According to VidyaMandir experts, candidates can analyse their NTA scores and calculate their approximate JEE Main rank through their percentile scores. Here’s how.

So, the total number of candidates who appeared for the JEE Main 2019 January exam was around 8.74 lakhs. The exam was conducted on four days in two slots (4 days x 2 slots = 8 sessions in total). So, approximately 1.09 lakh students took the exam in each session. Note that the JEE Main 2019 results/scorecard mentions only the NTA score for each subject and not the subject-wise marks secured by candidates. The percentile score displayed against each of the subjects and the overall percentile score is calculated up to 7 decimals using a pre-determined formula.

Candidates who fall among the top 2,50,000 rankers in JEE Main 2020 across all categories are eligible to appear for JEE Advanced. Out of this number, the percentage of candidates eligible for admissions under different categories are as follows:

• General category candidates (from the common merit list) – 46.5%
• General EWS – 4%
• OBC-NCL candidates – 27%
• SC candidates – 15%
• ST candidates – 7.5%
• PWD candidates - Horizontal reservation of 5% in all the above categories

Hello aspirant

If you are from general category then you are not qualified or not can't be estimated. Yet Jee main session 2 is to be commenced and then after results , Jee main 2020 cut offs will be announced . It depends on previous year's cut off. 2019 Jee main cut off was 89.75 This year's cut off may increase or decrease, can't say because it is said cut off will be reduced by 15 or 20 %.  So it difficult to predict 2020 cut offs.
For College prediction following link may help you
https:// engineering.careers360.com/ jee-main-college-predictor

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