RRB JE 2024 answer key objection deadline today; calculate probable scores

NEW DELHI: The Railway Recruitment Boards will close the answer key objection facility for RRB Junior Engineer (JE), Depot Material Superintendent (DMS), Chemical & Metallurgical Assistant (CMA) and other posts today December 28. Candidates who appeared in the recruitment exams can raise objections against the RRB JE 2024 answer key through the official website of RRBs.

As per the schedule, candidates will be able to raise objections against the RRB JE answer key 2024 up to 11:55 pm. To raise objections against the RRB JE answer key 2024, candidates must use their login credentials such as registration number and date of birth along with supporting documents.

The RRB JE recruitment drive 2024 aims to fill up a total of 7,951 vacancies in the department. Out of which 17 vacancies are for chemical supervisor, research and metallurgical supervisor and 7,934 posts will be filled for junior engineer, depot material superintendent and chemical and metallurgical assistant.

Candidates shortlisted from CBT 1 will progress to subsequent stages, including CBT 2, document verification, and medical examination round.

Also read RRB SI 2024 answer key objection window closes today; calculate probable scores

RRB JE Answer Key 2024 objection fee

Candidates who are not satisfied with the solutions have to pay an online fee of Rs 50, along with any applicable bank service charges per each objection raised.

if an objection is found valid, the fee paid for such objections will be refunded to the candidate after deducting the applicable bank charges, the notice further informed. The refund amount will be processed to the same account used for the original payment.

RRB JE 2024 Answer Key: Calculate probable scores

The RRB JE exam consisted of 100 questions, each carrying one mark, with a penalty of one-third mark for each incorrect answer. To estimate your score:

Marks Scored=(Number of Correct Answers×1)−(Number of Incorrect Answers×13)\text{Marks Scored} = (\text{Number of Correct Answers} \times 1) - (\text{Number of Incorrect Answers} \times \frac{1}{3})