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What is a Good Rank for JEE Main 2025?: Various factors like academic goals, target institution and branch of study will contribute to a good rank in JEE Main 2025. The top 1000 ranks are considered exceptional and ensures admission into premier IIIT's, NIT's and other top institutions with higher chances for CSE and ECE as a branch of study. A rank of 10,000 or less allows admission into IIIT's, NIT's and government-funded technical institutions (GFTI's) with popular branches, while CSE may require a better rank. The JEE Main 2025 exam for session 2 was conducted on April 2, 3, 4, 7, 8 & 9, 2025.
Also Check: JEE Main 2025 City Intimation Slip | JEE Main 2025 Exam Date
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A rank ranging between 10,000 to 15,000 opens opportunities in mid tier IIIT's and NIT's and state level government and private engineering colleges. The JEE Main rank of 50,000 might give less probability for admission in NIT's and IIIT's, but state government and private institutions remain viable options for this category of candidates. Hence, a proper study schedule and clear career goals will help candidates secure a good rank in JEE Main 2025 securing a seat in a top NIT or IIIT. For more details, candidates can refer to the article below.
Related: JEE Main Marks vs Rank 2025 | What is a Good Score for JEE Main 2025
Candidates with JEE Mains percentiles between 90 and 95, or those holding ranks between 14,000 and 45,000, have a higher chance of securing admission to one of the IIITs.
Rank Type | Rank Range |
Good | Up to 25000 |
Above average | 25000 to 50000 |
Average | 50000 |
Candidates can refer to the table below for the expected JEE Main 2025 Marks vs Ranks. Additionally, the Previous Year's Marks vs Rank data for JEE Main is available on this page.
S.No | JEE Main 2025 Score out of 300 (Expected) | JEE Main Rank 2025 (Expected) |
1 | 288- 294 | 20-11 |
2 | 280-284 | 44-22 |
3 | 270- 279 | 107-63 |
4 | 252- 268 | 522-106 |
5 | 231-249 | 1385-546 |
6 | 215-230 | 2798-1421 |
7 | 202-214 | 4666-2862 |
8 | 190-200 | 6664- 4830 |
9 | 175-189 | 10746-7151 |
10 | 161-174 | 16163-11018 |
11 | 149-159 | 21145-16495 |
12 | 132-148 | 32826-22238 |
13 | 120-131 | 43174-33636 |
14 | 110-119 | 54293-44115 |
15 | 102-109 | 65758-55269 |
16 | 95-101 | 76260-66999 |
17 | 89-94 | 87219-78111 |
18 | 79-88 | 109329-90144 |
19 | 64-87 | 169542-92303 |
20 | 44-62 | 326517-173239 |
21 | 1-42 | 1025009-334080 |
S.No | JEE Main 2024 Score out of 300 | JEE Main Rank 2024 |
1 | 286- 292 | 19-12 |
2 | 280-284 | 42-23 |
3 | 268- 279 | 106-64 |
4 | 250- 267 | 524-108 |
5 | 231-249 | 1385-546 |
6 | 215-230 | 2798-1421 |
7 | 200-214 | 4667-2863 |
8 | 189-199 | 6664- 4830 |
9 | 175-188 | 10746-7152 |
10 | 160-174 | 16163-11018 |
11 | 149-159 | 21145-16495 |
12 | 132-148 | 32826-22238 |
13 | 120-131 | 43174-33636 |
14 | 110-119 | 54293-44115 |
15 | 102-109 | 65758-55269 |
16 | 95-101 | 76260-66999 |
17 | 89-94 | 87219-78111 |
18 | 79-88 | 109329-90144 |
19 | 62-87 | 169542-92303 |
20 | 41-61 | 326517-173239 |
21 | 1-40 | 1025009-334080 |
The JEE Main percentile score is calculated using the following formula:
Percentile Score of a Candidate = 100 × [(Number of candidates with raw score less than or equal to the given candidate score )/(Total number of candidates appeared in that particular session)]
The formula plays a useful role for estimating the candidate’s performance in comparison to other candidates in the same session.
The authority has announced the JEE Main result 2025 dates for session 1 on the official website. According to the notification, the JEE Main 2025 result will be announced on February 12, 2025. To check the result, JEE Main 2025 aspirants will require login credentials like application number and date of birth.
Create a Study Plan
Solve as many JEE Main mock tests as possible
Refer to NCERT Books
Revision is Key
Let’s Have a look at the JEE Exam pattern
Prepare Notes
Study Resources and Material
Understand the JEE Main Conceptual Depth
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Frequently Asked Questions (FAQs)
A good rank depends on the institute and branch. For top NITs (CSE/IT), a rank below 5,000 is ideal, while other core branches require a rank below 15,000. IIITs and GFTIs accept ranks up to 25,000, and state-level colleges consider ranks below 50,000.
For the newer IITs, a rank higher than 4500 is generally sufficient. To be considered for admission to any IIT, candidates must score at least 250 or higher and fall within the 85-95 percentile in JEE Main. However, to secure a seat in Computer Science Engineering (CSE) at an IIT, a score between 280 and 300+ in JEE Advanced is typically required.
A score of 250 or higher is considered strong. A percentile range of 85–95 is ideal for candidates aiming for admission to NITs and IITs.
The qualifying marks for JEE Main 2025 are as follows: 90.78 for General, 0.001 for General-PwD, 75.62 for EWS, 73.61 for OBC-NCL, 51.98 for SC, and 37.23 for ST.
To register to sit for JEE Main in 2025, candidates will require a minimum of 95 percentile equivalent to scores 170-180.
For JEE Main 2025, it is desirable to be in the first thousand ranks. The ranks below 5000 are also not bad for getting a good chance to enter the recognized colleges.
On Question asked by student community
Here’s a plan for JEE Mains 2026 in 4 months:
1. Divide time: 2 months for Class 12 syllabus, 1 month for Class 11, 1 month for full revision & mock tests.
2. Daily schedule: 6–7 hours study; 50% for theory & problem-solving, 50% for practice & revision.
3. Topic-wise focus: Prioritize high-weightage chapters and weak areas first.
4. Daily problem practice: Solve previous year questions and chapter-wise exercises.
5. Weekly tests: Take 1 full-length test weekly, analyze mistakes, and revise weak concepts.
6. Consistency: Avoid skipping days; maintain notes and formula sheets for quick revision.
If you want to crack JEE exam you read to dedicatedly prepared for that from the scratch to the advance focus on high weightage topic and prepare question in the time based and continuously practice the previous question this will help to know the pattern of JEE exam questions
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With your JEE Main CRL 476199 and OBC rank 173160, admission to Civil Engineering at Indira Gandhi Institute of Technology (IGIT), Sarang under JoSAA counseling is highly unlikely, since even civil branch closing ranks in IGIT usually remain within 50k–80k CRL (general) and much lower for OBC. However, you may try through state counseling (Odisha JEE) where cutoffs are sometimes more flexible, but chances are still very low. You should keep other private/state colleges as backup.
Hey! With a JEE Main rank of around 1 lakh, you still have a good chance to get admission in several engineering colleges that provide quality education at a reasonable fee. For government colleges, you can aim for NITs or GFTIs in branches like Civil, Mechanical, or Chemical Engineering, as these usually have slightly higher closing ranks. For example, colleges like NIT Goa, NIT Puducherry, and NIT Srinagar might have seats available in certain branches.
Apart from NITs, state government engineering colleges are also a good option, especially under home state quota. They often have lower tuition fees and good faculty, so you can get a solid education without spending much.
If you are open to private colleges, there are options like VIT Vellore, KIIT Bhubaneswar, or Amity University that provide decent infrastructure and placements. However, the fees may be higher than government colleges.
My advice is to focus on branches that are less competitive, check all counseling rounds, and make use of state quotas if applicable. With proper planning, a rank of 1 lakh can still help you get a good college and start your engineering career success
fully.
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