Classical

Database MCQ - Database design

11:  

If every non-key attribute is functionally depedent primary key, then the relation will be in

A.

First normal form

B.

Second normal form

C.

Third form

D.

Fourth normal form

 
 

Option: B

Explanation :

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Amit said: (11:15pm on Friday 6th September 2013)
in 2nf partial dependency should not exist den how it can be 2nf
Atul said: (7:56am on Thursday 12th January 2017)
I think it should be A. First normal form , i.e for 2nf there would be FULLY functionally dependent..

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12:  

 If a relation is in BCNF, then it is also in

A.

1 NF

B.

2 NF

C.

3 NF

D.

All of the above

 
 

Option: D

Explanation :

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13:  

Consider the relation given below and ind the maximum normal form applicable to them

1. R(A, B) with productions { A --> B }

2. R(A, B) with productions  { B --> A }

3 R(A, B) with productions {A —> B, B --> A }

4  R(A, B, C) with productions {A -->B, B --> A, AB --> C }

A.

1, 2 and 3 are in 3NF and 4 is in BCNF

B.

1 and 2 are in BCNF and 3 and 4 are in 3NF

C.

All are in 3NF

D.

All are in BCNF

 
 

Option: D

Explanation :

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Naeem Ahmad said: (10:45am on Wednesday 17th June 2015)
Answer should be A because 4 is not in BCNF.In 4, FDs A-->B and B-->A do not hold property of neither trivial nor super key.A and B are not super key since AB is primary key.FDs are nontrivial FDs.
naresh said: (11:14pm on Friday 4th September 2015)
if single relation is there then it will be in bcnf so a and b are in bcnf but c and d are in 3nf becoz determinant is not always key or super key. so answer should be b.
Prashant said: (12:57pm on Friday 1st September 2017)
4 will be in bcnf because the closure of a,b and ab will be {a,b,c} and all will become a key.

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14:  

Consider the following

S1: Foreign key allows null values.

S2: Every binary tables is in BCNF

Which of the following is true 

A.

Both S1 and S2 are true

B.

S1 is true

C.

S2 is true

D.

None of these

 
 

Option: B

Explanation :

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15:  

Consider the following functional dependencies in a database:

Date_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

A.

in second normal form but not in third normal form

B.

in third normal form but not in BCNF

C.

in BCNF

D.

in none of the above

 
 

Option: D

Explanation :

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