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I got 145 Mark's in jee advance 2020, what would be my rank?


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hanubee 1st Oct, 2020
Answers (2)
Naman Raturi 1st Oct, 2020

Hello There ,

Your score is very good , but see rank , varies every year according to the toughness of the paper . Although according to last year cutoffs your rank would be around 3000-7000 , in all india category . With this rank you can get ECE and EE and other core branch in many decent IIT'S , although Cs can be little tricky as the cutoffs for cs are increasing every year , but EE and ECE , mech , chem , electronics , civil etc and other good branch you can get in  decent IIT going by last year cutoff ,

For more details you can refer the following page. -

https://engineering.careers360.com/articles/jee-advanced-cutoff-for-iits

https://engineering.careers360.com/articles/jee-advanced-cutoff

Hope it Helps  .

Use careers360 JEE Advanced college predictor to predict colleges like iit , iipe , iiser , iist etc .

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


ARYAN SAGAR 1st Oct, 2020
Dear student,

As you have not mentioned your category, so,  assuming that you are from general category, then you got a good marks in jee advanced.

With 145 marks you can easily get into some IIT. As per the cutoff of JEE advanced 2020, if you belong from  general category then you have to score at least 35% of the 396 marks to get qualified in JEE advanced and also you have to clear every individual subject cutoff that is 10% of 132 marks you have to secure.
Now, assuming that you have cleared the individual Subject cut off and also got 145 marks, then you are qualified in JEE advanced 2020.

And as per the previous years stats, your rank with 145 marks must lie between 6500 to 7000 .
So, with this much rank in general category you can get an IIT.

Also check out our JEE advanced college predictor tool to get an appropriate idea of the IIT that you can get on the basis of JEE advanced marks :-
https://engineering.careers360.com/jee-advanced-college-predictor?icn=QnA&ici=qna_answer
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