Tuesday, September 15, 2020

Database management system ER model, normalization quiz with answers

 Solved quiz questions in RDBMS, ER model quiz questions, normalization quiz questions, questions and answers explained in dbms

Quiz Questions in DBMS - TRUE / FALSE


1. A many-to-many relationship R connects entity sets E and F. In an ER diagram, we can use an arrow directed towards F to indicate that R has referential integrity from E to F.

(a) TRUE                                          (b) FALSE

Answer: FALSE

Referential integrity is a constraint only added to a many-one relationship in the “one direction.”


2. R is a relation in 4NF. If the MD A →→ BC (A multi-determines BC) holds in R and is non-trivial, then the FD A BC holds in R.

(a) TRUE                                          (b) FALSE

Answer: TRUE

Since R is in 4NF, the MD does not violate 4NF. Therefore, A must be a superkey, from which the FD A → BC follows.


3. If relation R has n tuples and relation S has m tuples, min(m, n) is the maximum number of tuples that R ∩ S can contain. (Assume set-theoretic semantics.)

(a) TRUE                                          (b) FALSE

Answer: TRUE

We try to find the number of common tuples (records) between R and S. The maximum number of tuples is based on the number of tuples in each relation and the smallest relation among them. Hence, the result is the minimum(m, n).


4. Number of possible non-trivial multi-valued dependencies in a relation with two attributes will be 1.

(a) TRUE                                          (b) FALSE

Answer: FALSE

Zero. Any multi-valued dependency in this relation must involve both two attributes. Hence, it is trivial.


5. In a relation R(X, Y, Z, T), if XY Y Z then X Z.

(a) TRUE                                          (b) FALSE

Answer: FALSE

As per the augmentation rule, if we like to add an attribute or set of attributes with a functional dependency, we can do so by adding the said attribute on both sides of the given functional dependency. But, we cannot remove an attribute (or set of attributes) that exists already in a FD on both sides always. Because, in the given FD, the attribute (or set of attributes) that appear on both sides may not be redundant on LHS to determine the rest of the attributes on the RHS.



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