Hi

To get 10,000 obc rank you need at least around 40,000 crl rank/general rank ,which is around  96 percentile as per our predictor at

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

Now very limited time is there for jee main remaining sessions so to prepare well

• Revise the concepts

• practice  chapterwise important mcqs/ questions which you you would have marked for revision during your preparation.

• solve at least previous 10 year jee main question papers as concept behind the question often gets repeated in the examination so it will be of great help in your preparation  ,

• go through mock test papers as it will let you know your preparation level and and weak areas so that you can work on them and perform well in your examination.

To boost your performance we have :- https://learn.careers360.com/jee-main-rank-booster

Thank you

Dear Aspirant,

In my opinion 140+ marks will fetch you obc rank under 10,000. Your All India Rank should be around 50,000. I have stated the marks for you, so that you could aim high. May be these numbers vary but you should atleast target 140+ marks in JEE Mains exam to grab a good college and a good branch.
Check out these links for more reference:

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

Hope this Helps!
Good Luck

For a OBC Rank of 10k, your Corresponding CRL in JEE Mains would expectedly be between 40-45k. This is perhaps a Good Rank, and you do have fair share of prospects during Counselling in both Home and Other State with this OBC Rank.

For more information and college prediction do follow the link below

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

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