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Exam Code: 1Z0-051
Exam Name: Oracle Database: SQL Fundamentals I
Updated: Apr 13, 2017
Q&As: 292

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Teststarter Latest and Most Accurate Oracle 1Z0-051 Dumps Exam Q&As 

QUESTION 3
Here is the structure and data of the CUST_TRANS table:
Exhibit:

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Dates are stored in the default date format dd-mm-rr in the CUST_TRANS table. Which three SQL
statements would execute successfully? (Choose three.)
A. SELECT transdate + ’10’ FROM cust_trans;
B. SELECT * FROM cust_trans WHERE transdate = ’01-01-07′;
C. SELECT transamt FROM cust_trans WHERE custno > ’11’;
D. SELECT * FROM cust_trans WHERE transdate=’01-JANUARY-07′;
E. SELECT custno + ‘A’ FROM cust_trans WHERE transamt > 2000;
Correct Answer: ACD
Explanation
Explanation/Reference:

 

 

QUESTION 4
See the Exhibit and examine the structure and data in the INVOICE table:
Exhibit:

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Which two SQL statements would executes successfully? (Choose two.)

A. SELECT MAX(inv_date),MIN(cust_id) FROM invoice;
B. SELECT MAX(AVG(SYSDATE – inv_date)) FROM invoice;
C. SELECT (AVG(inv_date) FROM invoice;
D. SELECT AVG(inv_date – SYSDATE),AVG(inv_amt) FROM invoice;
Correct Answer: AD
Explanation
Explanation/Reference:

 

 

QUESTION 5
See the Exhibit and examine the structure of the CUSTOMERS table:

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Using the CUSTOMERS table, you need to generate a report that shown the average credit limit for
customers in WASHINGTON and NEW YORK.
Which SQL statement would produce the required result?
A. SELECT cust_city, AVG(cust_credit_limit)
FROM customers
WHERE cust_city IN (‘WASHINGTON’,’NEW YORK’)
GROUP BY cust_credit_limit, cust_city;
B. SELECT cust_city, AVG(cust_credit_limit)
FROM customers
WHERE cust_city IN (‘WASHINGTON’,’NEW YORK’)
GROUP BY cust_city,cust_credit_limit;
C. SELECT cust_city, AVG(cust_credit_limit)
FROM customers
WHERE cust_city IN (‘WASHINGTON’,’NEW YORK’)
GROUP BY cust_city;

D. SELECT cust_city, AVG(NVL(cust_credit_limit,0))
FROM customers
WHERE cust_city IN (‘WASHINGTON’,’NEW YORK’);
Correct Answer: C
Explanation
Explanation/Reference:
Explanation:1Z0-051 dumps
Creating Groups of Data: GROUP BY Clause Syntax
You can use the GROUP BY clause to divide the rows in a table into groups. You can then use the group
functions to return summary information for each group.
In the syntax:
group_by_expression Specifies the columns whose values determine the basis for grouping rows
Guidelines
· If you include a group function in a SELECT clause, you cannot select individual results as well, unless
the individual column appears in the GROUP BY clause. You receive an error message if you fail to
include the column list in the GROUP BY clause. · Using a WHERE clause, you can exclude rows before
dividing them into groups.
· You must include the columns in the GROUP BY clause.
· You cannot use a column alias in the GROUP BY clause.

 

 

QUESTION 6
Where can sub queries be used? (Choose all that apply)
A. field names in the SELECT statement
B. the FROM clause in the SELECT statement
C. the HAVING clause in the SELECT statement
D. the GROUP BY clause in the SELECT statement
E. the WHERE clause in only the SELECT statement
F. the WHERE clause in SELECT as well as all DML statements
Correct Answer: ABCF
Explanation
Explanation/Reference:
Explanation:
SUBQUERIES can be used in the SELECT list and in the FROM, WHERE, and HAVING clauses of a
query.
A subquery can have any of the usual clauses for selection and projection. The following are required
clauses:
A SELECT list
A FROM clause
The following are optional clauses:
WHERE
GROUP BY
HAVING
The subquery (or subqueries) within a statement must be executed before the parent query that calls it, in
order that the results of the subquery can be passed to the parent.

 

 

QUESTION 7
Evaluate the following SQL statement:

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Which statement is true regarding the outcome of the above query?
A. It produces an error because the ORDER BY clause should appear only at the end of a compound
query-that is, with the last SELECT statement
B. It executes successfully and displays rows in the descending order of PROMO_CATEGORY
C. It executes successfully but ignores the ORDER BY clause because it is not located at the end of the
compound statement

D. It produces an error because positional notation cannot be used in the ORDER BY clause with SET
operators
Correct Answer: A
Explanation
Explanation/Reference:
Explanation:
Using the ORDER BY Clause in Set Operations
The ORDER BY clause can appear only once at the end of the compound query. Component queries
cannot have individual ORDER BY clauses. The ORDER BY clause recognizes only the columns of the
first SELECT query. By default, the first column of the first SELECT query is used to sort the output in an
ascending order.

 

 

QUESTION 8
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:

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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
B. name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT
value
S(e.employee_id, e.first_name ||’,
‘||e.last_name);
C. MERGE new_employees c
USING employees e
ON (c.employee_id = e.employee_id)
WHEN EXISTS THEN
UPDATE SET
D. name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN INSERT
valueS(e.employee_id, e.first_name ||’,
‘||e.last_name);
E. MERGE INTO new_employees cUSING employees e
ON (c.employee_id = e.employee_id)
WHEN EXISTS THEN
UPDATE SET
F. name = e.first_name ||’,’|| e.last_name
WHEN NOT MATCHED THEN
INSERT
value
S(e.employee_id, e.first_name ||’,
‘||e.last_name);
G. MERGE new_employees c
FROM employees e ON (c.employee_id = e.employee_id)
WHEN MATCHED THEN
UPDATE SET
H. 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);
Correct Answer: A
Explanation
Explanation/Reference:
Explanation: this is the correct MERGE statement syntax
Incorrect answer:
B it should MERGE INTO table_name
C it should be WHEN MATCHED THEN
D it should MERGE INTO table_name
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 8-29

 

 

 

QUESTION 9
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
Correct Answer: D
Explanation
Explanation/Reference:
Explanation: view the columns associated with the constraint names in the USER_CONS_COLUMNS
view.
Incorrect answer:
A table to view all constraints definition and names
B show all object name belong to user
C does not display column associated
E no such view
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 10-25

 

 

QUESTION 15
The COMMISSION column shows the monthly commission earned by the employee.
Exhibit

C2090-930 exam

Which two tasks would require sub queries 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
Correct Answer: AC
Explanation  1Z0-051 dumps

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