I got 99.0293652 percentile in JEE MAIN paper 2 January 2020. Can I get B Arch in NIT calicut?
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Hello,
Your expected rank will be around 1093 and the formula to calculate it is (100-your total percentile)*112679/100
112679 candidates appeared in January session in Paper 2. You have not mentioned your category and state. Check the cut off in 2019 for NIT, Calicut, the ranks are for other state and home state respectively
- General: 1362, 1668
- EWS: 230, 333
- OBC-NCL: 804, 693
- SC: 358,676
- ST: 81, 1108
Considering this you may get seat irrespective of your category. But cut off is subjected to change depending on number of candidates appeared, number of seats etc. To know the other college in which you may get seat you can use our JEE main college predictor tool
Reply
100 - (your total percentile) 112679 /100
In your case your rank must be around 1104
Cut off vary every year on various factors like the level of question in the entrance examination and total number of applicants in that examination.
As per previous year records its expected that yes you'll get admission in nit calicut with this rank as last year the cut off was 1362.
For more updates refer to the link mentioned below
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-nit-calicut/amp
Feel free to comment if you have any doubt
Good luck
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