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UGC NET Computer Science Previous Year Questions (PYQs)

UGC NET Computer Science Database Management Systems PYQ

UGC NET Computer Science PYQ
Consider relation R(A, B, C, D, E) with the following functional dependencies:
C → F, E → A, EC → D, A → B. 
Which of the following is a key for R?





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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Goal: Find a set of attributes whose closure covers all attributes of R = {A, B, C, D, E}.

Check (1) CD+:
From C → F we get F. D gives nothing further. So CD+ = {C, D, F}.
Missing A, B, E ⇒ not a key.

Check (2) EC+:
Start {E, C}. Using E → A ⇒ add A. Using A → B ⇒ add B. Using EC → D ⇒ add D.
Therefore EC+ = {E, C, A, B, D} = {A, B, C, D, E}. Covers all attributes of R ⇒ key.

Check (3) AE+:
A → B gives B; E → A adds nothing new. No way to reach C or D ⇒ not a key.

Check (4) AC+:
A → B gives B; C → F gives F. No E or D ⇒ not a key.

Answer: (2) EC

Note: F is outside R, so while C → F is valid, a key must cover only attributes of R. EC+ covers {A, B, C, D, E}.

UGC NET Computer Science PYQ
Consider the following three relations:

Employee(eid, eName), Comp(cid, cName), Own(eid, cid).
Which of the following relational algebra expressions return the set of eids who own all brands?

(1) π_eid( π_eid,cid(Own) ÷ π_cid(Comp) )
(2) π_eid( π_eid(Own) × π_cid(Comp) )
(3) π_eid( π_eid,cid(Own) × π_cid(Comp) )
(4) π_eid( π_eid(Own) × (π_cid,cName(Own) ÷ π_cid(Comp)) )





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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Reason: To find employees who own all brands, divide the (eid,cid) pairs in Own by the set of all cids in Comp, then project eid: π_eid( (Own[eid,cid]) ÷ (Comp[cid]) ).
UGC NET Computer Science PYQ
In a relational database, which one of the following is CORRECT:





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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

A prime attribute is one that is part of a candidate key.
In BCNF (Boyce-Codd Normal Form), for every functional dependency $X \to Y$, $X$ must be a super key.
If all attributes are prime, there cannot exist a dependency where a non-superkey determines another attribute.
Hence, the relation is automatically in BCNF.
UGC NET Computer Science PYQ
Consider T1(A,B,C,D,E) with FDs { EB→C, D→E, EA→B } and T2(A,B,C,D) with FDs { C→A, A→B, A→D }. Which is TRUE?





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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Candidate key for T1 is AD (A,D are prime; B,C,E non-prime). FDs EB→C, D→E, EA→B have LHS not a superkey and RHS non-prime → violates 3NF (and even 2NF due to D→E partial on key).
UGC NET Computer Science PYQ
Considering the following statements:
A. A non-serial schedule is said to be conflict-serializable, if it is conflict-equivalent to some serial schedule.
B. A non-serial schedule is said to be view serializable if it is view-equivalent to some serial schedule.
C. A schedule is said to be serial, if instructions of participating transactions are chronologically interleaved with each other.
D. A conflict-serializable schedule will be view-serializable also but vice-versa may not be true.





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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

A True (definition of conflict serializability).
B True (definition of view serializability).
C False — a serial schedule has no interleaving; one transaction completes before the next starts.
D True — conflict-serializable ⇒ view-serializable, but not every view-serializable schedule is conflict-serializable.

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