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JEE Main 2026 Percentile Calculator: Aspirants can use the JEE Main 2026 percentile calculator based on the marks in the exam. The rank of the candidates is based on their performance compared to other candidates in the exam. NTA employs a normalisation formula to derive the percentile from raw marks. Candidates can use the JEE Main 2026 percentile predictor for free to get the JEE Main exam score. To calculate their marks, candidates can refer to the JEE Main 2026 final answer key to estimate their probable percentile. Also, check the JEE Main 2026 marks vs percentile for test performance. The percentile score is calculated using a normalisation process. The score helps in assessing a candidate's overall standing.
Also Check: JEE Main 2026 Score Calculator Link | How to Calculate JEE Main Marks From Answer Key?
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Please note that the JEE Main 2026 ranking will be based on the percentile scores, rather than just raw marks. The authority will release the JEE Main 2026 cutoff online. The JEE Mains 2026 percentile release date for session 2 is April 20, 2026. Check JEE Main 2026 marks vs percentile vs rank with expected trends, cutoff analysis, shift-wise data, below.
To get into the calculation of the percentile, it is important to understand what a percentile is before you get into the calculation. A percentile is a statistical measurement that is applied to show the performance of a student compared to other students. It is a figure between 0 and 100, where:
The percentile score is calculated using the following formula

This formula indicates the percentage of students who scored lower than you.
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The normalisation of scoring is used to give the percentile of the JEE Main. NTA employs a method called normalisation to account for the difference in the level of difficulty in different shifts.
| Particulars | Details |
|---|---|
Total Percentile (T1P) | (100 x No. of candidates from the session with a raw score equal to or less than T1 score) / Total No. of candidates who appeared in the session |
Mathematics Percentile (M1P) | (100 x No. of candidates appeared from the session with a raw score equal to or less than the M1 score in Mathematics)/ Total No. of candidates who appeared in the session |
Chemistry Percentile (C1P): | (100 x No. of candidates appeared from the session with a raw score equal to or less than C1 score in Chemistry) / Total No. of candidates who appeared in the session |
Physics Percentile (P1P) | (100 x No. of candidates appeared from the session with a raw score equal to or less than P1 score in Physics) / Total No. of candidates who appeared in the session |
| Percentile | 2 Apr S1 | 2 Apr S2 | 4 Apr S1 | 4 Apr S2 | 5 Apr S1 | 5 Apr S2 | 6 Apr S1 | 6 Apr S2 | 8 Apr S2 |
|---|---|---|---|---|---|---|---|---|---|
99 | 176–180 | 181–185 | 186–190 | 191–195 | 171–175 | 166–170 | 151–155 | 161–165 | 156–160 |
98 | 156–160 | 161–165 | 166–170 | 171–175 | 151–155 | 146–150 | 131–135 | 141–145 | 136–140 |
95 | 121–125 | 126–130 | 131–135 | 136–140 | 116–120 | 111–115 | 96–100 | 106–110 | 101–105 |
90 | 86–90 | 91–95 | 96–100 | 101–105 | 81–85 | 76–80 | 61–65 | 71–75 | 66–70 |
| Percentile | 21S1 | 21S2 | 22S1 | 22S2 | 23S1 | 23S2 | 24S1 | 24S2 | 28S1 | 28S2 |
|---|---|---|---|---|---|---|---|---|---|---|
| 99 | 167 | 171 | 158 | 155 | 158 | 163 | 162 | 160 | 161 | 162 |
| 98.5 | 154 | 157 | 144 | 143 | 145 | 150 | 150 | 146 | 149 | 150 |
| 98 | 144 | 147 | 134 | 131 | 135 | 139 | 141 | 137 | 138 | 140 |
| 97.5 | 137 | 139 | 126 | 126 | 127 | 132 | 134 | 128 | 132 | 133 |
| 97 | 130 | 132 | 120 | 120 | 121 | 126 | 127 | 121 | 126 | 126 |
| 96.5 | 124 | 127 | 115 | 115 | 116 | 120 | 123 | 115 | 121 | 123 |
| 96 | 120 | 122 | 110 | 111 | 111 | 115 | 118 | 110 | 114 | 116 |
| 95.5 | 115 | 117 | 106 | 107 | 107 | 111 | 113 | 104 | 112 | 112 |
| 95 | 110 | 113 | 101 | 104 | 103 | 106 | 109 | 101 | 109 | 108 |
| 94 | 105 | 106 | 93 | 97 | 96 | 100 | 102 | 93 | 102 | 103 |
| 93 | 99 | 100 | 88 | 92 | 89 | 94 | 96 | 88 | 96 | 95 |
| 92 | 94 | 93 | 84 | 88 | 86 | 88 | 92 | 81 | 91 | 91 |
| 91 | 90 | 89 | 79 | 84 | 81 | 83 | 86 | 77 | 86 | 86 |
| 90 | 85 | 84 | 74 | 81 | 77 | 79 | 80 | 73 | 82 | 85 |
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Candidates can check the expected marks vs percentile for JEE Main 2026 in the table below.
| Marks | Percentile |
|---|---|
292 - 281 | 99.99890732-99.99635773 |
280-271 | 99.99617561-99.99153171 |
268-261 | 99.99034797-99.98087806 |
259-241 | 99.97687156-99.92078054 |
240-231 | 99.91549924-99.87388626 |
230-221 | 99.87060821-99.793392 |
220-211 | 99.78191884-99.70206198 |
210-201 | 99.69159044-99.57959052 |
200-181 | 99.57503767-99.19970133 |
180-161 | 99.17311273-98.563305 |
160-141 | 98.52824811-97.59527961 |
140-121 | 97.54301298-96.12871797 |
120-101 | 96.0687115-93.89928202 |
100-81 | 93.8020333-90.4734501 |
80-61 | 90.27631202-84.22540213 |
60-41 | 83.89085926-70.26839007 |
40-21 | 69.5797271- 38.687975234 |
20-11 | 36.58463962-19.33774352 |
Thus, the formula to calculate the probable JEE Main 2026 rank using the percentile for the January session is as follows:
JEE Main 2026 rank (probable) = (100 – NTA percentile score) × 823,967 / 100
For instance, if your NTA percentile score is 99.78, your JEE Main rank would be (100 – 99.78) × 823,967 / 100 = 1,812,76.
In case of a tie in percentile score, the merit of two or more candidates will be decided by the following criteria, in any order:
Applicants who score higher percentile in Mathematics will have the upper hand.
In case of the tie, those candidates who have scored high percentile scores in Physics will be taken into account.
In case the tie is not broken, candidates who have scored higher percentile marks in Chemistry will be selected.
In case the tie has not been broken, preference will be given to older candidates.
To obtain an estimated rank on the basis of the raw scores, the Candidates can use JEE Main 2026 Rank Predictor and Percentile Calculator which are free to use. Rank Predictor is a formula that involves the calculation of approximate ranks through the marking scheme (4 marks per correct answer, -1 per incorrect answer), and is applied to assist students in estimating their performance.
Frequently Asked Questions (FAQs)
To calculate your JEE Main percentile, use the formula: (100 × Number of students who scored less than you) / Total number of students in your session
To estimate your JEE Main rank, subtract your percentile score from 100 to determine the percentage of candidates who scored higher than you. Then, multiply this value by the total number of students who appeared in the January session (13,00,368) and divide the result by 100.
NTA calculates raw marks into percentile marks by eliminating marks of students who scored equally or less during the same shift.
The JEE Main 2026 Percentile result date for session 2 is April 20, 2026.
On Question asked by student community
Hello Aspirant,
With JEE main rank of 8,19,756 General (Delhi) Category and 68% in class 12 getting CSC or it in top IPU colleges like BPIT, VIPS, BVCOE, Akhilesh Das, or HMRITM is unlikely through the regular counselling rounds.
However, you should still participate in all IPU counselling rounds including
Hello,
Based on your JEE Main Paper 2 percentile (96.3), CRL 2566, NATA percentile (88.69), and Class 12 score (88.8%), you have a good academic profile for B.Arch admissions.
Since COMEDK has its own merit preparation for B.Arch admissions, your exact rank cannot be predicted before the merit list is
Hello Student,
With your rank og 129537 as an OBC candidate, it is highly unlikely that you will get a seat in core branches. There is a possibility, though, that you might get a seat in lower-tier NITs or GFTIs, in branches such as Biotechnology , Mining, Metallurgy and Civil
Hello Dear Student,
Neither branch is universally "better"; it depends entirely on your interests.
Computer Science Engineering (CSE) focuses on software, coding, and algorithms, and generally yields higher immediate IT salaries. Electronics and Communication Engineering (ECE) focuses on hardware, microchips, and circuits, offering a dual-advantage of core hardware jobs +
Hey there,
With 94.95 percentile in JEE Main, Rajasthan home state, and REAP OBC-NCL merit rank 202, you have a very good chance of getting B.Tech CSE at MBM University, Jodhpur. While admission cannot be guaranteed until the official seat allotment is announced, your profile is competitive based on previous
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