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Use the following tables (with partial data shown) for a car rental database to answer questions 20-25: CUSTOMER In table CUSTOMER, CID is the...
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Use the following tables (with partial data shown) for a car rental database to answer questions 20-25:
CUSTOMER In table CUSTOMER, CID is the primary key (Customer ID). RENTALS In the table RENTALS, RTN provides the rental number (the primary key), CID is the customer's unique id, PICKUP is the city where the car was picked up, and Return is the city where the car was returned. RENTCOST RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ If the return city of table RENTALS is the one listed in table CITYADJ, the cost of the rental is multiplied by FACTOR and by DAYS shown in table RENTLENGTH below. RENTLENGTH RENTLENGTH shows the number of days for the rental number (RTN) shown in table RENTALS. In a database used in reality, this table would be merged with the RENTALS table.
20. SELECT DISTINCT CID, CNAME FROM CUSTOMER WHERE CID IN (SELECT CID FROM RENTALS WHERE MAKE IN ('FORD', 'TOYOTA')) The CNAMEs shown by the execution of this query are: A. BLACK B. BLACK, JONES C. BLACK, JONES, MARTIN D. BLACK, JONES, MARTIN, VERNON
21. SELECT DISTINCT CUSTOMER.CID, CNAME FROM CUSTOMER, RENTALS, RENTCOST WHERE CUSTOMER.CID = RENTALS.CID AND RENTALS.MAKE = RENTCOST.MAKE AND NOT EXISTS (SELECT * FROM RENTALS R, RENTCOST C WHERE R.MAKE = C.MAKE AND RENTALS.CID = R.CID AND RENTCOST.COST <> C.COST) The meaning of this query is: A. List all customers with more than one car make rented B. List all customers with one or more rentals for which the cost of each car make rented is the same C. List all customers who have only rented one make D. None of the above
22. SELECT MAKE FROM RENTALS, CUSTOMER WHERE RENTALS.CID = CUSTOMER.CID AND RESID_CITY = 'HEMET' GROUP BY MAKE HAVING COUNT (DISTINCT RENTALS.CID) = (SELECT COUNT(*) FROM CUSTOMER WHERE RESID_CITY = 'HEMET') The execution of this query produces the following number of rows: A. 0 B. 1 C. 2 D. 3
23. SELECT MAKE FROM RENTALS, CUSTOMER WHERE RENTALS.CID = CUSTOMER.CID AND RESID_CITY = 'HEMET' GROUP BY MAKE HAVING COUNT (DISTINCT RENTALS.CID) = (SELECT COUNT(*) FROM CUSTOMER WHERE RESID_CITY = 'HEMET') The meaning of this query is the following: A. List all makes of cars rented to customers residing in Hemet B. List all makes of cars rented to at least one customer residing in Hemet C. List all makes of cars rented to all customers residing in Hemet D. None of the above
24. SELECT CID, CNAME FROM CUSTOMER WHERE 0 = (SELECT COUNT(*) FROM RENTALS WHERE CUSTOMER.CID = RENTALS.CID) What is the interpretation of this query? A. List the customers who do not have rentals B. List the customers who have one rental C. List the customers who have 0 or more rentals D. List the customers who have 1 or more rentals
25. SELECT CNAME, DATE_OUT, RTN FROM CUSTOMER, RENTALS WHERE CUSTOMER.CID = RENTALS.CID AND BIRTHPLACE IN ('ERIE', 'CARY') AND EXISTS (SELECT * FROM RENTCOST WHERE COST < 40 AND RENTALS.MAKE= RENTCOST.MAKE) The CNAMEs shown by the execution of this query are: A. SIMON B. GREEN, BLACK, SIMON C. GREEN D. GREEN, SIMON
