[Jun 05, 2022] Get New 1z0-071 Certification Practice Test Questions Exam Dumps [Q140-Q161]

[Jun 05, 2022] Get New 1z0-071 Certification Practice Test Questions Exam Dumps

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Exam Details

The Oracle 1Z0-071 exam is made up of 78 multiple-choice questions that are to be completed within the allocated timeframe of 120 minutes. The test is proctored by Pearson VUE, the official partner of Oracle. To schedule your session, you need to pay the fee of $245. But before that, you have to create an account with CertView, a web-based portal for all the Oracle certification activity. After you complete the exam, you will receive a notification from Oracle to your email address that your results are available on CertView. Log into the portal and click “See My New Exam Result Now” to find out whether you passed your 1Z0-071 test or not. To ace this exam, you need to give no less than 63% of correct answers.

It is strongly recommended that you undertake the relevant training before attempting the certification test. You can opt for one of the official courses offered by Oracle on its website. The examinees are encouraged to use Oracle training in combo with hands-on experience gained with the help of practical labs.

NO.140 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
Identify three ORDERBYclauses either one of which can complete the query.

 ORDER BY “Last name”

 ORDER BY 2, cust_id

 ORDER BY CUST_NO

 ORDER BY 2, 1

 ORDER BY “CUST_NO”

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.

NO.141 Which three are true about the MERGEstatement?

 It can combine rows from multiple tables conditionally to insert into a single table.

 It can merge rows only from tables.

 It can use subqueries to produce source rows.

 It can update, insert, or delete rows conditionally in multiple tables.

 It can update the same row of the target table multiple times.

 It can use views to produce source rows.

Explanation/Reference: https://www.oracletutorial.com/oracle-basics/oracle-merge/

NO.142 Examine the structure of the INVOICE table.

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

 SELECT inv_no,NVL2(inv_date,’Pending’,’Incomplete’)FROM invoice;

 SELECT inv_no,NVL2(inv_amt,inv_date,’Not Available’)FROM invoice;

 SELECT inv_no,NVL2(inv_date,sysdate-inv_date,sysdate)FROM invoice;

 SELECT inv_no,NVL2(inv_amt,inv_amt*.25,’Not Available’)FROM invoice;

NO.143 Which two tasks require subqueries?

 Display the total number of products supplied by supplier 102 which have a product status of obsolete.

 Display suppliers whose PROD_LIST_PRICE is less than 1000.

 Display the number of products whose PROD_LIST_PRICE is more than the average PROD_LIST_PRICE.

 Display the minimum PROD_LIST_PRICE for each product status.

 Display products whose PROD_MIN_PRICE is more than the average PROD_LIST_PRICE of all products, and whose status is orderable.

NO.144 Examine the structure of the employees table.

There is a parent/child relationship betweenEMPLOYEE_IDandMANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager asthe employee whoseEMPLOYEE_ID123.
Which query provides the correct output? A)

B)

C)

D)

 Option A

 Option B

 Option C

 Option D

NO.145 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

NO.146 View the exhibit and examine the description of the EMPLOYEES table. (Choose two.) You executed this SQL statement:
SELECT first_name, department_id, salary
FROM employees
ORDER BY department_id, first_name, salary desc;
Which two statements are true regarding the result?

 The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID and FIRST_NAME column.

 The values in the FIRST_NAME column would be returned in ascending order for all employees having the same value in the DEPARTMENT_ID column.

 The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.

 The values in the all columns would be returned in descending order.

 The values in the FIRST_NAME column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.

NO.147 View the exhibit and examine the description of the PRODUCT_INFORMATIONtable.

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

 SELECT COUNT (DISTINCT list_price)
FROM product_information
WHERE list_price is NULL

 SELECT COUNT (NVL(list_price, 0))
FROM product_information
WHERE list_price is NULL

 SELECT COUNT (list_price)
FROM product_information
WHERE list_price i= NULL

 SELECT COUNT (list_price)
FROM product_information
WHERE list_price is NULL

NO.148 View the exhibit and examine the structure of the EMPLOYEES table.
You want to display all employees and their managers having 100 as the MANAGER_ID.
You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

 SELECT m.last_name “Manager”, e.last_name “Employee”FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id = 100;

 SELECT m.last_name “Manager”, e.last_name “Employee”FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id = 100;

 SELECT m.last_name “Manager”, e.last_name “Employee”FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id = 100;

 SELECT m.last_name “Manager”, e.last_name “Employee”FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100

NO.149 View the Exhibit and examine the structure of the ORDER_ITEMS table.

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_id

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

 FROM order_items
WHERE(unit_price*quantity) = (SELECT MAX(unit_price*quantity)
FROM order_items)
GROUP BY order_id;
SELECT order_id

 FROM order_items
WHERE(unit_price*quantity) = MAX(unit_price*quantity)
GROUP BY order_id;
SELECT order_id

 FROM order_items
WHERE (unit_price*quantity) = (SELECT MAX(unit_price*quantity)
FROM order_items
GROUP BY order_id)

NO.150 Which two statements are true about Oracle synonyms? (Choose two.)

 A synonym can have a synonym.

 All private synonym names must be unique in the database.

 Any user can create a PUBLICsynonym.

 A synonym can be created on an object in a package.

 A synonym has an object number.

Explanation/Reference: https://docs.oracle.com/cd/B19306_01/server.102/b14200/statements_7001.htm

NO.151 You issue the following command to drop the PRODUCTS table:
SQL>DROP TABLE products;
What is the implication of this command? (Choose all that apply.)

 All data in the table are deleted but the table structure will remain

 All data along with the table structure is deleted

 All viewsand synonyms will remain but they are invalidated

 The pending transaction in the session is committed

 All indexes on the table will remain but they are invalidated

NO.152 Examine the description of the PROMOTIONS TABLE:
You want to display the unique is promotion costs in each promotion category.
Which two queries can be used?

 SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;

 SELECT promo_cost, promo_category FROM promotions ORDER BY 1

 SELECT promo_category, DISTINCT promo_cost FROM promotiong ORDER BY 2:

 select DISTINCT promo_categoryIl ‘has’||promol_cost as COSTS FROM promotions ORDER BY 1:

 SELECT DISTINCT promo_cost ||’in’IIDISTINCT promo_category promotions ORDER BY1;

NO.153 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, whose 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’.

NO.154 Evaluate the following two queries:

Which statement is true regarding the above two queries?

 Performance would improve in query 2 only if there are null values in the CUST_CREDIT_LIMIT column.

 There would be no change in performance.

 Performance would degrade in query 2.

 Performance would improve in query 2.

NO.155 View the exhibit and examine the description of the PRODUCT_INFORMATIONtable.

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

 SELECT COUNT (DISTINCT list_price)
FROM product_information
WHERE list_price is NULL

 SELECT COUNT (NVL(list_price, 0))
FROM product_information
WHERE list_price is NULL

 SELECT COUNT (list_price)
FROM product_information
WHERE list_price i= NULL

 SELECT COUNT (list_price)
FROM product_information
WHERE list_price is NULL

NO.156 Which two are true about dropping columns from a table?

 A column drop is implicitly committed.

 A column that is referenced by another column in any other table cannot be dropped.

 A column can be removed only if it contains no data.

 Multiple columns can be dropped simultaneously using the ALTER TABLEcommand.

 A column must be set as unused before it is dropped from a table.

 A primary key column cannot be dropped.

Explanation/Reference: https://oracle-base.com/articles/8i/dropping-columns

NO.157 View the Exhibit and examine the structure of the CUSTOMERStable.

CUSTOMER_VUis a view based on CUSTOMERS_BR1 table which has the same structure as CUSTOMERStable.
CUSTOMERSneed to be updated to reflect the latest information about the customers.
What is the error in the following MERGEstatement?

 The CUSTOMER_IDcolumn cannot be updated.

 The INTOclause is misplaced in the command.

 The WHEREclause cannot be used with INSERT.

 CUSTOMER_VUcannot be used as a data source.

Explanation/Reference:

NO.158 The first DROP operation is performed on PRODUCTS table using the following command:
DROP TABLE products PURGE;
Then you performed the FLASHBACK operation by using the following command:
FLASHBACK TABLE products TO BEFORE DROP;
Which statement describes the outcome of the FLASHBACK command?

 It recovers only the table structure.

 It recovers the table structure, data, and the indexes.

 It recovers the table structure and data but not the related indexes.

 It is not possible to recover the table structure, data, or the related indexes.

NO.159 Which three statements are true regarding group functions? (Choose three.)

 They can be used on columns or expressions.

 They can be passed as an argument to another group function.

 They can be used only with a SQL statement that has the GROUP BY clause.

 They can be used on only one column in the SELECT clause of a SQL statement.

 They can be used along with the single-row function in the SELECT clause of a SQL statement.

NO.160 You must create a table EMPLOYEES in which the values in the columns EMPLOYEES_ID and LOGIN_ID must be unique and not null.
Which two SQL statements would create the required table? (Choose two.)

 CREATE TABLE employees(employee_id NUMBER,login_id NUMBER,employee_name VARCHAR2(100),hire_date DATE,CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id));

 CREATE TABLE employees(employee_id NUMBER,login_id NUMBER,employee_name VARCHAR2(25),hire_date DATE,CONSTRAINT emp_id_pk PRIMARY KEY (employee_id, login_id));

 CREATE TABLE employees(employee_id NUMBER CONSTRAINT emp_id_pk PRIMARY KEY,login_id NUMBER UNIQUE,employee_name VARCHAR2(25),hire_date DATE);

 CREATE TABLE employees(employee_id NUMBER,login_id NUMBER,employee_name VARCHAR2(100),hire_date DATE,CONSTRAINT emp_id_uk UNIQUE (employee_id, login_id);CONSTRAINT emp_id_nn NOT NULL (employee_id, login_id));

 CREATE TABLE employees(employee_id NUMBER CONSTRAINT emp_id_nn NOT
NULL,login_id NUMBER CONSTRAINT login_id_nn NOT NULL,employee_name
VARCHAR2(100),hire_date DATE,CONSTRAINT emp_num_id_uk UNIQUE (employee_id, login_id));

NO.161 Which two are true about dropping columns from a table? (Choose two.)

 A column drop is implicitly committed.

 A column that is referenced by another column in any other table cannot be dropped.

 A column can be removed only if it contains no data.

 Multiple columns can be dropped simultaneously using the ALTER TABLEcommand.

 A column must be set as unused before it is dropped from a table.

 A primary key column cannot be dropped.

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