what is the qualifying marks in Advance for St girl
Dear aspirant!
Hope you are doing well !
The scores released by the NTA in percentile show that for general category candidates, the cut off for JEE (Advanced)-2022 is 88.4, which was 87.9 in 2021. In this category, the qualifying marks were 90.3 and 89.7 in 2020 and 2019 , respectively..
While for St category it is just only 26.7 percentile ,which is very easy to get . So may be any average student can cut the cutoff.
Hope it helps you !
Thanks for asking query
Hi
There is no qualifying cut off marks rather there is qualifying cut off percentile .
Also it doesn't matter whether you are a girl or a boy , cut off percentile is same for both the genders.
Minimum percentile in jee main 2022 to qualify it to be eligible for jee advanced is given below category wise :-
--------) General :- 88.4121383 percentile
--------) EWS:- 63.1114141 percentile
--------) OBC-NCL :- 67.0090297 percentile
--------) SC :- 43.0820954 percentile
--------) ST:- 26.7771328 percentile
--------) PWD :- 0.0031029 percentile
You can check the same at our page the link is provided below
https://engineering.careers360.com/articles/jee-main-cutoff
- So,if your percentile is more than or equal to 26.7771328 percentile then you have qualified to be eligible for JEE advanced otherwise you are not eligible
To have a personalised report on your chances of getting a college you can go through our college predictor, it will give you complete list of colleges which you can get on the basis of the category you belong to, your home state, your gender etc, the link for jee main college predictor is provided below
https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer
We also have Btech companion for you in which you get 15+ college predictors, you also get information regarding various counselling i.e Josaa, MHTCET etc and you get to attend live webinars with experts to guide you and get all your doubts cleared. The link is given below:-
https://www.careers360.com/btech-companion??icn=QnA&ici=qna_answer
Regards
ADITYA KUMAR




