Exam answers IT-6202 Database Management System 1 (Oracle)

ABELELLEN SA_REPTALORJONATHAN SA_REPGRANTKIMBERLY SA_REPFAYPAT MK_REP Correct
ABELELLEN SA_REPTALORJONATHAN SA_REPGRANTKIMBERLY SA_REP
ABELELLEN SA_REPGRANTKIMBERLY SA_REPFAYPAT MK_REP
TALORJONATHAN SA_REPGRANTKIMBERLY SA_REPFAYPAT MK_REP

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT FIRSTNAME, SALARY, LPAD(SALARY * 0.10 + SALARY =E2=80=93 100, 8, =E2=80=98$=E2=80=99) AS BONUSFROM EMPLOYEESWHERE DEPARTMENT_ID NOT IN (10, 110, 50AND SALARY=17000;

NENA 17000 $$$18600LEX 17000 $$$18600 Correct
NENA 17000 $$$17000LEX 17000 $$$17000
NENA 17000 17000$$$LEX 17000 17000$$$
NENA 17000 18600$$$LEX 17000 18600$$$

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT INITCAP(LASTNAME||=E2=80=99,=E2=80=99||FIRSTNAME) AS NAMEFROM EMPLOYEESWHERE JOB_ID LIKE =E2=80=99%PR%=E2=80=99;

KING,STEVENHUNOLD,ALEXANDERERNST,BRUCELORENTZ,DIANA
King,StevenHunold,AlexanderErnst,BruceLorentz,Diana Correct
king,stevenhunold,alexanderernst,brucelorentz,Diana
King=E2=80=99,=E2=80=99StevenHunold=E2=80=99,=E2=80=99AlexanderErnst=E2=80=99,=E2=80=99BruceLorentz=E2=80=99,=E2=80=99Diana

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT LASTNAME,SALARY, RPAD(SALARY,4,=E2=80=99@=E2=80=99)FROM EMPLOYEESWHERE SALARY BETWEEN 4000 AND 9000LASTNAME LIKE =E2=80=98%S=E2=80=99;

DAVIES 3100 3100MATOS 2600 2600VARGAS 2500 2500HIGGINS 12000 12000
MOURGOS 5800 5800 Correct
MOURGOS 5800 5800@@@@RAJS 3500 3500@@@@DAVIES 3100 3100@@@@MATOS 2600 2600@@@@VARGAS 2500 2500@@@@HIGGINS 12000 12000@@@
MOURGOS 5800 @@@@5800

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT LOWER(LASTNAME||=E2=80=99with a salary of=E2=80=99||SALARY) as Record, LENGTH(LASTNAME) as LnameFROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY IS NULL;

king with a salary of 24000 4hunold with a salary of 2400 6
kingwith a salary of 24000 4hunoldwith a salary of 2400 6 Correct
KINGWITH A SALARY OF 24000 4HUNOLDWITH A SALARY OF 2400 6
KING WITH A SALARY OF 24000 4HUNOLD WITH A SALARY OF 2400 6

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT MOD(SALARY,10) AS =E2=80=9CEXCESS SALARY=E2=80=9DFROM EMPLOYEESWHERE SALARY < 5000;

33333
00000
333333
000000 Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT TRIM(=E2=80=98K=E2=80=99 FROM LASTNAME)AS TRIM, LASTNAME, DEPARTMENT_ID, MANAGER_IDFROM EMPLOYEESWHERE DEPARTMENT_ID = 50AND MANAGER_ID = 100;

RAJS RAJS 50 124
MOURGOS MOURGOS 50 100 Correct
ZLOTKEY ZLOTKEY 80 100
MOURGOS ZLOTKEY 80 100

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT TRUNC(563.396,-2) FROM DUAL;

563
500 Correct
56339
5633

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT TRUNC(563.396,1)FROM DUAL;

56339
5633 Correct
563
500

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT TRUNC(563.396,2)FROM DUAL;

5633
563
56339 Correct
500

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order to display the employees lastname concatenated to salary. Format the salary column to 6 character long right padded with =E2=80=98*=E2=80=99 as special character for all employees whose manager_id is null or salary between 4000 and 6000 Rename the column as employees and their Salaries

SELECT (FIRSTNAME||RPAD(SALARY,6,'*')) AS "EMPLOYEES AND THEIR SALARIES"FROM EMPLOYEESWHERE MANAGER_ID = NULLOR SALARY BETWEEN 4000 AND 6000;
SELECT (FIRSTNAME||RPAD(SALARY,6,'*')) AS "EMPLOYEES AND THEIR SALARIES"FROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY BETWEEN 4000 OR 6000;
SELECT (FIRSTNAME||RPAD(SALARY,6,'*')) AS "EMPLOYEES AND THEIR SALARIES"FROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY BETWEEN 4000 AND 6000; Correct
SELECT (FIRSTNAME||(SALARY,6,'*')) AS "EMPLOYEES AND THEIR SALARIES"FROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY BETWEEN 4000 AND 6000;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order to display the employees lastname concatenated with firstname and with a =E2=80=98, (comma)=E2=80=99 in between. A rename on the concatenated column as Complete Name. Note all values in Complete Name column should be in lowercase plus display the length of employees lastname for all employees whose lastname starts with letter M sort the lastname in its default order.

SELECT LOWER(LASTNAME||','||FIRSTNAME) AS "COMPLETE NAME",LENGTH(LASTNAME)FROM EMPLOYEESWHERE LASTNAME LIKE 'M%'; Correct
SELECT LOWER(LASTNAME||','||FIRSTNAME) AS COMPLETE NAME,LENGTH(LASTNAME)FROM EMPLOYEESWHERE LASTNAME LIKE '%M';
SELECT LOWER(LASTNAME||','||FIRSTNAME) AS COMPLETE NAME,LENGTH(LASTNAME)FROM EMPLOYEESWHERE LASTNAME LIKE 'M%';
SELECT LOWER(LASTNAME,FIRSTNAME) AS COMPLETE NAME,LENGTH(LASTNAME)FROM EMPLOYEESWHERE LASTNAME LIKE 'M%';

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order to display the firstname concatenated to salary with additional column salary that provides a computation of salary * 2. Rename the column as Increase of all employees whose lastname ends with N.

SELECT (FIRSTNAME || 'SALARY OF' || SALARY || 'IF MULITPLY BY TWO THEN HE/SHE WLL GOT A NEW SALARY OF' || SALARY * 2)AS INCREASE FROM EMPLOYEES WHERE LASTNAME LIKE 'N%';
SELECT (FIRSTNAME || 'SALARY OF' SALARY 'IF MULITPLY BY TWO THEN HE/SHE WLL GOT A NEW SALARY OF' || SALARY * 2)AS INCREASE FROM EMPLOYEES WHERE LASTNAME LIKE 'N%';
SELECT (FIRSTNAME || 'SALARY OF' || SALARY || 'IF MULITPLY BY TWO THEN HE/SHE WLL GOT A NEW SALARY OF' || SALARY * 2)AS INCREASE FROM EMPLOYEES WHERE LASTNAME LIKE '%N'; Correct
SELECT lower(FIRSTNAME || 'SALARY OF' || SALARY || 'IF MULITPLY BY TWO THEN HE/SHE WLL GOT A NEW SALARY OF' || SALARY * 2)AS INCREASE FROM EMPLOYEES WHERE LASTNAME LIKE 'N%';

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order to display the lastname and salary of all employees whose department_id = 60 or job_id like =E2=80=98_T%=E2=80=99. Format the salary to be 15 character long, left padded with =E2=80=98$=E2=80=99 as special character. Label the column Salary.

SELECT LASTNAME, RPAD(SALARY,15,'$') AS SALARYFROM EMPLOYEESWHERE DEPARTMENT_ID = 60 OR JOB_ID LIKE '_T';
SELECT LASTNAME, LPAD(SALARY,15,'$') AS SALARYFROM EMPLOYEESWHERE DEPARTMENT_ID = 60 OR JOB_ID LIKE '_T'; Correct
SELECT LASTNAME, LPAD(SALARY,15,'$')FROM EMPLOYEESWHERE DEPARTMENT_ID = 60 OR JOB_ID LIKE '_T';
SELECT LASTNAME, LPAD(SALARY,15,'$') AS SALARYFROM EMPLOYEESWHERE DEPARTMENT_ID = 60 AND JOB_ID LIKE 'T';

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order to display the salary leftpadded with 15 character long and =E2=80=98$=E2=80=99 as special character and another column salary right padded with 10 character long with =E2=80=98@=E2=80=99 as special character used of all employees in 201, 176 and 144.

SELECT LPAD(SALARY,15,'$'), RPAD(SALARY,10,'@')FROM EMPLOYEESWHERE EMPLOYEE_ID = (201,176,144);
SELECT RPAD(SALARY,15,'$'), LPAD(SALARY,10,'@')FROM EMPLOYEESWHERE EMPLOYEE_ID IN (201,176,144);
SELECT LPAD(SALARY,15,'$'), RPAD(SALARY,10,'@')FROM EMPLOYEESWHERE EMPLOYEE_ID IN (201,176,144); Correct
SELECT LPAD(SALARY,15,'@'), RPAD(SALARY,10,'$')FROM EMPLOYEESWHERE EMPLOYEE_ID IN (201,176,144);

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order todisplay a report that trim the letter =E2=80=98A=E2=80=99 from lastname of all employees whose department_id between 60 and 90.

SELECT TRIM('A' FROM LASTNAME)FROM EMPLOYEESWHERE DEPARTMENT_ID BETWEEN 60 AND 90; Correct
SELECT TRIM(LASTNAME,=E2=80=99A=E2=80=99)FROM EMPLOYEESWHERE DEPARTMENT_ID BETWEEN 60 AND 90;
SELECT TRIM('A' FROM LASTNAME)FROM EMPLOYEESWHERE DEPARTMENT_ID BETWEEN 90 AND 60;
SELECT TRIM('A' FROM LASTNAME)FROM EMPLOYEESWHERE DEPARTMENT_ID BETWEEN 60 OR 90;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order todisplay the firstname and length of firstname rename the column length of firstname as Number of Character of all employees whose salary is between 4400 and 8300

SELECT FIRSTNAME, LENGTH(FIRSTNAME) AS NUMBER OF CHARACTERFROM EMPLOYEESWHERE SALARY BETWEEN 4400 AND 8300;
SELECT FIRSTNAME, LENGTH(FIRSTNAME) AS "NUMBER OF CHARACTER"FROM EMPLOYEESWHERE SALARY BETWEEN 8300 AND 4400;
SELECT FIRSTNAME, LENGTH(FIRSTNAME) AS "NUMBER OF CHARACTER"FROM EMPLOYEESWHERE SALARY BETWEEN 4400 AND 8300; Correct
SELECT FIRSTNAME, INSTR(FIRSTNAME) AS "NUMBER OF CHARACTER"FROM EMPLOYEESWHERE SALARY BETWEEN 4400 AND 8300;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order todisplay the Firstname concatenated to employees original salary plus concatenate again a new column salary that multiplies the original salary into three. Rename the column as Dream Salaries.Note sort the salary in descending order.

SELECT (FIRSTNAME||' EARNS '|| SALARY || 'MONTHLY BUT WANTS' || SALARY * 3)AS "DREAM SALARIES"FROM EMPLOYEESORDER BY SALARY;
SELECT (FIRSTNAME||' EARNS MONTHLY BUT WANTS' || SALARY * 3)AS "DREAM SALARIES"FROM EMPLOYEESORDER BY SALARY DESC;
SELECT (FIRSTNAME||' EARNS '|| SALARY || 'MONTHLY BUT WANTS' || SALARY * 3)AS "DREAM SALARIES"FROM EMPLOYEESORDER BY SALARY DESC; Correct
SELECT (FIRSTNAME||' EARNS MONTHLY BUT WANTS' || SALARY * 3)AS "DREAM SALARIES"FROM EMPLOYEESORDER BY SALARY;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order todisplay the firstname in capitalized format rename the column as pangalan whose job_id is equal to =E2=80=98SA_REP=E2=80=99.

SELECT UPPER(FIRSTNAME) PANGALANFROM EMPLOYEESWHERE JOB_ID = 'SA_REP';
SELECT FIRSTNAME AS PANGALANFROM EMPLOYEESWHERE JOB_ID = 'SA_REP';
SELECT UPPER(FIRSTNAME) AS PANGALANFROM EMPLOYEESWHERE JOB_ID IS 'SA_REP';
SELECT UPPER(FIRSTNAME) AS PANGALANFROM EMPLOYEESWHERE JOB_ID = 'SA_REP'; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhat query should be used in order toget theSUBSTR function that returns the job_id = =E2=80=98REP=E2=80=99.

SELECT JOB_ID FROM EMPLOYEESWHERE JOB_ID LIKE =E2=80=98%REP%=E2=80=99;
SELECT JOB_ID FROM EMPLOYEESWHERE SUBSTR(JOB_ID,4)='%REP%';
SELECT JOB_ID FROM EMPLOYEESWHERE SUBSTR(JOB_ID,4)='REP'; Correct
SELECT JOB_ID FROM EMPLOYEESWHERE SUBSTR='REP';

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that counts the number of ST_CLERK job_id? Rename the column as =E2=80=9CTotal no. of ST_CLERK=E2=80=9D.

SELECT COUNT(JOB_ID), AS TOTAL NO OF ST_CLERKFROM EMPLOYEESWHERE JOB_ID = =E2=80=98ST_CLERK=E2=80=99;
SELECT COUNT(JOB_ID), AS =E2=80=9CTOTAL NO OF ST_CLERK=E2=80=9DFROM EMPLOYEESWHERE JOB_ID != =E2=80=98ST_CLERK=E2=80=99;
SELECT COUNT * (JOB_ID), AS =E2=80=9CTOTAL NO OF ST_CLERK=E2=80=9DFROM EMPLOYEESWHERE JOB_ID = =E2=80=98ST_CLERK=E2=80=99;
SELECT COUNT(JOB_ID), AS =E2=80=9CTOTAL NO OF ST_CLERK=E2=80=9DFROM EMPLOYEESWHERE JOB_ID = =E2=80=98ST_CLERK=E2=80=99; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that display the Job_id and add the ff: function:Count the total number of job_id per distinct values; Compute for the summary of salary per job_id; and Compute for the average salary per job_id

SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS =E2=80=9CNO OF JOB_ID=E2=80=9D, SUM(SALARY) AS =E2=80=9CTOTAL SALARY=E2=80=9D, AVG(SALARY) AS =E2=80=9CAVERAGE SALARY=E2=80=9D FROM EMPLOYEES GROUP BY JOB_ID; Correct
SELECT COUNT (DISTINCT(JOB_ID) AS =E2=80=9CNO OF JOB_ID=E2=80=9D, SUM(SALARY) AS =E2=80=9CTOTAL SALARY=E2=80=9D, AVG(SALARY) AS =E2=80=9CAVERAGE SALARY=E2=80=9D FROM EMPLOYEES GROUP BY JOB_ID;
SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS =E2=80=9CNO OF JOB_ID=E2=80=9D, SUM(SALARY) AS =E2=80=9CTOTAL SALARY=E2=80=9D, AVG(SALARY) AS =E2=80=9CAVERAGE SALARY=E2=80=9D FROM EMPLOYEES;
SELECT DISTINCT(JOB_ID), COUNT * (JOB_ID) AS =E2=80=9CNO OF JOB_ID=E2=80=9D, SUM(SALARY) AS =E2=80=9CTOTAL SALARY=E2=80=9D, AVG(SALARY) AS =E2=80=9CAVERAGE SALARY=E2=80=9D FROM EMPLOYEES GROUP BY JOB_ID;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that display the maximum lastname concatenated to firstname and rename the column as =E2=80=9CEmployees Name=E2=80=9D, Job_id and apply the ff. functions:Count the Job_id; Apply where condition whose lastname ends with letter =E2=80=98N=E2=80=99; Group the job_id; and Apply having clause of employees having average salary that is greater than 10000.

SELECT MAX(LASTNAME||FIRSTNAME) AS "EMPLOYEES NAME", JOB_ID, COUNT(JOB_ID)FROM EMPLOYEESWHERE LASTNAME LIKE '%N'GROUP BY JOB_IDHAVING AVG(SALARY)<10000;
SELECT MAX(LASTNAME||FIRSTNAME) AS "EMPLOYEES NAME", JOB_ID, COUNT(JOB_ID)FROM EMPLOYEESWHERE LASTNAME LIKE '%N'GROUP BY JOB_IDHAVING AVG(SALARY)>10000; Correct
SELECT MAX(LASTNAME||FIRSTNAME) AS EMPLOYEES NAME, JOB_ID, COUNT(JOB_ID)FROM EMPLOYEESWHERE LASTNAME LIKE '%N'GROUP BY JOB_IDHAVING AVG(SALARY)>10000;
SELECT MAX(LASTNAME||FIRSTNAME) AS "EMPLOYEES NAME", JOB_ID, COUNT(JOB_ID)FROM EMPLOYEESGROUP BY JOB_IDWHERE LASTNAME LIKE '%N'HAVING AVG(SALARY)>10000;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that display the Minimum firstname concatenated to lastname? Look for the maximum salary that is less than 10000 per department_id.

SELECT MIN(FIRSTNAME,LASTNAME)FROM EMPLOYEESGROUP BY DEPARTMENT_IDHAVING MAX(SALARY)<10000;
SELECT MIN(FIRSTNAME||LASTNAME)FROM EMPLOYEESGROUP BY DEPARTMENT_IDHAVING MAX(SALARY)<10000; Correct
SELECT MIN(FIRSTNAME||LASTNAME)FROM EMPLOYEESGROUP BY DEPARTMENT_IDHAVING MAX(SALARY);
SELECT MIN(FIRSTNAME||LASTNAME)FROM EMPLOYEESHAVING MAX(SALARY)<10000GROUP BY DEPARTMENT_ID;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that displays the lowest salary rename as =E2=80=9CLOWEST SALARY=E2=80=9D, maximum salary rename as =E2=80=9CHIGHEST SALARY=E2=80=9D and department_id concatenated to job_id?

SELECT CONCAT(DEPARTMENT_ID,JOB_ID),MIN(SALARY), MAX(SALARY)FROM EMPLOYEESGROUP BY DEPARTMENT_ID;
SELECT CONCAT(DEPARTMENT_ID,JOB_ID),MIN(SALARY), MAX(SALARY)FROM EMPLOYEESGROUP BY JOB_ID;
SELECT CONCAT(DEPARTMENT_ID,JOB_ID),MIN(SALARY), MAX(SALARY)FROM EMPLOYEESGROUP BY JOB_ID,DEPARTMENT_ID; Correct
SELECT CONCAT(DEPARTMENT_ID,JOB_ID),MIN(SALARY), MAX(SALARY)FROM EMPLOYEES;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that displays the maximum salary?

SELECT MAX SALARY FROM EMPLOYEES;
SELECT COUNT (MAX(SALARY)) FROM EMPLOYEES;
SELECT MAX(SALARY) FROM EMPLOYEES; Correct
SELECT COUNT (DISCTINCT MAX(SALARY)) FROM EMPLOYEES;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that displays the minimum lastname?

SELECT MIN LASTNAME FROM EMPLOYEES;
SELECT MIN(LASTNAME) FROM EMPLOYEES; Correct
SELECT COUNT (MIN(LASTNAME)) FROM EMPLOYEES;
SELECT COUNT (DISCTINCT MIN(LASTNAME)) FROM EMPLOYEES;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct query that displays the MINIMUM salary of employees per job_id? Note job_id should be in lowercase.

SELECT MIN(SALARY) AS LOWEST SALARY, LOWER(JOB_ID)FROM EMPLOYEES;
SELECT MIN(SALARY) AS =E2=80=9CLOWEST SALARY=E2=80=9D, LOWER(JOB_ID)FROM EMPLOYEESGROUP BY JOB_ID; Correct
SELECT MIN(SALARY) AS =E2=80=9CLOWEST SALARY=E2=80=9D, LOWER(JOB_ID)FROM EMPLOYEES;
SELECT MIN(SALARY) AS LOWEST SALARY, LOWER(JOB_ID)FROM EMPLOYEESGROUP BY JOB_ID;

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct report that display the smallest (minimum) Surname.. and apply the ff. functions:Get the average salary; Group the data per job_id; Get only job_id with a keyword =E2=80=9CREP=E2=80=9D; and Apply having clause, which the max salary per job_id is greater than 5000.

SELECT CONCAT(DEPARTMENT_ID,JOB_ID),MIN(SALARY), MAX(SALARY)FROM EMPLOYEESGROUP BY JOB_ID;
SELECT MIN(LASTNAME), AVG(SALARY)FROM EMPLOYEESWHERE JOB_ID NOT LIKE =E2=80=98%REP%=E2=80=99GROUP BY JOB_IDHAVING MAX(SALARY)>500;
SELECT MIN(LASTNAME), AVG(SALARY)FROM EMPLOYEESGROUP BY JOB_IDWHERE JOB_ID LIKE =E2=80=98%REP%=E2=80=99HAVING MAX(SALARY)>500;
SELECT MIN(LASTNAME), AVG(SALARY)FROM EMPLOYEESWHERE JOB_ID LIKE =E2=80=98%REP%=E2=80=99GROUP BY JOB_IDHAVING MAX(SALARY)>500; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown below.Table 1.0 EMPLOYEESWhich of the following is the correct that display distinct job_id and the the total number per distinct (unique) job_id.

SELECT DISTICT(JOB_ID), COUNT*(JOB_ID) AS =E2=80=9CTOTAL NUMBER OF JOB_ID=E2=80=9DFROM EMPLOYEESGROUP BY JOB_ID;
SELECT DISTICT(JOB_ID), COUNT(JOB_ID) AS =E2=80=9CTOTAL NUMBER OF JOB_ID=E2=80=9DFROM EMPLOYEES;
SELECT DISTICT(JOB_ID), COUNT(JOB_ID) AS =E2=80=9CTOTAL NUMBER OF JOB_ID=E2=80=9DFROM EMPLOYEESGROUP BY JOB_ID; Correct
SELECT DISTICT((JOB_ID), COUNT(JOB_ID)) AS =E2=80=9CTOTAL NUMBER OF JOB_ID=E2=80=9DFROM EMPLOYEESGROUP BY JOB_ID;

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT FIRSTNAME, SALARY, LPAD(SALARY * 010 + SALARY =E2=80=93 100, 8, =E2=80=98$') AS BONUSFROM EMPLOYEESWHERE DEPARTMENT_ID NOT IN (10, 110, 50AND SALARY=17000;

NENA 17000 $$$18600LEX 17000 $$$18600 Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT INITCAP(LASTNAME||','||FIRSTNAME) AS NAMEFROM EMPLOYEESWHERE JOB_ID LIKE '%PR%';

King,StevenHunold,AlexanderErnst,BruceLorentz,Diana Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT LASTNAME,SALARY, RPAD(SALARY,4,'@')FROM EMPLOYEESWHERE SALARY BETWEEN 4000 AND 9000LASTNAME LIKE =E2=80=98%S';

MOURGOS 5800 5800 Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT LOWER(LASTNAME||'with a salary of'||SALARY) as Record, LENGTH(LASTNAME) as LnameFROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY IS NULL;

kingwith a salary of 24000 4hunoldwith a salary of 2400 6 Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESBased on the given SELECT statement below what would be the possible output?SELECT TRIM(=E2=80=98K' FROM LASTNAME)AS TRIM, LASTNAME, DEPARTMENT_ID, MANAGER_IDFROM EMPLOYEESWHERE DEPARTMENT_ID = 50AND MANAGER_ID = 100;

MOURGOS MOURGOS 50 100 Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhat query should be used in order to display the employees lastname concatenated to salary Format the salary column to 6 character long right padded with =E2=80=98*' as special character for all employees whose manager_id is null or salary between 4000 and 6000 Rename the column as employees and their Salaries

SELECT (FIRSTNAME||RPAD(SALARY,6,'*')) AS "EMPLOYEES AND THEIR SALARIES"FROM EMPLOYEESWHERE MANAGER_ID IS NULLOR SALARY BETWEEN 4000 AND 6000; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhat query should be used in order to display the employees lastname concatenated with firstname and with a =E2=80=98, (comma)' in between A rename on the concatenated column as Complete Name Note all values in Complete Name column should be in lowercase plus display the length of employees lastname for all employees whose lastname starts with letter M sort the lastname in its default order

SELECT LOWER(LASTNAME||','||FIRSTNAME) AS "COMPLETE NAME",LENGTH(LASTNAME)FROM EMPLOYEESWHERE LASTNAME LIKE 'M%'; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhat query should be used in order to display the lastname and salary of all employees whose department_id = 60 or job_id like =E2=80=98_T%' Format the salary to be 15 character long, left padded with =E2=80=98$' as special character Label the column Salary

SELECT LASTNAME, LPAD(SALARY,15,'$') AS SALARYFROM EMPLOYEESWHERE DEPARTMENT_ID = 60 OR JOB_ID LIKE '_T'; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhat query should be used in order to display the salary leftpadded with 15 character long and =E2=80=98$' as special character and another column salary right padded with 10 character long with=E2=80=98@' as special character used of all employees in 201, 176 and 144

SELECT LPAD(SALARY,15,'$'), RPAD(SALARY,10,'@')FROM EMPLOYEESWHERE EMPLOYEE_ID IN (201,176,144); Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhat query should be used in order todisplay a report that trim the letter =E2=80=98A' from lastname of all employees whose department_id between 60 and 90

SELECT TRIM('A' FROM LASTNAME)FROM EMPLOYEESWHERE DEPARTMENT_ID BETWEEN 60 AND 90; Correct

Choose the letter of the correct answer based on the table EMPLOYEES as shown belowTable 10 EMPLOYEESWhich of the following is the correct query that display the maximum lastname concatenated to firstname and rename the column as =E2=80=9CEmployees Name=E2=80=9D, Job_id and apply the ff functions:Count the Job_id; Apply where condition whose lastname ends with letter =E2=80=98N'; Group the job_id; and Apply having clause of employees having average salary that is greater than 10000

SELECT MAX(LASTNAME||FIRSTNAME) AS "EMPLOYEES NAME", JOB_ID, COUNT(JOB_ID)FROM EMPLOYEESWHERE LASTNAME LIKE '%N'GROUP BY JOB_IDHAVING AVG(SALARY)>10000; Correct

Consists of a collection of DML statements that form a logical unit of work.

Select
Transaction Correct
DDL
DML

Extracts a string of determined length.

Trim
Length
SUBSTR Correct
INSTR

Given the output below. Which of the following is the correct PL/SQL to be used

SELECT WAREHOUSE, CLASS FROM PARTS WHERE CLASS IN (=E2=80=98AP=E2=80=99,=E2=80=99SG=E2=80=99,=E2=80=99HW=E2=80=99);
SELECT WAREHOUSE, CLASS FROM PARTS;
SELECT DISTINCT WAREHOUSE FROM PARTS; Correct
SELECT DICTINCT WAREHOUSE, CLASS FROM PARTS WHERE WAREHOUSE >=1;

Given the output below. Which of the following is the correct PL/SQL to be used?

SELECT DESCRIPTION, ONHAND,CLASSFROM PARTSWHERE CLASS = =E2=80=98HW=E2=80=99; Correct
SELECT DESCRIPTION, ONHAND,CLASSFROM PARTSWHERE CLASS = HW;
SELECT DESCRIPTION, ONHAND,CLASSFROM PARTSWHERE ONHAND >=21;
SELECT DESCRIPTION, ONHAND,CLASSFROM PARTSWHERE ONHAND IN(50,21,22);