Board Weightage in JEE Main

Can you tell me about the weightage given to 12th Class Board Examination Marks/Score in Joint Entrance Exam Main (JEE Main)?

[pawan]

To qualify for the admission in the IITs/NITs/IIITs and such other CFTIs whose admission is based on the JEE (Joint Entrance Exam) (Advanced)/JEE(Main) ranks, need to have scored minimum 75% marks in the 12th class exam, or be in the top 20 percentile in the 12th class exam conducted by the respective Boards.

For SC/ST students the qualifying marks would be 65% in the 12th class exam.

The marks scored in JEE Mains paper are allotted 60% weightage in the final percentile score. The rest 40% is allotted for board marks.

The percentile is calculated relative to the number of students who attempted the exam. This final percentile score made up from JEE Main score and Board marks is used to calculate the All India Rank. Moreover a percentile score (around 110/360 for Open category) was set as a benchmark to attempt JEE Advance.

According to the existing policy, 60% weightage of JEE Mains score + 40% of Normalized Board exam marks is equal to Final AIR in JEE Main.

Thus, only scoring high in JEE Main is not enough, one need to have Class XII board marks good, so that JEE Main AIR can go higher.

To calculate JEE Main rank, follow these steps:

1. JEE Main marks scored by a candidate = A
2. Aggregate marks in Class XII board (basis a list of five subjects meeting the JEE Main Eligibility Criteria) = B
3. Convert marks into percentile by formula JEE Main aggregate marks corresponding to Percentile (P) at the All-India level = B1
4. JEE Main aggregate marks corresponding to Percentile (P) at particular board level = B2
5. Computation of normalized Class XII board score of the candidate:
6. Bfinal = 0.5 * (B1 + B2)
7. Calculation of the composite score for JEE Main Merit List:
8. C = 0.6 * A + 0.4 * Bfinal

This composite score forms the basis for the JEE Main Merit Ranks.