Konkan Gyanpeeth College of Engineering Raigarh(Mh) News & Updates 2025: Notifications, Notice, Result

List of Questions Expected to be Dropped from JEE Main 2025 Provisional Key Session 2: The JEE Main 2025 Provisional Key for Session 2 has sparked discussions regarding the removal of certain questions due to incorrect options, errors in wording, or misalignment with the syllabus. After a thorough discussion, it has been determined that 9 questions are likely to be dropped from the JEE Main 2025 Provisional Key for Session 2. For Shift 1 on April 3, a bonus will be awarded for 1 Physics question due to a wrongly framed question. Similarly, for the Chemistry question in Shift 1 on April 4, 2025, a bonus is also likely to be given. In addition, if the NTA considers the correct answer for the Mathematics question from April 2, 2025 (Shift 1) to be 2477 instead of 4949, all students who chose 4949 will receive 4 marks, while others will have their marks deducted. In this case, NTA's answer will be regarded as correct for valuation. Additionally, if the authorities decide to drop a question due to an incorrect option, all candidates who attempted that question will receive bonus marks.

45,000 Rank in JEE Mains 2025 Percentile and Marks: Below is an estimate of the JEE Mains 2025 percentile and marks for a 45000 rank, based on the previous year's trend. According to our analysis, a 45000 rank may correspond to a 99.965+ percentile, necessitating scores of 151.38+ for an easy paper, 139.415+ for a medium paper, and 116.65+ for a tough paper. Please note that this is merely an estimate and actual conversion data may vary due to additional factors such as candidate performance, previous year's trends, and total seat availability. Nonetheless, candidates can check out the table for reference purposes. Attaining a 45000 rank, equivalent to a 96+ percentile, is a significant achievement. As one of the top JEE ranks, it presents a favourable opportunity for admission to elite engineering colleges.

1000 Rank in JEE Mains 2025 Percentile and Marks: 1000 rank in JEE Mains 2025 exam is considered an extraordinarily good rank, with which the candidates can expect to secure a seat at any of the top-listed NITs or IITs. Also, the candidates can expect to take admission to CSE, ECE or EE programs with 1000 rank in the JEE Mains 2025 exam. Based on the previous years’ analysis, it can be assumed that to secure 1000 rank in the JEE Mains 2025 exam, the candidates should get 247.82+, 236.85+ and 223.06+ marks if the papers are easy, moderate, and tough. Whereas the expected percentile for 1000 rank in the JEE Mains 2025 exam is tentatively 99.933+. Candidates who are aiming to secure a seat to any of the core subjects in any of the top listed NITs or IITs should be ranked between 1000 in the JEE Main 2025 exam. 

2000 Rank in JEE Mains 2025 Percentile and Marks: According to the analysis, All India Rank (AIR) of 2000 is expected to translate to an impressive 99.867+ percentile and 229.72+ marks if the level of paper is considered to be medium. However, it's essential to note that the marks required to achieve this rank can vary significantly depending on the difficulty level of the paper. For instance, in an easy paper, students may need to score around 241.55+ to achieve an AIR of 2000, whereas in a tough paper, the required marks may be slightly lower, above 212.73. Here's a detailed breakdown of the expected marks and percentiles for AIR 1550 to 2000 for easy, moderate, and tough papers to help students anticipate their selection chances in top engineering institutions. 

90,000 Rank in JEE Mains 2025 Percentile and Marks: As per the experts, 90,000 rank in the JEE Mains 2025 exam is a moderate rank and the candidates who will score 94.3+ percentile, can expect to secure the specific rank. This implies that the candidates can hardly get admission to the top ranked NITs or IIITs with 90000 rank in the JEE Mains 2025 exam. For instance, the candidates can take admission to Mizoram University, Shri Mata Vaishno Devi University, and some NITs (National Institutes of Technology) or IIITs (Indian Institutes of Technology) with 90000 ranks. According to the analysis, 90,000 rank is equivalent to 125+, 112.5+, 92+ marks of easy, medium and tough level papers respectively.