Massive Update on JEE Main 2022 Topper Selection Criteria
NTA devises a Tie-breaker formula to decide AIR Rank
The National Testing Agency (NTA) has devised a tie-breaker formula on the basis of which there will be only 1 student on AIR-1 this year. Taking a cue from the JEE Main 2021 results, in which there were 18 students on AIR-1, NTA has set such criteria as per which in case of a tie in the NTA score of two students, there will be only one student on every AIR Rank. Amit Ahuja, career counseling expert, Allen Career Institute, said that there are some students who are getting perfect scores of 300 marks out of 300 in JEE-Main Session 1 (June Attempt). Similarly, in the JEE Mains 2022 Session 2 (July attempt) also, there may be some students who may have 300 marks out of 300. In order to calculate the All India Rank of such students in JEE Main 2022, the ascending order of age and JEE-Main application number will be determined as the last criterion.
In a situation of a tie of NTA scores among the students, the top All India Rank (AIR) ranks of the students, who have 300 marks as well as 100 percentile, will be released on the basis of age and JEE-Main application number. In such a situation, the possibility becomes absolutely minimal that with the age of two students, the application number is also the same.
Along with the NTA score and AIR Rank, the eligibility for JEE-Advanced will also be released. JEE-Main was conducted from June 24 to 30 and the July session from July 25 to 30, in which more than 9 lakh candidates appeared.
9 Criteria for determining AIR after a tie in NTA score
Amit Ahuja said that “For the first time this year, 9 criteria have been set for determining the rank in case of tie in the higher NTA score of the students”. If the total NTA score of two students is the same, then the NTA score for Mathematics will be seen first in the All India Rank determination. If the NTA score in Mathematics will also be the same then the NTA score of Physics, followed by Chemistry, will be seen. If the NTA score of all the 3 subjects is the same then the ratio of the number of correct and incorrect answers will be seen. In this case, the ratio of the number of correct and wrong answers in the Mathematics subject will be considered first. If there will be tie in the ratio of correct and wrong answers in Mathematics, then the ratio of the number of correct and wrong answers in Physics will be checked. In the case of tie in Physics answers, the ratio of a number of correct and incorrect answers in Chemistry will be seen. In case of tie in all the above criteria, the student whose age is more will be given preference in All India Rank. At the level of age criterion also, if a tie situation is formed, priority will be given to the ascending order of application number of the aspirant. Since the application of 2 aspirants cannot be the same so topper rank will be decided.
JEE-Advanced application process from August 7
Ahuja informed that this year’s JEE-Advanced exam is being conducted by IIT Bombay on August 28. The top 2.5 lakh students selected on the basis of JEE-Main will be declared eligible for the advanced examination. These students can apply for Advanced on the JEE-Advanced website from August 7 to 11. During the application, the students have to upload the 10th and 12th mark sheet certificate and category certificate by filling in their required information and examination center. The students were given the opportunity to choose from 8 examination centres during the application. During the application, students will have to give OBC and EWS certificates after April 1, 2022. Such students who do not have OBC and EWS certificates available within the given time, then they can apply through declaration. The declaration format can be downloaded from the information bulletin of JEE-Advanced. The result of the advanced examination will be released on September 11.
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