1z1-071 Test Papers - Valid 1z1-071 Test Questions

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Oracle 1z1-071 Exam Syllabus Topics:

Section	Weight	Objectives
Relational Database Concepts	10%	- ERD and SQL statement mapping - Relationship between database and SQL - Theoretical and physical aspects of relational databases
Working with Time Zones and Date/Time Data		- Time zone conversion functions - TIMESTAMP, INTERVAL data types
Retrieving Data Using SQL SELECT Statement	20%	- Arithmetic operations and NULL handling - Column aliases, concatenation, literals and DISTINCT - SELECT statement syntax and usage
Managing Indexes, Synonyms and Sequences		- Create and maintain indexes - Create and use sequences - Create and use synonyms
Aggregating Data Using Group Functions	10%	- SUM, AVG, COUNT, MAX, MIN functions - GROUP BY and HAVING clauses - Handling NULLs in group functions
Manipulating Data and Managing Transactions		- Transaction control: COMMIT, ROLLBACK, SAVEPOINT - Multi-table inserts - INSERT, UPDATE, DELETE, MERGE statements
Using Data Dictionary Views		- Retrieve metadata and object information
Controlling User Access		- System and object privileges - Roles, granting and revoking privileges
Using Subqueries to Solve Queries		- Single-row and multi-row subqueries - Correlated subqueries for update/delete
Using SET Operators		- UNION, UNION ALL, INTERSECT, MINUS operators - Ordering results with set operations
Conversion Functions and Conditional Expressions	10%	- TO_CHAR, TO_NUMBER, TO_DATE functions - NVL, NULLIF, COALESCE and CASE expressions - Implicit and explicit data type conversion
Using Single-Row Functions to Customize Output	20%	- Nested functions usage - Character, number, and date functions
Restricting and Sorting Data	15%	- Substitution variables, DEFINE and VERIFY commands - Operator precedence - WHERE clause and filtering conditions - ORDER BY clause and sorting rules
Managing Views		- Create, modify and drop views - Updatable and read-only views
Managing Tables and Constraints		- Constraints: primary key, foreign key, unique, check, not null - Temporary and external tables - Data types, creating/altering/dropping tables
Displaying Data from Multiple Tables	15%	- Cartesian products - Equijoins and non-equijoins - Self-joins and outer joins

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Oracle Database SQL Sample Questions (Q270-Q275):

NEW QUESTION # 270
Which two statements are true about * TABLES views?

A. You must have SELECT privileges on a table to view it in ALL TABLES.
B. You must have SELECT privileges on a table to view it in DBA TABLES.
C. USER TABLES displays all tables owned by the current user.
D. All users can query DBA TABLES successfully.
E. All TABLES displays all tables owned by the current user.
F. You must have SELECT privileges on a table to view it in USER TABLES.

Answer: A,C

NEW QUESTION # 271
Examine the description of the BOOKS_TRANSACTIONS table:

Examine this partial SQL statement:
SELECT * FROM books_transactions
Which two WHERE conditions give the same result?

A. WHERE borrowed_date = SYSDATE AND (transaction_type = 'RM' AND member_id = 'A101' OR member_id = 'A102');
B. WHERE borrowed_date = SYSDATE AND (transaction_type = 'RM' OR member_id IN ('A101',
'A102'));
C. WHERE borrowed_date = SYSDATE AND (transaction_type = 'RM' AND (member_id = A101' OR member_id = 'A102'));
D. WHERE (borrowed_date = SYSDATE AND transaction_type = 'RM') OR member_id IN ('A101',
'A102');
E. WHERE borrowed_date = SYSDATE AND transaction_type = 'RM' OR member_id IN ('A101',
'A102');

Answer: C,D

NEW QUESTION # 272
View the Exhibit and examine the description of the tables.

You execute this SQL statement:

Which three statements are true?

A. A product can have a different unit price at different times.
B. The statement will fail because a subquery may not be contained in a VALUES clause.
C. The statement will fail if a row already exists in the SALES table for product 23.
D. The statement will execute successfully and a new row will be inserted into the SALES table.
E. The SALES table has five foreign keys.
F. A customer can exist in many countries.

Answer: D,E,F

NEW QUESTION # 273
Examine the description of the ENPLYEES table:

Which two queries return all rows for employees whose salary is greater than the average salary in their department?

A. SELECT "
FROM employees
WHERE salary > ANY
SELECT AVG (salary)
EROM employees
GROUP BY department_ id);
B. SELECT"
FROM employees e1
WHERE salary >!
SELECT AVG (salary)
FROM employees e2
WHERE e1. Department _id = e2, department_ id
C. SELECT
FROM employees
WHERE salary > AVG (salary) OVER (PARTITION BY department _ id);
D. SELECT"
FROM employees
WHERE salary >
( SELECT AVG
(salary) FROM
employees
E. SELECT.
FROM
SELECT e.", AVG (salary) OVER (PARTITION BY department id) avg_ sal
FROM employees e
WHERE salary > avg_ sal;

Answer: B,C

Explanation:
GROUP BY department _ id
Explanation:
To return all rows for employees whose salary is greater than the average salary in their department, you would use either a subquery or an analytic function:
Option B:
SELECT ... FROM employees WHERE salary > AVG(salary) OVER (PARTITION BY department_id); This uses the window function AVG with PARTITION BY to calculate the average salary per department, and it compares each employee's salary to this average.
Option C:
SELECT ... FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e1.department_id = e2.department_id); This correlated subquery compares each employee's salary to the average salary in their department using a subquery to calculate the average salary for that department.
Options A, D, and E are incorrect because:
Option A: The use of ANY with the subquery does not ensure comparison with the average salary of their respective department.
Option D: This is syntactically incorrect; the subquery alias avg_sal is not accessible outside the subquery.
Option E: The subquery does not correlate with the outer query to ensure that each employee's salary is compared to the average salary of their respective department.

NEW QUESTION # 274
You must find the number of employees whose salary is lower than employee 110.
Which statement fails to do this?

A. SELECT COUNT (*)
FROM employees
JOIN employees a
ON e. salary< a. salary
WHERE a. employee_id= 110;
B. SELECT COUNT (*)
FROM employees e
JOIN (SELECT salary FROM employees WHERE employee_id= 110) a
ON e. salary< a. salary;
C. SELECT COUNT (* )
FROM employees e
WHERE e. salary < (SELECT a. salary FROM employees a WHERE e. employee_id = 110);
D. SELECT COUNT (* )
FROM employees
WHERE salary < (SELECT salary FROM employees WHERE employee_id =
110) ;

Answer: C

Explanation:
In this context, the correct answers will provide a count of employees with a salary less than that of employee with ID 110. Let's evaluate the provided statements:
A: This statement is incorrect due to a syntax error; it uses alias 'e' without defining it, and 'a' is also used incorrectly in the JOIN clause. It should have been written with proper aliases.
B: This statement is correct as it uses a subquery to find the salary of employee ID 110 and compares it to the salaries of all employees to get the count.
C: This statement is also correct as it creates a derived table with the salary of employee ID 110 and compares it with the salaries of the 'employees' table using a JOIN.
D: This is the statement that fails. It is incorrect because the subquery references 'e.employee_id = 110', which will compare the salary of each employee to themselves if their ID is 110, not to the salary of employee 110 for all employees.

NEW QUESTION # 275
......

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