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Oracle Database SQL Sample Questions (Q190-Q195):

NEW QUESTION # 190
Examine this query:
SELECT INTERVAL ‘100’ MONTH DURATION FROM DUAL;
What will be the output?

  • A. DURATION
    +08
  • B. DUFATION
    +100
  • C. an error
  • D. DURATION
    +08-04

Answer: D

NEW QUESTION # 191
View the Exhibit and examine the structure of ORDERS and CUSTOMERS tables.

There is only one customer with the cust_last_name column having value Roberts. Which INSERT statement should be used to add a row into the ORDERS table for the customer whose CUST_LAST_NAME is Roberts and CREDIT_LIMIT is 600?

  • A. INSERT INTO orders (order_id, order_data, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name=’Roberts’ ANDcredit_limit=600), order_total)VALUES (1,
    ’10-mar-2007′, ‘direct’, &customer_id, 1000).
  • B. INSERT INTO(SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_totalFROM orders o, customers cWHERE o.customer_id = c.customer_idAND c.cust_last_name=’Roberts’ AND c.credit_limit=600 )VALUES (1, ’10-mar-2007′, ‘direct’, (SELECT customer_idFROM customersWHERE cust_last_name=’Roberts’ ANDcredit_limit=600), 1000);
  • C. INSERT INTO orders (order_id, order_data, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name=’Roberts’ ANDcredit_limit=600), order_total)VALUES(1,
    ’10-mar-2007′, ‘direct’, &&customer_id, 1000);
  • D. INSERT INTO ordersVALUES(1, ’10-mar-2007′, ‘direct’,(SELECT customer_idFROM customersWHERE cust_last_name=’Roberts’ ANDcredit_limit=600), 1000);

Answer: D

NEW QUESTION # 192
Examine the business rule:
Each student can work on multiple projects and each project can have multiple students.
You need to design an Entity Relationship Model (ERD) for optimal data storage and allow for generating reports in this format:
STUDENT_ID FIRST_NAME LAST_NAME PROJECT_ID PROJECT_NAME PROJECT_TASK
Which two statements are true in this scenario?

  • A. The ERD must have a M:M relationship between the STUDENTSand PROJECTSentities that must be resolved into 1:M relationships.
  • B. STUDENT_IDmust be the primary key in the STUDENTSentity and foreign key in the PROJECTSentity.
  • C. The ERD must have a 1:M relationship between the STUDENTSand PROJECTSentities.
  • D. PROJECT_IDmust be the primary key in the PROJECTSentity and foreign key in the STUDENTSentity.
  • E. An associative table must be created with a composite key of STUDENT_ID and PROJECT_ID, which is the foreign key linked to the STUDENTS and PROJECTS entities.

Answer: A,E

Explanation:
Explanation/Reference:
References:
http://www.oracle.com/technetwork/issue-archive/2011/11-nov/o61sql-512018.html

NEW QUESTION # 193
View the exhibit and examine the structure and data in the INVOICE table.

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

  • A. SELECT MAX(AVG(SYSDATE -inv_date))
    FROM invoice
  • B. SELECT AVG( inv_date -SYSDATE), AVG(inv_amt)
    FROM invoice
  • C. SELECT MAX(inv_date), MIN(cust_id)
    FROM invoice
  • D. SELECT AVG(inv_date)
    FROM invoice

Answer: B,C

NEW QUESTION # 194
View the Exhibit and examine the structure of the CUSTOMERS table.

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

  • A. listing of customers who do not have a credit limit and were born before 1980
  • B. finding the number of customers, in each city, whose marital status is ‘married’
  • C. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
  • D. listing of those customers whose credit limit is the same as the credit limit of customers residing in the city ‘Tokyo’
  • E. finding the average credit limit of male customers residing in ‘Tokyo’ or ‘Sydney’

Answer: C,D

Explanation:
Describe the Types of Problems That the Subqueries Can Solve
There are many situations where you will need the result of one query as the input for
another.
Use of a Subquery Result Set for Comparison Purposes
Which employees have a salary that is less than the average salary? This could be
answered by two statements, or by a single statement with a subquery. The following
example uses two statements:
select avg(salary) from employees;
select last_name from employees where salary < result_of_previous_query ;
Alternatively, this example uses one statement with a subquery:
select last_name from employees where salary < (select avg(salary)from employees);
In this example, the subquery is used to substitute a value into the WHERE clause of the
parent query: it is returning a single value, used for comparison with the rows retrieved by
the parent query.
The subquery could return a set of rows. For example, you could use the following to find
all departments that do actually have one or more employees assigned to them:
select department_name from departments where department_id in
(select distinct(department_id) from employees);

NEW QUESTION # 195
……

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