diff --git a/solutions.sql b/solutions.sql
index d0eddcc..708c1d9 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,90 @@
 -- Add you solution queries below:
+#How many copies of the film Hunchback Impossible exist in the inventory system?
+
+SELECT COUNT(i.inventory_id) AS num_hunch
+FROM sakila.inventory i
+WHERE i.film_id = (SELECT film_id FROM sakila.film WHERE 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 length > (select avg(length) from sakila.film)
+order by length;
+
+#Use subqueries to display all actors who appear in the film Alone Trip.
+select  concat(first_name,' ',last_name) as actor_name
+from sakila.actor
+where actor_id in (select ft.actor_id from sakila.film f inner join film_actor ft
+on f.film_id = ft.film_id and 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 distinct title from film
+where film_id in (
+select film_id from film_category
+where category_id in (
+select category_id from category 
+where 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.
+select concat(cust.first_name,' ',cust.last_name) as customer_name,
+       cust.email
+from customer as cust
+inner join store as s
+on cust.store_id = s.store_id
+inner join address a
+on s.address_id = a.address_id
+inner join city c
+on a.city_id = c.city_id
+inner join country ct
+on c.country_id = ct.country_id
+and ct.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.film_id, f.title
+from film as f
+where film_id in (select film_id
+from film_actor as fa
+where fa.actor_id in 
+(select actor_id as prolific_actor
+from film_actor
+group by actor_id
+having count(film_id) = (select max(num_films) from 
+(select count(film_id) as num_films
+from film_actor
+group by actor_id)as tab )));
+
+
+#Films rented by most profitable customer. 
+#You can use the customer table and payment table to find the most profitable customer 
+#i.e. the customer that has made the largest sum of payments
+with cust_rank as 
+(select customer_id , dense_rank() over (order by total_revenue) as rnk
+from (
+select customer_id, 
+	   sum(amount) total_revenue
+from payment
+group by customer_id) as tbl)
+select f.title
+from film as f
+inner join inventory i
+on f.film_id = i.film_id
+inner join rental r
+on i.inventory_id = r.inventory_id
+and r.customer_id in (select customer_id from cust_rank
+where rnk = 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_spent
+from payment
+group by customer_id
+having sum(amount) > (select avg(total_spent)
+from (select sum(amount) as total_spent from payment group by customer_id) as tbl);
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