National Testing Agency (NTA) has released the preliminary answer key for JEE Main 2022 session 1. Students who appeared for the engineering entrance exam from June 23 to June 29 can download their answer keys from nta.ac.in or jeemain.nta. .nic.in. The answer key consists of the question and its corresponding correct answer.
If students feel that the marked answer is wrong or find any error in the answer key, they can raise an objection against it. A fee of Rs 200 will be applicable per objection. Students can raise the challenge only till July 4, 5 PM. No challenge will be entertained without receipt of processing fee. Challenges through any other means will not be accepted as per NTA.
JEE Main 2022: How to raise objections
Step 1: Visit jeemain.nta.nic.in
Step 2: Click on the answer key link
Step 3: Click on the answer which you think is wrong
Step 4: Select correction option, explain your reasons, attach supporting documents
Step 5: Pay the Fee, Submit
“The challenges raised by the candidates will be verified by the panel of subject experts. If the challenge of any candidate is found correct, the answer key will be revised and accordingly the response of all the candidates will be applied. On the basis of the revised final answer key, the result will be prepared and declared. No individual candidate will be informed about the acceptance/rejection of his/her challenge. The key to be finalized by the experts after the challenge will be final,” the NTA said in an official notice.
JEE Main 2022: How to estimate marks
Students can also use the answer key to estimate the marks obtained in the examination. Students can give themselves 4 marks for each correct answer or if the answer written by them matches with the answer given in the answer key. The final total will be their raw score. NTA result is given in percentile score.
The highest RAW marks obtained will be given 100 percentile while comparative marking will be given in all the results. To calculate the rank, the marks obtained by the students in all the four sessions will be normalized and the rank will be calculated as per the rules.
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