National Testing Agency has announced the results today for JEE mains 2019 January attempt. A lot of students are curious about the ranks and percentile calculation based on the normalized score provided by NTA.
Percentile scores will be used by NTA to rank students, which will further depend on the number of students appeared for the exam.
As per our expertise and understanding, we have analyzed how the rank will be calculated by NTA. Below is the formula for calculation of your rank for JEE mains 2019:
The exact percentage of people ahead of you is - 100 minus your total percentile score
Now, 874469 are the number of candidates appeared for JEE mains 2019 January attempt
Rank Formula = (100 - your total score) X 874469 /100.
The interesting question here is what if the number of students appearing for Jee Mains April 2019 attempt is much more ( which will obviously be the case), what will happen to rank? How will it be affected?
Suppose - The number of candidates for the April exam increased to 1400000, then your final rank now will be (100 - your total score) X 1400000/100
Let’s take an example here:
Suppose your total Percentile is 90.70
Your Rank will be ( 100-90.70 ) X 874469/100, which is 81325.
Now if the candidates increase to 1400000 in April, your rank will be (100 - 90.70) x 1400000/100, which is 1,30,200.
As per previous year JEE mains, Any rank below 2,00,000 qualify for JEE advanced.
By this way, you can have quite an accurate idea about the rank and your qualification for JEE advanced 2019. The exact rank for JEE mains 2019 will be announced after both the January and April 2019 result compilation on 30th April 2019
We hope that this article helps you with the prediction of your JEE mains 2019 rank and helps you prepare better for JEE mains April 2019 attempt.
Do you know that 60% of JEE mains 2019 January attempt came from our Most Asked Concepts section of JEE Knockout? Prepare Smart for Jee Mains 2019 April attempt. Begin your preparation NOW - JEE mains 2019 April - Knockout
Disclaimer - that this is our understanding actual rank may vary and we will not be responsible for your decision
For a train engine moving with speed of , the driver must apply brakes at a distance of before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed The value of is ______________(Assuming same retardation is produced by brakes)
200
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A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The energy of the block at . The spring constant of the spring is ________
50
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Given below are two statements :
Statement I : Sulphanilic acid gives esterification test for carboxyl group.
Statement II : Sulphanilic acid gives red colour in Lassigne's test for extra element detection.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I is incorrect but Statement II is correct
Both Statement I and Statement II are incorrect
Statement I is correct but Statement II is incorrect
Both Statement I and Statement II are correct