diff --git a/solutions.sql b/solutions.sql
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+-- Query 1 : Get the `id` values of the first 5 clients from `district_id` with a value equals to 1.
+
+SELECT client_id
+FROM client
+WHERE district_id = 1
+LIMIT 5
+
+-- Query 2 : In the `client` table, get an `id` value of the last client where the `district_id` equals to 72.
+
+SELECT client_id
+FROM client
+WHERE district_id = 72
+ORDER BY client_id DESC
+LIMIT 1
+
+-- Query 3 : Get the 3 lowest amounts in the `loan` table.
+
+SELECT amount
+FROM loan
+ORDER BY 1
+LIMIT 3
+
+-- Query 4 : What are the possible values for `status`, ordered alphabetically in ascending order in the `loan` table?
+
+SELECT DISTINCT status 
+FROM loan 
+ORDER BY status ASC;
+
+-- Query 5 : What is the `loan_id` of the highest payment received in the `loan` table?
+SELECT loan_id 
+FROM loan 
+ORDER BY payments DESC 
+LIMIT 1;
+
+-- Query 6 : What is the loan `amount` of the lowest 5 `account_id`s in the `loan` table? Show the `account_id` and the corresponding `amount`
+
+SELECT account_id, amount 
+FROM loan 
+ORDER BY account_id ASC 
+LIMIT 5;
+
+-- Query 7 : What are the `account_id`s with the lowest loan `amount` that have a loan `duration` of 60 in the `loan` table?
+
+SELECT account_id 
+FROM loan 
+WHERE duration = 60 
+ORDER BY amount ASC 
+LIMIT 5;
+
+-- Query 8 : What are the unique values of `k_symbol` in the `order` table?
+-- Note: There shouldn't be a table name `order`, since `order` is reserved from the `ORDER BY` clause. You have to use backticks to escape the `order` table name.
+
+SELECT DISTINCT k_symbol 
+FROM `order` 
+WHERE k_symbol <> ' ';
+
+-- Query 9 : In the `order` table, what are the `order_id`s of the client with the `account_id` 34?
+
+SELECT order_id 
+FROM `order` 
+WHERE account_id = 34;
+
+-- Query 10 : In the `order` table, which `account_id`s were responsible for orders between `order_id` 29540 and `order_id` 29560 (inclusive)?
+
+SELECT DISTINCT account_id 
+FROM `order` 
+WHERE order_id BETWEEN 29540 AND 29560;
+
+-- Query 11 : In the `order` table, what are the individual amounts that were sent to (`account_to`) id 30067122?
+
+SELECT amount 
+FROM `order` 
+WHERE account_to = 30067122;
+
+-- Query 12 : In the `trans` table, show the `trans_id`, `date`, `type` and `amount` of the 10 first transactions from `account_id` 793 in chronological order, from newest to oldest.
+
+SELECT trans_id, date, type, amount 
+FROM trans 
+WHERE account_id = 793 
+ORDER BY date DESC 
+LIMIT 10;
+
+-- Query 13 : In the `client` table, of all districts with a `district_id` lower than 10, how many clients are from each `district_id`? Show the results sorted by the `district_id` in ascending order.
+
+SELECT district_id, COUNT(client_id) 
+FROM client 
+WHERE district_id < 10 
+GROUP BY district_id 
+ORDER BY district_id ASC;
+
+-- Query 14 : In the `card` table, how many cards exist for each `type`? Rank the result starting with the most frequent `type`.
+
+SELECT type, COUNT(card_id) 
+FROM card 
+GROUP BY type 
+ORDER BY COUNT(card_id) DESC;
+
+-- Query 15 : Using the `loan` table, print the top 10 `account_id`s based on the sum of all of their loan amounts.
+
+SELECT account_id, SUM(amount) 
+FROM loan 
+GROUP BY account_id 
+ORDER BY SUM(amount) DESC 
+LIMIT 10;
+
+-- Query 16 : In the `loan` table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
+
+SELECT date, COUNT(loan_id) 
+FROM loan 
+WHERE date < 930907 
+GROUP BY date 
+ORDER BY date DESC;
+
+-- Query 17 : In the `loan` table, for each day in December 1997, count the number of loans issued for each unique loan duration, ordered by date and duration, both in ascending order. You can ignore days without any loans in your output.
+
+SELECT date, duration, COUNT(loan_id) 
+FROM loan 
+WHERE date LIKE '9712%' 
+GROUP BY date, duration 
+ORDER BY date ASC, duration ASC;
+
+-- Query 18 : In the `trans` table, for `account_id` 396, sum the amount of transactions for each type (`VYDAJ` = Outgoing, `PRIJEM` = Incoming). Your output should have the `account_id`, the `type` and the sum of amount, named as `total_amount`. Sort alphabetically by type.
+
+SELECT account_id, type, SUM(amount) AS total_amount 
+FROM trans 
+WHERE account_id = 396 
+GROUP BY type 
+ORDER BY type ASC;
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