what is the minimum rank in jee mains for get into nit and IIT?
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Hello aspirant
Minimum rank to qualify for jee main 2020 is difficult to estimate. But we can assume based on previous year's cut off.
For general category expected Jee main cut off 2020 is above 89.7548849 percentile. That means minimum 90 percentile. Now for 90 percentile you need to score minimum 80 marks.
With 80 marks, your expected all India rank will
be 1,74,900 rank. This calculation is expected .
For detailed information following links will help you.
https://engineering.careers360.com/article/jee-main-marks-vs-percentile
https://engineering.careers360.com/article/jee-main-marks-vs-rank
JEE Mains is an examination through which one can get admission in NITs, IIITs, GFITs. It doesn't offer the gateway to IITs. In order to get into IITs, a candidate should have qualified the JEE Advanced examination with good score.
The opening and closing ranks or percentile depends upon the NIT and also the branch you wish to choose. For NITs the opening percentile is 99.9. a candidates rank should be under approx 30k to get admission in NITs.
https://engineering.careers360.com/jee-main-college-predictor
https://engineering.careers360.com/jee-advanced-college-predictor
Hi there,
To get into NIT your rank should be under 30k
And to get into the top Nit then your rank should be under 15k
And to get into too branch of the top Nit your rank should be under 5k.
For IIT your rank should be under 10k in the jee advance paper to get into IIT, And to get into top branch of top IIT your rank should be under 3000.
I hope this help
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