I am jeemain 60.42percentile obc ncl iam attend+2 exam 2019 I get eligibility for JEE advance exam

Hello Gokulranjan,

As per our rank predictor your rank will be 336430.

Our rank predictor formula = ((100 - your percentile)/100)*Total number of candidate.

Total number of candidate in this jee 8.5 lack.

Your rank range will be 330000 - 340000 .

Based on this percentile you not able to clear Jee.

To clear Jee for general candidate 85 percentile is require as per our expert.

This percentile is not for get into NIT, for NIT percentile should be more than this.

Previous year top 224000 clear the Jee main exam.



You should go for April Jee main to clear the Jee main.

Best of both Jee marks will conclude during admission in institution.



Note:All this information is based on our rank predictor,Final rank may vary .We not take responsibility for vary in your rank by NTA and our prediction.

To know the percentile system follow the link below:https://engineering.careers360.com/articles/jee-main-normalisation-process-how-scores-are-calculated

To know details through video follow the link below:https://youtu.be/lXb58NIhSRE

If you have any questions remain than comment below.

Good luck.

hii

you need to find rank first

The official Jee rankings will be released after the examination in April. You cod use the formula to predict your rank:

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

However,this rank may change as only the best of the two scores will be considered for the final rank list.

please refer the below link

https://engineering.careers360.com/percentile-predictor-result/1094?s=39600026441548134962

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

https://engineering.careers360.com/articles/jee-main-cutoff

for eligibilty of the JEE adavance you need get rank below 2,28,000 rank then you will be eligible

regards