The National Testing Agency (NTA) will announce JEE Main 2025 Result in the form of percentile score. NTA uses the normalization formula to calculate the JEE Main percentile score of candidates. JEE Main percentile and normalization formula used by NTA (National Testing Agency) is provided here. To determine JEE Main 2025 results and rank list, NTA will adopt the process of normalization which is based on percentile calculation. The rank list of JEE Main 2025 will be prepared by the NTA based on percentile scores and not on the raw marks secured by aspirants in the exam.
Also Check: JEE Main Rank Predictor 2025
| JEE Main Result 2025 | JEE Main 2025 Marks vs Percentile |
| JEE Main Question Paper PDF 2024-2013 | JEE Main Cutoff |
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Predict JEE Main CollegesHow To Calculate JEE Main 2025 Percentile Score
JEE Main 2025 exam will be held on multiple days and sessions with the difficulty level of the exam varying from day to day and from session to session. Thus, to ensure equity in assessing the performance of all the test-takers, the exam conducting authority introduced the percentile system based on the normalization process. NTA uses the given below formula to calculate the percentile score of a candidate
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Key Terms For Calculating JEE Main 2025 Percentile Score
Before we delve deeper into how will NTA calculate the JEE Main 2025 Percentile Score through the Normalisation Process, there are some key terms you should know.
| Key terms |
What they mean |
|---|---|
| Percentile/ Percentile Score |
Percentile Score is the score based on the relative performance of candidates who have appeared for the exam. To calculate a percentile score, the marks secured are converted into a scale ranging from 100 to 0 for each session of examinees. NOTE: Percentile score is not the same as the percentage of marks secured. |
| Normalisation |
It is the practice of comparing candidates' scores who appeared in different shifts of the exam. The normalization process will be similar to the ones currently being adopted for other national level competitive exams like CAT. |
| Highest Raw Score and Percentile Score |
All the highest raw scores will have a normalised Percentile Score of 100 for their respective session. This means, that the highest raw scores secured by candidates would be normalised to a 100 Percentile Score for their respective session. |
| Lowest Raw Score and Percentile Score |
Percentile Score of all the lowest raw scores is different. In this case, the percentile score will depend on the total number of candidates who have taken the examination for their respective sessions. |
Highlights of JEE Main 2025 Percentile Score
- JEE Main percentile score is the normalized score for the exam and not the raw marks of the test taker.
- The percentile score specifies the percentage of candidates who have scored equal to below (raw scores) a particular percentile in JEE Main. Thus, the topper of each session of JEE Main will get the same percentile i.e. 100.
- Percentiles will also be calculated and converted for marks obtained in between the highest and lowest scores
- The percentile score will be used to prepare the JEE Main merit list
- The percentile scores for JEE Main will be calculated up to 7 decimal places to reduce tie-breaking scenarios and avoid the bunching effect
- The overall percentile is not the aggregate of the percentile of the individual subject
To help you understand how the agency will calculate JEE Main percentile score, here’s an example.
Q: Is 75 percent marks in class 12 required for JEE Mains?
Candidates need not have 75 percent marks in class 12 for appearing in JEE Main. As per the official brochure released by the NTA there is no minimum percentage required for appearing in the JEE Main. The requirement of 75 percent marks need not be associated with JEE Main. It is to be understood that 75 percent marks is required for admission to institutes which include NITs, IIITs, GFTIs and IITs. There are many institutes which offer admission on 60% and even less along with JEE Main. Candidates must clearly understand that the requirement of 75 percent marks in class 12 is only for taking admission in the institutes.
Q: What is the expected NIT Calicut JEE Main cutoff 2024?
Going as per the past three year trends candidates can expect the closing ranks for NIT Calicut JEE Main cutoff 2024 Round 1 Seat Allotment for top specialisations between the following range:
- B.Tech. in Computer Science and Engineering must range between 2717 and 4455
- B.Tech. in Electrical and Electronics Engineering must range between 10612 and 12119
- B.Tech. in Electronics and Communication Engineering must range between 7127 and 7563
- Bachelor of Architecture (B.Arch.) must range between 269 and 316
- B.Tech. in Engineering Physics must range between 16153 and 16430
Q: What is NIT Rourkela cutoff for BTech?
NIT Rourkela cutoff 2024 has been released across various specialisations in opening and closing ranks. NIT Rourkela BTech cutoff 2024 ranged between 2680 and 35593 for the General AI category candidates, wherein BTech in CSE is the most competitive course. The easiest BTech specialisation to secure admission in is B.Tech. in Ceramic Engineering + M.Tech. in Industrial Ceramic Engineering. For a detailed NIT Rourkela branch-wise cutoff 2024, candidates can refer to the table below for the Round 1 cutoff. The cutoff mentioned below is for the General category under the AI Quota.
Example: In 2024, JEE Main was held in two sessions of test-takers where the days and shifts were allotted randomly to candidates.
- Session-1: Day-1 Shift-1
- Session-2: Day-1 Shift-2
The JEE Main test takers were distributed in the following manner:
| Distribution of candidates |
||||||
|---|---|---|---|---|---|---|
| Session |
Day/Shift |
No of Candidates |
|
Marks |
|
|
| Absent |
Appeared |
Total |
Highest |
Lowest |
||
| Session-1 |
Day-1 Shift-1 |
3974 |
28012 |
31986 |
335 |
-39 |
| Session-2 |
Day-1 Shift-2 |
6189 |
32541 |
38730 |
346 |
-38 |
| Total (Session-1 to Session-4) |
25273 |
142482 |
167755 |
346 |
-49 |
|
The above example of the distribution of candidates, the method to calculate the Highest Raw Score in each paper will be normalized to 100, thus indicating that 100% of candidates have obtained scores that are equal to or lesser than the highest scorer/ topper for that session.
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Example: Highest Raw Score and Percentile Score
According to the official website of JEE Main, the Percentile Score of all the lowest raw scores will be different and will be based on the total number of test-takers for their respective sessions.
Highest Raw Score and Percentile Score
| Session |
Total Candidates Appeared |
Highest Raw Score |
Candidates who scored EQUAL OR LESS THAN Highest Raw Score |
Percentile Score |
|---|---|---|---|---|
| Session-1 |
28012 |
335 |
28012 |
100.0000000 [(28012/28012)*100] |
| Session-2 |
32541 |
346 |
32541 |
100.0000000 [(32541/32541)*100] |
| Session-3 |
41326 |
331 |
41326 |
100.0000000 [(41326/41326)*100] |
Lowest Raw Score and Percentile Score
| Session |
Total Candidates Appeared |
Lowest Raw Score |
Candidates who scored EQUAL OR LESS THAN Lowest Raw Score |
Percentile Score |
|---|---|---|---|---|
| Session-1 |
28012 |
-39 |
1 |
0.0035699 [(1/28012)*100] |
| Session-2 |
32541 |
-38 |
1 |
0.0030730 [(1/32541)*100] |
| Session-3 |
41326 |
-49 |
1 |
0.0024198 [(1/41326)*100] |
JEE Main 2025 Tie Breaker Rules
JEE Main merit list/rank list will be prepared based on the Percentile scores of the Total Raw Scores. In case, where two or more test takers secure equal Percentile Scores in the exam, the inter-se merit will be determined in the following order:
Q: What is the cutoff for BTech at NIT Andhra Pradesh for OBC category?
NIT Andhra Pradesh BTech cutoff 2024 for BTech courses ranged between 25704 and 91199 for OBC category. Wherein, the most competitive course was BTech in Computer Science and Engineering and the least competitive course was BTech in Biotechnology. While the admissions are based on the scores/rank obtained in JEE Main, the last cutoffs also change a lot depending on All India quota or Home State. It also varies if the applicant is a female.
Q: What is the cutoff for BTech in Computer Science and Engineering at NIT Hamirpur for other than General categories?
The last year 2022 cutoff ranks for BTech in Computer Science and Engineering at NIT Hamirpur - National Institute of Technology other than General category are - 17445 for OBC, 1663 for SC, 578 for ST, 16770 for EWS, 187395 for PWD. While the admissions are based on the scores/rank obtained in JEE exam, the last cut offs also change a lot depending on All India quota or Home State. It also varies if the applicant is a female.
Q: How to calculate percentile in JEE?
NTA uses a formula to calculate NTA score in JEE Main. The formula used is
Q: How much is 120 marks in JEE Mains percentile?
The National Testing Agency (NTA) will declare the JEE Mains result 2025 in the form of percentile scores. Candidates can check here the JEE Main 2025 marks vs percentile data to understand the 120 marks in JEE Main percentile.
As per the last year JEE Main marks vs percentile analysis, 120 marks would be around 95 to 97 percentile. However, the exact JEE Main 2025 percentile marks 120 would vary depending upon the highest number scored by the candidate in a particula shift and the total number candidates appeared in that particular shift. More details about JEE Main marks vs percentile vs rank can be checked to understand in details.
Q: What is JEE Main cutoff for IIITs?
- Candidates obtaining higher Percentile Score in Mathematics in the Test.
- Candidates obtaining higher Percentile Score in Physics in the Test.
- Candidates obtaining higher Percentile Score in Chemistry in the Test.
- Candidates older in age are preferred.
Normalization Process To Prepare JEE Main 2025 Ranklist
NTA’s normalization process for the preparation of the JEE Main merit/rank list will comprise of the following steps:
Step-1: Refers to the conduct of the exam. JEE Main is conducted in two shifts over four days.
Step-2: Once the examination is over, the results for each session will be prepared using raw scores and percentile scores for each subject.
Step-3: The third and last step would be the compilation of the NTA score and preparation of JEE Main merit list. Remember, the Final Ranking/ Merit would be prepared only after resolving ties (if any).
To compile JEE Main 2024 result, the four NTA scores (Physics, Chemistry, Mathematics, and overall score) of the first attempt and the second attempt will be merged for the compilation of results and preparation of the overall Merit List/Ranking. Candidates who appear in both the attempts; their best of the two NTA scores will be considered further for the preparation of Merit List /Ranking.
Read Important FAQs Here
Q. What is JEE Main Percentile Score?
A. JEE Main Percentile Score is the score based on the relative performance of candidates who have appeared for the JEE Main entrance exam and it is not the same as the percentage of marks secured.
Q. What is Normalisation in JEE Main?
A. It is the practice of comparing candidates' scores who appeared in different shifts of the JEE Main exam.
Q. What is Highest Raw Score and Percentile Score in JEE Main?
A. It is the practice of comparing candidates' scores who appeared in different shifts of the JEE Main exam.
Q. What is Highest Raw Score and Percentile Score in JEE Main?
A. All the highest raw scores will have a normalised Percentile Score of 100 for their respective session. This means, that the highest raw scores secured by candidates would be normalised to a 100 Percentile Score for their respective session.
Q. How JEE Main merit list is prepared?
A. The percentile score will be used to prepare JEE Main merit list.
Q. What are the JEE Main 2025 Tie Breaker Rules?
A. In case, where two or more test takers secure equal Percentile Scores in the exam, the inter-se merit will be determined in the following order: Candidates obtaining higher Percentile Score in Mathematics in the Test, their higher Percentile Score in Physics in the Test, higher Percentile Score in Chemistry in the Test and Candidates older in age to be preferred.
Q. How final JEE Main result will be prepared?
A. To compile the final JEE Main 2025 result, the four NTA scores (Physics, Chemistry, Mathematics, and overall score) of the first attempt and the second attempt will be merged for the compilation of results and preparation of the overall Merit List/Ranking.
Q. How JEE Main rank list is prepared?
A. The rank list of JEE Main 2025 will be prepared by the NTA based on percentile scores and not on the raw marks secured by aspirants in the exam.
Q. How will NTA calculate JEE Main 2025 Percentile Score?
A. The National Testing Agency will calculate JEE Main 2025 Percentile Score through the Normalisation Process.
Q. Will NTA release the JEE Main result along with the percentile score?
A. NTA is responsible for releasing JEE Main 2025 result along with the percentile score. To determine JEE Main 2025 results and rank list, NTA adopted the process of normalization which was based on percentile calculation.
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Student Forum
Anything you would want to ask experts?Answered 5 hours ago
Of course you can get nit
M
Beginner-Level 2
Answered 7 hours ago
Considering the JEE Main cutoff 2024, yes, it is possible to get IT at Shree LR Tiwari College with a percentile of 88.144. The Shree L.R. Tiwari College of Engineering accepts JEE Main entrance exam scores, followed by the JoSAA counselling process for entry into the BTech programs. UNIRAJ offers a
T
Beginner-Level 5
Answered 9 hours ago
There are about 120 Engineering colleges in India that accept JEE Main score. Some of them are mentioned below along with their total tuition fees:
| College Name | Total Tuition Fee |
|---|---|
| IIT Madras | INR 10 lakh |
| IIT Delhi | INR 8 lakh |
| IIT Bombay | INR 8 lakh |
| IIT Kanpur | INR 8 lakh |
| IIT Roorkee | INR 8 lakh |
Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.
T
Contributor-Level 10
Answered 18 hours ago
i don't think so brother
M
Beginner-Level 1
Answered Yesterday
Yes, whomever have completed their class 12 with the Science stream can opt for a degree in BTech without JEE Main or any other entrance exam scores with the criteria that you must have passed the 10+2 exam or equivalent with Physics and Mathematics as compulsory subjects. But the chances to get
J
Contributor-Level 7
Answered Yesterday
NTA has announced the JEE Main 2025 exam date session 1. The session 1 exam will be conducted on January 22, 23, 24 28 and 29, 2025 for Paper 1. For Paper 2, the exam date is January 30, 2025. JEE Main 2025 session 2 will be held from April 1 to 8, 2025. NTA has ended JEE Main 2025 session 1 re
V
Contributor-Level 10
Answered Yesterday
V
Beginner-Level 5
Answered Yesterday
Hello I hope you are doing fine
As you are asking for the chapter wise weightage of physics in JEE Mains so here is information for you:
From electrostatic, capacitors, simple harmonic motion, sound waves, elasticity, error in measurements, circular motion, electromagnetic waves, semi conductor, the
M
Contributor-Level 7
Answered Yesterday
Yes, admissions are possible in Atria Institute of Technology without JEE Main. The accepted entrance exams are COMEDK UGET, KCET, DCET, and others. Candidates applying through the COMEDK UGET must first register for the accepted entrance exam. Then appear for the accepted entrance exam. Then with
R
Beginner-Level 5
Answered Yesterday
Hello, I hope you are doing fine
Every year at least 10000 to 12000 student score 99 percentile in the JEE Mains exam. In India exams are very competitive and also very stressful. So many students fight for a rank and there's only one who can achieve it. Students get the dream college or the best c
M
Contributor-Level 7
Can I get a seat for B.Tech in NIT after getting OBC NCL PwD rank 955 in JEE Mains?