A table ‘student’ with schema (roll, name, hostel, marks), and another table ‘hobby’ with schema (roll, hobbyname) contains records as shown below:
Roll | Name | Hostel | Marks |
---|---|---|---|
1798 | Manoj Rathod | 7 | 95 |
2154 | Soumic Banerjee | 5 | 68 |
2369 | Gumma Reddy | 7 | 86 |
2581 | Pradeep Pendse | 6 | 92 |
2643 | Suhas Kulkarni | 5 | 78 |
2711 | Nitin Kadam | 8 | 72 |
2872 | Kiran Vora | 5 | 92 |
2926 | Manoj Kunkalikar | 5 | 94 |
2959 | Hemant Karkhanis | 7 | 88 |
3125 | Rajesh Doshi | 5 | 82 |
Roll | Hobbyname |
---|---|
1798 | chess |
1798 | music |
2154 | music |
2369 | swimming |
2581 | cricket |
2643 | chess |
2643 | hockey |
2711 | volleyball |
2872 | football |
2926 | cricket |
2959 | photography |
3125 | music |
3125 | chess |
The following SQL query is executed on the above tables:
select hostel from student natural join hobby where marks > = 75 and roll between 2000 and 3000;
Relations S and H with the same schema as those of these two tables respectively contain the same information as tuples. A new relation S’ is obtained by the following relational algebra operation:
S’ = ∏hostel ((σs.roll = H.roll (σmarks > 75 and roll > 2000 and roll < 3000 (S)) X (H))
The difference between the number of rows output by the SQL statement and the number of tuples in S’ is
(A) 6
(B) 4
(C) 2
(D) 0
Answer: (B)
Explanation: Output of above Query
Roll | Hostel |
---|---|
2369 | 7 |
2581 | 6 |
2643 | 5 |
2643 | 5 |
2872 | 5 |
2926 | 5 |
2959 | 7 |
Total rows selected by running SQL Query: 7
Total rows by Relation Algebra : 4 i.e 5,6,7 (Unique values only)
7-3=4
Answer is B
Quiz of this Question
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