diff --git a/week-1/mandatory/2-classes-db/task.md b/week-1/mandatory/2-classes-db/task.md
index 735de6ae..795348d1 100644
--- a/week-1/mandatory/2-classes-db/task.md
+++ b/week-1/mandatory/2-classes-db/task.md
@@ -7,6 +7,32 @@ Below you will find a set of tasks for you to complete to consolidate and extend
 To submit this homework write the correct commands for each question here:
 
 ```sql
+1.  Which rooms have a rate of more than 100.00?
+    cyf_hotel=# SELECT room_no, rate, room_type FROM rooms WHERE rate > 100;
+2.  List the reservations that have a checkin date this month and are for more than three nights.
+    cyf_hotel=# SELECT id, cust_id, room_no, checkin_date, checkout_date, no_guests, booking_date, checkout_date - checkin_date AS duration_of_stay FROM reservations WHERE EXTRACT(MONTH FROM checkin_date) = EXTRACT(MONTH FROM current_date) AND checkout_date - checkin_date > 3;
+3.  List all customers from cities that begin with the letter 'M'.
+    SELECT * FROM Customers WHERE City LIKE 'M%';
+
+    Insert some new data into the room_types and rooms tables, querying after each stage to check the data, as follows:
+4. Make a new room type of PENTHOUSE with a default rate of £185.00
+    cyf_hotel=# INSERT INTO room_types (room_type, def_rate) VALUES ('PENTHOUSE', 185.00);
+5. Add new rooms, 501 and 502 as room type PENTHOUSE and set the room rate of each to the default value (as in the new room type).
+    INSERT INTO rooms (room_no, rate, room_type) VALUES (501, (SELECT def_rate FROM room_types WHERE room_type = 'PENTHOUSE'), 'PENTHOUSE'),(502, (SELECT def_rate FROM room_types WHERE room_type='PENTHOUSE'), 'PENTHOUSE');
+6. Add a new room 503 as a PREMIER PLUS type similar to the other PREMIER PLUS rooms in the hotel but with a room rate of 143.00 to reflect its improved views over the city.
+    INSERT INTO rooms (room_no, rate, room_type) VALUES (503, 143.00, 'PREMIER PLUS');
+    To check if room no. 503 been added or not i have checked using SELECT room_no, rate, room_type FROM rooms WHERE room_no = 503;yes its worked.
+7. The hotel manager wishes to know how many rooms were occupied any time during the previous month - find that information.
+    SELECT COUNT(DISTINCT room_no) AS occupied_rooms FROM reservations WHERE (checkin_date <=date_trunc('month',current_date) - interval '1 day') AND (checkout_date >= date_trunc('month', current_date - interval '1 month'));
+8.  Get the total number of nights that customers stayed in rooms on the second floor (rooms 201 - 299).
+    SELECT SUM (checkout_date - checkin_date) AS total_nights FROM reservations WHERE room_no BETWEEN 201 AND 299;
+9. How many invoices are for more than £300.00 and what is their grand total and average amount?
+    SELECT COUNT(*) AS num_invoices, SUM(total) AS grand_total, AVG(total) AS average_total FROM invoices WHERE total > 300.00;
+10.  Bonus Question: list the number of nights stay for each floor of the hotel (floor no is the hundreds part of room number, e.g. room **3**12 is on floor **3**)
+    SELECT FLOOR(room_no / 100) AS floor_number, SUM(checkout_date - checkin_date) AS total_nights FROM reservations WHERE room_no BETWEEN 101 AND 499 GROUP BY FLOOR(room_no / 100) ORDER BY floor_number;
+
+
+
 
 
 ```