i got 99.5 percentile in jee mains 2020 which college can i expect for cse branch ??
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it is difficult to evaluate the rank from the percentile as of now but to calculate the rank you could use the formula given here (100-your percentile)x869010/100
by this formula your rank would likely be less than 5000 and with that rank you can apply to the good NITs, it will be difficult to get admission in the older NITs which are on the top but you can get a good NIT with less than 5000.
Here's a list of an NITs that you can apply to
- National Institute of Technology Delhi
- Visvesvaraya National Institute of Technology, Nagpur
- Maulana Azad National Institute of Technology Bhopal
- National Institute of Technology Durgapur
- National Institute of Technology, Kurukshetra
- Dr. B R Ambedkar National Institute of Technology, Jalandhar
- National Institute of Technology Goa
- National Institute of Technology Puducherry
- National Institute of Technology Hamirpur
99.5–99.99 then it would be under 5000 rank. 99–99.5 then expected rank b/w 5000–8567. 98–99 then tough competition it would be somewhat 8567–15786. After this percentile it would be less chance to get core branch in good NITs {MNIT Jaipur,Trichi,Warangal,etc.}
You have to go down to less good NITs for getting in CSE branch.
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