JKCET 2024 Answer key out: The Jammu and Kashmir Board of Professional Entrance Examinations (JKBOPEE) has published the answer key for the Jammu and Kashmir Common Entrance Test (JKCET) 2024, which was held on June 9. Initially scheduled for June 5, the exam was rescheduled to June 9. The state-level Common Entrance Test (CET) is conducted for admission to undergraduate engineering courses in various government and private institutes across Jammu and Kashmir.
The provisional answer key is now available on the official JKBOPEE website, jkbopee.gov.in. The JKCET 2024 exam included three sections: Physics, Chemistry, and Mathematics, with a total of 180 multiple-choice questions, each worth one mark. Candidates were penalized 0.25 marks for each incorrect answer. The exam was held offline and lasted for three hours.
Here’s the direct link to download the JKCET Answer Key
Candidates can download the answer key to estimate their scores. The board allows candidates to challenge the answer key by submitting representations, along with a payment of Rs 1000 per paper, at the BOPEE offices in Srinagar or Jammu. These challenges must be submitted within three working days after the exam.
How to calculate probable scores
To compute their scores in the JKCET 2024 exam, candidates need to tally their correct and incorrect answers. According to the exam pattern, candidates receive one mark for every correct response and face a penalty of 0.25 marks for each incorrect answer. Using the JKCET 2024 answer key, candidates can apply the following formula:
The provisional answer key is now available on the official JKBOPEE website, jkbopee.gov.in. The JKCET 2024 exam included three sections: Physics, Chemistry, and Mathematics, with a total of 180 multiple-choice questions, each worth one mark. Candidates were penalized 0.25 marks for each incorrect answer. The exam was held offline and lasted for three hours.
Here’s the direct link to download the JKCET Answer Key
Candidates can download the answer key to estimate their scores. The board allows candidates to challenge the answer key by submitting representations, along with a payment of Rs 1000 per paper, at the BOPEE offices in Srinagar or Jammu. These challenges must be submitted within three working days after the exam.
How to calculate probable scores
To compute their scores in the JKCET 2024 exam, candidates need to tally their correct and incorrect answers. According to the exam pattern, candidates receive one mark for every correct response and face a penalty of 0.25 marks for each incorrect answer. Using the JKCET 2024 answer key, candidates can apply the following formula:
- Multiply the number of correct answers by 1.
- Multiply the number of incorrect answers by 0.25.
- Subtract the total incorrect answers from the total correct answers to determine the final score.
- JKCET Score = [1 x number of correct answers] – [0.25 x number of incorrect answers]
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