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Evaluate the following query:
SQL> SELECT TRUNC (ROUND (156.00, -1),-1)
What would be the outcome?
Correct Answer: C
Which three statements are true regarding the data types?
A. The minimum column width that can be specified for a VARCHAR2 data type column is one.
B. Only one LONG column can be used per table.
C. A TIMESTAMP data type column stores only time values with fractional seconds.
D. The BLOB data type column is used to store binary data in an operating system file.
E. The value for a CHAR data type column is blank-padded to the maximum defined column width.
Correct Answer: ABE
Which three arithmetic operations can be performed on a column by using a SQL function that is built into Oracle database? (Choose three.)
A. Finding the lowest value
B. Finding the quotient
C. Raising to a power
Correct Answer: ACE
Which two statements are true regarding the COUNT function?
A. A SELECT statement using the COUNT function with a DISTINCT keyword cannot have a WHERE clause.
B. COUNT (DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULL values in the INV_AMT column.
C. COUNT (cust_id) returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_ID column.
D. COUNT (*) returns the number of rows including duplicate rows and rows containing NULL value in any of the columns.
E. The COUNT function can be used only for CHAR, VARCHAR2, and NUMBER data types.
Correct Answer: BD
Examine the structure of the PROGRAMS table:
Which two SQL statements would execute successfully?
A. SELECT NVL (ADD_MONTHS (END_DATE,1) SYSDATE) FROM programs;
B. SELECT TO_DATE (NVL (SYSDATE-END_DATE, SYSDATE)) FROM programs;
C. SELECT NVL (MONTHS_BETWEEN (start_date, end_date), `Ongoing\’) FROM programs;
D. SELECT NVL (TO_CHAR (MONTHS_BETWEEN (start-date, end_date)), `Ongoing\’) FROM programs
Correct Answer: AD
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Evaluate the following CRTEATE TABLE commands:
(ord_no NUMBER (2) CONSTRAINT ord_pk PRIMARY KEY,
cust_id NUMBER (4) );
CREATE TABLE ord_items
(ord _no NUMBER (2),
item_no NUMBER(3), qty NUMBER (3) CHECK (qty BETWEEEN 100 AND 200),
expiry_date date CHECK (expiry_date> SYSDATE),
CONSTRAINT it_pk PRIMARY KEY (ord_no, item_no),
CONSTARAINT ord_fk FOREIGN KEY (ord_no) REFERENCES orders (ord_no) );
Why would the ORD_ITEMS table not get created?
A. SYSDATE cannot be used with the CHECK constraint.
B. The BETWEEN clause cannot be used for the CHECK constraint.
C. The CHECK constraint cannot be placed on columns having the DATE data type.
D. ORD_NO and ITEM_NO cannot be used as a composite primary key because ORD_NO is also the FOREIGN KEY.
Correct Answer: A
Examine the commands used to create DEPARTMENT_DETAILS and COURSE_DETAILS:
You want to generate a report that shows all course IDs irrespective of whether they have corresponding department IDs or not but no department IDs if they do not have any courses. Which SQL statement must you use?
A. SELECT course_id, department_id, FROM department_details d RIGHT OUTER JOIN course_details c USING (department_id)
B. SELECT c.course_id, d.department_id FROM course_details c RIGHT OUTER JOIN .department_details d ON (c.depatrment_id=d.department_id)
C. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_id=d. department_id)
D. SELECT c.course_id, d.department_id FROM course_details c FULL OUTER JOIN department_details d ON (c.department_idd. department_id)
Correct Answer: C
Using the CUSTOMERS table, you need to generate a report that shows 50% of each credit amount in each income level. The report should NOT show any repeated credit amounts in each income level. Which query would give the required result?
A. SELECT cust_income_level || ` \’ || cust_credit_limit * 0.50 AS “50% Credit Limit” FROM customers.
B. SELECT DISTINCT cust_income_level || ` \’ || cust_credit_limit * 0.50 AS “50% Credit Limit” FROM customers.
C. SELECT DISTINCT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS “50% Credit Limit” FROM customers.
D. SELECT cust_income_level, DISTINCT cust_credit_limit * 0.50 AS “50% Credit Limit” FROM customers
Correct Answer: B
View the Exhibit and examine PRODUCTS and ORDER_ITEMS tables.
You executed the following query to display PRODUCT_NAME and the number of times the product has been ordered: SQL>SELECT p.product_name, i.item_cnt FROM (SELECT product_id, COUNT (*) item_cnt FROM order_items GROUP BY product_id) i RIGHT OUTER JOIN products p ON i.product_id = p.product_id; What would happen when the above statement is executed?
A. The statement would execute successfully to produce the required output.
B. The statement would not execute because inline views and outer joins cannot be used together.
C. The statement would not execute because the ITEM_CNT alias cannot be displayed in the outer query.
D. The statement would not execute because the GROUP BY clause cannot be used in the inline.
Correct Answer: A
Examine the structure of the Books_TRANSACTIONS table:
Examine the SQl statement:
SQL> SELECT * FROM books_transactions WHERE brrrowed_data<SYSDATE AND transaction_type=\'RM\' OR MEMBER_ID IN
Which statement is true about the outcome?
A. It displays details only for the members who have borrowed before today with RM as TRANSACTION_TYPE.
B. It displays details for members who have borrowed before today\’s with either RM as TRANSACTION_TYPE or MEMBER_ID as A101 and A102.
C. It displays details for only member A101 and A102 have borrowed before today with RM as TRANSACTION-TYPE.
D. It displays details for members who have borrowed today with RM as TRANSACTION_TYPE and details for member A101 or A102.
Correct Answer: A
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