Normalization
Question 1 
Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).
Consider the two statements given below.

 I. Both Y and Z are in BCNF
 II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?
I only  
Neither I nor II  
II only  
Both I and II 
R → P
R^{+} = RP
* In R → P, 'R' is a super key. So, Y is in BCNF.
Z = (QRS)
QR → S
S → Q
CK's = QR, RS
* In, S → Q, 'S' is not a super key. So, Z is not in BCNF.
* Y is in BCNF and Z is not in BCNF.
* 'R' is common attribute in the relations Y and Z. and R is candidate key for Y. So, the decomposition is lossless.
* The FD, R → P is applicable on Y and QR → S, S → Q are applicablein 2.
So, the decomposition is dependency preserving.
* Hence, Statement II is correct.
Question 2 
Consider the following four relational schemas. For each schema, all nontrivial functional dependencies are listed. The underlined attributes are the respective primary keys.
Schema I: Registration(rollno, courses) Field ‘courses’ is a setvalued attribute containing the set of courses a student has registered for. Nontrivial functional dependency rollno → courses Schema II: Registration (rollno, coursid, email) Nontrivial functional dependencies: rollno, courseid → email email → rollno Schema III: Registration (rollno, courseid, marks, grade) Nontrivial functional dependencies: rollno, courseid, → marks, grade marks → grade Schema IV: Registration (rollno, courseid, credit) Nontrivial functional dependencies: rollno, courseid → credit courseid → credit
Which one of the relational schemas above is in 3NF but not in BCNF?
Schema I  
Schema II  
Schema III  
Schema IV 
Registration (rollno, courses) rollno → courses
For the given schema Registration ‘rollno’ is a primary key.
Leftside of the functional dependency is a superkey so, Registration is in BCNF.
Schema II:
Registrstion (rollno, courseid, email)
rollno, courseid → email
email → rollno
From the given schema the candidate key is (rollno + courseid).
There is no part of the key in the left hand of the FD’s so, it is in 2NF.
In the FD email→rollno, email is nonprime attribute but rollno is a prime attribute.
So, it is not a transitive dependency.
No transitive dependencies so, the schema is in 3NF.
But in the second FD email→rollno, email is not a superkey.
So, it is violating BCNF.
Hence, the schema Registration is in 3NF but not in BCNF.
Schema III:
Registration (rollno, courseid, marks, grade)
rollno, courseid → marks, grade
marks → grade
For the schema the candidate key is (rollno + courseid).
There are no part of the keys are determining nonprime attributes.
So, the schema is in 2NF.
In the FD marks → grade, both the attributes marks and grade are nonprime.
So, it is a transitive dependency.
The FD is violating 3NF.
The schema Registration is in 2NF but not in 3NF.
Schema IV:
Registration (rollno, courseid, credit)
rollno, courseid → credit
courseid → credit
The candidate key is (rollno + courseid).
In the FD, courseid → credit, courseid is part of the key (prime attribute) and credit is nonprime.
So, it is a partial dependency.
The schema is violating 2NF.
Question 3 
Which of the following is NOT a superkey in a relational schema with attributes V, W, X, Y, Z and primary key VY?
VXYZ  
VWXZ  
VWXY  
VWXYZ 
Any superset of “VY” is a super key. So, option (B) does not contain “Y”.
Question 4 
The primary key is (VOLUME, NUMBER, STARTPAGE, ENDPAGE) and the following functional dependencies exist in the schema. (VOLUME, NUMBER, STARTPAGE, ENDPAGE) → TITLE (VOLUME, NUMBER) → YEAR (VOLUME, NUMBER, STARTPAGE, ENDPAGE) → PRICE The database is redesigned to use the following schemas. (VOLUME, NUMBER, STARTPAGE, ENDPAGE, TITLE, PRICE) (VOLUME, NUMBER, YEAR) Which is the weakest normal form that the new database satisﬁes, but the old one does not?
1NF  
2NF  
3NF  
BCNF 
V – VOLUME
N – NUMBER
S – STARTPAGE
E – ENDPAGE
T – TITLE
Y – YEAR
P – PRICE
Primary key: (V, N, S, E)
FD set:
(V, N, S, E) → T
(V, N) → Y
(V, N, S, E) → P
In (V, N) → Y; V, N is a part of the key and Y is nonprime attribute.
So, it is a partial dependency.
Now, the schema “Journal” is in 1NF but not in 2NF.
The database is redesigned as follows:
Both R_{1} and R_{2} are in BCNF.
Therefore, 2NF is the weakest normal form that the new database satisfies, but the old one does not.
Question 5 
Given the following two statements:
S1: Every table with two singlevalued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB>C, D>E, E>C is a minimal cover for the set of functional dependencies AB>C, D>E, AB>E, E>C.
Which one of the following is CORRECT?
S1 is TRUE and S2 is FALSE.  
Both S1 and S2 are TRUE.  
S1 is FALSE and S2 is TRUE.  
Both S1 and S2 are FALSE. 
If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also.
Let R(AB) be a two attribute relation, then
If {A→B} exists then BCNF since {A}^{+} = AB = R
If {B→A} exists then BCNF since {B}^{+} = AB = R
If {A→B, B→A} exists then BCNF since A and B both are Super Key now.
If {No nontrivial Functional Dependency} then default BCNF.
Hence it’s proved that a Relation with two singlevalued attributes is in BCNF hence it’s also in 1NF, 2NF, 3NF.
S2: False
The canonical cover for the given FD set is {AB→C, D→E, AB→E, E→C}. As we can see AB→E is not covered in minimal cover since {AB}^{+} = ABC in the given cover {AB→C, D→E, E→C}
Question 6 
Relation R has eight attributes ABCDEFGH . Fields of R contain only atomic values.
F = {CH→G, A→BC, B→CFG, E→A, F→EG}, is a set of functional dependencies (FDs) so that F^{+} is exactly the set of FDs that hold for R.
How many candidate keys does the relation R have?
3  
4  
5  
6 
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD^{+} = {ABCDEFGH}
BD^{+} = {ABCDEFGH}
ED^{+} = {ABCDEFGH}
FD^{+} = {ABCDEFGH}
But CD^{+}, GD^{+} and HD^{+} does not give all the attributes hence CD, GD and HD are not candidate keys.
Hence no. of candidate keys are 4: AD, BD, ED, FD.
Question 7 
Relation R has eight attributes ABCDEFGH . Fields of R contain only atomic values.
F = {CH→G, A→BC, B→CFG, E→A, F→EG}, is a set of functional dependencies (FDs) so that F^{+} is exactly the set of FDs that hold for R.
The relation R is
in 1NF, but not in 2NF.  
in 2NF, but not in 3NF.  
in 3NF, but not in BCNF.  
in BCNF. 
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD^{+} = {ABCDEFGH}
BD^{+} = {ABCDEFGH}
ED^{+} = {ABCDEFGH}
FD^{+}= {ABCDEFGH}
But CD^{+}, GD^{+} and HD^{+} does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.
Question 8 
Which of the following is TRUE?
Every relation in 3NF is also in BCNF  
A relation R is in 3NF if every nonprime attribute of R is fully functionally dependent on every key of R  
Every relation in BCNF is also in 3NF  
No relation can be in both BCNF and 3NF 
Question 9 
Consider the following relational schema:
Suppliers(sid:integer, sname:string, city:string, street:string) Parts(pid:integer, pname:string, color:string) Catalog(sid:integer, pid:integer, cost:real)
Assume that, in the suppliers relation above, each supplier and each street within a city has a unique name, and (sname, city) forms a candidate key. No other functional dependencies are implied other than those implied by primary and candidate keys. Which one of the following is TRUE about the above schema??
The schema is in BCNF  
The schema is in 3NF but not in BCNF  
The schema is in 2NF but not in 3NF  
The schema is not in 2NF 
(Sid, Street) → Sname
As Sid is a primary key, then
(Sid, Street) will be super key.
Hence, it is in BCNF.
Question 10 
Consider the following relational schemes for a library database:

Book(Title, Author, Catalog_ no, Publisher, Year, Price)
Collection (Title, Author, Catalog_no)
with in the following functional dependencies:

I. Title Author → Catalog_no
II. Catalog_no → Title Author Publisher Year
III. Publisher Title Year → Price
Assume {Author, Title} is the key for both schemes. Which of the following statements is true?
Both Book and Collection are in BCNF  
Both Book and Collection are in 3NF only
 
Book is in 2NF and Collection is in 3NF  
Both Book and Collection are in 2NF only 
Book(Title, Author, Catalog_no, Publisher, Year, Price)
Collection(Title, Author, Catalog_no)
I) Title Author ⟶ Catalog_no ⟶ BCNR
II) Catalog_no ⟶ Title, Author, Publisher, Year ⟶ 3NF
III) Publisher Title Year ⟶ Price ⟶ 2NF Book’s in 2NF
Collection is in 3NF.
Question 11 
Which one of the following statements if FALSE?
Any relation with two attributes is in BCNF  
A relation in which every key has only one attribute is in 2NF  
A prime attribute can be transitively dependent on a key in a 3NF relation  
A prime attribute can be transitively dependent on a key in a BCNF relation 
i) It is in 3NF.
ii) For any dependency X→ Y
where X is a super key.
iii) Functional dependency has been removed.
Option D is false.
→ Because a prime attribute can’t be transitive dependent on a key in a BCNF relation.
Question 12 
Which one of the following statements about normal forms is FALSE?
BCNF is stricter than 3NF  
Lossless, dependencypreserving decomposition into 3NF is always possible
 
Lossless, dependencypreserving decomposition into BCNF is always possible  
Any relation with two attributes is in BCNF 
Option B: Lossless, dependency preserving decomposition into 3NF is always possible.
Option C: It is false.
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can't be decomposed into BCNF.
Option D: It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and viceversa.
Question 13 
The relation scheme Student Performance (name, courseNo, rollNo, grade) has the following functional dependencies:
name, courseNo → grade rollNo, courseNo → grade name → rollNo rollNo → name
The highest normal form of this relation scheme is
2 NF  
3 NF  
BCNF  
4NF 
name, courseNo → grade →(I)
rollNo, courseNo → grade →(II)
name → rollNo →(III)
rollNo → name →(IV)
Candidate keys: name, courseNo (or) rollNo
Its is not BCNF, because the relation III, there is no relationship from super key.
name → rollNo
It is not BCNF, name is not super key.
It belongs to 3NF, because if X→Y, Y is prime then it is in 3NF.
Question 14 
Consider the following functional dependencies in a database:
Data_of_Birth → Age Age → Eligibility Name → Roll_number Roll_number → Name Course_number → Course_name Course_number → Instructor (Roll_number, Course_number) → Grade
The relation (Roll_number, Name, Date_of_birth, Age) is:
in second normal form but not in third normal form  
in third normal form but not in BCNF  
in BCNF  
in none of the above 
Date_of_Birth → Age
Name → Roll_number
Roll_number → Name
Candidate keys for the above are:
(Date_of_Birth, Name) and (Date_of_Birth, Roll_number)
Clearly, there is a partial dependency,
Date_of_Birth → Age
So, it is only in 1NF.
Question 15 
Relation R with an associated set of functional dependencies, F, is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set of relations is
Zero  
More than zero but less than that of an equivalent 3NF decomposition  
Proportional to the size of F^{+}  
Indetermine 
Question 16 
For relation R = (L, M, N , O, P), the following dependencies hold:
M → O NO → P P → L and L → MN
R is decomposed into R_{1} = (L, M, N , P) and R_{2} = (M, O).
 (a) Is the above decomposition a losslessjoin decomposition? Explain.
(b) Is the above decomposition dependencypreserving? If not, list all the dependencies that are not preserved.
(c) What is the highest normal form satisfied by the above decomposition?
Theory Explanation is given below. 
Question 17 
R(A,B,C,D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition?
A → B, B → CD  
A → B, B → C, C → D  
AB → C, C → AD  
A → BCD 
Question 18 
Consider the schema R = (S T U V) and the dependencies S → T, T → U, U → V and V → S. Let R = (R1 and R2) be a decomposition such that R1 ∩ R2 ≠ ∅ . The decomposition is
not in 2NF  
in 2NF but not 3NF  
in 3NF but not in 2NF  
in both 2NF and 3NF 
And since every attribute is key so the decomposed relation will be in BCNF and hence in 3NF.
Question 19 
Which normal form is considered adequate for normal relational database design?
2 NF  
5 NF  
4 NF  
3 NF 
Question 20 
For a database relation R(a,b,c,d), where the domains a, b, c, d include only atomic values, only the following functional dependencies and those that can be inferred from them hold:
a → c b → d
This relation is
in first normal form but not in second normal form  
in second normal form but not in third normal form  
in third normal form  
None of the above 
Since all a, b, c, d are atomic. So the relation is in 1NF.
Checking the FD's
a → c
b → d
We can see that there is partial dependencies. So it is not 2NF.
So answer is option (A).
Question 21 
(a) Consider the relation scheme R(A, B, C) with the following functional dependencies:
A, B → C, C → A
Show that the scheme R is the Third Normal Form (3NF) but not in BoyceCode Normal Form (BCNF).
(b) Determine the minimal keys of relation R.
Theory Explanation. 
Question 22 
Consider a relational table R that is in 3NF, but not in BCNF. Which one of the following statements is TRUE?
A cell in R holds a set instead of an atomic value.  
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a nonprime attribute and X is not a proper subset of any key.
 
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a nonprime attribute and X is a proper subset of some key.  
R has a nontrivial functional dependency X→A, where X is not a superkey and A is a prime attribute. 
FDs:
AB → C
BC → A
(BD)^{+} = BD ✖
(ABD)^{+} = ABDC ✔
(CBD)^{+} = CBDA ✔
Candidate keys = {ABD, CBD}
• The relation R is in 3NF, as there are no transitive dependencies.
• The relation R is not in BCNF, because the left side of both the FD’s are not Super keys.
• In R, BC → A is a nontrivial FD and in which BC is not a Super key and A is a prime attribute.
Question 23 
Choose the correct alternatives (More than one may be correct).
Indicate which of the following statements are true: A relational database which is in 3NF may still have undesirable data redundancy because there may exist:
Transitive functional dependencies.  
Nontrivial functional dependencies involving prime attributes on the rightside.
 
Nontrivial functional dependencies involving prime attributes only on the leftside.
 
Nontrivial functional dependencies involving only prime attributes.  
Both (B) and (D). 
B) 3NF because right side is prime attribute.
C) Not in 3NF, because lets suppose ABC is a candidate key. Now consider
AB → Nonprime attribute
which show it is not in 3NF
D) Involves only prime attribute, so right side should definitely contain only prime attribute. So in 3NF.
Question 24 
Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold:
AB → CD, DE → P, C → E, P → C and B → G.The relational schema R is
in BCNF  
in 3NF, but not in BCNF  
in 2NF, but not in 3NF  
not in 2NF 
Since there is a partial dependency B→G.
So the relational schema R is Not in 2NF.
Question 25 
A table has fields Fl, F2, F3, F4, F5 with the following functional dependencies
F1 → F3, F2→ F4, (F1.F2) → F5
In terms of Normalization, this table is in
1 NF  
2 NF  
3 NF  
None 
F2 → F4 ......(ii)
(F1⋅F2) → F5 .....(iii)
F1F2 is the candidate key.
F1 and F2 are the prime key.
In (i) and (ii) we can observe that the relation from P → NP which is partial dependency. So this is in 1NF.
Question 26 
Consider the decomposition of the relation R into the consistent relations according to the following two decomposition schemes.
D_{1}: R=[(P,Q,S,T); (P,T,X); (Q,Y); (Y,Z,W)]
D_{2}: R=[(P,Q,S);(T,X);(Q,Y);(Y,Z,W)]
Which one of the following options is correct?
D^{1}is a lossy decomposition, but D^{2}is a lossless decomposition.
 
Both D^{1}and D^{2}are lossy decompositions.  
Both D^{1}and D^{2}are lossless decompositions.  
D^{1}is a lossless decomposition, but D^{2}is a lossy decomposition. 
Given functional dependencies set:
PQ>X
P>YX
Q>Y
Y>ZW
 While merging the tables there should be some common attribute(s) and it should be a candidate key of one of the tables.
 R1 should be merged with R2 because PT is a key of R2.
 R3 should be merged with PQSTX because Q is a key of R3.
 R4 should be merged with PQSTXY because Y is a key of R4.
 R1 should be merged with R3 because Q is a key of R3.
 R4 should be merged with PQSY because Y is a key of R4.
 Now, there is no common attribute in between R2(TX) and PQSYZW.
 Hence, D2 is lossy decomposition.
Question 27 
A relation with only two attributes is always in BCNF.  
If all attributes of a relation are prime attributes, then the relation is in BCNF.  
Every relation has at least one nonprime attribute.  
BCNF decompositions preserve functional dependencies. 
Example:
R(A, B).
Two functional dependencies possible for the relation: (1) A>B and (2) B>A
If there is no functional dependency, we can assume trivial functional dependencies like AB>A and AB>B.
In all cases, functional dependencies like A>B, A must be a key.
So they all will be in BCNF irrespective of the functional depencies set.
Question 28 
Two (or more) candidate keys  
Two candidate keys and composite  
The candidate key overlap  
Two mutually exclusive foreign keys 
Transformation into BoyceCodd normal form deals with the problem of overlapping keys.
Question 29 
First Normal Form  
Second Normal Form
 
Third Normal Form  
Third Normal Form 
Second normal form (2NF) is a normal form used in database normalization.
To qualify for second normal form a relation must:
→be in first normal form (1NF)
→not have any nonprime attribute that is dependent on any proper subset of any candidate key of the relation.
A nonprime attribute of a relation is an attribute that is not a part of any candidate key of the relation.
Question 30 
ISBN → Title
ISBN → Publisher
Publisher → Address
First Normal Form  
Second Normal Form  
Third Normal Form  
BCNF 
→In order to check whether it is in Second normal form or not,fist we need to find the candidate key.
→After finding closure of attribute of ISBN from the above dependencies,the candidate key is ISBN.
→A relation is in 2NF if it is in 1NF and every nonprime attribute of the relation is dependent on the whole of every candidate key.
→The given dependencies satisfies second normal form rules as as non prime attribute of the relation is dependent on the whole of every candidate key.
→So the above dependencies are in Second normal form.
Question 31 
The table is in which normal form?
First Normal Form  
Second Normal Form  
Third Normal Form but not BCNF  
Third Normal Form but BCNF 
AB→ CDE
C→ B
The candidate keys are "AB" and "AC".
The relation is not in BCNF because the functional dependencies C→ B do not contains the key AB or AC
So, the given relation is in 3NF but not in BCNF
Question 32 
For a database relation R(a, b, c, d) where the domains of a, b, c and d include only atomic values, and only the following functional dependencies and those that can be inferred from them hold :

a → c
b → d
The relation is in
First normal form but not in second normal form
 
Second normal form but not in third normal form  
Third normal form
 
BCNF 
And there is a partial dependency exist in given FD’s so the given relation is in 1NF but not in second normal form.
Question 33 
A manytoone relationship exists between entity sets r1 and r2. How will it be represented using functional dependencies if Pk(r) denotes the primary key attribute of relation r?
Pk(r1) → Pk(r2)  
Pk(r2) → Pk(r1)  
Pk(r2) → Pk(r1) and Pk(r1) → Pk(r2)  
Pk(r2) → Pk(r1) or Pk(r1) → Pk(r2)

Here, we have a many to one relationship between between Set(r1) and Set(r2).
→ Elements of Set(r2) can’t identify elements of Set(r1) because one value element in Set(r2) is pointing to more than one element of Set(r1).
→ So, we can’t say Pk(r2) → Pk(r1) but elements of Set(r1) are pointing to exactly one element of Set(r2) so we can say that Pk(r2) → Pk(r1) because r1 is uniquely identifying r2.
Question 34 
First normal form  
Second normal form  
Fourth normal form  
Third normal form 
1. The relation R (table) is in second normal form (2NF)
2. Every nonprime attribute of R is nontransitively dependent on every key of R.
Question 35 
Composite keys  
Determinants  
Candidate keys  
Foreign keys 
● Each table can have one or more candidate keys, but one candidate key is unique, and it is called the primary key.
Question 36 
BCNF is stricter than 3NF  
Lossless,dependency preserving decomposition into BCNF is always possible  
Lossless,dependency preserving decomposition into 3NF is always possible  
Any relation with two attributes is BCNF 
It is not possible to have dependency preserving in BCNF decomposition.
→ Let take an example, 3NF can't be decomposed into BCNF. Option D : It is true.
Let consider two attributes (X, Y).
If (X→Y), X is a candidate key. It is in BCNF and viceversa.
Question 37 
normalized of data  
denomination of data  
isolation of data  
denormalized of data 
Most commonly used normal forms
First normal form(1NF)
Second normal form(2NF)
Third normal form(3NF)
Boyce & Codd normal form (BCNF)
Question 38 
(Assume that no faculty member within a single department has same name. Each faculty member has only one office identified in office. 3NF refers to third normal form and BCNF refers to BoyceCodd Normal Form. Then Faculty is
Not in 3NF, in BCNF  
In 3NF, not in BCNF  
In 3NF, in BCNF  
Not in 3NF, not in BCNF 
FacName ➝ dept,office
Question 39 
Which of the following relation schemas is definitely in BCNF?
R1(A,B)  
R4(A,B,C,D,E)
 
R3(A,B,C,D)
 
R2(A,B,C) 
1. BCNF is free from redundancy.
2. If a relation is in BCNF, then 3NF is also also satisfied.
3. Every Binary Relation ( a Relation with only 2 attributes ) is always in BCNF.
4. Sometimes going for BCNF form may not preserve functional dependency.
Question 40 
atomic values, only the following functional dependencies and those that can be inferred
from them hold
A → c
B → d
The relation is in
1 normal form but not in 2 normal form  
2 normal form but not in 3 normal form  
3 normal form  
None of these 
Since all a, b, c, d are atomic. So the relation is in 1NF.
Checking the FD's
a → c
b → d
We can see that there is partial dependencies. So it is not 2NF.
Question 41 
3NF  
2NF  
4NF  
5NF 
● A relation R is in BCNF if and only if it is in 3NF and no any prime attribute is transitively dependent on the primary key.
● An attribute C is transitively dependent on attribute A if there exist an attribute B such that A>B and B>C.
4NF
● A table is in the 4NF if it is in BCNF and has no multivalued dependencies.
Question 42 
A 4NF is more immune against logical inconsistencies than a 3NF table.  
A 3NF table will have fewer anomalies than a 2NF table  
A 3NF table is more vulnerable than a 2NF table  
A database is said to be in 3NF if all its tables are in 3NF 
1. It should be in the BoyceCodd Normal Form (BCNF).
2. The table should not have any Multivalued Dependency.
Question 43 
(i) To achieve physical data independence
(ii) To remove data anomalies (insertion,update,delete,anomalies)
(iii) To save space on disk.
(i),(ii) and (iii)  
(i) and (ii)  
(i) and (iii)  
(ii) and (iii) 
1. To remove data anomalies (insertion,update,delete,anomalies)
2. To save space on disk(It happens because of avoiding redundancy)
Question 44 
ia, iic, iiib, ivd  
id, iic, iiia, ivb  
id, iic, iiib, iva  
ia, iib, iiic, ivd  
None of the above 
→ A relation is in 3NF if it is in eliminates transitive dependency from the relation.
→ 4NF eliminates multivalued dependency from a relation.
→ 5NF eliminates join dependency from a table.
Question 45 
2 NF  
3 NF  
4 NF  
5 NF 
→ A relation is in 4NF if it is in BCNF and has no multivalued dependency.
→ A relation will contain Multivalued dependency if there exists a dependency A → B such that for a single value of A, multiple values of B exists.
Question 46 
BoyceCodd Normal form  
Third Normal form  
Second Normal form  
First Normal form 
AB → D
C → A
D → B
A + ={A}
B + ={B}
C + ={C,A}
D + ={D,B}
Since single attributes can't determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.
AB + ={A,B,C,D}
AC + ={A,C}
AD + ={A,D,B,C}
BC + ={B,C,A,D}
AB + ={A,B,C,D}
AC + ={A,C}
AD + ={A,D,B,C}
BC + ={B,C,A,D}
BD + ={B,D}
CD + ={C,D,B,A}
So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional dependency should be a super key) so the given relation is in 3NF only.
Question 47 
Any relation with two attributes is in BCNF.  
A relation in which every key has only one attribute is in 2NF.  
A prime attribute can be transitively dependent on a key in 3NF relation.  
A prime attribute can be transitively dependent on a key in BCNF relation. 
i) It is in 3NF.
ii) For any dependency X→ Y
where X is a super key.
iii) Functional dependency has been removed.
Option D is false.
→ Because a prime attribute can’t be transitive dependent on a key in a BCNF relation.
Question 48 
A relation in BCNF is always in 3NF.  
A relation in 3NF is always in BCNF.  
BCNF and 3NF are same.  
A relation in BCNF is not in 3NF. 
Question 49 
Redundancy  
Inconsistencies  
Anomalies  
All of the above 
Question 50 
2NF  
3NF  
4NF  
5NF 
Question 51 
(a)1, (b)5, (c)4, (d)2, (e)3  
(a)1, (b)5, (c)2, (d)3, (e)4  
(a)5, (b)1, (c)2, (d)3, (e)4  
(a)5, (b)4, (c)3, (d)1, (e)2 
Nameless object→ Within a statement
Template support→ Generic programming
A forward reference→ defines a class
Derived class inherits from base class→ Member function
Question 52 
I(a), II(c), III(b), IV(d)
 
I(d), II(a), III(b), IV(c)  
I(c), II(d), III(a), IV(b)  
I(d), II(b), III(a), IV(C)  
I(c), II(a), III(b), IV(d) 
→ 3NF eliminates transitive dependency from the relation. A relation can be in 3NF if it is in 2NF and have no transitive dependency.
→ 4NF eliminates multivalued dependency from the relation. A relation is in 4NF if it is in BCNF and have no multivalued dependency.
→ 5NF ensures that the relation contains no join dependency. A relation is in 5NF if it is 4NF and ave no join dependency.
Note: Given question is wrong. We are added correct extra option. Excluded for correction.
Question 53 
A relation in 3NF is always in BCNF  
A relation in BCNF is always in 3NF  
BCNF and 3NF are totally different  
A relation in BCNF is in 2NF but not in 3NF 
Question 54 
no nonprime attribute is functionally dependent on other nonprime attributes  
no nonprime attribute is functionally dependent on prime attributes  
all attributes are functionally independent  
prime attribute is functionally independent of all nonprime attributes 
Example: Let us consider a relation R(A, B, C, D) and let’s take ABC is the primary key of the given relation R then A, B, C are the prime attributes and D is a nonkey prime attribute.
Now for 2NF, B → D should not exist but B → A can exist.
3NF: A relation is said to be in 3NF if no transitive dependency exist in the relation i.e. either the LHS of a functional dependency should be a super key or the RHS of a functional dependency should be a Prime attribute.
Option(A) is correct it means the the LHS will always be a super key or both LHS or RHS of a functional dependency will be a prime key attribute. In both the cases the relation will be in 3NF and when the relation will be in 3NF it will automatically be in 2NF.
Option (B) is incorrect because according to this if there could be a case when in a functional dependency its LHS will not have prime key attribute and RHS will also not be a prime key attribute. So in that case the relation can't be in 3NF. so this option is not correct.
Option(C) is incorrect, if all attributes will be functionally independent then the conditions of being in 2NF and 3NF will not be satisfied. SSo this option is incorrect.
Question 55 
First normal form  
Second normal form  
Third normal form  
Fourth normal form 
2NF: A relation is in 2NF is there is no partial dependency exist in the relation. i.e.
primary_key → nonkey attribute
prime_attribute → nonkey attribute.
3NF : A relation is in 3NF if there is no transitive dependency or we can say that a relation is in 3NF if either LHS of a functional dependency is a super key or the RHS of the functional dependency is a prime key attribute.
4NF : A relation is in 4NF if there is no multivalued dependency exist in the relation.
Hence the correct answer is option (C)
NOTE: A prime key attribute is a attribute which is the part of primary key attributes. For example: If ABC is the the primary key then A,B,C, AB, AC, BC are the prime key attributes.
Question 56 
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .
in 1NF but not in 2NF  
in 2NF but not in 3NF
 
in 3NF but not in BCNF  
in BCNF 
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.
Question 57 
Question 58 
A → C
B → D
The relation R is in _______.
First normal form but not in second normal form.  
Both in first normal form as well as in second normal form.  
Second normal form but not in third normal form.  
Both in second normal form as well as in third normal form. 
Since all a, b, c, d are atomic. So the relation is in 1NF.
Checking the FD's
a → c
b → d
We can see that there is partial dependencies. So it is not 2NF.
So answer is option (A).
Question 59 
R is in 4NF  
R is not in 1NF  
R is in 2Nf and not in 3NF  
R is in 2NF and 3NF 
Question 60 
Lossless preserving decomposition into 3NF is always possible  
Lossless preserving decomposition into BCNF is always possible  
Any Relation with two attributes is in BCNF  
BCNF is stronger than 3NF 
Option(B) is false because BCNF ensures lossless decomposition but do not preserves functional dependency.
Option(C) is true because Any Relation with two attributes is in BCNF
Option(D) is true because BCNF ensure less number of duplicate values in a relation than 3NF.
Question 61 
Non_key attribute V_name is dependent on V_no which is part of composite key  
Non_key attribute V_name is dependent on Qty_sup  
Key attribute Qty_sup is dependent on primary_key unit price  
Key attribute V_ord_no is dependent on primary_key unit price  
None of the above 
Option(A) is not correct because Non_key attribute V_name is dependent on V_no which is part of composite key i.e there is a partial dependency exists because of V_no.
Option(B) is also not correct because there are chances of having a partial dependency because Qty_sup could be a prime attribute.
Option(C) and Option(D) are correct because in both the options Key attributes depend upon primary key.
Question 62 
R( A,B,C,D)  
R( A,B,C)  
R( A,B,C,D,E)  
R( A,B) 
1. BCNF is the advance version of 3NF. It is stricter than 3NF.
2. A table is in BCNF if every functional dependency X → Y, X is the super key of the table.
3. For BCNF, the table should be in 3NF, and for every FD, LHS is super key.
4. Two (or) Binary tuples are always be in BCNF
Question 63 
First normal form  
Second normal form  
Third normal form  
Fourth normal form 
2. In the question it is not mentioned that nonprime attribute is only dependent on primary key.
3. So, the functional dependency B → C, is perfectly valid.
4. This relation is in 2NF but not in 3NF because of every nonkey attribute is transitively dependent on the primary key. Here “A” will be a candidate key.
Question 64 
1NF but not in 2NF  
2 NF but not in 3NF  
3NF but not in BCNF  
BCNF but not in 4NF 
And clearly there is partial functional dependency which is
A → C ,in which non prime attribute C is partially dependent on A
Question 65 
1NF  
2NF  
4NF  
BCNF 
Question 66 
If a relation is an 2NF and 3NF forms, then
no nonprime attribute is functionally dependent on other nonprime attributes  
no nonprime attribute is functionally depend on the prime attributes  
all attributes are functionally independent  
prime attribute is functionally independent of all nonprime attributes

Question 67 
Which normal form is considered adequate for relational database design?
2 NF  
3 NF  
4 NF
 
BCNF

Question 68 
A relation is in ____ form if every field consists only of atomic values, that is, not lists or sets.
First normal
 
Third normal
 
Second normal
 
Fourth normal

Question 69 
Partial dependencies are removed to achieve which normal form?
First normal form  
Second normal form
 
BCNF
 
Third normal form

Question 70 
Zero  
More than Zero but less than that of an equivalent 3NF  
Proportional to the size of F^{+}  
Indeterminate 
Question 71 
1 NF  
2 NF  
3 NF  
BCNF 
Let there be relation R(C,D).
Possible functional dependencies are,
C>D (Candidate key C)
or
D>C (Candidate key is D)
or
CD>CD (Candidate key is CD)
No violation is there in any of the above three cases for BCNF.
Question 72 
2NF  
4NF  
3NF  
5NF 
Question 73 
Partial dependency  
Total dependency  
Transitive dependency  
None of the above 
Question 74 
A relation with two attributes is in BCNF  
Lossless dependency preserving decomposition into BCNF is always possible  
BCNF is stricter than 3NF  
Lossless, dependency preserving decomposition into 3NF is always possible

Question 75 
Consider the following relational schemas for a library database : Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection(Title, Author, Catalog_no) with the following functional dependencies :
I. Title, Author → Catalog_no
II. Catalog_no → Title, Author, Publisher, Year
III. Publisher, Title, Year → Price Assume (Author, Title) is the key for both schemas. Which one of the following is true ?
Both Book and Collection are in BCNF.  
Both Book and Collection are in 3NF.  
Book is in 2NF and Collection in 3NF.  
Both Book and Collection are in 2NF. 
(Author, Title) is the key .
Title, Author → Catalog_no
Catalog_no →Publisher
Here we are having a transitive dependency
(Key→nonkey
Nonkey→ nonKey)
Hence this relation is not in 3NF.
Collection(Title, Author, Catalog_no)
(Author, Title) is the key
Catalog_no → Title, Author
Since here LHS is not a Key so it is not in BCNF but since the RHS having prime key attribute so it is in 3NF.
Question 76 
Consider the schema R = {S, T, U, V} and the dependencies
S → T, T → U, U → V and V → S If R = (R1 and R2)
be a decomposition such that R1 ∩ R2 = φ then the decomposition is
not in 2NF  
in 2NF but not in 3NF  
in 3NF but not in 2NF  
in both 2NF and 3NF 
A relation which is in 2NF is always lossless.
In question, it is given that R1 ∩ R2 = φ which is violating lossless decomposition condition. Hence the given relation R is not in 2NF.
Question 77 
Every binary relation is never be in BCNF.  
Every BCNF relation is in 3NF.  
1 NF, 2 NF, 3 NF and BCNF are based on functional dependencies.  
Multivalued Dependency (MVD) is a special case of Join Dependency (JD). 
Question 78 
What is the highest normal form of a relation R(A, B, C, D, E) with FD set?
{B → A, A → C, BC → D, AC→ BE}
2NF  
3NF  
BCNF  
4NF 
A → C
BC → D
AC → BE
B^{+} = BACDE
A^{+} = ACBED
So A & B are Candidate key.
There is no partial dependency, so in 2NF.
But in the BC → D, neither BC is key nor D is prime attribute, hence not in 3NF.
Note: Official Key given optionC is correct.
Question 79 
The relation scheme student performance(name, courseno, rollNo, grade) has the following functional dependencies:
Name, courseNo → grade
rollNo, courseNo → grade
Name → rollNo
rollNo → name
The highest normal form of this relation scheme is
2NF  
BCNF
 
4NF  
3NF 
name, courseNo → grade →(I)
rollNo, courseNo → grade →(II)
name → rollNo →(III)
rollNo → name →(IV)
Candidate keys: name, courseNo (or) rollNo
Its is not BCNF, because the relation III, there is no relationship from super key.
name → rollNo
It is not BCNF, name is not super key.
It belongs to 3NF, because if X→Y, Y is prime then it is in 3NF.
Question 80 
Any relation with two attributes is in BCNF  
A relation in which every key has one attribute is in 2NF  
A prime attribute can be transitively dependent on a key in 3NF relation  
A prime attribute can be transitively dependent on a key in BCNF relation 
Option B is true because there cant be any partial functional dependencies.
Option C is true because the condition for 3NF is either left side of functional dependencies is super key or right side of functional dependencies is prime attribute.
Option D is false because the condition for BCNF is that left side of functional dependencies must be super key.
Question 81 
2nd Normal form  
3rd Normal form  
PJNF  
BCNF 
Question 82 
Functional dependency  
Database modelling  
Normalization  
Decomposition 
Question 83 
1NF oniy  
2NF and hence in lNF  
3NF and hence also in 2NF and 1NF  
BCNF and hence also in 3NF and 2NF and 1NF 
Pincode → city, state
Street, city, state → pincode
The candidate keys are
(empcode, name, pincode)
and
(empcode, name, street, city, state)
Since in both the functional dependencies right side is prime attributes but left side is not super keys. So the given relation is in 3NF but not in BCNF.
Question 84 
Relation R is decomposed using a set of functional dependencies, F, and relation S Is decomposed using another set of functional dependencies, G. One decomposition is definitely BCNF, the other is definitely 3NF, but it is not known which is which.
To make a guaranteed identification, which one of the following tests should be used on the decompositions? (Assume that the closure of F and G are available).
Losslessjoin  
BCNF definition  
3NF definition  
DependencyPreservation 
Question 85 
Second Normal Form  
Boyce Codd Normal Form  
Fourth Normal Form  
First Normal Form 
Question 86 
Suppose relation R(A,B,C,D,E) has the functional dependencies
A→ B
BC → D
BE → C
AD → E
CE → A
What are all the keys of R? Of the five given FDs, which ones violate the BCNF condition?
Descriptive solution 