Oracle Midterm 2

February 25, 2010

Section 1 Lesson 1

(Answer all questions in this section)

1.  Evaluate this SELECT statement:

SELECT LENGTH(email)

FROM employee;

What will this SELECT statement display?

Mark for Review

(1) Points

The longest e-mail address in the EMPLOYEE table.

The email address of each employee in the EMPLOYEE table.

The number of characters for each value in the EMAIL column in the employees table. (*)

The maximum number of characters allowed in the EMAIL column.

Correct

2.  Which SQL function can be used to remove heading or trailing characters (or both) from a character string?  Mark for Review

(1) Points

LPAD

CUT

NVL2

TRIM (*)

Correct

3.  Which SQL function is used to return the position where a specific character string begins within a larger character string?  Mark for Review

(1) Points

CONCAT

INSTR (*)

LENGTH

SUBSTR

Correct

4.  You need to display the number of characters in each customer’s last name. Which function should you use?  Mark for Review

(1) Points

LENGTH (*)

LPAD

COUNT

SUBSTR

Correct

5.  What will the following SQL statemtent display?

SELECT last_name, LPAD(salary, 15, ‘$’)SALARY

FROM employees;

Mark for Review

(1) Points

The last name of employees that have a salary that includes a $ in the value, size of 15 and the column labeled SALARY.

The last name and the format of the salary limited to 15 digits to the left of the decimal and the column labeled SALARY.

The last name and salary for all employees with the format of the salary 15 characters long, left-padded with the $ and the column labeled SALARY. (*)

The query will result in an error: “ORA-00923: FROM keyword not found where expected.”

Correct

6.  Which functions can be used to manipulate character, number, and date column values?  Mark for Review

(1) Points

CONCAT, RPAD, and TRIM (*)

UPPER, LOWER, and INITCAP

ROUND, TRUNC, and MOD

ROUND, TRUNC, and ADD_MONTHS

Incorrect. Refer to Section 1

7.  Which three statements about functions are true? (Choose three.)  Mark for Review

(1) Points

(Choose all correct answers)

The SYSDATE function returns the Oracle Server date and time. (*)

The ROUND number function rounds a value to a specified decimal place or the nearest whole number. (*)

The CONCAT function can only be used on character strings, not on numbers.

The SUBSTR character function returns a portion of a string beginning at a defined character position to a specified length. (*)

Incorrect. Refer to Section 1

Section 1 Lesson 2

(Answer all questions in this section)

8.  You issue this SQL statement:

SELECT ROUND (1282.248, -2)

FROM dual;

What value does this statement produce?

Mark for Review

(1) Points

1200

1282

1282.25

1300 (*)

Correct

9.  Which comparison operator retrieves a list of values?  Mark for Review

(1) Points

IN (*)

LIKE

BETWEEN…IN…

IS NULL

Correct

10.  Evaluate this function: MOD (25, 2) Which value is returned?  Mark for Review

(1) Points

1 (*)

2

25

0

Correct

Section 1 Lesson 3

(Answer all questions in this section)

11.  Which SELECT statement will return a numeric value?  Mark for Review

(1) Points

SELECT SYSDATE + 600 / 24

FROM employee;

SELECT ROUND(hire_date, DAY)

FROM employee;

SELECT (SYSDATE – hire_date) / 7

FROM employee;

(*)

SELECT SYSDATE – 7

FROM employee;

Incorrect. Refer to Section 1

12.  Evaluate this SELECT statement:

SELECT SYSDATE + 30

FROM dual;

Which value is returned by the query?

Mark for Review

(1) Points

the current date plus 30 hours

the current date plus 30 days (*)

the current date plus 30 months

No value is returned because the SELECT statement generates an error.

Incorrect. Refer to Section 1

13.  Which of the following SQL statements will correctly display the last name and the number of weeks employed for all employees in department 90?  Mark for Review

(1) Points

SELECT last_name, (SYSDATE-hire_date)/7 AS WEEKS

FROM employees

WHERE department_id = 90;

(*)

SELECT last name, (SYSDATE-hire_date)/7 DISPLAY WEEKS

FROM employees

WHERE department id = 90;

SELECT last_name, # of WEEKS

FROM employees

WHERE department_id = 90;

SELECT last_name, (SYSDATE-hire_date)AS WEEK

FROM employees

WHERE department_id = 90;

Correct

14.  You need to display the current year as a character value (for example: Two Thousand and One). Which element would you use?  Mark for Review

(1) Points

RR

YY

YYYY

YEAR (*)

Incorrect. Refer to Section 1

15.  You want to create a report that displays all orders and their amounts that were placed during the month of January. You want the orders with the highest amounts to appear first. Which query should you issue?  Mark for Review

(1) Points

SELECT orderid, total

FROM orders

WHERE order_date LIKE ’01-jan-02′ AND ’31-jan-02′

ORDER BY total DESC;

SELECT orderid, total

FROM orders

WHERE order_date IN ( 01-jan-02 , 31-jan-02 )

ORDER BY total;

SELECT orderid, total

FROM orders

WHERE order_date BETWEEN ’01-jan-02′ AND ’31-jan-02′

ORDER BY total DESC;

(*)

SELECT orderid, total

FROM orders

WHERE order_date BETWEEN ’31-jan-02′ AND ’01-jan-02′

ORDER BY total DESC;

Correct

Section 2 Lesson 1

(Answer all questions in this section)

16.  The EMPLOYEES table contains these columns:

EMPLOYEE_ID NUMBER(9)

LAST_NAME VARCHAR2 (25)

FIRST_NAME VARCHAR2 (25)

HIRE_DATE DATE

You need to display HIRE_DATE values in this format:

January 28, 2000

Which SELECT statement could you use?

Mark for Review

(1) Points

SELECT TO_CHAR(hire_date, Month DD, YYYY)

FROM employees;

SELECT TO_CHAR(hire_date, ‘Month DD, YYYY’)

FROM employees;

(*)

SELECT hire_date(TO_CHAR ‘Month DD’, ‘ YYYY’)

FROM employees;

SELECT TO_CHAR(hire_date, ‘Month DD’, ‘ YYYY’)

FROM employees;

Incorrect. Refer to Section 2

17.  You have been asked to create a report that lists all customers who have placed orders of at least $2,500. The report’s date should be displayed in the Day, Date Month, Year format (For example, Tuesday, 13 April, 2004 ). Which statement should you issue?  Mark for Review

(1) Points

SELECT companyname, TO_CHAR (sysdate, ‘fmdd, dy month, yyyy’), total

FROM customers NATURAL JOIN orders

WHERE total >= 2500;

SELECT companyname, TO_DATE (date, ‘day, dd month, yyyy’), total

FROM customers NATURAL JOIN orders

WHERE total >= 2500;

SELECT companyname, TO_DATE (sysdate, ‘dd, dy month, yyyy’), total

FROM customers NATURAL JOIN orders

WHERE total >= 2500;

SELECT companyname, TO_CHAR (sysdate, ‘fmDay, dd Month, yyyy’), total

FROM customers NATURAL JOIN orders

WHERE total >= 2500;

(*)

Incorrect. Refer to Section 2

18.  If you use the RR format when writing a query using the date 27-OCT-17 and the year is 2001, what year would be the result?  Mark for Review

(1) Points

2001

1901

2017 (*)

1917

Correct

19.  Which two statements concerning SQL functions are true? (Choose two.)  Mark for Review

(1) Points

(Choose all correct answers)

Character functions can accept numeric input.

Not all date functions return date values. (*)

Number functions can return number or character values.

Conversion functions convert a value from one data type to another data type. (*)

Single-row functions manipulate groups of rows to return one result per group of rows.

Correct

20.  The EMPLOYEES table contains these columns:

EMPLOYEE_ID NUMBER(9)

LAST_NAME VARCHAR2 (25)

FIRST_NAME VARCHAR2 (25)

SALARY NUMBER(6)

You need to create a report to display the salaries of all employees. Which script should you use to display the salaries in format: “$45,000.00″?

Mark for Review

(1) Points

SELECT TO_CHAR(salary, ‘$999,999′)

FROM employees;

SELECT TO_NUM(salary, ‘$999,990.99′)

FROM employees;

SELECT TO_NUM(salary, ‘$999,999.00′)

FROM employees;

SELECT TO_CHAR(salary, ‘$999,999.00′)

FROM employees;

(*)

Correct

Section 2 Lesson 1

(Answer all questions in this section)

21.  Which statement concerning single row functions is true?  Mark for Review

(1) Points

Single row functions can accept only one argument, but can return multiple values.

Single row functions cannot modify a data type.

Single row functions can be nested. (*)

Single row functions return one or more results per row.

Incorrect. Refer to Section 2

Section 2 Lesson 2

(Answer all questions in this section)

22.  Which statement about group functions is true?  Mark for Review

(1) Points

NVL and NVL2, but not COALESCE, can be used with group functions to replace null values.

NVL and COALESCE, but not NVL2, can be used with group functions to replace null values.

NVL, NVL2, and COALESCE can be used with group functions to replace null values. (*)

COALESCE, but not NVL and NVL2, can be used with group functions to replace null values.

Incorrect. Refer to Section 2

23.  The STYLES table contains this data:

STYLE_ID  STYLE_NAME  CATEGORY  COST

895840  SANDAL  85940  12.00

968950  SANDAL  85909  10.00

869506  SANDAL  89690  15.00

809090  LOAFER  89098  10.00

890890  LOAFER  89789  14.00

857689  HEEL  85940  11.00

758960  SANDAL  86979

Evaluate this SELECT statement:

SELECT style_id, style_name, category, cost

FROM styles

WHERE style_name LIKE ‘SANDAL’ AND NVL(cost, 0) < 15.00

ORDER BY category, cost;

Which result will the query provide?

Mark for Review

(1) Points

STYLE_ID  STYLE_NAME  CATEGORY  COST

895840  SANDAL  85940  12.00

968950  SANDAL  85909  10.00

758960  SANDAL  86979

STYLE_ID  STYLE_NAME  CATEGORY  COST

895840  SANDAL  85909  12.00

968950 SANDAL  85909  10.00

869506  SANDAL  89690  15.00

758960  SANDAL  86979

STYLE_ID  STYLE_NAME  CATEGORY  COST

895840  SANDAL  85909  12.00

968950  SANDAL  85909  10.00

758960  SANDAL  86979

869506  SANDAL  89690  15.00

STYLE_ID  STYLE_NAME  CATEGORY  COST

968950  SANDAL  85909  10.00

895840  SANDAL  85940  12.00

758960  SANDAL  86979

(*)

Correct

24.  Which of the following General Functions will return the first non-null expression in the expression list?  Mark for Review

(1) Points

NVL

NVL2

NULLIF

COALESCE (*)

Correct

Section 3 Lesson 2

(Answer all questions in this section)

25.  The CUSTOMERS and SALES tables contain these columns:

CUSTOMERS

CUST_ID NUMBER(10) PRIMARY KEY

COMPANY VARCHAR2(30)

LOCATION VARCHAR2(20)

SALES

SALES_ID NUMBER(5) PRIMARY KEY

CUST_ID NUMBER(10) FOREIGN KEY

TOTAL_SALES NUMBER(30)

Which SELECT statement will return the customer ID, the company and the total sales?

Mark for Review

(1) Points

SELECT c.cust_id, c.company, s.total_sales

FROM customers c, sales s

WHERE c.cust_id = s.cust_id (+);

SELECT cust_id, company, total_sales

FROM customers, sales

WHERE cust_id = cust_id;

SELECT c.cust_id, c.company, s.total_sales

FROM customers c, sales s

WHERE c.cust_id = s.cust_id;

(*)

SELECT cust_id, company, total_sales

FROM customers c, sales s

WHERE c.cust_id = s.cust_id;

Correct

26.  You have been asked to create a report that lists all corporate customers and all orders that they have placed. The customers should be listed alphabetically beginning with the letter ‘A’, and their corresponding order totals should be sorted from the highest amount to the lowest amount.

Which of the following statements should you issue?  Mark for Review

(1) Points

SELECT c.custid, c.companyname, o.orderdate, o. custid, o.amount

FROM customers c, orders o

WHERE c.custid = o.custid

ORDER BY amount DESC, companyname;

SELECT c.custid, c.companyname, o.orderdate, o. custid, o.amount

FROM customers c, orders o

WHERE c.custid = o.custid

ORDER BY companyname, amount DESC;

(*)

SELECT c.custid, c.companyname, o.orderdate, o. custid, o.amount

FROM customers c, orders o

WHERE c.custid = o.custid

ORDER BY companyname, amount;

SELECT c.custid, c.companyname, o.orderdate, o. custid, o.amount

Q FROM customers c, orders o

WHERE c.custid = o.custid

ORDER BY companyname ASC, amount ASC;

Correct

27.  What happens when you create a Cartesian product?  Mark for Review

(1) Points

All rows from one table are joined to all rows of another table (*)

The table is joined to itself, one column to the next column, exhausting all possibilities

The table is joined to another equal table

All rows that do not match in the WHERE clause are displayed

Correct

28.  Evaluate this SQL statement:

SELECT e.employee_id, e.last_name, e.first_name, d.department_name

FROM employees e, departments d

WHERE e.department_id = d.department_id AND employees.department_id > 5000

ORDER BY 4;

Which clause contains a syntax error?

Mark for Review

(1) Points

SELECT e.employee_id, e.last_name, e.first_name, d.department_name

FROM employees e, departments d

WHERE e.department_id = d.department_id

AND employees.department_id > 5000 (*)

ORDER BY 4;

Incorrect. Refer to Section 3

29.  What is produced when a join condition is not specified in a multiple-table query?  Mark for Review

(1) Points

a self-join

an outer join

an equijoin

a Cartesian product (*)

Incorrect. Refer to Section 3

30.  Your have two tables named EMPLOYEES and SALES. You want to identify the sales representatives who have generated at least $100,000 in revenue.

Which query should you issue?  Mark for Review

(1) Points

SELECT e.fname, e.lname, s.sales

FROM employees e, sales s

WHERE e.emp_id = s.emp_id AND revenue > 100000;

SELECT e.fname, e.lname, s.sales

FROM employees e, sales s

WHERE e.emp_id = s.emp_id AND revenue >= 100000;

(*)

SELECT e.fname, e.lname, s.sales

FROM employees, sales

WHERE e.emp_id = s.emp_id AND revenue >= 100000;

SELECT fname, lname, sales

Q FROM employees e, sales s

WHERE e.emp_id = s.emp_id AND revenue > 100000;

Incorrect. Refer to Section 3

Section 3 Lesson 4

(Answer all questions in this section)

31.  Evaluate this SELECT statement:

SELECT p.player_id, m.last_name, m.first_name, t.team_name

FROM player p

LEFT OUTER JOIN player m ON (p.manager_id = m.player_id)

LEFT OUTER JOIN team t ON (p.team_id = t.team_id);

Which join is evaluated first?

Mark for Review

(1) Points

the self-join of the player table (*)

the join between the player table and the team table on TEAM_ID

the join between the player table and the team table on MANAGER_ID

the join between the player table and the team table on PLAYER_ID

Incorrect. Refer to Section 3

32.  The EMPLOYEE_ID column in the EMPLOYEE table corresponds to the EMPLOYEE_ID column of the ORDER table. The EMPLOYEE_ID column in the ORDER table contains null values for rows that you need to display.

Which type of join should you use to display the data?  Mark for Review

(1) Points

natural join

self-join

outer join (*)

equijoin

Incorrect. Refer to Section 3

33.  Which operator would you use after one of the column names in the WHERE clause when creating an outer join?  Mark for Review

(1) Points

(+) (*)

*

+

=

Correct

Section 4 Lesson 2

(Answer all questions in this section)

34.  You need to join two tables that have two columns with the same name and compatible data types. Which type of join would you create to join the tables on both of the columns?  Mark for Review

(1) Points

Natural join (*)

Cross join

Outer join

Self-join

Correct

35.  A join between tables where the result set includes matching values from both tables but does NOT return any unmatched rows could be called which of the following? (Choose three)  Mark for Review

(1) Points

(Choose all correct answers)

Equijoin (*)

Self join (*)

Nonequijoin

Simple join (*)

full outer join

Incorrect. Refer to Section 4

36.  Which of the following best describes a natural join?  Mark for Review

(1) Points

A join between two tables that includes columns that share the same name, datatypes and lengths (*)

A join that produces a Cartesian product

A join between tables where matching fields do not exist

A join that uses only one table

Correct

Section 4 Lesson 3

(Answer all questions in this section)

37.  Which SELECT clause creates an equijoin by specifying a column name common to both tables?  Mark for Review

(1) Points

A HAVING clause

The FROM clause

The SELECT clause

A USING clause (*)

Correct

38.  Below find the structures of the PRODUCTS and VENDORS tables:

PRODUCTS

PRODUCT_ID NUMBER

PRODUCT_NAME VARCHAR2 (25)

VENDOR_ID NUMBER

CATEGORY_ID NUMBER

VENDORS

VENDOR_ID NUMBER

VENDOR_NAME VARCHAR2 (25)

ADDRESS VARCHAR2 (30)

CITY VARCHAR2 (25)

REGION VARCHAR2 (10)

POSTAL_CODE VARCHAR2 (11)

You want to create a query that will return an alphabetical list of products, including the product name and associated vendor name, for all products that have a vendor assigned.

Which two queries could you use?

Mark for Review

(1) Points

(Choose all correct answers)

SELECT p.product_name, v.vendor_name

FROM products p

LEFT OUTER JOIN vendors v ON p.vendor_id = v.vendor_id

ORDER BY p.product_name;

SELECT p.product_name, v.vendor_name

FROM products p

JOIN vendors v ON (vendor_id)

ORDER BY p.product_name;

SELECT p.product_name, v.vendor_name

FROM products p NATURAL JOIN vendors v

ORDER BY p.product_name;

(*)

SELECT p.product_name, v.vendor_name

FROM products p

JOIN vendors v USING (p.vendor_id)

ORDER BY p.product_name;

SELECT p.product_name, v.vendor_name

FROM products p

JOIN vendors v USING (vendor_id)

ORDER BY p.product_name;

(*)

Incorrect. Refer to Section 4

39.  The primary advantage of using JOIN ON is:  Mark for Review

(1) Points

The join happens automatically based on matching column names and data types

It will display rows that do not meet the join condition

It permits columns with different names to be joined (*)

It permits columns that don’t have matching data types to be joined

Incorrect. Refer to Section 4

40.  Below find the structure of the CUSTOMERS and SALES_ORDER tables:

CUSTOMERS

CUSTOMER_ID NUMBER NOT NULL, Primary Key

CUSTOMER_NAME VARCHAR2 (30)

CONTACT_NAME VARCHAR2 (30)

CONTACT_TITLE VARCHAR2 (20)

ADDRESS VARCHAR2 (30)

CITY VARCHAR2 (25)

REGION VARCHAR2 (10)

POSTAL_CODE VARCHAR2 (20)

COUNTRY_ID NUMBER Foreign key to COUNTRY_ID column of the COUNTRY table

PHONE VARCHAR2 (20)

FAX VARCHAR2 (20)

CREDIT_LIMIT NUMBER(7,2)

SALES_ORDER

ORDER_ID NUMBER NOT NULL, Primary Key

CUSTOMER_ID NUMBER Foreign key to CUSTOMER_ID column of the CUSTOMER table

ORDER_DT DATE

ORDER_AMT NUMBER (7,2)

SHIP_METHOD VARCHAR2 (5)

You need to create a report that displays customers without a sales order. Which statement could you use?

Mark for Review

(1) Points

SELECT c.customer_name

FROM customers c

WHERE c.customer_id not in (SELECT s.customer_id FROM sales_order s);

(*)

SELECT c.customer_name

FROM customers c, sales_order s

WHERE c.customer_id = s.customer_id(+);

SELECT c.customer_name

FROM customers c, sales_order s

WHERE c.customer_id (+) = s.customer_id;

SELECT c.customer_name

FROM customers c

RIGHT OUTER JOIN sales_order s ON (c.customer_id = s.customer_id);

Correct

Section 4 Lesson 4

(Answer all questions in this section)

41.  Which two sets of join keywords create a join that will include unmatched rows from the first table specified in the SELECT statement?  Mark for Review

(1) Points

LEFT OUTER JOIN and FULL OUTER JOIN (*)

RIGHT OUTER JOIN and LEFT OUTER JOIN

USING and HAVING

OUTER JOIN and USING

Incorrect. Refer to Section 4

42.  Which query represents the correct syntax for a left outer join?  Mark for Review

(1) Points

SELECT companyname, orderdate, total

FROM customers c

LEFT JOIN orders o

ON c.cust_id = o.cust_id;

SELECT companyname, orderdate, total

FROM customers c

OUTER JOIN orders o

ON c.cust_id = o.cust_id;

SELECT companyname, orderdate, total

FROM customers c

LEFT OUTER JOIN orders o ON c.cust_id = o.cust_id;

(*)

SELECT companyname, orderdate, total

FROM customers c

LEFT OUTER JOIN orders o

ON c.cust_id = o.cust_id;

Incorrect. Refer to Section 4

43.  You need to display all the rows from both the EMPLOYEE and EMPLOYEE_HIST tables. Which type of join would you use?  Mark for Review

(1) Points

a right outer join

a left outer join

a full outer join (*)

an inner join

Correct

Section 5 Lesson 1

(Answer all questions in this section)

44.  Evaluate this SELECT statement:

SELECT MIN(hire_date), dept_id

FROM employee

GROUP BY dept_id;

Which values are displayed?

Mark for Review

(1) Points

The earliest hire date in each department. (*)

The the earliest hire date in the EMPLOYEE table.

The latest hire date in the EMPLOYEE table.

The hire dates in the EMPLOYEE table that contain NULL values.

Correct

45.  What is the best explanation as to why this SQL statement will NOT execute?

SELECT department_id “Department”, AVG (salary)”Average”

FROM employees

GROUP BY Department;

Mark for Review

(1) Points

Salaries cannot be averaged as not all the numbers will divide evenly.

You cannot use a column alias in the GROUP BY clause. (*)

The GROUP BY clause must have something to GROUP.

The department id is not listed in the departments table.

Incorrect. Refer to Section 5

46.  What will the following SQL Statement do?

SELECT job_id, COUNT(*)

FROM employees

GROUP BY job_id;

Mark for Review

(1) Points

Displays all the employees and groups them by job.

Displays each job id and the number of people assigned to that job id. (*)

Displays only the number of job_ids.

Displays all the jobs with as many people as there are jobs.

Incorrect. Refer to Section 5

47.  If a select list contains both a column as well as a group function then what clause is required?  Mark for Review

(1) Points

having clause

join clause

order by clause

group by clause (*)

Correct

Section 5 Lesson 2

(Answer all questions in this section)

48.  You need to calculate the average salary of employees in each department. Which group function will you use?  Mark for Review

(1) Points

AVG (*)

MEAN

MEDIAN

AVERAGE

Incorrect. Refer to Section 5

49.  Which aggregate function can be used on a column of the DATE data type?  Mark for Review

(1) Points

AVG

MAX (*)

STDDEV

SUM

Correct

50.  The CUSTOMER table contains these columns:

CUSTOMER_ID NUMBER(9)

FNAME VARCHAR2(25)

LNAME VARCHAR2(30)

CREDIT_LIMIT NUMBER (7,2)

CATEGORY VARCHAR2(20)

You need to calculate the average credit limit for all the customers in each category. The average should be calculated based on all the rows in the table excluding any customers who have not yet been assigned a credit limit value. Which group function should you use to calculate this value?

Mark for Review

(1) Points

AVG (*)

SUM

COUNT

STDDEV

Correct

Section 5 Lesson 2

(Answer all questions in this section)

51.  You need to calculate the standard deviation for the cost of products produced in the Birmingham facility. Which group function will you use?  Mark for Review

(1) Points

STDEV

STDDEV (*)

VAR_SAMP

VARIANCE

Incorrect. Refer to Section 5

52.  The TRUCKS table contains these columns:

TRUCKS

TYPE VARCHAR2(30)

YEAR DATE

MODEL VARCHAR2(20)

PRICE NUMBER(10)

Which SELECT statement will return the average price for the 4×4 model?

Mark for Review

(1) Points

SELECT AVG (price) FROM trucks WHERE model = ’4×4′; (*)

SELECT AVG (price) FROM trucks WHERE model IS ’4×4′;

SELECT AVG(price) FROM trucks WHERE model IS 4×4;

SELECT AVG(price), model FROM trucks WHERE model IS ’4×4′;

Correct

53.  The EMPLOYEES table contains these columns:

EMPLOYEE_ID NUMBER(9)

LAST_NAME VARCHAR2(20)

FIRST_NAME VARCHAR2(20)

SALARY NUMBER(9,2)

HIRE_DATE DATE

BONUS NUMBER(7,2)

COMM_PCT NUMBER(4,2)

Which three functions could be used with the HIRE_DATE, LAST_NAME, or SALARY columns? (Choose three.)

Mark for Review

(1) Points

(Choose all correct answers)

MAX (*)

SUM

AVG

MIN (*)

COUNT (*)

Incorrect. Refer to Section 5

54.  The VENDORS table contains these columns:

VENDOR_ID NUMBER Primary Key

NAME VARCHAR2(30)

LOCATION_ID NUMBER

ORDER_DT DATE

ORDER_AMOUNT NUMBER(8,2)

Which two clauses represent valid uses of aggregate functions for this table?

Mark for Review

(1) Points

(Choose all correct answers)

FROM MAX(order_dt)

SELECT SUM(order_dt)

SELECT SUM(order_amount) (*)

WHERE MAX(order_dt) = order_dt

SELECT location_id, MIN(AVG(order_amount)) (*)

Incorrect. Refer to Section 5

55.  Which group function would you use to display the highest salary value in the EMPLOYEE table?  Mark for Review

(1) Points

AVG

COUNT

MAX (*)

MIN

Correct

Section 5 Lesson 3

(Answer all questions in this section)

56.  Evaluate this SELECT statement:

SELECT COUNT(*)

FROM products;

Which statement is true?

Mark for Review

(1) Points

The number of rows in the table is displayed. (*)

The number of unique PRODUCT_IDs in the table is displayed.

An error occurs due to an error in the SELECT clause.

An error occurs because no WHERE clause is included in the SELECT statement.

Correct

57.  Group functions can avoid computations involving duplicate values by including which keyword?  Mark for Review

(1) Points

NULL

DISTINCT (*)

SELECT

UNLIKE

Correct

58.  Which SELECT statement will calculate the number of rows in the PRODUCTS table?  Mark for Review

(1) Points

SELECT COUNT(products);

SELECT COUNT FROM products;

SELECT COUNT (*) FROM products; (*)

SELECT ROWCOUNT FROM products;

Correct

59.  The EMPLOYEES table contains these columns:

EMPLOYEE_ID NUMBER(9)

LAST_NAME VARCHAR2(20)

FIRST_NAME VARCHAR2(20)

SALARY NUMBER(7,2)

DEPARTMENT_ID NUMBER(9)

You need to display the number of employees whose salary is greater than $50,000? Which SELECT would you use?

Mark for Review

(1) Points

SELECT * FROM employees

WHERE salary > 50000;

SELECT * FROM employees

WHERE salary < 50000;

SELECT COUNT(*) FROM employees

WHERE salary < 50000;

SELECT COUNT(*) FROM employees

WHERE salary > 50000;

(*)

SELECT COUNT(*) FROM employees

WHERE salary > 50000

GROUP BY employee_id, last_name, first_name, salary, department_id;

Correct

Section 6 Lesson 1

(Answer all questions in this section)

60.  What is the correct order of clauses in a SELECT statement?  Mark for Review

(1) Points

SELECT

FROM

WHERE

ORDER BY

HAVING

SELECT

FROM

HAVING

GROUP BY

WHERE

ORDER BY

SELECT

FROM

WHERE

GROUP BY

HAVING

ORDER BY

(*)

SELECT

FROM

WHERE

HAVING

ORDER BY

GROUP BY

Correct

Section 6 Lesson 1

(Answer all questions in this section)

61.  The EMPLOYEES table contains these columns:

ID_NUMBER NUMBER Primary Key

NAME VARCHAR2 (30)

DEPARTMENT_ID NUMBER

SALARY NUMBER (7,2)

HIRE_DATE DATE

Evaluate this SQL statement:

SELECT id_number, name, department_id, SUM(salary)

FROM employees

WHERE salary > 25000

GROUP BY department_id, id_number, name

ORDER BY hire_date;

Why will this statement cause an error?

Mark for Review

(1) Points

The HAVING clause is missing.

The WHERE clause contains a syntax error.

The SALARY column is NOT included in the GROUP BY clause.

The HIRE_DATE column is NOT included in the GROUP BY clause. (*)

Correct

62.  The PAYMENT table contains these columns:

PAYMENT_ID NUMBER(9) PK

PAYMENT_DATE DATE

CUSTOMER_ID NUMBER(9)

Which SELECT statement could you use to display the number of times each customer made a payment between January 1, 2003 and June 30, 2003 ?

Mark for Review

(1) Points

SELECT customer_id, COUNT(payment_id)

FROM payment

WHERE payment_date BETWEEN ’01-JAN-2003′ AND ’30-JUN-2003′

GROUP BY customer_id;

(*)

SELECT COUNT(payment_id)

FROM payment

WHERE payment_date BETWEEN ’01-JAN-2003′ AND ’30-JUN-2003′;

SELECT customer_id, COUNT(payment_id)

FROM payment

WHERE payment_date BETWEEN ’01-JAN-2003′ AND ’30-JUN-2003′;

SELECT COUNT(payment_id)

FROM payment

WHERE payment_date BETWEEN ’01-JAN-2003′ AND ’30-JUN-2003′

GROUP BY customer_id;

Incorrect. Refer to Section 6

63.  You want to write a report that returns the average salary of all employees in the company, sorted by departments. The EMPLOYEES table contains the following columns:

EMPLOYEES:

EMP_ID NUMBER(10) PRIMARY KEY

LNAME VARCHAR2(20)

FNAME VARCHAR2(20)

DEPT VARCHAR2(20)

HIRE_DATE DATE

SALARY NUMBER(10)

Which SELECT statement will return the information that you require?

Mark for Review

(1) Points

SELECT salary (AVG)

FROM employees

GROUP BY dept;

SELECT AVG (salary)

FROM employees

GROUP BY dept;

(*)

SELECT AVG (salary)

FROM employees

BY dept;

SELECT AVG salary

FROM employees

BY dept;

Correct

64.  Evaluate this SELECT statement:

SELECT COUNT(emp_id), mgr_id, dept_id

FROM employee

WHERE status = ‘I’

GROUP BY dept_id

HAVING salary > 30000

ORDER BY 2;

Why does this statement return a syntax error?

Mark for Review

(1) Points

MGR_ID must be included in the GROUP BY clause. (*)

The HAVING clause must specify an aggregate function.

A single query cannot contain a WHERE clause and a HAVING clause.

The ORDER BY clause must specify a column name in the EMPLOYEE table.

Incorrect. Refer to Section 6

65.  Evaluate this statement:

SELECT department_id, AVG(salary)

FROM employees

WHERE job_id <> 69879

GROUP BY job_id, department_id

HAVING AVG(salary) > 35000

ORDER BY department_id;

Which clauses restricts the result? Choose two.

Mark for Review

(1) Points

(Choose all correct answers)

SELECT department_id, AVG(salary)

WHERE job_id <> 69879 (*)

GROUP BY job_id, department_id

HAVING AVG(salary) > 35000 (*)

Correct

66.  The MANUFACTURER table contains these columns:

MANUFACTURER_ID NUMBER

MANUFACTURER_NAME VARCHAR2(30)

TYPE VARCHAR2(25)

LOCATION_ID NUMBER

You need to display the number of unique types of manufacturers at each location. Which SELECT statement should you use?

Mark for Review

(1) Points

SELECT location_id, COUNT(DISTINCT type)

FROM manufacturer

GROUP BY location_id;

(*)

SELECT location_id, COUNT(DISTINCT type)

FROM manufacturer;

SELECT location_id, COUNT(type)

FROM manufacturer

GROUP BY location_id;

SELECT location_id, COUNT(DISTINCT type)

FROM manufacturer

GROUP BY type;

Correct

67.  Which statement about the GROUP BY clause is true?  Mark for Review

(1) Points

To exclude rows before dividing them into groups using the GROUP BY clause, you use should a WHERE clause. (*)

You can use a column alias in a GROUP BY clause.

By default, rows are not sorted when a GROUP BY clause is used.

You must use the HAVING clause with the GROUP BY clause.

Incorrect. Refer to Section 6

Section 6 Lesson 2

(Answer all questions in this section)

68.  Which operator can be used with a multiple-row subquery?  Mark for Review

(1) Points

IN (*)

<>

=

LIKE

Correct

69.  You need to create a report to display the names of products with a cost value greater than the average cost of all products. Which SELECT statement should you use?  Mark for Review

(1) Points

SELECT product_name

FROM products

WHERE cost > (SELECT AVG(cost) FROM product);

(*)

SELECT product_name

FROM products

WHERE cost > AVG(cost);

SELECT AVG(cost), product_name

FROM products

WHERE cost > AVG(cost)

GROUP by product_name;

SELECT product_name

FROM (SELECT AVG(cost) FROM product)

WHERE cost > AVG(cost);

Incorrect. Refer to Section 6

70.  Examine the structures of the CUSTOMER and ORDER_HISTORY tables:

CUSTOMER

CUSTOMER_ID NUMBER(5)

NAME VARCHAR2(25)

CREDIT_LIMIT NUMBER(8,2)

OPEN_DATE DATE

ORDER_HISTORY

ORDER_ID NUMBER(5)

CUSTOMER_ID NUMBER(5)

ORDER_DATE DATE

TOTAL NUMBER(8,2)

Which of the following scenarios would require a subquery to return the desired results?

Mark for Review

(1) Points

You need to display the date each customer account was opened.

You need to display each date that a customer placed an order.

You need to display all the orders that were placed on a certain date.

You need to display all the orders that were placed on the same day as order number 25950. (*)

Incorrect. Refer to Section 6

Section 6 Lesson 2

(Answer all questions in this section)

71.  Which operator can be used with subqueries that return only one row?  Mark for Review

(1) Points

LIKE (*)

ANY

ALL

IN

Incorrect. Refer to Section 6

72.  Which statement about subqueries is true?  Mark for Review

(1) Points

Subqueries should be enclosed in double quotation marks.

Subqueries cannot contain group functions.

Subqueries are often used in a WHERE clause to return values for an unknown conditional value. (*)

Subqueries generally execute last, after the main or outer query executes.

Correct

Section 6 Lesson 3

(Answer all questions in this section)

73.  Which statement about the <> operator is true?  Mark for Review

(1) Points

The <> operator is NOT a valid SQL operator.

The <> operator CANNOT be used in a single-row subquery.

The <> operator returns the same result as the ANY operator in a subquery.

The <> operator can be used when a single-row subquery returns only one row. (*)

Incorrect. Refer to Section 6

74.  Which best describes a single-row subquery?  Mark for Review

(1) Points

a query that returns only one row from the inner SELECT statement (*)

a query that returns one or more rows from the inner SELECT statement

a query that returns only one column value from the inner SELECT statement

a query that returns one or more column values from the inner SELECT statement

Incorrect. Refer to Section 6

75.  If a single-row subquery returns a null value and uses the equality comparison operator, what will the outer query return?  Mark for Review

(1) Points

no rows (*)

all the rows in the table

a null value

an error

Incorrect. Refer to Section 6

Section 6 Lesson 4

(Answer all questions in this section)

76.  You need to create a SELECT statement that contains a multiple-row subquery, which comparison operator(s) can you use?  Mark for Review

(1) Points

IN, ANY, and ALL (*)

LIKE

BETWEEN…AND…

=, <, and >

Correct

77.  Which operator or keyword cannot be used with a multiple-row subquery?  Mark for Review

(1) Points

ALL

ANY

= (*)

>

Correct

78.  Examine the data in the PAYMENT table:

PAYMENT_ID  CUSTOMER_ID  PAYMENT_DATE  PAYMENT_TYPE  PAYMENT_AMOUNT

86590586  8908090  10-JUN-03  BASIC  859.00

89453485  8549038  15-FEB-03  INTEREST  596.00

85490345  5489304  20-MAR-03  BASIC  568.00

This statement fails when executed:

SELECT customer_id, payment_type

FROM payment

WHERE payment_id =

(SELECT payment_id

FROM payment

WHERE payment_amount = 596.00 OR payment_date = ’20-MAR-2003′);

Which change could correct the problem?

Mark for Review

(1) Points

Change the outer query WHERE clause to ‘WHERE payment_id IN’. (*)

Remove the quotes surrounding the date value in the OR clause.

Remove the parentheses surrounding the nested SELECT statement.

Change the comparison operator to a single-row operator.

Incorrect. Refer to Section 6

79.  Which of the following best describes the meaning of the ANY operator?  Mark for Review

(1) Points

Equal to any member in the list

Compare value to each value returned by the subquery (*)

Compare value to every value returned by the subquery

Equal to each value in the list

Incorrect. Refer to Section 6

80.  What would happen if you attempted to use a single-row operator with a multiple-row subquery?  Mark for Review

(1) Points

An error would be returned. (*)

No rows will be selected.

All the rows will be selected.

The data returned may or may not be correct.

Incorrect. Refer to Section 6

Section 6 Lesson 4

(Answer all questions in this section)

81.  Examine the structures of the PARTS and MANUFACTURERS tables:

PARTS:

PARTS_ID VARCHAR2(25)

PK PARTS_NAME VARCHAR2(50)

MANUFACTURERS_ID NUMBER

COST NUMBER(5,2)

PRICE NUMBER(5,2)

MANUFACTURERS:

ID NUMBER

PK NAME VARCHAR2(30)

LOCATION VARCHAR2(20)

Which SQL statement correctly uses a subquery?

Mark for Review

(1) Points

UPDATE parts SET price = price * 1.15

WHERE manufacturers_id =

(SELECT id

FROM manufacturers

WHERE UPPER(location) IN(‘ATLANTA ‘, ‘BOSTON ‘, ‘DALLAS ‘));

SELECT parts_name, price, cost

FROM parts

WHERE manufacturers_id !=

(SELECT id

FROM manufacturers

WHERE LOWER(name) = ‘cost plus’);

SELECT parts_name, price, cost

FROM parts

WHERE manufacturers_id IN

(SELECT id

FROM manufacturers m

JOIN part p ON (m.id = p.manufacturers_id));

(*)

SELECT parts_name

FROM

(SELECT AVG(cost)

FROM manufacturers)

WHERE cost > AVG(cost);

Incorrect. Refer to Section 6

82.  You are looking for Executive information using a subquery. What will the following SQL statement display?

SELECT department_id, last_name, job_id

FROM employees

WHERE department_id IN

(SELECT department_id

FROM departments

WHERE department_name = ‘Executive’);

Mark for Review

(1) Points

The department ID, department name and last name for every employee in the Executive department.

The department ID, last name, department name for every Executive in the employees table.

The department ID, last name, job ID from departments for Executive employees.

The department ID, last name, job ID for every employee in the Executive department. (*)

Correct

83.  Examine the data in the PAYMENT table:

PAYMENT_ID  CUSTOMER_ID  PAYMENT_DATE  PAYMENT_TYPE  PAYMENT_AMOUNT

86590586  8908090  10-JUN-03  BASIC  859.00

89453485  8549038  15-FEB-03  INTEREST  596.00

85490345  5489304  20-MAR-03  BASIC  568.00

This statement fails when executed:

SELECT payment_date, customer_id, payment_amount

FROM payment

WHERE payment_id =

(SELECT payment_id

FROM payment

WHERE payment_date >= ’05-JAN-2002′ OR payment_amount > 500.00);

Which change could correct the problem?

Mark for Review

(1) Points

Remove the subquery WHERE clause.

Change the outer query WHERE clause to ‘WHERE payment_id IN’. (*)

Include the PAYMENT_ID column in the select list of the outer query.

Remove the single quotes around the date value in the inner query WHERE clause.

Correct

84.  You need to display all the products that cost more than the maximum cost of every product produced in Japan. Which multiple-row comparison operator could you use?  Mark for Review

(1) Points

>ANY (*)

NOT=ALL

IN

>IN

Incorrect. Refer to Section 6

85.  Which statement about the ANY operator when used with a multiple-row subquery is true?  Mark for Review

(1) Points

The ANY operator compares every value returned by the subquery. (*)

The ANY operator can be used with the DISTINCT keyword.

The ANY operator is a synonym for the ALL operator.

The ANY operator can be used with the LIKE and IN operators.

Incorrect. Refer to Section 6

86.  Evaluate this SELECT statement:

SELECT player_id, name

FROM players

WHERE team_id IN

(SELECT team_id

FROM teams

WHERE team_id > 300 AND salary_cap > 400000);

What would happen if the inner query returned a NULL value?

Mark for Review

(1) Points

No rows would be returned by the outer query. (*)

A syntax error in the outer query would be returned.

A syntax error in the inner query would be returned.

All the rows in the PLAYER table would be returned by the outer query.

Correct

Section 7 Lesson 1

(Answer all questions in this section)

87.  The STUDENTS table contains these columns:

STU_ID NUMBER(9) NOT NULL

LAST_NAME VARCHAR2 (30) NOT NULL

FIRST_NAME VARCHAR2 (25) NOT NULL

DOB DATE

STU_TYPE_ID VARCHAR2(1) NOT NULL

ENROLL_DATE DATE

You create another table, named FT_STUDENTS, with an identical structure.You want to insert all full-time students, who have a STU_TYPE_ID value of “F”, into the new table. You execute this INSERT statement:

INSERT INTO ft_students

(SELECT stu_id, last_name, first_name, dob, stu_type_id, enroll_date

FROM students

WHERE UPPER(stu_type_id) = ‘F’);

What is the result of executing this INSERT statement?

Mark for Review

(1) Points

All full-time students are inserted into the FT_STUDENTS table. (*)

An error occurs because the FT_STUDENTS table already exists.

An error occurs because you CANNOT use a subquery in an INSERT statement.

An error occurs because the INSERT statement does NOT contain a VALUES clause.

Incorrect. Refer to Section 7

88.  Which statement about the VALUES clause of an INSERT statement is true?  Mark for Review

(1) Points

If no column list is specified, then the values must be in the order the columns are specified in the table. (*)

The VALUES clause in an INSERT statement is optional.

Character, date, and numeric data must be enclosed within single quotes in the VALUES clause.

To specify a null value in the VALUES clause, use an empty string (‘ ‘).

Incorrect. Refer to Section 7

89.  You need to copy rows from the EMPLOYEE table to the EMPLOYEE_HIST table. What could you use in the INSERT statement to accomplish this task?  Mark for Review

(1) Points

an ON clause

a SET clause

a subquery (*)

a function

Incorrect. Refer to Section 7

90.  The PRODUCTS table contains these columns:

PROD_ID NUMBER(4)

PROD_NAME VARCHAR2(25)

PROD_PRICE NUMBER(3)

You want to add the following row data to the PRODUCTS table:

(1) a NULL value in the PROD_ID column

(2) “6-foot nylon leash” in the PROD_NAME column

(3) “10″ in the PROD_PRICE column

You issue this statement:

INSERT INTO products

VALUES (null,’6-foot nylon leash’, 10);

What row data did you add to the table?

Mark for Review

(1) Points

The row was created with the correct data in all three columns. (*)

The row was created with the correct data in two of three columns.

The row was created with the correct data in one of the three columns.

The row was created completely wrong. No data ended up in the correct columns.

Incorrect. Refer to Section 7

Section 7 Lesson 2

(Answer all questions in this section)

91.  One of the sales representatives, Janet Roper, has informed you that she was recently married, and she has requested that you update her name in the employee database. Her new last name is Cooper. Janet is the only person with the last name of Roper that is employed by the company. The EMPLOYEES table contains these columns and all data is stored in lowercase:

EMP_ID NUMBER(10) PRIMARY KEY

LNAME VARCHAR2(20)

FNAME VARCHAR2(20)

DEPT VARCHAR2 (20)

HIRE_DATE DATE

SALARY NUMBER(10)

Which UPDATE statement will accomplish your objective?

Mark for Review

(1) Points

UPDATE employees

SET lname = ‘cooper’

WHERE lname = ‘roper’;

(*)

UPDATE employees lname = ‘cooper’

WHERE lname = ‘roper’;

UPDATE employees

SET lname = ‘roper’

WHERE lname = ‘cooper’;

UPDATE employees

SET cooper = ‘lname’

WHERE lname = ‘roper’;

Correct

92.  The EMPLOYEES table contains the following columns:

EMP_ID NUMBER(10) PRIMARY KEY

LNAME VARCHAR2(20)

FNAME VARCHAR2(20)

DEPT VARCHAR2(20)

HIRE_DATE DATE

SALARY NUMBER(9,2)

BONUS NUMBER(9,2)

You want to execute one DML statement to change the salary of all employees in department 10 to equal the new salary of employee number 89898. Currently, all employees in department 10 have the same salary value. Which statement should you execute?

Mark for Review

(1) Points

UPDATE employee

SET salary = SELECT salary

FROM employee

WHERE emp_id = 89898;

UPDATE employee

SET salary = (SELECT salary FROM employee WHERE emp_id = 89898);

UPDATE employee

SET salary = (SELECT salary FROM employee WHERE emp_id = 89898)

WHERE dept = 10;

(*)

UPDATE employee

SET salary = (SELECT salary FROM employee WHERE emp_id = 89898 AND dept = 10);

Correct

93.  The PLAYERS table contains these columns:

PLAYER_ID NUMBER NOT NULL

PLAYER_LNAME VARCHAR2(20) NOT NULL

PLAYER_FNAME VARCHAR2(10) NOT NULL

TEAM_ID NUMBER

SALARY NUMBER(9,2)

You need to increase the salary of each player for all players on the Tiger team by 12.5 percent. The TEAM_ID value for the Tiger team is 5960. Which statement should you use?

Mark for Review

(1) Points

UPDATE players (salary) SET salary = salary * 1.125;

UPDATE players SET salary = salary * .125 WHERE team_id = 5960;

UPDATE players SET salary = salary * 1.125 WHERE team_id = 5960; (*)

UPDATE players (salary) VALUES(salary * 1.125) WHERE team_id = 5960;

Correct

94.  You need to delete a record in the EMPLOYEES table for Tim Jones, whose unique employee identification number is 348. The EMPLOYEES table contains these columns:

ID_NUM NUMBER(5) PRIMARY KEY

LNAME VARCHAR2(20)

FNAME VARCHAR2(20)

ADDRESS VARCHAR2(30)

PHONE NUMBER(10)

Which DELETE statement will delete the appropriate record without deleting any additional records?

Mark for Review

(1) Points

DELETE FROM employees WHERE id_num = 348; (*)

DELETE FROM employees WHERE lname = jones;

DELETE * FROM employees WHERE id_num = 348;

DELETE ‘jones’ FROM employees;

Incorrect. Refer to Section 7

95.  You need to update the expiration date of products manufactured before June 30th . In which clause of the UPDATE statement will you specify this condition?  Mark for Review

(1) Points

the ON clause

the WHERE clause (*)

the SET clause

the USING clause

Correct

96.  Which of the following represents the correct syntax for an INSERT statement?  Mark for Review

(1) Points

INSERT VALUES INTO customers (3178 J. Smith 123 Main Street Nashville TN 37777;

INSERT INTO customers VALUES ’3178′ ‘J.’ ‘Smith’ ’123 Main Street’ ‘Nashville’ ‘TN’ ’37777′;

INSERT INTO customers VALUES (’3178′, ‘J.’, ‘Smith’, ’123 Main Street’, ‘Nashville’, ‘TN’, ’37777′); (*)

INSERT customers VALUES 3178, J., Smith, 123 Main Street, Nashville, TN, 37777;

Incorrect. Refer to Section 7

97.  One of your employees was recently married. Her employee ID is still 189, however, her last name is now Rockefeller. Which SQL statement will allow you to reflect this change?  Mark for Review

(1) Points

INSERT INTO my_employees SET last_name = ‘Rockefeller’ WHERE employee_ID = 189;

INSERT my_employees SET last_name = ‘Rockefeller’ WHERE employee_ID = 189;

UPDATE INTO my_employees SET last_name = ‘Rockefeller’ WHERE employee_ID = 189;

UPDATE my_employees SET last_name = ‘Rockefeller’ WHERE employee_ID = 189; (*)

Incorrect. Refer to Section 7

98.  When the WHERE clause is missing in a DELETE statement, what is the result?  Mark for Review

(1) Points

All rows are deleted from the table. (*)

The table is removed from the database.

An error message is displayed indicating incorrect syntax.

Nothing. The statement will not execute.

Correct

99.  What keyword in an UPDATE statement speficies the columns you want to change? Mark for Review

(1) Points

SELECT

WHERE

SET (*)

HAVING

Correct

100.  You need to update both the DEPARTMENT_ID and LOCATION_ID columns in the EMPLOYEE table using one UPDATE statement. Which clause should you include in the UPDATE statement to update multiple columns?  Mark for Review

(1) Points

the USING clause

the ON clause

the WHERE clause

the SET clause (*)

Incorrect. Refer to Section 7

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February 25, 2010

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