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It would be difficult to provide you the exact information about the qualifying score or admission cutoff of NITs 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 18,129. You can go through with the given below link to check the previous year 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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