User:Ranjithraj.vasam/EAMCET world

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EAMCET WORLD

EAMCET WORLD(eamcetworld.com)is a public community portal of EAMCET aspirants and an encyclopedia of EAMCET.

EAMCET

EAMCET, which stands for Engineering Agricultural and Medical Common Entrance Test, is an Entrance examination required for admission to various Engineering and Medical colleges in the state of Andhra Pradesh, India.

About

The EAMCET test is a yearly examination taken by a large number of Engineering and Medical aspirants in the state. Presently the JNTU located in Hyderabad conducts the examination. This is the entrance exam for Engineering ('E' Category)and Medicine ('AM' Category- Agriculture and Medicine).From 2009 onwards 25% weightage is being given to 10+2 results. The rank will be based on 75% EAMCET marks and 25% Intermediate marks (ICSE,CBSE or INTERMEDIATE).

Universities Under EAMCET

Universities (& affilated colleges) that admit students under the counseling based on EAMCET ranks fall under these University Regions:

Andhra University (AU Region)
Osmania University (OU Region)
Sri Venkateswara University (SVU Region)

Format

The format of the test had to be changed many times to make the test fair for all students. An expert committee constituted by the Educational Department of the state government decides how the test is conducted. The test is of 160 questions with 40 marks for each subject( Maths A/Botany, Maths B/Zoology, Physics and Chemistry). Physics and Chemistry question papers are different for E and AM categories though the syllabus is the same. The questions are of multiple-choice type with four options. Many models of questions may be given like Match the following, choose the correct statement, Assertion-Reason and Multiple-correct type questions. Students have to fill the circles given in the OMR sheet with the correct option. Two or more answers or blank will also be regarded as incorrect. There is no negative marking.

Weightage

From 2009 onwards 25% weightage is being given to 10+2 results. The rank will be based on 75% EAMCET marks and 25% marks secured in groups(maths/botony,zoology+physics+chemistry) of Intermediate. (ICSE,CBSE or INTERMEDIATE).

The calculation would be

for 75% (EAMCET)

((secured marks)/(total marks))*75 =a

For 25% (GROUPS MARKS)

((secured marks)/(total marks))*25 =b

Aggregate will be 'sum of above two calculations(a+b)'

ex: if you got 96 in eamcet & 524 in groups

a=((92)/(160))*75 =43.125
b=((524)/(600))*25 =21.833
Total aggregate =43.125=21.833=64.958

References

External links