I got 47.033 as nta score for 7th Jan shift 2 paper 2020 then what is my percentile and is the score enough to take admission in any NIT
Baidurja, NTA score is the same as percentile. So your percentile will be 47.033; this means that your performance was better that 47.033% of the total candidates.
You can calculate your approximate rank for JEE Mains using your percentile through the following formula:
(100-your total percentile) x 869010/100
So with 47.033 percentile, your approximate rank in CRL will be 460288.
Also, if the number of students increases to 12 lakh in the April session, the formula for rank calculation would change to:
(100-your total percentile) x 1200000/100
Then your rank would be around 635604.
While it is difficult to predict the exact qualifying cutoff, a percentile of 85 or above is decent enough to qualify for JEE Advanced for the unreserved students. In 2019, the NTA score cutoff for the Common Rank List was 89.7548849
According to your percentile, it will be difficult to clear the cutoffs for JEE Advanced. Also you won't be able to get a seat in the top NITs or IIITs with this rank. You need to have at least 97-98 percentile to get a seat in the top colleges. Although you can try for colleges through state level counselling.
I would suggest you to sit for the April session and give your best in that exam so that your chance of qualifying JEE Mains and getting a top college increases. Solve the previous years' papers, brush up your concepts and revise thoroughly.
You can also go through our college predictor to check which colleges you are eligible for through your JEE Mains score. The link for the college predictor is:
All the best.
