diff --git a/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/readers-like-you.md b/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/readers-like-you.md
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+---
+title: Readers Like You
+description: Practice using joins to find customers who share reading interests with a specific customer
+order: 160
+type: challenge
+setup: |
+  ```sql
+  CREATE TABLE customer (
+      id INT PRIMARY KEY,
+      name VARCHAR(255),
+      email VARCHAR(255)
+  );
+
+  CREATE TABLE sale (
+    id INT PRIMARY KEY,
+    customer_id INT,
+    book_id INT,
+    order_date DATE,
+    total_amount DECIMAL(10,2)
+  );
+
+  CREATE TABLE book (
+    id INT PRIMARY KEY,
+    title VARCHAR(255),
+    author VARCHAR(255)
+  );
+
+  INSERT INTO book (id, title, author)
+  VALUES 
+      (1, 'The Great Gatsby', 'F. Scott Fitzgerald'),
+      (2, 'To Kill a Mockingbird', 'Harper Lee'),
+      (3, '1984', 'George Orwell'),
+      (4, 'Pride and Prejudice', 'Jane Austen'),
+      (5, 'The Catcher in the Rye', 'J.D. Salinger');
+
+  INSERT INTO customer (id, name, email)
+  VALUES 
+      (1, 'Alice Smith', 'alice@example.com'),
+      (2, 'Bob Johnson', 'bob@example.com'),
+      (3, 'Carol White', 'carol@example.com'),
+      (4, 'David Brown', 'david@example.com'),
+      (5, 'Emily Davis', 'emily@example.com'),
+      (6, 'Frank Wilson', 'frank@example.com'),
+      (7, 'Grace Taylor', 'grace@example.com');
+
+  INSERT INTO sale (id, customer_id, book_id, order_date, total_amount)
+  VALUES 
+      (1, 1, 1, '2024-01-01', 10.00),  -- Alice: Great Gatsby
+      (2, 1, 2, '2024-01-15', 15.00),  -- Alice: Mockingbird
+      (3, 1, 3, '2024-02-01', 20.00),  -- Alice: 1984
+      (4, 2, 1, '2024-02-15', 10.00),  -- Bob: Great Gatsby
+      (5, 2, 3, '2024-03-01', 20.00),  -- Bob: 1984
+      (6, 3, 2, '2024-03-15', 15.00),  -- Carol: Mockingbird
+      (7, 3, 3, '2024-04-01', 20.00),  -- Carol: 1984
+      (8, 4, 1, '2024-04-15', 10.00),  -- David: Great Gatsby
+      (9, 5, 4, '2024-05-01', 25.00),  -- Emily: Pride and Prejudice
+      (10, 6, 3, '2024-05-15', 20.00); -- Frank: 1984
+  ```
+---
+
+The bookstore wants to implement a "Readers Like You" feature that shows customers who have similar reading interests. Given a specific customer ID, we want to find all other customers who have purchased at least one of the same books.
+
+Given the following data in table `customer`
+
+| id  | name         | email             |
+| --- | ------------ | ----------------- |
+| 1   | Alice Smith  | alice@example.com |
+| 2   | Bob Johnson  | bob@example.com   |
+| 3   | Carol White  | carol@example.com |
+| 4   | David Brown  | david@example.com |
+| 5   | Emily Davis  | emily@example.com |
+| 6   | Frank Wilson | frank@example.com |
+| 7   | Grace Taylor | grace@example.com |
+
+The following data in table `sale`
+
+| id  | customer_id | book_id | order_date | total_amount |
+| --- | ----------- | ------- | ---------- | ------------ |
+| 1   | 1           | 1       | 2024-01-01 | 10.00        |
+| 2   | 1           | 2       | 2024-01-15 | 15.00        |
+| 3   | 1           | 3       | 2024-02-01 | 20.00        |
+| 4   | 2           | 1       | 2024-02-15 | 10.00        |
+| 5   | 2           | 3       | 2024-03-01 | 20.00        |
+| 6   | 3           | 2       | 2024-03-15 | 15.00        |
+| 7   | 3           | 3       | 2024-04-01 | 20.00        |
+| 8   | 4           | 1       | 2024-04-15 | 10.00        |
+| 9   | 5           | 4       | 2024-05-01 | 25.00        |
+| 10  | 6           | 3       | 2024-05-15 | 20.00        |
+
+And the following data in table `book`
+
+| id  | title                  | author              |
+| --- | ---------------------- | ------------------- |
+| 1   | The Great Gatsby       | F. Scott Fitzgerald |
+| 2   | To Kill a Mockingbird  | Harper Lee          |
+| 3   | 1984                   | George Orwell       |
+| 4   | Pride and Prejudice    | Jane Austen         |
+| 5   | The Catcher in the Rye | J.D. Salinger       |
+
+Write a query that takes customer ID `1` (Alice Smith) and finds all other customers who have purchased any of the same books as Alice, along with the books they have in common.
+
+Your output should contain the following columns:
+
+- `customer_name` - name of the other customer
+- `common_book` - title of the book they both purchased
+- `purchase_date` - when the other customer purchased the book
+
+## Expected Output
+
+| customer_name | common_book           | purchase_date            |
+| ------------- | --------------------- | ------------------------ |
+| Bob Johnson   | 1984                  | 2024-03-01T00:00:00.000Z |
+| Bob Johnson   | The Great Gatsby      | 2024-02-15T00:00:00.000Z |
+| Carol White   | 1984                  | 2024-04-01T00:00:00.000Z |
+| Carol White   | To Kill a Mockingbird | 2024-03-15T00:00:00.000Z |
+| David Brown   | The Great Gatsby      | 2024-04-15T00:00:00.000Z |
+| Frank Wilson  | 1984                  | 2024-05-15T00:00:00.000Z |
+
+## Solution
+
+```sql
+SELECT
+    c2.name AS customer_name,
+    b.title AS common_book,
+    s2.order_date AS purchase_date
+FROM sale s1
+    INNER JOIN sale s2
+        ON s1.book_id = s2.book_id
+        AND s1.customer_id = 1
+        AND s2.customer_id != 1
+    INNER JOIN customer c2
+        ON s2.customer_id = c2.id
+    INNER JOIN book b
+        ON s1.book_id = b.id
+ORDER BY c2.name, b.title;
+```
+
+## Explanation
+
+Let's break down how this query works:
+
+We first start with all the purchases of Alice (customer_id = 1) to find the books she has purchased i.e.
+
+```sql
+FROM sale s1
+```
+
+Now, because we want to find all the other sales that have the same book_id as Alice's sales, we need to join the `sale` table again (aliased as `s2`) to compare purchases:
+
+```sql
+INNER JOIN sale s2 ON s1.book_id = s2.book_id
+```
+
+This will give us all the sales that have the same book_id as Alice's sales. But we are only interested in the sales to other customers, so we need to add a condition to exclude Alice's sales:
+
+```sql
+AND s2.customer_id != 1
+```
+
+Now we need to join the `customer` table to get the name of the other customer:
+
+```sql
+INNER JOIN customer c2 ON s2.customer_id = c2.id
+```
+
+Finally, we need to join the `book` table to get the title of the book:
+
+```sql
+INNER JOIN book b ON s1.book_id = b.id
+```
+
+Finally, we order the results by customer name and book title for easy reading:
+
+```sql
+ORDER BY c2.name, b.title;
+```
+
+This query efficiently finds all customers who share at least one book purchase with Alice, showing which books they have in common and when they made their purchase.
diff --git a/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/referred-customers.md b/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/referred-customers.md
index 8b22a3809..aaf5f11d0 100644
--- a/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/referred-customers.md
+++ b/src/data/courses/sql-mastery/chapters/multi-table-queries/lessons/referred-customers.md
@@ -40,7 +40,7 @@ Your output should contain the following columns:
 
 - `name`
 - `email`
-- `referred_by` (name of the customer or '- None -')
+- `referred_by` — name of the customer or `- None -`
 
 > **Tip:** You can use `COALESCE` function for `referred_by` column.