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Exam
: Oracle 1Z0-007
Title
:
Introduction to Oracle9i:
SQL
Version : Demo
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1. 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. What are two reasons to create synonyms? (Choose two.)
A.You have too many tables.
B.Your tables are too long.
C.Your tables have difficult names.
D.You want to work on your own tables.
E.You want to use another schema's tables.
F.You have too many columns in your tables.
Answer: CE
3. 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
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FROM student_grades
SORT ORDER BY gpa DESC;
F.SELECT student_id, gpa
FROM student_grades
ORDER BY gpa DESC;
Answer: F
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. 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
6. Click the Exhibit button and examine the data in the EMPLOYEES table.
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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));
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Answer: CDE
7. A SELECT statement can be used to perform these three functions:
1. Choose rows from a table.
2. Choose columns from a table.
3. Bring together data that is stored in different tables by creating a link between them.
Which set of keywords describes these capabilities?
A.difference, projection, join
B.selection, projection, join
C.selection, intersection, join
D.intersection, projection, join
E.difference, projection, product
Answer: B
8. 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
9. Evaluate this SQL statement:
SELECT e.employee_id, (.15* e.salary) + (.5 * e.commission_pct)
+ (s.sales_amount * (.35 * e.bonus)) AS CALC_VALUE
FROM employees e, sales s
WHERE e.employee_id = s.emp_id;
What will happen if you remove all the parentheses from the calculation?
A.The value displayed in the CALC_VALUE column will be lower.
B.The value displayed in the CALC_VALUE column will be higher.
C.There will be no difference in the value displayed in the CALC_VALUE column.
D.An error will be reported.
Answer: C
10. Which SQL statement generates the alias Annual Salary for the calculated column SALARY*12?
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A.SELECT ename, salary*12 'Annual Salary'
FROM employees;
B.SELECT ename, salary*12 "Annual Salary"
FROM employees;
C.SELECT ename, salary*12 AS Annual Salary
FROM employees;
D.SELECT ename, salary*12 AS INITCAP("ANNUAL SALARY")
FROM employees
Answer: B
11. Evaluate this SQL statement:
SELECT ename, sal, 12*sal+100
FROM emp;
The SAL column stores the monthly salary of the employee. Which change must be made to the above
syntax to calculate the annual compensation as "monthly salary plus a monthly bonus of $100, multiplied by
12"?
A.No change is required to achieve the desired results.
B.SELECT ename, sal, 12*(sal+100)
FROM emp;
C.SELECT ename, sal, (12*sal)+100
FROM emp;
D.SELECT ename, sal+100,*12
FROM emp;
Answer: B
12. The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER(4) NOT NULL
CUSTOMER_NAME VARCHAR2(100) NOT NULL
CUSTOMER_ADDRESS VARCHAR2(150)
CUSTOMER_PHONE VARCHAR2(20)
You need to produce output that states "Dear Customer customer_name, ".
The customer_name data values come from the CUSTOMER_NAME column in the CUSTOMERS table.
Which statement produces this output?
A.SELECT dear customer, customer_name,
B.SELECT "Dear Customer", customer_name || ','
FROM customers;
C.SELECT 'Dear Customer ' || customer_name ','
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FROM customers;
D.SELECT 'Dear Customer ' || customer_name || ','
FROM customers;
E.SELECT "Dear Customer " || customer_name || ","
FROM customers;
F.SELECT 'Dear Customer ' || customer_name || ',' ||
FROM customers;
Answer: D
13. Which two are attributes of iSQL*Plus? (Choose two.)
A.iSQL*Plus commands cannot be abbreviated.
B.iSQL*Plus commands are accessed from a browser.
C.iSQL*Plus commands are used to manipulate data in tables.
D.iSQL*Plus commands manipulate table definitions in the database.
E.iSQL*Plus is the Oracle proprietary interface for executing SQL statements.
Answer: BE
14. Which is an iSQL*Plus command?
A.INSERT
B.UPDATE
C.SELECT
D.DESCRIBE
E.DELETE
F.RENAME
Answer: D
15. Which are iSQL*Plus commands? (Choose all that apply.)
A.INSERT
B.UPDATE
C.SELECT
D.DESCRIBE
E.DELETE
F.RENAME
Answer: D
16. Which two statements are true about constraints? (Choose two.)
A.The UNIQUE constraint does not permit a null value for the column.
B.A UNIQUE index gets created for columns with PRIMARY KEY and UNIQUE constraints.
C.The PRIMARY KEY and FOREIGN KEY constraints create a UNIQUE index.
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D.The NOT NULL constraint ensures that null values are not permitted for the column.
Answer: BD
17. Which three statements correctly describe the functions and use of constraints? (Choose three.)
A.Constraints provide data independence.
B.Constraints make complex queries easy.
C.Constraints enforce rules at the view level.
D.Constraints enforce rules at the table level.
E.Constraints prevent the deletion of a table if there are dependencies.
F.Constraints prevent the deletion of an index if there are dependencies.
Answer: CDE
18. Which SQL statement defines a FOREIGN KEY constraint on the DEPTNO column of the EMP table?
A.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) NOT NULL,
CONSTRAINT emp_deptno_fk FOREIGN KEY deptno
REFERENCES dept deptno);
B.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2)
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
C.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) NOT NULL,
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno)
FOREIGN KEY (deptno));
D.CREATE TABLE EMP
(empno NUMBER(4),
ename VARCHAR2(35),
deptno NUMBER(7,2) FOREIGN KEY
CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
Answer: B
19. Which view should a user query to display the columns associated with the constraints on a table
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owned by the user?
A.USER_CONSTRAINTS
B.USER_OBJECTS
C.ALL_CONSTRAINTS
D.USER_CONS_COLUMNS
E.USER_COLUMNS
Answer: D
20. You need to design a student registration database that contains several tables storing academic
information.
The STUDENTS table stores information about a student. The STUDENT_GRADES table stores
information about the student's grades. Both of the tables have a column named STUDENT_ID. The
STUDENT_ID column in the STUDENTS table is a primary key.
You need to create a foreign key on the STUDENT_ID column of the STUDENT_GRADES table that points
to the STUDENT_ID column of the STUDENTS table. Which statement creates the foreign key?
A.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
CONSTRAINT student_id_fk REFERENCES (student_id)
FOREIGN KEY students(student_id));
B.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
student_id_fk FOREIGN KEY (student_id)
REFERENCES students(student_id));
C.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
gpa NUMBER(4,3),
CONSTRAINT FOREIGN KEY (student_id)
REFERENCES students(student_id));
D.CREATE TABLE student_grades
(student_id NUMBER(12),
semester_end DATE,
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gpa NUMBER(4,3),
CONSTRAINT student_id_fk FOREIGN KEY (student_id)
REFERENCES students(student_id));
Answer: D
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