Tag archives for JEE Main


What is the cut-offs of JEE Main Exam

  Updated on : 22/11/2017

 by : Admin

JEE or Joint  Entrance Exam Main is the most popular Engineering Entrance Exam. It is conducted to offer admissions to candidates in Undergraduate Engineering programmes in all the Engineering Universities and Colleges except NITs (National Institutes of Technology), IIITs (Indian Institutes of Information Technology). JEE Main exam is also the eligibility test for JEE Advanced which is the entrance exam to secure admissions in programmes which is offer by 16 IITs NITs and Indian School of Mines (ISM). The top 224,000 score holders in JEE Main are only eligible to apply for JEE Advanced exam. Every year, over 12 lakh candidates appear in this national level engineering entrance test. JEE Main Exam cut-offs are released after the JEE Main exam results are out. Cut-offs are made according to toughness of the paper, number of Candidates appeared for JEE Mains, and it varies from low to high. Cut-offs are also made to limit the admission. JEE exam generally are of moderate level still it is the toughest one in comparison to other Engineering Entrance Exam. Though the question pattern derives from 11- 12 level syllabus of Physics, Chemistry and Mathematics, still the questions goes upto the toughest  level. Nearly 10 to 11 lakh students appeared for the exam (though vary every year) almost every year where only nearly 20,000 candidates could clear the examination. As per the new eligibility, in order to be considered for admissions, candidates must secure either 75% aggregate marks (General Category) and 65% percent (Reserved Category) in the qualifying examination or they have to be in top 20 percentile in their respective class 12th/equivalent qualifying examination board.JEE Main Exam cutoffs for appearing in JEE Advanced, for the admission of NIT, IIT was for General category(81) students while the cutoff for OBC(49), SC(32), ST(27) and PwD (1)students, respectively. For further information, please contact our Experts.demo

Is JEE the toughest engineering exam in India

  Updated on : 18/11/2017

 by : Admin

Engineering as a career has attracted students in India in a large way and a large number of aspirants taking the engineering exams are testimony to this fact. Engineering is a multi-disciplined field with new disciplines and branches being added with the advancement of science and technology. Though in the modern days a number of new fields has been added, but the basic engineering disciplines remains in the field of  Mechanical, Chemical, Civil, Electrical and Aerospace engineering.The students gets admission into the different engineering courses through  a number of national/state/university level entrance examination. Among all the examination JEE Exam (Joint Entrance Test) is the most important and challenging examination. JEE exam is of two types; JEE Main and JEE Advanced. Almost every engineering college and institute accepts the scores of JEE mains conducted by CBSE. But to get admission into the IIts and NITs  the candidates needs to clear JEE Advanced conducted by any one of  IIT for which to clear JEE main is mandatory. The JEE Main exams can be attempted for three succeeding years. The JEE Main exam is on the same lines of the CBSE Class 11 and 12 Physics, Chemistry and Math syllabus. Generally the difficulty level of the test is of  medium level.Normally Physics and mathematics section are little tough than the Chemistry, though may vary year to year. In the last exam in 2017 also these two section was tough than chemistry. Nearly 10 t 11 lakh students appeared for the exam (though vary every year) almost every year where only nearly 20,000 candidates could clear the examination. When we come to  JEE Advanced exam it can be attempted for only two consecutive years. The syllabus of the two exams also differs from JEE Main.  The JEE Advanced requires a few extra topics to be covered for the exam along with 5 the JEE Main syllabus. To answer the JEE Advanced questions, the candidates must have in-depth knowledge of the concepts and their applications. Speed and accuracy is focused on in JEE Advanced exams. Though the JEE Advanced exam are generally of moderate level, but the difficulty level of the JEE advanced exam is little higher than the JEE mains. Approximately  15 lakh students appears for the exam every year, only 1.5 lakh could clear the exam. For more information regarding the details of the examination please contact our experts. - Ask Our Expertsdemo

How is the JEE Mains rank calculated?

  Updated on : 27/01/2017

 by : Admin

As the JEE Main 2017 is drawing near you are being more tensed and you are looking for all possible ways to stay on track. Among all the various details that you have searched about JEE entrance exam, one more thing that is keeping you busy is the way JEE entrance exam ranks are calculated.Candidates who are appearing for JEE Main 2017 must stay aware of the fact that admissions to NITs, IIITs and GFTIs and that are done based on their All India Rank. However, most of the JEE entrance exam aspirants may not know the fact that while calculating the JEE Main rank Class 12th marks are converted to normalized scores in order to calculate their composite score and award the desirable ranking against it.How is JEE entrance exam list prepared?If we look deep into the fact then you will come to know that JEE Main Rank List is ready with all the JEE Main 2017 composite scores & Normalized Marks of Class 12th. The ratios in which they are calculated are 60:40.To calculate the composite score they follow a simple numerical method, as follows –Composite Score = (JEE Main Marks)*60+ (Normalized score of Class 12th marks)*40 / 100How they calculate the normalized score of Class 12? Well, as per our view we have seen that normalized score is the average of 2 JEE Main scores. To get this score the examiner need to follow four easy steps –·         As per the class 12 result, the percentile is calculated suing the formula:Number of candidates in board with total marks less than you X 100 / Total no. of candidates in the entire group·         The score of all the JEE Main 2017 is arranged in a merit and as per percentile, they are chosen by various reputed colleges. Let this be A1.·         In step 3 of the process the JEE entrance exam score belonging to a corresponding JEE Main percentile is something that is next up the row to get figured out. Let this be A2·         Now in the fourth step you need to average A1 and A2 and get the composite score against which the rank for the JEE entrance exam rank will be decided.After all the scores in hands, the merit list is determined and the ranks are announced to the world. You must have gained much knowledge by this time, right? demo

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