1Z0-051 | Refresh 1Z0-051 Exam Study Guides With New Update Exam Questions
Q51. - (Topic 2)
Examine this statement:
SELECT student_id, gpa FROM student_grades WHERE gpa > &&value;
You run the statement once, and when prompted you enter a value of 2.0. A report is produced. What happens when you run the statement a second time?
A. An error is returned.
B. You are prompted to enter a new value.
C. A report is produced that matches the first report produced.
D. You are asked whether you want a new value or if you want to run the report based on the previous value.
Answer: C
Explanation:
use the double-ampersand if you want to reuse the variable value without prompting the user each time.
Incorrect Answer: Ais not an error
B&& will not prompt user for second time D&& will not ask the user for new value
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 7-13
Q52. - (Topic 2)
Which two statements are true regarding constraints? (Choose two.)
A. A foreign key cannot contain NULL values.
B. The column with a UNIQUE constraint can store NULLS .
C. A constraint is enforced only for an INSERT operation on a table.
D. You can have more than one column in a table as part of a primary key.
Answer: B,D
Q53. - (Topic 2)
View the Exhibit and examine the structure of the PRODUCT, COMPONENT, and PDT_COMP tables.
In PRODUCT table, PDTNO is the primary key.
In COMPONENT table, COMPNO is the primary key.
In PDT_COMP table, (PDTNO,COMPNO) is the primary key, PDTNO is the foreign key referencing PDTNO in PRODUCT table and COMPNO is the foreign key referencing the COMPNO in COMPONENT table.
You want to generate a report listing the product names and their corresponding component names, if the component names and product names exist.
Evaluate the following query:
SQL>SELECT pdtno,pdtname, compno,compname FROM product _____________ pdt_comp USING (pdtno) ____________ component USING(compno)
WHERE compname IS NOT NULL;
Which combination of joins used in the blanks in the above query gives the correct output?
A. JOIN; JOIN
B. FULL OUTER JOIN; FULL OUTER JOIN
C. RIGHT OUTER JOIN; LEFT OUTER JOIN
D. LEFT OUTER JOIN; RIGHT OUTER JOIN
Answer: C
Q54. - (Topic 1)
Which two statements are true regarding the COUNT function? (Choose two.)
A. COUNT(*) returns the number of rows including duplicate rows and rows containing NULL value in any of the columns
B. COUNT(cust_id) returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_ID column
C. COUNT(DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULL values in the INV_AMT column
D. A SELECT statement using COUNT function with a DISTINCT keyword cannot have a WHERE clause
E. The COUNT function can be used only for CHAR, VARCHAR2 and NUMBER data types
Answer: A,C
Explanation:
Using the COUNT Function
The COUNT function has three formats:
COUNT(*)
COUNT(expr)
COUNT(DISTINCT expr)
COUNT(*) returns the number of rows in a table that satisfy the criteria of the SELECT
statement, including duplicate rows and rows containing null values in any of the columns.
If a WHERE clause is included in the SELECT statement, COUNT(*) returns the number of
rows that satisfy the condition in the WHERE clause.
In contrast,
COUNT(expr) returns the number of non-null values that are in the column identified by
expr.
COUNT(DISTINCT expr) returns the number of unique, non-null values that are in the
column identified by expr.
Q55. - (Topic 1)
Which object privileges can be granted on a view?
A. none
B. DELETE, INSERT,SELECT
C. ALTER, DELETE, INSERT, SELECT
D. DELETE, INSERT, SELECT, UPDATE
Answer: D
Explanation: Object privilege on VIEW is DELETE, INSERT, REFERENCES, SELECT and UPDATE.
Incorrect Answer: AObject privilege on VIEW is DELETE, INSERT, REFERENCES, SELECT and UPDATE BObject privilege on VIEW is DELETE, INSERT, REFERENCES, SELECT and UPDATE CObject privilege on VIEW is DELETE, INSERT, REFERENCES, SELECT and UPDATE
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 13-12
Q56. - (Topic 1)
See the exhibit and examine the structure of the CUSTOMERS and GRADES tables:
You need to display names and grades of customers who have the highest credit limit.
Which two SQL statements would accomplish the task? (Choose two.)
A.
SELECT custname, grade
FROM customers, grades
WHERE (SELECT MAX(cust_credit_limit)
FROM customers) BETWEEN startval and endval;
B.
SELECT custname, grade FROM customers, grades WHERE (SELECT MAX(cust_credit_limit) FROM customers) BETWEEN startval and endval AND cust_credit_limit BETWEEN startval AND endval;
C.
SELECT custname, grade
FROM customers, grades
WHERE cust_credit_limit = (SELECT MAX(cust_credit_limit)
FROM customers)
AND cust_credit_limit BETWEEN startval AND endval;
D.
SELECT custname, grade
FROM customers , grades
WHERE cust_credit_limit IN (SELECT MAX(cust_credit_limit)
FROM customers)
AND MAX(cust_credit_limit) BETWEEN startval AND endval;
Answer: B,C
Q57. - (Topic 1)
Which view should a user query to display the columns associated with the constraints on a table owned by the user?
A. USER_CONSTRAINTS
B. USER_OBJECTS
C. ALL_CONSTRAINTS
D. USER_CONS_COLUMNS
E. USER_COLUMNS
Answer: D
Explanation: view the columns associated with the constraint names in the USER_CONS_COLUMNS view. Incorrect Answer: Atable to view all constraints definition and names Bshow all object name belong to user Cdoes not display column associated Eno such view
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 10-25
Q58. - (Topic 2)
Evaluate the SQL statement:
SELECT ROUND(45.953, -1), TRUNC(45.936, 2)
FROM dual;
Which values are displayed?
A. 46 and 45
B. 46 and 45.93
C. 50 and 45.93
D. 50 and 45.9
E. 45 and 45.93
F. 45.95 and 45.93
Answer: C
Explanation:
ROUND (45.953,-1) will round value to 1 decimal places to the left. TRUNC (45.936,2) will truncate value to 2 decimal The answer will be 50 and 45.93
Incorrect Answers :
A. Does not meet round and truncate functions
B. Does not meet round functions
D. Does not meet truncate functions
E. Does not meet round functions
F. Does not meet round functions
Refer: Introduction to Oracle9i: SQL, Oracle University Student Guide, Single-Row functions, p. 3-13
Q59. - (Topic 2)
View the Exhibit and examine the structure of the PRODUCTS tables.
You want to generate a report that displays the average list price of product categories where the average list price is less than half the maximum in each category.
Which query would give the correct output?
A.
SELECT prod_category,avg(prod_list_price) FROM products GROUP BY prod_category HAVING avg(prod_list_price) < ALL (SELECT max(prod_list_price)/2 FROM products GROUP BY prod_category);
B.
SELECT prod_category,avg(prod_list_price) FROM products GROUP BY prod_category HAVING avg(prod_list_price) > ANY (SELECT max(prod_list_price)/2 FROM products GROUP BY prod_category);
C.
SELECT prod_category,avg(prod_list_price) FROM products HAVING avg(prod_list_price) < ALL (SELECT max(prod_list_price)/2 FROM products GROUP BY prod_category);
D.
SELECT prod_category,avg(prod_list_price) FROM products GROUP BY prod_category HAVING avg(prod_list_price) > ANY (SELECT max(prod_list_price)/2 FROM products);
Answer: A
Explanation:
Using the ANY Operator in Multiple-Row Subqueries
The ANY operator (and its synonym, the SOME operator) compares a value to each value
returned by a subquery.
<ANY means less than the maximum.
>ANY means more than the minimum.
=ANY is equivalent to IN
Using the ALL Operator in Multiple-Row Subqueries
The ALL operator compares a value to every value returned by a subquery.
>ALL means more than the maximum and
<ALL means less than the minimum.
The NOT operator can be used with IN, ANY, and ALL operators.
Q60. - (Topic 2)
View the Exhibit and examine the data in the PROJ_TASK_DETAILS table.
The PROJ_TASK_DETAILS table stores information about tasks involved in a project and the relation between them.
The BASED_ON column indicates dependencies between tasks. Some tasks do not depend on the completion of any other tasks.
You need to generate a report showing all task IDs, the corresponding task ID they are dependent on, and the name of the employee in charge of the task it depends on.
Which query would give the required result?
A.
SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.based_on = d.task_id);
B.
SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p LEFT OUTER JOIN proj_task_details d ON (p.based_on =
d.task_id);
C.
SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p FULL OUTER JOIN proj_task_details d ON (p.based_on =
d.task_id);
D.
SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.task_id = d.task_id);
Answer: B