, A, B, C}$ never contain $D$ .

KEY IDEA

First classify attributes: one that never appears on a right-hand side must be in every key, and one that never appears on a left-hand side is in no key. Then take attribute closures of the remaining combinations and keep the minimal ones.

WATCH OUT FOR

Treating $A$ , $E$ or $C D$ as keys because $A \to B C$ and $E \to A$ look like they reach everything. These never derive $D$ , which appears on no right-hand side, so $A$ and $E$ alone cannot be keys. Another slip is stopping after finding $A D$ and $C D$ and missing $E D$ .

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GATE 2026 DA Q.17: solution and answer

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