i am from bihar belongs to OBC ncl. how much score(marks/300) did i need for cse in nit patna?

Hello Aspirant,

As we go through the latest cut-off of 2020 then it shows that the closing rank for NIT Patna was 4959 for CSE branch in NIT Patna under Home state quota for OBC-NCL category. Now to secure such rank you should have your JEE-Main 2021 score at least 170 marks out of 300 and percentile against this is expected to be in the range of 99.4569399985455 - 99.573193698637 .

You can check out marks Vs ranks Vs Percentile from the link given below :-

https://engineering.careers360.com/articles/jee-main-marks-vs-percentile

Or

You can also use JEE-Main Rank Predictor to predict your rank :-

JEE Main Rank Predictor 2020 - Predict your JEE Main Rank - JEE Main Percentile Calculator

JEE Main Rank Predictor - Know your expected IIT JEE Main 2020 Rank by using Careers360 JEE Main rank predictor tool based on JEE Mains scor ...

You can check out the latest cut-offs from the link given below :-

https://josaa.nic.in/Result/Result/currentorcr.aspx

Note that the above prediction is made on the basis of latest cut-offs available on official website of JOSAA or CSAB and the cut-offs are bound to change every year depending upon various factors such as the number of candidates appearing in an examination, the number of candidates qualifying an examination, difficulty level of the paper, and so on.

If you are preparing for JEE-Main 2021 or 2022 then I would recommend you to check the resources made by our experts to help aspirants like you in the best and smart preparation of JEE-Main 2021 or 2022 :-

https://learn.careers360.com/jee-main-coaching/?utm_source=%20QnA

I hope this information helps you.

Good Luck!!