FunctionalDependency
Question 1 
V → W
VW → X
Y → VX
Y → Z
Which of the following is irreducible equivalent for this set of functional dependencies?
V→W V→X Y→V Y→Z  
V→W W→X Y→V Y→Z  
V→W V→X Y→V Y→X Y→Z  
V→W W→X Y→V Y→X Y→Z 
V → W, VW → X, Y → V, Y → X, Y→ Z
Step 2:
V → W, VW → X, Y → V, Y → X, Y→ Z
(V)^{+} = V ×
(VW)^{+} = VW ×
(Y)^{+} = YXZ
(Y)^{+} = YVW ×
(Y)^{+} = YVWX
Without Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.
So, we can remove Y → X as its redundant.
Step 3:
V → W, VW → X, Y → V, Y → Z
(V)^{+} = VW, the closure of V is deriving W from the remaining FD’s.
So, W is redundant. We can remove it.
So, the final canonical form is
V→W, V→X, Y→V, Y→Z
⇾ So, option (A) is correct.
Question 2 
Consider the relation X(P, Q, R, S, T, U) with the following set of functional dependencies
F = { {P, R} → {S,T}, {P, S, U} → {Q, R} }
Which of the following is the trivial functional dependency in F+ is closure of F?
{P,R}→{S,T}  
{P,R}→{R,T}  
{P,S}→{S}  
{P,S,U}→{Q} 
Question 3 
Consider the relation scheme R = (E, F, G, H, I, J, K, L, M, N) and the set of functional dependencies {{E,F} → {G}, {F} → {I,J}, {E,H} → {K,L}, {K} → {M}, {L} → {N}} on R. What is the key for R?
{E,F}  
{E,F,H}  
{E,F,H,K,L}  
{E} 
B) (EFH)^{+} = EFHGIJKLMN, EFH is deriving all the attributes of R. So, EFH is a key.
C) If EFH is a key, then EFHKL will become the super key.
D) (E)^{+} = E
Question 4 
The following functional dependencies are given:
AB → CD, AF → D, DE → F, C → G, F → E, G → A.
Which one of the following options is false?
{CF}^{+} = {ACDEFG}  
{BG}^{+} = {ABCDG}
 
{AF}^{+} = {ACDEFG}  
{AB}^{+} = {ABCDFG} 
AF → D
DE → F
C → G
F → E
G → A
CF^{+} = {G,E,A,D,C,F} = {A,C,D,E,F,G} (✔️)
BG^{+} = {B,G,A,C,D} = {A,B,C,D,G} (✔️)
AF^{+} = {D,E,A,F} = {A,D,E,F} (❌)
AB^{+} = {A,B,C,D,G} (❌)
Question 5 
Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold: {A → B, BC → D, E → C, D → A}. What are the candidate keys of R?
AE, BE  
AE, BE, DE  
AEH, BEH, BCH  
AEH, BEH, DEH 
So only option (D) contains E and H as part of candidate key.
Question 6 
From the following instance of a relation scheme R(A,B,C), we can conclude that:
A functionally determines B and B functionally determines C  
A functionally determines B and B does not functionally determines C  
B does not functionally determines C  
A does not functionally determines B and B does not functionally determines C 
But for the given instance it can be seen that B does not functionally determines C, and it can be concluded for entire relation.
Question 7 
Consider a schema R(A,B,C,D) and functional dependencies A → B and C → D. Then the decomposition of R into R_{1}(AB) and R_{2}(CD) is
dependency preserving and lossless join  
lossless join but not dependency preserving  
dependency preserving but not lossless join  
not dependency preserving and not lossless join 
R_{1}∩R_{2} ≠ 0
Given R_{1}(A,B), R_{2}
R_{1}∩R_{2} = 0
Not lossless.
The given relation decomposed into R_{1}(A,B) and R_{2}(C,D) and there are only two functional dependencies A→B and C→D. So the given decomposition is dependency preserving.
Question 8 
Given the following relation instance.
x y z 1 4 2 1 5 3 1 6 3 3 2 2
Which of the following functional dependencies are satisfied by the instance?
XY → Z and Z → Y  
YZ → X and Y → Z  
YZ → X and X → Z  
XZ → Y and Y → X 
If for t1[A] = t2[A] then t1[Y] = t2[Y].
Question 9 
Let R = (A, B, C, D, E, F) be a relation scheme with the following dependencies: C→F, E→A, EC→D, A→B. Which of the following is a key of R?
CD  
EC  
AE  
AC 
A) (CD)^{+} = cdf
Not a key.
B) (EC)^{+} = ecdabf
Yes, it is a key.
C) (AE)^{+} = aeb
Not a key. D) (AC)^{+} = abcf
Not a key.
Question 10 
Let R (A, B, C, D) be a relational schema with the following functional dependencies: A → B, B → C, C → D and D → B.
The decomposition of R into (A, B), (B, C), (B, D)
gives a lossless join, and is dependency preserving  
gives a lossless join, but is not dependency preserving  
does not give a lossless join, but is dependency preserving  
does not give a lossless join and is not dependency preserving

(A, B, C) (B, D)  common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
Question 11 
Consider a relation R with five attributes V, W, X, Y, and Z. The following functional dependencies hold: VY → W, WX → Z, and ZY → V.
Which of the following is a candidate key for R?
VXZ  
VXY  
VWXY  
VWXYZ 
Candidate keys are
VXY, WXY, ZXY
Question 12 
P⟶ QR
RS⟶ T
Which of the following functional dependencies can be inferred from the above functional dependencies?
PS ⟶ Q  
PS ⟶ T  
R ⟶ T  
P ⟶ R 
Question 13 
AB → C; BC → D; C → E;
The number of superkeys in the relation R is _____________.
8 
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
Question 14 
X is functionally dependent on Y  
X is not functionally dependent on any subset of Y  
Both (a) and (b)  
None of these 
→ An attribute is fully functional dependent on another attribute, if it is Functionally Dependent on that attribute and not on any of its proper subset.
For example, an attribute X is fully functional dependent on another attribute Y, if it is Functionally Dependent on X and not on any of the proper subset of Y.
Must satisfied conditions:
i) X is functionally dependent on Y and
ii) X is not functionally dependent on any subset of Y.
Question 15 
Reflexive, Augmentation and Decomposition  
Transitive, Augmentation and Reflexive  
Augmentation, Transitive, Reflexive and Decomposition  
Reflexive, Transitive and Decomposition 
The axioms are sound in generating only functional dependencies in the closure of a set of functional dependencies when applied to that set
Axiom of Reflexivity:
→If X is a set of attributes and Y is a subset of X, then X holds Y. Hereby, X holds Y ( X > Y) means that X functionally determines Y.
Axiom of Augmentation:
→If X holds Y and Z is a set of attributes, then XZ holds YZ. It means that attribute in dependencies does not change the basic dependencies.
Axiom of Transitivity:
The axiom of transitivity says if X holds Y, and Y holds Z, then X must also hold Z.
Question 16 
Preserves dependency but cannot perform lossless join  
Preserves dependency and performs lossless join  
Does not perform dependency and cannot perform lossless join  
Does not preserve dependency but perform lossless join 
Schema R(A, B, C, D)
Functional dependencies are A→ B and C→ D
Decomposed Schema is R1(A,B) and R2(C,D)
Dependency preservation decomposition:
It is another property of decomposed relational database schema D in which each functional dependency X → Y specified in F either appeared directly in one of the relation schemas Ri in the decomposed D or could be inferred from the dependencies that appear in some Ri.
Decomposition D={ R1 , R2, R3,,.., ,Rm} of R is said to be dependencypreserving with respect to F if the union of the projections of F on each Ri , in D is equivalent to F.
In other words, R ⊂ join of R1, R1 over X. The dependencies are preserved because each dependency in F represents a constraint on the database. If decomposition is not dependencypreserving, some dependency is lost in the decomposition.
→ R1(A,B) is covered A→ B
→ R2(C,D) is covered C→ D
It is Functional Dependency preserving because both the functional dependencies are covered.
Lossless join:
The decomposition is a losslessjoin decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
According to functional dependency R1(A,B) ∩ R2(C,D) = null. There is no common key in both the tables. So, it is not lossless join.
Question 17 
A→ (B,C)  
A → B, A→→ C  
A→ B, C→→A  
A→→B, B→ C 
→→ is multivalued dependency.
Functional Dependency:
A → B means that the values of B are determined by the values of A. Two tuples sharing the same values of A will necessarily have the same values of B.
Multivalued dependency: It is a special case of a join dependency, with only two sets of values involved, i.e. it is a binary join dependency.
A multivalued dependency exists when there are at least three attributes (like A,B and C) in a relation and for a value of A there is a well defined set of values of B and a well defined set of values of C. However, the set of values of B is independent of set C and vice versa.
Question 18 
Top right side  
Down left side  
Left hand side  
Right hand side 
This relationship is indicated by the representation below :
X ———–> Y
The left side of the above FD diagram is called the determinant, and the right side is the dependent.
Question 19 
dependency preserving  
not dependency preserving  
need not be dependency preserving  
none of these 
Question 20 
y ⊆ x  
y ⊂ x  
x ⊆ y  
x ⊂ y and y ⊂ x 
Question 21 
Which of the following functional dependencies are satisfied by the instance?
XY→ Z and Z → Y  
YZ →X and Y → Z  
YZ → X and X→ Z  
XZ→ Y and Y → X 
The lefthand side attributes determine the values of attributes on the righthand side.
Question 22 
not in 2 NF  
in 2 NF but not in 3 NF  
in 3 NF not in 2 NF  
in both 2NF and 3 NF 
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
Question 23 
CD  
EC  
AE  
AC 
1. First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
2. In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE)^{ +} =(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
Question 24 
first normal form but not in second normal form  
Second normal form but not in third normal form  
Third normal form  
None of these 
● The Second Normal Form states that it should meet all the rules for 1NF and there must be no partial dependencies of any of the columns on the primary key.
● The questions states that relation consists of automatic values but there no information about prime key.So we don’t about partial dependency.
Question 25 
referential rule  
inferential rule  
augmentation rule  
reflexive rule 
Question 26 
FD is covered by E  
E is covered by F  
F is covered by E  
F plus is covered by E 
A covers B if A ^{+} ⊆ B^{ +} (+ means closure ).
Every set of functional dependencies has a canonical cover.
Question 27 
A> B  
B< A  
AB> R  
R< AB 
Question 28 
Primary key  
Super key  
candidate key  
Primary key 
→ In the relational model of databases, a candidate key of a relation is a minimal superkey for that relation; that is, a set of attributes such that:
1.The relation does not have two distinct tuples (i.e. rows or records in common database language) with the same values for these attributes (which means that the set of attributes is a superkey)
2.There is no proper subset of these attributes for which (1) holds (which means that the set is minimal).
→ Candidate keys are also variously referred to as primary keys, secondary keys or alternate keys. The constituent attributes are called prime attributes. Conversely, an attribute that does not occur in ANY candidate key is called a nonprime attribute.
→ Since a relation contains no duplicate tuples, the set of all its attributes is a superkey if NULL values are not used. It follows that every relation will have at least one candidate key. The candidate keys of a relation tell us all the possible ways we can identify its tuples. As such they are an important concept for the design of database schema.
Question 29 
CD  
EC  
AE  
AC 
First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
Question 30 
AC→ H  
C→ H  
G→ H  
A→ H 
Question 31 
Which of the following functional dependencies hold?
{AB} → D and D → A  
{AB → D and B → D  
{AB} → C and B → C  
{AB} → D and A → D 
→ If LHS of a functional dependency is repeating then see the value of RHS. If values of RHS are same for repeated value of LHS then you can say that LHS of a functional dependency is determining RHS of a functional dependency else not determining.
Now if you see the given options then only option(C) is satisfying above two rules.
Question 32 
If R(A,B,C,D) is a relation schema, which is decomposed into R1(A,B,C) and R2(C,D), which of the following ensures that the given decomposition is non additive(or lossless)?
R1 → R2
 
R2 → R1
 
R1 ∩ R2 → R1 or R1 ∩ R2 → R2
 
R1 U R2 → R1 R1 U R2 → R2

For lossless decomposition:
R_{1} ∩ R_{2} → R_{1} (OR) R_{1} ∩ R_{2} → R_{1}  R_{2}
R_{1} ∩ R_{2} → R_{2} (OR) R_{1} ∩ R_{2} → R_{2}  R_{1}
Question 33 
a and b  
Only a  
Only b  
Neither a nor b 
→If R is a relation with attributes X and Y, a functional dependency between the attributes is represented as X→Y, which specifies Y is functionally dependent on X. Here X is a determinant set and Y is a dependent attribute. Each value of X is associated with precisely one Y value.
Question 34 
Dependency preserving but not lossless join  
Dependency preserving and lossless join  
Lossless join but not dependency preserving  
Neither dependency preserving nor lossless join. 
and R _{2} . Given decomposition is lossless only if
1. R _{1} U R_{ 2} =R
2. R _{1} U R_{ 2} =φ
3. R _{1} ∩ R_{ 2} → R_{ 1} (or) R_{ 1} ∩ R_{ 2} → R_{ 2}
→ In this schema, there is no common key attribute between R _{1} and R _{2} . So, this relation is lossy relation.
Definition of Functional dependency preserving:
→ Let R be the relational schema with functional dependency set F is decomposed into R _{1} , R _{2} ,..,R _{n} with functional dependency sets F _{1} ,F_{ 2} ,...,F_{ n} respectively. In general F _{1} ,F_{ 2} ,...,F_{ n} can be ⊆ F.
So, dependencies are preserved in the given decomposition.
Question 35 
A  
AB  
ABC  
ABD 
Now check whether ABD is itself a candidate key or not by finding the closure.
(ABD) ^{+} = { ABDCEFG }
since ABD can identify each attribute of the given relation uniquely.
So, ABD is the candidate key.
Question 36 
Which of the following is candidate key?
A  
AC  
AD  
None of these 
Step1: Check whether AD could be a candidate key or not.
(AD) + = {A,D,C,B,F,E}
Step2: AD can uniquely identifies each attribute. So, AD is the candidate key for given relation.
Question 37 
A  
AE  
BE  
CE 
Step1: Attribute A and E are not present in the right hand side of any production. So, every key should include AE to be a candidate key.
Step2: To check whether AE could be a candidate key or not.
(AE) + = {A,B,C,D,E}
Since AE can uniquely identify each attribute and we can say this is the candidate key for given relation.
Question 38 
A➝B
B,C➝D
E➝C
D➝A
What are the candidate keys of R ?
AEF, BEF and DEF  
AEF, BEF and BCF  
AE and BE  
AE, BE and DE 
EFB ^{+} = {EFABCD}
EFC ^{+} = {EFC}
EFD ^{+} = {EFDCAB}
So EFA, EFB, EFD are the keys for the given relation R=(A, B, C, D, E, F).
Question 39 
CD  
EC  
AE  
AC 
→First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
→In right hand side, CE are missing. It means definitely EC should be there in candidate key.
(EC)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
Question 40 
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 
Question 41 
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 sert(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 42 
flight number  
flight number + place of departure  
flight number + date  
flight number + destination 
Question 43 
appeared directly in one of the relation schemas R_{i} in the decomposed D.  
could be inferred from dependencies that appear in some R_{i}.  
both (A) and (B)  
None of these 
Appeared directly in one of the relation schemas R_{i} in the decomposed D.
Could be inferred from dependencies that appear in some R_{i}.
Question 44 
1 – d, 2 – b, 3 – a, 4 – c  
2 – d, 3 – a, 1 – b, 4 – c  
4 – a, 3 – b, 2 – c, 1 – d  
3 – a, 4 – b, 1 – c, 2 – d 
Candidate key→ Uniquely identified a row. Candidate key is minimal super key.
Nonredundancy→ No attribute can be added
Functional dependency→ A constraint between two attribute
Question 45 
3 NF  
2 NF  
4 NF  
BCNF 
(1) be in first normal form (1NF)
(2) 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 46 
R_{1} ∩ R_{2} → R_{1} is in F^{+}  
R_{1} ∩ R_{2} → R_{2} is in F^{+}  
both R_{1} ∩ R_{2} → R_{1} and R_{1} ∩ R_{2} → R_{2} functional dependencies are in F^{+}  
at least one from R_{1} ∩ R_{2} → R_{1} and R_{1} ∩ R_{2} → R_{2} is in F^{+} 
R_{1} ∩ R_{2 }→ R_{1} and R_{1} ∩ R_{2} → R_{2} is in F+
Lossless join:
The decomposition is a losslessjoin decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R_{1} ∩ R_{2 }→ R_{1}
R_{1} ∩ R_{2} → R_{2}
Question 47 
F={A →B, AD→C, B→F, A→ E}.
Which of the following is Candidate Key ?
A
 
AC  
AD  
None of these 
Question 48 
A → B
B → C
C → D
D → B
Then decomposition of R into R1(A, B), R2(B, C) and R3(B, D) is __________ .
Dependency preserving and lossless join.  
Lossless join but not dependency preserving.  
Dependency preserving but not lossless join.  
Not dependency preserving and not lossless join. 
(A, B, C) (B, D)  common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
Question 49 
D_{1} : The decomposition of the schema R(A, B, C) into R_{1}(A, B) and R_{2} (A, C) is always lossless.
D_{2} : The decomposition of the schema R(A, B, C, D, E) having AD → B, C → DE, B → AE and AE → C, into R_{1} (A, B, D) and R_{2} (A, C, D, E) is lossless.
Both D1 and D2  
Neither D1 nor D2  
Only D1  
Only D2 
Question 50 
I. The decomposition of R into R1(C, A) and R2(A, B) is lossless.
II. The decomposition of R into R1(A, B) and R2(B, C) is lossy.
Only I  
Only II  
Both I and II  
Neither I nor II 
Here ‘B’ is common attribute but ‘B’ is not the primary key in any of the decomposed relations (R1 or R2). Hence this decomposition is not a lossless decomposition i.e., it is a lossy decomposition.
Hence, option (C) is correct.
Question 51 
R _{1} ∩ R_{ 2} ↠ (R_{ 1}  R_{ 2} )
R _{1} → R_{ 2}
R _{1} ∩ R_{ 2} ↠ (R_{ 2}  R_{ 1} )
(R _{2} → R_{ 1} ) ∩ R_{ 2}
(a) and (b) happens  
(a) and (d) happens  
(a) and (c) happens  
(b) and (c) happens 
Question 52 
Comprehension:
Answer question (96100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
:→ Minimal cover F’ of functional dependency set F is
F’ = {A → B, A → C, BC → D, D →E}  
F’ = {A → BC, B → D, D → E}  
F’ = {A → B, A → C, A → D, D → E}  
F’ = {A → B, A → C, B → D, C → D.D → E} 
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure. Above BC are two attributes in LHS.
B^{+} = {B}
C^{+} = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D
Question 53 
Comprehension:
Answer question (96100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
→ Assume that given table R is decomposed in two tables
Which of the following option is true w.r.t. given decomposition?
Dependency preservation property is followed  
R_{1} and R_{2} are both in 2 NF  
R_{2} is in 2 NF and R_{3} is in 3 NF  
R_{1} is in 3 NF and R_{2} is in 2 NF 
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.
Question 54 
Comprehension:
Answer question (96100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
→ Identify the redundant functional dependency in F
BC→D  
D→E  
A→D  
A→BC 
Question 55 
Comprehension:
Answer question (96100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
→ Identify primary key of table R with functional dependency set F
BC  
AD  
A  
AB 
A^{+} = { A,B,C,D,E}. Hence A is the primary key of relation R
Question 56 
Comprehension:
Answer question (96100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
→ Identify the normal form in which relation R belong to
1 NF  
2 NF  
3 NF  
BCNF 
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct
Question 57 
((R1∩ R2) →(R1R2)) ∈ F^{+}  
((R1 ∩R2) → (R1  R2)) ∉ F^{+}  
((R1 ∪ R2) →(R1  R2)) ∈ F^{+}  
((R1∩R2) →R1 ) ∈ F^{+} 
Question 58 
971658  
971659  
860547  
860548 
→Subtract each and every digit of given number by digit “9”.
999999
139452
________
860547
Question 59 
X  Y  Z 
X1  Y1  Z1 
X1  Y1  Z2 
X2  Y1  Z1 
X2  Y1  Z3 
R: Z → Y, X → Y, and X → Z  
R: Z →Y, X → Y, and XZ → Y  
R: X →Y, X → Z, and XZ → Y  
R: Z →Y, X → Z, and XZ → Y 
Option 3 is wrong because of X → Z.
Option 4 is wrong because of X → Z
Question 60 
Let the functional dependencies, (A → B, B → C) hold for relation P(A,B,C). If it is decomposed in R(A,B) and S(B,C), then the decomposition will be
Lossy
 
Lossless
 
NonLossless  
Strongly

Question 61 
Dependency preserving and lossless Join  
Lossless Joint but not dependency preserving  
Dependency preserving but not lossless Join  
Not dependency preserving and not lossless Join 
Question 62 
Key dependencies  
Relation dependencies
 
Database dependencies  
None of these 
Question 63 
C → F,E → A, EC → D, A → BWhich of the following is a key for R?
CD  
EC  
AE  
AC 
(CD)^{+} = CDF X
(EC)^{+} = ECFDAB ✓
(AE)^{+} = AEB X
(AC)^{+} = ACBF X
Question 64 
Tables  
Attributes  
Rows  
Relations 
Question 65 
AB → CD, AF → D, DE → F, C → G, F → E, G → AWhich one of the following options is false ?
BG^{+}={ABCDG}  
CF^{+}={ACDEFG}  
AB^{+}={ABCDG}  
AF^{+}={ACDEFG} 
(CF)^{+} = CFGEAD
(AB)^{+} = ABCDG
(AF)^{+} = AFDE
Hence option (D) is False.
Question 66 
ABCD,ɸ  
ABCD, E  
ABC, E 
E^{+} = {E}
Question 67 
dependency preserving and lossless join
 
lossless join but not dependency preserving  
dependency preserving but not lossless join  
neither dependency·preserving nor lossless join 
R_{2}(C,D) contains FD C→D
So, yes dependency preserving.
But there is no common attribute between R_{1} and R_{2}, hence not lossless join.
Question 68 
From the following instance of a relation scheme R(X,Y,Z), we can conclude that:
X Y Z
1 1 1
1 1 0
2 3 2
2 3 2
X functionally determines Y and Y functionally determines Z  
Y does not functionally determine Z  
X does not functionally determine Y and Y does not functionally determine Z  
None of the above 
But for the given instance it can be seen that Y does not functionally determines Z, and it can be concluded for the entire relation
Question 69 
Dependency preserving but not lossless join  
Dependency preserving and lossless join  
Lossless join but not dependency preserving  
Neither lossless join nor dependency preserving 
The relation R1 covers FD A → B and relation R2 covers FD C → D.Hence the decomposition is functional dependency preserving.
Question 70 
lossless join but not dependency preserving  
dependency preserving but not lossless join  
not dependency preserving and not lossless join  
part dependency preserving and lossless join 
Question 71 
Super Key  
Candidate Key  
Primary Key  
Candidate and Primary Key  
B and D 
Question 72 
A  
B  
E  
BC 
Closure(A)+ = { A, B, C, D, E }
Closure(B)+ = { B, D }
Closure(E)+ = { A, B, C, D, E }
Closure(BC)+ = { A, B, C, D, E }
Question 73 
Foreign key  
Primary key  
Assertion  
Referential constraint 