JEE Main Percentile Calculator 2025: How to Calculate Percentile using Answer Key

JEE Main Percentile Calculator 2025: How to Calculate Percentile using Answer Key

Edited By Samridhi Mishra | Updated on Apr 19, 2025 01:35 AM IST | #JEE Main

JEE Main Percentile Calculator 2025 - NTA uses a specific formula to calculate a candidate's JEE Mains percentile 2025. Those who took the JEE Mains 2025 exam must check the JEE Main 2025 percentile calculator details provided in this article. Using the JEE Mains percentile calculator, students can estimate their percentile as per their scores in the JEE Mains exam. The complete procedure for using the JEE Main percentile calculator 2025 is explained in this article in a simplified manner.

This Story also Contains
  1. JEE Main Result 2025 - Percentile Calculation
  2. JEE Main 2025 Percentile
  3. JEE Main Percentile vs Marks 2025
  4. JEE Main 2025 Tie-Breaking Rule
  5. JEE Main Tie-Breaking Rule for Paper 1 (B.Tech)
  6. JEE Main Tie-Breaking Rule for Paper 2 A (B.Arch)
  7. JEE Main Tie-Breaking Rule for Paper 2 B (B.Planning)
JEE Main Percentile Calculator 2025: How to Calculate Percentile using Answer Key
JEE Main Percentile Calculator 2025: How to Calculate Percentile using Answer Key

Using the JEE Main percentile calculator, candidates can assess their percentile and standing among all the other candidates who appeared for the JEE Main 2025 exam. Please note that the JEE Main rank list 2025 will be prepared based on a candidate's JEE Main percentile and not just raw scores.
Must Check: JEE Main Rank Predictor 2025 | JEE Main College Predictor 2025

JEE Main Result 2025 - Percentile Calculation

The National Testing Agency calculates the scores of JEE Main 2025 for percentile for individual subjects separately (Physics, Chemistry, and Mathematics) as well as in the combined format using the normalization formula as shown below.

Calculating JEE Main Percentile

Total Percentile (T1P)

(100 x No. of candidates from the session with a raw score equal to or less than T1 score) / Total No. of candidates who appeared in the session

Mathematics Percentile (M1P)

(100 x No. of candidates appeared from the session with a raw score equal to or less than than M1 score in Mathematics)

/ Total No. of candidates who appeared in the session

Chemistry Percentile (C1P):

(100 x No. of candidates appeared from the session with a raw score equal to or less than C1 score in Chemistry) / Total No. of candidates who appeared in the session

Physics Percentile (P1P)

(100 x No. of candidates appeared from the session with raw score equal to or less than P1 score in Physics) / Total No. of candidates who appeared in the session

JEE Main percentile calculation

JEE Main 2025 Percentile

NTA is expected to release a official notice consisting of a table to classify the range of percentages as secured by students who appeared in the JEE Main 2025 paper. Students will be able to check their official percentile as per their appeared shift from the NTA press release when uploaded. The same will be provided below.

JEE Main Percentile vs Marks 2025

NTA calculates the JEE Main 2025 percentile using the process of normalization of marks. There is a certain number of marks against which the percentile score is allotted, which in turn can be called the JEE Main 2025 marks vs percentile. Aspirants may refer to the table below to know the expected range of marks for which a certain percentile is allotted to the respective candidates.

JEE Main 2025 Expected Percentile vs Marks 2025

Marks out of 300Percentile
288- 29499.99826992- 99.99890732
280-28499.99617561 - 99.99790569
270-27999.99034797 - 99.99417236
252- 26899.95228621- 99.99016586
231-24999.87388626-99.95028296
215-23099.74522293-99.87060821
202-21499.57503767- 99.73930423
190-20099.39319714- 99.56019541
175-18999.02150308 - 99.3487614
161-17498.52824811-98.99673561
149-15998.07460288-98.49801724
132-14897.0109678-97.97507774
120-13196.0687115-96.93721175
110-11995.05625037-95.983027
102-10994.01228357-94.96737888
95-10193.05600452 -93.89928202
89-9492.05811248 -92.88745828
79-8890.0448455 -91.79177119
62-8784.56203931-91.59517945
41-6170.26839007-84.22540213
1-4266.6590786-69.5797271
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JEE Main 2025 Tie-Breaking Rule

NTA has changed the JEE Main tie-breaking rule 2025. Candidates can check the updated JEE Main 2025 tie-breaking policy below. The tie-breaking rules apply when two candidates obtain the same JEE Main scores. However, to allot the JEE Main ranks, the authorities follow the tie-breaking policy JEE Main 2025. The tie-breaking for JEE Main 2025 will apply in the following order.

JEE Main Tie-Breaking Rule for Paper 1 (B.Tech)

  • NTA score in Mathematics, followed by
  • NTA score in Physics, followed by
  • NTA score in Chemistry, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Physics in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Chemistry in the Test followed by
  • If there is a tie even after this, candidates will be assigned the same rank.
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JEE Main Tie-Breaking Rule for Paper 2 A (B.Arch)

The tie between candidates obtaining the same Total NTA scores will be resolved in the following manner:

  • NTA score in Mathematics, followed by
  • NTA score in Aptitude Test, followed by
  • NTA score in Drawing Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics (Part-I) in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Aptitude Test (Part-II) in the Test followed by
  • If there is a tie even after this, candidates will be assigned the same rank.
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JEE Main Tie-Breaking Rule for Paper 2 B (B.Planning)

In case two candidates obtain the same marks in the JEE Main 2025 B.Planning exam the tie-breaking will apply in the following order:

  • NTA score in Mathematics, followed by
  • NTA score in Aptitude Test, followed by
  • NTA score in Planning Based Questions, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in all the subjects in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Mathematics (Part- I) in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Aptitude Test (Part- II) in the Test, followed by
  • Candidate with less proportion of a number of attempted incorrect answers and correct answers in Planning Based Questions (Part-III) in the Test followed by
  • If there is a tie even after this, candidates will be assigned the same rank.

Frequently Asked Questions (FAQs)

1. How to calculate JEE Main percentile?

To calculate the JEE Mains percentile 2025, candidates have to (100 x No. of candidates from the session with a raw score equal to or less than T1 score) / Total No. of candidates who appeared in the session.

2. How much percentile is 70 marks in JEE Mains?

70 marks in JEE Mains translates to a percentile of 90.

3. What is the 99.5 percentile in JEE Mains?

99.5 percentile in JEE Mains means the candidate has obtained approximately 200 marks.

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Questions related to JEE Main

Have a question related to JEE Main ?

HEY THERE!!

With 967 IPE marks and 90 percentile in JEE Mains, you have a very high probability of getting a CSE core seat in SASTRA University through the usual merit-based admission process.

Admission Process at SASTRA

  1. Stream 1 (70% seats): It is conducted on a combined score: 75% weightage to your Class 12 (IPE) marks and 25% weightage to your JEE Main score.
  2. Stream 2 (30% seats): Only on Class 12 (IPE) marks.

Your Chances

  • Recent times student reports and expert answers indicate that above 900 IPE marks and a good JEE Main percentile (generally above 85–90) are generally sufficient to obtain a CSE seat in SASTRA.
  • SASTRA's CSE cutoff is likely to be tough, but your 967/1000 in IPE and 90 percentile in JEE Main are strong for the CSE core, especially in Stream 1.
  • Last year's cutoff every year is based on the set of candidates, but your profile is on or above the average admitted student profile for CSE.

Conclusion:

With your marks, you have a strong possibility of securing a CSE core seat at SASTRA University, though allotment will finally be made depending on the competition that year and the availability of seats. Attempt to apply through both streams to maximize chances.

Dear candidate,

You have 93 percentile in AEEE and 92.9 percentile in JEE Main, so there is a strong chance of getting admission in several branches such as :

  1. ECE
  2. EEE
  3. CSE in Amrita Vishwa Vidyapeetham, especially under Group B (higher fee).

Admission for CSE at coimbatore is competitive but there is a chance if you are from home state.

All the best!


HEY THERE!!!

On your marks-56% in JEE Main and 88% in CBSE Board (Tamil Nadu)-you are eligible to apply for the B.Tech. in Electronics Engineering (VLSI Design and Technology) at SASTRA University, Thanjavur, but a seat for the course will be allotted based on the combined merit and competition.

VLSI Course Prospects

  1. Admission is merit-based; there is no cut-off marks-it depends on the pool of candidates each year.
  2. For competitive streams like VLSI, the combined marks in boards are typically higher.
  3. With 88% boards, you have a good opportunity, especially through Stream 2 (on the basis of marks in boards only).
  4. In case you have a poor JEE percentile, you have a better chance through Stream 2.

Summary

You are eligible and stand a good chance for VLSI course in SASTRA Thanjavur, especially under Stream 2 with your good board marks. Admission is not guaranteed but depends on the merit list and competition annually. Apply early and monitor the official releases for counselling and rank lists.

HEY THERE!!!

With 978/1000 in Class 12th (98%) and 80.9 percentile in JEE Mains, you have a very high chance of getting Computer Science Engineering (CSE) with specialization at SASTRA University, Thanjavur.

How SASTRA Admission Works

  1. Stream 1: 70% of seats are filled based on a combined merit score (75% weightage to Class 12 marks + 25% weightage to JEE Main percentile).
  2. Stream 2: 30% of seats are filled based solely on Class 12 marks

Your Chances

  • Your Class 12 mark (978/1000) is good and will place you high in the merit list.
  • Your JEE Main percentile (80.9) is average and combined with your excellent board marks puts your overall merit score in competition.
  • Recent trends suggest that 95%+ board students and 75–85 percentile students in JEE Main have achieved CSE and its specializations at SASTRA, especially during the later rounds or through Stream 2

Conclusion

You are likely to receive CSE with specialization from SASTRA University through Stream 1 or Stream 2 because of your good board marks and average JEE Main percentile. Final allotment depends on competition and availability of seats, but your profile is very competitive to receive CSE and its specializations in SASTRA.

Hii,

A few key documents are required for JEE. Your admit card, which grants you entry to the exam room, comes first. To prove who you are, you also need identification, such as your school ID or Aadhar card. To prove your age, your birth certificate is required.

Your class 12 grade report is required as documentation of your education. A category certificate is necessary if you fall under one of the reserved categories, such as SC, ST, or OBC. Keep these with you on test day.

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