diff --git a/solutions.sql b/solutions.sql
index d0eddcc..04e9016 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,150 @@
--- Add you solution queries below:
+-- Start by selecting the Sakila database
+USE sakila;
+
+-- 1. How many copies of the film Hunchback Impossible exist in the inventory system?
+-- Explanation: The inventory table stores the copies of each film. 
+-- We use a subquery to fetch the film_id of "Hunchback Impossible" from the film table and count its occurrences in the inventory table.
+
+SELECT COUNT(*) AS number_of_copies
+FROM inventory
+WHERE film_id = (
+    SELECT film_id
+    FROM film
+    WHERE title = 'Hunchback Impossible'
+);
+
+-- 2. List all films whose length is longer than the average of all the films.
+-- Explanation: Calculate the average film length using a subquery and then filter films with length above this average.
+
+SELECT title, length
+FROM film
+WHERE length > (
+    SELECT AVG(length)
+    FROM film
+);
+
+-- 3. Use subqueries to display all actors who appear in the film Alone Trip.
+-- Explanation: Find the film_id of "Alone Trip," and use it to find the actors from the film_actor table.
+
+SELECT first_name, last_name
+FROM actor
+WHERE actor_id IN (
+    SELECT actor_id
+    FROM film_actor
+    WHERE film_id = (
+        SELECT film_id
+        FROM film
+        WHERE title = 'Alone Trip'
+    )
+);
+
+-- 4. Identify all movies categorized as family films.
+-- Explanation: Find all film titles associated with the "Family" category using the film_category and category tables.
+
+SELECT title
+FROM film
+WHERE film_id IN (
+    SELECT film_id
+    FROM film_category
+    WHERE category_id = (
+        SELECT category_id
+        FROM category
+        WHERE name = 'Family'
+    )
+);
+
+-- 5. Get name and email from customers from Canada using subqueries.
+-- Explanation: Find the country_id for "Canada" from the country table, 
+-- and then fetch the relevant customers based on their address_id linked through the address table.
+
+SELECT first_name, last_name, email
+FROM customer
+WHERE address_id IN (
+    SELECT address_id
+    FROM address
+    WHERE city_id IN (
+        SELECT city_id
+        FROM city
+        WHERE country_id = (
+            SELECT country_id
+            FROM country
+            WHERE country = 'Canada'
+        )
+    )
+);
+
+-- Same query using JOINs
+SELECT c.first_name, c.last_name, c.email
+FROM customer c
+JOIN address a ON c.address_id = a.address_id
+JOIN city ci ON a.city_id = ci.city_id
+JOIN country co ON ci.country_id = co.country_id
+WHERE co.country = 'Canada';
+
+-- 6. Which are films starred by the most prolific actor?
+-- Explanation: First, identify the actor who has acted in the most films using GROUP BY and ORDER BY. 
+-- Then, find all films they starred in using their actor_id.
+
+-- Step 1: Find the most prolific actor
+SELECT actor_id
+FROM film_actor
+GROUP BY actor_id
+ORDER BY COUNT(film_id) DESC
+LIMIT 1;
+
+-- Step 2: Find films starred by the most prolific actor
+SELECT title
+FROM film
+WHERE film_id IN (
+    SELECT film_id
+    FROM film_actor
+    WHERE actor_id = (
+        SELECT actor_id
+        FROM film_actor
+        GROUP BY actor_id
+        ORDER BY COUNT(film_id) DESC
+        LIMIT 1
+    )
+);
+
+-- 7. Films rented by the most profitable customer.
+-- Explanation: Find the customer who made the largest total payments using the payment table, 
+-- then list the films they rented using the rental table.
+
+-- Step 1: Find the most profitable customer
+SELECT customer_id
+FROM payment
+GROUP BY customer_id
+ORDER BY SUM(amount) DESC
+LIMIT 1;
+
+-- Step 2: Find films rented by the most profitable customer
+SELECT DISTINCT f.title
+FROM film f
+JOIN inventory i ON f.film_id = i.film_id
+JOIN rental r ON i.inventory_id = r.inventory_id
+WHERE r.customer_id = (
+    SELECT customer_id
+    FROM payment
+    GROUP BY customer_id
+    ORDER BY SUM(amount) DESC
+    LIMIT 1
+);
+
+-- 8. Get the client_id and the total_amount_spent of those clients who spent more than the average total amount spent by each client.
+-- Explanation: First calculate the average total amount spent using GROUP BY. Then filter clients who exceed this average.
+
+SELECT customer_id, total_spent
+FROM (
+    SELECT customer_id, SUM(amount) AS total_spent
+    FROM payment
+    GROUP BY customer_id
+) AS customer_totals
+WHERE total_spent > (
+    SELECT AVG(total_spent)
+    FROM (
+        SELECT SUM(amount) AS total_spent
+        FROM payment
+        GROUP BY customer_id
+    ) AS avg_spent
+);