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Delhi CET 2019 online application form released; eligibility, fees, exam date

Delhi CET 2019: The application process has begun today, April 1, 2019 and will conclude on April 30, 2019. Four different tests will be conducted for admission to diploma courses and lateral entry admissions.

By: Education Desk | New Delhi |
April 1, 2019 3:38:35 pm
CET, cet 2019, delhi cet 2019, common entrance test, college admission, diploma admission, lateral diploma admission, cet 2019 online form, form delhi cet 2019, cetdelhi.nic.in, cetdelhi.nic.in, and tte.delhigovt.nic.in, education news Delhi CET 2019: The application process will close on March 30, 2019. (Representational Image)

Delhi CET 2019: The online application process for the Common Entrance Test (CET) 2019 for admission to diploma programmes has begun on the official websites — cetdelhi.nic.in and tte.delhigovt.nic.in. The application process has begun today, April 1 and will conclude on April 30, 2019.

A total of 4,335 seats across colleges affiliated to Department of training and technical education (DTTE) are on offer through the entrance exam. There are four tests under CET. Test 1 will be for engineering/ technology-based diploma courses and occupational courses, test 2 for diploma course in modern office practice (Hindi and English). test 3 is for diploma courses in Pharmacy and test 4 is for lateral entry (for admission in the second year directly).

Delhi CET 2019: How to apply

Step 1: Visit the official website, cetdelhi.nic.in
Step 2: Click on the link ‘Apply CET 2019 for admission in diploma courses’
Step 3: Click on ‘Apply’ under Test 1, 2, 3 or 4
Step 4: Click on ‘apply’ under ‘new registration’
Step 5: Fill details to register
Step 6: Login- using the registration number
Step 7: Fill form upload documents
Step 8: Make payment

Delhi CET 2019: Fee

Candidates will have to pay application fee of Rs 400 and bank charges, if applicable.

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Delhi CET 2019: Exam pattern

In test 1 and test 4, candidates will be asked to answer 150 objective-type questions. In test-2, 310 questions will be asked out of which the candidate has to attempt 150 questions. In test-3, there will be a total of 210 questions, out of which the students will have to attempt 150 questions by opting 60 questions either from Mathematics or from Biology.

Candidates will be given 2 hours 30 minutes to solve the exam, an additional 50 minutes will be given to candidates belonging to PwD category. The test will carry 600 marks. The exam will be conducted in Hindi and English.

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