diff --git a/setoperations/sqlquery10.sql b/setoperations/sqlquery10.sql
new file mode 100644
index 0000000..e099a06
--- /dev/null
+++ b/setoperations/sqlquery10.sql
@@ -0,0 +1,42 @@
+Problem Statement:
+Write a SQL query to print the item name, item type of only the items which are available in shop 1 but not in shop 2 in the ascending order of item name.
+
+Information about the table:
+Table shop_1:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | pencil     | stationery |    10 |
+|       2 | soap       | toiletries |    25 |
+|       3 | eraser     | stationery |     5 |
+|       4 | toothbrush | toiletries |    30 |
+|       5 | toothpaste | toiletries |    50 |
++---------+------------+------------+-------+
+
+Table shop_2:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | facewash   | toiletries |    70 |
+|       2 | soap       | toiletries |    25 |
+|       3 | pencil     | stationery |    15 |
+|       4 | paintbrush | stationery |    30 |
+|       5 | shampoo    | toiletries |   100 |
++---------+------------+------------+-------+
+
+Note: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT item_name, item_type FROM shop_1 LEFT JOIN shop_2
+USING(item_name, item_type)
+WHERE shop_2.item_id IS NULL
+ORDER BY item_name;
+
+Output:
++------------+------------+
+| item_name  | item_type  |
++------------+------------+
+| eraser     | stationery |
+| toothbrush | toiletries |
+| toothpaste | toiletries |
++------------+------------+
\ No newline at end of file
diff --git a/setoperations/sqlquery11.sql b/setoperations/sqlquery11.sql
new file mode 100644
index 0000000..f0cfec0
--- /dev/null
+++ b/setoperations/sqlquery11.sql
@@ -0,0 +1,40 @@
+Problem Statement:
+Write a SQL query to print the item name, price of only the items which are available in shop 2 but not in shop 1 whose price is greater than 50.
+
+Information about the table:
+Table shop_1:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | pencil     | stationery |    10 |
+|       2 | soap       | toiletries |    25 |
+|       3 | eraser     | stationery |     5 |
+|       4 | toothbrush | toiletries |    30 |
+|       5 | toothpaste | toiletries |    50 |
++---------+------------+------------+-------+
+
+Table shop_2:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | facewash   | toiletries |    70 |
+|       2 | soap       | toiletries |    25 |
+|       3 | pencil     | stationery |    15 |
+|       4 | paintbrush | stationery |    30 |
+|       5 | shampoo    | toiletries |   100 |
++---------+------------+------------+-------+
+
+Note: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT item_name, price FROM shop_1 RIGHT JOIN shop_2
+USING(item_name, price)
+WHERE shop_2.price>50;
+
+Output:
++-----------+-------+
+| item_name | price |
++-----------+-------+
+| facewash  |    70 |
+| shampoo   |   100 |
++-----------+-------+
\ No newline at end of file
diff --git a/setoperations/sqlquery6.sql b/setoperations/sqlquery6.sql
new file mode 100644
index 0000000..2afbbd6
--- /dev/null
+++ b/setoperations/sqlquery6.sql
@@ -0,0 +1,36 @@
+Problem Statement:
+Find out all the details of employees that work for both the departments.
+
+Information about the table:
+Table Empdept1:
++---------+----------+----------+-------------------+
+| EmpCode | EmpFName | EmpLName | Job               |
++---------+----------+----------+-------------------+
+|    9369 | TONY     | STARK    | SOFTWARE ENGINEER |
+|    9499 | TIM      | ADOLF    | SALESMAN          |
+|    9566 | KIM      | JARVIS   | MANAGER           |
+|    9654 | SAM      | MILES    | SALESMAN          |
++---------+----------+----------+-------------------+
+
+Table Empdept2:
++---------+----------+----------+----------+
+| EmpCode | EmpFName | EmpLName | Job      |
++---------+----------+----------+----------+
+|    9566 | KIM      | JARVIS   | MANAGER  |
+|    9902 | ANDREW   | FAULKNER | ANALYST  |
+|    9685 | SAMAY    | DAGA     | SALESMAN |
++---------+----------+----------+----------+
+
+Note-1: The data should not contain duplicate rows of employees.
+Note-2: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT DISTINCT * FROM empdept1 INNER JOIN empdept2
+USING (empcode, empfname, emplname, job);
+
+Output:
++---------+----------+----------+---------+
+| EmpCode | EmpFName | EmpLName | Job     |
++---------+----------+----------+---------+
+|    9566 | KIM      | JARVIS   | MANAGER |
++---------+----------+----------+---------+
\ No newline at end of file
diff --git a/setoperations/sqlquery7.sql b/setoperations/sqlquery7.sql
new file mode 100644
index 0000000..39c2587
--- /dev/null
+++ b/setoperations/sqlquery7.sql
@@ -0,0 +1,38 @@
+Problem Statement:
+Write a SQL query to find the item name along with its type of stationery item which is available in both the shops.
+
+Information about the table:
+Table shop_1:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | pencil     | stationery |    10 |
+|       2 | soap       | toiletries |    25 |
+|       3 | eraser     | stationery |     5 |
+|       4 | toothbrush | toiletries |    30 |
+|       5 | toothpaste | toiletries |    50 |
++---------+------------+------------+-------+
+
+Table shop_2:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | facewash   | toiletries |    70 |
+|       2 | soap       | toiletries |    25 |
+|       3 | pencil     | stationery |    15 |
+|       4 | paintbrush | stationery |    30 |
+|       5 | shampoo    | toiletries |   100 |
++---------+------------+------------+-------+
+
+Note: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT DISTINCT item_name, item_type FROM shop_1 INNER JOIN shop_2
+USING (item_name, item_type) WHERE item_type = 'stationery';
+
+Output:
++-----------+------------+
+| item_name | item_type  |
++-----------+------------+
+| pencil    | stationery |
++-----------+------------+
\ No newline at end of file
diff --git a/setoperations/sqlquery8.sql b/setoperations/sqlquery8.sql
new file mode 100644
index 0000000..f62d50b
--- /dev/null
+++ b/setoperations/sqlquery8.sql
@@ -0,0 +1,38 @@
+Problem Statement:
+Write a SQL query to find the name and price of items whose price is greater than 20 and available in both the shops.
+
+Information about the table:
+Table shop_1:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | pencil     | stationery |    10 |
+|       2 | soap       | toiletries |    25 |
+|       3 | eraser     | stationery |     5 |
+|       4 | toothbrush | toiletries |    30 |
+|       5 | toothpaste | toiletries |    50 |
++---------+------------+------------+-------+
+
+Table shop_2:
++---------+------------+------------+-------+
+| item_id | item_name  | item_type  | price |
++---------+------------+------------+-------+
+|       1 | facewash   | toiletries |    70 |
+|       2 | soap       | toiletries |    25 |
+|       3 | pencil     | stationery |    15 |
+|       4 | paintbrush | stationery |    30 |
+|       5 | shampoo    | toiletries |   100 |
++---------+------------+------------+-------+
+
+Note: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT DISTINCT item_name, price FROM shop_1 INNER JOIN shop_2
+USING(item_name, price) WHERE price >20;
+
+Output:
++-----------+-------+
+| item_name | price |
++-----------+-------+
+| soap      |    25 |
++-----------+-------+
diff --git a/setoperations/sqlquery9.sql b/setoperations/sqlquery9.sql
new file mode 100644
index 0000000..585b123
--- /dev/null
+++ b/setoperations/sqlquery9.sql
@@ -0,0 +1,38 @@
+Problem Statement:
+List down all the details of employees working in dept1 but not in Dept2.
+
+Information about the table:
+Table Empdept1:
++---------+----------+----------+-------------------+
+| EmpCode | EmpFName | EmpLName | Job               |
++---------+----------+----------+-------------------+
+|    9369 | TONY     | STARK    | SOFTWARE ENGINEER |
+|    9499 | TIM      | ADOLF    | SALESMAN          |
+|    9566 | KIM      | JARVIS   | MANAGER           |
+|    9654 | SAM      | MILES    | SALESMAN          |
++---------+----------+----------+-------------------+
+
+Table Empdept2:
++---------+----------+----------+----------+
+| EmpCode | EmpFName | EmpLName | Job      |
++---------+----------+----------+----------+
+|    9566 | KIM      | JARVIS   | MANAGER  |
+|    9902 | ANDREW   | FAULKNER | ANALYST  |
+|    9685 | SAMAY    | DAGA     | SALESMAN |
++---------+----------+----------+----------+
+
+Note: Write keywords of syntax in uppercase alphabets.
+
+Solution:
+SELECT * FROM empdept1 LEFT JOIN empdept2 
+USING (empcode, empfname, emplname, job)
+WHERE empdept2.empcode IS NULL;
+
+Output:
++---------+----------+----------+-------------------+
+| EmpCode | EmpFName | EmpLName | Job               |
++---------+----------+----------+-------------------+
+|    9369 | TONY     | STARK    | SOFTWARE ENGINEER |
+|    9499 | TIM      | ADOLF    | SALESMAN          |
+|    9654 | SAM      | MILES    | SALESMAN          |
++---------+----------+----------+-------------------+