The JEE (Main) results, which were announced on Tuesday, will be used for two purposes.
The result will be used to decide the eligibility of a student to appear in JEE (Advanced). 1.5 lakh candidates, proportionately from all the categories, namely, general, OBC, SC, ST, etc, would be selected in a straightforward manner based on the results.
The result will be used to decide the eligibility of a student to appear in JEE (Advanced). 1.5 lakh candidates, proportionately from all the categories, namely, general, OBC, SC, ST, etc, would be selected in a straightforward manner based on the results.
The second use of this result would be to determine a place (rank) of students in the merit list.
This is prepared based on the JEE (Main) and Standard XII results by a weighted formula, as illustrated below.
The merit list for admissions to NITs and many other engineering colleges is based on the following 3 factors:
(a) 0.6 multiplied by the marks obtained in JEE (Main), (b) 20% weightage of the percentile of a student calculated from his XII Board marks, and (c) 20% weightage of the percentile of a student calculated among the candidates who appeared in JEE (Main) 2013 from the Board of the candidate. This will give the relative standing of the candidate among the total candidates who appeared in JEE (Main).
A percentile score simply gives the position of a candidate among all the students who appeared in the examination. For example, 90 percentile means that 90% of the students are placed below him, or alternatively 100-90 = 10% students are above him.
Since the last two scores are percentile scores, they need to be rescaled to the JEE (Main) marks for additions of the similar units in the above three components.
This is prepared based on the JEE (Main) and Standard XII results by a weighted formula, as illustrated below.
The merit list for admissions to NITs and many other engineering colleges is based on the following 3 factors:
(a) 0.6 multiplied by the marks obtained in JEE (Main), (b) 20% weightage of the percentile of a student calculated from his XII Board marks, and (c) 20% weightage of the percentile of a student calculated among the candidates who appeared in JEE (Main) 2013 from the Board of the candidate. This will give the relative standing of the candidate among the total candidates who appeared in JEE (Main).
A percentile score simply gives the position of a candidate among all the students who appeared in the examination. For example, 90 percentile means that 90% of the students are placed below him, or alternatively 100-90 = 10% students are above him.
Since the last two scores are percentile scores, they need to be rescaled to the JEE (Main) marks for additions of the similar units in the above three components.
Therefore, after getting the percentile of a candidate in the above two cases (b) and (c), one has to find the corresponding JEE (Main) scores for calculating the final score.
| XII Marks & Percentile | Marks in JEE (Main) (X) | Weight (Y) | Weighted sum X * Y | ||
|---|---|---|---|---|---|
| JEE Score | 200 | 0.6 | 120 | ||
| XII Result (based on 5 subjects) | Marks 400 Percentile 90 | Look at 90 percentile of JEE Total List, say 240 | 240 | 0.2 | 48 |
| Contribution from JEE of the same Board | -- | Look at 90 percentile of JEE Board-wise list, say 180 | 180 | 0.2 | 36 |
| TOTAL | 204 | ||||
sir
ReplyDeletea student has scored 150 out of 360 in jeemain.He secured 475 out of 500 in cbse board exams.How the normalisation works here....?
please illustrate.