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JEE main 2017: Cut-off goes low, know what is required to qualify for JEE advanced

JEE Main 2017 results: The percentage is calculated from the marks scored divided by the total marks and multiplied by 100

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JEE Main 2017 results: Students were greeted with a pleasant surprise as the Central Board of Secondary Education (CBSE) declared the JEE Main 2017 results today and the cut-off for JEE advanced was unexpectedly low this year. The cut-off for the general category is 81, which brings a twist in prior predictions which believed the cut-off would be between 95 to 105 this year.

More than one lakh students are eligible for JEE Advanced from the common rank list. The Joint Entrance Exam 2017 main were held in April 2, 8 and 9, this year.

Read | JEE Advanced 2017 result out. Check here

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“The cut off was low this year due to the level of difficulty of the mathematics paper,” says Venkat Ramana, JEE exam head of TIME Hyderabad. This year, the JEE main score will not be determined on the basis of the class 12 board marks. The eligibility for JEE advanced will completely be dependent on the JEE main score and rank.

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Last year, the JEE main cut-off was 100 and OBC, SC and ST categories was at 70, 52 and 48 respectively. The cut-off in 2015 was 105 for general category candidates and for OBC, SC and ST category candidates it was 70, 50 and 44 respectively.

About 10.20 lakh students participated in the JEE main exam this year. On the basis of JEE Main rankings, about 2.20 lakh top scorers will be given chance to appear in JEE Advanced 2017. The qualified candidates have to get registered in JEE Advanced.

D Ankita, an aspirant who wasn’t expecting to score well is surprised with the low cut-off. “I felt maths would bring down my numbers but I have made it to the JEE Advanced,” said she.

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Read | JEE main 2017 results out: Kalpit Veerwal scores perfect 360 by 360

The JEE Advanced paper will be conducted by the Indian Institute of Technology (IIT), Madras on May 21, 2017. To be eligible for advanced candidates will be required to have scored at least 75 per cent in their class 12 board examinations (65 per cent for SC and ST students).

Candidates can calculate their score based on their percentage or their percentile. The percentage is calculated from the marks scored divided by the total marks and multiplied by 100. The percentile is calculated with the total marks of all five subjects to determine the candidates position in the board. The percentile is used to calculate the normalised score.

Read | IIT JEE Mains 2017: Results declared on April 27

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While selecting the candidates from the final list after advanced, the following order will be followed:

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Only the top 2,20,000 scorers will be allowed to appear for the advanced paper and candidates should have scored at least 50.5 per cent for the same. This percentage varies for reserved categories — 27 per cent for OBC and NCL, 15 per cent for SC and 7.5 per cent for ST.

For more stories on JEE main 2017, click here

 

First published on: 27-04-2017 at 15:29 IST
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