Any person having passed SSC (Std X) or equivalent examination and above can appear for NATA. However, only candidates who have qualified an aptitude test in architecture and have secured at least 50% marks in 10+2 or equivalent examination with Mathematics as one of the subjects, shall be eligible for admission to B.Arch. Course. No direct lateral admission is allowed at any year/stage of B.Arch. course based on any qualification.
NATA-2017 score shall be valid only for admission in the session 2017-18.
Who Will Fit ?
Prospective applicants desirous of taking admission to First year of undergraduate course in Architecture (Bachelor of Architecture) in India take NATA. NATA scores are used by admissions authorities of different Government, Govt. Aided & unaided schools / colleges of Architecture, to provide common measure for comparing the qualifications of applicants for admission in addition to their scholastic performance in 10+2 or PUC equivalent examination.
Where to Appear?
NATA is offered over major part of admission season at designated Test Centers located at colleges / schools of architecture in India. Updated list of Test centers is available at the website www.nata.nic.in
Who Accepts NATA?
All schools / colleges of Architecture, Government, Government aided, University
Departments, private unaided, including colleges affiliated to self-financed Deemed Universities and Private Universities, or any department of Architecture within a college, requires that its applicants take the NATA exam.
How to Apply NATA & Download NATA Admit card?
Register NATA 2017
Download NATA2017 Admit Card
NATA 2017 Exam Date : 16.04.2017 (11AM TO 2 PM)
Result Date : 10TH JUNE, 2017
|NATA 2017 - EXAM PATTERN
|DATE OF EXAM||PART||SUBJECT||QUESTIONS||MARKS||TOTAL MARKS||TIMINGS||MODE OF EXAMINATION
|16/04/2017 (SUNDAY) Time: |
11:00am to 2.00pm
|B||Drawing||2||40||80||90Mins||Paper and Pencil
NATA SYLLABUS - Refer Here
Tie Breaking Rules for Overall Score
Unified merit list of candidates will be published in NATA-2017. During merit listing, all candidates would get unique rank.
Therefore, if more than one candidate gets the same overall marks, tie breaking logic would be applied in the following order:
- More overall marks obtained in Mathematics test component
- Less wrong answers in attempted ones of Mathematics test component
- More overall marks in problems of Aptitude test component
- Less wrong answers in attempted ones of aptitude test component
- Earlier date of birth