Maximum Grades By Making ready With 1z1-071 Dumps UPDATED 2026 [Q190-Q213]

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Maximum Grades By Making ready With 1z1-071 Dumps UPDATED 2026

Prepare 1z1-071 Exam Questions [2026] Recently Updated Questions

Oracle 1z1-071 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Aggregating Data Using Group Functions	10%	– Handling NULLs in group functions – GROUP BY and HAVING clauses – SUM, AVG, COUNT, MAX, MIN functions
Topic 2: Using SET Operators		– UNION, UNION ALL, INTERSECT, MINUS operators – Ordering results with set operations
Topic 3: Working with Time Zones and Date/Time Data		– TIMESTAMP, INTERVAL data types – Time zone conversion functions
Topic 4: Controlling User Access		– System and object privileges – Roles, granting and revoking privileges
Topic 5: Displaying Data from Multiple Tables	15%	– Self-joins and outer joins – Equijoins and non-equijoins – Cartesian products
Topic 6: Managing Tables and Constraints		– Data types, creating/altering/dropping tables – Temporary and external tables – Constraints: primary key, foreign key, unique, check, not null
Topic 7: Relational Database Concepts	10%	– ERD and SQL statement mapping – Relationship between database and SQL – Theoretical and physical aspects of relational databases
Topic 8: Using Data Dictionary Views		– Retrieve metadata and object information
Topic 9: Managing Indexes, Synonyms and Sequences		– Create and use synonyms – Create and maintain indexes – Create and use sequences
Topic 10: Retrieving Data Using SQL SELECT Statement	20%	– Column aliases, concatenation, literals and DISTINCT – SELECT statement syntax and usage – Arithmetic operations and NULL handling
Topic 11: Restricting and Sorting Data	15%	– Operator precedence – ORDER BY clause and sorting rules – WHERE clause and filtering conditions – Substitution variables, DEFINE and VERIFY commands
Topic 12: Manipulating Data and Managing Transactions		– Multi-table inserts – INSERT, UPDATE, DELETE, MERGE statements – Transaction control: COMMIT, ROLLBACK, SAVEPOINT
Topic 13: Conversion Functions and Conditional Expressions	10%	– Implicit and explicit data type conversion – NVL, NULLIF, COALESCE and CASE expressions – TO_CHAR, TO_NUMBER, TO_DATE functions
Topic 14: Managing Views		– Create, modify and drop views – Updatable and read-only views
Topic 15: Using Subqueries to Solve Queries		– Single-row and multi-row subqueries – Correlated subqueries for update/delete
Topic 16: Using Single-Row Functions to Customize Output	20%	– Character, number, and date functions – Nested functions usage

NEW QUESTION 190
Which two statements are true regarding a SAVEPOINT? (Choose two.)

A SAVEPOINTdoes not issue a COMMIT

Only one SAVEPOINTmay be issued in a transaction

Rolling back to a SAVEPOINT can undo a TRUNCATEstatement

Rolling back to a SAVEPOINTcan undo a CREATE INDEX statement

Rolling back to a SAVEPOINTcan undo a DELETEstatement

NEW QUESTION 191
Which three arithmetic operations can be performed on a column by using a SQL function that is built into Oracle database? (Choose three.)

Finding the lowest value

Finding the quotient

Raising to a power

Subtraction

Addition

NEW QUESTION 192
In the PROMOTIONS table, the PROMO_ BEGIN_DATE column is of data type and the default date format is DD-MON-RR Which two statements are true about expressions using PROMO_ BEGIN_DATE in a query?

TONUMBER (PROMO BEGIN_DATE) – 5 will return a number

PROMO_ BEGIN_DATE – 5 will return a date

PROMO_ BEGIN_DATE – SYSDATE will return a number

PROMO_ BEGIN_DATE – SYSDATE will return an error

TODATE(PROMO BEGIN_DATE *5) will return a date

NEW QUESTION 193
Which two are true about virtual columns?

They can be referenced In the where clause of an update or debete statement.

They can be referenced in the set clause of an update statement as the name of the column To be updated.

They can be indexed.

They cannot have a data type explicitly specified.

They can be referenced in the column expression of another virtxial column.

NEW QUESTION 194
Examine the structure of the MEMBERS table: (Choose the best answer.)

Examine the SQL statement:
SQL > SELECT city, last_name LNAME FROM MEMBERS ORDER BY 1, LNAME DESC; What would be the result execution?

It displays all cities in descending order, within which the last names are further sorted in descending order.

It fails because a column alias cannot be used in the ORDER BY clause.

It fails because a column number and a column alias cannot be used together in the ORDER BY clause.

It displays all cities in ascending order, within which the last names are further sorted in descending order.

NEW QUESTION 195
View the exhibit and examine the structure of the PROMOTIONS table.

You have to generate a report that displays the promo name and start date for all promos that started after the last promo in the ‘INTERNET’ category.
Which query would give you the required output?

SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date> ALL (SELECT MAX (promo_begin_date)FROM promotions) ANDpromo_category= ‘INTERNET’;

SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date IN (SELECT promo_begin_dateFROM promotionsWHERE promo_category= ‘INTERNET’);

SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date > ALL (SELECT promo_begin_dateFROM promotionsWHERE promo_category = ‘INTERNET’);

SELECT promo_name, promo_begin_date FROM promotionsWHERE promo_begin_date> ANY (SELECT promo_begin_dateFROM promotionsWHERE promo_category= ‘INTERNET’);

NEW QUESTION 196
Examine the description of the PROMTIONS table:

You want to display the unique promotion costs in each promotion category.
Which two queries can be used?

SELECT promo_cost, | pxomo_category FROM promotions ORDER BY 1;

SELECT promo_category, DISTINCT promo_cost PROM promotions ORDER BY 2:

SELECT DISTINCT promo_category ||’has’|| promo_cost AS COSTS FROM promotions ORDER BY 1;

SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;

SELECT DISTINCT promo_cost ||’ in’ II DISTINCT promo_category FROM promotions ORDER BY 1;

NEW QUESTION 197
What is true about non-equijoin statement performance?

The between condition always performs less well than using the >= and <= conditions.

The Oracle join syntax performs better than the SQL: 1999 compliant ANSI join syntax.

The join syntax used makes no difference to performance.

The between condition always performs better than using the >= and <= conditions.

Table aliases can improve performance.

NEW QUESTION 198
View the Exhibit and examine the details of PRODUCT_INFORMATIONtable.
PRODUCT_NAME
CATEGORY_ID SUPPLIER_ID
Inkjet C/8/HQ 12 102094
Inkjet C/4 12 102090
LaserPro 600/6/BW 12 102087
LaserPro 1200/8/BW 12 102099
Inkjet B/6 12 102096
Industrial 700/ID 12 102086
Industrial 600/DQ 12 102088
Compact 400/LQ 12 102087
Compact 400/DQ 12 102088
HD 12GB /R 13 102090
HD 10GB /I 13 102071
HD 12GB @7200 /SE 13 102057
HD 18.2GB @10000 /E 13 102078
HD 18.2GB @10000 /I 13 102050
HD 18GB /SE 13 102083
HD 6GB /I 13 102072
HD 8.2GB@5400 13 102093
You have the requirement to display PRODUCT_NAMEfrom the table where the CATEGORY_IDcolumn has values 12or 13, and the SUPPLIER_IDcolumn has the value 102088. You executed the following SQL statement:
SELECT product_name
FROM product_information
WHERE (category_id = 12 AND category_id = 13) AND supplier_id = 102088; Which statement is true regarding the execution of the query?

It would not execute because the same column has been used in both sides of the ANDlogical operator to form the condition.

It would not execute because the entire WHEREclause condition is not enclosed within the parentheses.

It would execute and the output would display the desired result.

It would execute but the output would return no rows.

NEW QUESTION 199
Which three statements are correct regarding indexes? (Choose three.)

A non-deferrable PRIMARY KEYor UNIQUE KEYconstraint in a table automatically attempts to create a unique index.

Indexes should be created on columns that are frequently referenced as part of any expression.

When a table is dropped, the corresponding indexes are automatically dropped.

For each DML operation performed, the corresponding indexes are automatically updated.

NEW QUESTION 200
Which two statements are true regarding working with dates? (Choose two.)

The RR date format automatically calculates the century from the SYSDATE function but allows the session user to enter the century.

The RR date format automatically calculates the century from the SYSDATE function and does not allow a session user to enter the century.

The default internal storage of dates is in character format.

The default internal storage of dates is in numeric format.

NEW QUESTION 201
You execute these commands:
SQL> DEFINE hiredate = ’01-APR -2011′;
SQL> SELECT employee_id, first_name, salary FROM employees WHERE hire date > &hiredate AND manager_id >&mgr_id; For which substitution variables will you be prompted?

none

&hiredate and &mgr_id

only &hiredate

only &mgr_id

NEW QUESTION 202
View the exhibit and examine the structure in ORDERS and ORDER_ITEMS tables.

You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATE VIEW statement would create the view successfully?

CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;

CREATE OR REPLACE VIEW ord_vu (order_id, order_date)AS SELECT o.order_id, o.order_date, COUNT(i.line_item_id)”NO OF ITEMS”FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;

CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)”NO OF ITEMS”FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;

CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)||’NO OF ITEMS’FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_dateWHITH CHECK OPTION;

NEW QUESTION 203
Examine the structure of the MARKStable:

Which two statements would execute successfully? (Choose two.)
SELECT SUM(DISTINCT NVL(subject1,0)), MAX(subject1)

FROM marks
WHERE subject1 > subject2;
SELECT student_name subject1

FROM marks
WHERE subject1 > AVG(subject1);
SELECT SUM(subject1+subject2+subject3)

FROM marks
WHERE student_name IS NULL;
SELECT student_name,SUM(subject1)

FROM marks
WHERE student_name LIKE ‘R%’;

NEW QUESTION 204
Which three are true about system and object privileges

WITH GRANT OPTION can be used when granting an object privilege to both users and roles

WITH GRANT OPTION cannot be used when granting an object privilege to PUBLIC

Revoking a system privilege that was granted with the WITH ADMIN OPTION has a cascading effect.

Revoking an object privilege that was granted with the WITH GRANT OPTION clause has a cascading effect

Adding a primary key constraint to an existing table in another schema requires a system privilege

Adding a foreign key constraint pointing to a table in another schema requires the REFERENCEs object privilege

NEW QUESTION 205
Which two queries return the string Hello! we’re ready?

SELECT q’! Hello! We’re ready! ‘FROM DUAL;

SELECT “Hello! We’re ready “FROM |DUAL;

SELECT q'[Hello! We’re ready]’FROM DUAL;

SELECT ‘Hello! we re ready’ ESCAPE’N’FROMDUAL:

SELECT ‘Hello! We’re ready’ FROM DUAL;

NEW QUESTION 206
View the exhibit and examine the description of the PRODUCT_INFORMATION table.

Which SQL statement would retrieve from the table the number of products having LIST_PRICE as NULL?

SELECT COUNT (DISTINCT list_price)FROM product_informationWHERE list_price is NULL

SELECT COUNT (NVL(list_price, 0))FROM product_informationWHERE list_price is NULL

SELECT COUNT (list_price)FROM product_informationWHERE list_price i= NULL

SELECT COUNT (list_price)FROM product_informationWHERE list_price is NULL

NEW QUESTION 207
View the Exhibit and examine the data in the PRODUCT_INFORMATIONtable.

Which two tasks would require subqueries? (Choose two.)

displaying all the products whose minimum list prices are more than average list price of products having the status orderable

displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE

displaying the number of products whose list prices are more than the average list price

displaying all supplier IDs whose average list price is more than 500

displaying the minimum list price for each product status

NEW QUESTION 208
Examine this SELECT statement and view the Exhibit to see its output:

SELECT constraints_name, constraints_type, search_condition, r_constraints_name, delete_rule, status, FROM user_constraints WHERE table_name = ‘ORDERS’; Which two statements are true about the output? (Choose two.)

The DELETE_RULE column indicates the desired state of related rows in the child table when the corresponding row is deleted from the parent table.

The R_CONSTRAINT_NAME column contains an alternative name for the constraint.

In the second column, ‘c’ indicates a check constraint.

The STATUS column indicates whether the table is currently in use.

NEW QUESTION 209
View the exhibit and examine the structure of the CUSTOMERStable.

Which two tasks would require subqueries or joins to be executed in a single statement?

finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers

finding the average credit limit of male customers residing in ‘Tokyo’or ‘Sydney’

listing of customers who do not have a credit limit and were born before 1980

finding the number of customers, in each city, who’s marital status is ‘married’.

listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city ‘Tokyo’.

NEW QUESTION 210
View the Exhibit and examine the structure in the EMPLOYEES tables.

Evaluate the following SQL statement:
SELECT employee_id, department_id
FROM employees
WHERE department_id= 50 ORDER BY department_id
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=90
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=10;
What would be the outcome of the above SQL statement?

The statement would not execute because the positional notation instead of the column name should be used with the ORDER BY clause.

The statement would execute successfully and display all the rows in the ascending order of DEPARTMENT_ID.

The statement would execute successfully but it will ignore the ORDER BY clause and display the rows in random order.

The statement would not execute because the ORDER BY clause should appear only at the end of the SQL statement, that is, in the last SELECT statement.

NEW QUESTION 211
Examine the structure of the MEMBERStable:
Name Null? Type
—————— ————— ——————————
MEMBER_ID NOT NULL VARCHAR2 (6)
FIRST_NAME VARCHAR2 (50)
LAST_NAME NOT NULL VARCHAR2 (50)
ADDRESS VARCHAR2 (50)
You execute the SQL statement:
SQL > SELECT member_id, ‘ ‘ , first_name, ‘ ‘ , last_name “ID FIRSTNAME LASTNAME
” FROM members;
What is the outcome?

It fails because the alias name specified after the column names is invalid.

It fails because the space specified in single quotation marks after the first two column names is invalid.

It executes successfully and displays the column details in a single column with only the alias column heading.

It executes successfully and displays the column details in three separate columns and replaces only the last column heading with the alias.

NEW QUESTION 212
View the exhibits and examine the structures of the COSTSand PROMOTIONStables.

Evaluate the following SQL statement:

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 213
Examine the create table statements for the stores and sales tables.
SQL> CREATE TABLE stores(store_id NUMBER(4) CONSTRAINT store_id_pk PRIMARY KEY, store_name VARCHAR2(12), store_address VARCHAR2(20), start_date DATE); SQL> CREATE TABLE sales(sales_id NUMBER(4) CONSTRAINT sales_id_pk PRIMARY KEY, item_id NUMBER(4), quantity NUMBER(10), sales_date DATE, store_id NUMBER(4), CONSTRAINT store_id_fk FOREIGN KEY(store_id) REFERENCES stores(store_id)); You executed the following statement:
SQL> DELETE from stores
WHERE store_id=900;
The statement fails due to the integrity constraint error:
ORA-02292: integrity constraint (HR.STORE_ID_FK) violated
Which three options ensure that the statement will execute successfully?

Disable the primary key in the STORES table.

Use CASCADE keyword with DELETE statement.

DELETE the rows with STORE_ID = 900 from the SALES table and then delete rows from STORES table.

Disable the FOREIGN KEY in SALES table and then delete the rows.

Create the foreign key in the SALES table on SALES_ID column with on DELETE CASCADE option.

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Maximum Grades By Making ready With 1z1-071 Dumps UPDATED 2026 [Q190-Q213]

Oracle 1z1-071 Exam Syllabus Topics:

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