In the year 2017 November, the National Testing Agency was set up. This was formed majorly to conduct entrance exams such as JEE Main, GPAT and also CMAT.
JEE Main exam is conducted twice a year in online mode. It is a computer-based test. Earlier, JEE Main was conducted only once a year. Due to the total number of students who are applying for the exam is increasing gradually, because of which the conducting board has decided to conduct it twice.
The exam conducting board NTA also releases mock tests on their official website to make the students understand the examination pattern and get familiar with the exam environment.
Council of Architecture has suggested NTA regarding the cut off percentile of JEE Main paper II. The students who want to pursue Architecture, they’ll have to secure at least 50% of marks in any of the qualifying examinations with physics, chemistry and mathematics as their primary subjects.
The official board has also set up (TPC’s) Test practice centres for taking up free mock tests, and these centres are basically schools and engineering colleges. The number of TPCs is approximately 3400.
Expert teachers will be preparing a huge data bank from which unique papers will be assigned to every student.
The JEE Main Rank List for JoSAA counselling will be prepared in May.
The rank list of JEE Main will be released by the National Testing Agency after the announcement of results. This is because, as the exam is conducted twice, many candidates would have applied for JEE Main January as well as JEE Main May. Based on the candidates’ percentile or score, the rank or the merit list of JEE Main will be produced. As previously mentioned, to calculate the merit list or the rank list, the best of the two will be considered. The lowest score secured will be ignored.
For both the papers 1 and 2 the JEE Main score will be calculated differently. The procedure for calculation has been mentioned below.
60 MCQ Questions Marking Scheme
For every correct answer, the student will be awarded 4 marks, and for every incorrect answer, 1 negative mark will be deducted. There will be no deduction of marks if you do not answer a question.
Numerical Type Questions Marking Scheme
For every correct answer, the students will be awarded 4 marks. There will be no deduction of marks if they do not answer the question.
The NTA will use the following method to calculate the percentile score of a candidate:
100 X Total number of students appeared for that particular session along with the raw score ‘Equal To Or Less’ than the students / the total number of the candidates appeared in that particular session.
Note: The percentile of JEE Main will be calculated up to 7 decimal places.
Ranks of the candidates are prepared on the basis of their JEE Main 2020 percentile score as mentioned above. In cases where more than two candidates obtain the same marks then in such cases, inter-se-merit guidelines shall be followed to allot a rank. In such cases, the ranks will be allotted based on the below-given criteria.
The student who secured a higher percentile in Mathematics will be allocated a higher rank.
If the tie still remains, the candidate who scored higher percentile in Physics shall be allocated higher rank.
If the tie still remains unresolved, then the candidate who scored higher percentile in Chemistry shall be allocated higher rank.
If the tie still persists then the candidate who has procured less negative marks will be allocated higher rank.
If the tie still remains, the applicant who is older shall get a higher rank.
Paper 2 of JEE Main will be directed in two different shifts. Initially, the candidates who have appeared for the examination in the first attempt will be receiving the raw marks. These raw of all candidates shall be declared. This list will have the accurate raw marks of the candidates which have been released for paper 2 JEE Main 2020 for the first attempt.
The rank list of paper 2 shall be drafted only after combining the first and the second attempt raw scores. If in case, the candidate has appeared in both the attempts examination, then the best of two exam marks will be taken into consideration. The JEE Main Paper 2 Rank List will be prepared after taking into consideration the marks of both attempts, the All India Category rank as well as All India Rank. Suppose, if there is a tie between two or more candidates, the below-mentioned procedure would be followed to come to a conclusion.
The student who secured the highest marks in the aptitude test in Paper 2 will be allotted a higher rank.
If the tie still persists the candidate who scored higher marks in the drawing test.
If the tie still remains, the applicant who is older will be allocated a higher rank.
Even after considering all the above mentioned, then the candidates will be allotted with the same rank.
JEE Main Common Rank List Important Points
The students are categorised under General, SC, ST, OBC – NCL, General - PwD, OBC - NCL - PwD, ST – PwD and SC - PwD. The candidate’s category, along with his/her performance, will be considered before the preparation of the common rank list.
The common rank list or the CRL will be the basis for the allotment process.
JEE Main will have two different rank lists. The first one will be the common rank list, and the second is the category rank list. In the category rank lists, each list will comprise of candidates with the same category and their ranks based on their exam performance.