GATE CS 2025 Set 2 — Question 54

MSQ+2 / -0MediumSQL JoinsSQLDatabasesSubqueriesAggregate Functions

Databases → SQL → Aggregate Functions

Last updated 5 September 2026

Question

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?

Your answer

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SELECT S.name FROM Students S WHERE EXISTS (SELECT * FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno);

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

SELECT S.name FROM Students S WHERE 0 < (SELECT COUNT(*) FROM Enrolled E WHERE E.courseno = 1470 AND E.rollno = S.rollno);

SELECT S.name FROM Students S NATURAL JOIN Enrolled E WHERE E.courseno = 1470;

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Correct answer (A) [code]; (C) [code]; (D) [code] Solution The goal is to find the names of students enrolled in course 1470. 1. Option (A) uses a correlated subquery with EXISTS . It selects a student if there exists a record in the Enrolled table with the same rollno and courseno = 1470 . This is a standard and correct SQL approach. 2. Option (B) uses SIZEOF, which is not a standard SQL operator/function. The correct function to count rows is COUNT . Therefore, this option is syntactically incorrect. 3. Option (C) uses a subquery that counts the number of matching enrollment records. If COUNT(*) is greater than 0, it means the student is enrolled. This is logically equivalent to EXISTS and is a correct SQL query. 4. Option (D) uses NATURAL JOIN . A natural join automatically joins tables based on columns with the same name and data type. In the given schema, Students and Enrolled share only the rollno attribute. Thus, NATURAL JOIN performs an inner join on rollno, correctly linking students to their enrollments. The WHERE clause filters for course 1470. This is a correct query. Therefore, options (A), (C), and (D) are correct. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (A) [code]; (C) [code]; (D) [code] Solution The goal is to find the names of students enrolled in course 1470. 1. Option (A) uses a correlated subquery with EXISTS . It selects a student if there exists a record in the Enrolled table with the same rollno and courseno = 1470 . This is a standard and correct SQL approach. 2. Option (B) uses SIZEOF, which is not a standard SQL operator/function. The correct function to count rows is COUNT . Therefore, this option is syntactically incorrect. 3. Option (C) uses a subquery that counts the number of matching enrollment records. If COUNT(*) is greater than 0, it means the student is enrolled. This is logically equivalent to EXISTS and is a correct SQL query. 4. Option (D) uses NATURAL JOIN . A natural join automatically joins tables based on columns with the same name and data type. In the given schema, Students and Enrolled share only the rollno attribute. Thus, NATURAL JOIN performs an inner join on rollno, correctly linking students to their enrollments. The WHERE clause filters for course 1470. This is a correct query. Therefore, options (A), (C), and (D) are correct.

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