**JEE Main Normalisation:** JEE, aka Joint Entrance Examination, is conducted on two levels. JEE Main is the first level, and those who clear the Mains will advance to attempt the JEE Advanced exam. The number of students who attempt the JEE Advanced changes drastically every year. Only 2 to 3 lakh students passed the Mains in the last few years.

However, to get admitted into India’s Ivy League colleges, such as IIT and NITs, the students should clear the JEE Advanced examination. It has been estimated that only 40 to 50 thousand candidates clear this exam each year.

This article will discuss everything related to the marking scheme of the JEE Main exam while understanding how to calculate the percentage score.

## Overview of the Percentile-Based Normalisation Procedure

This procedure enables the National Testing Agency (NTA) to prepare ranks for the JEE Main. This method has been incorporated for the fourth time in the JEE Main exam. It will help the students calculate their scores on the exam.

The normalisation procedure is a method that helps the students calculate and compare their scores across various question paper sessions. Percentile equivalence is majorly utilised for normalisation across all the sections. This technique has been implemented in all the popular state-level and national-level entrance tests.

Furthermore, the normalisation process helps derive a solution affected by multiple shifts and various question papers. Due to that, the candidates could be assured of experiencing no partiality based on the difficulty level of the examination. The main purpose is to uphold the rule of law and identify the actual merit.

In addition, this technique is well-established and could help the students compare scores over various examination sessions. Apart from that, it is also comparable to other major educational selection tests across the nation.

**Also Read: **

## Importance of Normalisation Method

The National Testing Agency has decided to conduct the JEE examination on different dates with two sessions per day. In every session, a different set of questions will be provided to the candidates, so there is a possibility that the difficulty levels might vary from one session to another.

Despite the efforts by NTA to maintain a similar level of difficulty, many aspirants are still experiencing tougher question sessions which make them lose marks. In many cases, the students who attended the easier sessions got more marks and benefits than those who attempted more difficult ones.

Therefore, the NTA introduced a normalization procedure to provide justice to the students who have attempted tougher question papers. It is based on the percentile scores, implying that no one would benefit or be conned. A candidate’s true merit could be identified through the normalization technique.

Hence, NTA has compiled JEE Main scores for various session question papers by normalizing procedure techniques based on percentile scores.

The percentile scores of JEE Main define the relative performance of all the candidates who attempted the JEE examination. The marks obtained by the aspirants in the JEE exam will be transformed into a scale that ranges from 100 to 0 for every exam session.

For instance, percentile scores indicate the percentage of candidates who have secured equal or lower marks than those who obtained higher marks in a particular JEE Main exam session. Later on, the marks of these candidates will be transformed into a percentile. This score will be denoted for all the JEE Main sessions, and the topper of every session will get 100 percentile flat.

Moreover, it is considered the normalised score for the JEE Main exam. It can also be used for preparing the merit or rank list for JEE. Hence, the percentile score will be calculated up to 7 decimal digits to avoid the bouncing effect or inequality and reduce ties.

**Also Read:**

JEE Main Session 2 Result Declared, Here’s how to Download your scorecard

JEE Main Session 2 Topper’s List

## Raw Score and the Percentile score

For every session the candidates attended, their highest raw score will be considered a 100 percentile score, and it will be converted into normalised percentile scores of 100.

Similarly, the percentile score will be different from calculating the lowest raw score. Here, the total number of students appearing for each session can be used to calculate it.

## Calculating Procedure for JEE Main Percentile Score

To calculate the percentile score of the JEE exam, the following details are required:

- A total number of applicants who have appeared in a particular session of the JEE exam.
- A highest raw score of a student in that specific session.

The percentile score acquired by a student will be different from the raw score. This score is also not similar to the percentage of marks secured by a student.

Provided below is the formula used for calculating the JEE Main percentile:

Percentile score = 100 * (The no. of candidates who have obtained a score equal to or less than the student) / (the total no. of candidates who have appeared in that particular session of the examination).

**JEE Main Related Links**

Let us see some examples to understand this concept better:

**Example 1:**

For each correct answer in the JEE main examination, 4 marks will be awarded. Similarly, for every wrong answer, 1 mark will be deducted. In total, the JEE exam has 90 questions for 360 marks. The marks secured out of 360 will not be considered for preparing ranks. It will be transformed into a scale that ranges from 100 to 0 through the normalisation procedure. The following is an example of that:

Let us consider the JEE Main was conducted in 4 sessions, and the following is the highest and lowest score of the candidates:

Session |
The highest raw score secured by a candidate |
The total no. of candidates who appeared for the exam |
Total no. of candidates who scored equal to or less than the highest raw score of a candidate |
Percentile of the candidate who scored the highest marks |
Explanation (apply the above formula) |

Session 1 | 335 | 28,012 | 28,012 | 100 Percentile | (28012 / 28012) X 100 = 100 |

Session 2 | 346 | 32,541 | 32,541 | 100 Percentile | (32541 / 32541) X 100 = 100 |

Session 3 | 331 | 41,326 | 41,326 | 100 Percentile | (41326 / 41326) X 100 = 100 |

Session 4 | 332 | 40,603 | 40,603 | 100 Percentile | (40603 / 40603) X 100 = 100 |

From the above tabulation, one can clearly understand that a student who has secured the highest raw score in every session will obtain the 100 percentile.

**Example 2:**

Let us take an instance where we can check the raw score and percentile of candidates in a specific JEE Main session:

Name of the student |
Raw Score secured by the student |
Students scored less raw scores than the candidate |
The total no. of students who attempted the exam |
Percentile of the student |
Explanation |

A | 335 out of 360 | 28,000 | 28,012 | 99.9571612 | (28000 / 28012) X 100 = 99.9571612 |

B | 330 out of 360 | 27,012 | 28,012 | 96.4301013 | (27012 / 28012) X 100 = 96.4301013 |

C | 310 out of 360 | 20,000 | 28,012 | 71.3979722 | (20000 / 28012) X 100 = 71.3979722 |

D | 270 out of 360 | 16,588 | 28,012 | 59.2174782 | (16588 / 28012) X 100 = 59.2174782 |

E | 230 out of 360 | 13,999 | 28,012 | 49.9750107 | (13999 / 28012) X 100 = 49.9750107 |

F | 99 out of 360 | 6,700 | 28,012 | 23.9183207 | (6700 / 28012) X 100 = 23.9183207 |

G | 58 out of 360 | 400 | 28,012 | 1.42795944 | (400 / 28012) X 100 = 1.42795944 |

## Tie-breaking Rules of JEE Main

In case two or more students secured the same percentile score in a JEE Main exam session, the below-mentioned rules can be applied to identify the order of the students in the merit list:

- The candidates who have scored the highest percentile in the Mathematics section will be considered first.
- Even if there is a tie after incorporating the rule mentioned above, the candidates who have secured the highest scores in Physics shall be taken into account.
- If the tie is still present, the students who have the highest percentile in Chemistry will be considered.
- Even after all that, if the tie is still present, the older students shall be considered.

These are some of the jee main tie-breaking rules that will be considered in case of a tie in the percentile.

## JEE Main Result Preparation

The National Testing Agency will prepare each session’s results while having the following components into consideration:

- The students scored total marks out of 360, also referred to as the raw score.
- The students who obtained percentile scores. A percentile score is nothing but the overall percentile obtained in all three subjects of the JEE Main exam, i.e., Maths, Physics, and Chemistry.

The following are illustrations of how a percentile score is being prepared. A similar formula is applied to denote the percentile score for each subject.

## The Ranking System of JEE Main

The percentile score of every session of the JEE exam can be calculated separately. While doing so, the raw scores of the students in all three subjects (Physics, Chemistry, and Maths) and overall raw scores can also be considered for preparing the overall rank list (or merit list).

Therefore, the final NTA score will be prepared on the basis of the overall raw score as well as the raw score in all three subjects.

## Conclusion

The whole of the JEE Main normalisation has been explained thoroughly in this article. The main purpose of this procedure is to provide equal opportunities to the candidates who appear for the exam in various sessions. As it contains multiple sessions a year, there are many chances that the students might face varied difficulties in the question paper.

**Also See: **

How to Calculate Your Rank in JEE Main Using NTA Percentile?

Moreover, with the assistance of this method, the ranking technique has become a lot more unique and easier for awarding the best-scoring students without any hassle.

## FAQs about JEE Main Normalisation

### How is the JEE Main cutoff being calculated?

There are various factors which are being considered before calculating the cutoff score for JEE Main. They are:

- The difficulty level of the examination

- The total number of candidates who attempted the exam

- Seat availability for each session

- Total number of sessions per year

### How good is the 90 percentile in JEE Mains?

In general, candidates who score around the 90 percentile might be ranked between 1,50,000 and 3,00,000. It clearly indicates that the applicants who obtain that percentile in the JEE Main exams might get accepted into some of the NITs and IITs. However, not every institution accepts that. In most cases, the candidates are obliged to appear for JEE Advanced in order to get a seat in India’s top public universities.

### How many percentiles is 120 in JEE?

On a medium-difficulty paper, scoring 120 marks out of 300 might obtain a percentile of 97 or 98 depending on the session. One can expect a rank of about 3,00,000; however, it is not accurate.

### How many attempts can a candidate take in the JEE exam?

The candidates can take up to a maximum of two attempts in a year. Moreover, they can appear for the JEE examination in three consecutive years, starting from the year they passed their 12th standard. Hence, NTA has not mandated any age criteria, so that anyone can appear for the JEE Mains.

### What is my percentile if my NTA score is 40?

40 marks in the NTA JEE Main 2023 exam will let a students get a percentile around 70.