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Simulation Exam Bible 1Z0-007 Q&a
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Web Demo:
1.Evaluate this SQL statement:
SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME
FROM EMPLOYEES e, DEPARTMENTS d
WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID;
In the statement, which capabilities of a SELECT statement are performed?
A.selection, projection, join
B.difference, projection, join
C.selection, intersection, join
D.intersection, projection, join
E.difference, projection, product
Answer: A
2. What does the FORCE option for creating a view do?
A.creates a view with constraints
B.creates a view even if the underlying parent table has constraints
C.creates a view in another schema even if you don't have privileges
D.creates a view regardless of whether or not the base tables exist
Answer: D
2. The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
The registrar requested a report listing the students' grade point averages (GPA) sorted from highest grade point average to lowest.
Which statement produces a report that displays the student ID and GPA in the sorted order requested by the registrar?
A.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa ASC;
B.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa ASC;
C.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa;
D.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa;
E.SELECT student_id, gpa
FROM student_grades
SORT ORDER BY gpa DESC;
F.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa DESC;
Answer: F
3. The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
STREET_ADDRESS VARCHAR2(150)
CITY_ADDRESS VARCHAR2(50)
STATE_ADDRESS VARCHAR2(50)
PROVINCE_ADDRESS VARCHAR2(50)
COUNTRY_ADDRESS VARCHAR2(50)
POSTAL_CODE VARCHAR2(12)
CUSTOMER_PHONE VARCHAR2(20)
The CUSTOMER_ID column is the primary key for the table.
You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?
A.COUNT(UPPER(country_address))
B.COUNT(DIFF(UPPER(country_address)))
C.COUNT(UNIQUE(UPPER(country_address)))
D.COUNT DISTINCT UPPER(country_address)
E.COUNT(DISTINCT (UPPER(country_address)))
Answer: E
4. In which three cases would you use the USING clause? (Choose three.)
A.You want to create a nonequijoin.
B.The tables to be joined have multiple NULL columns.
C.The tables to be joined have columns of the same name and different data types.
D.The tables to be joined have columns with the same name and compatible data types.
E.You want to use a NATURAL join, but you want to restrict the number of columns in the join condition.
Answer: CDE
5. Evaluate this SQL statement:
SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME
FROM EMPLOYEES e, DEPARTMENTS d
WHERE e.DEPARTMENT_ID = d.DEPARTMENT_ID;
In the statement, which capabilities of a SELECT statement are performed?
A.selection, projection, join
B.difference, projection, join
C.selection, intersection, join
D.intersection, projection, join
E.difference, projection, product
Answer: A
6. Click the Exhibit button and examine the data in the EMPLOYEES table.
Which three subqueries work? (Choose three.)
A.SELECT *
FROM employees
where salary > (SELECT MIN(salary)
FROM employees
GROUP BY department_id);
B.SELECT *
FROM employees
WHERE salary = (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
C.SELECT distinct department_id
FROM employees
WHERE salary > ANY (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
D.SELECT department_id
FROM employees
WHERE salary > ALL (SELECT AVG(salary)
FROM employees
GROUP BY department_id);
E.SELECT last_name
FROM employees
WHERE salary > ANY (SELECT MAX(salary)
FROM employees
GROUP BY department_id);
F.SELECT department_id
FROM employees
WHERE salary > ALL (SELECT AVG(salary)
FROM employees
GROUP BY AVG(SALARY));
Answer: CDE
7. A SELECT statement can be used to perform these three functions:
1. Choose rows from a table.
3.The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
STREET_ADDRESS VARCHAR2(150)
CITY_ADDRESS VARCHAR2(50)
STATE_ADDRESS VARCHAR2(50)
PROVINCE_ADDRESS VARCHAR2(50)
COUNTRY_ADDRESS VARCHAR2(50)
POSTAL_CODE VARCHAR2(12)
CUSTOMER_PHONE VARCHAR2(20)
The CUSTOMER_ID column is the primary key for the table.
You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?
A.COUNT(UPPER(country_address))
B.COUNT(DIFF(UPPER(country_address)))
C.COUNT(UNIQUE(UPPER(country_address)))
D.COUNT DISTINCT UPPER(country_address)
E.COUNT(DISTINCT (UPPER(country_address)))
Answer: E
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1z0-001 - 9i-Internet-Application-Developer-Introduction-to-Oracle:-SQL-and-PL/SQL
1z0-271 - Hyperion-Financial-Management-9.3-Administrator-I&II
1z0-402 - Oracle-Enterprise-Linux:-Fundamentals
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