Register

Let a, b and c be the number of entities in entity sets A, B, and C, respectively

2,466 views
0 votes
0 votes

Let a, b and c be the number of entities in entity sets A, B, and C, respectively. Let t be the maximum number of triples in the relationship set for R. Which of the following is a possible combination of values for a, b, c, and t
image:MST2/1Qa.png

  1.   a =100, b=10, c=1000, t=1000
  2.   a =1000, b=1, c=200, t=1200
  3.   a =5, b=3, c=4, t=20
  4.   a =100, b=10, c=20, t=100

--could'nt understand the question.what are triples??

User Avatar

1 Answer

Best answer
0 votes
0 votes

Relation between A and B = R(A,B) = Many to One.

Relation between A and C = R(A,C) = Many to One.

Relation between B and C = R(B,C) = One to One.

Foreign Key of B and C will be added in A to mantain A's Relationship with B and C .

So now , A will be the Deciding Factor. 

And Total Tuples in the Relation R will be total Tuples Carried by A.

So, here you can easily decide the Correct one as , Option D contains a=100=t. And that is TRUE.

Hope This will Clear Your Doubt !!!

• selected by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

6 votes
6 votes
5 answers
2
learncp asked Sep 5, 2015
5,242 views
GATE2004_13 Let R1 (A, B, C) and R2 (D, E) be two relation schema, where the primary key...
----SPOILERS------ My doubt is that why option "c" shouldnt be an answer, as join is performed on the C not equal to D, and there will be no case where C is not equal to d as it satisfies referential integrity constraints. . ?
SPOILERS My doubt is that why option "c" shouldnt be an answer, as join is performed on the C not equal to D, and there will be no case where C is not equal to d as i...
User Avatar
Link Copied!