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# Database management systems DBMS TRUE or FALSE questions with answers explained for competitive and entrance exams.

## TRUE or FALSE questions in DBMS

1. The functional dependency (FD), XW → Z follows from set of FDs {XY → Z, Y → W, WY → Z}
(a) TRUE                                          (b) FALSE

 Answer: (b) FALSE The closure can be used to check whether the FD holds in the relation as per the given set of functional dependencies.             XW+    = XW only No functional dependencies have either of these of both on their LHS. Hence, the FD XW → Z does not hold.

2. For a relation R(A, B, C, D) with set of functional dependencies, F = {B → C, AD → C}, the lossless decomposition is R1(A, B) and R2(B, C, D).
(a) TRUE                                          (b) FALSE

 Answer: (b) FALSE To check whether the given decomposition is lossless or not, we can find the intersection between the two decomposed relations. If the resultant set of attributes determine all the attribute of one of these relation, then we would say the decomposition is lossless, otherwise not.             R1 ∩ R2 = (AB ∩ BCD) = B. There are no FD such that B → A (to represent R1) or B → CD (to represent R2). Hence, the decomposition is not lossless.

3. If an attribute in a table is a foreign key, then the table cannot contain two tuples with the same value of that attribute.
(a) TRUE                                          (b) FALSE

 Answer: FALSE Foreign key attributes can have duplicate values, that is, any number of records can have the same value for that attribute. Only primary key and unique attributes cannot have duplicate values.

4. A relation R(A, B, C, D, E) with set of functional dependencies F = {AB → C, CD → E, C → A, C → D, D → B} is in BCNF.
(a) TRUE                                          (b) FALSE

 Answer: FALSE For a relation to be in BCNF, all the functional dependencies have the super keys on their left hand side. We can check that by finding the keys for the given relation. Let us start with first FD; (AB)+ = ABCDE through FDs AB → C, C → D, and CD → E AB is a candidate key since (AB)+ = R. (CD)+ = CDEAB through FDs CD → E, C → A, and D → B CD is a super key since (CD)+ = R. C+ = CADBE through FDs C → A, C → D, D → B, and CD → E C is a candidate key since C+ = R D+ = DB through FD D → B. and, D is a not a key. The FD D → B violates the rule for BCNF relation since D is not a key.

5. The relation R given in question (4) is in 3NF.
(a) TRUE                                          (b) FALSE

 Answer: TRUE 3NF - A relation is in third normal form (3NF) if it is in 2NF and no non-key attribute is transitively dependent on a candidate key. The candidate keys for R are AB, AD, and C. In R, there is no non-key attributes depends transitively on a candidate key. Hence, the table is in 3NF.

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