JEE Main 2019 Normalisation

National Testing Agency (NTA) has revived normalisation in Joint Entrance Examination (JEE) Main. It released information brochure and application form for JEE Main on September 01 and with it, the notice for normalisation is also issued. Central Board of Secondary Education (CBSE) removed normalization of class 12th marks in JEE after realizing that the impact of boards marks in JEE Main AIR was not fulfilling the determined agenda. Now, NTA has again introduced normalisation but for a different purpose.

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Earlier, the normalization was done to equate the class 12th marks in order to bring the level of all the Boards on the same line. However, now NTA will do normalisation to bring the difficulty level of exam that will be held in multiple sessions and days at the same benchmark.

Some of the candidates may end up attempting a relatively tougher set of questions when compared to other sets. The candidates who attempt the comparatively tougher examination are likely to get lower marks as compared to those who attempt the easier one. In order to overcome such a situation,“Normalization procedure based on Percentile Score” will be used for ensuring that candidates are neither benefitted nor disadvantaged due to the difficulty level of the examination. – By NTA (As mentioned in the official pdf)

Now, the questions that arises in the minds of candidates is what is JEE Main 2019 normalization? – Well it is a process to compare candidate’s score in multiple session. It basically ensures that there is no uneven level of difficulty in the exam. NTA will use percentile equivalence for JEE Main 2019. Normalisation is also done in other examinations, like GATE, where exam is held in multiple sessions.

How JEE Main 2019 normalization will be done is simple. NTA will use a predefined formula for normalization in order to prepare the merit list of JEE Main. To understand it, candidates first need to know few terminologies, like raw score, percentile score, subject percentile score, overall percentile score, and merit list.

NTA will calculate raw score of JEE Main 2019 on the basis of the marking scheme of the examination. For each correct answer marked by the candidates 4 marks will be added and for each incorrect answer, negative marking of 1 marks will be implemented.

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Overall percentile score of JEE Main is the calculate by comparing the raw score of the candidate, with the raw score secured by all the candidates in the respective session. The important point to note is that NTA will calculate the percentile score until 7 decimal places. To get this, NTA will use the formula below.

NTA has also decided to calculate percentile score for each subject, i.e. Mathematics, Physics, and Chemistry. This score will be used to break tie in the examination. NTA will calculate it for all the candidates and will use it when needed.

One the basis of overall percentile score, the conducting body will prepare JEE Main 2019 merit list. The candidate at the top of the list will be allocated JEE Main all India rank (AIR) 1. On the basis of AIR, the candidates will be called for counselling and will be allocated seat, in the courses offered by NITs, IIITs, GFTIs.

How JEE Main 2019 merit list will be prepared for two attempts? Well, this year, candidates can appear for JEE Main twice. The merit list for both January and April paper will be prepared combine. The Percentile score of the candidates who is appearing for both the attempts will be prepared separately for JEE Main January and JEE Main April. But to determine their final AIR, the best of both the attempts will be used.

JEE Main 2019 tie breaking will be done if the Overall Percentile Score of two or more candidates will come out to be the same. For this, NTA will first check the Percentile Score in JEE Main 2019 Mathematics section. However, if this does not work then the Percentile Score in Physics section will be considered. If the tie prevails still then the Percentile Score in Chemistry will be used. The higher the score, the better rank will be allocated to the candidates. If all of this fails, then NTA will allocate higher rank to the candidates who is elder in age.

JEE Main 2019 will be held from January 06 to 20, 2019. The registration for JEE Main 2019 is open right now. Candidates can apply until September 30, 2018. This year, JEE is only being held in online mode. JEE Main 2019 will also be held in the month of April, from 06 to 20. The details regarding the April exam will release in the month of February.

You may also have a look at the official pdf by NTA on JEE Main 2019 normalisation. We hope most of your doubts must have been cleared. However, if there is still any confusion regarding normalisation in JEE Main 2019 then comment below. 

2 COMMENTS

  1. Sir, I am in class 12 this year.
    Can I do the following in jee mains:
    I give jee in April only this year, I.e. give jee only in April in 2019.
    Then I drop, and give jee in both(January and April) in 2020….???
    Please clear my doubt…
    Will my 3 attempts exhaust in this method?
    How will my rank be calculated( which best of two will be taken?)

  2. How will the normalisation benefit when the final list of jee (after the April exam) will be prepared on the basis of raw marks of the student?

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