Dear Aspirant

Based on your score/percentile your rank  will be nearly equal to (100-87)*145386/100=18900.

To know previous year cutoff of NITs and GFITs you can visit below link-

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

For further query feel free to ask!

Hope it helps! GOOD LUCK!

Hey there!

Kcet , NATA and jee main paper 2 based admissions are not related. So, you just have to produce documents from the examination you are using for your admission procedure. You don't need documents from other architectural examinations in the counselling procedure.

Hello student,

First of all I want to congratulate you for getting such rank. This is really a good rank to get national Institute of technologies with course B architecture.

Although you belongs to general category you can get various NIT with this rank. You may get those NITs

• NIT

Hello there,

See, since 2019 NTA decided to conduct examinations of JEE twice in months of January as well as in april.

Now, as per your query and as per previous years, it is seen that the paper 2 of JEE Main is conducted just before the start date of

Hello aspirant,

You got 84 percentile in Jee paper 2 & 75% in your board examination.

According to the rank formula i. e. rank = (100-percentile) * (no. of students) /100.

Now, as the total no.  of students is 145386, you rank should be around 23.2k.

Now, look, you can't

Yes you can get the seats in Andhra pradesh

As your marks is quite appropriate for it..

Here the link is provided below .so that you can predict the college suitable for you

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer

Hope this will help you..

Better luck!!

As per the previous year cutoff, you can get admission in MNIT, Jaipur  as the closing rank for OBC is 1688.  The previous year's closing rank for obc category at National Institute of Technology, Calicut is  393. National Institute of Technology, Raipur closing rank for obc category is 13542 You