Figuer:
CUSTOMER In Table CUSTOMER, CID Is the Primary Key (Customer ID)

question 25

Multiple Choice

Figuer:
CUSTOMER $$\begin{array} { | l | l | l | l | l | } \hline \text { CID } & \text { CNAME } & \text { AGE } & \text { RESID\_CITY } & \text { BIRTHPLACE } \\\hline 10 & \text { BLACK } & 40 & \text { ERIE } & \text { TAMPA } \\\hline 20 & \text { GREEN } & 25 & \text { CARY } & \text { ERIE } \\\hline 30 & \text { JONES } & 30 & \text { HEMET } & \text { TAMPA } \\\hline 40 & \text { MARTIN } & 35 & \text { HEMET } & \text { TAMPA } \\\hline 50 & \text { SIMON } & 22 & \text { ERIE } & \text { ERIE } \\\hline 60 & \text { VERNON } & 60 & \text { CARY } & \text { CARY } \\\hline\end{array}$$ In table CUSTOMER, CID is the primary key (Customer ID) .
RENTALS $$\begin{array} { | l | l | l | l | l | l | } \hline \text { CID } & \text { MAKE } & \text { DATE\_OUT } & \text { PICKUP } & \text { RETURN } & \text { RTN } \\\hline 10 & \text { FORD } & 10 - 0 c t - 1994 & \text { CARY } & \text { CARY } & 1 \\\hline 10 & \text { GM } & 01 - N o v - 1995 & \text { TAMPA } & \text { CARY } & 2 \\\hline 10 & \text { FORD } & 01 - J a n - 1995 & \text { ERIE } & \text { ERIE } & 3 \\\hline 20 & \text { NISSAN } & 07 - J u 1 - 1994 & \text { TAMPA } & \text { TAMPA } & 4 \\\hline 30 & \text { FORD } & 01 - J u l - 1995 & \text { CARY } & \text { ERIE } & 5 \\\hline 30 & \text { GM } & 01 - A u g - 1995 & \text { ERIE } & \text { ERIE } & 6 \\\hline 40 & \text { FORD } & 01 - A u g - 1994 & \text { CARY } & \text { ERIE } & 7 \\\hline 50 & \text { GM } & 01 - S e p - 19 95 & \text { ERIE } & \text { CARY } & 8 \\\hline 70 & \text { TOYOTA } & 02 - S e p - 1995 & \text { RENO } & \text { RENO } & 9 \\\hline\end{array}$$ 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 $$\begin{array} { | l | l | } \hline \text { MAKE } & \text { COST } \\\hline \text { FORD } & 30 \\\hline \text { GM } & 40 \\\hline \text { NISSAN } & 30 \\\hline \text { TOYOTA } & 20 \\\hline \text { VOLVO } & 50 \\\hline\end{array}$$ RENTCOST shows the base cost of renting a given MAKE for one day.
CITYADJ $$\begin{array} { | l | l | } \hline \text { CITY } & \text { FACTOR } \\\hline \text { CARY } & 1 \\\hline \text { ERIE } & 1.1 \\\hline \text { RENO } & 0.9 \\\hline \text { TAMPA } & 0.8 \\\hline\end{array}$$ 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 $$\begin{array} { | l | l | } \hline \text { RTN } & \text { DAYS } \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 2 \\\hline 4 & 2 \\\hline 5 & 4 \\\hline 6 & 2 \\\hline 7 & 3 \\\hline 8 & 1 \\\hline\end{array}$$ 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.
-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

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