ERModel
Question 1 
Ovals that contain underlined identifiers
 
Rectangles with double/bold border
 
Diamonds with double/bold border
 
Ovals with double/bold border

Question 2 
In an EntityRelationship (ER) model, suppose R is a manytoone relationship from entity set E1 to entity set E2. Assume that E1 and E2 participate totally in R and that the cardinality of E1 is greater than the cardinality of E2.
Which one of the following is true about R?
Every entity in E1 is associated with exactly one entity in E2.
 
Some entity in E1 is associated with more than one entity in E2.
 
Every entity in E2 is associated with exactly one entity in E1.
 
Every entity in E2 is associated with at most one entity in E1.

The M : 1 relationship holds between two entities E1 and E2, in which each tuple from E2 is in relation with many tuples of E1. One tuple from E1 is in relation with only one tuple of E2. It is given that participation from both the sides is total and the cardinality of E1 is greater than E2.
Therefore, every entity E1 is associated with exactly one entity in E2.
Question 3 
An ER model of a database consists of entity types A and B. These are connected by a relationship R which does not have its own attribute. Under which one of the following conditions, can the relational table for R be merged with that of A?
Relationship R is onetomany and the participation of A in R is total.  
Relationship R is onetomany and the participation of A in R is partial.  
Relationship R is manytoone and the participation of A in R is total.  
Relationship R is onetomany and the participation of A in R is partial. 
The relational table for R be merged that of A, if the relationship R is Manytoone and the participation of A in R is total.
Question 4 
4  
6  
7  
8 
Question 5 
π_{A1} (σ_{(F1∧F2)} (r))  
π_{A1} (σ_{(F1∨F2)} (r))  
π_{A2} (σ_{(F1∧F2)} (r))  
π_{A2} (σ_{(F1∨F2)} (r)) 
Two Selects with Boolean expression can be combined into one select with AND of two Boolean expressions.
Question 6 
in all candidate keys of R.  
in some candidate key of R.  
in a foreign key of R.  
only in the primary key of R . 
Ex: AB, BC, CD are candidate keys of R(ABCD). In the FDs set one attribute may not be part of all the FDs.
Question 7 
some dependent.  
all dependents.  
some of his/her dependents.  
all of his/her dependents. 
Question 8 
I, II, III and IV  
I, II and III only  
I, II and IV only  
II, III and IV only 
Question 9 
An attribute of an entity can have more than one value  
An attribute of an entity can be composite  
In a row of a relational table, an attribute can have more than one value  
In a row of a relational table, an attribute can have exactly one value or a NULL value 
Option (B): In ER model, the attribute which can be further broken down into some other attributes is called composite attribute.
Option (C): In Relational model, the intersection of one row and column should contain only one value. So, option (C) is INCORRECT.
Option (D): In Relational model, the intersection of one row and column should contain either exactly one value or NULL.
Question 10 
2  
3  
4  
5 
➝ Here N is a Weak entity, but it need to modify the primary key of P such as P1
M = {M1, M2, M3, P1}
P = {P1, P2}
N = {N1, N2, P1}
➝ Relationship set has its own attribute, then no need to create a separate table.
➝ Finally we require 3 minimum tables to represent M,P,N,R1,R2.
Question 11 
{M1, M2, M3, P1}  
{M1, P1, N1, N2}
 
{M1, P1, N1}  
{M1, P1}

For M = {M1, M2, M3, P1}
P = {P1, P2}
N = {N1, N2, P1}
Question 12 
2  
3  
4  
5 
R_{1} is one to many.
R_{2} is many to many.
→ E_{1} and E_{2} have separate table because they need to store multiple values.
→ R_{2} also have separate table by considering Primary keys E_{1} and E_{2} as foreign keys.
→ R_{1} is converted to many side table i.e., E_{2} as Primary key and E_{1} as Foreign key.
So, totally we need 3 tables to store the value.
Question 13 
Person  
Hotel Room  
Lodging  
None of these 
Question 14 
CD → AC  
BD → CD  
BC → CD  
AC → BC 
Option (B):
BD → CD
BD^{+} = BD
i.e., BD cannot derive CD and hence is not implied.
Question 15 
Consider the following entity relationship diagram (ERD), where two entities E1 and E2 have a relation R of cardinality 1 : m.
The attributes of E1 are A11, A12 and A13 where A11 is the key attribute. The attributes of E2 are A21, A22 and A23 where A21 is the key attribute and A23 is a multivalued attribute. Relation R does not have any attribute. A relational database containing minimum number of tables with each table satisfying the requirements of the third normal form (3NF) is designed from the above ERD. The number of tables in the database is
2  
3  
5  
4 
Then, we get
T1: {A11, A12, A13}  key is A11
T2: {A21, A22, A11}  key is A21
T3: {A21, A23}  key is {A21, A23}
Question 16 
ROLL_NO  NAME  DEPT_ID 

1  ABC  1 
2  DEF  1 
3  GHI  2 
4  JKL  3 
DEPT_ID  DEPT_NAME 

1  A 
2  B 
3  C 
 update Student set Dept_id = Null where Roll_on = 1
 update Department set Dept_id = Null where Dept_id = 1
Both (i) and (ii) will fail  
(i) will fail but (ii) will succeed  
(i) will succeed but (ii) will fail  
Both (i) and (ii) will succeed 
But in (ii) if we set in Department table, Dept_id = Null, then it will produce inconsistency because in Student table we will still have the tuples containing the Dept_id = 1.