GO Classes Test Series 2024 | Mock GATE | Test 12 | Question: 23

Answer 1

Ans C

for option A:-“Domain of each attribute is an elementary type; that is, not a set or a record structure” it means that there atomic condition is satisfied which is clearly 1NF.

for option B :- A relation is in 3NF if at least one of the following conditions holds in every non-trivial function dependency X –> Y. 

• X is a super key.
• Y is a prime attribute (each element of Y is part of some candidate key)
• There should be no transitive dependency (Non Prime → Non Prime)

for option C :- It is in 2NF and condition is stating that there should be no non trivial dependency which is partial dependency.

for option D :- It is condition for BCNF “X should be a superkey/candidate key for every functional dependency (FD) X−>Y in a given relation”. 

  

 

Answer 2

(B)
X->A
Here,either
1)A is prime, so here X->A can be of type
  (i) ck->prime ,or    -->in 3NF
  (ii)prime->prime    -->in 3NF
or-
2) X is ck or sk, so here X→A can be of type
   (i) ck->prime ,or      -->in 3NF
   (ii)ck->non prime     -->in 3NF
so B is in 3nf(basically the cases don’t provide any counter eg )

 

(C)
X->A
Here,either
1)A is prime, so here X->A can be of type
  (i) ck->prime ,or    -->in 3NF
  (ii)prime->prime    -->in 3NF
or-
2) X is not proper subset of any key of R,
so here X is either non prime, then  X->A can be of type
    (i)np→ np, only  -->in 3NF violation!!!
   or
    X is candidate key itself
   (i) ck->prime ,or      -->in 3NF
   (ii)ck->non prime     -->in 3NF
so C  is not in 3nf, BUT is in 2NF