JEE Main Normalization Procedure 2025: NTA (National Testing Agency) conductsJEE Main 2025. With the introduction of the newly formed exam conducting body, major changes have been done at the procedure of JEE Mains. Every year, around 14 lakh candidates appear for the exam from various State and Central Boards across the country. The selection of the candidate is based on JEE Main marks and counselling. NTA has released the result online on the official site. Candidates can click on the link below to check.
Check Eligibility For NTA JEE Main Exam
What Is JEE Main Normalization Procedure 2025?
The often conflicted issue is the difficulty level of the exam which varies in each session which results in some of the candidates ending up with a difficult set of papers. Naturally, the candidate with a more difficult set of exams scores less in the exam in comparison to candidates getting an easier set of the question paper. The JEE Main 2025 will take place in February, March, April, and May. The exam of all four sessions has been over.
How To Calculate Percentile In JEE Main 2025?
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After the declaration of the JEE Main 2025 February exam, candidates can check their expected rank from the formula provided below. The candidate will have to multiply their percentile to 100 which will be divided by 100. The provided score will be multiplied again with the total number of candidates who appeared for the exam. The candidate must note that the below-provided formula will only provide the estimated rank of the respective session 2025. The rank might vary after the April exam. The declaration of the final rank will be released by the National Testing Agency (NTA) after JEE Main Feb/March/April/May 2025 exam.
Details Of JEE Main Normalization Procedure 2025
The National Testing Agency (NTA) will release the JEE Main result for each session in the form of a normalized score. This will help in resolving issues of discrepancy in the difficulty level of each session. To understand how the normalization procedure takes place, read further.
For Example, there two students, candidate A, and candidate B.
- Candidate A has taken JEE Main 2025 in shift 1 one and scored a raw mark of 340 out of 360. He is the highest scorer among all the shift 1 candidates where 56789 candidates appeared.
- Candidate B has taken JEE Main 2025 in shift 2 and scored 310 out of 260. She is the highest scorer among all the shift 2 candidates where 54689 candidates appeared.
Now, considering only the raw scores of these two candidates to prepare the merit list, candidate A will be the topper with a higher rank. However, if shift 2 is more difficult than shift 1 then this will be unfair for candidate B who took a more difficult paper and got fewer scores.
This is where JEE Main Normalization Procedure comes in. Let us see how the ranks change with the normalization procedure.
Shift 1: Total candidates who appeared = 56789
Candidates who scored equal to or less than the topper of shift 1 = 56789
Percentile Obtained = 100 (56789/56789)
Percentile Obtained = 100
Shift 2: Total candidates who appeared = 54689
Candidates who scored equal to or less than the topper of shift 2 = 54689
Percentile Obtained = 100 (54689/54689)
Percentile Obtained = 100
This shows us that although both candidates had a different score, based on their difference of difficulty level, both got the same percentile which is 100.
Now let us take 4 more candidates C, D, E, and F.
- Candidate C & D have a raw total score of 330 and secured the 2nd position in shift 1.
- Candidate E has a raw total score of 304 and secured the 3rd position in shift 1.
- And candidate F has a raw total score of 304 and secured the 2nd position in shift 2.
If the raw score of candidates is taken into consideration, the merit list will be something like this:
| Candidate A 340 | 1st Position |
| Candidate B 310 | 3rd Position |
| Candidate C 330 | 2nd Position |
| Candidate D 330 | 2nd Position |
| Candidate E 304 | 4th Position |
| Candidate F 304 | 4th Position |
If we consider these ranks of the candidate, then it will be unfair for Candidate F to be in the 4th position as shift 2 was difficult than shift 1 and also his rank is equal to candidate E. This happens because of the difference in the difficulty level of the various shifts held.
If we change these scores to percentile scores, the following will be the result:
Percentile of candidate A = 100 (56789/56789) = 100
Percentile of candidate B = 100 (54689/54689) = 100
Candidate C percentile= 100 (56788/56789) = 99.99823909
Percentile of candidate D = 100 (56788/56789) = 99.99823909
Percentile of candidate E = 100 (56786/56789) = 99.99471728
Candidate F percentile= 100 (54688/54689) = 99.99817147
Based on this percentile score, the ranks awarded will be something like this:
| Candidate A 340 | 1st Position |
| Candidate B 310 | 1st Position |
| Candidate C 330 | 2nd Position |
| Candidate D 330 | 2nd Position |
| Candidate F 304 | 3rd Position |
| Candidate E 304 | 4th Position |
This is how the normalization procedure will take place and candidates will be awarded a rank after their score is normalized and changed into a percentile score.
National Testing Agency is using the policy of ‘JEE Main Normalization Procedure based on percentile Score’ to overcome such problem of the unequal gain of marks. The Normalization Procedure was utilized earlier also for equalization of class 12th marks to bring all the boards on the same level. Now the normalization Procedure will be used for the JEE Main score which will be based on Percentile Score to ensure that the level of exam is equal for every candidate appearing for JEE Main 2025. Normalization is done to make sure there is no uneven difficulty level in the exam.
NTA’s New Percentile Score Based On JEE Main Normalization
Percentile Score: Percentile Score is the marks obtained by the candidate and is transformed into a scale ranging from 0 to 100. The purpose is to compare the scores of the candidates in multiple sessions. Percentile Score will be the normalized score of the examination.
JEE Main Normalization Procedure 2025 as per NTA
How Will The Procedure Of JEE Main 2025 Normalization Work?
- Candidates should know that the raw scores of candidate secured in JEE Main 2025 based on the marking scheme will be calculated.
- Each correct answer will add up the 4 marks and a negative marking of 1 marks will be for each incorrect answer.
- The calculation of the overall score of JEE Main 2025 is by comparing the raw score of the candidate in comparison to all the candidates.
- The percentile secured by the candidate means the percentage of the candidates who have secured less than you in the JEE Main exam or the percentage of candidates from whom you have secured above in the exam.
- The preparation of the result of JEE Main 2025 is based on a raw score and the percentile score of Total and all the three subjects (Physics, Chemistry, Mathematics).
Total Percentile (T1P)
Let T1, M1, P1, C1 be the raw scores in Total, Mathematics, Physics, Chemistry of a candidate and T1P, M1P, P1P, C1 be the Percentile Scores of Total, Mathematics, Physics, Chemistry of that candidate.
Compilation & Display of Result For Multi-Session Papers of JEE Main 2025
Since the examination of JEE Main 2025 will be held twice this year by the National Testing Agency, it is important for the aspirants that they are aware of the procedure and the calculation of the Result by the examination authorities. Given the difficulty level of this exam, the scores of the candidates from both the attempts will be taken for the final result and the preparation of the Merit List. Explaining this procedure further, we are providing a detailed explanation of the Result calculation and the preparation of the Merit List of JEE Main 2025 in the points below.
Compilation of Result for First attempt:
Since the first attempt will be conducted in multi-session, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per the above procedure. The calculated NTA scores for all the sessions will be merged for the declaration of result. The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the first attempt.
Compilation of Result for Second attempt:
Similarly, the second attempt will be conducted in multi-session, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per the above procedure. The calculated NTA scores for all the sessions will be merged for the declaration of result. The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the second attempt.
Compilation of Result and Preparation of Merit List / Ranking:
The four NTA scores for each of the candidates for the first attempt as well as for the second attempt will be merged for the compilation of results and preparation of the overall Merit List / Ranking. Those who appear in both the attempts; their best of the two NTA scores will be taken into consideration for the preparation of Merit List /Ranking. The example is given below.
Expected JEE Main 2025 Marks Vs Percentile
Expected JEE Main 2025 Marks Vs Percentile| Percentile | Score |
| 0.843517743614459 – 9.69540662201048 | 0 – 10 |
| 13.4958497103427 – 33.2291283360524 | 11 – 20 |
| 37.6945295632834 – 56.5693109770195 | 21 – 30 |
| 58.1514901857346 – 71.3020522957121 | 31 – 40 |
| 73.2878087751462 – 80.9821538087469 | 41 – 50 |
| 82.0160627661434 – 86.9079446541208 | 51 – 60 |
| 87.5122250914779 – 90.7022005707394 | 61 – 70 |
| 91.0721283110867 – 93.1529718505396 | 71 – 80 |
| 93.4712312797351 – 94.7494792463808 | 81 – 90 |
| 94.9985943180054 – 96.0648502433078 | 91 – 100 |
| 96.2045500677875 – 96.9782721725982 | 101 – 110 |
| 97.1429377776765 – 97.6856721385145 | 111 – 120 |
| 97.8112608696124 – 98.2541321080562 | 121 – 130 |
| 98.3174149345299 – 98.6669358629096 | 131 – 140 |
| 98.7323896268267 – 98.9902969950969 | 141 – 150 |
| 99.0286140409721 – 99.2397377073381 | 151 – 160 |
| 99.272084675244 – 99.4312143898418 | 161 – 170 |
| 99.4569399985455 – 99.573193698637 | 171 – 180 |
| 99.5973996511304 – 99.6885790237511 | 181 – 190 |
| 99.7108311325455 – 99.7824720681761 | 191 – 200 |
| 99.7950635053476 – 99.845212160289 | 201 – 210 |
| 99.8516164257469 – 99.8937326121479 | 211 – 220 |
| 99.9011137994553 – 99.9289017987302 | 221 – 230 |
| 99.9349804235716 – 99.9563641573886 | 231 – 240 |
| 99.9601632979145 – 99.9750342194015 | 241 – 250 |
| 99.9772051568448 – 99.9888196721667 | 250 – 262 |
| 99.9909906096101 – 99.9940299220308 | 263 – 270 |
| 99.9946812032638 – 99.997394875068 | 271 – 280 |
| 99.99989145 | 300 |
JEE Main 2020 Marks Vs Percentile
Candidates can refer below to check the approximate percentile based on the jee main scores.
| JEE Main Marks | JEE Main Percentile |
| 360 to 310 | 100 – 99.9872475 |
| 309 to 299 | 99.9855721 – 99.9740296 |
| 298 to 288 | 99.9758912 – 99.9625802 |
| 287 to 278 | 99.9614632 – 99.9442427 |
| 277 to 268 | 99.9409848 – 99.9192962 |
| 267 to 258 | 99.9183654 – 99.8908126 |
| 257 to 248 | 99.8864376 – 99.8484594 |
| 247 to 238 | 99.8451084 – 99.7995904 |
| 237 to 228 | 99.7967979 – 99.7329423 |
| 227 to 218 | 99.7300567 – 99.6547519 |
| 217 to 208 | 99.6488876 – 99.5527320 |
| 207 to 198 | 99.5491948 – 99.4360979 |
| 197 to 188 | 99.4200875 – 99.2895839 |
| 187 to 178 | 99.2647305 – 99.0778181 |
| 177 to 168 | 99.0632970 – 98.8568370 |
| 167 to 158 | 98.8250954 – 98.5571069 |
| 157 to 148 | 98.5310434 – 98.2004095 |
| 147 to 138 | 98.1493996 – 97.7568649 |
| 137 to 128 | 97.6947779 – 97.1764870 |
| 127 to 118 | 97.1384157 – 96.4199013 |
| 117 to 108 | 96.3551149 – 95.4974401 |
| 107 to 98 | 95.4267895 – 94.4012845 |
| 97 to 88 | 94.1947314 – 92.7470911 |
| 87 to 78 | 92.6189146 – 90.6422786 |
| 77 to 68 | 90.4290235 – 87.3418039 |
| 67 to 58 | 87.3418039 – 82.6801638 |
| 57 to 48 | 81.9257190 – 75.3902075 |
| 47 to 38 | 74.1345061 – 64.1130038 |
| 37 to 28 | 62.8883924 – 49.2876291 |
| 27 to 18 | 49.0122870 – 33.0881504 |
| 17 to 1 | 31.3257004 – 9.0775388 |
Expected JEE Main Percentile Vs Rank 2025
Candidates must note that the rank is obtained as per the percentile scores. Hence, the applicants can check the expected JEE Main rank 2025 as per the percentile scores.
| NTA JEE Main Scores 2025 | Expected ranks |
| 100 | 1 |
| 99 | 8750 |
| 98 | 17,500 |
| 97 | 26,200 |
| 96 | 35,000 |
| 95 | 35,000 |
| 94 | 52,400 |
| 93 | 61,200 |
| 92 | 70,000 |
| 91 | 78,700 |
| 90 | 87,450 |
| 85 | 1,31,100 |
| 80 | 1,74,900 |
| 70 | 2,62,350 |
| 60 | 3,49,800 |
| 50 | 4,37,240 |
| 40 | 5,24,700 |
| 30 | 6,12,150 |
| 20 | 6,99,600 |
| 10 | 7,87,000 |
| 05 | 8,30,750 |
| 0 | 8,74,470 |
JEE Main Answer Key 2025
After the completion of the JEE Mains exam 2025, NTA will release the official JEE Main Answer Key for the same. The Answer Key will contain correct responses to the questions asked in the exam. So, the candidates can check their estimated scores based on the NTA JEE Main Answer Key. Also, NTA provides the opportunity to challenge the official answer key. Candidates who feel that the answers given in the answer key are wrong can raise their objections by writing an application and submitting the required fee. All those who are going to appear for JEE Main 2025 can download the answer key.
JEE Main Previous Years Cutoff (2020, 2019, 2018, 2017, 2016, 2015, 2014)
Candidates can check the JEE Main Cutoff to prepare efficiently.
JEE Main 2020 Cutoff
| Category | JEE Main Paper 1 Cut Off 2020 |
| PwD | 0.0618524 |
| Common Rank List (CRL) | 90.3765335 |
| General-EWS | 70.2435518 |
| Other Backward Classes (OBC-NCL) | 72.8887969 |
| Scheduled Tribe (ST) | 39.0696101 |
| Scheduled Caste (SC) | 50.1760245 |
JEE Main 2019 Cutoff
| Category | JEE Main Paper 1 Cut Off 2019 (Percentile) |
| PwD | 0.11371730 |
| Common Rank List (CRL) | 89.7548849 |
| General-EWS | 78.2174869 |
| Other Backward Classes (OBC-NCL) | 74.3166557 |
| Scheduled Tribe (ST) | 44.3345172 |
| Scheduled Caste (SC) | 54.0128155 |
JEE Main Cutoff (2018, 2017, 2016, 2015, 2014)
| Category | 2018 | 2017 | 2016 | 2015 | 2014 |
| UR | 74 | 81 | 100 | 105 | 49 |
| OBC | 45 | 49 | 70 | 47 | 74 |
| SC | 29 | 32 | 52 | 50 | 70 |
| ST | 24 | 27 | 48 | 44 | 70 |
Comment:hello sir,i am till reading in 12th can i apply for jee mains 2019
Hi Sunny,
Yes, you can apply for JEE Main 2019.