diff --git a/solutions.sql b/solutions.sql
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+--- Query1: 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 ORDER BY client_id LIMIT 5;
+
+--- Query2: 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;
+
+--- Query3: Get the 3 lowest amounts in the `loan` table.
+SELECT amount FROM loan ORDER BY amount ASC LIMIT 3;
+
+--- Query4: 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 ; 
+
+--- Query5: 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;
+
+--- Query6: 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;
+
+--- Query7: 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;
+
+--- Query8: 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 != " " ORDER BY k_symbol ASC;
+
+--- Query9: 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 ORDER BY order_id ASC;
+
+--- Query10: 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 ORDER BY account_id ASC;
+
+--- Query11: In the `order` table, what are the individual amounts that were sent to (`account_to`) id 30067122?
+SELECT DISTINCT amount FROM `order` WHERE account_to = 30067122 ;
+
+--- Query12: 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;
+
+--- Query13: 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(*) AS client_count  FROM client WHERE district_id < 10 GROUP BY district_id ORDER BY district_id ASC;
+
+--- Query14: In the `card` table, how many cards exist for each `type`? Rank the result starting with the most frequent `type`.
+SELECT type, COUNT(*) AS card_count FROM card GROUP BY type ORDER BY  card_count DESC;
+
+--- Query15: 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) AS total_loan_amount  FROM loan GROUP BY account_id ORDER BY total_loan_amount DESC LIMIT 10;
+
+--- Query16: 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(*) AS loan_count FROM loan  WHERE date < 930907 GROUP BY date  ORDER BY date DESC ;
+
+--- Query17: 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(*) AS loan_count  FROM loan WHERE date LIKE '9712%' GROUP BY date, duration ORDER BY date ASC, duration ASC;
+
+--- Query18: 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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