i have got 87.886 percentile in jeemains2019 and i belong to O.C .Can i be eligible for NIT's and jee Advanced

#JEE Advanced #Narula Institute of Technology, Agarpara #Eligibility #Result

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i have got 87.886 percentile in jeemains2019 and i belong to O.C .Can i be eligible for NIT's and jee Advanced

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Navyasri Sirigineedi 11th Dec, 2018

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Sai 20th Jan, 2019

Hello Navyasri..

Percentile= (marks you got*100)/(toppers marks in that session).

According to careers360 rank formula,

Rank formula= [(100-total score) * (no of students wrote exam)] /100

No of students attended are 874469.

So your expected rank will be nearly around 1,05,933.

As per previous year jee mains, any rank below 2,00,000 will be qualified for jee advanced.

So according to expected rank of yours, so you are eligible for jee advanced.

Hope this helps you. All the best

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Sai

25th Jan, 2019

Need a small correction for my answer,percentile=(100*no of candidates who scored equal or less than you in that session)/(Total no of candidates appeared in that session).

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