1z0-071 – Premium VCE Dumps

5 Oct, 2026 By admin

Updated Oct 05, 2026 1z0-071 Exam Dumps – PDF Questions and Testing Engine [Q74-Q93]

Updated Oct 05, 2026 1z0-071 Exam Dumps – PDF Questions and Testing Engine

New (2026) Oracle 1z0-071 Exam Dumps

Oracle 1z0-071 Exam Syllabus Topics:

Section	Objectives
Topic 1: Relational Database Concepts	– Tables, rows, columns, and relational model basics – Data types and database objects overview
Topic 2: SQL Functions	– Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions – Single-row functions 1. Character, numeric, date functions
Topic 3: Data Manipulation and Schema Objects	– DDL operations 1. CREATE, ALTER, DROP tables and constraints – DML operations 1. INSERT, UPDATE, DELETE, MERGE
Topic 4: SQL Retrieval	– Joins and set operations 1. UNION, INTERSECT, MINUS 2. INNER, OUTER, CROSS joins – SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting
Topic 5: Subqueries and Advanced SQL	– Correlated subqueries and nested queries – Single-row and multi-row subqueries

NEW QUESTION 74
Which three statements are true about Oracle synonyms?

A synonym cannot be created for a PL /SQL package.

A SEQUENCE can have a synonym.

A synonym can be available to all users .

A synonym created by one user can refer to an object belonging to another user.

Any user can drop a PUBLIC synonym.

NEW QUESTION 75
View the Exhibit and examine the structure of the PRODUCT_INFORMATION table.

You want to see the product names and the date of expiration of warranty for all the products, if the product is purchased today. The products that have no warranty should be displayed at the top and the products with maximum warranty period should be displayed at the bottom.
Which SQL statement would you execute to fulfill this requirement?

SELECT product_name, SYSDATE+warranty_period AS “Warranty expire date”FROM product_informationORDER BY SYSDATE-warranty_period

SELECT product_name, SYSDATE+warranty_period AS “Warranty expire date”FROM product_informationORDER BY SYSDATE+warranty_period

SELECT product_name, SYSDATE+warranty_period AS “Warranty expire date”FROM product_informationORDER BY SYSDATE

SELECT product_name, SYSDATE+warranty_period AS “Warranty expire date”FROM product_informationWHERE warranty_period > SYSDATE

NEW QUESTION 76
Examine the structure of the MEMBERS table:

You want to display details of all members who reside in states starting with the letter A followed by exactly one character.
Which SQL statement must you execute?

SELECT * FROM MEMBERS WHERE state LIKE ‘%A_’;

SELECT * FROM MEMBERS WHERE state LIKE ‘A_’;

SELECT * FROM MEMBERS WHERE state LIKE ‘A_%’;

SELECT * FROM MEMBERS WHERE state LIKE ‘A%’;

NEW QUESTION 77
View the exhibits and examine the structures of the COSTS and PROMOTIONS tables.

Evaluate the following SQL statement:
SQL> SELECT prod_id FROM costs
WHERE promo_id IN (SELECT promo_id FROM promotions
WHERE promo_cost < ALL
(SELECT MAX(promo_cost) FROM promotions
GROUP BY (promo_end_date-
promo_begin_date)));
What would be the outcome of the above SQL statement?

It displays prod IDs in the promo with the lowest cost.

It displays prod IDs in the promos with the lowest cost in the same time interval.

It displays prod IDs in the promos with the highest cost in the same time interval.

It displays prod IDs in the promos which cost less than the highest cost in the same time interval.

NEW QUESTION 78
Examine the structure of the EMPLOYEES table.
You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.
Which query would retrieve the required result?

SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100)INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;

SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100)UNION ALL(SELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000);

SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = ‘100’)UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000;

(SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100))UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;

NEW QUESTION 79
The EMPLOYEES table contains columns EMP_ID of data type NUMBER and HIRE_DATE of data type DATE You want to display the date of the first Monday after the completion of six months since hiring.
The NLS_TERRITORY parameter is set to AMERICA in the session and, therefore, Sunday is the first day of the week Which query can be used?

SELECT emp_id,NEXT_DAY(ADD_MONTHS(hite_date,6),’MONDAY’) FROM employees;

SELECT emp_id,ADD_MONTHS(hire_date,6), NEXT_DAY(‘MONDAY’) FROM employees;

SELECT emp_id,NEXT_DAY(MONTHS_BETWEEN(hire_date,SYSDATE),6) FROM employees;

SELECT emp_id,NEXT_DAY(ADD_MONTHS(hire_date,6),1) FROM employees;

NEW QUESTION 80
You need to display the first names of all customers from the CUSTOMERS table that contain the character
‘e’ and have the character ‘a’ in the second last position.
Which query would give the required output?

SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, ‘e’)<>0 ANDSUBSTR(cust_first_name, -2, 1)=’a’;

SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, ‘e’)<>” ANDSUBSTR(cust_first_name, -2, 1)=’a’;

SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, ‘e’)IS NOT NULL ANDSUBSTR(cust_first_name, 1, -2)=’a’;

SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, ‘e’)<>0 ANDSUBSTR(cust_first_name, LENGTH(cust_first_name), -2)=’a’;

NEW QUESTION 81
Which two are true about transactions in the Oracle Database?

DDL statements automatically commit only data dictionary updates caused by executing the DDL.

A DDL statement issued by a session with an uncommitted transation automaticall commits that transaction.

An uncommitted transaction is automatically committed when the user exits SQL*PLUS

DML statements always start new transactions.

A session can see uncommitted updates made by the same user in a different session

NEW QUESTION 82
Which two statements are true regarding single row functions?

MOD: returns the quotient of a division

TRUNC: can be used with number and date values

CONCAT: can be used to combine any number of values

SYSDATE: returns the database server current date and time

INSTR: can be used to find only the first occurrence of a character in a string

TRIM: can be used to remove all the occurrences of a character from a string

NEW QUESTION 83
Which two are true about the precedence of opertors and condtions

+ (addition) has a higher order of precedence than * (mliplpition)

NOT has a higher order of precedence than AND and OR in a condition.

AND and OR have the same order of precedence in a condition

Operators are evaluated before conditions.

|| has a higher order of precedence than +(addition)

NEW QUESTION 84
Which three statements are true about time zones, date data types, and timestamp data types in an Oracle database?

The DBTIMEZONE function can return an offset from Universal Coordinated Time (UTC)

A TIMESTAMP WITH LOCAL TIMEZONE data type column is stored in the database using the time zone of the session that inserted the row

A TIMESTAMP data type column contains information about year, month, and day

The SESSIONTIMEZONE function can return an offset from Universal Coordinated Time (UTC)

The CURRENT_TIMESTAMP function returns data without time zone information

NEW QUESTION 85
Evaluate the following SQL commands:

The command to create a table fails. Identify the reason for the SQL statement failure.

You cannot use ORD_NOand ITEM_NOcolumns as a composite primary key because ORD_NOis also the FOREIGN KEY.

You cannot use the BETWEENclause in the condition of a CHECKconstraint.

You cannot use the NEXTVALsequence value as a DEFAULTvalue for a column.

You cannot use SYSDATEin the condition of a CHECKconstraint.

NEW QUESTION 86
Examine the description of the EMPLOYEES table:

NLS_DATE FORMAT is DD-MON-RR.
Which two queries will execute successfully?

SELECT dept_ id, AVG (MAX(salary)) FROM employees GROUP By dept_id HAVING hire_date> ‘ O1-JAN-19’;

SELECT dept_ id, AVG(MAX(salary)) FROM employees GROUP BY dept_id, salary;

SELECT dept id, MAX (SUM(salary)) FROM employees GROUP BY dept_id;

SELECT dept_ iD, sum(salary) FROM employees WHERE hire_date > ’01-JAN-9′ GROUP BY dept_id;

SELECT AVG(MAX(salary)) FROM employees GROUP BY salary;

NEW QUESTION 87
Examine the data in the COLORStable:

Examine the data in the BRICKStable:

Which two queries return all the rows from COLORS? (Choose two.)

NEW QUESTION 88
View the exhibit and examine the structure and data in the invoice table. (Choose two.)

Which two SQL statements would execute successfully?

SELECT MAX(AVG(SYSDATE -inv_date)) FROM invoice

SELECT AVG(inv_date) FROM invoice

SELECT MAX(inv_date), MIN(cust_id) FROM invoice

SELECT AVG( inv_date -SYSDATE), AVG(inv_amt) FROM invoice

NEW QUESTION 89
Examine the structure of the ORDERS table: (Choose the best answer.)

You want to find the total value of all the orders for each year and issue this command:
SQL> SELECT TO_CHAR(order_date,’rr’), SUM(order_total) FROM orders
GROUP BY TO_CHAR(order_date, ‘yyyy’);
Which statement is true regarding the result?

It executes successfully but does not give the correct output.

It executes successfully and gives the correct output.

It returns an error because the TO_CHAR function is not valid.

It return an error because the datatype conversion in the SELECT list does not match the data type conversion in the GROUP BY clause.

NEW QUESTION 90
View the Exhibits and examine PRODUCTSand SALES tables.
Exhibit 1

Exhibit 2

You issue the following query to display product name the number of times the product has been sold:

What happens when the above statement is executed?

The statement executes successfully and produces the required output.

The statement produces an error because a subquery in the FROMclause and outer-joins cannot be used together.

The statement produces an error because the GROUP BYclause cannot be used in a subquery in the FROM clause.

The statement produces an error because ITEM_CNTcannot be displayed in the outer query.

NEW QUESTION 91
Examine the description of the EMPLOYEES table:

Which statement will fail?

SELECT department_id, COUNT (*)
FROM employees
HAVING department_ id <> 90 AND COUNT(*) >= 3
GROUP BY department_id;

SELECT department_id, COUNT (*)
FROM employees
WHERE department_ id <> 90 AND COUNT(*) >= 3
GROUP BY department_id;

SELECT department_id, COUNT(*)
FROM employees
WHERE department_id <> 90 HAVING COUNT(*) >= 3
GROUP BY department_id;

SELECT department_id, COUNT(*)
FROM employees
WHERE department_id <> 90 GROUP BY department_id
HAVING COUNT(*) >= 3;

NEW QUESTION 92
Evaluate the following SQL statement:
SQL> SELECT cust_id, cust_last_name “Last Name” FROM customers WHERE country_id = 10 UNION
SELECT cust_id CUST_NO, cust_last_name
FROM customers
WHERE country_id = 30;
Which ORDER BY clause are valid for the above query? (Choose all that apply.)

ORDER BY 2,1

ORDER BY CUST_NO

ORDER BY 2,cust_id

ORDER BY “CUST_NO”

ORDER BY “Last Name”

NEW QUESTION 93
View the Exhibit and examine the structure of the ORDER_ITEMS table. (Choose the best answer.)

You must select the ORDER_ID of the order that has the highest total value among all the orders in the ORDER_ITEMS table.
Which query would produce the desired result?

SELECT order_idFROM order_itemsGROUP BY order_idHAVING
SUM(unit_price*quantity) = (SELECT MAX (SUM(unit_price*quantity))FROM order_items GROUP BY order_id);

SELECT order_idFROM order_itemsWHERE(unit_price*quantity) = (SELECT MAX (SUM(unit_price*quantity)FROM order_items) GROUP BY order_id);

SELECT order_idFROM order_itemsWHERE(unit_price*quantity)
MAX(unit_price*quantity)GROUP BY order_id);

SELECT order_idFROM order_itemsWHERE (unit_price*quantity) = (SELECT MAX(unit_price*quantity)FROM order_itemsGROUP BY order_id)

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Updated Oct 05, 2026 1z0-071 Exam Dumps – PDF Questions and Testing Engine [Q74-Q93]

Oracle 1z0-071 Exam Syllabus Topics: