CET Delhi 2019 (Delhi Polytechnic): Application Form, Eligibility, Exam Dates. Delhi CET 2019 Application Form can be filled till 19th April 2019. CET Delhi is the pre-requisite test for admission to engineering and non-engineering diploma programmes in the government and private polytechnic institutions of Delhi. It is conducted by the Department of Training and Technical Education, Government of National Capital Territory of Delhi. CET Delhi 2019 is also known as the Delhi Polytechnic entrance exam. This examination also provides entry to second year diploma courses. In addition to this, candidates can also take Delhi Common Entrance Test for admission to one year certificate courses in World Class Skill Center (WCSC). Candidates desiring for Delhi CET 2019 can check details about application form, eligibility, exam dates, pattern, admit card, etc.
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CET Delhi 2019 Exam Date
Here, we are mentioning all the important dates related to CET Delhi 2019.
| Events | Dates |
| Release of application form | 20th March 2019 |
| Last date for submission of application form | 19th April 2019 (5: 00 PM) |
| Application correction | 20th – 23rd April 2019 |
| Issue of Admit Card | 3rd week of May 2019 |
| Exam Date | 26th & 27th May 2019 |
| Declaration of Results | 14th June 2019 (4: 00 PM) |
| Counselling starts | 15th June 2019 |
CET Delhi Application Form 2019
- CET Delhi application form has been filled & submitted online only.
- No other mode is entertained for filling & submitting the application form.
- The application form is available from 20th March to 19th April 2019.
- Candidates have to apply separately for each test.
- The application will be rejected, in case of submitting more than one application form for the same test.
- Recent colour photograph (size 7.00 cm x 5.5 cm) and signature must be uploaded in the application form.
- The authority may also provide the correction facility in the application form.
- At last, candidates must retain the printout of filled-in application form for further use.
Application Fee
Non-refundable application fee of Rs. 400/- (Bank charges Rs. 18/- + GST Rs. 3/-) will be charged from every applicant. Fee can be deposited through credit card net-banking/debit-card.
Eligibility Criteria
The eligibility criteria of CET Delhi for all of its 5 tests is different. Here, we are mentioning the eligibility conditions test-wise:
Qualification Eligibility
| Test No. | Courses | Eligibility |
| Test – 1 | Engineering/Technology based Diploma Courses & Occupational Courses |
|
| Test – 2 | (i) Diploma Course in Modern Office Practice – English(ii) One year certificate course in Hospitality Operations, Retail Services, Finance Executive in World Class Skill Centre (WCSC) |
|
| Test – 3 | Diploma Course in Modern Office Practice – Hindi |
|
| Test – 4 |
(i) Diploma Course in Pharmacy
(ii) one year certificate course in Software Testing Assistant in World Class Skill Centre (WCSC) |
|
| Test – 5 | Lateral entry admission against 10% supernumerary seats in analogous diploma courses |
|
Age Criteria
The maximum age limit for various categories as on 1st August 2019 is given in the below table:
| Test No. | Gen (Boys) | Gen (Girls) | OBC (Boys) | OBC (Girls) | SC/ST/PWD (Boys) | SC/ST/PWD (Girls) |
| 1 | 21 | 23 | 24 | 26 | 26 | 28 |
| 2 | 24 | 30 | 27 | 30 | 29 | 35 |
| 3 | 24 | 30 | 27 | 30 | 29 | 35 |
| 4 | 24 | 30 | 27 | 30 | 29 | 35 |
| 5 | 24 | 30 | 27 | 30 | 29 | 35 |
CET Delhi Exam Pattern 2019
Aspirants can check details about the exam pattern to know how the question paper of CET Delhi is going to set for the academic year 2019-2024:
- Exam Mode: CET Delhi will be conducted as pen and paper based test.
- Number of Questions: The question paper will consist total 150 questions.
- Type of Questions: Only objective type questions will be asked in the examination.
- Language: For test 1, 4 and 5, the medium of question paper will be English and Hindi(Bilingual). For test 2, the question paper medium will be English only and for test 3, the medium will be Hindi only.
- Exam Duration: The duration of the examination will be 2 hours 30 minutes.
- Marking Scheme: Candidate will receive 4 marks for each correct answer.
- Negative Marking: For each incorrect answer, 1 mark will be deducted.
- Exam Centers: The exam centres will be only in Delhi.
| Tests | Subjects | No. of Questions |
| Test-1: | Mathematics | 50 |
| Physics | 40 | |
| Chemistry | 30 | |
| Biology | 15 | |
| English | 15 | |
| Test-2: | English Comprehension including Grammar | 60 |
| Current Affairs | 45 | |
| Logical & Analytical Ability | 45 | |
| Test-3: | Hindi Comprehension including Grammar | 60 |
| Current Affairs | 45 | |
| Logical & Analytical Ability | 45 | |
| Test-4: | Mathematics/Biology | 60 |
| Physics | 45 | |
| Chemistry | 45 | |
| Biology | 15 | |
| Test-5: | Mathematics | 60 |
| Physics | 60 | |
| Chemistry | 30 |
CET Delhi 2019 Syllabus
CET Delhi 2019 syllabus is different for different tests. The syllabus is of basically 10th or 12th class standard depending upon the test for which you are appearing. Syllabus of Physics, Chemistry, Mathematics and Biology is based on 12th standard.
CET Delhi Admit Card 2019
All the registered candidates can download their CET Delhi 2019 Admit Card by logging into the official website of CET Delhi. The admit card will be available for downloading before one week of the examination. Admit card must be retained safe till the admission process gets completed.
CET Delhi Result
CET 2019 result will be displayed on the official website of CET Delhi. The result will be declared on 14th June 2019. Admissions will be done on the basis of merit prepared as per the candidate’s performance in entrance test.
CET Delhi Counselling 2019
A web counselling will be conducted for CET Delhi. It will be statred from 15th June 2019. It is expected that there will be 4 rounds in the counselling process. At the time of counselling, candidates must bring all documents (in original) and one set of attested copies for verification purpose. To appear the counselling round, all the qualified candidates have to pay Rs. 700/-.
