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At Exampdf, people can get all the required exam information for Oracle 1Z0-051 test, which cover all the knowledge points of the actual test.

 

 
 
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Published:  February 03, 2012
 
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Slide 1: Many new exams are available at Exampdf.com, such as I40-420, 00M-646, 000-M224, and so on. With Exampdf real Q&As of any IT certification exam, all the examinees can clear the exams easily. Take charge of your time now.
Slide 2: The safer , easier way to help you pass any IT exams. Exam : 1Z0-051 Title : Oracle Database: SQL Fundamentals I Version : DEMO 1 / 11
Slide 3: The safer , easier way to help you pass any IT exams. 1. View the Exhibit and examine the structure of the SALES, CUSTOMERS, PRODUCTS, and TIMES tables. The PROD_ID column is the foreign key in the SALES table, which references the PRODUCTS table. Similarly, the CUST_ID and TIME_ID columns are also foreign keys in the SALES table referencing the CUSTOMERS and TIMES tables, respectively. Evaluate the following CREATE TABLE command: CREATE TABLE new_sales(prod_id, cust_id, order_date DEFAULT SYSDATE) AS SELECT prod_id, cust_id, time_id FROM sales; Which statement is true regarding the above command? A. The NEW_SALES table would not get created because the DEFAULT value cannot be specified in the column definition. B. The NEW_SALES table would get created and all the NOT NULL constraints defined on the specified columns would be passed to the new table. 2 / 11
Slide 4: The safer , easier way to help you pass any IT exams. C. The NEW_SALES table would not get created because the column names in the CREATE TABLE command and the SELECT clause do not match. D. The NEW_SALES table would get created and all the FOREIGN KEY constraints defined on the specified columns would be passed to the new table. Answer: B 2. View the Exhibit to examine the description for the SALES table. Which views can have all DML operations performed on it? (Choose all that apply.) A. CREATE VIEW v3 AS SELECT * FROM SALES WHERE cust_id = 2034 WITH CHECK OPTION; B. CREATE VIEW v1 AS SELECT * FROM SALES WHERE time_id <= SYSDATE - 2*365 WITH CHECK OPTION; C. CREATE VIEW v2 AS SELECT prod_id, cust_id, time_id FROM SALES WHERE time_id <= SYSDATE - 2*365 WITH CHECK OPTION; D. CREATE VIEW v4 AS SELECT prod_id, cust_id, SUM(quantity_sold) FROM SALES WHERE time_id <= SYSDATE - 2*365 GROUP BY prod_id, cust_id WITH CHECK OPTION; Answer: AB 3. You need to extract details of those products in the SALES table where the PROD_ID column contains the string '_D123'. Which WHERE clause could be used in the SELECT statement to get the required output? A. WHERE prod_id LIKE '%_D123%' ESCAPE '_' B. WHERE prod_id LIKE '%\_D123%' ESCAPE '\' 3 / 11
Slide 5: The safer , easier way to help you pass any IT exams. C. WHERE prod_id LIKE '%_D123%' ESCAPE '%_' D. WHERE prod_id LIKE '%\_D123%' ESCAPE '\_' Answer: B 4. Which two statements are true regarding single row functions? (Choose two.) A. They a ccept only a single argument. B. They c an be nested only to two levels. C. Arguments can only be column values or constants. D. They a lways return a single result row for every row of a queried table. E. They c an return a data type Answer: DE 5. Which SQL statements would display the value 1890.55 as $1,890.55? (Choose three .) A. SELECT TO_CHAR(1890.55,'$0G000D00') FROM DUAL; B. SELECT TO_CHAR(1890.55,'$9,999V99') FROM DUAL; C. SELECT TO_CHAR(1890.55,'$99,999D99') FROM DUAL; D. SELECT TO_CHAR(1890.55,'$99G999D00') FROM DUAL; E. SELECT TO_CHAR(1890.55,'$99G999D99') FROM DUAL; Answer: ADE 6. Examine the structure of the SHIPMENTS table: name PO_ID PO_DATE Null Type NUMBER(3) DATE DATE VARCHAR2(30) NUMBER(8,2) value different from the one that is referenced. NOT NULL NOT NULL SHIPMENT_DATE NOT NULL SHIPMENT_MODE SHIPMENT_COST You want to generate a report that displays the PO_ID and the penalty amount to be paid if the SHIPMENT_DATE is later than one month from the PO_DATE. The penalty is $20 per day. Evaluate the following two queries: SQL> SELECT po_id, CASE WHEN MONTHS_BETWEEN (shipment_date,po_date)>1 THEN 4 / 11
Slide 6: The safer , easier way to help you pass any IT exams. TO_CHAR((shipment_date - po_date) * 20) ELSE 'No Penalty' END PENALTY FROM shipments; SQL>SELECT po_id, DECODE (MONTHS_BETWEEN (po_date,shipment_date)>1, TO_CHAR((shipment_date - po_date) * 20), 'No Penalty') PENALTY FROM shipments; Which statement is true regarding the above commands? A. Both execute successfully and give correct results. B. Only the first query executes successfully but gives a wrong result. C. Only the first query executes successfully and gives the correct result. D. Only the second query executes successfully but gives a wrong result. E. Only the second query executes successfully and gives the correct result. Answer: C 7. Which two statements are true regarding the USING and ON clauses in table joins? (Choose two.) A. Both USING and ON clauses can be used for equijoins and nonequijoins. B. A maximum of one pair of columns can be joined between two tables using the ON clause. C. The types. D. The WHERE clause can be used to apply additional conditions in SELECT statements containing the USING clause. ON clause can be used to join tables on columns that have different names but compatible data ON or the Answer: CD 8. View the Exhibit and examine the structure of the CUSTOMERS table. Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.) A. listing of customers who do not have a credit limit and were born before 1980 B. finding the number of customers, in each city, whose marital status is 'married' C. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney' D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the 5 / 11
Slide 7: The safer , easier way to help you pass any IT exams. city 'Tokyo' E. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers Answer: DE 9. Which statement is true regarding the INTERSECT operator? A. It ignores NULL values. B. Reversing the order of the intersected tables alters the result. C. The names of columns in all SELECT statements must be identical. D. The number of columns and data types must be identical for all SELECT statements in the query. Answer: D 10. View the Exhibit; e xamine the structure of the PROMOTIONS table. Each promotion has a duration of at least seven days . Your manager has asked you to generate a report, done to l date. Which query would achieve the required result? which provides the weekly cost for each promotion A. SELECT promo_name, promo_cost/promo_end_date-promo_begin_date/7 FROM promotions; B. SELECT promo_name,(promo_cost/promo_end_date-promo_begin_date)/7 FROM promotions; C. SELECT promo_name, promo_cost/(promo_end_date-promo_begin_date/7) FROM promotions; D. SELECT promo_name, promo_cost/((promo_end_date-promo_begin_date)/7) FROM promotions; Answer: D 11. View the Exhibit and examine the structure of the PRODUCTS table. All products have a list price. You issue the following command to display the total price of each product after a discount of 25% and a 6 / 11
Slide 8: The safer , easier way to help you pass any IT exams. tax of 15% are applied on it. Freight charges of $100 have to be applied to all the products. SQL>SELECT prod_name, prod_list_price -(prod_list_price*(25/100)) +(prod_list_price -(prod_list_price*(25/100))*(15/100))+100 AS "TOTAL PRICE" FROM products; What would be the outcome if all the parenthese s are removed from the above statement? A. It produces a syntax error. B. The result remains unchanged. C. The total price value would be lower than the correct value. D. The total price value would be higher than the correct value. Answer: B 12. You need to produce a report where each customer's credit limit has been incremented by $1000. In the output, t he customer's last name should have the heading Name and the incremented credit limit should be labeled New Credit Limit. The column headings should have only the first letter of each word in uppercase . Which statement would accomplish this requirement? A. SELECT cust_last_name Name, cust_credit_limit + 1000 "New Credit Limit" FROM customers; B. SELECT cust_last_name AS Name, cust_credit_limit + 1000 AS New Credit Limit FROM customers; C. SELECT cust_last_name AS "Name", cust_credit_limit + 1000 AS "New Credit Limit" FROM customers; D. SELECT INITCAP(cust_last_name) "Name", cust_credit_limit + 1000 INITCAP("NEW CREDIT LIMIT") FROM customers; 7 / 11
Slide 9: The safer , easier way to help you pass any IT exams. Answer: C 13. View the Exhibit and examine the structure of the PRODUCTS table. You need to generate a report in the following format: CATEGORIES 5MP Digital Photo Camera's category is Photo Y Box's category is Electronics Envoy Ambassador's category is Hardware Which two queries would give the required output? (Choose two.) A. SELECT prod_name FROM products; B. SELECT prod_name FROM products; C. SELECT prod_name FROM products; D. SELECT prod_name FROM products; Answer: CD q'''s category is ' prod_category CATEGORIES q'['s ]'category is ' prod_category CATEGORIES q'\'s\' ' category is ' prod_category CATEGORIES q'<'s >' 'category is ' prod_category CATEGORIES 14. Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit amount in each income level. The report should NOT show any repeated credit amounts in each income level. Which query would give the required result? A. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50% Credit Limit" FROM customers; B. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS "50% Credit Limit" FROM customers; C. SELECT DISTINCT cust_income_level ' ' cust_credit_limit * 0.50 8 / 11
Slide 10: The safer , easier way to help you pass any IT exams. AS "50% Credit Limit" FROM customers; D. SELECT cust_income_level ' ' cust_credit_limit * 0.50 AS "50% Credit Limit" FROM customers; Answer: C 15. View the Exhibit and examine the data in the CUSTOMERS table. Evaluate the following query: SQL> SELECT cust_name AS "NAME", cust_credit_limit/2 AS MIDPOINT,MIDPOINT+100 AS "MAX LOWER LIMIT" FROM customers; The above query produces an error on execution. What is the reason for the error? A. An alias cannot be used in an expression. B. The a lias NAME should not be enclosed with in double quotation marks . C. The MIDPOINT+100 expression values. D. The a lias MIDPOINT should be enclosed with in double quotation marks CUST_CREDIT_LIMIT/2 expression . Answer: A 16. Evaluate the following query: SQL> SELECT promo_name q'{'s start date was }' promo_begin_date for the gives an error because CUST_CREDIT_LIMIT contains NULL AS "Promotion Launches" FROM promotions; What would be the outcome of the above query? A. It produces an error because flower braces have been used. B. It produces an error because the data types are not matching. 's at the end of each promo_name in the output. C. It executes successfully and introduces an D. It executes successfully and displays the literal " {'s start date was } " for each row in the output. Answer: C 9 / 11
Slide 11: The safer , easier way to help you pass any IT exams. 17. View the E xhibit and examine the data in the EMPLOYEES table. You want to generate a report showing the total compensation paid to each employee to You issue the following query: SQL>SELECT ename ' joined on ' hiredate ', the total compensation paid is ' TO_CHAR(ROUND(ROUND(SYSDATE-hiredate)/365) * sal + comm) "COMPENSATION UNTIL DATE" FROM employees; What is the outcome? date. A. It generates an error because the alias is not valid. B. It executes successfully and gives the correct output. C. It executes successfully but does not give the correct output. D. It generates an error because the usage of the ROUND function in the expression is not valid. E. It generates an error because the concatenation operator can be used to combine only two items. Answer: C 18. Examine the structure of the PROMOTIONS table: name PROMO_ID PROMO_NAME Null Type NUMBER(6) VARCHAR2(30) VARCHAR2(30) NUMBER(10,2) NOT NULL NOT NULL PROMO_CATEGORY NOT NULL PROMO_COST NOT NULL The management wants to see a report of unique promotion costs in each promotion category. Which query would achieve the required result? A. SELECT DISTINCT promo_cost, promo_category FROM promotions; B. SELECT promo_category, DISTINCT promo_cost FROM promotions; C. SELECT DISTINCT promo_cost, DISTINCT promo_category FROM promotions; D. SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1; 10 / 11
Slide 12: The safer , easier way to help you pass any IT exams. Answer: D 19. Evaluate the following query: SELECT INTERVAL '300' MONTH, INTERVAL '54-2' YEAR TO MONTH, INTERVAL '11:12:10.1234567' HOUR TO SECOND FROM dual; What is the correct output of the above query? A. +25-00 , +54-02, +00 11:12:10.123457 B. +00-300, +54-02, +00 11:12:10.123457 C. +25-00 , +00-650, +00 11:12:10.123457 D. +00-300 , +00-650, +00 11:12:10.123457 Answer: A 20. Which three statements are true regarding the data types in Oracle Database 10g/11g? (Choose three.) A. Only one LONG column can be used per table. B. A TIMESTAMP data type column stores only time values with fractional seconds. C. The BLOB data type column is used to store binary data in an operating system file. D. The minimum column width that can be specified for a VARCHAR2 data type column is one. E. The value for a CHAR data type column is blank-padded to the maximum defined column width. Answer: ADE 11 / 11

   
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