Oracle Database SQL — Question 244

In your session, the NLS_DATE_FORMAT is DD-MM-YYYY.
There are 86400 seconds in a day.
Examine this result:

DATE -
-----------
02-JAN-2020
Which statement returns this?

Answer options

• A. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '4' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
• B. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '3' MONTH + INTERVAL '7' DAY - INTERVAL '360' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
• C. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
• D. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '86410' SECOND, 'DD-MON- YYYY') AS "date" FROM DUAL;
• E. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '6' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;

Correct answer: C

Explanation

The correct answer is C, as it accurately adds 2 months and 5 days to the date '29-10-2019' and then subtracts 120 seconds, resulting in '02-JAN-2020'. The other options either add incorrect intervals or subtract the wrong amount of seconds, leading to different dates that do not match the specified result.