How to calculate overall rank in jee paper1

Hello Sarthak,

The overall rank as well as percentile is calculated in the follwing ways.

How to calculate my JEE Main 2019 rank based on my NTA Percentile score?

Let’s take this for the January session of JEE Main 2019 first. As per the NTA declared result statistics, 8,74,469 students appeared for the exam across four days with two sessions each making that a total of eight sessions. There are 15 students who have scored 100 percentile in these sessions.

To calculate the JEE Main 2019 rank, you first need to know the percentage of people ahead of you with higher percentile scores. To calculate that, the formula would be 100 which is the highest score minus your percentile score.

Since the percentile is calculated from a scale ranging from 100 to 0 for each session of examinees, the formula applies the total number of students in the January session which is 874469 /100 multiplied by the percentage of people who are ahead of you to arrive at the rank.

Therefore, the formula to calculate the JEE Main 2019 rank for January 2019 is

JEE Main 2019 rank (probable) = (100- NTA percentile score ) X 874469 /100

If NTA percentile score is 90.70, JEE Main rank will be (100-90.70 ) X 874469/100 = 81325

Consider the NTA score of 80.60 then JEE Main 2019 rank = (100-80.60) X 874469 /100. The rank will be 169646.

** This formula would not be for a 100 percentile score as those are the top ranks.

Refer to the follwing artcle for more help.

https://engineering.careers360.com/articles/how-calculate-jee-main-2019-rank-based-on-nta-percentile-score

Hello,

Simple formula of getting a great rank is that score more than others. But IIT is a little complex and so is it's ranking procedure.

first of all let's check the procedure for jee main. It's even more complex to understand due to introduction of normalisation method .

The details for normalization of Qualifying examination marks is as follows:-

i. Note down the aggregate marks (A0) obtained by each student in JEE- Main.

ii. Compute the percentile (P) of each student on the basis of aggregate marks in his/her own board (B0) computed from the list of five subjects specified (each marked out of 100). The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE-Main. The variable B0 is only a base for calculating percentile (P), which is further used to get corresponding JEE (Main) marks. This is used in computing JEE Main rank.

iii. Determine the JEE- Main aggregate marks corresponding to percentile (P) at the All-India level. Regard this as B1.

iv. Also, determine the JEE-Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE-Main by the students of that board.Regard this as B2.

The normalized board score of the candidate was computed as:

Bfinal = 0.5 * (B1 + B2)

For the purpose of admission to CFTIs where it has been decided to use the JEE Main performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list was computed as:

C = 0.6 * AO + 0.4 *Bfinal

Now considering C as the score for comparison ranks are alloted  to students.

In case of jee advance method is quite simple because IIT is only interested in your score in jee advence. So , according to total score (paper1 +paper2) ranks are alloted to students.

yet,there are few more things to know for a condition when two students have same score

In such a case 2 solutions  are used:

1> Now, score of both the students in the toughest paper among PCM is considered(usually score of math is considerd) , and the student who has more score in this paper is awarded a better rank.

2> Now if score of both students are same in this paper also then both are awarded the same rank..

.

Hope it helps.

Thank you.