The cutoff for the B.Des course at IIT Delhi in 2025 (Round 1) was a closing rank of 35 for the General (AI) category. For other categories, the cutoff ranks were 21 for OBC and 11 for SC candidates. Admission is based on UCEED scores followed by a counselling process.
B.Des Cut-off
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New answer posted
a year ago
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Rashmi Sinha
Contributor-Level 10
The NIFT Cut-off 2026 has been declared for various categories of students who fall under the General AI quota. The closing rank for the round four was 2707, and the most popular course with the lowest cut-off rank, which was 4, is the Master of Fashion Technology (M.FTech).
For other courses the cutoff ranks are below:
| Course | NIFT Round 4 Cutoff 2026 |
|---|---|
| B.Des. in Accessory Design | 458 |
| B.Des. in Textile Design | 1350 |
| B.Des. in Knitwear Design | 1890 |
| B.FTech. in Apparel Production | 482 |
| Master of Fashion Management (MFM) | 221 |
| B.Des. in Fashion Design | 735 (R2) |
| B.Des. in Leather Design | 2707 |
| B.Des. in Fashion Communication | 365 (R2) |
| Master of Fashion Technology (M.FTech) | 4 |
New answer posted
a year ago
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Rashmi Sinha
Contributor-Level 10
IIT Roorkee B.Des cutoff 2026 has been released for multiple rounds and categories. For the General AI category students, the maximum rank required in UCEED was 90 for admission to the B.Des course. To check out the category-wise break-up, refer to the table below:
| Category | UCEED Cutoff 2026 |
|---|---|
| General | 90 |
| OBC | 51 |
| SC | 24 |
| ST | 17 |
| EWS | 20 |
New answer posted
a year ago
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New answer posted
a year ago
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Neha Arora
Contributor-Level 10
IIT Roorkee accepts the UCEED entrance exam scores, followed by its counselling process for admission to BDes courses. So, to get a seat at the IIT Roorkee B.Des course, candidates must have a valid score to be eligible for the counselling process. For the General AI category students, the maximum UCEED cutoff rank required for admission was 90.
New answer posted
2 years ago
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Virajita Sharma
Contributor-Level 10
The NIFT Cutoff 2026 has been announced for Round 2 for different categories. During Round 2 cutoffs, the last ranks were 212 - 2430 where Master of Design (M.Des.) was found to be the most difficult course among all courses at National Institute of Fashion Technology, Kannur for the students of General AI category students. In order to take admission in the institute, the students have to score a valid score. For other categories of students, you can visit this page.
New answer posted
2 years ago
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Neha Arora
Contributor-Level 10
NIFT Kannur Cutoff 2026 Round 4 was released for different M.Des, MFM and B.Des courses. The Cutoff was released for different categories under the AI quota. For the General AI category candidates, the Cutoff ranks ranged between 354 and 3462 .
The course with highest rank for the General AI category candidates was B.Des. in Knitwear Design at 3462. The most competitive course was Master of Design (M.Des), with closing cutoff of 354 .
Candidates can refer to the table below to check the closing Cutoff ranks for different courses offered at NIFT Kannur:
| Course | 2026 |
|---|---|
| Master of Design (M.Des.) | 354 |
| Master of Fashion Management (MFM) | 471 |
| B.Des. in Fashion Communication | 733 |
| B.FTech. in Apparel Production | 991 |
| B.Des. in Fashion Design | 1595 |
| B. Des. in Textile Design | 2932 |
| B.Des. in Knitwear Design | 3462 |
New answer posted
2 years ago
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Neha Arora
Contributor-Level 10
According to the NIFT Jodhpur cutoff 2026 round 4 lies in the range from 361 to 1944 for BDes, B.FTech and MFM courses Therefore, those students who have a rank within 1919 can qualify for admission in NIFT. This is the cutoff range for the General AI category .
The cutoff ranks for other categories are provided below :
| Category | NIFT Jodhpur cutoff round 4 |
|---|---|
| OBC | 2880-5997 |
| SC | 6075-10222 |
| ST | 8533 (R3) |
New answer posted
2 years ago
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Rashmi Sinha
Contributor-Level 10
Yes, with a cutoff of 36, you can get into IIT Hyderabad for B.Des admission in Round 2 of the UCEED Seat Allotment process. The Round 2 closing rank for B.Des was 36 for OBC category students, so a cutoff of 36 should be sufficient for admission.
Candidates can refer to the table below to check out the round 1 and the round 2 B.Des cutoff 2024 for the OBC category candidates:
| Rounds | UCEED Cutoff Rank for OBC |
|---|---|
| 1 | 36 |
| 2 | 36 |
New answer posted
2 years ago
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Janki H Gajjar
Contributor-Level 10
For the students belonging to the General AI quota, the IIT Bombay UCEED cutoff 2025 was 14 in the last round. The cutoff rank varies for different rounds and categories. Candidates can refer to the table below to view the category-wise UCEED cutoff 2025 for the last round.
| Categories | UCEED Last Round Cutoff 2025 |
|---|---|
| General | 14 |
| OBC | 13 |
| SC | 7 |
| ST | 5 |
| EWS | 4 |
New answer posted
2 years ago
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Janki H Gajjar
Contributor-Level 10
In order for the OBC AI candidates to qualify for the BDes program, they should have a rank of 3836 or below .
Candidates may check out the following table that shows the closing ranks for different programs in Round 4.
| Course | Round 4 (Closing Rank) |
|---|---|
| B.Des. in Accessory Design | 1456 |
| B.Des. in Textile Design | 2496 |
| B.Des. in Knitwear Design | 3836 |
| B.Des. in Fashion Design | 1174 |
| B.Des. in Fashion Communication | 321 |
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