I have scored 85.19 in JEE main Jan 2019 (9th jan- 1st shift). What will my rank be? Will I be eligible for JEE advanced?
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Answer (1)
20th Jan, 2019
Hello Candidate,
Your rank will be 1,41,170 as per our rank predictor formula.
Rank predictor formula = ((100 - your percentile)/100)*total number of candidate
This year 9.5 lacks candidate given the Jee main
As per view previous year 2,24,000 candidate are eligible for Jee Advance previous year.
As this year categories cutoff are not released by NTA .
After release of categories cutoff we can comment more about Advance eligibility more clearly but as per viewing previous year trends your seems eligible for advanced
If any other queries remain than comment below.
Good luck...
Your rank will be 1,41,170 as per our rank predictor formula.
Rank predictor formula = ((100 - your percentile)/100)*total number of candidate
This year 9.5 lacks candidate given the Jee main
As per view previous year 2,24,000 candidate are eligible for Jee Advance previous year.
As this year categories cutoff are not released by NTA .
After release of categories cutoff we can comment more about Advance eligibility more clearly but as per viewing previous year trends your seems eligible for advanced
If any other queries remain than comment below.
Good luck...
3 Comments
Comments (3)
Reply
If a candidate gets 100 percentile then your formula isn't valid
For 100 percentile we take candidate percentile as 99.99 for the sake of calculating approximate rank
In case of 99.99 percentile rank of a candidate will be 95 .
Means if candidate score 99.99 percentile than candidate will have rank range from 75 - 125.
Thank you...
In case of 99.99 percentile rank of a candidate will be 95 .
Means if candidate score 99.99 percentile than candidate will have rank range from 75 - 125.
Thank you...
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