Hello,

You have not clearly specified if you belong to OBC-NCL or OBC-CL. If you belong to OBC-CL then you are considered as general category candidate. According to our JEE Main Rank Predictor Tool you can expect overall rank to be between 21171 - 25321 and category rank between 4675 - 6325 with 120 marks. This analysis is based on previous year data. In 2021 the rank corresponding to the score may change depending on factors like number of candidates appeared, top score in the exam, normalized score etc. In 2020 the cut off in round 6 to get admission in NIT, Patna in Computer Science Engineering was as follows for home state candidate

• General:15617
• OBC-NCL:4959

Based on the previous year cut off rank you have chances of getting seat in the preferred NIT if you belong to OBC-NCL but it is difficult to get seat if you belong to OBC-CL as you will be considered as general category candidate. However, cut off changes every year depending on number of candidates qualified, difficulty of exam and number of seats available etc. So you should aim to score minimum 140-150 marks to increase the chance if getting seat in NIT, Patna.

We have JEE Main coaching program which you may consider to enroll to help you with your preparation. It comes with E-Lectures, smart study material, performance analysis, structured time table, test series and more. To check the details and enroll click the given link

https://learn.careers360.com/jee-main-coaching/?icn=jee-main_coaching&ici=exit-intent

Hello,

The JEE Main cut off depends upon various factors such as number of candidates appeared for the exam, highest marks obtained, question paper difficulty level etc. So, the cut off varies each year. Since, this year a lot of changes have been done in JEE Main, so it is tough to come to any disclosure.

JEE Main 2019 home state cut off for CSE in NIT Patna  for general category candidates was 16347 and in JEE Main 2018 the cut off rank was 29652.

So, with 120 marks your chance is quite low to get CSE in NIT Patna if you are general category candidate. But, if you have any reservation(SC/ST/OBC/EWS) then your chance of getting a seat will increase.

You can check the previous year cut off from the link below

https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-patna/amp

good luck!

Hello Aspirant,

Since you have not mentioned your category so considering it as General or Open category.

If we go through the analysis of marks Vs Percentile Vs rank for JEE-Main 2020 then according to the marks mentioned by you in the question your rank and percentile is expected to be in the range of 21171 - 25321 and 97.49 - 97.69 respectively, 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 :-

https://engineering.careers360.com/jee-main-rank-predictor

As we go through the latest cut-off of 2020 then it shows that the closing rank for NIT Patna was 15617 for CSE branch in NIT Patna under Home state quota for Open category or General category. So there are quite less chances for you to get seat for CSE branch in NIT Patna

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!!