Which JEE Main 2025 score will NTA use for common rank list? - NTA conducts the most important entrance examination JEE Main for admission into undergraduate engineering colleges in India. The National Testing Agency administers the exam twice a year: January and April sessions. For JEE Main 2025 exam the January session will take place from January 22 to 31, while the April session will be conducted from April 1 to 8, 2025. Candidates' best scores from the two sessions will be used for ranking purposes, which plays a crucial role in securing admissions. Despite NTA outlining the criteria for selecting the best score, some confusion remains among students regarding the preparation of the common rank list.
Aspirants who are planning to appear for the JEE Main 2025 exam should keep a close eye on the official website of NTA to get the latest updates about the exam dates and other details. Understanding the selection criteria and preparing accordingly can significantly increase the chances of success in the exam and securing admission into the top engineering colleges in India.
The National Testing Agency will release the JEE Main 2025 rank list after the JEE Main result. Due to the large number of candidates appearing for the JEE Main exam, the rank/merit list is prepared based on the JEE Mains cutoff 2025 percentile score of the candidate. To calculate the percentile score, if a candidate has appeared for both exam sessions, the better of the two scores will be considered. The score used by NTA to prepare the JEE Main common rank list will prioritize the better score obtained by the candidate from the two sessions.
Appearing in JEE Main 2025? Scores of the respective test will be used.
Appearing in both JEE Main 2025? The better of the two scores will be used. The lower score will be ignored.
| Marks out of 300 | Rank |
| 288- 294 | 20-11 |
| 280-284 | 44-22 |
| 270- 279 | 107-63 |
| 252- 268 | 522-106 |
| 231-249 | 1385-546 |
| 215-230 | 2798-1421 |
| 202-214 | 4666-2862 |
| 190-200 | 6664- 4830 |
| 175-189 | 10746-7151 |
| 161-174 | 16163-11018 |
| 149-159 | 21145-16495 |
| 132-148 | 32826-22238 |
| 120-131 | 43174-33636 |
| 110-119 | 54293-44115 |
| 102-109 | 65758-55269 |
| 95-101 | 76260-66999 |
| 89-94 | 87219-78111 |
| 79-88 | 109329-90144 |
| 64-87 | 169542-92303 |
| 44-62 | 326517-173239 |
| 1-42 | 1025009-334080 |
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The JEE Main score 2025 for Paper-1 and Paper-2 will be calculated differently. Mentioned below is the procedure for the calculation for both the papers.
Score of JEE Main = Number of correct answers x 4 – number of incorrect answers x 1
The percentile score will be the normalized score for JEE Main exam 2025 (instead of the raw marks of the candidate). The JEE Main 2025 percentile scores will be calculated up to 7 decimal places. NTA will use the following method to calculate the percentile score of a candidate:
100 * Number of candidates appeared in the ‘Session’ with raw score ‘Equal To Or Less’ than the candidate/Total number of the candidates appeared in the ‘Session'.
Ranks will be prepared based on the percentile scores of the candidates in JEE Main 2025 as mentioned above. However, there may be instances where more than two candidates may score the same marks. In such a case the inter-se-merit guidelines will be enforced to allot the rank.
JEE Main paper-2 is conducted in two shifts, the raw marks obtained by all the candidates who appeared in the first attempt will be declared. This shall comprise the actual marks obtained in Paper 2 of JEE (Main) 2025 for the first attempt.
The rank list will be prepared by merging the raw scores of the first and the second attempt. Candidates who have appeared in both the attempts; their best of the two raw marks will be considered. The total raw marks of both the attempts and rank of Paper-2 for all candidates will be declared. This shall comprise the total raw marks obtained by the candidate in the first attempt as well as in the second attempt along with All India rank and All India category rank. In case of a tie, inter-se merit will be determined in the following order.
Important Points of JEE Main Common Rank List
The common rank list of JEE Main will include candidates given ranks as per their performance but with a tag of GEN, GEN-PwD, OBC-NCL, OBC-NCL-PwD, SC, SC-PwD, ST or ST-PwD.
The category ranks will be separate from the JEE Main 2025 common rank list where the list will consist of the same category candidates assigned ranks based on their performance in JEE Main 2025.
Allotment will be based on CRL which is also called as the common rank list.
JEE Main 2025 new official website is jeemain.nta.nic.in.
As with the previous year, Section B does not contain any more optional questions. Applicants must sit for all 5 numerical questions in each subject.
The details of JEE Main 2025 exam pattern for Mathematics: Section A consists of 20 MCQs, Section B has 10 numerical problems. As mentioned earlier five numerical questions will be asked in section B wherein none of the questions will be covered in other questions of the same type. This pattern holds good for both Paper 1 (B.E./B.Tech) and Paper 2 (B.Arch/B.Planning).
JEE Main is held by the National Testing Agency which was set up in November 2017 exclusively for conducting entrance exams pertaining to higher education.
JEE Main 2025 will be a computer-based test and will be held twice a year
JEE Main Mock tests will be available on the official website.
As recommended by the Council of Architecture to NTA regarding JEE Main Paper II, candidates must secure at least 50% marks in the qualifying examination and 50% marks in PCM individually.
3400 Schools/ Engineering Colleges as Test Practice Centres (TPCs) are open for practice on the mock tests for free.
Each student will be assigned a unique paper from a large data bank prepared by experts.
Rank List will be prepared in May for JoSAA Counselling.
On Question asked by student community
Hello,
The link to the question paper is attached here. Careers360 help students in their exam preparation. students can find all the required study materials and exam materials, exam materials, results and cutoffs on the website.
Thank you.
Hello,
Despite of the prior backlogs, you can appear for JEE Mains in 2026 because you passed class 12; that marks the basic eligibility.
Anyways, this would be regarded as the third JEE Mains attempt, because there's a limit that a candidate can attempt only a limited number of times;
You can get free mock tests for JEE 2026 on the Careers360 website. The mock tests are based on the latest exam pattern and help in understanding the level of questions and time management. You just need to register and start attempting the tests online.
Direct link for JEE mock
Hello,
If you passed 12th in 2025 and did not appear in JEE till now, then this is your eligibility:
JEE Main can be given for 3 consecutive years after 12th.
Each year has 2 sessions .
You are eligible in 2025, 2026 and 2027 .
Since you
Hello,
For JEE (for IT / engineering) :
10th percentage
There is no fixed minimum percentage needed in Class 10 to prepare for JEE.
You just need to pass Class 10 .
Important part is Class 12
You must study Physics, Chemistry, and Mathematics in Class 11 and 12 .
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