How will CBSE calculate JEE Main 2016 Final Rankings and Class XII Board Exam Marks?

CBSE  has announced the standardization/normalization  procedure  (page 5) for class 12 board exam marks in preparing all India JEE Main 2016 merit list. Here is what the CBSE Notice says

The JEE Interface Group (JIG) has decided that 50% of Boards marks be standardized by equating percentile amongst different boards/examining bodies and anchoring them to All India JEE-Main percentiles, and 50% be normalized by equating each Board’s/examining Body’s percentile with JEE(Main) percentile marks of respective Boards /examining bodies.

The detailed procedure for normalization of Board marks will be as follows:

Note down the aggregate marks (A0) obtained by each student in JEE- Main.

Compute the percentile (P) of each student on the basis of aggregate marks in his/her own board (B0) computed from the list of five subjects specified (each marked out of 100). The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE-Main.

Determine the JEE- Main aggregate marks corresponding to percentile (P) at the All- India level. Regard this as B1.

Also, determine the JEE- Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE- Main by the students of that board. Regard this as B2.

The normalized board score of the candidate will be computed as:

B (final) = 0.5 * (B1 + B2)

For the purpose of admission to CFTIs where it has been decided to use the JEE Mains performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list will be computed as:

C = 0.6 * AO + 0.4 * B (final),

Five subjects to be used for normalization:

  • Physics
  • Mathematics
  • Any one of the subjects Chemistry, Biology, Biotechnology and Computer Science
  • One language
  • Any subject other than the above four subjects.

In respect of 3, 4 and 5, the best mark in a given category will be chosen.