Exam Number/Code:
1Z0-051
Exam Name: Oracle Database: SQL Fundamentals I
Questions and Answers: 175 Q&As
Price:$ 100.00
Update Time: 2009-12-20
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Oracle Database: SQL Fundamentals I : 1Z0-051 Exam
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1Z0-051 practice test Oracle Database: SQL Fundamentals I * Exam Number/Code : 1Z0-051 * Exam Name : Oracle Database: SQL Fundamentals I * Questions and Answers : 175 Q&As * Update Time: 2009-12-28 Exam 1Z0-051 covers topics such as: * retrieving data using SQL select statements * reporting aggregated data using group functions * displaying data from multiple tables * using subqueries to solve queries * using set operator * manipulating data * using DDL statements to create… [ more.. ]
1Z0-051 practice test Oracle Database: SQL Fundamentals I * Exam Number/Code : 1Z0-051 * Exam Name : Oracle Database: SQL Fundamentals I * Questions and Answers : 175 Q&As * Update Time: 2009-12-28 Exam 1Z0-051 covers topics such as: * retrieving data using SQL select statements * reporting aggregated data using group functions * displaying data from multiple tables * using subqueries to solve queries * using set operator * manipulating data * using DDL statements to create… [ more.. ]
Exam : Oracle 1Z0-051
Title : Oracle Database: SQL Fundamentals I
1. 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 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
2. Examine the structure of the SHIPMENTS table:
name Null Type
PO_ID NOT NULL NUMBER(3)
PO_DATE NOT NULL DATE
SHIPMENT_DATE NOT NULL DATE
SHIPMENT_MODE VARCHAR2(30)
SHIPMENT_COST NUMBER(8,2)
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
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
3. 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 ON clause can be used to join tables on columns that have different names but compatible data types.
D. The WHERE clause can be used to apply additional conditions in SELECT statements containing the ON or the USING clause.
Answer: CD
4. 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 ''
C. WHERE prod_id LIKE '%_D123%' ESCAPE '%_'
D. WHERE prod_id LIKE '%_D123%' ESCAPE '_'
Answer: B
5. 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 value different from the one that is referenced.
Answer: DE
6. 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.
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
7. 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
8. 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