Which JEE Main score will NTA use for common rank list?
Prabha Dhavala, 19 Jan 2019, #JEE Main
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Which JEE Main score will NTA use for common rank list? - With the announcement of two tests per year and better of the two scores along with a host of other changes by the National Testing Agency aka NTA, there is still some ambiguity in the minds of students about which JEE Main score will be used by NTA for the common merit rank list. From 2019, NTA is conducting the exam two times – once in January and once in April. Candidates can attempt one of the exams or both. So if you are one of those students who is attempting the exam twice – both in January and April 2019 then you should read the article below to know which JEE Main score will be used by NTA to determine the rank list.

Latest: JEE Main 2019 Result announced on January 19!

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Salient Facts about New NTA JEE Main

  • JEE Main will be held by the National Testing Agency which was set up in November 2017 exclusively for conducting entrance exams pertaining to higher education.

  • JEE Main 2019 will be a computer-based test and will be held twice a year

  • The first test will be on January 9, 10, 11, and 12, 2019 for Paper 1, while for Paper 2, it will be conducted on January 8 while the second attempt will be from April 6 to 20, 2019.

  • JEE Main Mock tests are available from September 1, 2018.

  • As recommended by Council of Architecture to NTA regarding JEE Main Paper II, candidates must secure at least 50% marks in the qualifying examination and 50% marks in PCM individually.

  • 3400 Schools/ Engineering Colleges as Test Practise Centres (TPCs) will be opened for practice on the mock tests for free.

  • Each student will be assigned a unique paper from a large databank prepared by experts.

  • Result for the first test will be declared on January 31, 2019, while results for the second test will be announced on April 30, 2019.

  • Rank List will be prepared in May for JoSAA Counselling.

Procedure for JEE Main Ranks

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National Testing Agency will announce the rank list after the result for JEE Main 2019 April is announced. The reason is that many candidates will have appeared for both the exams – JEE Main January and April. The JEE Main 2019 rank/ merit list will be prepared based on the percentile score of the candidate. So, to calculate the percentile score, in the case where the candidate has appeared for both the exam, the better of the two scores will be considered. So how will the score be used by NTA to prepare the JEE Main common rank list? The details are given below.

Appeared in JEE Main 2019 January OR April? Scores of the respective January OR April test will be used

Appeared in both JEE Main 2019 January and JEE Main April 2019? The better of the two scores will be used. So it means that if the score for JEE Main April is more than January or vice versa, it will be used. The lower score will be ignored.

How is the JEE Main 2019 score calculated?

The JEE Main score 2019 for Paper-1 and Paper-2 will be calculated differently. Mentioned below is the procedure for the calculation for both the papers.

  • Score of JEE Main = Number of correct answers x 4 – number of incorrect answers x 1

How is Candidate's Percentile Score for JEE Main paper-1 Will be Calculated?

The percentile score will be the normalized score for JEE Main 2019 (instead of the raw marks of the candidate). The JEE Main 2019 percentile scores will be calculated up to 7 decimal places. NTA will use the following method to calculate the percentile score of a candidate:

100 * Number of candidates appeared in the ‘Session’ with raw score ‘Equal To Or Less’ than the candidate/Total number of the candidates appeared in the ‘Session'.

Rank List Preparation for JEE Main 2019 for Paper-1

Ranks will be prepared on the basis of the percentile scores of the candidates in JEE Main 2019 as mentioned above. However, there may be instances where more than two candidates may score the same marks. In such a case the inter-se-merit guidelines will be enforced to allot the rank.

  • Candidates with the higher percentile score obtained in Mathematics will be given a better rank.
  • If the tie still remains unresolved the candidates with higher percentile score in Physics will be awarded a better rank.
  • If the tie still remains unresolved the candidates with the higher percentile score in Chemistry will be awarded a better rank.
  • In case the tie is still not resolved, then the candidates older in age to be preferred.

Raw marks for JEE Main paper-2

JEE Main paper-2 will be conducted in two shifts, the raw marks obtained by all the candidates who appeared in the first attempt will be declared by 31st January 2019. This shall comprise the actual marks obtained in Paper-2 of JEE (Main)– 2019 for the first attempt.

Rank List Preparation for JEE Main 2019 for Paper-2

The rank list will be prepared by merging the raw scores of the first and the second attempt. Candidates who have appeared in both the attempts; their best of the two raw marks will be considered. The total raw marks of both the attempts and rank of Paper-2 for all candidates will be declared by 30th April 2019. This shall comprise the total raw marks obtained by the candidate in the first attempt as well as in the second attempt along with All India rank and All India category rank. In case of a tie, inter-se merit will be determined in the following order.

  • Candidates obtaining higher marks in Aptitude Test in Paper-2
  • Candidates obtaining higher marks in Drawing Test in Paper-2
  • Candidates older in age
  • If the resolution is not possible after this criterion, candidates will be given the same rank.

Important Points of JEE Main Common Rank List

  • The common rank list of JEE Main will include candidates given ranks as per their performance but with a tag of GEN, GEN-PwD, OBC-NCL, OBC-NCL-PwD, SC, SC-PwD, ST or ST-PwD.

  • The category ranks will be separate from the JEE Main 2019 common rank list where the list will consist of the same category candidates assigned ranks on the basis of their performance in JEE Main 2019.

  • Allotment will be on the basis of CRL which is also called as the common rank list.

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Aditya kumar Student Expert 23rd Oct, 2019
Hello there!


Iiser Aptitude test can be considered somewhat similar to JEE Main exam , but surely its level is below JEE Advanced.One thing which makes iiser aptitude test tougher is that it also asks questions from biology as jee doesnt include questions from biology , so it becomes difficult for pcm students. The iiser aptitude test consists of 4 subjects; Physics, Chemistry, Maths & Biology. Each section has 15 Multiple Choice Questions and the time duration for all the section is 180 minutes.Every correct response fetches (+3) marks. (-1) mark is deducted for an incorrect response. No marks are deducted for unattempted responses.


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Dear Aspirant,

No good college will give you admission in their CSE branch as the minimum requirement of all the good colleges is at least 60% in their +2. Moreover, all the good colleges needs a minimum of 75% in the +2 for their engineering courses including IITs and NITs.

If you are going to take admission in any normal below average private college then you might not get a package above 2 lakh per annum. CSE is a top trade only if done from some top college, otherwise you'll struggle for job.

Hope this helps. Thank you.


sir how to apply jee mains application for telangana open school

Aditi Amrendra Student Expert 23rd Oct, 2019

Its is very simple.
Fill the Application Form.
Upload Scanned Images.
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I am providing you a link where you can fill jee mains application form.


If you face any problem while filling the form, just refer to the below link for help in filling your jee application form:


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How can I crack jee mains 2020(both the exams) By working from now.. And what kind of timetable should I stick to?

Aditi Amrendra Student Expert 23rd Oct, 2019

There is no ideal time table.Everyone has a different time table.And also there is no stanadard startegy that will syrely help you to crcak jee.All you have to do is that have you own strategy.Make a time table always stick to it.Be consistent and have determination.Have right sources and PYQ papers for practice.Join online mock test.Work on you waek areas and strenthen you strong areas.
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Anupam Kumar Student Expert 23rd Oct, 2019
Hey there

Getting 150+ is a little tough you start preparing from now honestly. See if you have atleast prepared good for boards and know basic PCM then you can get 150+ which requires hard work. First you need to memorize the whole NCERT of chemistry thoroughly. If you had done coachings then revise the notes properly and solve as many questions as you can from chemistry make it your strong subject because it's scoring and can give you 60+ easily if you'll be honest with the subject. Then maths is also scoring for that solve rd Sharma, coaching notes and materials and advance previous year questions. Same for physics. In physics 12th portion is important as it will come in boards also. Rotation is important. But Careers360 has launched a wonderful platform which you can use to get 150+ in April exam if you follow the platform whole heartedly.
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