JEE Main 2023 Toughest vs Easiest Shift: National Testing Agency successfully concluded the Session 2 exam on 15th April 2023. The session 2 exams were conducted in 2 different shifts and the difficulty levels of all 7 days and shifts were different. To determine fair rank and percentile for all the candidate NTA make use of the Normalization process. In the session 2 exam, the 11th April Shift 1 exam was the most difficult shift, whereas the 8th April Shift 1 exam was the easiest shift. As per the analysis by resonance, physics was the easiest subject in the JEE Main and Mathematics was the most difficult. Below candidate can check the JEE Main Toughest vs Easiest Shift along with how NTA normalizes the score in JEE Main.
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Toughest to Easiest Shift in JEE Main 2023 Session 2
Below candidates can check the list of the top 3 Toughest to Easiest Shifts in JEE Main 2023 Session 2 Exam:
S.No | Toughest Shift Details | Difficulty Level out of 3 | Easiest Shift Details | Difficulty Level out of 3 |
---|---|---|---|---|
1. | 11th April 2023 Shift 1 | 1.89 | 8th April 2023 Shift 1 | 1.40 |
2. | 6th April 2023 Shift 1 | 1.81 | 15th April 2023 Shift 1 | 1.42 |
3. | 6th April 2023 Shift 1 | 1.78 | 10th April 2023 | 1.43 |
By now you must be aware of the most difficult and easier subject in the JEE Main Session 2 Exam. Next, let's take a look at how NTA normalizes the JEE Main Score.
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How will NTA Normalize the JEE Main Scores?
As mentioned above the difficulty level of all the days and shifts in the JEE Main Session 2 exam was different. So to determine a fair percentile for all the candidates, NTA makes use of the normalization process. Authorities make use of the following details while calculating the percentile score:
- Total Candidte appeared for the exam
- Highest raw score in the JEE Main Session 2 2023
The below candidate can check two different cases to understand how NTA normalizes the JEE Mian score:
Case 1: Let's consider that the JEE exam is conducted in two different sessions and the lowest & highest score of the candidate is as follows:
Session | Candidate's Highest Raw Score | Total Candidates Appeared for the Exam | Total aspirants who secured marks Equal to or Less than the highest raw score | Percentile of the candidate who secured the Highest Marks | Explanation (Apply Formula) |
---|---|---|---|---|---|
JEE Main Session 1 | 335 | 28,012 | 28,012 | 100 Percentile | (28012/28012) X 100 = 100 |
JEE Main Session 2 | 346 | 32,541 | 32,541 | 100 Percentile | (32541/32541) X 100 = 100 |
From the above one can easily determine that the candidate must have secured 100 marks
Case 2: Below candidate can check the percentile of the students in a particular session
Candidate Name | Candidate Raw Score | Aprirants Scored Less raw score than the candidate | Total Candidates Appeared for the JEE Exam | Candidte Percentile | Detailed Explanation |
---|---|---|---|---|---|
A | 335 out of 360 | 28,000 | 28,012 | 99.9571 | (28000/28012) X 100 = 99.9571 |
B | 330 out of 360 | 27,012 | 28,012 | 96.4301 | (27012/28012) X 100 = 96.4301 |
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