The Bitsat examination form has been released  You can apply by going on the link provided on the page of careers 360. The form is available till 20th March 2019 till 5:00 pm. So apply before the last date and I wish you all the best for your examination. 

Hello varshini

86.9 percentile is not enough for nit trichy.

And also you may qualify or not it is also a doubt.So lets wait for the results .

You should start preparing for the next attempt and score atleast 99.5 percentile for NIT trichy.( May vary de pending upon other

I think you are not.

Hii

No site or app can provide you with free mocktests for JEE and BITSAT. 
You can use entrance360. This platform provides you with a lot of study materials, sample papers, previous papers, a proper timetable for your help, guidance, and is pretty much affordable.
Here i am providing a

Hello aspirant, 

In BITS, admissions are through BITS entrance, and there is direct entry for students possessing 1st ranks in respective board,If you are a topper in your board, you can go for direct entry to BITSAT. Otherwise you have to go for BITSAT entrance and qualify.

Good luck

Hello,

According to th eligibility criteria for bitsat2019

The candidate must scored a minimum of 75% aggregate marks in Physics, Chemistry and Mathematics subjects. In case candidate has opt for Biology in BITSAT 2019, he/she should have scored a minimum of aggregate 75% marks in Physics, Chemistry and Biology subjects.

hello varun srivastava,

please refer this link:

https://engineering.careers360.com/articles/bitsat-fee-structure

copy the above link and paste it in the address bar of your browser.

all the best!

I Hope this information will be useful to you.

if you need additional information,let us know to help you.

Thank you!!!:)

Hi Nisarg,

Getting admission in BITS,PILANI is quite difficult because you need to score well enough marks ,If you are preparing for JEE exams you don't need to do much things only you need to increase your problem solving speed as  much you can and you need to solve mock