Use the following tables (with partial data shown) for a car rental database to answer questions 20-25: SEE QUESTIONS BELOW.

CUSTOMER

CID	CNAME	AGE	RESID_CITY	BIRTHPLACE
10	BLACK	40	ERIE	TAMPA
20	GREEN	25	CARY	ERIE
30	JONES	30	HEMET	TAMPA
40	MARTIN	35	HEMET	TAMPA
50	SIMON	22	ERIE	ERIE
60	VERNON	60	CARY	CARY

In table CUSTOMER, CID is the primary key (Customer ID).

RENTALS

CID	MAKE	DATE_OUT	PICKUP	RETURN	RTN
10	FORD	10-OCT-1994	CARY	CARY	1
10	GM	01-NOV-1995	TAMPA	CARY	2
10	FORD	01-JAN-1995	ERIE	ERIE	3
20	NISSAN	07-JUL-1994	TAMPA	TAMPA	4
30	FORD	01-JUL-1995	CARY	ERIE	5
30	GM	01-AUG-1995	ERIE	ERIE	6
40	FORD	01-AUG-1994	CARY	ERIE	7
50	GM	01-SEP-1995	ERIE	CARY	8
70	TOYOTA	02-SEP-1995	RENO	RENO	9

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

MAKE	COST
FORD	30
GM	40
NISSAN	30
TOYOTA	20
VOLVO	50

RENTCOST shows the base cost of renting a given MAKE for one day.

CITYADJ

CITY	FACTOR
CARY	1
ERIE	1.1
RENO	0.9
TAMPA	0.8

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

RTN	DAYS
1	1
2	3
3	2
4	2
5	4
6	2
7	3
8	1

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