Gate cse Ace test series

Question

Answer 1

8 tuples are returned.

R=A $\cup$ B

A.ID	A.Name	A.Age
5	Bala	19
10	Akash	21
15	Raj	18
20	Babbar	16
25	Monty	20
3	Rohan	12
8	Rahul	22

 

R $\times$ C

A.Id	A.Name	A.Age	C.ID	C.Name	C.Age
5	Bala	19	9	Manish	19
5	Bala	19	18	Billu	22
10	Akash	21	9	Manish	19
10	Akash	21	18	Billu	22
...	...
...	...
25	Monty	20	9	Manish	19
25	Monty	20	18	Billu	22

8 Tuples returned after filtering.

Comments

I cant seem to understand  it.

Please correct me wherever Im taking a wrong turn

We are taking Join of A U B with C

Taking Cartesian should give 8*2 = 16 rows right?

Then filtering

For all the rows where C.ID = 9 (8 tuples)
and all the rows where A.ID = 25 (2 tuples)
We have 10 rows. One of this is common to both conditions i.e A=25, C=9
So effectively we should get 9 tuples no?

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8 tuples are returned.

I wrote as final answer. Didn’t mean for A $\cup$ B

You guys thought as for union…

A $\cup$ B returning 7 tuples, After crossing with 2 tuples of C returning 14 tuples. Now condition applicable on 8 tuples. That’s why final answer as 8.

Hope it’s clear now...

@dheerajgudipati @Mrityudoot

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@minimalist yep man all good now. Thanks

Answer 2

A U B has 7 tuples (ID = 3,5,8,10,15,20,25) 
It is joined to C on the condition : EITHER A.id > 20 in A U B (Which is A = 25) OR C.id < 10 in C (Which is C =9)
2 tuples are returned. 

 

ID	Name	Age
25	Monty	20
9	Manish	19

 

Comments

If the equality condition had been then 3(common columns) or 6(All columns) could have been based on the convention followed (author dependent).

But other than equality conditions irrespective of column name(same or different) all columns of join tables have to be included.