JEE Main 2019: Merit List to be based on Normalized/Percentile Score

The National Testing Agency (NTA) will check each and every objection submitted before 17 January (11:50 PM) by students who appeared for JEE Main 2019 (Paper I, II) in any shift between 8 January and 12 January. In some shifts for JEE Main 2019, students found the paper to be more difficult than in other shifts. For this reason, they are very nervous regarding their scores in JEE Main 2019. But students need not worry about this as the National Testing Agency will follow normalization process to prepare JEE Main 2019 Merit List.  In this article, we are going to explain you the complete normalization process to be followed by the National Testing Agency in detail.

JEE Main 2019: Expected Cut-off to Qualify the Exam - Check This Link!

About the exam:

JEE Main Exam 2019 was conducted between Jan 8 and 12 in two shifts i.e., First Shift (9:30 AM to 12:30 PM) and Second Shift (2:30 PM to 5:30 PM). A total of 11,09,250 candidates registered (9,29,198 for Paper-I & 1,80,052 for Paper II) for JEE Main 2019. This year, the national testing agency (NTA) has conducted JEE Main in fully computer based test mode (CBT) for the very first time. Students will also get another chance to appear for JEE Main 2019 in the month of April.

The candidates who will qualify JEE Main will be eligible for appearing in JEE Advanced 2019, which is the next level for admissions to Indian Institute of Technology (IITs). Only candidates who will qualify JEE Advanced 2019 will be shortlisted for final admissions to IITs. However, the successful candidates of JEE Main 2019 can take admission in different NITs, GFTIs and IIITs through JoSAA counselling.

JEE Main 2019: Detailed Analysis and Review of Paper I (9, 10, 11 and 12 January, All Shifts)

JEE Main 2019: Analysis and Review 8 Jan (Paper II - All Shifts)

The National Testing Agency (NTA) will calculate marks of qualified candidates based on a normalisation process to prepare the rank list for JEE Main 2019. In this article, we are going to explain you the complete normalisation process to be followed by the National Testing Agency.

Basic Terminologies:


Raw score:

The row score will be calculated by NTA on the basis of the marking scheme of the examination. For each correct answer marked by the candidates 4 marks will be added, while for each incorrect answer, 1 mark will be deducted from the total score.

Percentile Scores:

The percentile scores will be obtained by converting the raw marks into a scale ranging from 100 to 0 for each session of examinees.

The Percentile score will be the Normalized Score for the examination (instead of the raw marks of the candidate) and shall be used for preparation of the merit lists.

Note:- The Percentile Scores will be calculated up to 7 decimal places to avoid bunching effect and reduce ties.

Steps to be followed for getting JEE Main 2019 Normalised Score:

Step 1:

The row score of each candidate for each subject (Maths, Physics and Chemistry) will be calculated by the National Testing Agency.

Step 2:

The percentile scores for each of the subject and total percentile score will be calculated using the below formula:

Here, T1,M1, P1, C1 are the raw scores in Total, Mathematics, Physics, Chemistry of a candidate and T1P, M1P, P1P, C1P are the percentile scores of Total, Mathematics, Physics, Chemistry of that candidate.

Subjects to be used for normalization:

  1. Physics
  2. Mathematics
  3. Chemistry


Tie-breaking rules:

(a) Candidates obtaining higher Percentile Score in Mathematics will get higher rank.

(b) Candidates obtaining higher Percentile Score in Physics will get higher rank.

(c) Candidates obtaining higher Percentile Score in Chemistry will get higher rank.

(d) Candidates older in age will get higher rank.

JEE Main 2019 Question Paper 1: Memory Based Questions and Topics

Advertisement

Related Categories

Advertisement