Log In
2 votes

in Databases 164 views
R is not the primary key in relation r2 therefore redundant data will result maximum selection of tuples.
b is correct answer.

1 Answer

1 vote
Best answer
Given R1(P, Q, R) and R2(R ,U, V) where R is primary key in R1 and U is primary key in R2.
R in R1 is primary key so it should contain unique values assume 1,2,3,......1500(1500tuples)
There is no constraint on R in R2 so it can contain duplicate values assume 1,1,1,...1(2000 tuples)
Natural join is performed on the Common attribute i.e R
1 in R( R1 table) can map with 2000 1's in R( R2 table).
Hence maximum 2000 tuples are possible in natural join

selected by

when 1500 wud be the answer then???

plz check this too. . .

Now I got it clearly....  Thank u so much

Related questions

–1 vote
1 answer
83 views asked Sep 26, 2018 in Databases aoao 83 views
0 votes
0 answers
Consider the relation schema: Student(roll no, name course no) Enroll(roll no, course no,course name) The number of tuples in the student and enroll table is 30 and 40 respectively The natural join is performed on roll no., what is max. and min. no. of tuples that are possible?
asked Dec 16, 2017 in Databases srestha 379 views
2 votes
2 answers
Say we have two relations R (a,b,c) and S (b,d,e). Now, R has 200 tuples and S has 300 tuples. What will be Minimum number of tuples when we do R ⋈ S ( ⋈ = Natural Join)?
asked Dec 6, 2017 in Databases iarnav 388 views
2 votes
1 answer
R(A, B) foreign key B refers S. S(A, B) foreign key A refers R. (primary keys are bold) None of the attributes is nullable. R contains 75 tuples and S contains 25 tuples. What is the least and most number of tuples that R⋈S contains? a) 0, 25 b) 0, 75 c) 25, 75 d) 1, 25
asked Nov 9, 2017 in Databases techbrk3 420 views