diff --git a/README.md b/README.md
index 4c8b1a5..f563f0b 100644
--- a/README.md
+++ b/README.md
@@ -35,22 +35,36 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 
 ### find all customers that live in London. Returns 6 records.
 > This can be done with SELECT and WHERE clauses
-
+SELECT *
+FROM customers
+WHERE city = 'London'
 
 ### find all customers with postal code 1010. Returns 3 customers.
 > This can be done with SELECT and WHERE clauses
-
+SELECT *
+FROM customers
+WHERE postal_code = '1010'
 
 ### find the phone number for the supplier with the id 11. Should be (010) 9984510.
 > This can be done with SELECT and WHERE clauses
+SELECT phone
+FROM suppliers
+WHERE supplier_id = '11'
 
 
 ### list orders descending by the order date. The order with date 1998-05-06 should be at the top.
 > This can be done with SELECT, WHERE, and ORDER BY clauses
 
+SELECT *
+From orders
+ORDER BY order_date DESC
+
 
 ### find all suppliers who have names longer than 20 characters. You can use `length(company_name)` to get the length of the name. Returns 11 records.
 > This can be done with SELECT and WHERE clauses
+SELECT *
+FROM suppliers
+WHERE length(company_name) > '20'
 
 
 ### find all customers that include the word 'MARKET' in the contact title. Should return 19 records.
@@ -59,6 +73,9 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 > Don't forget the wildcard '%' symbols at the beginning and end of your substring to denote it can appear anywhere in the string in question
 
 > Remember to convert your contact title to all upper case for case insenstive comparing so upper(contact_title)
+SELECT *
+FROM customers
+WHERE contact_title LIKE '%Market%'
 
 
 ### add a customer record for   
@@ -70,7 +87,8 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 * the postal code is '111'
 * the country is 'Middle Earth'
 > This can be done with the INSERT INTO clause
-
+INSERT INTO customers(customer_id, company_name, contact_name, address, city, postal_code, country)
+VALUES ('SHIRE', 'The Shire', 'Bilbo Baggins', '1 Hobbit-Hole', 'Bag End', 111, 'Middle Earth' )
 
 ### update _Bilbo Baggins_ record so that the postal code changes to _"11122"_.
 > This can be done with UPDATE and WHERE clauses
@@ -80,15 +98,29 @@ Answer the following data queries. Keep track of the SQL you write by pasting it
 > This can be done with SELECT, COUNT, JOIN and GROUP BY clauses. Your count should focus on a field in the Orders table, not the Customer table
 
 > There is more information about the COUNT clause on [W3 Schools](https://www.w3schools.com/sql/sql_count_avg_sum.asp)
-
+SELECT c.company_name, COUNT(*) AS numOfOrders
+FROM orders o
+LEFT JOIN customers c
+ON o.customer_id = c.customer_id
+GROUP BY c.company_name
 
 ### list customers names and the number of orders per customer. Sort the list by number of orders in descending order. _Save-a-lot Markets should be at the top with 31 orders followed by _Ernst Handle_ with 30 orders. Last should be _Centro comercial Moctezuma_ with 1 order.
 > This can be done by adding an ORDER BY clause to the previous answer
-
+SELECT c.company_name, COUNT(*) AS numOfOrders
+FROM orders o
+LEFT JOIN customers c
+ON o.customer_id = c.customer_id
+GROUP BY c.company_name
+ORDER BY numOfOrders DESC
 
 ### list orders grouped by customer's city showing number of orders per city. Returns 69 Records with _Aachen_ showing 6 orders and _Albuquerque_ showing 18 orders.
 > This is very similar to the previous two queries, however, it focuses on the City rather than the CustomerName
-
+SELECT c.city, COUNT(*) AS numOfOrders
+FROM orders o
+LEFT JOIN customers c
+ON o.customer_id = c.customer_id
+GROUP BY c.city
+ORDER BY numOfOrders DESC
 
 ## Data Normalization