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Database Management System 1 (Oracle)

Showing 1-75 of 144 answers

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||FIRSTNAME), JOB_IDFROM EMPLOYEESWHERE SUBSTR (JOB_ID,4)=E2=80=99REP=E2=80=99;
  • 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 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 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;
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);
He proposed the relational model for database systems in 1970.
  • Dr EF Codd Correct
  • DrcLF Codd
  • Dr Lee Codd
  • Dr JF Codd
It is a collection of relations or two-dimensional tables controlled by the Oracle server.
  • Database management system
  • Relational database Correct
  • Relational server
  • Relational model
It is use to accept numeric input and return numeric values.
  • Multiple row function
  • Character function
  • Number function Correct
  • Case manipulation function
Refer to table 1.0 Using the SELECT statement below what would be the possible output:SELECT NAME, ORG, POSITION, ID * (100 + 3)FROM ORGCHARWHERE ORG = =E2=80=98AMATHS=E2=80=99
  • It will retrieve the record of ANNA with new ID of 10
  • It will retrieve the record of ANNA with new ID of 1003
  • It will retrieve the record of ANNA with new ID of 1030 Correct
  • 0 rows selected
Refer to table 1.0 Using the SELECT statement below what would be the possible output:SELECT NAME, ORG, POSITIONFROM ORGCHARWHERE NAME LIKE =E2=80=98%A=E2=80=99;
  • 0 rows selected / no rows selected
  • Incorrect SELECT statement
  • It retrieves the record of AGAPITO
  • It retrieves the record of ANNA and MOJICA Correct
Refer to table 1.0 Using the SELECT statement below what would be the possible output:SELECT NAME, ORG, POSITIONFROM ORGCHARWHERE NAME LIKE =E2=80=98A%=E2=80=99AND POSITION = =E2=80=98SEC=E2=80=99;
  • It retrieves the record of AGAPITO Correct
  • Incorrect SELECT statement
  • It retrieves the record of ANNA and MOJICA
  • 0 rows selected / no rows selected
Refer to table 1.0 Using the SELECT statement below what would be the possible output:SELECT NAME, ORG, POSITIONFROM ORGCHARWHERE ORG =E2=80=99JPCS=E2=80=99OR POSITION = =E2=80=98SEC=E2=80=99;
  • Incorrect SELECT statement
  • It will retrieve the record of JOAN andROAN
  • 0 rows selected
  • It will retrieve the record of JOAN,ROAN and JUN Correct
Refer to table 1.0, suppose that user insert the following values using the statement below what would be the possible output?INSERT INTO ORGCHART (ID, NAME)VALUES (11,=E2=80=99YAMBAO,JUN=E2=80=99); 
  • The ORG and POSITION of YAMBAO will automatically sets to NULL Correct
  • The ORG and POSITION of YAMBAO will automatically sets to None
  • Cannot insert the record because of missing values for ORG and POSITION
  • The ORG and POSITION of YAMBAO will automatically sets to N/A
Refer to table 1.0, suppose that user insert the following values using the statement below what would be the possible output?INSERT INTO ORGCHART VALUES (11,=E2=80=99YAMBAO,JUN=E2=80=99, NULL, NULL); 
  • The ORG and POSITION of YAMBAO will automatically sets to N/A
  • The ORG and POSITION of YAMBAO will automatically sets to None
  • The ORG and POSITION of YAMBAO will automatically sets to NULL Correct
  • Cannot insert the record because of missing values for ORG and POSITION
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