How to calculate JEE Main 2026 rank based on Percentile score? - Candidates can check this page for the way to calculate the JEE Main rank from percentile score. Careers360 has derived the JEE Main 2026 rank from the percentile based on the previous year's statistics. The percentile score is a measure to know the percentage of candidates who have marks equal to or less than. NTA uses a JEE normalisation method to get the percentile from raw marks. Those with higher percentile are assigned higher rank in JEE Main. By knowing how to calculate the JEE Main rank based on percentile, candidates get a chance to judge their test performance. Also, they can use the JEE Main rank predictor tool. Note that the final JEE Main result 2026 will mention the rank.
Also Check: JEE Main marks vs percentile
With the help of expected rank, aspirants can also check the chances of admission to desired colleges through JEE Main college predictor. Candidates can check this article to know how to calculate JEE Main 2026 rank based on percentile score.
JEE Main 2026 result date will be released online. Since the authority conducts two sessions of JEE Main exam, the final ranks announced through JEE Main 2026 session will be final. The institutes consider this rank during the JoSAA counselling. Below is the process of calculating rank from percentile using a formula provided by coaching institutes and an example. As per the previous year's NTA JEE Main statistics, 14,75,103 students appeared for JEE Main in the January session. NTA had announced that 24 students have scored 100 percentile in 2025 and have declared them as JEE Main Toppers.
To calculate the JEE Main rank, you first need to know the percentage of candidates with higher percentile scores than you. To calculate that, the formula would be-
100 (highest score) - your percentile score.
Since the percentile is calculated from 100 to 0 for each session of examinees, the formula applies the total number of students in the session which is-
14,75,103 /100 * percentage of people ahead of you to arrive at the rank.
Therefore, the formula to calculate the JEE Main rank through percentile is-
JEE Main rank (probable) = (100- NTA percentile score ) X 14,75,103 /100
*This formula would not be for a 100 percentile score as those are the top ranks.
Suppose the total number of candidates who appeared for JEE Main exam are 823967, and then the approximate JEE Main rank can be-
If the NTA percentile score is 99.999, JEE Main rank will be (100-99.999 ) X 823967/100 = 914 (approx.)
Consider the NTA score of 80.60 then JEE Main rank = (100-80.60) X 823967/100. The rank will be 159849
Note: It should be undertood that the formula given for the calculation of percentile in JEE Main is indicative and should not be taken as accurate as the actual ranks will vary on account of many factors. For the actual rank, candidates should check the alloted rank on the JEE Main 2026 scorecard.
Until the result is declared, aspirants can go through the expected JEE Mains percentile vs rank based on the previous year statistics to get an idea of how to calculate the JEE Main rank from the percentile.
JEE Main Percentile | JEE Main Rank |
100 | 1 |
99.994681 - 99.997394 | 56-25 |
99.988780 - 99.994681 | 115-55 |
99.956404 - 99.988551 | 402-115 |
99.901123 - 99.956364 | 978-401 |
99.795063 - 99.901123 | 2001 — 978 |
99.573193 - 99.782472 | 3901 - 2001 |
99.239937 - 99.782472 | 7003 - 3901 |
98.732389 - 99.239937 | 12200 - 7003 |
96.978108 - 98.7322 | 210010 -12200 |
96.064850 - 96.978108 | 35000 - 21010 |
Candidates can take a probable calculation of percentile in JEE Main January 2026 session and check the rank. Based on your performance in the JEE Main January session, you can decide if you want to attempt the April session or not.
According to the total number of registrations in JEE Main 2026 session 1, the expected number of candidates who may appeared for the JEE Main January session are around 14 Lakhs. Using this as a basis, use the NTA score and see what is the JEE Main rank.
The reason is that the authority will consider the better of the two session's scores is to provide benefit to aspirants. So if any candidate performs badly in January, they can take another attempt in April. The January NTA score will be counted for determiing the rank if they would have performed better in that session or vice-versa.
General category candidates with an NTA percentile score of 89 and above are expected to have better chances to qualify for JEE Advanced if one takes a look at the JEE Main cutoff from last year. It is a fact that IITs are dream institutions for most engineering aspirants. So a student who has done exceptionally well in January session of JEE Main may wish to skip the JEE Main April session to concentrate on JEE Advanced preparation. Another student may have scored well enough and feels that another session is too strenuous so may wish to skip it if only he/she has a fair chance of a seat in one good NIT or IIIT. Others may wish to know if they have the chance for any admissions based on their January JEE Main percentile scores or if should they appear for the April exam to better this.
We hope that the decision to write or skip can be made a bit easier using the calculation methods mentioned in this article.
Frequently Asked Questions (FAQs)
JEE Main 2026 normalisation process will be done by the authority to ensure all candidates' scores are at an equal level, regardless of the difficulty level of the session.
The JEE Main 2026 result date will be announced.
Aspirants can use a formula to convert a percentile to rank in JEE Mains as given in the article. However, the said formula provides only a probable rank.
On Question asked by student community
Hello
Yes, it creates a problem if you're 12th LC(state OBC) and JEE(Central OBC/EWS) categories differ.
JoSAA requires a central OBC-NCL certificate for OBC reservation; since Karnataka OBC isn't central, you will be treated as general, or you can use a Declaration for OBC-to-General conversions from during counseling, but switching to EWS needs you to have applied as EWS initially. Your EWS certificate works if you meet the income criteria, but yes the important thing is Central OBC list for OBC, not state list.
Hope it helps you, in case of any doubts you can directly drop your query or you can visit to Careers360.com
Hello,
Yes, in JEE Mains, 95 percentile and above is good, and you can get admission in mid to upper-tier NITs. Here is the list of some NITs where you can get admission.
1. NIT Agartala
2. NIT Raipur
3. NIT Durgapur
4. NIT Puducherry
Thank you.
hello,
The link to the most relevant chapter of JEE Mains is attached herewith. You can also find the sample papers with an answer key, which will help you analyse your in-depth performance. Careers360 gives every aspirant an opportunity for a free mock test. the registration is going on. The last date of registration on 8th January.
https://engineering.careers360.com/articles/most-important-chapters-of-jee-main
Thank you.
Hello,
The exam date of the All India JEE Mains mock test season one is in mid to late January, and season two will be held in early to mid-April. You can enroll for the Careers360 free mock test. The last date on 8th January, 2026. The mock test question papers are set chapter-wise wise which can help you analysing your in-depth performance.
Best Regards.
Hello,
You passed your Class 12 (Intermediate) from UP Board in
2024
.
As per
JEE Advanced rules
, a candidate can appear
only in the year of Class 12 passing and the next consecutive year
Eligible years for you were 2024 and 2025
You already appeared in 2025
2026 is not allowed
Your health issue does not change this rule.
You can still take JEE Main again in 2026 , but not JEE Advanced.
Hope it helps !
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