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I have got 88.57 percentile in jee mains paper 2 2019 ..What would be my expected rank?


Ashi sharma 1st Feb, 2019
Answers (2)
Anurag Patel 1st Feb, 2019

Hi student,

Approximately 13,00,000 students appeared for jee mains paper 2 in 2019.

Your percentile is 88.57 and this means that 88.57 percent students of the total students appeared for jee mains examination paper 2 are behind you. Your approximate rank can be calculated as follows:

Rank(approximate)=(100-NTA score)*(number of students appeared)÷100

Therefore your approximate rank is 14859.

However this is not your final rank and this will vary with the official rank list due to various reasons. So it is better to wait for the official rank list.

Good luck

2 Comments
Comments (2)
1st Feb, 2019
it is wrong because 145386 students appeared only.... 
Reply
1st Feb, 2019
THANKS  
Reply
Yogita Punjabi 1st Feb, 2019
Hello Ashi



You can estimate your rank by the below given formula:

(100 - your percentile) * (874469/100)



Here, 874469 is the total number of students who appeared in the January session of JEE Mains.



Accordingly, your rank comes around- 1 lakh. However, this rank may vary with the actual one depending upon the number of candidates appearing in the April session of JEE Mains exam.

Well, based on your percentile, you have chances to qualify for JEE Advanced level.

The final cutoff percentile along with the result will be declared after the April session and so you will have to wait for now. Hope for the best.!



Good Luck.!



For any further information-queries, you can ask here at CAREERS360 anytime .!












1 Comment
Comments (1)
1st Feb, 2019
this is also wrong because ...no of student appeared for jee paper 2 is jst 145368
Reply
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