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JEE Main rank predictor 2025 is the best tool to calculate JEE Main rank for session 2 before the final result. Careers360's JEE Mains rank predictor 2025 helps in estimating your rank based on the marks obtained in the JEE Main 2025 exam. It’s a great way to know if you have a chance to get into colleges like NITs or IITs. JEE Main Rank Predictor uses smart AI technology to give you an accurate JEE Main marks vs rank comparison.
To use the JEE Main rank predictor 2025 tool, just enter your JEE Main score, exam shift, and the difficulty level of your paper. JEE Mains rank predictor also checks past cutoffs, total seats, and category-wise seat distribution to predict your rank. Remember, the JEE Main rank predictor 2025 gives an estimate. The actual rank will be announced with the official JEE Main results.
Scroll up and enter the following details of JEE Main exam
Careers360's JEE Main Rank Predictor tool provides a most accurate rank and percentile based on the scores obtained in the JEE Main exam. However, for a more accurate results, candidates should wait for the official NTA JEE results. The rank calculator for JEE Main 2025 is considered most reliable tool and its accuracy depends on several factors.
The JEE Main 2025 Rank Predictor tool by Careers360 uses comprehensive data from various colleges to provide precise predictions. The tool is designed to help students anticipate their potential ranks and explore college details, including cutoffs, with accuracy. Here are the key benefits of using the JEE Main Rank Predictor for 2025:
Candidates can evaluate their standing among other JEE Main aspirants, gaining insight into the competition level.
Students can pre-select potential colleges based on predicted results and assess their eligibility for admission.
By using the JEE Mains rank predictor tool, candidates can stay ahead of the competition, making early, informed decisions about their future.
It’s important to remember that the rank shown by the tool is based on the information provided by the candidates, not the actual rank. Nevertheless, using the JEE Main Rank Predictor from Percentile will give users valuable insights into different aspects of their results.
Yes, JEE Mains rank predictor is free to use.
Admission into most of the NITs closes at 20000 rank.
In the above provided field, just select the JEE Main percentile instead of scores to know the expected rank.
Yes, the Careers360's JEE Rank predictor also provide the tentative percentile based on the score.
Yes, Careers360 JEE Mains Rank predictor is available fee.
The JEE Main rank with 99.5 marks will be around 61 to 80K while with 99.5 percentile, you will get a rank between around 5320 - 7354.
With 75 marks wyou will get a percentile around 85 or 90. It is a good score but possibility to get a good college with this marks is tough.
Careers360's JEE Main rank predictor is the best.
I think you want to say 86 percentile because JEE Mains releases merit lists based on percentile, not percentages. So, with this percentile, you can get into several government colleges or private colleges in the city of Hyderabad. There many colleges in hyderabad that accept mains score, some popular ones arfe
But there are several colleges in this rank, which you can check with our college predictor in this link from Career360:
https://engineering.careers360.com/jee-main-college-predictor?utm_source=qna&utm_medium=jee_cp
Hello,
There is good news for you. You can get admission to the JECRC Foundation for the B.Tech program with just 12th marks. They can give you admission without a JEE Mains score; they just need your 12th marks, and based on this merit, you can obtain a seat in this institute. While you have 90.20% in your 12th, so it's a strong merit from your end. There is a high possibility that you will get admission to JECRC Foundation.
I hope it's clear your query!!
With 88 percentile in JEE Mains, chances for CS/ECE/EEE in top NITs or IIITs are low. However, you may get these branches in private universities like Amity, Sharda, Galgotias, or lower-tier state government colleges through state counselling.
Hello aspirant,
The syllabus of JEE MAINS PAPER 2A for degree B.Arch inculdes three subjects: maths, aptitude test, and drawing.
The exam is of 400 marks with 77 questions and of 3 hours. -1 negative marking will be given for every wrong answer of maths or numerical value question. For drawing test, there will be no negative marking.
There are many platforms online providing sample papers, one of which is careers360. The link of which, I am attaching here,
https://engineering.careers360.com/articles/jee-main-syllabus-2026
Best of luck for your preparation.
Hello
NIT Kurukshetra – Civil Engineering (JEE Main Cut-offs)
2024 Round 3 and 4 Final Closing Ranks
All India (AI Other State) Category – Gender Neutral
Opening Rank: Approximately 15,535
Closing Rank: Around 29,292 in Round 1, and up to 33,266 by Round 4
Home State (HS Haryana) Category – Gender Neutral
Opening Rank: Approximately 9,853
Closing Rank: Around 38,638 across various rounds
Expected Cut-Offs for 2025 (Projected)
The estimated closing ranks for Civil Engineering in the early rounds are as follows:
All India (Open): Between 15,520 and 34,638
EWS (All India): Between 4,090 and 46,628
OBC NCL (All India): Between 7,946 and 12,000
SC (All India): Between 1,445 and 5,155
ST (All India): Between 960 and 1,136
Note: These expected cutoffs are only estimates based on previous year trends and are subject to change depending on this year's exam and counselling dynamics.
Dear candidate , Cutoffs for JEE Main 2024 for Tatyasaheb Kore Engineering college ( TKIET ) . Warananagar : - CSE : 65,000 ( All India ) , 36,000 ( Home state ) . - Electronics and Telecom : 40,000 - Chemical Engineering : 47,000 . - Cyber security : up to - 85,000 .
Hello Srinidhi,
In order to acquire Mechanical Engineering at NIT Kurukshetra, generally the required JEE Main rank will be in the range of:
The cutoffs that I have provided above change from year to year, as well as category and gender. For reserved categories (SC/ST/OBC/EWS), the ranks are much more relaxed.
For JEE Main 2024, the closing ranks at Rajarambapu Institute of Technology, Islampur (General category, All India) were approximately:
Computer Engineering: 2,10,000 – 2,40,000
CSE (AI/ML or IT): 2,30,000 – 2,80,000
Electronics & Telecommunication: 3,60,000 – 3,85,000
Electrical Engineering: 4,70,000 – 4,80,000
Civil Engineering: 4,15,000 – 4,85,000
Mechanical Engineering: 6,05,000 – 6,15,000
Robotics & Automation: 5,50,000 – 6,00,000
These are approximate ranges based on Round 3 or final round cutoffs. Some variation can occur based on seat availability and category.
As per your query, the JEE rank required for NIT Kurukshetra in Computer Science Engineering branch is,
All India Closing Rank (General category) : 6562
Home state Closing Rank (General category) : 8198
For more information, click on the link given below,
https://www.careers360.com/university/national-institute-of-technology-kurukshetra/cut-off
Regards
Yes, you are eligible for JEE Advanced 2026.
You passed Class 12 in 2024 (CBSE) but didn’t meet the 75% requirement, so you reappeared through NIOS in 2025 and scored 75%. Since NIOS is a recognized board and you now meet the 75% eligibility, you’re allowed to appear for JEE Advanced in 2026. Also, JEE Advanced allows attempts in the year of passing 12th and the following year. So as long as you qualify JEE Main 2026 and are in the top 2.5 lakh candidates, you’re eligible.
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