1. 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, FROM customers;
B. SELECT "Dear Customer", customer_name || ',' FROM customers;
C. SELECT 'Dear Customer ' || customer_name ',' 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
2. 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
3. Which statement correctly describes SQL and iSQL*Plus Commands?
A. Both SQL and iSQL*plus allow manipulation of values in the database.
B. iSQL*Plus recognizes SQL statements and sends them to the server; SQL is the
Oracle proprietary interface for executing SQL statements.
C. iSQL*Plus is a language for communicating with the Oracle server to access data;
SQL recognizes SQL statements and sends them to the server.
D. SQL manipulates data and table definitions in the database; iSQL*Plus does not allow
manipulation of values in the database.
Answer: D
4. 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
5. From SQL*Plus, you issue this SELECT statement:
SELECT* From orders;
You use this statement to retrieve data from a data table for __________. (Choose all
that apply)
A. Updating
B. Viewing
C. Deleting
D. Inserting
E. Truncating
Answer: B, D
6. Which two are attributes of iSQL*Plus? (Choose two)
A. iSQL*Plus commands cannot be abbreviated.
B. iSQL*Plus commands are accesses 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: B, E
7. Which SQL statement generates the alias Annual Salary for the calculated column
SALARY*12?
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
8. Evaluate this SQL statement:
SELECT e.EMPLOYEE_ID,e.LAST_NAME,e.DEPARTMENT_ID, d.DEPARTMENT_NAME
FROM EMP e, DEPARTMENT 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. Which is an iSQL*Plus command?
A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME
Answer: D
10. You need to produce a report for mailing labels for all customers. The mailing label
must have only the customer name and address. The CUSTOMERS table has these
columns:
CUST_ID NUMBER(4) NOT NULL
CUST_NAME VARCHAR2(100) NOT NULL
CUST_ADDRESS VARCHAR2(150)
CUST_PHONE VARCHAR2(20)
Which SELECT statement accomplishes this task?
A. SELECT * FROM customers;
B. SELECT name, address FROM customers;
C. SELECT id, name, address, phone FROM customers;
D. SELECT cust_name, cust_address FROM customers;
E. SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;
Answer: D
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
13. The STUDENT_GRADES table has these columns
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
Which statement finds students who have a grade point average (GPA) greater than 3.0
for the calendar year 2001?
A. SELECT student_id, gpa
FROM student_grades
WHERE semester_end BETWEEN ’01-JAN-2001’ AND ’31-DEC-2001’
OR gpa > 3.;
B. SELECT student_id, gpa
FROM student_grades
WHERE semester_end BETWEEN ’01-JAN-2001’ AND ’31-DEC-2001’
AND gpa gt 3.0;
C. SELECT student_id, gpa
FROM student_grades
WHERE semester_end BETWEEN ’01-JAN-2001’ AND ’31-DEC-2001’
AND gpa > 3.0;
D. SELECT student_id, gpa
FROM student_grades
WHERE semester_end BETWEEN ’01-JAN-2001’ AND ’31-DEC-2001’
AND gpa > 3.0;
E. SELECT student_id, gpa
FROM student_grades
WHERE semester_end > ’01-JAN-2001’ OR semester_end < ’31-DEC-2001’
AND gpa >= 3.0;
Answer: C
Lesson 2: Restricting and Sorting Data
1. 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
2. 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)
Which statement finds the rows in the CUSTOMERS table that do not have a postal
code?
A. SELECT customer_id, customer_name
FROM customers
WHERE postal_code CONTAINS NULL;
B. SELECT customer_id, customer_name
FROM customers
WHERE postal_code = '________';
C. SELECT customer_id, customer_name
FROM customers
WHERE postal_code IS NULL;
D. SELECT customer_id, customer_name
FROM customers
WHERE postal code IS NVL;
E. SELECT customer_id, customer_name
FROM customers
WHERE postal_code = NULL;
Answer: C
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 FROM student_grades SORT ORDER BY gpa DESC;
F. SELECT student_id, gpa FROM student_grades ORDER BY gpa DESC;
Answer: F
4. The EMPLOYEES table contains these columns:
EMPLOYEE_ID NUMBER(4)
LAST_NAME VARCHAR2 (25)
JOB_ID VARCHAR2(10)
You want to search for strings that contain 'SA_' in the JOB_ID column. Which SQL
statement do you use?
A. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE
'%SA\_%' ESCAPE '\';
B. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE
'%SA_';
C. SELECT employee_id, last_name, job_id FROM employees WHERE job_id LIKE '%SA_'
ESCAPE "\";
D. SELECT employee_id, last_name, job_id FROM employees WHERE job_id = '%SA_';
Answer : A
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)
A promotional sale is being advertised to the customers in France. Which WHERE
clause identifies customers that are located in France?
A. WHERE lower(country_address) = "france"
B. WHERE lower(country_address) = 'france'
C. WHERE lower(country_address) IS 'france'
D. WHERE lower(country_address) = '%france%'
E. WHERE lower(country_address) LIKE %france%
Answer: B
6. The PRODUCTS table has these columns:
PRODUCT_ID NUMBER(4)
PRODUCT_NAME VARCHAR2(45)
PRICE NUMBER(8,2)
Evaluate this SQL statement:
SELECT * FROM PRODUCTS ORDER BY price, product_name;
What is true about the SQL statement?
A. The results are not sorted.
B. The results are sorted numerically.
C. The results are sorted alphabetically.
D. The results are sorted numerically and then alphabetically.
Answer: D
7. Evaluate these two SQL statements:
SELECT last_name, salary, hire_dateFROM EMPLOYEES ORDER BY salary DESC;
SELECT last_name, salary, hire_dateFROM EMPLOYEES ORDER BY 2 DESC;
What is true about them?
A. The two statements produce identical results.
B. The second statement returns a syntax error.
C. There is no need to specify DESC because the results are sorted in descending order by
default.
D. The two statements can be made to produce identical results by adding a column alias for
the salary column in the second SQL statement.
Answer: A
8. Examine the structure of the STUDENTS table:
STUDENT_ID NUMBER NOT NULL, Primary Key
STUDENT_NAME VARCHAR2(30)
COURSE_ID VARCHAR2(10) NOT NULL
MARKS NUMBER
START_DATE DATE
FINISH_DATE DATE
You need to create a report of the 10 students who achieved the highest ranking in the
course INT SQL and who completed the course in the year 1999.
Which SQL statement accomplishes this task?
A. SELECT student_ id, marks, ROWNUM "Rank"
FROM students
WHERE ROWNUM <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99
AND course_id = 'INT_SQL'
ORDER BY marks DESC;
B. SELECT student_id, marks, ROWID "Rank"
FROM students
WHERE ROWID <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99'
AND course_id = 'INT_SQL'
ORDER BY marks;
C. SELECT student_id, marks, ROWNUM "Rank"
FROM (SELECT student_id, marks
FROM students
WHERE ROWNUM <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99'
AND course_id = 'INT_SQL'
ORDER BY marks DESC);
D. SELECT student_id, marks, ROWNUM "Rank”
FROM (SELECT student_id, marks
FROM students
WHERE (finish_date BETWEEN ’01-JAN-99 AND ’31-DEC-99’
AND course_id = ‘INT_SQL’
ORDER BY marks DESC)
WHERE ROWNUM <= 10 ;
E. SELECT student id, marks, ROWNUM “Rank”
FROM (SELECT student_id, marks
FROM students
ORDER BY marks)
WHERE ROWNUM <= 10
AND finish date BETWEEN ’01-JAN-99’ AND ’31-DEC-99’
AND course_id = ‘INT_SQL’;
Answer: D
9. 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
10. 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
11. In a SELECT statement that includes a WHERE clause, where is the GROUP BY clause
placed in the SELECT statement?
A. Immediately after the SELECT clause
B. Before the WHERE clause
C. Before the FROM clause
D. After the ORDER BY clause
E. After the WHERE clause
Answer: E
12. The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
The register has requested a report listing the students' grade point averages (GPA),
sorted from highest grade point average to lowest within each semester, starting from
the earliest date. Which statement accomplishes this?
A. SELECT student_id, semester_end, gpa
FROM student_grades
ORDER BY semester_end DESC, gpa DESC;
B. SELECT student_id, semester_end, gpa
FROM student_grades
ORDER BY semester_end ASC, gpa ASC;
C. SELECT student_id, semester_end, gpa
FROM student_grades
ORDER BY semester_end, gpa DESC;
D. SELECT student_id, semester_end, gpa
FROM student_grades
ORDER BY gpa DESC, semester_end DESC;
E. SELECT student_id, semester_end, gpa
FROM student_grades
ORDER BY gpa DESC, semester_end ASC;
Answer: C
13. The ORDERS table has these columns:
ORDER_ID NUMBER(4) NOT NULL
CUSTOMER_ID NUMBER(12) NOT NULL
ORDER_TOTAL NUMBER(10,2)
The ORDERS table tracks the Order number, the order total, and the customer to
whom the Order belongs. Which two statements retrieve orders with an inclusive total
that ranges between 100.00 and 2000.00 dollars? (Choose two.)
A. SELECT customer_id, order_id, order_total
FROM orders
RANGE ON order_total (100 AND 2000) INCLUSIVE;
B. SELECT customer_id, order_id, order_total
FROM orders
HAVING order_total BETWEEN 100 and 2000;
C. SELECT customer_id, order_id, order_total
FROM orders
WHERE order_total BETWEEN 100 and 2000;
D. SELECT customer_id, order_id, order_total
FROM orders
WHERE order_total >= 100 and <= 2000;
E. SELECT customer_id, order_id, order_total
FROM orders
WHERE order_total >= 100 and order_total <= 2000;
Answer: C, E
14. The EMP table has these columns:
ENAME VARCHAR2(35)
SALARY NUMBER(8,2)
HIRE_DATE DATE
Management wants a list of names of employees who have been with the company for
more than five years. Which SQL statement displays the required results?
A. SELECT ENAME
FROM EMP
WHERE SYSDATE-HIRE_DATE > 5;
B. SELECT ENAME
FROM EMP
WHERE HIRE_DATE-SYSDATE > 5;
C. SELECT ENAME
FROM EMP
WHERE (SYSDATE-HIRE_DATE)/365 > 5;
D. SELECT ENAME
FROM EMP
WHERE (SYSDATE-HIRE_DATE)* 365 > 5;
Answer: C
15. Evaluate these two SQL statements:
SELECT last_name, salary , hire_date
FROM EMPLOYEES
ORDER BY salary DESC;
SELECT last_name, salary, hire_date
FROM EMPLOYEES
ORDER BY 2 DESC;
What is true about them?
A. The two statements produce identical results.
B. The second statement returns a syntax error.
C. There is no need to specify DESC because the results are sorted in descending order
by default.
D. The two statements can be made to produce identical results by adding a column alias
for the salary column in the second SQL statement.
Answer: A
16. Which two statements are true regarding the ORDER BY clause? (Choose two)
A. The sort is in ascending by order by default.
B. The sort is in descending order by default.
C. The ORDER BY clause must precede the WHERE clause.
D. The ORDER BY clause is executed on the client side.
E. The ORDER BY clause comes last in the SELECT statement.
F. The ORDER BY clause is executed first in the query execution.
Answer: A, E
17. You need to display the last names of those employees who have the letter “A” as the
second character in their names. Which SQL statement displays the required results?
A. SELECT last_name FROM EMP
WHERE last_name LIKE ‘_A%’;
B. SELECT last_name FROM EMP
WHERE last name =’*A%’
C. SELECT last_name FROM EMP
WHERE last name =’_A%’;
D. SELECT last_name FROM EMP
WHERE last name LIKE ‘*A%’
Answer: A
Lesson 3: Single-Row Functions
1. 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
2. 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
3. The EMPLOYEES table contains these columns:
LAST_NAME VARCHAR2 (25)
SALARY NUMBER (6,2)
COMMISSION_PCT NUMBER (6)
You need to write a query that will produce these results:
1. Display the salary multiplied by the commission_pct.
2. Exclude employees with a zero commission_pct.
3. Display a zero for employees with a null commission value.
Evaluate the SQL statement:
SELECT LAST_NAME, SALARY*COMMISSION_PCT
FROM EMPLOYEES
WHERE COMMISSION_PCT IS NOT NULL;
What does the statement provide?
A. All of the desired results
B. Two of the desired results
C. One of the desired results
D. An error statement
Answer: C
4. Evaluate the SQL statement:
SELECT ROUND(45.953, -1), TRUNC(45.936, 2)
FROM dual;
Which values are displayed?
A. 46 and 45
B. 46 and 45.93
C. 50 and 45.93
D. 50 and 45.9
E. 45 and 45.93
F. 45.95 and 45.93
Answer: C
5. The EMPLOYEES table contains these columns:
EMPLOYEE_ID NUMBER(4)
ENAME VARCHAR2 (25)
JOB_ID VARCHAR2(10)
Which SQL statement will return the ENAME, length of the ENAME, and the numeric
position of the letter "a" in the ENAME column, for those employees whose ENAME
ends with a the letter "n"?
A. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES
WHERE SUBSTR(ENAME, -1, 1) = 'n';
B. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES
WHERE SUBSTR(ENAME, -1, 1) = 'n';
C. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES
WHERE INSTR(ENAME, 1, 1) = 'n';
D. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES
WHERE INSTR(ENAME, -1, 1) = 'n';
Answer: A
6. Which four are attributes of single row functions? (Choose four.)
A. cannot be nested
B. manipulate data items
C. act on each row returned
D. return one result per row
E. accept only one argument and return only one value
F. accept arguments which can be a column or an expression
Answer: B, C, D, F
7. Which SQL statement returns a numeric value?
A. SELECT ADD_MONTHS(MAX(hire_Date), 6) FROM EMP;
B. SELECT ROUND(hire_date) FROM EMP;
C. SELECT sysdate-hire_date FROM EMP;
D. SELECT TO_NUMBER(hire_date + 7) FROM EMP;
Answer: C
8. Which two tasks can you perform using only the TO_CHAR function? (Choose two.)
A. convert 10 to 'TEN'
B. convert '10' to 10
C. convert 10 to '10'
D. convert 'TEN' to 10
E. convert a date to a character expression
F. convert a character expression to a date
Answer: C, E
9. Which four are types of functions available in SQL? (Choose 4)
A. string
B. character
C. integer
D. calendar
E. numeric
F. translation
G. date
H. conversion
Answer: B, E, G, H
10. Which SQL statement displays the date March 19, 2001 in a format that appears as
“Nineteenth of March 2001 12:00:00 AM”?
A. SELECT TO_CHAR(TO_DATE('19-Mar-2001’, ‘DD-Mon-YYYY’), ‘fmDdspth
“of” Month YYYY fmHH:MI:SS AM’) NEW_DATE
FROM dual;
B. SELECT TO_CHAR(TO_DATE(’19-Mar-2001’, ‘DD-Mon-YYYY’), ‘Ddspth
“of” Month YYYY fmHH:MI:SS AM’) NEW_DATE
FROM dual;
C. SELECT TO_CHAR(TO_DATE(’19-Mar-2001’, ‘DD-Mon-YYYY’), ‘fmDdspth “of” Month
YYYY HH:MI:SS AM’) NEW_DATE FROM dual;
D. SELECT TO_CHAR(TO_DATE(’19-Mar-2001’, ‘DD-Mon-YYYY), ‘fmDdspth “of” Month
YYYYfmtHH:HI:SS AM') NEW_DATE FROM dual;
Answer: A
11. You would like to display the system date in the format "Monday, 01 June, 2001".
Which SELECT statement should you use?
A. SELECT TO_DATE(SYSDATE, 'FMDAY, DD Month, YYYY') FROM dual;
B. SELECT TO_CHAR(SYSDATE, 'FMDD, DY Month, 'YYYY') FROM dual;
C. SELECT TO_CHAR(SYSDATE, 'FMDay, DD Month, YYYY') FROM dual;
D. SELECT TO_CHAR(SYSDATE, 'FMDY, DDD Month, YYYY') FROM dual;
E. SELECT TO_DATE(SYSDATE, 'FMDY, DDD Month, YYYY') FROM dual;
Answer: C
12. Which SELECT statement will the result ‘ello world’ from the string ‘Hello World’?
A. SELECT SUBSTR( ‘Hello World’,1) FROM dual;
B. SELECT INITCAP(TRIM (‘Hello World’, 1,1)) FROM dual;
C. SELECT LOWER(SUBSTR(‘Hello World’, 1, 1) FROM dual;
D. SELECT LOWER(SUBSTR(‘Hello World’, 2, 1) FROM dual;
E. SELECT LOWER(TRIM (‘H’ FROM ‘Hello World’)) FROM dual;
Answer: E
13. Which four statements correctly describe functions that are available in SQL? (Choose
four)
A. INSTR returns the numeric position of a named character.
B. NVL2 returns the first non-null expression in the expression list.
C. TRUNCATE rounds the column, expression, or value to n decimal places.
D. DECODE translates an expression after comparing it to each search value.
E. TRIM trims the heading of trailing characters (or both) from a character string.
F. NVL compares two expressions and returns null if they are equal, or the first
expression of they are not equal.
G. NULLIF compares twp expressions and returns null if they are equal, or the first
expression if they are not equal.
Answer: A, D, E, G
14. Which two tasks can you perform by using the TO_CHAR function? (Choose two)
A. Convert 10 to ‘TEN’
B. Convert ‘10’ to 10
C. Convert 10 to ‘10’
D. Convert ‘TEN’ to 10
E. Convert a date to a character expression
F. Convert a character expression to a date
Answer: C, E
15. Which two are character manipulation functions? (Choose two.)
A. TRIM
B. REPLACE
C. TRUNC
D. TO_DATE
E. MOD
F. CASE
Answer: A, B
16. Which SELECT statement should you use to extract the year from the system date and
display it in the format "1998"?
A. SELECT TO_CHAR(SYSDATE,'yyyy') FROM dual;
B. SELECT TO_DATE(SYSDATE,'yyyy') FROM dual;
C. SELECT DECODE(SUBSTR(SYSDATE, 8), 'YYYY') FROM dual;
D. SELECT DECODE(SUBSTR(SYSDATE, 8), 'year') FROM dual;
E. SELECT TO_CHAR(SUBSTR(SYSDATE, 8,2),'yyyy') FROM dual;
Answer: A
Which three SELECT statements displays 2000 in the format “$2,000.00”? (Choose three)
A. SELECT TO_CHAR(2000, ‘$#,###.##’) FROM dual;
B. SELECT TO_CHAR(2000, ‘$0,000.00’) FROM dual;
C. SELECT TO_CHAR(2000, ‘$9,999.00’) FROM dual;
D. SELECT TO_CHAR(2000, ‘$9,999.99’) FROM dual;
E. SELECT TO_CHAR(2000, ‘$2,000.00’) FROM dual;
F. SELECT TO_CHAR(2000, ‘$N,NNN.NN’) FROM dual;
Answer: B, C, D
Lesson 4: Joining Tables
1. Examine the structure of the EMPLOYEES and DEPARTMENTS tables:
EMPLOYEES
EMPLOYEE_ID NUMBER
DEPARTMENT_ID NUMBER
MANAGER_ID NUMBER
LAST_NAME VARCHAR2(25)
DEPARTMENTS
DEPARTMENT_ID NUMBER
MANAGER_ID NUMBER
DEPARTMENT_NAME VARCHAR2(35)
LOCATION_ID NUMBER
You want to create a report displaying employee last names, department names, and
locations. Which query should you use to create an equi-join?
A. SELECT last_name, department_name, location_id
FROM employees , departments ;
B. SELECT employees.last_name, departments.department_name, departments.location_id
FROM employees e, departments D
WHERE e.department_id =d.department_id;
C. SELECT e.last_name, d.department_name, d.location_id
FROM employees e, departments D
WHERE manager_id =manager_id;
D. SELECT e.last_name, d.department_name, d.location_id
FROM employees e, departments D
WHERE e.department_id =d.department_id;
Answer: D
2. In which two cases would you use an outer join? (Choose two.)
A. The tables being joined have NOT NULL columns.
B. The tables being joined have only matched data.
C. The columns being joined have NULL values.
D. The tables being joined have only unmatched data.
E. The tables being joined have both matched and unmatched data.
F. Only when the tables have a primary key/foreign key relationship.
Answer: C, E
3. You want to retrieve all employees’ last names, along with their manager’s last names
and their department names. Which query would you use?
A. SELECT last_name, manager_id, department_name
FROM employees e
FULL OUTER JOIN departments 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 departments 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)
LEFT OUTER JOIN departments 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 departments 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 departments d ON (e.department_id = d.department_id) ;
Answer: B
4.
The COMMISSION column shows the monthly commission earned by the employee.
Which two tasks would require subqueries or joins in order to be performed in a single
step? (Choose two.)
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
Answer: A, C
5. 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
6. Examine the structure of the EMPLOYEES and DEPARTMENTS tables:
EMPLOYEES
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
USING (e.department_id, d.department_id);
Answer: C
7. 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
8. 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
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: A, D
9. The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
The registrar has asked for a report on the average grade point average (GPA) for
students enrolled during semesters that end in the year 2000. Which statement
accomplish this?
A. SELECT AVERAGE(gpa)
FROM student_grades
WHERE semester_end > ’01-JAN-2000’ and semester end < 31-DEC-2000’;
B. SELECT COUNT(gpa)
FROM student grades
WHERE semester_end > ’01-JAN-2000’ and semester end < ’31-DEC-2000’;
C. SELECT MIN(gpa)
FROM student grades
WHERE semester_end > ’01-JAN-2000’ and semester end < ’31-DEC-2000’;
D. SELECT AVG(gpa)
FROM student_grades
WHERE semester_end BETWEEN ’01-JAN-2000’ and ’31.DEC.2000’;
E. SELECT SUM(gpa)
FROM student grades
WHERE semester_end > ’01-JAN-2000’ and semester end < ’31-DEC-2000’;
F. SELECT MEDIAN(gpa)
FROM student_grades
WHERE semester end > ’01-JAN-2000’ and semester end < ’31-DEC-2000’;
Answer: D
10. Examine the data of the EMPLOYEES table.
EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers
and refers to the EMPLOYEE_ID)
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 HR_MGR 5000
106 Bryan 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
Which statement lists the ID, name, and salary of the employee, and the ID and name of
the employee's manager, for all the employees who have a manager and earn more than
4000?
A. SELECT employee_id "Emp_id", emp_name "Employee",
salary, employee_id "Mgr_id", emp_name "Manager"
FROM employees WHERE salary > 4000;
B. SELECT e.employee_id "Emp_id", e.emp_name "Employee",
e.salary, m.employee_id "Mgr_id", m.emp_name "Manager"
FROM employees e, employees m
WHERE e.mgr_id = m.mgr_id
AND e.salary > 4000;
C. SELECT e.employee_id "Emp_id", e.emp_name "Employee",
e.salary, m.employee_id "Mgr_id", m.emp_name "Manager"
FROM employees e, employees m
WHERE e.mgr_id = m.employee_id
AND e.salary > 4000;
D. SELECT e.employee_id "Emp_id", e.emp_name "Employee",
e.salary, m.mgr_id "Mgr_id", m.emp_name "manager"
FROM employees e, employees m
WHERE e.mgr_id = m.employee_id
AND e.salary > 4000;
E. SELECT e.employee_id "Emp_id", e.emp_name "Employee",
e.salary, m.mgr_id "Mgr_id", m.emp_name "Manager"
FROM employees e, employees m
WHERE e.employee_id = m.employee_id
AND e.salary > 4000;
Answer: C
11. Examine the data of the EMPLOYEES table.
EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers
and refers to the EMPLOYEE_ID)
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 HR_MGR 5000
106 Bryan 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
Evaluate this SQL statement:
SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary,
m.employee_id "Mgr_id", m.emp_name "Manager"
FROM employees e, employees m
WHERE e.mgr_id = m.employee_id
AND e.salary > 4000;
What is its output?
A.
EMP_id EMPLOYEE SALARY Mgr_id Manager
110 Bob 8000 Bob
120 Ravi 6500 110 Ravi
108 Jennifer 6500 110 Jennifer
103 Chris 4200 120 Chris
105 Diana 5000 108 Diana
B.
EMP_id EMPLOYEE SALARY Mgr_id Manager
120 Ravi 6500 110 Bob
108 Jennifer 6500 110 Bob
103 Chris 4200 120 Ravi
105 Diana 5000 108 Jennifer
C.
EMP_id EMPLOYEE SALARY Mgr_id Manager
110 Bob 8000
120 Ravi 6500 110 Bob
108 Jennifer 6500 110 Bob
103 Chris 4200 120 Ravi
105 Diana 5000 108 Jennifer
D
EMP_id EMPLOYEE SALARY Mgr_id Manager
110 Bob 8000 110 Bob
120 Ravi 6500 120 Ravi
108 Jennifer 6500 108 Jennifer
103 Chris 4200 103 Chris
105 Diana 5000 105 Dina
E. The SQL statement produces an error.
Answer: B
12. In which case would you use a FULL OUTER JOIN?
A. Both tables have NULL values.
B. You want all unmatched data from one table.
C. You want all matched data from both tables.
D. You want all unmatched data from both tables.
E. One of the tables has more data than the other.
F. You want all matched and unmatched data from only one table.
Answer: D
13. Examine the structures of the EMPLOYEES and TAX 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
TAX
MIN_SALARY NUMBER
MAX_SALARY NUMBER
TAX_PERCENT NUMBER Percentage tax for given salary range
You need to find the percentage tax applicable for each employee. Which SQL statement
would you use?
A. SELECT employee_id, salary, tax_percent
FROM employees e, tax t
WHERE e.salary BETWEEN t.min_salary AND t.max_salary;
B. SELECT employee_id, salary, tax_percent
FROM employees e, tax t
WHERE e.salary > t.min_salary, tax_percent;
C. SELECT employee_id, salary, tax_percent
FROM employees e, tax t
WHERE MIN(e.salary) = t.min_salary
AND MAX(e.salary) = t.max_salary
D. You cannot find the information because there is no common column between the two
tables.
Answer: A
14. Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX 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
DEPARTMENTS
DEPARTMENT_ID NUMBER NOT NULL, Primary Key
DEPARTMENT_NAME VARCHAR2(30)
MGR_ID NUMBER References MGR_ID column of the
EMPLOYEES table
TAX
MIN_SALARY NUMBER
MAX_SALARY NUMBER
TAX_PERCENT NUMBER
For which situation would you use a nonequijoin query?
A. To find the tax percentage for each of the employees.
B. To list the name, job id, and manager name for all the employees.
C. To find the name, salary, and department name of employees who are not working
with Smith.
D. To find the number of employees working for the Administrative department and
earning less then 4000.
E. To display name, salary, manager ID, and department name of all the employees, even
if the employees do not have a department ID assigned.
Answer: A
15. What is true about joining tables through an equijoin?
A. You can join a maximum of two tables through an equijoin.
B. You can join a maximum of two columns through an equijoin.
C. You specify an equijoin condition in the SELECT or FROM clauses of a SELECT
statement.
D. To join two tables through an equijoin, the columns in the join condition must be
primary key and foreign key columns.
E. You can join n tables (all having single column primary keys) in a SQL statement by
specifying a minimum of n-1 join conditions.
Answer: E
16. Which three are true regarding the use of outer joins? (Choose three.)
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 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.
Answer: D, E, F
Lesson 5: Group Functions
1. 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
2. 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 VARCHAR2(30)
SALARY NUMBER(8,2)
Which statement shows the department ID, minimum salary, and maximum salary paid
in that department, only of the minimum salary is less then 5000 and the maximum
salary is more than 15000?
A. SELECT dept_id, MIN(salary(, MAX(salary)
FROM employees
WHERE MIN(salary) < 5000 AND MAX(salary) > 15000;
B. SELECT dept_id, MIN(salary), MAX(salary)
FROM employees
WHERE MIN(salary) < 5000 AND MAX(salary) > 15000
GROUP BY dept_id;
C. SELECT dept_id, MIN(salary), MAX(salary)
FROM employees
HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
D. SELECT dept_id, MIN(salary), MAX(salary)
FROM employees
GROUP BY dept_id
HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
E. SELECT dept_id, MIN(salary), MAX(salary)
FROM employees
GROUP BY dept_id, salary
HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;
Answer: D
3. Which two are true about aggregate functions? (Choose two.)
A. You can use aggregate functions in any clause of a SELECT statement.
B. You can use aggregate functions only in the column list of the SELECT clause and in
the WHERE clause of a SELECT statement.
C. You can mix single row columns with aggregate functions in the column list of a
SELECT statement by grouping on the single row columns.
D. You can pass column names, expressions, constants, or functions as parameters to an
aggregate function.
E. You can use aggregate functions on a table, only by grouping the whole table as one
single group.
F. You cannot group the rows of a table by more than one column while using aggregate
functions.
Answer: C, D
4. 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.
Which two statements find the number of customers? (Choose two.)
A. SELECT TOTAL(*) FROM customers;
B. SELECT COUNT(*) FROM customers;
C. SELECT TOTAL(customer_id) FROM customers;
D. SELECT COUNT(customer_id) FROM customers;
E. SELECT COUNT(customers) FROM customers;
F. SELECT TOTAL(customer_name) FROM customers;
Answer: B, D
5. You need to calculate the total of all salaries in the accounting department. Which group
function should you use?
A. MAX
B. MIN
C. SUM
D. COUNT
E. TOTAL
F. LARGEST
Answer: C
6. Which clause should you use to exclude group results?
A. WHERE
B. HAVING
C. RESTRICT
D. GROUP BY
E. ORDER BY
Answer: B
7. Which two statements are true about WHERE and HAVING clauses? (Choose two)
A. A WHERE clause can be used to restrict both rows and groups.
B. A WHERE clause can be used to restrict rows only.
C. A HAVING clause can be used to restrict both rows and groups.
D. A HAVING clause can be used to restrict groups only.
E. A WHERE clause CANNOT be used in a query of the query uses a HAVING clause.
F. A HAVING clause CANNOT be used in subqueries.
Answer: B, C
8. 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)
Which statement produces the number of different departments that have employees
with last name Smith?
A. SELECT COUNT(*) FROM employees WHERE last_name='Smith';
B. SELECT COUNT (dept_id) FROM employees WHERE last_name='Smith';
C. SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name='Smith';
D. SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name='Smith';
E. SELECT UNIQUE(dept_id) FROM employees WHERE last_name='Smith';
Answer: D
9. What is true of using group functions on columns that contain NULL values?
A. Group functions on columns ignore NULL values.
B. Group functions on columns returning dates include NULL values.
C. Group functions on columns returning numbers include NULL values.
D. Group functions on columns cannot be accurately used on columns that contain NULL
values.
E. Group functions on columns include NULL values in calculations if you use the keyword
INC_NULLS.
Answer: A
10. The STUDENT_GRADES table has these columns:
STUDENT_ID NUMBER(12)
SEMESTER_END DATE
GPA NUMBER(4,3)
Which statement finds the highest grade point average (GPA) per semester?
A. SELECT MAX(gpa) FROM student_grades WHERE gpa IS NOT NULL;
B. SELECT (gpa) FROM student_grades GROUP BY semester_end WHERE gpa IS NOT
NULL;
C. SELECT MAX(gpa) FROM student_grades WHERE gpa IS NOT NULL GROUP BY
semester_end;
D. SELECT MAX(gpa) GROUP BY semester_end WHERE gpa IS NOT NULL FROM
student_grades;
E. SELECT MAX(gpa) FROM student_grades GROUP BY semester_end WHERE gpa IS
NOT NULL;
Answer: C
11. You need to write a SQL statement that returns 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.
Which statement accomplishes this task?
A. SELECT a.emp_name, a.sal, b.dept_id, MAX(sal) FROM employees a, departments b
WHERE a.dept_id = b.dept_id AND a.sal < MAX(sal) GROUP BY b.dept_id;
B. SELECT a.emp_name, a.sal, a.dept_id, b.maxsal FROM employees a, (SELECT dept_id,
MAX(sal) maxsal FROM employees GROUP BY dept_id) b WHERE a.dept_id = b.dept_id
AND a.sal < b.maxsal;
C. SELECT a.emp_name, a.sal, a.dept_id, b.maxsal FROM employees a WHERE a.sal <
(SELECT MAX(sal) maxsal FROM employees b GROUP BY dept_id);
D. SELECT emp_name, sal, dept_id, maxsal FROM employees, (SELECT dept_id,
MAX(sal) maxsal FROM employees GROUP BY dept_id) WHERE a.sal < maxsal;
Answer: B
13. Which clause would you use in a SELECT statement to limit the display to those
employees whose salary is greater then 5000?
A. ORDER BY SALARY > 5000
B. GROUP BY SALARY > 5000
C. HAVING SALARY > 5000
D. WHERE SALARY > 5000
Answer: D
14. The CUSTOMERS table has these columns:
CUSTOMER_ID NUMBER (4) NOT NULL
CUSTOMER_NAME VARCHAR2 (100) NOT NULL
STREET_ADDRESS VARCHAR2 (150)
CITY_ADDRESS VARHCAR2 (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
15. Examine the description of the CUSTOMERS table:
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.
Which statement returns the city address and the number of customers in the cities Los
Angeles or San Francisco?
A. SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los
Angeles', 'San Francisco');
B. SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los
Angeles', 'San Francisco') GROUP BY city_address;
C. SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN
('Los Angeles', 'San Francisco') GROUP BY city_address, customer_id;
D. SELECT city_address, COUNT(customer_id) FROM customers GROUP BY city_address
IN ('Los Angeles', 'San Francisco');
Answer: B
Lesson 6: Subqueries
1. 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
2. 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
3. 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
4. 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 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
5. Which two statements about subqueries are true? (Choose two.)
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 table 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 a condition where the LIKE operator is
used for comparison.
Answer: B, D
6. Examine the structure of the STUDENTS table:
STUDENT_ID NUMBER NOT NULL, Primary Key
STUDENT_NAME VARCHAR2(30)
COURSE_ID VARCHAR2(10) NOT NULL
MARKS NUMBER
START_DATE DATE
FINISH_DATE DATE
You need to create a report of the 10 students who achieved the highest ranking in the
course INT SQL and who completed the course in the year 1999.
Which SQL statement accomplishes this task?
A. SELECT student_ id, marks, ROWNUM "Rank"
FROM students
WHERE ROWNUM <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99'
AND course_id = 'INT_SQL'
ORDER BY marks DESC;
B. SELECT student_id, marks, ROWID "Rank"
FROM students
WHERE ROWID <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99'
AND course_id = 'INT_SQL'
ORDER BY marks;
C. SELECT student_id, marks, ROWNUM "Rank"
FROM (SELECT student_id, marks
FROM students
WHERE ROWNUM <= 10
AND finish_date BETWEEN '01-JAN-99' AND
'31-DEC-99'
AND course_id = 'INT_SQL'
ORDER BY marks DESC);
D. SELECT student_id, marks, ROWNUM "Rank”
FROM (SELECT student_id, marks
FROM students
ORDER BY marks DESC)
WHERE ROWNUM <= 10
AND finish_date BETWEEN '01-JAN-99' AND '31-DEC-99'
AND course_id = 'INT_SQL';
Answer: D
7. Examine the data from the EMP table:
EMP_ID DEPT_ID COMMISSION
1 10 500
2 20 1000
3 10
4 10 600
5 30 800
6 30 200
7 10
8 20 300
The COMMISSION column shows the monthly commission earned by the employee.
Which three tasks would require subqueries or joins in order to perform in a single
step? (Choose three)
A. Deleting the records of employees who do not earn commission.
B. Increasing the commission of employee 3 by the average commission earned in
department 20.
C. Finding the number of employees who do NOT earn commission and are working for
department 20.
D. Inserting into the table a new employee 10 who works for department 20 and earns a
commission that is equal to the commission earned by employee 3.
E. Creating a table called COMMISSION that has the same structure and data as the
columns EMP_ID and COMMISSIONS of the EMP table.
F. Decreasing the commission by 150 for the employees who are working in department
30 and earning a commission of more then 800.
Answer: B, D, E
8. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
DEPARTMENT_ID NUMBER
SALARY NUMBER
What is the correct syntax for an inline view?
A. SELECT a.last_name, a.salary, a.department_id,
b.maxsal
FROM employees a,
(SELECT department_id, max(salary)maxsal
FROM employees
GROUP BY department_id) b
WHERE a.department_id = b.department_id
AND a.salary < b.maxsal;
B. SELECT a.last name, a.salary, a.department_id
FROM employees a
WHERE a.department_id IN
(SELECT department_id
FROM employees b
GROUP BY department_id having salary =
(SELECT max(salary) from employees))
C. SELECT a.last_name, a.salary, a.department_id
FROM employees a
WHERE a.salary =
(SELECT max(salary)
FROM employees b
WHERE a.department_id = b.department_id);
D. SELECT a.last_name, a.salary, a.department_id
FROM employees a
WHERE (a.department_id, a.salary) IN
(SELECT department_id, a.salary) IN
(SELECT department_id max(salary)
FROM employees b
GROUP BY department_id
ORDER BY department_id);
Answer: A
9. Which three statements about subqueries are true? (Choose three)
A. A single row subquery can retrieve only one column and one row.
B. A single row subquery can retrieve only one row but many columns.
C. A multiple row subquery can retrieve multiple rows and multiple columns.
D. A multiple row subquery can be compared by using the “>” operator.
E. A single row subquery can use the IN operator.
F. A multiple row subquery can use the “=” operator.
Answer: B, C, D
10. Which operator can be used with a multiple-row subquery?
A. =
B. LIKE
C. BETWEEN
D. NOT IN
E. IS
F. <>
Answer: D
11. Examine the data from the ORDERS and CUSTOMERS tables.
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
109 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 Lind New York
Evaluate the SQL statement:
SELECT *
FROM orders
WHERE cust_id = (SELECT cust_id
FROM customers
WHERE cust_name = 'Smith');
What is the result when the query is executed?
A.
ORD_ID ORD_DATE CUST_ID ORD_TOTAL
102 09-MAR-200 35 12500
106 18-JUL-2000 35 7000
108 04-AUG-2000 10 8000
B.
ORD_ID ORD_DATE CUST_ID ORD_TOTAL
102 09-MAR-200 35 12500
106 18-JUL-2000 35 7000
C.
ORD_ID ORD_DATE CUST_ID ORD_TOTAL
108 04-AUG-2000 10 8000
D. The query fails because the subquery returns more than one row.
E. The query fails because the outer query and the inner query are using different tables.
Answer: D
12. You define a multiple-row subquery in the WHERE clause of an SQL query with a
comparison operator "=". What happens when the main query is executed?
A. The main query executes with the first value returned by the subquery.
B. The main query executes with the last value returned by the subquery.
C. The main query executes with all the values returned by the subquery.
D. The main query fails because the multiple-row subquery cannot be used with the
comparison operator.
E. You cannot define a multiple-row subquery in the WHERE clause of a SQL query.
Answer: D
13. A subquery can be used to _________.
A. Create groups of data
B. Sort data in a specific order
C. Convert data to a different format
D. Retrieve data based on an unknown condition
Answer: D
14. In which scenario would TOP N analysis be the best solution?
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.
Answer: D
15. EMPLOYEES and DEPARTMENTS data:
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
On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID
managers and refers to the EMPLOYEE_ID.
On the DEPARTMENTS table DEPARTMENT_ID is the primary key.
Evaluate this UPDATE statement.
UPDATE employees SET mgr_id = (SELECT mgr_id
FROM employees
WHERE dept_id= (SELECT department_id
FROM departments
WHERE department_name = 'Administration')),
Salary = (SELECT salary FROM employees
WHERE emp_name = 'Smith') WHERE job_id = 'IT_ADMIN';
What happens when the statement is executed?
A. The statement executes successfully, leaves the manager ID as the existing value, and
changes the salary to 4000 for the employees with ID 103 and 105.
B. The statement executes successfully, changes the manager ID to NULL, and changes
the salary to 4000 for the employees with ID 103 and 105.
C. The statement executes successfully, changes the manager ID to NULL, and changes
the salary to 3000 for the employees with ID 103 and 105.
D. The statement fails because there is more than one row matching the employee name
Smith.
E. The statement fails because there is more than one row matching the IT_ADMIN job
ID in the EMPLOYEES table.
F. The statement fails because there is no 'Administration' department in the
DEPARTMENTS table.
Answer: D
16. Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:
EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2 (25)
LAST_NAME VARCHAR2 (25)
HIRE_DATE DATE
NEW EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
NAME VARCHAR2 (60)
Which DELETE statement is valid?
A. DELETE FROM employees
WHERE employee_id = (SELECT employee_id
FROM employees);
B. DELETE * FROM employees
WHERE employee_id = (SELECT employee_id
FROM new_employees);
C. DELETE FROM employees
WHERE employee_id IN (SELECT employee_id
FROM new_employees
WHERE name = 'Carrey');
D. DELETE * FROM employees
WHERE employee_id IN (SELECT employee_id
FROM new_employees
WHERE last_name = 'Carrey');
Answer: C
17. In which four clauses can a subquery be used? (Choose four.)
A. in the INTO clause of an INSERT statement
B. in the FROM clause of a SELECT statement
C. in the GROUP BY clause of a SELECT statement
D. in the WHERE clause of a SELECT statement
E. in the SET clause of an UPDATE statement
F. in the VALUES clause of an INSERT statement
Answer: B, D, E, F
18. Examine the data in the EMPLOYEES table.
LAST_NAME DEPARTMENT_ID SALARY
Getz 10 3000
Davis 20 1500
King 20 2200
Davis 30 6000
…
Examine the subquery:
SELECT last_name FROM employees
WHERE salary IN (SELECT MAX(salary)
FROM employees GROUP BY department_id);
Which statement is true?
A. The SELECT statement is syntactically accurate.
B. The SELECT statement does not work because there is no HAVING clause.
C. The SELECT statement does not work because the column specified in the GROUP BY
clause is not in the SELECT list.
D. The SELECT statement does not work because the GROUP BY clause should be in the
main query and not in the subquery.
Answer: A
19. Which two statements about subqueries are true? (Choose two.)
A. A subquery should retrieve only one row.
B. A subquery can retrieve zero or more rows.
C. A subquery can be used only in SQL query statements.
D. Subqueries CANNOT be nested by more than two levels.
E. A subquery CANNOT be used in an SQL query statement that uses group functions.
F. When a subquery is used with an inequality comparison operator in the outer SQL
statement, the column list in the SELECT clause of the subquery should contain only one
column.
Answer: B, F
20
ORDERS
ORD_ID ORD_DATE CUST_ID ORD_TOTAL
100 12-JAB-2000 15 10000
101 09-MAR-2000 40 8000
102 09-MAR-2000 35 12500
103 15-MAR-2000 15 12500
104 25-JUN-2000 15 6000
105 18-JUL-2000 20 5000
106 18-JUL-2000 35 7000
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
Evaluate this SQL statement:
SELECT cust_id, ord_total
FROM orders
WHERE ord_total > ANY (SELECT ord_total
FROM orders
WHERE cust_id IN (SELECT cust_id
FROM customers
WHERE city LIKE
‘New York’));
What is the result when the above query is executed?
A.
CUST_ID ORD_TOTAL
15 10000
40 8000
35 12500
15 12500
10 8000
B.
CUST_ID ORD_TOTAL
15 10000
35 12500
15 12000
ok
C.
CUST_ID ORD_TOTAL
15 10000
40 8000
15 12000
15 6000
20 5000
35 7000
20 6500
10 8000
D.
CUST_ID ORD_TOTAL
15 6000
20 5000
20 6500
E. The query returns no rows.
F. The query fails because ANY is not a valid operator with a subquery.
Answer: A
Lesson 7: Data Manipulation
1. You added a PHONE_NUMBER column of NUMBER data type to an existing
EMPLOYEES table. The EMPLOYEES table already contains records of 100
employees. Now, you want to enter the phone numbers of each of the 100 employees into
the table.
Some of the employees may not have a phone number available.
Which data manipulation operation do you perform?
A. MERGE
B. INSERT
C. UPDATE
D. ADD
E. ENTER
F. You cannot enter the phone numbers for the existing employee records.
Answer: C
2. Which are DML statements? (Choose all that apply)
A. COMMIT…
B. MERGE…
C. UPDATE…
D. DELETE…
E. CREATE…
F. DROP…
Answer: B, C, D
3. 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
4. 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.
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
5. Examine the structure if the EMPLOYEES and NEW EMPLOYEES tables:
EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
NEW EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
NAME VARCHAR2(60)
Which MERGE statement is valid?
A. MERGE INTO new_employees c
USING employees e
ON (c.employee_id = e.employee_id)
WHEN MATCHED THEN
UPDATE SET
c.name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT VALUES(e.employee_id, e.first_name ||’,
‘||e.last_name);
B. MERGE new_employees c
USING employees e
ON (c.employee_id = e.employee_id)
WHEN EXIST THEN
UPDATE SET
c.name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT VALUES(e.employee_id, e.first_name ||’,
‘||e.last_name);
C. MERGE INTO new employees c
USING employees e
ON (c.employee_id = e.employee_id)
WHEN EXISTS THEN
UPDATE SET
c.name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT VALUES(e.employee_id, e.first_name ||’,
‘||e.last_name);
D. MERGE new_employees c
FROM employees e
ON (c.employee_id = e.employee_id)
WHEN MATCHED THEN
UPDATE SET
c.name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT INTO new_employees VALUES(e.employee_id, e.first_name ||’.’||e.last_name);
Answer: A
6. Examine the data in the EMPLOYEES and EMP_HIST 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 HR_MGR 5000
106 Bryan 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
EMP_HIST
EMPLOYEE_ID NAME JOB_ID SALARY
101 Smith SA_CLERK 2000
103 Chris IT_CLERK 2200
104 John HR_CLERK 2000
106 Smith AD_ASST 3000
108 Jennifer HR_MGR 4500
The EMP_HIST table is updated at the end of every year. The employee ID, name, job
ID, and salary of each existing employee are modified with the latest data. New employee
details are added to the table.
Which statement accomplishes this task?
A. UPDATE emp_hist
SET employee_id, name, job_id, salary =
(SELECT employee_id, name, job_id, salary
FROM employees)
WHERE employee_id IN
(SELECT employee_id
FROM employees);
B. MERGE INTO emp_hist eh
USING employees e
ON (eh.employee_id = e.employee_id)
WHEN MATCHED THEN
UPDATE SET eh.name = e.name,
eh.job_id = e.job_id,
eh.salary = e.salary
WHEN NOT MATCHED THEN
INSERT VALUES (e.employee id, e.name,
e.job id, e.salary);
C. MERGE INTO emp_hist eh
USING employees e
ON (eh.employee_id = e.employee_id)
WHEN MATCHED THEN
UPDATE emp hist SET eh.name = e.name,
eh.job_id = e.job_id,
eh.salary = e.salary
WHEN NOT MATCHED THEN
INSERT INTO emp_hist
VALUES (e.employee_id, e.name, e.job_id, e.salary);
D. MERGE INTO emp_hist eh
USING employees e
WHEN MATCHED THEN
UPDATE emp_hist SET eh.name = e.name,
eh.job_id = e.job_id,
eh.salary = e.salary
WHEN NOT MATCHED THEN
INSERT INTO emp_hist
VALUES (e.employee_id, e.name, e.job_id, e.salary);
Answer: B
7. Examine the data in the EMPLOYEES table.
EMPLOYEES
EMPLOYEE_ID 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 Bryan 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
On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of
managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL
column.
Evaluate this DELETE statement:
DELETE employee_id, salary, job_id
FROM employees
WHERE dept_id = 90;
Why does the DELETE statement fail when you execute it?
A. There is no row with dept_id 90 in the EMPLOYEES table.
B. You cannot delete the JOB_ID column because it is a NOT NULL column.
C. You cannot specify column names in the DELETE clause of the DELETE statement.
D. You cannot delete the EMPLOYEE_ID column because it is the primary key of the
table.
Answer: C
8. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
You issue these statements:
CREATE table new_emp ( employee_id NUMBER, name VARCHAR2(30));
INSERT INTO new_emp SELECT employee_id , last_name from employees;
Savepoint s1;
UPDATE new_emp set name = UPPER(name);
Savepoint s2;
Delete from new_emp;
Rollback to s2;
Delete from new_emp where employee_id =180;
UPDATE new_emp set name = 'James';
Rollback to s2;
UPDATE new_emp set name = 'James' WHERE employee_id =180;
Rollback;
At the end of this transaction, what is true?
A. You have no rows in the table.
B. You have an employee with the name of James.
C. You cannot roll back to the same savepoint more than once.
D. Your last update fails to update any rows because employee ID 180 was already
deleted.
Answer: A
9. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER NOT NULL
EMP_NAME VARCHAR2(30)
JOB_ID VARCHAR2(20) DEFAULT 'SA_REP'
SAL NUMBER
COMM_PCT NUMBER
MGR_ID NUMBER
DEPARTMENT_ID NUMBER
You need to update the records of employees 103 and 115. The UPDATE statement you
specify should update the rows with the values specified below:
JOB_ID: Default value specified for this column definition.
SAL: Maximum salary earned for the job ID SA_REP.
COMM_PCT: Default value specified for this commission percentage column, if any.
If no default value is specified for the column, the value should be
NULL.
DEPARTMENT_ID: Supplied by the user during run time through substitution variable.
Which UPDATE statement meets the requirements?
A. UPDATE employees
SET job_id = DEFAULT
AND Sal = (SELECT MAX(sal)
FROM employees
WHERE job_id = 'SA_REP')
AND comm_pct = DEFAULT
AND department_id = &did
WHERE employee_id IN (103,115);
B. UPDATE employees
SET job_id = DEFAULT
AND Sal = MAX(sal)
AND comm_pct = DEFAULT OR NULL
AND department_id = &did
WHERE employee_id IN (103,115)
AND job_id = 'SA_REP';
C. UPDATE employees
SET job_id = DEFAULT,
Sal = (SELECT MAX(sal)
FROM employees
WHERE job_id = 'SA_REP'),
comm_pct = DEFAULT,
department_id = &did
WHERE employee_id IN (103,115);
D. UPDATE employees
SET job_id = DEFAULT,
Sal = MAX(sal),
comm_pct = DEFAULT,
department_id = &did
WHERE employee_id IN (103,115)
AND job_id = 'SA_REP';
E. UPDATE employees
SET job_id = DEFAULT,
Sal = (SELECT MAX(sal)
FROM employees
WHERE job_id = 'SA_REP')
comm_pct = DEFAULT OR NULL,
department_id = &did
WHERE employee_id IN (103,115);
Answer: C
10. Which two statements complete a transaction? (Choose two)
A. DELETE employees;
B. DESCRIBE employees;
C. ROLLBACK TO SAVEPOINT C;
D. GRANT SELECT ON employees TO SCOTT;
E. ALTER TABLE employees SET UNUSED COLUMN sal;
F. Select MAX(sal) FROM employees WHERE department_id = 20;
Answer: D,E
11. A data manipulation language statement _____.
A. completes a transaction on a table
B. modifies the structure and data in a table
C. modifies the data but not the structure of a table
D. modifies the structure but not the data of a table
Answer: C
12. You own a table called EMPLOYEES with this table structure:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
What happens when you execute this DELETE statement?
DELETE employees;
A. You get an error because of a primary key violation.
B. The data and structure of the EMPLOYEES table are deleted.
C. The data in the EMPLOYEES table is deleted but not the structure.
D. You get an error because the statement is not syntactically correct.
Answer : C
13. Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:
EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
NEW_EMPLOYEES
EMPLOYEE_ID NUMBER Primary Key
NAME VARCHAR2(60)
Which MERGE statement is valid?
A. MERGE INTO new_employees c USING employees e ON (c.employee_id =
e.employee_id) WHEN MATCHED THEN UPDATE SET c.name = e.first_name ||','||
e.last_name WHEN NOT MATCHED THEN INSERT VALUES (e.employee_id,
e.first_name ||', '||e.last_name);
B. MERGE new_employees c USING employees e ON (c.employee_id = e.employee_id)
WHEN EXISTS THEN UPDATE SET c.name = e.first_name ||','|| e.last_name WHEN NOT
MATCHED THEN INSERT VALUES (e.employee_id, e.first_name ||', '||e.last_name);
C. MERGE INTO new_employees c USING employees e ON (c.employee_id =
e.employee_id) WHEN EXISTS THEN UPDATE SET c.name = e.first_name ||','||
e.last_name WHEN NOT MATCHED THEN INSERT VALUES(e.employee_id,
e.first_name ||', '||e.last_name);
D. MERGE new_employees c FROM employees e ON (c.employee_id = e.employee_id)
WHEN MATCHED THEN UPDATE SET c.name = e.first_name ||','|| e.last_name WHEN
NOT MATCHED THEN INSERT INTO new_employees VALUES (e.employee_id,
e.first_name ||', '||e.last_name);
Answer: A
14. Which three are true? (Choose three.)
A. A MERGE statement is used to merge the data of one table with data from another.
B. A MERGE statement replaces the data of one table with that of another.
C. A MERGE statement can be used to insert new rows into a table.
D. A MERGE statement can be used to update existing rows in a table.
Answer: A, C, D
15. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
HIRE_DATE DATE
Which INSERT statement is valid?
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);
Answer: B
16. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2 (25)
LAST_NAME VARCHAR2 (25)
HIRE_DATE DATE
Which UPDATE statement is valid?
A. UPDATE employees
SET first_name = ‘John’
SET last_name = ‘Smith’
WHERE employee_id = 180;
B. UPDATE employees
SET first_name = ‘John’,
SET last_name = ‘Smoth’
WHERE employee_id = 180;
C. UPDATE employee
SET first_name = ‘John’
AND last_name = ‘Smith’
WHERE employee_id = 180;
D. UPDATE employee
SET first_name = ‘John’, last_name = ‘Smith’
WHERE employee_id = 180;
Answer: D
Lesson 8: Working with Tables.
1. Which describes the default behavior when you create a table?
A. The table is accessible to all users.
B. Tables are created in the public schema.
C. Tables are created in your schema.
D. Tables are created in the DBA schema.
E. You must specify the schema when the table is created.
Answer: C
2. You need to create a table named ORDERS that contains four columns:
1. an ORDER_ID column of number data type
2.- a CUSTOMER_ID column of number data type
3. an ORDER_STATUS column that contains a character data type
4. a DATE_ORDERED column to contain the date the order was placed
When a row is inserted into the table, if no value is provided for the status of the order,
the value PENDING should be used instead.
Which statement accomplishes this?
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 = SYSDATE);
D. CREATE TABLE orders (
order_id NUMBER (10),
customer_id NUMBER (8)
order_status NUMBER (10),
date_ordered DATE DEFAULT SYSDATE);
Answer: B
3. Which is a valid CREATE TABLE statement?
A. CREATE TABLE EMP9$# (empid number(2));
B. CREATE TABLE EMP*123 (empid number(2));
C. CREATE TABLE PACKAGE (packid number(2));
D. CREATE TABLE 1EMP_TEST (empid number(2));
Answer: A
4. 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
5. Which four are correct guidelines for naming database tables? (Choose four)
A. Must begin with either a number or a letter.
B. Must be 1-30 characters long.
C. Should not be an Oracle Server reserved word.
D. Must contain only A-Z, a-z, 0-+, _, *, and #.
E. Must contain only A-Z, a-z, 0-9, _, $, and #.
F. Must begin with a letter.
Answer: B, C, E, F
6. What does the TRUNCATE statement do?
A. Removes the table
B. Removes all rows from a table
C. Shortens the table to 10 rows
D. Removes all columns from a table
E. Removes foreign keys from a table
Answer: B
7. You need to change the definition of an existing table. The COMMERCIALS table needs
its DESCRIPTION column changed to hold varying length characters up to 2000 bytes.
The column can currently hold 1000 bytes per value. The table contains 20000 rows.
Which statement is valid?
A. ALTER TABLE commercials MODIFY (description CHAR2(2000));
B. ALTER TABLE commercials CHANGE (description CHAR2(2000));
C. ALTER TABLE commercials CHANGE (description VARCHAR2(2000));
D. ALTER TABLE commercials MODIFY (description VARCHAR2(2000));
E. You cannot increase the size of a column if the table has rows.
Answer: D
8. The EMPLOYEES table has these columns:
LAST NAME VARCHAR2(35)
SALARY NUMBER(8,2)
HIRE_DATE DATE
Management wants to add a default value to the SALARY column. You plan to alter the
table by using this SQL statement:
ALTER TABLE EMPLOYEES
MODIFY (SALARY DEFAULT 5000);
What is true about your ALTER statement?
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.
Answer: B
9. Which statement describes the ROWID data type?
A. Binary data up to 4 gigabytes.
B. Character data up to 4 gigabytes.
C. Raw binary data of variable length up to 2 gigabytes.
D. Binary data stored in an external file, up to 4 gigabytes.
E. A hexadecimal string representing the unique address of a row in its table.
Answer: E
10. Evaluate the SQL statement
DROP TABLE DEPT:
Which four statements are true of the SQL statement? (Choose four)
A. You cannot roll back this statement.
B. All pending transactions are committed.
C. All views based on the DEPT table are deleted.
D. All indexes based on the DEPT table are dropped.
E. All data in the table is deleted, and the table structure is also deleted.
F. All data in the table is deleted, but the structure of the table is retained.
G. All synonyms based on the DEPT table are deleted.
Answer: A, B, D, E
11. Evaluate the SQL statement:
TRUNCATE TABLE DEPT;
Which three are true about the SQL statement? (Choose three.)
A. It releases the storage space used by the table.
B. It does not release the storage space used by the table.
C. You can roll back the deletion of rows after the statement executes.
D. You can NOT roll back the deletion of rows after the statement executes.
E. An attempt to use DESCRIBE on the DEPT table after the TRUNCATE statement
executes will display an error.
F. You must be the owner of the table or have DELETE ANY TABLE system privileges to
truncate the DEPT table
Answer: A, D, F
Lesson 9: Constraints
1. Which four are valid Oracle constraint types? (Choose four.)
A. CASCADE
B. UNIQUE
C. NONUNIQUE
D. CHECK
E. PRIMARY KEY
F. CONSTANT
G. NOT NULL
Answer: B, D, E, G
2. Which statement adds a constraint that ensures the CUSTOMER_NAME column of the
CUSTOMERS table holds a value?
A. ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name
IS NOT NULL;
B. ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK
customer_name IS NOT NULL;
C. ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn
NOT NULL;
D. ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS
NOT NULL;
E. ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
F. ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name
NOT NULL;
Answer: C
3. 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 studentsMODIFY CONSTRAINT stud_id_pk PRIMARY KEY(student_id);
Answer: D
4. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key
FIRST_NAME VARCHAR2(25)
LAST_NAME VARCHAR2(25)
Which three statements insert 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
5. Examine the SQL statement that creates ORDERS table:
CREATE TABLE orders (SER_NO NUMBER UNIQUE, ORDER_ID NUMBER,
ORDER_DATE DATE NOT NULL, STATUS VARCHAR2(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
6. Which constraint can be defined only at the column level?
A. UNIQUE
B. NOT NULL
C. CHECK
D. PRIMARY KEY
E. FOREIGN KEY
Answer: B
7. Which view should a user query to display the columns associated with the constraints
on a table owned by the user?
A. USER_CONSTRAINTS
B. USER_OBJECTS
C. ALL_CONSTRAINTS
D. USER_CONS_COLUMNS
E. USER_COLUMNS
Answer : D
8. 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, gpa
NUMBER(4,3), CONSTRAINT student_id_fk FOREIGN KEY (student_id) REFERENCES
students(student_id));
Answer: D
9. 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.
D. The NOT NULL constraint ensures that null values are not permitted for the column.
Answer: B, D
10. 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
11. 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;
F. ALTER TABLE table_name TURN ON CONSTRAINT constraint_name;
Answer: C
12. 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
13. 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 employees(employee id), JOB_ID VARCHAR2(15).
SALARY NUMBER);
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
14. 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)
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
15. 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
16. 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: C, D, E
17. Which constraint can be defines only at the column level?
A. UNIQUE
B. NOT NULL
C. CHECK
D. PRIMARY KEY
E. FOREIGN KEY
Answer: B
18. Which two statements about creating constraints are true? (Choose two)
A. Constraint names must start with SYS_C.
B. All constraints must be defines at the column level.
C. Constraints can be created after the table is created.
D. Constraints can be created at the same time the table is created.
E. Information about constraints is found in the VIEW_CONSTRAINTS dictionary view.
Answer: C, D
Lesson 10: Views
1. Top N analysis requires _____ and _____. (Choose two.)
A. the use of rowid
B. a GROUP BY clause
C. an ORDER BY clause
D. only an inline view
E. an inline view and an outer query
Answer: C, E
2. Which object privileges can be granted on a view?
A. none
B. DELETE, INSERT,SELECT
C. ALTER, DELETE, INSERT, SELECT
D. DELETE, INSERT, SELECT, UPDATE
Answer: D
3. What is true about updates through a view?
A. You cannot update a view with group functions.
B. When you update a view group functions are automatically computed.
C. When you update a view only the constraints on the underlying table will be in effect.
D. When you update a view the constraints on the views always override the constraints on
the underlying tables.
Answer: A
4. 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
5. Which best describes an inline view?
A. a schema object
B. a subquery that can contain an ORDER BY clause
C. another name for a view that contains group functions
D. a subquery that is part of the FROM clause of another query
Answer: D
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
7. 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
8. Examine the structure if the EMPLOYEES table:
Column name Data Type Remarks
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2(30)
JOB_ID VARCHAR2(20) NOT NULL
SAL NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID
column of the DEPARTMENTS table
You need to create a view called EMP_VU that allows the user to insert rows through
the view. Which SQL statement, when used to create the EMP_VU view, allows the user
to insert rows?
A. CREATE VIEW emp_Vu AS
SELECT employee_id, emp_name,
department_id
FROM employees
WHERE mgr_id IN (102, 120);
B. CREATE VIEW emp_Vu AS
SELECT employee_id, emp_name, job_id
department_id
FROM employees
WHERE mgr_id IN (102, 120);
C. CREATE VIEW emp_Vu AS
SELECT department_id, SUM(sal) TOTALSAL
FROM employees
WHERE mgr_id IN (102, 120)
GROUP BY department_id;
D. CREATE VIEW emp_Vu AS
SELECT employee_id, emp_name, job_id,
DISTINCT department_id
FROM employees;
Answer: B
9. You need to create a view EMP_VU. The view should allow the users to manipulate the
records of only the employees that are working for departments 10 or 20.
Which SQL statement would you use to create the view EMP_VU?
A. CREATE VIEW emp_vu AS
SELECT * FROM employees
WHERE department_id IN (10,20);
B. CREATE VIEW emp_vu AS
SELECT * FROM employees
WHERE department_id IN (10,20)
WITH READ ONLY;
C. CREATE VIEW emp_vu AS
SELECT * FROM employees
WHERE department_id IN (10,20)
WITH CHECK OPTION;
D. CREATE FORCE VIEW emp_vu AS
SELECT * FROM employees
WHERE department_id IN (10,20);
E. CREATE FORCE VIEW emp_vu AS
SELECT * FROM employees
WHERE department_id IN (10,20)
NO UPDATE;
Answer: C
10. Mary has a view called EMP_DEPT_LOC_VU that was created based on the
EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. She granted SELECT
privilege to Scott on this view.
Which option enables Scott to eliminate the need to qualify the view with the name
MARY .EMP_DEP_LOC_VU each time the view is referenced?
A. Scott can create a synonym for the EMP_DEPT_LOC_VU bus using the command:
CREATE PRIVATE SYNONYM EDL_VU
FOR mary.EMP DEPT_LOC_VU;
then he can prefix the columns with this synonymn.
B. Scott can create a synonym for the EMP_DEPT_LOC_VU by using the command:
CREATE SYNONYM EDL_VU
FOR mary.EMP_DEPT_LOC_VU;
then he can prefix the columns with this synonym.
C. Scott can create a synonym for the EMP_DEPT_LOC_VU by using the command:
CREATE LOCAL SYNONYM EDL_VU
FOR mary.EMP DEPT_LOC_VU;
then he can prefix the columns with this synonym.
D. Scott can create a synonym for the EMP_DEPT_LOC_VU by using the command:
CREATE SYNONYM EDL_VU
ON mary(EMP_DEPT_LOC_VU);
then he can prefix the columns with this synonym.
E. Scott cannot create a synonym because synonyms can be created only for tables.
F. Scott cannot create any synonym for Mary’s view. Mary should create a private
synonym for the view and grant SELECT privilege on that synonym to Scott.
Answer: B
11. You need to perform certain data manipulation operations through a view called
EMP_DEPT_VU, which you previously created.
You want to look at the definition of the view (the SELECT statement on which the view
was create.)
How do you obtain the definition of the view?
A. Use the DESCRIBE command in the EMP_DEPT VU view.
B. Use the DEFINE VIEW command on the EMP_DEPT VU view.
C. Use the DESCRIBE VIEW command on the EMP_DEPT VU view.
D. Query the USER_VIEWS data dictionary view to search for the EMP_DEPT_VU
view.
E. Query the USER_SOURCE data dictionary view to search for the EMP_DEPT_VU
view.
F. Query the USER_OBJECTS data dictionary view to search for the EMP_DEPT_VU
view.
Answer: D
12. You are granted the CREATE VIEW privilege. What does this allow you to do?
A. Create a table view.
B. Create a view in any schema.
C. Create a view in your schema.
D. Create a sequence view in any schema.
E. Create a view that is accessible by everyone.
F. Create a view only of it is based on tables that you created.
Answer: C
13. Examine the structure of the EMP_DEPT_VU view:
Column Name Type Remarks
EMPLOYEE_ID NUMBER From the EMPLOYEES table
EMP_NAME VARCHAR2(30) From the EMPLOYEES table
JOB_ID VARCHAR2(20) From the EMPLOYEES table
SALARY NUMBER From the EMPLOYEES table
DEPARTMENT_ID NUMBER From the DEPARTMENTS table
DEPT_NAME VARCHAR2(30) From the DEPARTMENTS table
Which SQL statement produces an error?
A. SELECT * FROM emp_dept_vu;
B. SELECT department_id, SUM(salary) FROM emp_dept_vu
GROUP BY department_id;
C. SELECT department_id, job_id, AVG(salary) FROM emp_dept_vu
GROUP BY department_id, job_id;
D. SELECT job_id, SUM(salary) FROM emp_dept_vu
WHERE department_id IN (10,20)
GROUP BY job_id
HAVING SUM(salary) > 20000;
E. None of the statements produce an error; all are valid.
Answer: E
14. Which SQL statement would you use to remove a view called EMP_DEPT_VU from
your schema?
A. DROP emp_dept_vu;
B. DELETE emp_dept_vu;
C. REMOVE emp_dept_vu;
D. DROP VIEW emp_dept_vu;
E. DELETE VIEW emp_dept_vu;
F. REMOVE VIEW emp_dept_vu;
Answer: D
Lesson 11: Other Objects
1. Which two statements about sequences are true? (Choose two)
A. You use a NEXTVAL pseudo column to look at the next possible value that would be
generated from a sequence, without actually retrieving the value.
B. You use a CURRVAL pseudo column to look at the current value just generated from
a sequence, without affecting the further values to be generated from the sequence.
C. You use a NEXTVAL pseudo column to obtain the next possible value from a
sequence by actually retrieving the value from the sequence.
D. You use a CURRVAL pseudo column to generate a value from a sequence that would
be used for a specified database column.
E. If a sequence starting from a value 100 and incremented by 1 is used by more then one
application, then all of these applications could have a value of 105 assigned to their
column whose value is being generated by the sequence.
F. You use REUSE clause when creating a sequence to restart the sequence once it
generates the maximum value defined for the sequence.
Answer: B, C
2. 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
3. In which scenario would index be most useful?
A. The indexed column is declared as NOT NULL.
B. The indexed columns are used in the FROM clause.
C. The indexed columns are part of an expression.
D. The indexed column contains a wide range of values.
Answer: D
4. Examine the structure of the EMPLOYEES table:
Column name Data type Remarks
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
LAST_NAME VARCNAR2(30)
FIRST_NAME VARCNAR2(30)
JOB_ID NUMBER
SAL NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER
You need to create an index called NAME_IDX on the first name and last name fields of
the EMPLOYEES table. Which SQL statement would you use to perform this task?
A. CREATE INDEX NAME_IDX (first_name, last_name);
B. CREATE INDEX NAME_IDX (first_name AND last_name);
C. CREATE INDEX NAME_IDX ON (first_name, last_name);
D. CREATE INDEX NAME_IDX ON employees (first_name AND last_name);
E. CREATE INDEX NAME_IDX ON employees(first_name, last_name);
F. CREATE INDEX NAME_IDX FOR employees(first_name, last_name);
Answer: E
5. 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: B, E
6. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER NOT NULL, Primary Key
EMP_NAME VARCHAR2(30)
JOB_ID NUMBER
SAL NUMBER
MGR_ID NUMBER References EMPLOYEE_ID column
DEPARTMENT_ID NUMBER Foreign key to DEPARTMENT_ID column of the
DEPARTMENTS table
You created a sequence called EMP_ID_SEQ in order to populate sequential values for
the EMPLOYEE_ID column of the EMPLOYEES table.
Which two statements regarding the EMP_ID_SEQ sequence are true? (Choose two.)
A. You cannot use the EMP_ID_SEQ sequence to populate the JOB_ID column.
B. The EMP_ID_SEQ sequence is invalidated when you modify the EMPLOYEE_ID
column.
C. The EMP_ID_SEQ sequence is not affected by modifications to the EMPLOYEES table.
D. Any other column of NUMBER data type in your schema can use the EMP_ID_SEQ
sequence.
E. The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEES
table.
F. The EMP_ID_SEQ sequence is dropped automatically when you drop the EMPLOYEE_ID
column.
Answer: C, D
7. User Mary has a view called EMP_DEPT_LOC_VU that was created based on the
EMPLOYEES, DEPARTMENTS, and LOCATIONS tables. She has the privilege to
create a public synonym, and would like to create a synonym for this view that can be
used by all users of the database.
Which SQL statement can Mary use to accomplish that task?
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;
Answer: A
8. Examine the statement:
Create synonym emp for hr.employees;
What happens when you issue the statement?
A. An error is generated.
B. You will have two identical tables in the HR schema with different names.
C. You create a table called employees in the HR schema based on you EMP table.
D. You create an alternative name for the employees table in the HR schema in your own
schema.
Answer: D
9. What is true about sequences?
A. The start value of the sequence is always 1.
B. A sequence always increments by 1.
C. The minimum value of an ascending sequence defaults to 1.
D. The maximum value of descending sequence defaults to 1.
Answer: A
10. 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
11. What is true about sequences?
A. Once created, a sequence belongs to a specific schema.
B. Once created, a sequence is linked to a specific table.
C. Once created, a sequence is automatically available to all users.
D. Only the DBA can control which sequence is used by a certain table.
E. Once created, a sequence is automatically used in all INSERT and UPDATE statements.
Answer: A
12. Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER NOT NULL
EMP_NAME VARCHAR2(30)
JOB_ID VARCHAR2(20)
SAL NUMBER
MGR_ID NUMBER
DEPARTMENT_ID NUMBER
You want to create a SQL script file that contains an INSERT statement. When the
script is run, the INSERT statement should insert a row with the specified values into
the EMPLOYEES table. The INSERT statement should pass values to the table columns
as specified below:
EMPLOYEE_ID: Next value from the sequence
EMP_ID_SEQEMP_NAME and JOB_ID: As specified by the user during run time,
through substitution variables
SAL: 2000
MGR_ID: No value
DEPARTMENT_ID: Supplied by the user during run time through substitution variable.
The INSERT statement should fail if the user supplies a value other than 20 or 50.
Which INSERT statement meets the above requirements?
A. INSERT INTO employees
VALUES (emp_id_seq.NEXTVAL, '&ename', '&jobid', 2000, NULL, &did);
B. INSERT INTO employees
VALUES (emp_id_seq.NEXTVAL, '&ename', '&jobid',
2000, NULL, &did IN (20,50));
C. INSERT INTO (SELECT *
FROM employees
WHERE department_id IN (20,50))
VALUES (emp_id_seq.NEXTVAL, '&ename', '&jobid', 2000, NULL, &did);
D. INSERT INTO (SELECT *
FROM employees
WHERE department_id IN (20,50)
WITH CHECK OPTION)
VALUES (emp_id_seq.NEXTVAL, '&ename', '&jobid', 2000, NULL, &did);
E. INSERT INTO (SELECT *
FROM employees
WHERE (department_id = 20 AND
department_id = 50)
WITH CHECK OPTION )
VALUES (emp_id_seq.NEXTVAL, '&ename', '&jobid', 2000, NULL, &did);
Answer: D
Lesson 12: User Access
1. 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
2. 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
3. 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
4. Which two statements accurately describe a role? (Choose two.)
A. A role can be given to a maximum of 1000 users.
B. A user can have access to a maximum of 10 roles.
C. A role can have a maximum of 100 privileges contained in it.
D. Privileges are given to a role by using the CREATE ROLE statement.
E. A role is a named group of related privileges that can be granted to the user.
F. A user can have access to several roles, and several users can be assigned the same
role.
Answer: D, F
5. What is necessary for your query on an existing view to execute successfully?
A. The underlying tables must have data.
B. You need SELECT privileges on the view.
C. The underlying tables must be in the same schema.
D. You need SELECT privileges only on the underlying tables.
Answer: B
6. Scott issues the SQL statements:
CREATE TABLE dept (deptno NUMBER(2), dname VARCHAR2(14),
loc VARCHAR2(13)};
GRANT SELECT ON DEPT TO SUE;
If Sue needs to select from Scott's DEPT table, which command should she use?
A. SELECT * FROM DEPT;
B. SELECT * FROM SCOTT.DEPT;
C. SELECT * FROM DBA.SCOTT.DEPT;
D. SELECT * FROM ALL_USERS WHERE USER_NAME = 'SCOTT'
AND TABLE NAME = 'DEPT';
Answer: B
7. Which statement creates a new user?
A. CREATE USER susan;
B. CREATE OR REPLACE USER susan;
C. CREATE NEW USER susan DEFAULT;
D. CREATE USER susan IDENTIFIED BY blue;
E. CREATE NEW USER susan IDENTIFIED by blue;
F. CREATE OR REPLACE USER susan IDENTIFIED BY blue;
Answer: D
8. What is true about the WITH GRANT OPTION clause?
A. It allows a grantee DBA privileges.
B. It is required syntax for object privileges.
C. It allows privileges on specified columns of tables.
D. It is used to grant an object privilege on a foreign key column.
E. It allows the grantee to grant object privileges to other users and roles.
Answer: E
9. The DBA issues this SQL command:
CREATE USER scott
IDENTIFIES by tiger;
What privileges does the user Scott have at this point?
A. No privileges.
B. Only the SELECT privilege.
C. Only the CONNECT privilege.
D. All the privileges of a default user.
Answer: A
10. You are the DBA for an academic database. You need to create a role that allows a
group of users to modify existing rows in the STUDENT_GRADES table.
Which set of statements accomplishes this?
A. CREATE ROLE registrar; GRANT MODIFY ON student_grades TO registrar; GRANT
registrar to user1, user2, user3
B. CREATE NEW ROLE registrar; GRANT ALL ON student_grades TO registrar; GRANT
registrar to user1, user2, user3
C. CREATE ROLE registrar; GRANT UPDATE ON student_grades TO registrar; GRANT
ROLE registrar to user1, user2, user3
D. CREATE ROLE registrar; GRANT UPDATE ON student_grades TO registrar; GRANT
registrar to user1, user2, user3;
E. CREATE registrar; GRANT CHANGE ON student_grades TO registrar; GRANT
registrar;
Answer: D
11. The user Sue issues this SQL statement:
GRANT SELECT ON sue.EMP TO alice WITH GRANT OPTION;
The user Alice issues this SQL statement:
GRANT SELECT ON sue.EMP TO reena WITH GRANT OPTION;
The user Reena issues this SQL statement:
GRANT SELECT ON sue.EMP TO timber;
The user Sue issues this SQL statement:
REVOKE select on sue.EMP FROM alice;
For which users does the revoke command revoke SELECT privileges on the SUE.EMP
table?
A. Alice only
B. Alice and Reena
C. Alice, Reena, and Timber
D. Sue, Alice, Reena, and Timber
Answer: C
12. The DBA issues this SQL command:
CREATE USER scott IDENTIFIED by tiger;
What privileges does the user Scott have at this point?
A. no privileges
B. only the SELECT privilege
C. only the CONNECT privilege
D. all the privileges of a default user
Answer: A
13. The DBA issues this SQL command:
CREATE USER scott IDENTIFIED by tiger;
What privileges does the user Scott have at this point?
A. no privileges
B. only the SELECT privilege
C. only the CONNECT privilege
D. all the privileges of a default user
Answer: A
14. Examine the statement:
GRANT select, insert, update
ON student_grades
TO manager
WITH GRANT OPTION;
Which two are true? (Choose two.)
A. MANAGER must be a role.
B. It allows the MANAGER to pass the specified privileges on to other users.
C. It allows the MANAGER to create tables that refer to the STUDENT_GRADES table.
D. It allows the MANAGER to apply all DML statements on the STUDENT_GRADES table.
E. It allows the MANAGER the ability to select from, insert into, and update the
STUDENT_GRADES table.
F. It allows the MANAGER the ability to select from, delete from, and update the
STUDENT_GRADES table.
Answer: B, E
15. The user Alice wants to grant all users query privileges on her DEPT table.
Which SQL statement accomplishes this?
A. GRANT select ON dept
TO ALL_USERS;
B. GRANT select ON dept
TO ALL;
C. GRANT QUERY ON dept
TO ALL_USERS
D. GRANT select ON dept
TO PUBLIC;
Answer: D
16. Which object privileges can be granted on a view?
A. none
B. DELETE, INSERT,SELECT
C. ALTER, DELETE, INSERT, SELECT
D. DELETE, INSERT, SELECT, UPDATE
Answer: D
17. Which one is a system privilege?
A. SELECT
B. DELETE
C. EXECUTE
D. ALTER TABLE
E. CREATE TABLE
Answer: E
18. Examine these statements:
CREATE ROLE registrar;
GRANT UPDATE ON student_grades TO registrar;
GRANT registrar to user1, user2, user3;
What does this set of SQL statements do?
A. The set of statements contains an error and does not work.
B. It creates a role called REGISTRAR, adds the MODIFY privilege on the
STUDENT_GRADES object to the role, and gives the REGISTRAR role to three users.
C. It creates a role called REGISTRAR, adds the UPDATE privilege on the
STUDENT_GRADES object to the role, and gives the REGISTRAR role to three users.
D. It creates a role called REGISTRAR, adds the UPDATE privilege on the
STUDENT_GRADES object to the role, and creates three users with the role.
E. It creates a role called REGISTRAR, adds the UPDATE privilege on three users, and gives
the REGISTRAR role to the STUDENT_GRADES object.
F. It creates a role called STUDENT_GRADES, adds the UPDATE privilege on three users,
and gives the UPDATE role to the registrar.
Answer: C
Lesson 13: Producing Readable Output with iSQL*Plus
1. Which iSQL*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 iSQL*Plus.
Answer: A
2. Which substitution variable would you use if you want to reuse the variable without
prompting the user each time?
A. &
B. ACCEPT
C. PROMPT
D. &&
Answer: D
3. Which SQL statement accepts user input for the columns to be displayed, the table
name, and WHERE condition?
A. SELECT &1, "&2" FROM &3 WHERE last_name = '&4';
B. SELECT &1, '&2' FROM &3 WHERE '&last_name = '&4' ';
C. SELECT &1, &2 FROM &3 WHERE last_name = '&4';
D. SELECT &1, '&2' FROM EMP WHERE last_name = '&4';
Answer: C
4. Which is an iSQL*Plus command?
A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME
Answer: D
5. Examine this statement:
SELECT student_id, gpa FROM student_grades WHERE gpa > &&value;
You run the statement once, and when prompted you enter a value of 2.0. A report is
produced. What happens when you run the statement a second time?
A. An error is returned.
B. You are prompted to enter a new value.
C. A report is produced that matches the first report produced.
D. You are asked whether you want a new value or if you want to run the report based on the
previous value.
Answer: C
6. Which are iSQL*Plus commands? (Choose all that apply.)
A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME
Answer: D
7. Which is an iSQL*Plus command?
A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME
Answer: D
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