I have scored 92.85 percentile in mht cet 2019. 75% in HSC. I belong to open category. What will be my ranking?
Hello Vineet....
Your Score in mh cet is 92.85 in 2019 and in HSC is 75%....
So as per your score your rank will be 29,549 Rank...
To calculate the percent as per your score the these is the formula..
Since the percentile is determined from a scale running from 100 to 0, the recipe applies the all out number of understudies which is 413284/100
duplicated by the level of individuals who are in front of you to touch base at the rank.
In this manner, the equation to compute the rank is
Likely Rank = (100-percentile score) X 413284/100
In the event that percentile score is 92.85, position will be
(100-92.85) X 413284/100 = 29,549.
In this manner 29,549 will be the plausible position.
Like these the rank is calculated using the MH CET Percentile...
Hope these will be helpful..
All the best!!!
Hi Vineet,
Total number of students who applied for the Engineering field was-
1,20,000 (PCM) + 1,70,000 (PCMB) = 2,90,000
According to your percentile, your rank would be-
2,90,000 * 0.0715 = 20,735
With this rank, you can get a good college through MHT CET 2019 counselling.
You can check all details about MHT CET counselling from the given link-
https://engineering.careers360.com/articles/mht-cet-counselling
All the best!!!
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