i got 90.54 percentile in jee paper 2 2019 . Can i get NITs by this percentile?
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The exact prediction would be difficult as JEE Main is being conducted twice by NTA for the first time. As a total No. of 1,45,386 candidates had appeared for January Session, Using the Formula (100-your score)* 1,45,386 / 100- Your Expected rank will be 13753. With this rank the chances are less to get a seat in NITs, You can also prefer the previous year cutoff to take an Idea about the cutoff.
https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
NOTE: This is only a prediction. NTA will merge the scores obtained in first and second attempt by candidates of JEE Main Paper 2 2019 for the compilation and preparation of overall merit/rank list. The rank will be released on 30th April, 2019.
Good Luck!
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