diff --git a/Practical/Practical Exam/SQL/S6 - SQL Queries.md b/Practical/Practical Exam/SQL/S6 - SQL Queries.md
new file mode 100644
index 0000000..0653789
--- /dev/null
+++ b/Practical/Practical Exam/SQL/S6 - SQL Queries.md	
@@ -0,0 +1,149 @@
+# S6 - SQL Queries (in MySQL)
+
+**Problem Statement:**
+Consider following Relation
+Employee(emp_id,employee_name,street,city)
+Works(employee_name,company_name,salary)
+Company(company_name,city)
+Manages(employee_name,manager_name)
+Create above tables with appropriate constraints like primary key,
+foreign key, not null etc.
+1. Change the city of employee working with InfoSys to ‘Bangalore’
+2. Find the names of all employees who earn more than the average
+salary of all employees of their company. Assume that all people
+work for at most one company.
+3. Find the names, street address, and cities of residence for all
+employees who work for 'TechM' and earn more than $10,000.
+4. Change name of table Manages to Management.
+5. Create Simple and Unique index on employee table.
+6. Display index Information
+
+---
+
+## Creating the database
+```sql
+CREATE DATABASE Companies2;
+USE Companies2;
+
+```
+
+## Creating tables:
+
+```sql
+CREATE TABLE Employee (
+  emp_id INT UNIQUE NOT NULL, -- can be set to auto increment using AUTO_INCREMENT
+  employee_name VARCHAR(255),
+  street VARCHAR(255),
+  city VARCHAR(255),
+  PRIMARY KEY (employee_name)
+);
+
+CREATE TABLE Works (
+  employee_name VARCHAR(255),
+  company_name VARCHAR(255),
+  salary INT -- use FLOAT if you are feeling fancy and pay your employees in Paise
+);
+
+CREATE TABLE Company (
+  company_name VARCHAR(255),
+  city VARCHAR(255),
+  PRIMARY KEY (company_name)
+);
+
+CREATE TABLE Manages (
+  employee_name VARCHAR(255),
+  manager_name VARCHAR(255)
+);
+
+```
+
+## Declaring foreign keys
+
+```sql
+ALTER TABLE Works ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
+ALTER TABLE Works ADD FOREIGN KEY (company_name) REFERENCES Company (company_name);
+ALTER TABLE Manages ADD FOREIGN KEY (employee_name) REFERENCES Employee (employee_name);
+
+
+```
+
+## Inserting data
+
+```sql
+INSERT INTO Employee VALUES
+(1, 'Mehul', 'Street 42', 'Pune'),
+(2, 'Himanshu', 'Street 74', 'Mumbai'),
+(3, 'Gundeti', 'Street 14', 'Pune'),
+(4, 'Salvi', 'Street 38', 'Pune'),
+(5, 'Afan', 'Steet 98', 'Pune'),
+(6, 'Jambo', 'Street 23', 'Mumbai');
+
+INSERT INTO Company VALUES
+('TCS', 'Pune'),
+('Infosys', 'Mumbai'),
+('TechM', 'Pune'),
+('MEPA', 'Pune');
+
+INSERT INTO Works VALUES
+('Mehul', 'MEPA', 15000),
+('Himanshu', 'TCS', 25000),
+('Gundeti', 'TCS', 9000),
+('Salvi', 'TechM', 8000),
+('Afan', 'Infosys', 13000),
+('Jambo', 'MEPA', 28000);
+
+
+INSERT INTO Manages VALUES
+('Mehul', 'Kalas'),
+('Himanshu', 'Kshitij'),
+('Gundeti', 'Macho'),
+('Salvi', 'Kshitij'),
+('Afan', 'Kalas'),
+('Jambo', 'Macho');
+
+```
+
+## Queries
+
+1. Change the city of employee working with InfoSys to ‘Bangalore’
+```sql
+UPDATE Company SET city = "Bangalore" WHERE company_name = "Infosys";
+
+```
+
+2. Find the names of all employees who earn more than the average salary of all employees of their company. Assume that all people work for at most one company.
+```sql
+SELECT employee_name, salary, company_name FROM Works as W WHERE salary > (SELECT AVG(salary) FROM Works WHERE company_name = W.company_name);
+
+```
+
+3. Find the names, street address, and cities of residence for all employees who work for 'TechM' and earn more than $10,000.
+```sql
+SELECT Employee.employee_name, street, city FROM Employee INNER JOIN Works ON Employee.employee_name = Works.employee_name WHERE salary > 10000;
+
+```
+
+4. Change name of table Manages to Management.
+```sql
+ALTER TABLE Manages RENAME TO Management;
+SHOW TABLES;
+
+```
+
+5. Create Simple and Unique index on employee table.
+```sql
+-- Simple Index
+CREATE INDEX emp_index ON Employee(employee_name);
+
+-- Unique Index
+CREATE UNIQUE INDEX emp_uniqueIndex ON Employee(emp_id);
+
+```
+
+6. Display index Information
+```sql
+SHOW INDEX FROM Employee;
+
+```
+
+---