MikeXydas/qr2t_benchmark huggingface dataset:
Table: climber, mountain
Question: Show the height of the mountain climbed by the climber with the maximum points.
SQL query:
SELECT T2.Height AS "mountain Height", T1.Points AS "climber Points" 
 FROM climber AS T1 JOIN mountain AS T2 ON T1.Mountain_ID = T2.Mountain_ID ORDER BY T1.Points DESC
Result:
,climber Points,mountain Height
0,15.0,5895.0

Answer: The climber points are 15 and the mountain height is 5895.
Table: elimination, wrestler
Question: What is the time of elimination for the wrestler with the most days held?
SQL query:
SELECT T1.Time AS "elimination Time", T2.Days_held AS "wrestler Days_held" 
 FROM elimination AS T1 JOIN wrestler AS T2 ON T1.Wrestler_ID = T2.Wrestler_ID ORDER BY T2.Days_held DESC
Result:
,elimination Time,wrestler Days_held
0,10:45,960

Answer: The elimination time of the wrestler with 960 days held, is 10:45
Table: track
Question: Return the minimum, maximum, and average seating across all tracks.
SQL query:
SELECT MIN(seating) AS "minimum seating", MAX(seating) AS "maximum seating", AVG(seating) AS "average seating" 
 FROM track
Result:
,average seating,maximum seating,minimum seating
0,91526.88888888889,168000.0,63000.0

Answer: The average seating is 91526.88888888889, the maximum seating is 168000 and the minimum seating is 63000 across all tracks
Table: Person, PersonFriend
Question: Who are Bob's friends?
SQL query:
SELECT T2.friend AS "PersonFriend friend", T1.name AS "Person name" 
 FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name 
 WHERE T1.name = 'Bob'
Result:
,Person name,PersonFriend friend
0,Bob,Zach

Answer: Zach is Bob's friend
Table: Customers
Question: When are the birthdays of customer who are classified as 'Good Customer' status?
SQL query:
SELECT date_of_birth, customer_status_code 
 FROM Customers 
 WHERE customer_status_code = 'Good Customer'
Result:
,date_of_birth,customer_status_code
0,1994-06-21 01:34:56,Good Customer

Answer: The person who is classified as 'Good Customer' was born in 1994-06-21 time 01:34:56
Table: product
Question: What are the products with the maximum page size eqal to A4 or a pages per minute color less than 5?
SQL query:
SELECT product, max_page_size, pages_per_minute_color 
 FROM product 
 WHERE max_page_size = 'A4' OR pages_per_minute_color < 5
Result:
,product,max_page_size,pages_per_minute_color
0,Fujitsu fi-6130 A4 Series Scanner,A4,30.0

Answer: The product Fujitsu fi-6130 A4 Series Scanner has a maximum page size equal to A4 and a pages per minute color of 30.
Table: Companies
Question: Compute the average profits companies make.
SQL query:
SELECT AVG(Profits_billion) AS "average Profits_billion" 
 FROM Companies
Result:
,average Profits_billion
0,17.063157894736843

Answer: The average of profits is 17.063157894736843 billions.
Table: song
Question: What is the average rating and resolution of all bangla songs?
SQL query:
SELECT AVG(rating) AS "average rating", AVG(resolution) AS "average resolution", languages 
 FROM song 
 WHERE languages = 'bangla'
Result:
,languages,average rating,average resolution
0,bangla,7.5,558.0

Answer: The songs of language bangla have an average rating of 7.5 and an average resolution of 558.
Table: Apartments
Question: Show the most common apartment type code.
SQL query:
SELECT apt_type_code 
 FROM Apartments 
 GROUP BY apt_type_code ORDER BY COUNT(*) DESC
Result:
,apt_type_code
0,Flat

Answer: The most common apartments type is "Flat"
Table: airport
Question: What are the names of the airports which are not in the country 'Iceland'?
SQL query:
SELECT name, country 
 FROM airport 
 WHERE country <> 'Iceland'
Result:
,name,Country
0,Schiphol Airport,Netherlands

Answer: Schiphol Airport is in the country of Netherlands.
Table: budget
Question: Count the number of budgets in year 2001 or before whose budgeted amount is greater than 3000
SQL query:
SELECT COUNT(*) AS "count of budget", budgeted, YEAR 
 FROM budget 
 WHERE budgeted > 3000 AND YEAR <= 2001
Result:
,Year,Budgeted,count of budget
0,1999,4979,3

Answer: The number of budgets in 1999 whose budgeted is more than 4979 is 3.
Table: Third_Party_Companies, Maintenance_Contracts
Question: Which company started the earliest the maintenance contract? Show the company name.
SQL query:
SELECT T1.company_name AS "Third_Party_Companies company_name", T2.contract_start_date AS "Maintenance_Contracts contract_start_date" 
 FROM Third_Party_Companies AS T1 JOIN Maintenance_Contracts AS T2 ON T1.company_id = T2.maintenance_contract_company_id ORDER BY T2.contract_start_date ASC
Result:
,Third_Party_Companies company_name,Maintenance_Contracts contract_start_date
0,"Grimes, Mosciski and Renner",2017-04-21 00:03:07

Answer: The company name is Grimes, Mosciski and Renner and the contract start date is 2017-04-21 00:03:07
Table: captain
Question: Find the name of captains whose rank are either Midshipman or Lieutenant.
SQL query:
SELECT name, rank 
 FROM captain 
 WHERE rank = 'Midshipman' OR rank = 'Lieutenant'
Result:
,Name,Rank
0,Captain Sir Henry Langford,Midshipman

Answer: Captain Sir Henry Langford has a rank of Midshipman.
Table: city
Question: What are the census rankings of cities that do not have the status "Village"?
SQL query:
SELECT Census_Ranking, Status 
 FROM city 
 WHERE Status <> 'Village'
Result:
,Status,Census_Ranking
0,Town,636 of 5008

Answer: The cities that have the status "Town" have a census ranking of 636 of 5008
Table: station
Question: What is the name of every city that has at least 15 stations and how many stations does it have?
SQL query:
SELECT city, COUNT(*) AS "count of station" 
 FROM station 
 GROUP BY city 
 HAVING COUNT(*) >= 15
Result:
,city,count of station
0,San Francisco,35

Answer: The city of San Francisco has 35 stations.
Table: channel
Question: List the names of all the channels owned by either CCTV or HBS
SQL query:
SELECT name, OWNER 
 FROM channel 
 WHERE OWNER = 'CCTV' OR OWNER = 'HBS'
Result:
,Name,Owner
0,CCTV-1,CCTV

Answer: The name is CCTV-1 and the owner is CCTV.
Table: Invoices
Question: Check the invoices record and compute the average quantities ordered with the payment method "MasterCard".
SQL query:
SELECT AVG(Order_Quantity) AS "average Order_Quantity", payment_method_code 
 FROM Invoices 
 WHERE payment_method_code = 'MasterCard'
Result:
,payment_method_code,average Order_Quantity
0,MasterCard,1.8333333333333333

Answer: The payment method MasterCards has an average order quantity of 1.8333333333333333.
Table: ship
Question: What is the number of ships?
SQL query:
SELECT COUNT(*) AS "count of ship" 
 FROM ship
Result:
,count of ship
0,8

Answer: There are 8 ship.
Table: Customers
Question: When did customer with first name as Carole and last name as Bernhard became a customer?
SQL query:
SELECT date_became_customer, last_name, first_name 
 FROM Customers 
 WHERE first_name = 'Carole' AND last_name = 'Bernhard'
Result:
,last_name,first_name,date_became_customer
0,Bernhard,Carole,2016-05-11 17:03:48

Answer: Carole Bernhard became a customer in 	2016-05-11 17:03:48.
Table: county
Question: How many counties are there in total?
SQL query:
SELECT COUNT(*) AS "count of county" 
 FROM county
Result:
,count of county
0,5

Answer: there are 5 countries
Table: program
Question: Count the number of programs.
SQL query:
SELECT COUNT(*) AS "count of program" 
 FROM program
Result:
,count of program
0,5

Answer: There are 5 programs
Table: player, tryout
Question: Find the average hours for the students whose tryout decision is no.
SQL query:
SELECT AVG(T1.HS) AS "average HS", T2.decision AS "tryout decision" 
 FROM player AS T1 JOIN tryout AS T2 ON T1.pID = T2.pID 
 WHERE T2.decision = 'no'
Result:
,average HS,tryout decision
0,925.0,no

Answer: The average hours for the students whose tryout decision is no is 925
Table: employees
Question: What is the title, phone and hire date of Nancy Edwards?
SQL query:
SELECT title, phone, hire_date, last_name, first_name 
 FROM employees 
 WHERE first_name = 'Nancy' AND last_name = 'Edwards'
Result:
,phone,title,hire_date,last_name,first_name
0,+1 (403) 262-3443,Sales Manager,2002-05-01 00:00:00,Edwards,Nancy

Answer: The employee Nancy Edwards has the phone +1 (403) 262-3443, she is a Sales Manager and was hired in 2002-05-01 00:00:00.
Table: salary, team
Question: Compute the average salary of the players in the team called 'Boston Red Stockings'.
SQL query:
SELECT AVG(T1.salary) AS "average salary", T2.name AS "team name" 
 FROM salary AS T1 JOIN team AS T2 ON T1.team_id = T2.team_id_br 
 WHERE T2.name = 'Boston Red Stockings'
Result:
,team name,average salary
0,Boston Red Stockings,2856616.7595628416

Answer: The average salary of the players in the team called 'Boston Red Stockings' is 2856616.7595628416
Table: assignedto, projects, scientists
Question: Find the name of the project for which a scientist whose name contains ‘Smith’ is assigned to.
SQL query:
SELECT T2.name AS "projects name", T3.name AS "scientists name" 
 FROM assignedto AS T1 JOIN projects AS T2 ON T1.project = T2.code JOIN scientists AS T3 ON T1.scientist = T3.SSN 
 WHERE T3.name LIKE '%Smith%'
Result:
,projects name,scientists name
0,Build Your Own Telescope,Carol Smith

Answer: The scientist name is Carol Smith and the project name is Build Your Own Telescope
Table: Settlements
Question: Compute the total amount of settlement across all the settlements.
SQL query:
SELECT SUM(Amount_Settled) AS "sum of Amount_Settled" 
 FROM Settlements
Result:
,sum of Amount_Settled
0,21993

Answer: The sum of the all the settled amounts is 21993.
Table: pilot
Question: Show the names of pilots from team "Bradley" or "Fordham".
SQL query:
SELECT Pilot_name, Team 
 FROM pilot 
 WHERE Team = 'Bradley' OR Team = 'Fordham'
Result:
,Team,Pilot_name
0,Bradley,Patrick O'Bryant

Answer: The pilot Patrick O'Bryant is in team Bradley.
Table: employees
Question: Return the full names and salaries for employees with first names that end with the letter m.
SQL query:
SELECT first_name, last_name, salary 
 FROM employees 
 WHERE first_name LIKE '%m'
Result:
,SALARY,LAST_NAME,FIRST_NAME
0,8200,Fripp,Adam

Answer: The employee with first name Adam and last name Fripp has a salary of 8200.
Table: Allergy_type
Question: What is allergy type of a cat allergy?
SQL query:
SELECT allergytype, allergy 
 FROM Allergy_type 
 WHERE allergy = 'Cat'
Result:
,Allergy,AllergyType
0,Cat,animal

Answer: The allergy in cats is of type animal.
Table: customers
Question: What is the name of customers who do not use Cash as payment method.
SQL query:
SELECT customer_name, payment_method 
 FROM customers 
 WHERE payment_method <> 'Cash'
Result:
,customer_name,payment_method
0,Tillman Ernser,Cheque

Answer: The customer Tillman Ernser has the payment method cheque.
Table: physician, trained_in, procedures
Question: Which physicians are trained in procedures that are more expensive than 5000?
SQL query:
SELECT T1.name AS "physician name", T3.cost AS "procedures cost" 
 FROM physician AS T1 JOIN trained_in AS T2 ON T1.employeeid = T2.physician JOIN procedures AS T3 ON T3.code = T2.treatment 
 WHERE T3.cost > 5000
Result:
,physician name,procedures cost
0,Christopher Turk,5600.0

Answer: The physician Christopher Turk has a procedure with a cost of 5600.
Table: music_festival, volume
Question: Please show the date of ceremony of the volumes that last more than 2 weeks on top.
SQL query:
SELECT T1.Date_of_ceremony AS "music_festival Date_of_ceremony", T2.Weeks_on_Top AS "volume Weeks_on_Top" 
 FROM music_festival AS T1 JOIN volume AS T2 ON T1.Volume = T2.Volume_ID 
 WHERE T2.Weeks_on_Top > 2
Result:
,volume Weeks_on_Top,music_festival Date_of_ceremony
0,3.0,18 February 2011

Answer: The festival at 18 February 2011 was on top for 3 weeks
Table: track
Question: Count the number of tracks.
SQL query:
SELECT COUNT(*) AS "count of track" 
 FROM track
Result:
,count of track
0,9

Answer: There are 9 tracks
Table: College
Question: What is the number of colleges with a student population greater than 15000?
SQL query:
SELECT COUNT(*) AS "count of College", enr 
 FROM College 
 WHERE enr > 15000
Result:
,enr,count of College
0,18000,3

Answer: The number of colleges is 3
Table: station
Question: For each city, what is the highest latitude for its stations?
SQL query:
SELECT city, MAX(lat) AS "maximum lat" 
 FROM station 
 GROUP BY city
Result:
,city,maximum lat
0,Mountain View,37.406940000000006

Answer: The city of Mountain View has a maximum latitude of 37.406940000000006.
Table: Student
Question: What are the first names for all students who are from the major numbered 600?
SQL query:
SELECT Fname, Major 
 FROM Student 
 WHERE Major = 600
Result:
,Fname,Major
0,Linda,600

Answer: The student with first name Linda is from the major 600.
Table: student, department
Question: Which department has the highest average student GPA, and what is the average gpa?
SQL query:
SELECT T2.dept_name AS "department dept_name", AVG(T1.stu_gpa) AS "average stu_gpa", T1.dept_code AS "student dept_code" 
 FROM student AS T1 JOIN department AS T2 ON T1.dept_code = T2.dept_code 
 GROUP BY T1.dept_code ORDER BY AVG(T1.stu_gpa) DESC
Result:
,average stu_gpa,student dept_code,department dept_name
0,3.1799999999999997,CIS,Computer Info. Systems

Answer: The department with code CIS and name Computer Info. Systems has a student with an average gpa of 3.1799999999999997
Table: member
Question: What are the card numbers of members from Kentucky?
SQL query:
SELECT card_number, Hometown 
 FROM member 
 WHERE Hometown LIKE '%Kentucky%'
Result:
,Hometown,Card_Number
0,"Louisville, Kentucky",LT98

Answer: The member with hometown in Louisville, Kentucky had the card number LT98.
Table: submission
Question: What is the average score of submissions?
SQL query:
SELECT AVG(Scores) AS "average Scores" 
 FROM submission
Result:
,average Scores
0,82.9

Answer: The average submission score is 82.9.
Table: Customers
Question: What is the phone and email for customer with first name Aniyah and last name Feest?
SQL query:
SELECT customer_phone, customer_email, customer_first_name, customer_last_name 
 FROM Customers 
 WHERE customer_first_name = 'Aniyah' AND customer_last_name = 'Feest'
Result:
,customer_email,customer_phone,customer_last_name,customer_first_name
0,fahey.dorian@example.com,(673)872-5338,Feest,Aniyah

Answer: The customer Aniyah Feest has the email fahey.dorian@example.com and the phone (673)872-5338.
Table: accelerator_compatible_browser, browser, web_client_accelerator
Question: When did the web accelerator 'CACHEbox' and browser 'Internet Explorer' become compatible?
SQL query:
SELECT T1.compatible_since_year AS "accelerator_compatible_browser compatible_since_year", T3.name AS "web_client_accelerator name", T2.name AS "browser name" 
 FROM accelerator_compatible_browser AS T1 JOIN browser AS T2 ON T1.browser_id = T2.id JOIN web_client_accelerator AS T3 ON T1.accelerator_id = T3.id 
 WHERE T3.name = 'CACHEbox' AND T2.name = 'Internet Explorer'
Result:
,browser name,web_client_accelerator name,accelerator_compatible_browser compatible_since_year
0,Internet Explorer,CACHEbox,1995

Answer: The browser Internet Explorer and the accelerator CACHEbox are compatible since 1995.
Table: Sportsinfo
Question: Which sport has most number of students on scholarship?
SQL query:
SELECT sportname, onscholarship 
 FROM Sportsinfo 
 WHERE onscholarship = 'Y' 
 GROUP BY sportname ORDER BY COUNT(*) DESC
Result:
,SportName,OnScholarship
0,Football,Y

Answer: Football has students on scholarship.
Table: BOOKINGS
Question: List the order dates of all the bookings.
SQL query:
SELECT Order_Date 
 FROM BOOKINGS
Result:
,Order_Date
0,2016-12-12 10:43:01

Answer: There was a booking in 2016-12-12 10:43:01
Table: Ref_Shipping_Agents
Question: What are the shipping agent names?
SQL query:
SELECT shipping_agent_name 
 FROM Ref_Shipping_Agents
Result:
,shipping_agent_name
0,UPS

Answer: The shipping agent names are UPS
Table: user_profiles
Question: Find the name and email of the users who have more than 1000 followers.
SQL query:
SELECT name, email, followers 
 FROM user_profiles 
 WHERE followers > 1000
Result:
,name,email,followers
0,Iron Man,ts@richest.com,6662425

Answer: Iron Man has 6662425 followers and email ts@richest.com.
Table: accounts, checking, savings
Question: Find the name and total checking and savings balance of the accounts whose savings balance is lower than corresponding checking balance.
SQL query:
SELECT T1.name AS "accounts name", T3.balance + T2.balance 
 FROM accounts AS T1 JOIN checking AS T2 ON T1.custid = T2.custid JOIN savings AS T3 ON T1.custid = T3.custid 
 WHERE T3.balance < T2.balance
Result:
,accounts name,T3.balance + T2.balance
0,Weeks,7060.0

Answer: The account with name Weeks has a total checking and savings balance of 7060
Table: products
Question: What are the names and prices of all products in the store?
SQL query:
SELECT name, price 
 FROM products
Result:
,Name,Price
0,Hard drive,240

Answer: The product named Hard drive is priced 240.
Table: journal, journal_committee
Question: What is the average sales of the journals that have an editor whose work type is 'Photo'?
SQL query:
SELECT AVG(T1.sales) AS "average sales", T2.work_type AS "journal_committee work_type" 
 FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID 
 WHERE T2.work_type = 'Photo'
Result:
,average sales,journal_committee work_type
0,4660.25,Photo

Answer: The journal that has a committee with a work type of "Photo", has average sales of 4660.25
Table: Player
Question: What is the average weight of all players?
SQL query:
SELECT AVG(weight) AS "average weight" 
 FROM Player
Result:
,average weight
0,168.38028933092224

Answer: The average weight is 168.38028933092224.
Table: school
Question: Take the average of the school enrollment.
SQL query:
SELECT AVG(Enrollment) AS "average Enrollment" 
 FROM school
Result:
,average Enrollment
0,1663.3333333333333

Answer: The average enrollment is 1663.3333333333333.
Table: company
Question: What is the minimum, maximum, and average market value for every company?
SQL query:
SELECT MIN(market_value) AS "minimum market_value", MAX(market_value) AS "maximum market_value", AVG(market_value) AS "average market_value" 
 FROM company
Result:
,average market_value,maximum market_value,minimum market_value
0,225.34,407.4,147.4

Answer: The average market value is 225.34, the maximum value is 407.4, and the minimum value is 147.4.
Table: weather
Question: For each zip code, select all those that have an average mean visiblity below 10.
SQL query:
SELECT zip_code 
 FROM weather 
 GROUP BY zip_code 
 HAVING AVG(mean_visibility_miles) < 10
Result:
,zip_code
0,94041

Answer: The zip code 94041 has an average mean visibility below 10.
Table: Customers
Question: What are the email addresses and date of births for all customers who have a first name of Carole?
SQL query:
SELECT email_address, date_of_birth, first_name 
 FROM Customers 
 WHERE first_name = 'Carole'
Result:
,first_name,date_of_birth,email_address
0,Carole,1998-12-15 13:24:40,everette.goyette@example.org

Answer: The customer with first name "Carole" and email "everette.goyette@example.org" was born in 1998-12-15 time 13:24:40.
Table: BOOKINGS
Question: Count the total number of bookings made.
SQL query:
SELECT COUNT(*) AS "count of BOOKINGS" 
 FROM BOOKINGS
Result:
,count of BOOKINGS
0,15

Answer: The total number of bookings made is 15
Table: tracks
Question: What is the unit price of the tune "Fast As a Shark"?
SQL query:
SELECT unit_price, name 
 FROM tracks 
 WHERE name = 'Fast As a Shark'
Result:
,name,unit_price
0,Fast As a Shark,0.99

Answer: The unit price of the tune "Fast As a Shark" is 0.99.
Table: stock
Question: Give the average quantity of stocks.
SQL query:
SELECT AVG(Quantity) AS "average Quantity" 
 FROM stock
Result:
,average Quantity
0,110.875

Answer: The average quantity is 110.875.
Table: drivers
Question: Find the forename and surname of drivers whose nationality is German?
SQL query:
SELECT forename, surname, nationality 
 FROM drivers 
 WHERE nationality = 'German'
Result:
,surname,forename,nationality
0,Heidfeld,Nick,German

Answer: The forename is Nick and the surname is Heidfeld, with a German nationality.
Table: Apartments
Question: Which apartments have type code "Flat"? Give me their apartment numbers.
SQL query:
SELECT apt_number, apt_type_code 
 FROM Apartments 
 WHERE apt_type_code = 'Flat'
Result:
,apt_number,apt_type_code
0,Suite 645,Flat

Answer: The apartment with number Suite 645 is of type flat.
Table: Financial_transactions
Question: What is the average, minimum, maximum, and total transaction amount?
SQL query:
SELECT AVG(transaction_amount) AS "average transaction_amount", MIN(transaction_amount) AS "minimum transaction_amount", MAX(transaction_amount) AS "maximum transaction_amount", SUM(transaction_amount) AS "sum of transaction_amount" 
 FROM Financial_transactions
Result:
,sum of transaction_amount,average transaction_amount,maximum transaction_amount,minimum transaction_amount
0,19935.839999999997,1329.0559999999998,1979.66,368.46

Answer: The average is 1329.0559999999998, the minimum is 368.46, the maximum is 1979.66, and the total is 19935.839999999997 of the transaction amount
Table: Player
Question: What are the names and number of hours spent training for each player who trains for less than 1500 hours?
SQL query:
SELECT pName, HS 
 FROM Player 
 WHERE HS < 1500
Result:
,HS,pName
0,1200,Andrew

Answer: The player named Andrew has trained for 1200 hours
Table: available_policies
Question: What are the customer phone numbers under the policy "Life Insurance"?
SQL query:
SELECT customer_phone, policy_type_code 
 FROM available_policies 
 WHERE policy_type_code = 'Life Insurance'
Result:
,Customer_Phone,policy_type_code
0,+16(2)5838999222,Life Insurance

Answer: The customer phone number under the policy type Life Insurance is +16(2)5838999222
Table: COURSE
Question: What are the course names for courses taught on MTW?
SQL query:
SELECT CName, Days 
 FROM COURSE 
 WHERE Days = 'MTW'
Result:
,Days,CName
0,MTW,COMPUTER LITERACY

Answer: Course "COMPUTER LITERACY" was taught on MTW.
Table: technician
Question: What is the number of technicians?
SQL query:
SELECT COUNT(*) AS "count of technician" 
 FROM technician
Result:
,count of technician
0,10

Answer: There are 10 technicians.
Table: papers
Question: How many papers are published in total?
SQL query:
SELECT COUNT(*) AS "count of papers" 
 FROM papers
Result:
,count of papers
0,7

Answer: The number of published papers is 7
Table: company
Question: Show minimum, maximum, and average market value for all companies.
SQL query:
SELECT MIN(market_value) AS "minimum market_value", MAX(market_value) AS "maximum market_value", AVG(market_value) AS "average market_value" 
 FROM company
Result:
,average market_value,maximum market_value,minimum market_value
0,225.34,407.4,147.4

Answer: The average market value is 225.34, the maximum is 407.4 and the minimum is 147.4.
Table: Companies
Question: Count the number of companies.
SQL query:
SELECT COUNT(*) AS "count of Companies" 
 FROM Companies
Result:
,count of Companies
0,19

Answer: There are 19 companies
Table: Invoices
Question: How many invoices do we have?
SQL query:
SELECT COUNT(*) AS "count of Invoices" 
 FROM Invoices
Result:
,count of Invoices
0,15

Answer: We have 15 invoices.
Table: candidate
Question: Which poll resource provided the most number of candidate information?
SQL query:
SELECT poll_source 
 FROM candidate 
 GROUP BY poll_source ORDER BY COUNT(*) DESC
Result:
,Poll_Source
0,Newsweek Poll

Answer: The most number of candidate information came from the poll source named Newsweek Poll
Table: club, member_of_club, student
Question: Who is the president of the club "Bootup Baltimore"? Give me the first and last name.
SQL query:
SELECT t3.fname AS "student fname", t3.lname AS "student lname", t2.position AS "member_of_club position", t1.clubname AS "club clubname" 
 FROM club AS t1 JOIN member_of_club AS t2 ON t1.clubid = t2.clubid JOIN student AS t3 ON t2.stuid = t3.stuid 
 WHERE t1.clubname = 'Bootup Baltimore' AND t2.position = 'President'
Result:
,club clubname,student fname,student lname,member_of_club position
0,Bootup Baltimore,Linda,Smith,President

Answer: The name of President of the club Bootup Baltimore is Linda Smith