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Test: Functional Dependencies & Normalization- 2 - Computer Science Engineering (CSE) MCQ


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15 Questions MCQ Test - Test: Functional Dependencies & Normalization- 2

Test: Functional Dependencies & Normalization- 2 for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Test: Functional Dependencies & Normalization- 2 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Functional Dependencies & Normalization- 2 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Functional Dependencies & Normalization- 2 below.
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Test: Functional Dependencies & Normalization- 2 - Question 1

A functional dependency of the form X → Y is trivial if

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 1

A trivial functional dependency is a database dependency that occurs when describing a functional dependency of an attribute or of a collection of attributes that includes the original attribute.

So, Option c is correct answer.

Test: Functional Dependencies & Normalization- 2 - Question 2

In an entity relationship, y is the dominant entity and x is a subordinate entity. Which of the following is/are incorrect?

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 2

1. y is dominant entity.

2. x is subordinate entity..
Since, y is dominant entity. So does not depend on any other and x is subordinate. So x is existence dependent on y and deletion of x does not effect y, So, x is detected, so y is incorrect.

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Test: Functional Dependencies & Normalization- 2 - Question 3

Relations produced from an E-R model will always be in

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 3

Relations produced from E-R model generally in 1-NF form.

Test: Functional Dependencies & Normalization- 2 - Question 4

Let R be a relation. Which of the following com m ents about the relation R are correct?
1. R will necessarily have a composite key if R is in BCNF but not in 4NF.

2. If R is in 3NF and if every key of R is simple , then R is in BCNF.

3. If R is in BCNF and if R has at least one

4. If R is in 3NF and if its every key is simple, then R is in 5NF.

Test: Functional Dependencies & Normalization- 2 - Question 5

Consider the relation:
Employee (Emp-No, Emp-name, salary, project- no, due-date)
(Assuming an 1-1 relationship between project and employees)
Project-no is functionally dependent on

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 5

The relation EMPLOYEE has one to one relation b/w project and employee i.e. each employee has being assigned a project and each project is associated with only one employee, hence project- number is functionally dependent on Emp-no.

Test: Functional Dependencies & Normalization- 2 - Question 6

Let R ( a, b, c) and S (d, e, f) be two relations in which d is the foreign key of S that refers to the primary key of R. Consider the following four operations.
1. Insert into R
2. Insert into S
3. Delete from R
4. Delete from S

Which of the following is true about the referential integrity constraint about?

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 6

R (a, b, d, c) and S (e,f)
1. Insertion into R does not create any violation,
2. Insertion into S create violation.
3. Deletion in S does not create any violation.
4. Deletion in R can create violation.

Test: Functional Dependencies & Normalization- 2 - Question 7

Consider the following set of functional dependencies on the scheme (A, B, C).
A —> BC, B —> C, A —> B ,A B —> C
The canonical cover for this set is

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 7

 

In the given FDs,

1) Convert elements on LHS as singleton

           Thus, A -> BC can be written as A-> B and A->C

2) Remove composite attributes from LHS

           Here, AB -> C can be written as A -> C, because we have an FD {A -> B}.

3) Remove redundant attributes

        We have FD = {A -> B, A -> C , B -> C}

       this is a transitive depenency A -> B -> C. After remove redundancy, we get A->B, B->C.

Option A is the required answer

Test: Functional Dependencies & Normalization- 2 - Question 8

Which of the following statement is false?

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 8

BCNF relation may or may not preserve dependency also any relation cannot be necessarily decomposed into BCNF relation,

Test: Functional Dependencies & Normalization- 2 - Question 9

R(A, B, C, D) with FD’s
F(A →B, A →C, A → D) and  R1 (A, B, D) with FD’s
F1(A —> B, A → D) and R2(B, C) with FD's
F2 = { } 
​Then find which statements is true:

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 9


Clearly we can see that B is the common attribute and B is not candidate key of any of the decomposed relations. Hence, dependency is not preserved.

Test: Functional Dependencies & Normalization- 2 - Question 10

R = (A, B, C, D, E, F)
R1 - (A, B)
R2 = (C,D,E)
R3 = (A, C, F)
F= {A→B; C → DE ; AC → F;}
​Does the above FD’s has lossless join property?

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 10

R = {A, B, C, D, E, F}
F = {A → B , C → DE, AC → F)
R1 = ( A B )
R2 =.{C,D,E}
R3 = {A, C, F}


Convert b to a in 3rd row convert b to a in 3rd row. Since 3rd row contains all ‘a’s hence the relation has lossless join property.

Test: Functional Dependencies & Normalization- 2 - Question 11

Which of the following list represents the MVDs (Mean Value Dependencies), satisfied by a relation R(A, B, C) that has the following tuples: (a1, b1, c1),(a1, b1, c2), (a2, b1, c1), (a2, b1,c3).

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 11


Test: Functional Dependencies & Normalization- 2 - Question 12

From the following instance of a relation schema R(A, B, C), we can conclude that:

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 12

Generally Normalization is done on the schema itself. From the relational instance given,we may strike out FD s that do not hold. e.g.B does not functionally determine C(This is true). But we cannot say that A functionally determines B for the entire relation itself. This is because that, A->B holds for this instance, but in future there might be some tuples added to the instance that may violate A->B. So overall on the relation we cannot conclude that A->B, from the relational instance which is just a subset of an entire relation. 

Test: Functional Dependencies & Normalization- 2 - Question 13

Consider the following declaration:
F: Set of functional dependencies
R: Relation of functional dependencies . R : (A, B, C, D, E)
F: (A —> BC: CD —> E; B —> D; E —> A)
Then candidate keys for R is ___________

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 13

R = (A, B, C, D, E)
F = (A →BC, C D → E, D,B → D, E→A)
A+ = ABCDE so A is the candidate key.
A can be replaced by E
So, E also a candidate key.
Scan be replaced by CD.
Hence, (A, E, C, D) are the candidate keys.

Test: Functional Dependencies & Normalization- 2 - Question 14

A given relation is known to be in third normal form. Select the statement which can be inferred from this

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 14

A relation is in 3NF if for X →A
(i) X is a super key or candidate key
(ii) A is a prime attribute.
Hence, for A to be non-key attribute (non-prime attribute)
X must be satisfying (i)

Test: Functional Dependencies & Normalization- 2 - Question 15

A relation Empdtl is defined with attribute empcode (unique), name, street, city, state and pincode. For any pincode, there is only one city and state. Also, for any given street, city and state, there is just one pincode. in normalization terms, Empdtl is a relation in

Detailed Solution for Test: Functional Dependencies & Normalization- 2 - Question 15

1) Pin code →  city, state

2) Street, city, state →  pin code

The candidate keys of relation is empcode as it uniquely identifies the relation.

Hence, it is in 2NF and therefore, also in 1NF.

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