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How many marks are required to clear JEE Mains 2025?- Aspirants appearing for BTech admission through JEE Main 2025 must be curious to know the minimum marks required to clear the exam. Every year, the NTA releases a press note stating how many marks are required to clear JEE Mains. According to the previous year trends, candidates should score above 92 marks in the general category in the JEE Main exam to clear the JEE Main 2025 cutoff. The authority conducted the JEE Main exam for session 1 between January 22 to 30, 2025, while session 2 was conducted from April 2 to 9, 2025. Students can check this article for category-wise JEE Main 2025 marks vs percentile and year-wise comparison between the same.
Get latest updates on JEE Main cutoff, percentile
The last year's closing ranks at IIT Bombay CSE prorgamme were as follows.
Seat Type | Gender | 2025 closing rank | 2024 closing rank |
OPEN | Gender-Neutral | 66 | 68 |
OPEN | Female-only (including Supernumerary) | 369 | 421 |
OPEN (PwD) | Gender-Neutral | 4 | 3 |
OPEN (PwD) | Female-only (including Supernumerary) | 12 | 7 |
EWS | Gender-Neutral | 20 | 23 |
EWS | Female-only (including Supernumerary) | 53 | 79 |
EWS (PwD) | Gender-Neutral | 3 | 3 |
OBC-NCL | Gender-Neutral | 54 | 50 |
OBC-NCL | Female-only (including Supernumerary) | 331 | 274 |
OBC-NCL (PwD) | Gender-Neutral | 3 | 2 |
SC | Gender-Neutral | 31 | 20 |
SC | Female-only (including Supernumerary) | 270 | 32 |
SC (PwD) | Gender-Neutral | 1 | 71 |
ST | Gender-Neutral | 19 | 1 |
ST | Female-only (including Supernumerary) | 76 | 13 |
ST (PwD) | Gender-Neutral | 1 | 40 |
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The authority will release the mandated JEE Mains marks in the form of an NTA score. It is a percentile score that will indicate the number of candidates equal to or less than the candidate. Aspirants will be able to predict their results based on the passing marks of JEE Main 2025.
Last year the minimum marks required to clear the JEE Main was around 93 for general category candidates. However, due to changes in the syllabus and pattern this year, the expected marks required to clear JEE Main 2025 might go as high as 95 or above. For more details, check the following tables for the previous three years' JEE Main passing marks for various categories below.
Category | NTA Score |
|---|---|
General | 92 - 100 |
General-PwD | 0.0017 - 92 |
EWS | 83 - 92 |
OBC-NCL | 78 - 92 |
SC | 61 - 92 |
ST | 47 - 92 |
Category | NTA Score |
|---|---|
General | 93.2362181 |
General-PwD | 0.0018700 |
EWS | 81.3266412 |
OBC-NCL | 79.6757881 |
SC | 60.0923182 |
ST | 46.6975840 |
Category | NTA Score |
|---|---|
General | 90.78 |
General-PwD | 0.001 |
EWS | 75.62 |
OBC-NCL | 73.61 |
SC | 51.98 |
ST | 37.23 |
Category | NTA Score |
|---|---|
General | 88.41 |
General-PwD | 0.003 |
EWS | 63.11 |
OBC-NCL | 67.01 |
SC | 43.08 |
ST | 26.78 |
| Also Check: | |
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JEE Main result will consist of the NTA score for each subject and overall percentile as well. However, there will be no subject-wise cut off to clear the JEE Mains exam. Students will get to learn about their percentile only after downloading the scorecard of JEE Main.
The exam conducting authority has provided the formula to determine the minimum marks required to clear JEE Mains. The is given below.
Total Percentile (TPI) = 100 X Number of candidates who appeared in the ‘Session’ with a raw score equal to or less than the candidate / The total number of the candidates who appeared in the ‘Session’.
Candidates who score above the NTA marks will be eligible to apply for the JEE Advanced. Aspirants not only have to clear the first phase test, that is, JEE Main, but they have to be among the top 2.5 lakh rank holders as well. This year, IIT Kanpur will be conducting the exam for admission to IIT colleges in India.
| Know more about: | |
|---|---|
What will be the Expected Percentile for 180 Marks in JEE Main | |
NTA does not provide raw scores. Candidates will be able to check the normalized score, also called the NTA score.
NTA Score is the percentile of JEE Main. For instance, a 93 percentile means, 93% of the total test takers have marks less than or equal to those with that percentile.
The best NTA score out of both JEE Main attempts will be considered while preparing the final merit list.
A specific formula is used to convert the raw scores of Mathematics, Physics and Chemistry and total score into NTA scores.
JEE Main result is comparative in nature. A candidate's final result depends on the performance in the test as well as overall competition.
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Frequently Asked Questions (FAQs)
The expected marks required to clear JEE Main 2025 are approximately 90-95 percentile for the General category, 80-85 percentile for the EWS category, 75-80 percentile for the OBC-NCL category, 55-60 percentile for the SC category, and 45-50 percentile for the ST category.
The overall competition, difficulty level of the paper and candidate’s performance in the test determined whether a particular score is good or bad. As per previous year trends, 150 marks could fetch a rank between 21245 to 16495.
A score of 120 is not a bad score. As per JEE Main marks vs rank vs percentile data, 120 marks will be equal to 95 percentile.
Aspirants can check the above article to figure out the marks to clear JEE Mains.
As per previous year trends and expert analysis, this year, JEE Main 2025 difficulty level is expected to be easy to moderate. However, candidates must thoroughly prepare for the exam.
On Question asked by student community
Hello,
If you’re appearing for Class 12 (HSC) in 2026 and don’t have your board admit card or roll number yet, you can leave the “Registration No./Enrollment No./Roll No.” field blank or enter “NA” (Not Applicable) if the form allows.
Do not enter your school GR number, as it’s not recognized by the exam board or NTA.
Once your board issues the admit card or registration number, it can be updated later during JEE Main form correction or at the time of result verification.
So, for now, safely enter “NA” or leave blank — not your GR number.
Hope you understand.
Hello,
For JEE Main EWS certificate, your parent's income certificate is required, not yours.
The EWS category is based on the family’s annual income, which must be below 8 lakhs from all sources.
The certificate should be issued by a competent government authority and must be valid for the current financial year.
Hope you understand.
Hello,
To secure admission to the University of Hyderabad for the Integrated BTech + MTech in Computer Science and Engineering (CSE) course with an EWS certificate and home state quota, you would likely need a JEE Main percentile of approximately 95-97 percentile or higher. Also, you can check the JEE Main Cutoff , to know more details.
I hope it will clear your query!!
Hello,
Here’s how you can get JEE Main Previous Year by following these steps :
Go to the official JEE Main website.
Click on the “Previous Year Question Papers” or “Downloads” section.
Choose the year and session you want (for example, 2024, 2023, 2022, etc.).
Download the paper in PDF format.
You can also find solved papers and answer keys for the last 5 years on Careers360, where they are available year-wise and subject-wise for free. Click on this link to get papers : JEE Main Question Papers with solution
Hope it helps !
You can get the most repeated and comprehensive questions in the JEE MAINS for the JEE MAINS 2026 EXAMINATION. You can get their pdf from the article link given below by careers360 and download the topic wise questions from every subject to strengthen the preparation.
link- Most Repeated Questions in JEE Mains: Comprehensive Analysis
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