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  1. Home
  2. Oracle Certification
  3. 1Z1-051 Exam
  4. Oracle.1Z1-051.dumpsfiles Dumps

Free Oracle 1Z1-051 Exam Dumps Questions & Answers

Exam Code/Number:1Z1-051Join the discussion
Exam Name:Oracle Database: SQL Fundamentals I
Certification:Oracle
Question Number:254
Publish Date:Aug 29, 2026
Rating
100%
Page: 1 / 51
Total 254 questions
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Question 1

Examine the data in the CUSTOMERS table:

You want to list all cities that have more than one customer along with the customer details.
Evaluate the following query:
SQL>SELECT c1.custname, c1.city
FROM Customers c1 __________________ Customers c2
ON (c1.city=c2.city AND c1.custname<>c2.custname);
Which two JOIN options can be used in the blank in the above query to give the correct output? (Choose two.)

Correct Answer: C,D

Question 2

In the CUSTOMERS table, the CUST_CITY column contains the value 'Paris' for the
CUST_FIRST_NAME 'ABIGAIL'.
Evaluate the following query:

What would be the outcome?

Correct Answer: A

Question 3

Which three statements are true regarding the data types in Oracle Database 10g/11g? (Choose three.)

Correct Answer: A,D,E
Explanation: (Only visible for DumpsFiles members)

Question 4

You work as a database administrator at ABC.com. You study the exhibit carefully.
Exhibit:

You want to create a SALE_PROD view by executing the following SQL statements:

Which statement is true regarding the execution of the above statement?

Correct Answer: B
Explanation: (Only visible for DumpsFiles members)

Question 5

View the Exhibit and examine the structures of the EMPLOYEES and DEPARTMENTS tables.
You want to update the EMPLOYEES table as follows:4 ? 4;
-Update only those employees who work in Boston or Seattle (locations 2900 and 2700).
-Set department_id for these employees to the department_id corresponding to London
(location_id 2100).
-Set the employees' salary in location_id 2100 to 1.1 times the average salary of their department.
-Set the employees' commission in location_id 2100 to 1.5 times the average commission of their
department.
You issue the following command:
SQL>UPDATE employees
SET department_id =
(SELECT department_id
FROM departments
WHERE location_id = 2100),
(salary, commission) =
(SELECT 1.1*AVG(salary), 1.5*AVG(commission)
FROM employees, departments
WHERE departments.location_id IN(2900,2700,2100))
WHERE department_id IN
(SELECT department_id
FROM departments
WHERE location_id = 2900
OR location_id = 2700)
What is the outcome?

Correct Answer: D

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