96.20 percentile obc ncl in jee mains 2019 can i get nit please reply me

Hi student,

A percentile of 96.20 means that 96.20 percent of students who appeared for the examination are behind you. You approximate rank will be around 33,229. However this rank wil change to some extent due to another jee mains Examination in April.

According to 2017 cutoffs, you can get following NITs:

• National Institute of Technology, Andhra Pradesh
• National Institute of Technology Calicut
• National Institute of Technology Sikkim
• National Institute of Technology, Mizoram
• National Institute of Technology Durgapur
• National Institute of Technology, Srinagar
• National Institute of Technology Raipur
• National Institute of Technology, Uttarakhand

Hope it helps.

Good luck.

hiii sainath

your rank approximately would be 33,230

you could calculate using the 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-scoHttps://engineering.careers360.com/articles/jee-main-cutoff

while coming to admission in NITs you have quite chances of getting admission but you need wait , till the rank and marks are released

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

Regards