✅ SQL Practice Questions with Answers: Part-3 🧠🗃️

🔍 Q13. Find employees whose salary is above the company average
🧾 Table: employees
id | name | salary
1 | Arjun | 40000
2 | Riya | 55000
3 | Meena | 62000
4 | Kabir | 45000

📄 Query:

SELECT *  
FROM employees  
WHERE salary > (SELECT AVG(salary) FROM employees);

📌 Result: Riya, Meena

🔍 Q14. Get the 3 most recent joined employees
🗓 Table: employees
id | name | join_date
1 | Arjun | 2021-03-10
2 | Riya | 2022-05-12
3 | Meena | 2023-02-01
4 | Kabir | 2023-11-09

📄 Query:

SELECT *  
FROM employees  
ORDER BY join_date DESC  
LIMIT 3;

📌 Result: Kabir, Meena, Riya

🔍 Q15. Retrieve employees who don’t have a manager assigned
🧾 Table: employees
id | name | manager_id
1 | Arjun | NULL
2 | Riya | 1
3 | Meena | NULL
4 | Kabir | 2

📄 Query:

SELECT id, name  
FROM employees  
WHERE manager_id IS NULL;

📌 Result: Arjun, Meena

🔍 Q16. Find departments where more than 2 employees work
🧾 Table: employees
id | name | department
1 | Arjun | HR
2 | Riya | IT
3 | Meena | IT
4 | Kabir | HR
5 | John | IT

📄 Query:

SELECT department, COUNT(*) AS total  
FROM employees  
GROUP BY department  
HAVING COUNT(*) > 2;

📌 Result: IT → 3 employees

🔍 Q17. Select employees whose salary equals department average
🧾 Table: employees
id | name | dept | salary
1 | Arjun | HR | 40000
2 | Riya | IT | 55000
3 | Meena | IT | 62000
4 | Kabir | HR | 45000

📄 Query:

SELECT e.*  
FROM employees e  
JOIN (  
  SELECT dept, AVG(salary) AS avg_salary  
  FROM employees  
  GROUP BY dept  
) t ON e.dept = t.dept  
WHERE e.salary = t.avg_salary;

📌 Result: None (but logic works if matches exist)

🔍 Q18. Get employees who have at least one matching project
📂 Table: project_assignments
employee_id | project_id
1 | 101
2 | 101
2 | 102
3 | 103
4 | 101

📄 Query:

SELECT p1.employee_id, p2.employee_id AS colleague  
FROM project_assignments p1  
JOIN project_assignments p2  
ON p1.project_id = p2.project_id  
AND p1.employee_id <> p2.employee_id;

📌 Result:
Employees 1, 2, 4 share project 101

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