What percentile I will get with 89 marks in jee main January 9 shift 1, obc

hello aspirant,

It’s difficult to predict actual rank or JEE advanced qualify marks. you may calculate your expected rank by this formula

your percentile is to understand the number of student who are ahead of you. 869010 are the number of candidates appeared for Jan exam. So rank formula is (100 - your total score) X 869010 /100. If the number of candidates for April exam increased to 1200000 then your final rank shall be (100 - your total score) X 1200000/100

It’s difficult to tell exact qualifying marks for jee advanced eligibility but if you are getting more than 85 percentile score then you have decent chances to qualify for jee advanced

moreover the percentile of jee mains can be calculated by using the formula of the following.

total number of student below your score /total number of students who attended the exam X 100 gives your percentile. as per the last year result analysis your percentile score comes as 94-95 in between.

hope this helps..

all the best

You haven't mentioned your Category. Assuming your Category to be General, here you go:

After considering the Paper 1 inputs

Your Overall Percentile is: 94.47 - 94.67

Your General Rank is: 48980 - 59130

But if you belong to any other Category, please feel free to use our JEE rank predictor:

JEE Main Rank Predictor 2020 - Predict your JEE Main Rank - JEE Main Percentile Calculator

JEE Main Rank Predictor - Know your expected IIT JEE Main 2020 Rank by using Careers360 JEE Main rank predictor tool based on JEE Mains scor ...

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