i got 90.49 percentile in jee mains can i get nit Bhopal under state quota and I'm in general category
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Answers (3)
28th Jan, 2020
hi,
Hope you are doing great.
According to your percentile your rank will be as follows ,
100- percentile obtained total no. of students 100=
100-90.49869010100
=82642
According to previous year trends , you may not get into NIT Bhopal even under state quota.
But maybe it can vary this year.
For better understanding I am adding a link for clearance https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-manit-bhopal/amp
Hope you are doing great.
According to your percentile your rank will be as follows ,
100- percentile obtained total no. of students 100=
100-90.49869010100
=82642
According to previous year trends , you may not get into NIT Bhopal even under state quota.
But maybe it can vary this year.
For better understanding I am adding a link for clearance https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-manit-bhopal/amp
Comments (0)
28th Jan, 2020
no it will be very difficult for general category.but you can even get by filling right choices during counciling.you can easily get into get any govt college with such a percentile.so best of luck for your future.
Comments (0)
Student Expert
28th Jan, 2020
Hello,
Your expected rank will be around 82642.You can calculate your expected rank using the formula given below
(100-your total percentile)*869010/100.
869010 is the total number of candidates appeared in January session.
So it is further calculated as
(100-90.49)*869010/100=82642
It is expected that 1200000 candidates will appear in April session so it will be (100-your total percentile)*1200000/100.Your expected rank will change.
Now in 2019 the cut off were as follows for each branch for general category
You can predict the colleges in which you may get seat using our JEE main college predictor
Your expected rank will be around 82642.You can calculate your expected rank using the formula given below
(100-your total percentile)*869010/100.
869010 is the total number of candidates appeared in January session.
So it is further calculated as
(100-90.49)*869010/100=82642
It is expected that 1200000 candidates will appear in April session so it will be (100-your total percentile)*1200000/100.Your expected rank will change.
Now in 2019 the cut off were as follows for each branch for general category
- Chemical Engineering -24829
- Civil engineering- 30670
- Computer science engineering- 12229
- Electrical engineering- 22374
- Electronics and communication engineering- 18541
- Mechanical Engineering- 31269
- Materials Science and Metallurgical Engineering-35386
You can predict the colleges in which you may get seat using our JEE main college predictor
Comments (0)
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