MHT CET result 2024 on June 19; tie-breaking policy, score calculation formula

NEW DELHI: The Maharashtra CET Cell will be declaring the Maharashtra Common Entrance Test (MHT CET) result 2024 soon. Recently, the MHT CET 2024 result date has been revised. The MHT CET result 2024 for PCB and PCM will be declared on June 19. Earlier, the MHT CET result date was June 10. Candidates who appeared for the entrance exam can access and download the MHT CET result 2024 by visiting the official website, cetcell.mahacet.org.

MHT CET 2024 exam was conducted from April 22 to May 17. The MHT CET answer key 2024 was published on May 21. Candidates could raise objections against the MHT CET 2024 answer key till May 26. The final answer key will be out along with the MHT CET result 2024. Candidates will require their application number and date of birth to access and download the MHT CET result.

MHT CET Result 2024: Tie-breaking policy

Candidates can have a look at the MHT CET 2024 tie-breaking policy below.

• In case of a tie, candidates with the fewest incorrect attempts in Physics, chemistry, and mathematics or biology will be given preference.
• If the tie remains, those with the fewest incorrect attempts in Mathematics, Biology will be selected.
• If the tie continues, candidates with fewer incorrect answers in chemistry will have an edge.
• If the tie remains, those with higher scores on the Class 12 board exam will be given priority.
• If the tie remains, individuals older in age will be given preference.

MHT CET 2024 Result: Normalisation method and score calculation

The MHT CET 2024 exam was held across multiple shifts and days, potentially leading to variations in the difficulty level of question paper sets. The normalization method is used for MHT CET results. This method aims to prevent any advantage or disadvantage to candidates due to differences in question paper difficulty. Percentile scores of MHT CET candidates are computed accurately to five decimal places using a specific formula:

MHT CET percentile score = 100 * (Number of candidates in the exam with normalized marks ≤ the candidate) ÷ total number of candidates in the exam.