QUESTION NO: 1
Examine the data in the EMPLOYEES and DEPARTMENTS tables.
EMPLOYEES
LAST_NAME DEPARTMENT_ID SALARY
Getz 10 3000
Davis 20 1500
King 20 2200
Davis 30 5000
Kochhar 5000
DEPARTMENTS
DEPARTMENT_ID DEPARTMENT_NAME
10 Sales
20 Marketing
30 Accounts
40 Administration
You want to retrieve all employees, whether or not they have matching departments in
the departments table. Which query would you use?
A. SELECT last_name, department_name
FROM employees , departments(+);
B. SELECT last_name, department_name
FROM employees JOIN departments (+);
C. SELECT last_name, department_name
FROM employees(+) e JOIN departments d
ON (e.department_id = d.department_id);
D. SELECT last_name, department_name
FROM employees e
RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);
E. SELECT last_name, department_name
FROM employees(+) , departments
ON (e.department_id = d.department_id);
F. SELECT last_name, department_name
FROM employees e LEFT OUTER
JOIN departments d ON (e.department_id = d.department_id);
Answer: F
Explanation:
Answer F is correct. This query shows correct syntax to retrieve all employees, whether or not
they have matching departments in the department table. Oracle9i extends its compliance with
ANSI/ISO by supporting that standard’s requirements for outer join syntax and semantics.
Incorrect Answers
A: This query uses “+” to create outer join as it was in Oracle8i, but it requires also usage of
WHERE clause in SELECT statement.
B: The JOIN clause cannot be used with in conjunction with “+”: syntax is incorrect.
C: The JOIN clause cannot be used with in conjunction with “+”: syntax is incorrect.
D: This statement requires LEFT OUTER JOIN, not RIGHT OUTER JOIN.
E: This query uses incorrect syntax with “+” and ON to create outer join.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 112-114
Chapter 3: Advanced Data Selection in Oracle
QUESTION NO: 2
Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
Which three statements inserts a row into the table? (Choose three)
A. INSERT INTO employees
VALUES ( NULL, ‘John’,‘Smith’);
B. INSERT INTO employees( first_name, last_name)
VALUES(‘John’,‘Smith’);
C. INSERT INTO employees
VALUES (‘1000’,‘John’,NULL);
D. INSERT INTO employees(first_name,last_name, employee_id)
VALUES ( 1000, ‘John’,‘Smith’);
E. INSERT INTO employees (employee_id)
VALUES (1000);
F. INSERT INTO employees (employee_id, first_name, last_name)
VALUES ( 1000, ‘John’,‘’);
Answer: C, E, F
Explanation:
Since EMPLOYEE_ID column is used as primary key, it cannot be NULL, so only INSERT
statements in C, E and F are correct. You can insert the row with NULL LAST_NAME as in
answer C, or only the row with EMPLOYEE_ID as in answer E, or the row with empty
LAST_NAME column.
Incorrect Answers
A: This answer is incorrect because a primary key cannot be NULL.
B: INSERT statement does not contain primary key value at all, so this answer needs to be
eliminated as correct one.
D: This statement shows incorrect order of columns of row which needs to be inserted into
the table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 260-267
Chapter 6: Manipulating Oracle Data
QUESTION NO: 3
You need to give the MANAGER role the ability to select from, insert into, and modify
existing rows in the STUDENT_GRADES table. Anyone given this MANAGER role
should be able to pass those privileges on to others.
Which statement accomplishes this?
A. GRANT select, insert, update
ON student_grades
TO manager;
B. GRANT select, insert, update
ON student_grades
TO ROLE manager;
C. GRANT select, insert, modify
ON student_grades
TO manager
WITH GRANT OPTION;
D. GRANT select, insert, update
ON student_grades
TO manager
WITH GRANT OPTION;
E. GRANT select, insert, update
ON student_grades
TO ROLE manager
WITH GRANT OPTION;
F. F.GRANT select, insert, modify
ON student_grades
TO ROLE manager
WITH GRANT OPTION;
Answer: D
Explanation:
This answer provides correct syntax of GRANT command to give the MANAGER role all
asked privileges. Clause WITH GRANT OPTION will allow this role to pass those privileges
on to others.
Incorrect Answers
A: This statement would be correct if it included WITH GRANT OPTION clause to allow
this role to pass those privileges on to others.
B: This statement uses incorrect clause TO ROLE.
C: There is no option with name MODIFY in the GRANT command.
E: This statement uses incorrect clause TO ROLE.
F: There is no option with name MODIFY in the GRANT command. And this statement also
uses incorrect clause TO ROLE.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 356-365
Chapter 8: User Access Control in Oracle
QUESTION NO: 4
Examine the data in the EMPLOYEES table:
LAST_NAME DEPARTMENT_ID SALARY
Getz 10 3000
Davis 20 1500
King 20 2200
Davis 30 5000
…
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: C, D, E
Explanation:
These answers show correct syntax, because they use ANY and ALL keywords for convert
multi-row output of sub-query to one-row result.
Incorrect Answers
A: This SELECT statement is incorrect because of multi-row return of sub-query: it will
return minimal salary for EACH department.
B: This SELECT statement is incorrect because of multi-row return of sub-query: it will
return average salary for EACH department.
F: This SELECT statement is incorrect because GROUP BY clause cannot contain functions,
like AVG(), MIN(), MAX() and so on.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 145-156
Chapter 4: Subqueries
QUESTION NO: 5
The database administrator of your company created a public synonym called HR for
the HUMAN_RESOURCES table of the GENERAL schema, because many users
frequently use this table.
As a user of the database, you created a table called HR in your schema. What happens
when you execute this query?
SELECT *
FROM HR;
A. You obtain the results retrieved from the public synonym HR created by the database
administrator.
B. You obtain the results retrieved from the HR table that belongs to your schema.
C. You get an error message because you cannot retrieve from a table that has the same
name as a public synonym.
D. You obtain the results retrieved from both the public synonym HR and the HR table
that belongs to your schema, as a Cartesian product.
E. You obtain the results retrieved from both the public synonym HR and the HR table
that belongs to your schema, as a FULL JOIN.
Answer: B
Explanation:
By executing this query you will extract data from the HR table in your own schema, it will
not work with HR synonym for the HUMAN_RESOURCES table of the GENERAL schema.
Incorrect Answers
A: The results will be retrieved from the table in your own schema, not from the GENERAL
schema, using synonym HR.
C: There is no error: data from the table in your own schema will be retrieved by this query.
D: This query will not generate Cartesian product from both tables.
E: This query will not retrieve data from both tables as a FULL JOIN.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 331-335
Chapter 7: Creating Other Database Objects in Oracle
QUESTION NO: 6
Which two statements about views are true? (Choose two.)
A. A view can be created as read only.
B. A view can be created as a join on two or more tables.
C. A view cannot have an ORDER BY clause in the SELECT statement.
D. A view cannot be created with a GROUP BY clause in the SELECT statement.
E. A view must have aliases defined for the column names in the SELECT statement.
Answer: A, B
Explanation:
A view can be created as read only object. However, it is possible to change data in the
underlying table(s) with some restrictions.A view also can be created as a join on two or more
tables. This type of view is called complex view. Complex views provide complicated data
models where many base tables are drawn together into one virtual table.
Incorrect Answers
C: Query operations containing ORDER BY clause are also permitted, so long as the
ORDER BY clause appearsoutside the parentheses. The following is an example of what I
mean: CREATE VIEW my_view AS (SELECT*FROM emp) ORDER BYempno.
D: A view can be created with a GROUP BY clause in the SELECT statement.
E: It is not required to have aliases defined for the column names in the SELECT statement.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 292-309
Chapter 7: Creating Other Database Objects in Oracle
QUESTION NO: 7
Examine the description of the EMPLOYEES table:
EMP_ID NUMBER(4) NOT NULL
LAST_NAME VARCHAR2(30) NOT NULL
FIRST_NAME VARCHAR2(30)
DEPT_ID NUMBER(2)
JOB_CAT VARCHARD2(30)
SALARY NUMBER(8,2)
Which statement shows the maximum salary paid in each job category of each
department?
A. SELECT dept_id, job_cat, MAX(salary)
FROM employees
WHERE salary > MAX(salary);
B. SELECT dept_id, job_cat, MAX(salary)
FROM employees
GROUP BY dept_id, job_cat;
C. SELECT dept_id, job_cat, MAX(salary)
FROM employees;
D. SELECT dept_id, job_cat, MAX(salary)
FROM employees
GROUP BY dept_id;
E. SELECT dept_id, job_cat, MAX(salary)
FROM employees
GROUP BY dept_id, job_cat, salary;
Answer: B
Explanation:
This answer provides correct syntax and semantics to show the maximum salary paid in each
job category of each department.
Incorrect Answers
A: This query will not return any row because condition SALARY > MAX(SALARY) is
FALSE.
C: This query will return error because you cannot show maximum salary with DEPT_ID
and JOB_CAT without grouping by these columns.
D: The GROUP BY clause is missing JOB_ID column.
E: You don’t need to group results of query by SALARY in the GROUP BY column.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 356-365
Chapter 8: User Access Control in Oracle
QUESTION NO: 8
Management has asked you to calculate the value 12*salary* commission_pct for all the
employees in the EMP table. The EMP table contains these columns:
LAST NAME VARCNAR2(35) NOT NULL
SALARY NUMBER(9,2) NOT NULL
COMMISION_PCT NUMBER(4,2)
Which statement ensures that a value is displayed in the calculated columns for all
employees?
A. SELECT last_name, 12*salary* commission_pct
FROM emp;
B. SELECT last_name, 12*salary* (commission_pct,0)
FROM emp;
C. SELECT last_name, 12*salary*(nvl(commission_pct,0))
FROM emp;
D. SELECT last_name, 12*salary*(decode(commission_pct,0))
FROM emp;
Answer: C
Explanation:
This SELECT statement provides correct usage of NVL function to calculate columns for all
employees. Oracle give you possibility to substitute a value in place of NULL. The basic
syntax for NVL() is NVL(column_name, value_if_null). Notice that the column specified in
NVL() contains an actual value. That value is what Oracle returns; when the column is NULL,
the special string is returned. The value specified to be returned if the column value is NULL
must be the same datatype as the column specified.
Incorrect Answers
A: This SELECT statement will return NULL value for rows with NULL
COMMISION_PCT column.
B: It is incorrect syntax in this query: NVL function needs to be used for correct result.
D: The DECODE function is used as substitution of IF-THEN-ELSE PL/SQL construction in
SQL queries. The SELECT statement provides incorrect syntax of it cannot have only two
parameters.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 31-32
Chapter 1: Overview of Oracle Databases
QUESTION NO: 9
Which syntax turns an existing constraint on?
A. ALTER TABLE table_name
ENABLE constraint_name;
B. ALTER TABLE table_name
STATUS = ENABLE CONSTRAINT constraint_name;
C. ALTER TABLE table_name
ENABLE CONSTRAINT constraint_name;
D. ALTER TABLE table_name
STATUS ENABLE CONSTRAINT constraint_name;
E. ALTER TABLE table_name
TURN ON CONSTRAINT constraint_name;
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F. ALTER TABLE table_name
TURN ON CONSTRAINT constraint_name;
Answer: C
Explanation:
ALTER TABLE statement with ENABLE CONSTRAINT keywords is correct answer to
enable an existing constraint.
Incorrect Answers
A: This statement is missing CONSTRAINT keyword.
B: “STATUS =” is incorrect syntax to enable constraint for the table.
D: There is no STATUS keyword in the command to enable constraint.
E: There is no TURN ON keywords in the command to enable constraint.
F: There is no TURN ON keywords in the command to enable constraint.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 239-240
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 10
Examine the description of the STUDENTS table:
STD_ID NUMBER(4)
COURSE_ID VARCHARD2(10)
START_DATE DATE
END_DATE DATE
Which two aggregate functions are valid on the START_DATE column? (Choose two)
A. SUM(start_date)
B. AVG(start_date)
C. COUNT(start_date)
D. AVG(start_date, end_date)
E. MIN(start_date)
F. MAXIMUM(start_date)
Answer: C, E
Explanation:
It is possible to apply COUNT() and MIN() functions on the column with DATE data type.
Incorrect Answers
A: Function SUM() cannot be used with DATE data type column.
B: Function AVG() cannot be used with DATE data type column.
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D: Function AVG() cannot be used with DATE data type column. And function AVG() just
has one parameter X, not two. It averages all X column values returned by the SELECT
statement.
F: There is no MAXIMUM() function in Oracle, only MAX() function exists.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 81-85
Chapter 2: Limiting, Sorting, and Manipulating Return Data
QUESTION NO: 11
The EMPLOYEE tables has these columns:
LAST_NAME VARCHAR2(35)
SALARY NUMBER(8,2)
COMMISSION_PCT NUMBER(5,2)
You want to display the name and annual salary multiplied by the commission_pct for
all employees. For records that have a NULL commission_pct, a zero must be displayed
against the calculated column.
Which SQL statement displays the desired results?
A. SELECT last_name, (salary * 12) * commission_pct
FROM EMPLOYEES;
B. SELECT last_name, (salary * 12) * IFNULL(commission_pct, 0)
FROM EMPLOYEES;
C. SELECT last_name, (salary * 12) * NVL2(commission_pct, 0)
FROM EMPLOYEES;
D. SELECT last_name, (salary * 12) * NVL(commission_pct, 0)
FROM EMPLOYEES;
Answer: D
Explanation:
This SELECT statement provides correct usage of NVL function to calculate columns for all
employees. Oracle give you possibility to substitute a value in place of NULL. The basic
syntax for NVL() is NVL(column_name, value_if_null). Notice that the column specified in
NVL() contains an actual value. That value is what Oracle returns; when the column is NULL,
the special string isreturned. The value specified to be returned if the column value is NULL
must be the same datatype as the column specified.
Incorrect Answers
A: This SELECT statement will return NULL value for rows with NULL
COMMISION_PCT column.
B: There is no IFNULL() function in Oracle.
C: The NVL2() function requires 3 parameters, not 2. Function NVL2(expr1, expr2, expr3)
returns expr2 if expr1 is not NULL. If expr1 is NULL, it returns expr3.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 31-32
Chapter 1: Overview of Oracle Databases
QUESTION NO: 12
Examine the data from the ORDERS and CUSTOMERS table.
ORDERS
ORD_ID ORD_DATE CUST_ID ORD_TOTAL
100 12-JAN-2000 15 10000
101 09-MAR-2000 40 8000
102 09-MAR-2000 35 12500
103 15-MAR-2000 15 12000
104 25-JUN-2000 15 6000
105 18-JUL-2000 20 5000
106 18-JUL-2000 35 7000
107 21-JUL-2000 20 6500
108 04-AUG-2000 10 8000
CUSTOMERS
CUST_ID CUST_NAME CITY
10 Smith Los Angeles
15 Bob San Francisco
20 Martin Chicago
25 Mary New York
30 Rina Chicago
35 Smith New York
40 Linda New York
Which SQL statement retrieves the order ID, customer ID, and order total for the
orders that are placed on the same day that Martin places his orders?
A. SELECT ord_id, cust_id, ord_total
FROM orders, customers
WHERE cust_name=’Mating’
AND ord_date IN (’18-JUL-2000’,’21-JUL-2000’);
B. SELECT ord_id, cust_id, ord_total
FROM orders
Where ord_date IN (SELECT ord_date
FROM orders
WHERE cust_id = (SELECT cust_id
FROM customers
WHERE cust_name =
‘Martin’));
C. SELECT ord_id, cust_id, ord_total
FROM orders
Where ord_date IN (SELECT ord_date
FROM orders, customers
Where cust_name = ‘Martin’);
D. SELECT ord_id, cust_id, ord_total
FROM orders
WHERE cust_id IN (SELECT cust_id
FROM customers
WHERE cust name = ‘Martin’);
Answer: B
Explanation:
This query will return the order ID, customer ID, and order total for the orders that are placed
on the same day that Martin places his orders.
Incorrect Answers
A: This query returns only Martin’s orders for July 18, 2000 and July 21, 2002, not orders of
others that were placed on the same day that Martin placed his orders.
C: This query uses incorrect sub-query to extract dates when Martin placed his orders.
D: This query will return only Martin’s orders.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 145-156
Chapter 4: Subqueries
QUESTION NO: 13
You need to modify the STUDENTS table to add a primary key on the STUDENT_ID
column. The table is currently empty.
Which statement accomplishes this task?
A. ALTER TABLE students
ADD PRIMARY KEY student_id;
B. ALTER TABLE students
ADD CONSTRAINT PRIMARY KEY (student_id);
C. ALTER TABLE students
ADD CONSTRAINT stud_id_pk PRIMARY KEY student_id;
D. ALTER TABLE students
ADD CONSTRAINT stud_id_pk PRIMARY KEY (student_id);
E. ALTER TABLE students
MODIFY CONSTRAINT stud_id_pk PRIMARY KEY (student_id);
Answer: D
Explanation:
This statement provides correct syntax to add a primary key on the STUDENT_ID column of
the STUDENT table.
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Incorrect Answers
A: This ALTER TABLE statement is missing CONSTRAINT keyword and the name of the
constraint.
B: This ALTER TABLE statement is missing the name of the constraint.
C: It’s incorrect syntax in the ALTER TABLE command: STUDENT_ID must be used with
brackets.
E: We need to add constraint, not to modify existing one. Usage of the MODIFY keyword is
incorrect in this case.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 239-240
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 14
Evaluate the SQL statement:
1 SELECT a.emp_name, a.sal, a.dept_id, b.maxsal
2 FROM employees a,
3 (SELECT dept_id, MAX(sal) maxsal
4. FROM employees
5 GROUP BY dept_id) b
6 WHERE a.dept_id = b.dept_id
7 AND a.sal < b.maxsal;
What is the result of the statement?
A. The statement produces an error at line 1.
B. The statement produces an error at line 3.
C. The statement produces an error at line 6.
D. The statement returns the employee name, salary, department ID, and maximum salary
earned in the department of the employee for all departments that pay less salary then
the maximum salary paid in the company.
E. The statement returns the employee name, salary, department ID, and maximum salary
earned in the department of the employee for all employees who earn less than the
maximum salary in their department.
Answer: E
Explanation:
The statement returns the employee name, salary, department ID, and maximum salary earned
in the department of the employee for all employees who earn less than the maximum salary
in their department. This query is example of an inline view which is the sub-query in the
FROM clause of the main query. The sub-query can be a SELECT statement that utilizes
joins, the GROUP BY clause, or the ORDER BY clause.
Incorrect Answers
A: The statement does not produce an error at line 1.
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B: The statement does not produce an error at line 3.
C: The statement does not produce an error at line 6.
D: The statement returns the employee name, salary, department ID, and maximum salary
earned in the department of the employee for all EMPLOYEES, NOT DEPARTMENTS, who
earn less than the maximum salary in their department.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 161-165
Chapter 4: Subqueries
QUESTION NO: 15
Examine the data in the EMPLOYEES and DEPARTMENTS tables:
EMPLOYEES
EMPLOYEE_ID EMP_NAME DEPT_ID MGR_ID JOB_ID SALARY
101 Smith 20 120 SA_REP 4000
102 Martin 10 105 CLERK 2500
103 Chris 20 120 IT_ADMIN 4200
104 John 30 108 HR_CLERK 2500
105 Diana 30 108 IT_ADMIN 5000
106 Smith 40 110 AD_ASST 3000
108 Jennifer 30 110 HR_DIR 6500
110 Bob 40 EX_DIR 8000
120 Ravi 20 110 SA*DIR 6500
DEPARTMENTS
DEPARTMENT_ID DEPARTMENT_NAME
10 Admin
20 Education
30 IT
40 Human Resources
Also examine the SQL statements that create the EMPLOYEES and DEPARTMENTS
tables:
CREATE TABLE departments
(department_id NUMBER PRIMARY KEY,
department_name VARCHAR2(30));
CREATE TABLE employees
(EMPLOYEE_ID NUMBER PRIMARY KEY,
EMP_NAME VARCHAR2(20),
DEPT_ID NUMBER REFERENCES
departments(department_id),
MGR_ID NUMBER REFERENCES
employees(employee id),
MGR_ID NUMBER REFERENCES
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employees(employee id),
JOB_ID VARCHAR2(15).
SALARY NUMBER);
ON the EMPLOYEES,
On the EMPLOYEES table, EMPLOYEE_ID is the primary key.
MGR_ID is the ID of managers and refers to the EMPLOYEE_ID.
DEPT_ID is foreign key to DEPARTMENT_ID column of the DEPARTMENTS table.
On the DEPARTMENTS table, DEPARTMENT_ID is the primary key.
Examine this DELETE statement:
DELETE
FROM departments
WHERE department id = 40;
What happens when you execute the DELETE statement?
A. Only the row with department ID 40 is deleted in the DEPARTMENTS table.
B. The statement fails because there are child records in the EMPLOYEES table with
department ID 40.
C. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the
rows with employee IDs 110 and 106 are deleted from the EMPLOYEES table.
D. The row with department ID 40 is deleted in the DEPARTMENTS table. Also the
rows with employee IDs 106 and 110 and the employees working under employee 110
are deleted from the EMPLOYEES table.
E. The row with department ID 40 is deleted in the DEPARTMENTS table. Also all the
rows in the EMPLOYEES table are deleted.
F. The statement fails because there are no columns specifies in the DELETE clause of
the DELETE statement.
Answer: B
Explanation:
It will be error generated because there are 2 child records in the EMPLOYEES table with
department number you try to delete from the DEPARTMENTS table.
Incorrect Answers
A: The row with department ID 40 will not be deleted because of the child records in the
EMPLOYEES table.
C: Neither the row with department ID 40 will not be deleted not child records in the
EMPLOYEES table will be deleted.
D: It will be error when you try to execute the DELETE statement, no rows will be deleted in
the EMPLOYEES or the DEPARTMENTS tables.
E: It will be error when you try to execute the DELETE statement, no rows will be deleted in
the EMPLOYEES or the DEPARTMENTS tables.
F: The statement fails because of constraint violation not because there are no columns
specifies in the DELETE clause of the DELETE statement.
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OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 240-245
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 16
Which three are DATETIME data types that can be used when specifying column
definitions? (Choose three.)
A. TIMESTAMP
B. INTERVAL MONTH TO DAY
C. INTERVAL DAY TO SECOND
D. INTERVAL YEAR TO MONTH
E. TIMESTAMP WITH DATABASE TIMEZONE
Answer: A, C, D
Explanation:
TIMESTAMP, INTERVAL DAY TO SECOND and INTERVAL YEAR TO MONTH can be
used to specify column definition.
Incorrect Answers
B: The INTERVAL MONTH TO DAY data type cannot be used when specifying column
definitions there are only INTERVAL DAY TO SECOND and INTERVAL YEAR TO
MONTH data types.
E: The TIMESTAMP WITH DATABASE TIMEZONE data type cannot be used when
specifying column definitions, because there are only TIMESTAMP WITH TIME ZONE
and TIMESTAMP WITH LOCAL TIME ZONE data types.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 215-217
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 17
Which SQL statement defines the FOREIGN KEY constraint on the DEPTNO column
of the EMP table?
A. CREATE TABLE EMP
(empno NUMBER(4),
ename VARCNAR2(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 VARCNAR2(35),
deptno NUMBER(7,2)
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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 VARCNAR2(35),
deptno NUMBER(7,2) FOREIGN KEY
CONSTRAINT emp deptno fk REFERENCES dept (deptno));
Answer: B
Explanation:
This statement provides correct syntax to define the FOREIGN KEY constraint on the
DEPTNO column of the EMP table.
Incorrect Answers
A: There is incorrect syntax, because list of columns and column for the constraint need to be
surrounded with the brackets.
C: It is incorrect to use FOREIGN KEY keywords to define constraint on the table. It can be
used to add integrity constraint to existing table.
D: It is incorrect to use FOREIGN KEY keywords to define constraint on the table. It can be
used to add integrity constraint to existing table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 238-245
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 18
Evaluate the set of SQL statements:
CREATE TABLE dept
(deptno NUMBER(2),
dname VARCNAR2(14),
loc VARCNAR2(13));
ROLLBACK;
DESCRIBE DEPT
What is true about the set?
A. The DESCRIBE DEPT statement displays the structure of the DEPT table.
B. The ROLLBACK statement frees the storage space occupies by the DEPT table.
C. The DESCRIBE DEPT statement returns an error ORA-04043: object DEPT does not
exist.
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D. The DESCRIBE DEPT statement displays the structure of the DEPT table only if
there is a COMMIT statement introduced before the ROLLBACK statement.
Answer: A
Explanation:
The structure of the DEPT table will be displayed because the CREATE TABLE statement is
DDL operation and it cannot be rolled back because implicit commit occurs on the database
when a user exits SQL*Plus or issues a data-definition language (DDL) command such as a
create table statement, user to create a database object, or an alter table statement, used to
alter a database object.
Incorrect Answers
B: The ROLLBACK statement has nothing to do with the storage space of the DEPT table.
C: The DESCRIBE DEPT statement does not produce the error. It displays the structure of
the DEPT table.
D: The COMMIT statement does not need to be introduced because implicit commit occurs
on the database after creation of the table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 281-283
Chapter 6: Manipulating Oracle Data
QUESTION NO: 19
Which data dictionary table should you query to view the object privileges granted to
the user on specific columns?
A. USER_TAB_PRIVS_MADE
B. USER_TAB_PRIVS
C. USER_COL_PRIVS_MADE
D. USER_COL_PRIVS
Answer: D
Explanation:
The USER_COL_PRIVS data dictionary view will show the object privileges granted to the
user on specific columns.
Incorrect Answers
A: There is no USER_TAB_PRIVS_MADE view in Oracle.
B: The USER_TAB_PRIVS data dictionary view is used to show the object privileges
granted to the user on the tables, not specific columns.
C: There is no USER_COL_PRIVS_MADE view in Oracle.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 356-365
Chapter 8: User Access Control in Oracle
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QUESTION NO: 20
Examine the structure of the EMPLOYEES and DEPARTMENTS tables:
E
MPLOYEES
Column name Data type Remarks
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2 (30)
JOB_ID VARCHAR2 (20)
SALARY NUMBER
MGR_ID NUMBER References EMPLOYEE_ID COLUMN
DEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID
column of the DEPARTMENTS table
DEPARTMENTS
Column name Data type Remarks
DEPARTMENT_ID NUMBER NOT NULL, Primary Key
DEPARTMENT_NAME VARCHAR2(30)
MGR_ID NUMBER References MGR_ID column of the
EMPLOYEES table
Evaluate this SQL statement:
SELECT employee_id, e.department_id, department_name,
salary
FROM employees e, departments d
WHERE e.department_id = d.department_id;
Which SQL statement is equivalent to the above SQL statement?
A. SELECT employee_id, department_id, department_name,
salary
FROM employees
WHERE department_id IN (SELECT department_id
FROM departments);
B. SELECT employee_id, department_id, department_name,
salary
FROM employees
NATURAL JOIN departments;
C. SELECT employee_id, d.department_id, department_name,
salary
FROM employees e
JOIN departments d
ON e.department_id = d.department_id;
D. SELECT employee_id, department_id, department_name,
Salary
FROM employees
JOIN departments
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USING (e.department_id, d.department_id);
Answer: C
Explanation:
This query shows correct JOIN ON clause syntax and provides equivalent to the above SQL
statement.
Incorrect Answers
A: This statement will show data only for the EMPLOYEES table with records that have
department ID from DEPARTMENTS table, not join result of two tables.
B: NATURAL join selects rows from the tables that have equal values in all matched
columns (same column names). If the columns having the same names have different
datatypes, an error is returned.
D: There is incorrect usage of JOIN clause with USING keyword.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 112-114
Chapter 3: Advanced Data Selection in Oracle
QUESTION NO: 21
The EMP table contains these columns:
LAST NAME VARCHAR2(25)
SALARY NUMBER(6,2)
DEPARTMENT_ID NUMBER(6)
You need to display the employees who have not been assigned to any department.
You write the SELECT statement:
SELECT LAST_NAME, SALARY, DEPARTMENT_ID
FROM EMP
WHERE DEPARTMENT_ID = NULL;
What is true about this SQL statement?
A. The SQL statement displays the desired results.
B. The column in the WHERE clause should be changed to display the desired results.
C. The operator in the WHERE clause should be changed to display the desired results.
D. The WHERE clause should be changed to use an outer join to display the desired
results.
Answer: C
Explanation:
The operator in the WHERE clause should be changed to display the desired results. There are
times when you want to substitute a value in place of NULL. Oracle provides this
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functionality with a special function, called NVL(). You cannot use operation equal with
NULL, but you can achieve desired results using NVL() function after the WHERE clause.
Incorrect Answers
A: The SQL statement will generate an error because you cannot use operation equal with
NULL.
B: The column in the WHERE clause should not be changed to display the desired results.
D: Since there is only one table used in this query you don’t need to use outer join to display
the desired results.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 31-32
Chapter 1: Overview of Oracle Databases
QUESTION NO: 22
Evaluate the SQL statement:
SELECT ROUND(TRUNC(MOD(1600,10),-1),2)
FROM dual;
What will be displayed?
A. 0
B. 1
C. 0.00
D. An error statement
Answer: A
Explanation:
Result will be 0. MOD(x,y) function calculates the modulus of x, defined in long division as
the integer remainder when x is divided by y until no further whole number can be produced.
TRUNC() function truncates x to the decimal precision of y. ROUND(x,y) rounds x to the
decimal precision of y.
Incorrect Answers
B: Result will be 0, not 1.
C: Result will be 0, not 0.00 because MOD(1600,10) return 0 and all other functions
(TRUNC and ROUND) return 0 also.
D: There is no error in this statement.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 69-71
Chapter 2: Limiting, Sorting, and Manipulating Return Data
Examine the description of the MARKS table:
STD_ID NUMBER(4)
STUDENT_NAME VARCHAR2(30)
SUBJ1 NUMBER(3)
SUBJ2 NUMBER(3)
SUBJ1 and SUBJ2 indicate the marks obtained by a student in two subjects.
Examine this SELECT statement based on the MARKS table:
SELECT subj1+subj2 total_marks, std_id
FROM marks
WHERE subj1 > AVG(subj1) AND subj2 > AVG(subj2)
ORDER BY total_marks;
What is the result of the SELECT statement?
A. The statement executes successfully and returns the student ID and sum of all marks
for each student who obtained more than the average mark in each subject.
B. The statement returns an error at the SELECT clause.
C. The statement returns an error at the WHERE clause.
D. The statement returns an error at the ORDER BY clause.
Answer: C
Explanation:
The statement returns an error at the WHERE clause because group function AVG() cannot be
used in the WHERE clause. Group functions can be used in SELECT clause and GROUP BY
clause. They allow you to perform data operations on several values in a column of data as
though the column were one collective group of data.
Incorrect Answers
A: The statement does not execute successfully because an error will be generated.
B: The statement returns an error at the WHERE, not at the SELECT clause.
D: The statement returns an error at the WHERE, not at the ORDER BY clause.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 122-125
Chapter 3: Advanced Data Selection in Oracle
QUESTION NO: 24
Which /SQL*Plus feature can be used to replace values in the WHERE clause?
A. Substitution variables
B. Replacement variables
C. Prompt variables
D. Instead-of variables
E. This feature cannot be implemented through /SQL*Plus.
Answer: A
Explanation:
Lexical substitution variables can be used to replace values in the WHERE clause.
Incorrect Answers
B: There is no replacement variables SQL*Plus feature in Oracle.
C: There is no prompt variables SQL*Plus feature in Oracle.
D: There is no instead-of variables SQL*Plus feature in Oracle.
E: This feature is implemented in the SQL*Plus with lexical substitution variables.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 165-173
Chapter 4: Subqueries
QUESTION NO: 25
You want to display the titles of books that meet these criteria:
1. Purchased before January 21, 2001
2. Price is less then $500 or greater than $900
You want to sort the results by their data of purchase, starting with the most recently
bought book.
Which statement should you use?
A. SELECT book_title
FROM books
WHERE price between 500 and 900
AND purchase_date < ’21-JAN-2001’
ORDER BY purchase_date;
B. SELECT book_title
FROM books
WHERE price IN (500,900)
AND purchase_date < ’21-JAN-2001’
ORDER BY purchase date ASC;
C. SELECT book_title
FROM books
WHERE price < 500 or > 900
AND purchase_date < ’21-JAN-2001’
ORDER BY purchase date DESC;
D. SELECT book_title
FROM books
WHERE (price < 500 OR price > 900)
AND purchase_date < ’21-JAN-2001’
ORDER BY purchase date DESC;
Answer: D
Explanation:
This statement provides required results.
Incorrect Answers
A: This query will show books with price in range $500 and $900, not less then $500 or
greater than $900.
B: This query will show books with prices exactly $500 or $900, not less then $500 or
greater than $900.
C: This order will not show correct rows because of incorrect syntax in the WHERE clause..
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 56-66
Chapter 2: Limiting, Sorting, and Manipulating Return Data
QUESTION NO: 26
Which statement explicitly names a constraint?
A. ALTER TABLE student_grades
ADD
FOREIGN KEY (student_id) REFERENCES students(student_id);
B. ALTER TABLE student_grades
ADD CONSTRAINT NAME = student_id_fk
FOREIGN KEY (student_id) REFERENCES students(student_id);
C. ALTER TABLE student_grades
ADD CONSTRAINT student_id_fk
FOREIGN KEY (student_id) REFERENCES students(student_id);
D. ALTER TABLE student grades
ADD NAMED CONSTRAINT student_id_fk
FOREIGN KEY (student_id) REFERENCES students(student_id);
E. ALTER TABLE student grades
ADD NAME student_id_fk
FOREIGN KEY (student_id) REFERENCES students(student_id);
Answer: C
Explanation:
This statement provides correct syntax to add a foreign key constraint to the existing table.
Incorrect Answers
A: The ADD FOREIGN KEY is wrong construction to add a foreign key constraint to the
existing table.
B: The ADD CONSTRAINT NAME is wrong construction to add a foreign key constraint
to the existing table.
D: The ADD NAMED CONSTRAINT is wrong construction to add a foreign key constraint
to the existing table.
E: The ADD NAME is wrong construction to add a foreign key constraint to the existing
table.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 238-239
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 27
Examine the SQL statements that creates ORDERS table:
CREATE TABLE orders
(SER_NO NUMBER UNIQUE,
ORDER_ID NUMBER,
ORDER_DATE DATE NOT NULL
STATUS VARCHARD2(10)
CHECK (status IN (‘CREDIT’,’CASH’)),
PROD_ID_NUMBER
REFERENCES PRODUCTS(PRODUCT_ID),
ORD_TOTAL NUMBER,
PRIMARY KEY (order id, order date));
For which columns would an index be automatically created when you execute the above
SQL statement? (Choose two)
A. SER_NO
B. ORDER_ID
C. STATUS
D. PROD_ID
E. ORD_TOTAL
F. Composite index on ORDER_ID and ORDER_DATE
Answer: A, F
Explanation:
Indexes are created automatically by Oracle to support integrity constraints that enforce
uniqueness. The two types of integrity constraints that enforce uniqueness are PRIMARY
KEY and UNIQUE constraints. When the primary key or UNIQUE constraint is declared, a
unique index to support the column’s uniqueness is also created, and all values in all columns
that were defined as part of the primary key or UNIQUE constraint are placed into the index.
Incorrect Answers
B: There will not be index for ORDER_ID column.
C: There will not be index for STATUS column.
D: There will not be index for PROD_ID column.
E: There will not be index for ORD_TOTAL column.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 237-238
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 28
You created a view called EMP_DEPT_VU that contains three columns from the
EMPLOYEES and DEPARTMENTS tables:
EMPLOYEE_ID, EMPLOYEE_NAME AND DEPARTMENT_NAME.
The DEPARTMENT_ID column of the EMPLOYEES table is the foreign key to the
primary key DEPARTMENT_ID column of the DEPARTMENTS table.
You want to modify the view by adding a fourth column, MANAGER_ID of NUMBER
data type from the EMPLOYEES tables.
How can you accomplish this task?
A. ALTER VIEW emp_dept_vu (ADD manager_id NUMBER);
B. MODIFY VIEW emp_dept_vu (ADD manager_id NUMBER);
C. ALTER VIEW emp_dept_vu AS
SELECT employee_id, employee_name,
department_name, manager_id
FROM employee e, departments d
WHERE e.department_id = d.department_id;
D. MODIFY VIEW emp_dept_vu AS
SELECT employee_id, employee_name,
department_name, manager_id
FROM employees e, departments d
WHERE e.department_id = d.department_id;
E. CREATE OR REPLACE VIEW emp_dept_vu AS
SELECT employee_id, employee_name,
department_name, manager_id
FROM employees e, departments d
WHERE e.department_id = d.department_id;
F. You must remove the existing view first, and then run the CREATE VIEW command
with a new column list to modify a view.
Answer: E
Explanation:
When we want to alter the underlying data used in the definition of a view, we use the
CREATE OR REPLACE VIEW statement. When a CREATE OR REPLACE VIEW
statement is issued, Oracle will disregard the error that arises when it encounters the view that
already exists with that name, and it will overwrite the definition for the old view with the
definition for the new one.
Incorrect Answers
A: There is no ALTER VIEW command in Oracle.
B: There is no MODIFY VIEW command in Oracle.
C: There is no ALTER VIEW command in Oracle.
D: There is no MODIFY VIEW command in Oracle.
F: You don’t need to remove the existing view to create modified view. You are able to do
that with CREATE OR REPLACE command.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 310-313
Chapter 7: Creating Other Database Objects in Oracle
QUESTION NO: 29
For which two constraints does the Oracle Server implicitly create a unique index?
(Choose two.)
A. NOT NULL
B. PRIMARY KEY
C. FOREIGN KEY
D. CHECK
E. UNIQUE
Answer: B, E
Explanation:
Indexes are created automatically by Oracle to support integrity constraints that enforce
uniqueness. The two types of integrity constraints that enforce uniqueness are PRIMARY
KEY and UNIQUE constraints. When the primary key or UNIQUE constraint is declared, a
unique index to support the column’s uniqueness is also created, and all values in all columns
that were defined as part of the primary key or UNIQUE constraint are placed into the index.
Incorrect Answers
A: Oracle will not implicitly create an unique index for the NOT NULL constraint.
C: Oracle will not implicitly create an unique index for the FOREIGN KEY constraint.
D: Oracle will not implicitly create an unique index for the FOREIGN KEY constraint.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 237-238
Chapter 5: Creating Oracle Database Objects
QUESTION NO: 30
Which three SELECT statements displays 2000 in the format “$2,000.00”? (Choose
three)
A. SELECT TO CNAR(2000, ‘$#,###.##’)
FROM dual;
B. SELECT TO CNAR(2000, ‘$0,000.00’)
FROM dual;
C. SELECT TO CNAR(2000, ‘$9,999.00’)
FROM dual;
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D. SELECT TO CNAR(2000, ‘$9,999.99’)
FROM dual;
E. SELECT TO CNAR(2000, ‘$2,000.00’)
FROM dual;
F. SELECT TO CNAR(2000, ‘$N,NNN.NN’)
FROM dual;
Answer: B, C, D
Explanation:
Only queries in answers B, C and D will show result as in the format “$2,000.00”.
Incorrect Answers
A: Oracle error “ORA-01481: invalid number format model” will be generated.
E: Oracle error “ORA-01481: invalid number format model” will be generated.
F: Oracle error “ORA-01481: invalid number format model” will be generated.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 356-365
Chapter 8: User Access Control in Oracle
QUESTION NO: 31
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:
EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHARD2(25)
LAST_NAME VARCHARD2(25)
HIRE_DATE DATE
NEW EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
NAME VARCHAR2(60)
Which UPDATE statement is valid?
A. UPDATE new_employees SET name = (Select last_name||
first_name
FROM employees
Where employee_id
=180)
WHERE employee_id =180;
B. UPDATE new_employees SET name = (SELECT
last_name||first_name
FROM employees)
WHERE employee_id =180;
C. UPDATE new_employees SET name = (SELECT last_name||
first_name
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FROM employees
WHERE employee_id
=180)
WHERE employee_id =(SELECT employee_id
FROM new employees);
D. UPDATE new_employees SET name = (SELECT last name||
first_name
FROM employees
WHERE employee_id=
(SELECT employee_id
FROM new_employees))
WHERE employee_id
=180;
Answer: A
Explanation:
Sub-query in this answer will return one row value, concatenated first and last name for the
employee with ID 180, so update will be successful. When sub-queries are linked to the
parent by equality comparisons, the parent query expects only one row of data from the subquery.
Incorrect Answers
B: Sub-query will return concatenated first and last name for ALL records from the table
EMPLOYEES. It will cause an error for the parent query, because it expects one only one
row.
C: Last WHERE statement in this query will generate error because sub-query returns multirow
result.
D: SELECT statement for the NEW_EMPLOYEES table in this query will generate error
because sub-query returns multi-row result.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 150-156
Chapter 4: Subqueries
QUESTION NO: 32
Examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS
tables.
EMPLOYEES
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2 (30)
JOB_ID VARCHAR2 (20)
SALARY NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the
DEPARTMENTS table
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DEPARTMENTS
DEPARTMENT_ID NUMBER NOT NULL, Primary Key
DEPARTMENT_NAME VARCHAR2
(30)
MGR_ID NUMBER References NGR_ID column of the
EMPLOYEES table
LOCATION_ID NUMBER Foreign key to LOCATION_ID column of the
LOCATIONS table
LOCATIONS
LOCATION_ID NUMBER NOT NULL, Primary Key
CITY VARCHAR2 |30)
Which two SQL statements produce the name, department name, and the city of all the
employees who earn more then 10000? (Choose two)
A. SELECT emp_name, department_name, city
FROM employees e
JOIN departments d
USING (department_id)
JOIN locations 1
USING (location_id)
WHERE salary > 10000;
B. SELECT emp_name, department_name, city
FROM employees e, departments d, locations 1
JOIN ON (e.department_id = d.department id)
AND (d.location_id =1.location_id)
AND salary > 10000;
C. SELECT emp_name, department_name, city
FROM employees e, departments d, locations 1
WHERE salary > 10000;
D. SELECT emp_name, department_name, city
FROM employees e, departments d, locations 1
WHERE e.department_id = d.department_id
AND d.location_id = 1.location_id
AND salary > 10000;
E. SELECT emp_name, department_name, city
FROM employees e
NATURAL JOIN departments, locations
WHERE salary > 10000;
Answer: B, D<
