MHT CET 2021 answer key released at cetcell.mahacet.org; Download now

The State CET Cell, Maharashtra has released the MHT CET 2021 answer key on October 10. Get direct link to check the MHT CET answer key here.

MHT CET 2021 answer key released at cetcell.mahacet.org; Download now
mohit.dhyani | Oct 12, 2021 - 7:54 a.m. IST
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NEW DELHI: The State Common Entrance Test Cell, Maharashtra has released the MHT CET answer key 2021 on October 11. Applicants can check the MHT CET 2021 answer key from the official website - cetcell.mahacet.org. The login details required to check MHT CET answer key 2021 are application number and password. Candidates are also allowed to raise objections in MHT CET exam answer key 2021 from the official website.
Direct link to download MHT CET 2021 answer key

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MHT CET 2021 exam was conducted from September 20 to October 1. However, the MHT CET exam for candidates who missed the exam due to heavy rainings was conducted on October 9 and 10. Applicants who appeared for the exam can download the MHT CET 2021 answer key from cetcell.mahacet.org.
Also Check: MHT CET 2021 result date

How to check MHT CET answer key 2021?

  • Visit the MHT CET official website- cetcell.mahacet.org.

  • Click on the direct link to check the MHT CET answer key.

  • Enter the login credentials in the required fields.

  • Check details mentioned in the answer key.

  • Download the answer key of MHT CET 2021 for future reference.

Applicants are allowed to raise objections in MHT CET answer key 2021 from October 12 to 13. Candidates are required to submit the supporting documents for MHT CET 2021 answer key challenge from the candidate portal.

After considering all the objections raised by students, the exam authority will release the MHT CET 2021 final answer key at the official website. Along with the final answer key, the State CET Cell will also release the MHT CET 2021 result. The result of MHT CET 2021 will be released on the basis of the normalisation process as mentioned below.

MHT CET Percentile Score = 100 * (Number of candidates in exam with normalized marks ≤ the candidate) ÷ Total number of candidates in the exam


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