JEE Main 2025 Session 1 Result – Know the Formula for Calculating NTA Score

14 Students have Scored 100 NTA Score in JEE Main 2025 Session 1 Result

National Testing Agency has released the result of JEE Main 2025 Session 1. In the exam conducted in total 10 shifts, 14 students have scored 100 NTA score, which includes maximum 5 students from Rajasthan, 2 from Delhi, 2 from Uttar Pradesh, and 1 each from Maharashtra, Gujarat, Andhra Pradesh, Karnataka and Telangana.

This year, maximum numbers of students from Rajasthan have scored 100 NTA. 5 Students from ALLEN have scored 100 NTA scores. List of 44 state toppers has also been released in the declared result of whom 13 State Toppers are from ALLEN. Students will get the result only with their application number and password created during application.

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Here’s the Formula

In the result released for the first session of JEE Main, the total NTA score of the students along with separate NTA scores of Physics, Chemistry and Maths are released in 7 decimal percentile. JEE Main exam is conducted in different shifts. In such a situation, the difficulty level of the paper of each shift can also be different.

In such a situation, the raw scores of the students in all the different shifts are normalized on the basis of their respective shifts and released as percentile NTA score in 7 decimals. First of all, the raw score of each student is calculated, after that the number of students having equal and lesser raw score than that of the student is taken, this is multiplied by 100, divided by the total number of students giving the exam and the percentile is calculated in 7 decimal places, which is called the NTA score of that student.

This percentile formula is applicable on the total average marks of the students as well as on the marks obtained in Maths, Physics, and Chemistry. With this formula, the total NTA score of each student as well as the NTA score of each subject is calculated. The NTA score of the April session will also be calculated.

If the student takes both the JEE Main exams in January and April, then the All India Rank will be released on the basis of maximum NTA score of both the examinations.

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Which NIT, IIIT Available on Which Percentile

JEE-Main January results were released in the form of 7 decimal percentile. The curiosity of the students about the NITs, IIIT and GFTI available, on the basis of their respective percentile score, on NTA score is clearly visible.

The possibilities of colleges given below may change for General, OBC, EWS, SC-ST students as per the category. – Those who have more than 99 percentile, there are clear chances of getting admission in top NITs like Trichy, Warangal, Suratkal, Allahabad, Rourkela, Calicut and Jaipur, and in core branches of Kurukshetra NIT and IIIT Allahabad.

If they have 99 to 98 percentile, then there are chances of getting core branches in top 10 NITs as well as other branches as well as core branches in top 10-20 NITs and IIIT Jabalpur, Gwalior, Guwahati, Kota, Lucknow. These NITs include Bhopal, Surat, Nagpur, Jalandhar, Delhi, Hamirpur, Durgapur.

If they have 98 to 96 percentile, then apart from core branches in top 20 NITs, they can get admission in other branches and remaining NITs which include NITs of Northeast as well as NITs of Patna, Raipur, Agartala, Srinagar, Silchar, Uttarakhand NIT and institutions like BITS Misra, Punjab Engineering College Chandigarh, JNU, and Hyderabad University.

Also, students will have the possibility of getting core branches of new IIIT Vadodara, Pune, Sonipat, Surat, Nagpur, Bhopal, Trichy, Raichur, Kanchipuram, Ranchi, Dharwad, Agartala, Kalyani.

On having 96 to 94 percentile score, there can be chances of getting admission in other branches and GFTIs in addition to the core branches of top 25 to 31 NITs.

Who Should Do What?

This year students have 2 options to take JEE-Main exam 2025. Those students whose January JEE-Main percentile is more than 99.5 should prepare for JEE-Advanced exam with full focus since on this percentile, the option of getting core branches in good NIT has become secured.

Students with percentile between 99.5 to 98.5 can give JEE Main as per their convenience or start preparing for Advanced. Students whose percentile is less than 98.5 should pay full attention to the preparation of Advanced along with JEE Main Session 2. These suggestions can change according to the category.

13.11 Registered for Paper 1, 12.58 lakh took the Exam

Career Counseling Expert of ALLEN Career Institute told that JEE Main January session for B.E.-B.Tech. was held in 10 shifts in 5 days between 22 to 29 January. The final answer key was released on 10 February. According to the official press release of NTA, 13 lakh 11 thousand 544 students had applied for the exam and 12 lakh 58 thousand 136 students appeared in the exam.

This exam was conducted in 13 languages. This exam was conducted in 618 exam centers in 304 exam cities of the country as well as in 15 cities abroad. 8 lakh 33 thousand 325 boys and 4 lakh 24 thousand 810 girls appeared in this exam.

In this, 4 lakh 66 thousand 358 students of General category, 1 lakh 38 thousand 699 of EWS, 4 lakh 90 thousand 275 of OBC, 1 lakh 22 thousand 845 of SC and 39 thousand 959 of ST appeared in the exam. The category-wise list of students with 100 percentiles has also been released.

👇Watch JEE Main 2025 Session 1 Complete Paper Analysis👇

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