Special Report, New Delhi/Mumbai: The notification for the Joint Entrance Examination (Advanced) 2017 has been already released while official notification for Joint Entrance Examination (Main) 2017 is yet to be released. As per the notification on key dates of JEE Advanced 2017, the exam will be held on 21 May 2017 (Sunday) while JEE Main 2017 (offline exam) will be held on 02 April 2017 (Sunday). To get admission into Indian Institutes of technology (IITs) for B-Tech courses, candidates have to appear the above two entrance tests.
The top rank holders will apply for admission into all the Indian Institute of Technology (IITs) and the Indian School of Mines (ISM) for various B-Tech courses. In JEE Main 2016, nearly 12 lakh candidates had appeared, while out of those 1,98,000 students were declared eligible for appearing the JEE advanced examination 2016. Out of registered 155,948 candidates in JEE Main 2016, only 36,566 were qualified as rank holders of JEE Advanced 2016. Total number of seats for all the 23 IITs (including Indian School of Mining (ISM), Dhanbad) was 10,575 for the academic year 2016-17 (B-Tech courses).
On an average out of nearly 113 candidates applied for JEE Main 2016, only one candidate finally managed to get a seat in any IITs. Hence, IIT-JEE test is considered as one the toughest entrance tests in the world. Here, Skilloutlook presents the detailed procedure how the final rank list will be prepared in the JEE Advanced 2017.
20K more candidates will be qualified for appearing JEE Advanced 2017: For JEE Advanced 2017, 20,000 more candidates will be made eligible to write the paper after passing the JEE Main 2017, and thus getting the number to 2,20,000 (Two Lakh Twenty). For JEE Advanced 2016, there were provision to declare only 2,00,000 (Two Lakh) candidates eligible out of the total candidates appearing JEE Main 2016.
Rank lists will be based on aggregate marks: As per the information bulletin, only candidates who appear in both Paper 1 and Paper 2 will be considered for ranking. The marks obtained by a candidate in Physics in JEE (Advanced) 2017 will be equal to the marks scored in Physics part of Paper 1 plus the marks scored in Physics part of Paper 2. Marks obtained in Chemistry and Mathematics will be calculated in the same way. The aggregate mark obtained by a candidate in JEE (Advanced) 2017 is the sum of the marks awarded to him/her in Physics, Chemistry and Mathematics. Rank lists is prepared based on the aggregate marks in JEE (Advanced) 2017.
Minimum marks for each subject and aggregate mark to qualify for ranking: Only candidates who score the minimum prescribed marks in each subject and in aggregate will be included in the rank list. The minimum prescribed mark varies with the category as given in the below table. For general candidates (Common rank list), minimum % of marks in each subject is 10 while minimum % of aggregate marks is 35. For OBC- NCL category students, minimum % of marks in each subject is 9 while minimum % of aggregate marks is 31.5. For SC/ST category students, minimum % of marks in each subject is 5 while minimum % of aggregate marks is 17.5.
|Rank List||Minimum % of marks
in each subject
|Minimum % of aggregate marks
|Common rank list (CRL)||10.0||35.0|
|OBC-NCL rank list||9.0||31.5|
|SC rank list||5.0||17.5|
|ST rank list||5.0||17.5|
|Common-PwD rank list (CRL-PwD)||5.0||17.5|
|OBC-NCL-PwD rank list||5.0||17.5|
|SC-PwD rank list||5.0||17.5|
|ST-PwD rank list||5.0||17.5|
|Preparatory course rank lists||2.5||8.75|
Rank lists in case of same aggregate marks: If the aggregate marks scored by two or more candidates are the same, then the following tie-break policy will be used for awarding ranks: Higher rank will be assigned to the candidate who has obtained higher marks in Mathematics. If this does not break the tie, higher rank will be assigned to the candidate who has obtained higher marks in Physics. If there is a tie even after this, candidates will be assigned the same rank. There will be no waiting list for ranking.
Here, it can be mentioned that there will be no weightage for the 12th class marks in calculating the ranks in JEE Main Examination 2017. IIT Council had taken the decision earlier and issued circular in this regard. However, Class XII scores will be a key parameter only for determining the eligibility criteria for appearing the exam (JEE Main 2017). For candidates to qualify for admission in the IITs/NITs, IIITs and such other CFTIs, whose admission are based on the JEE Ranks, they should have secured at least 75% marks in the 12th class exam, or be in the top 20 percentile in the 12th class exam conducted by their Boards. For SC/ST students the qualifying marks will be 65% in the XII exam.
The JEE Advanced 2017 Results will be declared on 11 June 2017 (Sunday).