The JEE Main 2020 results are just around the corner and most students would try to understand how to calculate the percentile and rank for all of India.
The outcome of JEE Main 2020 would be determined through the normalization process. The exam was conducted on different days and sessions. It is quite possible that some candidates will get an easy job, while others will have to answer more difficult questions on the JEE Main 2020 exam. To ensure that no applicant benefits or is disadvantaged by this, the standardization process would be adopted for determine the main JEE results and ranges. NTA would rank students based on the percentile score obtained on the test. This percentile score would be calculated based on a predetermined formula.
What is the percentile score?
The percentile score would be based on the relative performance of all applicants appearing on the exam. The score would be obtained after transforming the scores on a scale ranging from 100 to 0 for each test session.
According to the official NTA website, “The percentile score indicates the percentage of candidates who scored at or below (raw scores equal to or below) that particular percentile on that exam. Therefore, the topper (highest score) of each session gets the same percentile of 100 that is desirable. Scores between the highest and lowest scores also become appropriate percentiles. “
Difference between percentage and percentile
The percentile and the percentage of grades are two totally different entities. Therefore, applicants should not confuse the two terms. This is the difference:
Percentage: The percentage is a number of 100
Percentile : A candidate’s percentile score would reflect how many candidates scored lower than that candidate on an exam. A percentile score is the value below which a certain percentage of observations fall.
Declaration of final result of JEE Main 2020 and all ranges of India
NTA ran JEE Main 2020 in two attempts: January and September. Since the exam was conducted in multiple shifts each day throughout the two sessions, the calculation of the score for a single percentile score is done using a normalization process.
What is normalization?
Normalization is the process of identifying a candidate’s actual merit / rank by creating the same playing field for all students. The same process is used in other exams across the country to solve similar problems.
The formula for calculating the JEE Main 2020 scores is:
NTA, to avoid the clustering effect and reduce the tiebreaker, computes the percentile scores to 7 decimal places.
Steps involved in standardizing and preparing the JEE 2020 core list of merits / ranks:
Step 1: Distribution of examinees over two days and two shifts per day
The candidates would be distributed over four sessions at random, so that each session has roughly the same number of candidates. These four sessions are as follows:
• Session 1: Turn 1 of day 1
• Session 2: Turn 2 of day 1
• Session 3: Turn 1 of day 2
• Session 4: Turn 2 of day 2
Step 2: Elaboration of results for each session
The JEE Main 2020 results for each session would be prepared in the form of:
• Raw scores
• Separate percentile scores for each of the three subjects (Mathematics, Physics, Chemistry) and the total score.
• The following four percentiles would be calculated for each candidate in one session:
Step 3: compiling the NTA score and preparing the overall list of merits / ranks
The percentile scores for the total raw score from the four sessions calculated in the previous step would be merged and known as NTA scores. This would then be used for the compilation of results and the preparation of the JEE 2020 main merit overall list / ranking.
Note that the percentile for the four sessions would be calculated separately to determine the total raw score and the raw scores in three subjects: Mathematics, Physics, and Chemistry.
Build and View JEE Result 2020 for Multisession
Compilation of the JEE Main 2020 output for the first attempt:
The first attempt of JEE Main 2020 took place in multiple sessions, so the NTA scores would be calculated corresponding to the raw grades obtained by a student in each session according to the above process. The calculated NTA scores for all sessions would be merged to report the JEE Main 2020 result. The result would comprise the four NTA scores for each of the three subjects: Mathematics, Physics, and Chemistry along with the total from the first attempt.
Compiling the output of JEE Main 2020 for the second attempt:
The same method given above would be followed to compile the main JEE 2020 results for the second attempt.
Compilation of results and preparation of list / ranking of merits
The four NTA scores from each candidate’s first and second attempt would be combined for the compilation of the main JEE result and the preparation of the overall merit list / ranking. The candidates who appeared in both attempts, their best score from the two NTAs would be considered for the preparation of the list of merits / rankings. The method is explained in the following example:
JEE 2020 Main Result: Tiebreaker Resolution
In the event that two or more candidates achieve equal percentile scores in the JEE Main 2020 result, the ranking will be determined as follows:
Tiebreaker for Test 1 (BTech):
• NTA score in Math: the candidate who scores higher than the percentile in Math would be ranked higher
• NTA Score in Physics: the candidate who achieves a higher percentile score in Physics would be ranked higher
• NTA score in Chemistry: the candidate with the highest percentile score in Chemistry would rank higher
• Date of birth: the oldest candidate would occupy a higher position
Note: In the event that the tie remains unresolved, the candidates are awarded the same rank.
JEE Main 2020 tiebreaker for paper 2 (BArch):
• NTA Mathematics Score: Candidate who scores higher in Mathematics will receive a higher rank
• NTA score on the aptitude test: then if the tie still persists, then the applicant with a higher score on the aptitude test will be assigned a higher rank
• NTA Score on Drawing Test: If the tie still persists, applicants with higher scores on the Drawing test will receive a higher rank.
• Fewer negative response: candidates have fewer negative responses at work.
• Date of birth: the oldest candidate would occupy a higher position
JEE Main 2020 Tiebreaker for Test 3 (B. Planning):
• NTA Score in Math: Applicants who score higher in Math will receive a higher rank
• NTA score on the aptitude test: if the tie persists, then the applicant with the highest scores on the aptitude test will receive a higher rank
• NTA score on planning-based questions: if the above criteria do not break the tie, then the candidate would score higher on planning-based questions would be given the highest ranking
• Fewer negative response: candidates have fewer negative responses at work.
• Date of birth: the oldest candidate would occupy a higher position
What happens after the JEE Main 2020 result?
• Cut for JEE Advanced
NTA would announce the cutoff score for JEE Advanced. In 2019, the cut-off score for students in the overall category was 89.75, while that of OBC – NCL students was 74.
Selection / advice procedure
After the JEE Main and JEE Advanced results are reported, JoSAA conducts a joint process of assigning positions for admissions to 100 institutes, including 23 IITs, 31 NITs, 25 IIITs, and 28 other GFTIs. Admission to all of these participating JEE Main and JEE Advanced institutes would be through a single advisory platform.
(The author of this article is a FITJEE expert. Opinions expressed here are personal).
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