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Oracle Database SQL Sample Questions (Q255-Q260):

NEW QUESTION # 255
You must find the number of employees whose salary is lower than employee 110.
Which statement fails to do this?

A. SELECT COUNT (*)
FROM employees
JOIN employees a
ON e. salary< a. salary
WHERE a. employee_ id= 110;
B. SELECT COUNT (*)
FROM employees e
JOIN (SELECT salary FROM employees WHERE employee_ id= 110) a
ON e. salary< a. salary;
C. SELECT COUNT (* )
FROM employees e
WHERE e. salary < (SELECT a. salary FROM employees a WHERE e. employee_ id = 110);
D. SELECT COUNT (* )
FROM employees
WHERE salary < (SELECT salary FROM employees WHERE employee.id =
110) ;

Answer: C

NEW QUESTION # 256
You execute this query:

What is the result?

A. It generates an error.
B. It executes successfully but does not return any result.
C. It returns the date for the first Monday of the next month.
D. It returns the date for the last Monday of the current month.

Answer: C

NEW QUESTION # 257
View the Exhibit and examine the description of the EMPLOYEES table.

You want to calculate the total remuneration for each employee. Total remuneration is the sum of the annual salary and the percentage commission earned for a year. Only a few employees earn commission.
Which SQL statement would you execute to get the desired output?

A. SELECT first_name, salary, salary*12 + NVL(salary,0)*commission_pct, "Total"FROM EMPLOYEES;
B. SELECT first_name, salary, salary*12+(salary*NVL2 (commission_pct,
salary,salary+commission_pct))"Total"FROM EMPLOYEES;
C. SELECT first_name, salary (salary + NVL (commission_pct, 0)*salary)*12 "Total"FROM EMPLOYEES;
D. SELECT first_name, salary, salary*12+salary*commission_pct "Total"FROM EMPLOYEES;

Answer: B

NEW QUESTION # 258
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.

The CUSTOMERS table contains the current location of all currently active customers. The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?

A. UNION ALL
B. INTERSECT
C. UNION
D. MINUS

Answer: D

NEW QUESTION # 259
View the Exhibit and examine the structure of the PRODUCT_INFORMATION table.

PRODUCT_ID column is the primary key.
You create an index using this command:
SQL > CREATE INDEX upper_name_idx
ON product_information(UPPER(product_name));
No other indexes exist on the PRODUCT_INFORMATION table.
Which query would use the UPPER_NAME_IDX index? (Choose the best answer.)

A. SELECT product_idFROM product_informationWHERE UPPER(product_name) IN ('LASERPRO',
'CABLE');
B. SELECT UPPER(product_name)FROM product_information;
C. SELECT product_id, UPPER(product_name)FROM product_informationWHERE
UPPER(product_name)='LASERPRO' OR list_price > 1000;
D. SELECT UPPER(product_name)FROM product_informationWHERE product_id = 2254;

Answer: A

NEW QUESTION # 260
......

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