Consider the following relational schema: Students ( rollno: integer, name: string, age: integer, cgpa: real) Courses ( courseno: integer, cname: string, credits: integer) Enrolled ( rollno: integer, courseno: integer, grade: string) Which of the following options is/are correct SQL query/queries to retrieve the names of the students enrolled in course number (i.e., courseno) 1470? A) SELECT S.name FROM Students S WHERE EXISTS (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno); B) SELECT S.name FROM Students S WHERE SIZEOF (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno) > 0; C) SELECT S.name FROM Students S WHERE 0 D) SELECT S.name FROM Students S NATURAL JOIN Enrolled E WHERE E.courseno = 1470;

Official answer: A) SELECT S.name FROM Students S WHERE EXISTS (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno);; C) SELECT S.name FROM Students S WHERE 0; D) SELECT S.name FROM Students S NATURAL JOIN Enrolled E WHERE E.courseno = 1470;. Explanation: To retrieve the names of students enrolled in `courseno = 1470`: 1. Option (A) uses a correlated subquery with the `EXISTS` predicate. For each student tuple in `Students`, it checks whether there exists at least one record in `Enrolled` with the matching `rollno` and `courseno = 1470`. If true, `S.name` is selected. This is valid and correct. 2. Option (B) uses `SIZEOF`, which is not a standard SQL keyword or aggregate function for computing the cardinality of a subquery result. Thus, this query is syntactically invalid. 3. Option (C) uses a scalar subquery `SELECT COUNT(*) FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno`. The condition `0 < (SELECT COUNT(*) ...)` correctly evaluates to true whenever the student has at least one enrollment record in course 1470. This is valid and correct. 4. Option (D) performs a `NATURAL JOIN` between `Students` and `Enrolled`. Since the only common attribute with the same name and type between the two tables is `rollno`, the join automatically equates `Students.rollno = Enrolled.rollno`. The `WHERE` clause then filters for `E.courseno = 1470` and projects `S.name`. This is valid and correct. Therefore, the correct options are A, C, and D.

B) SELECT S.name FROM Students S WHERE SIZEOF (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno) > 0;

SQL

GATE CSE 2025 Set 2 · Q54

MSQ2 marksMedium

Time00:00

Consider the following relational schema: Students ( $\underline{\text{rollno: integer}}$ , name: string, age: integer, cgpa: real) Courses ( $\underline{\text{courseno: integer}}$ , cname: string, credits: integer) Enrolled ( $\underline{\text{rollno: integer}}$ , $\underline{\text{courseno: integer}}$ , grade: string) Which of the following options is/are correct SQL query/queries to retrieve the names of the students enrolled in course number (i.e., courseno) 1470?

Select all that applyOne or more correct

Pick every correct option

Solution

Step-by-step verified solution

Official answer: A) SELECT S.name FROM Students S WHERE EXISTS (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno);; C) SELECT S.name FROM Students S WHERE 0; D) SELECT S.name FROM Students S NATURAL JOIN Enrolled E WHERE E.courseno = 1470;

SQL

To retrieve the names of students enrolled in courseno = 1470:

Option (A) uses a correlated subquery with the EXISTS predicate. For each student tuple in Students, it checks whether there exists at least one record in Enrolled with the matching rollno and courseno = 1470. If true, S.name is selected. This is valid and correct.
Option (B) uses SIZEOF, which is not a standard SQL keyword or aggregate function for computing the cardinality of a subquery result. Thus, this query is syntactically invalid.
Option (C) uses a scalar subquery SELECT COUNT(*) FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno. The condition 0 < (SELECT COUNT(*) ...) correctly evaluates to true whenever the student has at least one enrollment record in course 1470. This is valid and correct.
Option (D) performs a NATURAL JOIN between Students and Enrolled. Since the only common attribute with the same name and type between the two tables is rollno, the join automatically equates Students.rollno = Enrolled.rollno. The WHERE clause then filters for E.courseno = 1470 and projects S.name. This is valid and correct.

Therefore, the correct options are A, C, and D.

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This SQL previous year question is Q54 of the GATE CSE 2025 Set 2 paper. Find more GATE CSE PYQs on SQL in Databases, or browse all GATE CSE PYQs.

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SQL has 17 GATE CSE PYQs across 14 papers from 2011 to 2025. This is the only 2-mark MSQ among them, and it is the only one in GATE CSE 2025 Set 2.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.