luni, 21 martie 2011

1. You need to display each employee's name in all uppercase letters. Which function should you use? Mark for Review
(1) Points


CASE


UCASE


UPPER (*)


TOUPPER



Incorrect. Refer to Section 1 Lesson 1.


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 (*)



Incorrect. Refer to Section 1 Lesson 1.


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. 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. (*)



Correct


5. 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 12.00
You query the database and return the value 79. Which script did you use?

Mark for Review
(1) Points


SELECT INSTR(category, 2,2)
FROM styles
WHERE style_id = 895840;


SELECT INSTR(category, -2,2)
FROM styles
WHERE style_id = 895840;


SELECT SUBSTR(category, 2,2)
FROM styles
WHERE style_id = 895840;


SELECT SUBSTR(category, -2,2)
FROM styles
WHERE style_id = 758960;
(*)




Incorrect. Refer to Section 1 Lesson 1.


6. You query the database with this SQL statement:
SELECT LOWER(SUBSTR(CONCAT(last_name, first_name)), 1, 5) "ID"
FROM employee;

In which order are the functions evaluated?

Mark for Review
(1) Points


LOWER, SUBSTR, CONCAT


LOWER, CONCAT, SUBSTR


SUBSTR, CONCAT, LOWER


CONCAT, SUBSTR, LOWER (*)



Correct


7. You query the database with this SQL statement:
SELECT CONCAT(last_name, (SUBSTR(LOWER(first_name), 4))) "Default Password"
FROM employees;

Which function will be evaluated first?

Mark for Review
(1) Points


CONCAT


SUBSTR


LOWER (*)


All three will be evaluated simultaneously.



Correct


8. 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."



Incorrect. Refer to Section 1 Lesson 1.


9. You issue this SQL statement:
SELECT TRUNC(751.367,-1) FROM dual;
Which value does this statement display?

Mark for Review
(1) Points


700


750 (*)


751


751.3



Incorrect. Refer to Section 1 Lesson 2.


10. Which two functions can be used to manipulate number or date column values, but NOT character column values? (Choose two.) Mark for Review
(1) Points

(Choose all correct answers)


RPAD


TRUNC (*)


ROUND (*)


INSTR


CONCAT



Incorrect. Refer to Section 1 Lesson 2.

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


1 (*)


2


25


0



Incorrect. Refer to Section 1 Lesson 2.


12. 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;



Incorrect. Refer to Section 1 Lesson 3.


13. The EMPLOYEES table contains these columns:
LAST_NAME VARCHAR2(20)
FIRST_NAME VARCHAR2(20)
HIRE_DATE DATE
EVAL_MONTHS NUMBER(3)

Evaluate this SELECT statement:

SELECT hire_date + eval_months
FROM employees;

The values returned by this SELECT statement will be of which data type?

Mark for Review
(1) Points


DATE (*)


NUMBER


DATETIME


INTEGER



Incorrect. Refer to Section 1 Lesson 3.


14. You need to display the number of months between today's date and each employee's hiredate. Which function should you use? Mark for Review
(1) Points


ROUND


BETWEEN


ADD_MONTHS


MONTHS_BETWEEN (*)



Correct


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;



Incorrect. Refer to Section 1 Lesson 3.


16. Which function would you use to return the current database server date and time? Mark for Review
(1) Points


DATE


SYSDATE (*)


DATETIME


CURRENTDATE



Correct




Section 2
(Answer all questions in this section)

17. 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 Lesson 2.


18. When executed, which statement displays a zero if the TUITION_BALANCE value is zero and the HOUSING_BALANCE value is null? Mark for Review
(1) Points


SELECT NVL (tuition_balance + housing_balance, 0) "Balance Due"
FROM student_accounts;
(*)



SELECT NVL(tuition_balance, 0), NVL (housing_balance), tuition_balance + housing_balance "Balance Due"
FROM student_accounts;


SELECT tuition_balance + housing_balance
FROM student_accounts;


SELECT TO_NUMBER(tuition_balance, 0), TO_NUMBER (housing_balance, 0), tutition_balance + housing_balance "Balance Due"
FROM student_accounts;



Correct


19. 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 (*)



Incorrect. Refer to Section 2 Lesson 2.


20. The PRODUCT table contains this column: PRICE NUMBER(7,2)
Evaluate this statement:
SELECT NVL(10 / price, '0')
FROM PRODUCT;

What would happen if the PRICE column contains null values?

Mark for Review
(1) Points


The statement would fail because values cannot be divided by 0.


A value of 0 would be displayed. (*)


A value of 10 would be displayed.


The statement would fail because values cannot be divided by null.



Correct

21. Which SQL Statement should you use to display the prices in this format: "$00.30"? Mark for Review
(1) Points


SELECT TO_CHAR(price, '$99,900.99')
FROM product;
(*)



SELECT TO_CHAR(price, "$99,900.99")
FROM product;


SELECT TO_CHAR(price, '$99,990.99')
FROM product;


SELECT TO_NUMBER(price, '$99,900.99')
FROM product;



Incorrect. Refer to Section 2 Lesson 1.


22. Which best describes the TO_CHAR function? Mark for Review
(1) Points


The TO_CHAR function can be used to specify meaningful column names in an SQL statement's result set.


The TO_CHAR function can be used to remove text from column data that will be returned by the database.


The TO_CHAR function can be used to display dates and numbers according to formatting conventions that are supported by Oracle. (*)


The TO_CHAR function can only be used on Date columns.



Incorrect. Refer to Section 2 Lesson 1.


23. Which arithmetic operation will return a numeric value? Mark for Review
(1) Points


TO_DATE('01-JUN-2004') - TO_DATE('01-OCT-2004') (*)


NEXT_DAY(hire_date) + 5


SYSDATE - 6


SYSDATE + 30 / 24



Incorrect. Refer to Section 2 Lesson 1.


24. 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.



Correct


25. Which three statements concerning explicit data type conversions are true? (Choose three.) Mark for Review
(1) Points

(Choose all correct answers)


Use the TO_NUMBER function to convert a number to a character string.


Use the TO_DATE function to convert a character string to a date value. (*)


Use the TO_NUMBER function to convert a character string of digits to a number. (*)


Use the TO_DATE function to convert a date value to character string or number.


Use the TO_CHAR function to convert a number or date value to character string. (*)



Incorrect. Refer to Section 2 Lesson 1.


26. 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




Section 3
(Answer all questions in this section)

27. Which statement about a self join is true? Mark for Review
(1) Points


The NATURAL JOIN clause must be used.


Table aliases must be used to qualify table names. (*)


Table aliases cannot be used to qualify table names.


A self join must be implemented by defining a view.



Incorrect. Refer to Section 3 Lesson 4.


28. Evaluate this SELECT statement:
SELECT *
FROM employee e, employee m
WHERE e.mgr_id = m.emp_id;
Which type of join is created by this SELECT statement?

Mark for Review
(1) Points


a self join (*)


a cross join


a left outer join


a full outer join



Correct


29. Which SELECT statement implements a self join? Mark for Review
(1) Points


SELECT p.part_id, t.product_id
FROM part p, part t
WHERE p.part_id = t.product_id;
(*)



SELECT p.part_id, t.product_id
FROM part p, product t
WHERE p.part_id = t.product_id;


SELECT p.part_id, t.product_id
FROM part p, product t
WHERE p.part_id = t.product_id (+);


SELECT p.part_id, t.product_id
FROM part p, product t
WHERE p.part_id =! t.product_id;



Incorrect. Refer to Section 3 Lesson 4.


30. 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



Incorrect. Refer to Section 3 Lesson 3.
31. Which type of join returns rows from one table that have NO direct match in the other table? Mark for Review
(1) Points


Equijoin


Self join


Outer join (*)


Natural join



Incorrect. Refer to Section 3 Lesson 3.


32. 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 orders o
ON c.cust_id = o.cust_id;



Incorrect. Refer to Section 3 Lesson 3.


33. The primary advantages of using JOIN ON is: (Select two) Mark for Review
(1) Points

(Choose all correct answers)


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 3 Lesson 2.


34. 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 3 Lesson 2.


35. Which of the following statements is the simplest description of a nonequijoin? Mark for Review
(1) Points


A join condition containing something other than an equality operator (*)


A join condition that is not equal to other joins.


A join condition that includes the (+) on the left hand side.


A join that joins a table to itself



Incorrect. Refer to Section 3 Lesson 2.


36. You created the CUSTOMERS and ORDERS tables by issuing these CREATE TABLE statements in sequence:
CREATE TABLE customers
(custid varchar2(5),
companyname varchar2(30),
contactname varchar2(30),
address varchar2(30),
city varchar2(20),
state varchar2(30),
phone varchar2(20),
constraint pk_customers_01 primary key (custid));

CREATE TABLE orders
(orderid varchar2(5) constraint pk_orders_01 primary key,
orderdate date,
total number(15),
custid varchar2(5) references customers (custid));

You have been instructed to compile a report to present the information about orders placed by customers who reside in Nashville. Which query should you issue to achieve the desired results?

Mark for Review
(1) Points


SELECT custid, companyname
FROM customers
WHERE city = 'Nashville';


SELECT orderid, orderdate, total
FROM orders o
NATURAL JOIN customers c ON o.custid = c.custid
WHERE city = 'Nashville';


SELECT orderid, orderdate, total
FROM orders o
JOIN customers c ON o.custid = c.custid
WHERE city = 'Nashville';
(*)



SELECT orderid, orderdate, total
FROM orders
WHERE city = 'Nashville';



Incorrect. Refer to Section 3 Lesson 2.


37. Evaluate this SELECT statement:
SELECT a.lname || ', ' || a.fname as "Patient", b.lname || ', ' || b.fname as "Physician", c.admission
FROM patient a
JOIN physician b
ON (b.physician_id = c.physician_id)
JOIN admission c
ON (a.patient_id = c.patient_id);

Which clause generates an error?

Mark for Review
(1) Points


JOIN physician b


ON (b.physician_id = c.physician_id); (*)


JOIN admission c


ON (a.patient_id = c.patient_id)



Incorrect. Refer to Section 3 Lesson 2.


38. Which keyword in a SELECT statement 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 (*)



Incorrect. Refer to Section 3 Lesson 2.


39. For which condition would you use an equijoin query with the USING keyword? Mark for Review
(1) Points


You need to perform a join of the CUSTOMER and ORDER tables but limit the number of columns in the join condition. (*)


The ORDER table contains a column that has a referential constraint to a column in the PRODUCT table.


The CUSTOMER and ORDER tables have no columns with identical names.


The CUSTOMER and ORDER tables have a corresponding column, CUST_ID. The CUST_ID column in the ORDER table contains null values that need to be displayed.



Correct




Section 4
(Answer all questions in this section)

40. Which group functions below act on character, number and date data types? (Choose more than one answer) Mark for Review
(1) Points

(Choose all correct answers)


SUM


MAX (*)


MIN (*)


AVG


COUNT (*)



Incorrect. Refer to Section 4 Lesson 2.

41. Which group function would you use to display the lowest value in the SALES_AMOUNT column? Mark for Review
(1) Points


AVG


COUNT


MAX


MIN (*)



Incorrect. Refer to Section 4 Lesson 2.


42. You need to compute the total salary for all employees in department 10. Which group function will you use? Mark for Review
(1) Points


MAX


SUM (*)


VARIANCE


COUNT



Incorrect. Refer to Section 4 Lesson 2.


43. 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
You need to determine the average payment amount made by each customer in January, February and March of 2003.
Which SELECT statement should you use?

Mark for Review
(1) Points


SELECT AVG(payment_amount)
FROM payment
WHERE payment_date
BETWEEN '01-JAN-2003' AND '31-MAR-2003';
(*)



SELECT AVG(payment_amount)
FROM payment;


SELECT SUM(payment_amount)
FROM payment
WHERE payment_date BETWEEN '01-JAN-2003' and '31-MAR-2003';


SELECT AVG(payment_amount)
FROM payment
WHERE TO_CHAR(payment_date) IN (JAN, FEB, MAR);



Incorrect. Refer to Section 4 Lesson 2.


44. 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 4 Lesson 2.


45. The CUSTOMER table contains these columns:
CUSTOMER_ID NUMBER(9)
FIRST_NAME VARCHAR2(25)
LAST_NAME 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



Incorrect. Refer to Section 4 Lesson 2.


46. 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 4x4 model?

Mark for Review
(1) Points


SELECT AVG(price)
FROM trucks
WHERE model = '4x4';
(*)



SELECT AVG(price)
FROM trucks
WHERE model IS '4x4';


SELECT AVG(price)
FROM trucks
WHERE model IS 4x4;


SELECT AVG(price), model
FROM trucks
WHERE model IS '4x4';



Correct


47. Evaluate this SQL statement:
SELECT COUNT (amount)
FROM inventory;

What will occur when the statement is issued?

Mark for Review
(1) Points


The statement will return the greatest value in the INVENTORY table.


The statement will return the total number of rows in the AMOUNT column.


The statement will replace all NULL values that exist in the AMOUNT column.


The statement will count the number of rows in the INVENTORY table where the AMOUNT column is not null. (*)



Incorrect. Refer to Section 4 Lesson 3.


48. 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.



Incorrect. Refer to Section 4 Lesson 3.


49. Evaluate this SELECT statement:
SELECT COUNT(*)
FROM employees
WHERE salary > 30000;

Which results will the query display?

Mark for Review
(1) Points


The number of employees that have a salary less than 30000.


The total of the SALARY column for all employees that have a salary greater than 30000.


The number of rows in the EMPLOYEES table that have a salary greater than 30000. (*)


The query generates an error and returns no results.



Correct


50. 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