Your browser does not support iframes.
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!
Latest : [ IIT/JEE 2020 Online Preparation – Crack JEE 2020 with JEE Knockout Program ( AI-Based Coaching), If you Do Not Quality- Get 100% MONEY BACK. Know More
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
Crack JEE 2020 with JEE Knockout Program, If you Do Not Quality- Get 100% MONEY BACK
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.
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.
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.
JMI B.Tech Counselling 2020 - Candidates will be able to apply for JMI B.Tech 2020 counselling in...
DAIICT B.Tech Admission 2020 - Candidates will be able to apply for DAIICT B.Tech admission 2020 ...
JMI B.Tech Admission 2020 - Candidates will be able to apply for JMI B.Tech admission 2020 tentat...
IIIT Hyderabad B.Tech Counselling 2020 - IIIT Hyderabad will conduct the B.Tech counselling 2020 ...
Merit list of IIIT Hyderabad will be formulated on the basis of applicants marks in entrance exam...
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.
Your JEE Main brochure has been successfully mailed to your registered email id .
The Question containing Inaapropriate or Abusive Words
Question lacks the basic details making it difficult to answer
Topic Tagged to the Question are not relevant to Question
Question drives traffic to external sites for promotional or commercial purposes
The Question is not relevant to User
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile