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Exam : Oracle 1Z0-001
Title : INTRODUCTION TO ORACLE: SQL AND PL/SQL


1. Click on the EXHIBIT button and examine the table instance chart for the patient table.
You need to create the patient_id_seq sequence to be used with the patient table's primary key column. The sequence should begin at 1000, have a maximum value of 999999999, never reuse any numbers, and increment by 1.
Which statement would you use to complete this task?
A. CREATE SEQUENCE patient_id_seq
START WITH 1000
MAXVALUE 999999999
NOCYCLE;
B. CREATE SEQUENCE patient_id_seq
START WITH 1000
MAXVALUE 999999999
STEP BY 1;
C. CREATE SEQUENCE patient_id_seq
ON patient (patient_id)
MINVALUE 1000
MAXVALUE 999999999
INCREMENT BY 1
NOCYCLE;
D. This task cannot be accomplished.
Answer: A

2. Examine this block of code:
SET SERVEROUTPUT ON
DECLARE
x NUMBER;
v_sal NUMBER;
v_found VARCHAR2(10) := 'TRUE';
BEGIN
x := 1;
v_sal := 1000;
DECLARE
v_found VARCHAR2(10);
y NUMBER;
BEGIN
IF (v_sal > 500) THEN
v_found := 'YES';
END IF;
DBMS_OUTPUT.PUT_LINE ('Value of v_found is '|| v_found);
DBMS_OUTPUT.PUT_LINE ('Value of v_sal is '|| v_sal);
y := 20;
END;
DBMS_OUTPUT.PUT_LINE ('Value of v_found is '|| v_found);
DBMS_OUTPUT.PUT_LINE ('Value of Y is '|| TO_CHAR(y));
END;
SET SERVEROUTPUT OFF
What is the result of executing this block of code?
A. PLS-00201: identifier 'Y' must be declared
B. Value of v_found is YES
Value of v_sal is 1000
Value of v_found is TRUE
C. Value of v_found is YES
Value of v_sal is 1000
Value of v_found is YES
Value of Y is 20
D. PLS-00201: identifier 'v_sal' must be declared
PLS-00201: identifier 'Y' must be declared
E. Value of v_found is YES
Value of v_sal is 1000
Value of v_found is TRUE
Value of Y is 20
Answer: A

3. Click on the EXHIBIT button and examine the table instance chart for the cars table.
Which SELECT statement will display style, color, and lot number for all cars based on the model entered at the prompt, regardless of case?
A. SELECT style, color, lot_no
FROM cars
WHERE model = UPPER('&model');
B. SELECT style, color, lot_no
FROM cars
WHERE model = '&model';
C. SELECT style, color, lot_no
FROM cars
WHERE UPPER(model) = UPPER('&model');
D. SELECT style, color, lot_no
FROM cars
WHERE UPPER(model) = '&model';
Answer: C

4. The structure of the DEPT table is as follows:
Name Null? Type
------------------------------- -------- -------
DEPTNO NOT NULL NUMBER(2)
DNAME VARCHAR2(14)
LOC VARCHAR2(13)
Examine the code:
DECLARE
TYPE dept_record_type IS RECORD
(dno NUMBER,
name VARCHAR2(20));
dept_rec dept_record_type;
BEGIN
SELECT deptno, dname
INTO dept_rec
FROM dept
WHERE deptno = 10;
END;
Which statement displays the name of the selected department?
A. DBMS_OUTPUT.PUT_LINE(name);
B. DBMS_OUTPUT.PUT_LINE(dname);
C. DBMS_OUTPUT.PUT_LINE(dept_rec.name);
D. DBMS_OUTPUT.PUT_LINE(dept_rec.dname);
E. DBMS_OUTPUT.PUT_LINE(dept_rec(name));
Answer: C

5. Click on the EXHIBIT button and examine the structure of the DEPARTMENT and EMPLOYEE tables.
Evaluate this SQL statement:
CREATE INDEX emp_dept_id_idx
ON employee(dept_id);
Which result will the statement provide?
A. Store an index in the EMPLOYEE table.
B. Increase the chance of full table scans.
C. May reduce the amount of disk I/O for SELECT statements.
D. May reduce the amount of disk I/O for INSERT statements.
E. Override the unique index created when the FK relationship was defined.
Answer: C

6. Which statement is valid within the executable section of a PL/SQL block?
A. BEGIN
emp_rec emp%ROWTYPE;
END;
B. WHEN NO_DATA_FOUND THEN
DBMS_OUTPUT.PUT_LINE ('No records found');
C. SELECT ename, sal
INTO v_ename, v_sal
FROM emp
WHERE empno = 101;
D. PROCEDURE calc_max (n1 NUMBER,n2 NUMBER,p_max OUT NUMBER) IS
BEGIN
IF n1 > n2 THEN
p_max := n1;
ELSE
p_max := n2;
END;
Answer: C

7. You need to perform a major update on the EMPLOYEE table. You have decided to disable the PRIMARY KEY constraint on the empid column and the CHECK constraint on the job column.
What happens when you try to enable the constraints after the update is completed?
A. You need to recreate the constraints once they are disabled.
B. Any existing rows that do not confirm with the constraints are automatically deleted.
C. Only the future values are verified to confirm with the constraints, leaving the existing values unchecked.
D. The indexes on both the columns with the PRIMARY KEY constraint and the CHECK constraint are automatically re-created.
E. All the existing column values are verified to confirm with the constraints and an error message is generated if any existing values do not confirm.
Answer: E

8. Which ALTER command would you use to reinstate a disabled primary key constraint?
A. ALTER TABLE cars
ENABLE PRIMARY KEY (id);
B. ALTER TABLE cars
ENABLE CONSTRAINT cars_id_pk;
C. ALTER TABLE cars
ENABLE PRIMARY KEY (id) CASCADE;
D. ALTER TABLE cars
ADD CONSTRAINT cars_id_pk PRIMARY KEY (id);
Answer: B

9. Evaluate this IF statement:
IF v_value > 100 THEN
v_new_value := 2 * v_value;
ELSIF v_value > 200 THEN
v_new_value := 3 * v_value;
ELSIF v_value v_new_value := 4 * v_value;
ELSE
v_new_value := 5 * v_value;
END IF;
What would be assigned to V_NEW_VALUE if V_VALUE is 250?
A. 250
B. 500
C. 750
D. 1000
E. 1250
Answer: B

10. The EMPLOYEE table contains these columns:
BONUSNUMBER(7,2)
DEPT_ID NUMBER(9)
There are 10 departments and each department has at least 1 employee. Bonus values are greater than 500; not all employees receive a bonus.
Evaluate this PL/SQL block:
DECLARE
v_bonusemployee.bonus%TYPE := 300;
BEGIN
UPDATE employee
SET bonus = bonus + v_bonus
WHERE dept_id IN (10, 20, 30);
COMMIT;
END;
What will be the result?
A. All employees will be given a 300 bonus.
B. A subset of employees will be given a 300 bonus.
C. All employees will be given a 300 increase in bonus.
D. A subset of employees will be given a 300 increase in bonus.
Answer: D

11. Given this executable section of a PL/SQL block:
BEGIN
FOR employee_record IN salary_cursor LOOP
employee_id_table(employee_id) :=
employee_record.last_name;
END LOOP;
CLOSE salary_cursor;
END;
Why does this section cause an error?
A. The cursor needs to be opened.
B. No FETCH statements were issued.
C. Terminating conditions are missing.
D. The cursor does not need to be closed.
Answer: D

12. Examine this block of code:
SET SERVEROUTPUT ON
DECLARE
x NUMBER;
v_sal NUMBER;
v_found VARCHAR2(10) := 'TRUE';
BEGIN
x := 1;
v_sal := 1000;
DECLARE
v_found VARCHAR2(10);
y NUMBER;
BEGIN
IF (v_sal > 500) THEN
v_found := 'YES';
END IF;
DBMS_OUTPUT.PUT_LINE ('Value of v_found is '|| v_found);
DBMS_OUTPUT.PUT_LINE ('Value of v_sal is '|| v_sal);
y := 20;
END;
DBMS_OUTPUT.PUT_LINE ('Value of v_found is '|| v_found);
DBMS_OUTPUT.PUT_LINE ('Value of Y is '|| TO_CHAR(y));
END;
SET SERVEROUTPUT OFF
Why does this code produce an error when executed?
A. The value of V_FOUND cannot be 'YES'.
B. Variable V_FOUND is declared at more than one location.
C. Variable Y is declared in the inner block and referenced in the outer block.
D. Variable V_SAL is declared in the outer block and referenced in the inner block.
Answer: C

13. You issue this command:
CREATE SYNONYM emp
FOR ed.employee;
Which task has been accomplished?
A. The need to qualify an object name with its schema was eliminated for user Ed.
B. The need to qualify an object name with its schema was eliminated for only you.
C. The need to qualify an object name with its schema was eliminated for all users.
D. The need to qualify an object name with its schema was eliminated for users with access.
Answer: B

14. Which should you do after each FETCH statement in a PL/SQL block?
A. Open the cursor.
B. Close the cursor.
C. Initialize the loop.
D. Test for rows using a cursor attribute.
Answer: D