My daughter has scored 62.3 percentile.is she selected for advance??

Hi there.



The final cut off list will be declared by NTA after the April session of the JEE mains examination.



As for now, your probable rank can be calculated.



Now, A total of 874469 candidates wrote the exam. Since the percentile is calculated from a scale ranging from 100 to 0 for each session of examinees, the formula applies the total number of students in the January session which is 874469 /100 multiplied by the percentage of people who are ahead of you to arrive at the rank.



Therefore, the formula to calculate the JEE Main 2019 rank for January 2019 is



JEE Main 2019 rank (probable) = (100- 62.3 ) X 874469 /100 = 329675 (approximately)



This rank may inflate or deflate depending on the total number of candidates who write the examination. in the April session.



You could take a probable calculation and see where you stand. This means that you take a hypothetical number of the candidates who will appear in April. If you go by past trends, the figures are somewhat between 11 lakhs to 12 lakhs. Using this as a basis, use the NTA score and see which JEE Main rank you will get.



Using the same formula,



If NTA percentile score is 62.3, JEE Main rank will be (100-62.3) X 1200000/100 = 452400



As per experts from various leading coaching centres, an NTA percentile score of 85 will stand a chance at JEE Advanced but this is still not clear. Until April 30, there is also no hard fast means to know the minimum cutoff to qualify for JEE Advanced 2019.



Now, using this estimated calculation, you can check previous year cut off trends and think of an idea of the probable colleges.



Check the following links for previous year trends in cut off :



https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff/amp



For college predictor of JEE mains, check the link below:



https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-college-predictor/amp



Hope this helps.

Good luck!