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

Hello Aspirant,

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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