A. CREATE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status VARCHAR2 (10), date_ordered DATE = SYSDATE);
B. CREATE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status VARCHAR2 (10), date_ordered DATE DEFAULT SYSDATE);
C. CREATE OR REPLACE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status VARCHAR2 (10), date_ordered DATE DEFAULT SYSDATE);
D. CREATE OR REPLACE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status NUMBER (10), date_ordered DATE = SYSDATE);
E. CREATE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status NUMBER (10), date_ordered DATE = SYSDATE);
F. CREATE TABLE orders ( order_id NUMBER (10), customer_id NUMBER (8), order_status NUMBER (10), date_ordered DATE DEFAULT SYSDATE);

A. Column definitions cannot be altered to add DEFAULT values.
B. A change to the DEFAULT value affects only subsequent insertions to the table.
C. Column definitions cannot be altered at add DEFAULT values for columns with a NUMBER data type.
D. All the rows that have a NULL value for the SALARY column will be updated with the value 5000.

A. listing the employees who earn the same amount of commission as employee 3
B. finding the total commission earned by the employees in department 10
C. finding the number of employees who earn a commission that is higher than the average commission of the company
D. listing the departments whose average commission is more that 600
E. listing the employees who do not earn commission and who are working for department 20 in descending order of the employee ID 
F. listing the employees whose annual commission is more than 6000

A. CREATE PUBLIC SYNONYM EDL_VU ON emp_dept_loc_vu
B. CREATE PUBLIC SYNONYM EDL:VU FOR mary (emp_dept_loc_vu);
C. CREATE PUBLIC SYNONYM EDL_VU FOR emp _dept_loc_vu;
D. CREATE SYNONYM EDL_VU ON emp_dept_loc_vu FOR EACH USER;
E. CREATE SYNONYM EDL_VU FOR EACH USER ON emp_dept_loc_vu
F. CREATE PUBLIC SYNONYM EDL_VU ON emp_dept_loc_vu FOR ALL USERS;

A. A single row subquery can retrieve data from only one table.
B. A SQL query statement cannot display data from table B that is referred to in its subquery, unless table B is included in the main query's FROM clause.
C. A SQL query statement can display data from table B that is referred to in its subquery, without including B in its own FROM clause.
D. A single row subquery can retrieve data from more than one table.
E. A single row subquery cannot be used in a condition where the LIKE operator is used for comparison.
F. A multiple-row subquery cannot be used in an INSERT statement to insert multiple rows at a time.

A. &
B. ACCEPT
C. PROMPT
D. &&

A. UNIQUE
B. NOT NULL
C. CHECK
D. PRIMARY KEY
E. FOREIGN KEY

A. INSERT INTO employees (employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01/01/01);
B. INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01 january 01');
C. INSERT INTO employees(employee_id, first_name, last_name, Hire_date) VALUES (1000, 'John', 'smith', To_ date ('01/01/01));
D. INSERT INTO employees(employee_id, first_name, last_name, hire_date) VALUES (1000, 'John', 'smith','01-Jan-01');

A. You want to identify the most senior employee in the company.
B. You want to find the manager supervising the largest number of employees.
C. You want to identify the person who makes the highest salary for all employees.
D. You want to rank the top three sales representatives who have sold the maximum number of products.

A. You cannot use IN operator in a condition that involves an outerjoin.
B. You use (+) on both sides of the WHERE condition to perform an outerjoin.
C. You use (*) on both sides of the WHERE condition to perform an outerjoin.
D. You use an outerjoin to see only the rows that do not meet the join condition.
E. In the WHERE condition, you use (+) following the name of the column in the table without matching rows, to perform an outerjoin.
F. You cannot link a condition that is involved in an outerjoin to another condition by using the OR operator.

A. SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN department d ON (e.department_id = d.department_id);
B. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.managaer_id = m.employee_id) LEFT OUTER JOIN department d ON (e.department_id = d.department_id);
C. SELECT e.last_name, m.last_name, department_name FROM employees e RIGT OUTER JOIN employees m on ( e.manager_id = m.employee_id) FULL OUTER JOIN department d ON (e.department_id = d.department_id);
D. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGT OUTER JOIN department d ON (e.department_id = d.department_id);
E. SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id)
F. SELECT last_name, manager_id, department_name FROM employees e JOIN department d ON (e.department_id = d.department_id);