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Which JEE Main 2026 score will NTA use for common rank list? -National Testing Agency takes the candidate’s best scores from the two sessions, which will be used for ranking purposes. It plays an important role in securing admissions in top engineering colleges. Despite NTA outlining the criteria for selecting the best score, some confusion remains among students regarding the preparation of the common rank list. Understanding the JEE Main 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 JEE Main 2026 exam for session 1 is scheduled from January 21 to 29, 2026. The April session will be conducted from April 2 to 9, 2026. Read below for more details on the JEE Main exam and its common ranks list.
It is expected that the NTA will announce the JEE Mains 2026 response sheet release date soon on its official website. Based on the last year's trends, the JEE Main response sheet will be published in the first week of February.
The National Testing Agency will release the JEE Main 2026 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 2026 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 prioritise the better score obtained by the candidate from the two sessions.
Appearing in JEE Main 2026? Scores of the respective tests will be used.
Appearing in both JEE Main 2026? The better of the two scores will be used. The lower score will be ignored.
| JEE Main Marks | JEE Main Rank |
|---|---|
300-291 | 1 - 15 |
292-280 | 16 - 56 |
279-271 | 86 - 125 |
268-259 | 142 - 341 |
258-250 | 381 - 704 |
249-240 | 733 - 1246 |
239-230 | 1269 - 1909 |
229-220 | 2043 - 3217 |
219-210 | 3319 - 4549 |
209- 200 | 4647 - 6269 |
199-190 | 6488 - 8724 |
189- 180 | 8951 - 12197 |
179-170 | 12439 - 16524 |
169-160 | 16942 - 21710 |
159-150 | 22156 - 28132 |
149-140 | 28402 - 36243 |
139-130 | 36985 - 46047 |
129-120 | 47066 - 57991 |
119-110 | 59254 - 72925 |
109-100 | 74236 - 91426 |
99-90 | 93240 - 114780 |
89-80 | 117151 - 143434 |
79-70 | 1468491 - 184121 |
69-60 | 186799- 237626 |
59-50 | 243215 - 319343 |
49-40 | 327188 - 448730 |
39-30 | 460200 - 647703 |
29-20 | 678394 935442 |
19-10 | 954561 - 1207129 |
9-0 | 1222184 - 1390805 |
The JEE Main score 2026 for Paper-1 and Paper-2 will be calculated differently. Mentioned below is the procedure for the calculation for both papers.
Score of JEE Main = Number of correct answers x 4 – number of incorrect answers x 1
The percentile score will be the normalised score for the JEE Main exam 2026 (instead of the raw marks of the candidate). The JEE Main 2026 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 candidates appeared in the ‘Session'.
Ranks will be prepared based on the percentile scores of the candidates in JEE Main 2026 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.
Candidates with a higher percentile score obtained in Mathematics will be given a better rank.
If the tie remains unresolved, the candidates with higher percentile scores in Physics will be awarded a better rank.
If the tie still remains unresolved, the candidates with the higher percentile score in Chemistry will be awarded a better rank.
In case the tie is still not resolved, then the candidates older in age are preferred.
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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) 2026 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 attempts; their best of the two raw marks will be considered. The total raw marks of both attempts and the 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.
Candidates obtaining higher marks in the Aptitude Test in Paper 2
Candidates obtaining higher marks in the Drawing Test in Paper 2
Candidates older in age
If the resolution is not possible after this criterion, candidates will be given the same rank.
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 2026 common rank list where the list will consist of the same category candidates assigned ranks based on their performance in JEE Main 2026.
Allotment will be based onthe CRL, which is also called the common rank list.
JEE Main 2026 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 2026 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 2026 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.
The Rank List will be prepared in May for JoSAA Counselling.
On Question asked by student community
JEE Main 28 Jan shift 2 exam will end at 6 PM. The complete analysis and memory-based questions will solution will be updated in the below article. Keep checking the page-
Math and Chemistry was difficult and Physics was moderate. The complete analysis is available here- https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-1-question-paper-with-solutions-pdf
You can also check the memory-based questions and detailed solutions for JEE Main Jan 28 shift 1 paper.
Hi Lucky,
Please refer to this link and you can download the free pdf.
HI Manisha Maharana
You can download the JEE Mains 10 Free Mock Test with Detailed solutions. Its a feely downloadable pdf.
https://engineering.careers360.com/download/sample-papers/jee-main-10-full-mock-test-and-explanations-pdf
Also, you can check ad attemp the online mock test on our platform.
https://learn.careers360.com/test-series-jee-main-free-mock-test/
A general equation of a circle is
$
x^2+y^2+2 g x+2 f y+c=0
$
Since it passes through $(0,0)$,
$
c=0
$
So the equation becomes
$
x^2+y^2+2 g x+2 f y=0
$
It cuts the x -axis at ( $a, 0$ ).
Substituting:
$
a^2+2 g a=0
$
g=-a/2
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