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 
+ORDER BY client_id ASC 
+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 amount ASC 
+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_ids 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_ids 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'
+ORDER BY k_symbol ASC ;
+
+/*Query 9
+In the order table, what are the order_ids 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_ids were responsible for orders 
+between order_id 29540 and order_id 29560 (inclusive)?*/
+SELECT DISTINCT account_id 
+FROM 'order'
+WHERE order_id>=29540 and order_id<=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) AS client_count
+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) as card_freq
+FROM card
+GROUP BY type
+ORDER BY card_freq DESC;
+
+/*Query 15
+Using the loan table, print the top 10 account_ids 
+based on the sum of all of their loan amounts.*/
+SELECT account_id, sum(amount) as loan_amount
+FROM loan
+GROUP BY account_id
+ORDER BY loan_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(*) AS loan_count
+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(*) AS loan_count
+FROM loan
+WHERE date BETWEEN 971201 AND 971231
+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 account_id, type
+ORDER BY type ASC;
+
+