what should be my jee main score lies between...? to get NIT ROURKELA "CS" branch general category , with home state quota.
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Answer (1)
19th Jan, 2020
Hey Rashmi,
This year's cut-off are yet to release, but according to 2019 cut-off for CS branch, general category in female and home state quota, the closing cut-off rank was 10,746. To get this rank, you must have 98-99 percentile score. The results have already been released, so you can check your percentile and then calculate your rank using the formula:
(100- percentile)869010/100.
Here, 869010 is the number of candidates appeared in January session of JEE mains.
Kindly check the link below to see the previous year cut-offs:
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela
Also, you can use our College predictor for knowing the possibilities:
Hope this helps!
This year's cut-off are yet to release, but according to 2019 cut-off for CS branch, general category in female and home state quota, the closing cut-off rank was 10,746. To get this rank, you must have 98-99 percentile score. The results have already been released, so you can check your percentile and then calculate your rank using the formula:
(100- percentile)869010/100.
Here, 869010 is the number of candidates appeared in January session of JEE mains.
Kindly check the link below to see the previous year cut-offs:
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela
Also, you can use our College predictor for knowing the possibilities:
Hope this helps!
1 like
4 Comments
Comments (4)
thanks bro, but I'm Male🤞
Reply
19th Jan, 2020
R.Binod
okay, so for male category, you should have rank below 10,000 and expected percentile around 98-99 only. The percentile is kind of same for both.
Harshita Gupta
thanks 😊 guptajii
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