diff --git a/solutions.sql b/solutions.sql
index d0eddcc..a155afd 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,103 @@
 -- Add you solution queries below:
+/* How many copies of the film Hunchback Impossible exist in the inventory system? */
+SELECT COUNT(*) AS TOTAL_COPIES
+FROM sakila.film f
+JOIN sakila.inventory i ON f.film_id = i.film_id
+WHERE f.title = 'Hunchback Impossible';
+
+/* List all films whose length is longer than the average of all the films. */
+SELECT f.title, f.length
+FROM sakila.film f
+WHERE f.length > (SELECT AVG(length) FROM sakila.film);
+
+/* Use subqueries to display all actors who appear in the film Alone Trip. */
+SELECT first_name, last_name
+FROM sakila.actor
+WHERE actor_id IN (
+    SELECT actor_id
+    FROM sakila.film_actor fa
+    JOIN sakila.film f ON fa.film_id = f.film_id
+    WHERE f.title = 'Alone Trip'
+);
+
+/* Sales have been lagging among young families, 
+and you wish to target all family movies for a promotion. 
+Identify all movies categorized as family films. */
+SELECT f.title
+FROM sakila.film f
+JOIN sakila.film_category fc ON f.film_id = fc.film_id
+JOIN sakila.category c ON fc.category_id = c.category_id
+WHERE c.name = 'Family';
+
+/* Get name and email from customers from Canada using subqueries. 
+Do the same with joins. 
+Note that to create a join, you will have to identify the correct tables with their primary keys 
+and foreign keys, that will help you get the relevant information. */
+
+/* Subqueried */ 
+SELECT c.first_name, c.last_name, c.email
+FROM sakila.customer c
+WHERE c.address_id IN (
+    SELECT a.address_id
+    FROM sakila.address a
+    WHERE a.city_id IN (
+        SELECT ci.city_id
+        FROM sakila.city ci
+        WHERE ci.country_id = (
+            SELECT co.country_id
+            FROM sakila.country co
+            WHERE co.country = 'Canada'
+        )
+    )
+);
+
+/* Joined */
+SELECT c.first_name, c.last_name, c.email
+FROM sakila.customer c
+JOIN sakila.address a ON c.address_id = a.address_id
+JOIN sakila.city ci ON a.city_id = ci.city_id
+JOIN sakila.country co ON ci.country_id = co.country_id
+WHERE co.country = 'Canada';
+
+/* Which are films starred by the most prolific actor? 
+Most prolific actor is defined as the actor that has acted in the most number of films. 
+First you will have to find the most prolific actor 
+and then use that actor_id to find the different films that he/she starred. */
+SELECT f.title
+FROM sakila.film_actor fa
+JOIN sakila.film f ON fa.film_id = f.film_id
+WHERE fa.actor_id = (
+    SELECT fa.actor_id
+    FROM sakila.film_actor fa
+    GROUP BY fa.actor_id
+    ORDER BY COUNT(fa.film_id) DESC
+    LIMIT 1
+);
+
+/* Films rented by most profitable customer. You can use the customer table 
+and payment table to find the most profitable customer 
+ie the customer that has made the largest sum of payments */
+SELECT f.title
+FROM sakila.rental r
+JOIN sakila.inventory i ON r.inventory_id = i.inventory_id
+JOIN sakila.film f ON i.film_id = f.film_id
+WHERE r.customer_id = (
+    SELECT p.customer_id
+    FROM sakila.payment p
+    GROUP BY p.customer_id
+    ORDER BY SUM(p.amount) DESC
+    LIMIT 1
+);
+
+/* Get the client_id and the total_amount_spent of those clients who spent more than the average of the total_amount spent by each client. */
+SELECT customer_id, SUM(amount) AS total_amount_spent
+FROM sakila.payment
+GROUP BY customer_id
+HAVING SUM(amount) > (
+    SELECT AVG(total_spent)
+    FROM (
+        SELECT SUM(amount) AS total_spent
+        FROM sakila.payment
+        GROUP BY customer_id
+    ) AS avg_spent
+);
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