UPSSSC: Recruitment for 3446 posts of Technical Assistant, applications will start from May 1

UPSSSC: Recruitment for 3446 posts of Technical Assistant, applications will start from May 1

UPSSSC Technical Assistant exam 2024: Uttar Pradesh Subordinate Service Selection Commission (UPSSSC) has invited applications for recruitment to 3446 posts of Technical Assistant. Candidates who want to apply for this post can visit the official website for information related to recruitment. Before applying, candidates are advised to read the information related to recruitment, only then start the further process.

Know about category wise posts

3446 posts of Technical Assistant will be filled. In which 1813 posts of unreserved category, 509 posts of Scheduled Caste (SC), 151 posts of Scheduled Tribe (ST). Applications have been invited for 629 posts of OBC category and 344 posts for Economic Weaker Section category candidates.

application fee

To apply for the posts of Technical Assistant, the age of the candidates should be minimum 21 years and maximum 40 years.

UPPSC Technical Assistant exam 2024: You can apply through these steps

Step 1 – First of all you have to go to the official website

Step 2 – Click on 'Live Advertisements' link on the home page.

Step 3- Once the link for the application process is activated, one can apply for the posts of Technical Assistant.

Step 4 – After which candidates can register and proceed with the application process.

Step 5 – In which the information required in the application form will have to be filled and the necessary documents will have to be submitted.

Step 6 – After filling all the necessary information, the form will have to be submitted and the application fee will be paid. If you want, you can download the form and take its printout.

The exam pattern will be like this

The exam pattern will consist of 100 objective multiple choice questions. A total of two hours will be given to solve the exam. One mark will be given for each question. There will be negative marking of 0.25 marks for wrong answer.

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