JEE NEET EXAM 2025-26 | frequently Ask Question,Doubts
How many students appeared for jee mains 2025
Total unique candidates registered (either January or April 2025): 1,539,848
Total unique candidates who appeared in 2025: 1,475,103
In Session 1 (January 2025) — 13,11,544 registered; 12,58,136 appeared.
In Session 2 (April 2025) — 10,61,840 registered; 9,92,350 appeared
How many students appeared for jee mains 2024
Total appeared (both sessions, all papers combined): ≈ 1,415,110 candidates.
Breakdown by session:
- Session 1 (January 2024): ~ 1,170,048 appeared
- Session 2 (April 2024): ~ 1,067,959 appeared
JEE NEET EXAM 2025-26 | frequently Ask Question,Doubts
What is jee ?
JEE stands for Joint Entrance Examination.
It is a national-level engineering entrance exam in India conducted for admission to top engineering colleges.
Types of JEE
JEE is conducted in two stages:
1. JEE Main
- Conducted by NTA (National Testing Agency)
- For admission to:
- NITs (National Institutes of Technology)
- IIITs (Indian Institutes of Information Technology)
- GFTIs (Government-funded Technical Institutes)
- Also acts as eligibility test for JEE Advanced.
Exam Papers:
- Paper 1 → B.Tech/B.E.
- Paper 2A → B.Arch
- Paper 2B → B.Planning
2. JEE Advanced
- Conducted by IITs
- Only top 2.5 lakh rankers of JEE Main can appear.
- For admission into:
- IITs (Indian Institutes of Technology)
Why is JEE important?
- It is the gateway to the best engineering colleges in India.
- Very competitive—lakhs of students appear every year.
Who can apply?
- Students who have passed or are appearing in Class 12 with Physics, Chemistry, and Maths.
How Many Attempts for JEE Main?
1. Number of Years Allowed
You can attempt JEE Main for 3 consecutive years after Class 12.
2. Number of Attempts per Year
JEE Main is held in 2 sessions each year:
- Session 1 → January
- Session 2 → April
You can attempt both sessions.
Total Attempts
3 years × 2 sessions per year = 6 total attempts
So, maximum = 6 attempts for JEE Main.
Why 2 attempts per year?
It gives students:
- a chance to improve score
- reduce exam pressure
- use the best score for ranking
RESOURCE FOR JEE : Study Material JEE / NEET
How to calculate percentile in jee ?
JEE Percentile Formula
NTA uses the normalization formula based on relative performance, not the raw marks.
Percentile Score = (100 × Number of candidates who scored ≤ your score) ÷ Total number of candidates in that session
This is calculated separately for each subject:
- Physics percentile
- Chemistry percentile
- Maths percentile
- Overall percentile
Finally, the overall NTA Score is the average of the percentile in all three subjects.
Example
Suppose:
- Total candidates in your session = 100,000
- Number of students scoring equal or less than you = 85,000
Then,
Percentile = (100 × 85,000 / 100,000)
= 85 percentile
This means you scored better than 85% candidates in that session.
Important Points
✔ Percentile ≠ Percentage
Percentile is relative position; percentage is marks out of 100.
✔ Highest score = 100 percentile
Not 100 marks — it just means “top scorer”.
✔ Percentile depends on paper difficulty
That’s why NTA uses normalization.
✔ Rank is derived from combined percentile of all sessions.
when is jee mains 2025
JEE Main for 2025 was held in two sessions.
- Session 1 (2025): January 22–30, 2025
- Session 2 (2025): April 2–9, 2025
JEE NEET EXAM 2025-26 | frequently Ask Question,Doubts
99 percentile in jee mains means how many marks
JEE Main Percentile → Marks Conversion (Approx.)
These are based on recent trends from 2022–2025.
1. 99.5 Percentile = How Many Marks?
Approx. 195 – 210 marks
2. 95, 98, 99.8 Percentile = Marks
| Percentile | Approx. Marks (out of 300) |
|---|---|
| 95 percentile | 120 – 140 marks |
| 98 percentile | 150 – 170 marks |
| 99.8 percentile | 240 – 260 marks |
3. Rank for 99 Percentile
| Percentile | Approx. Rank |
|---|---|
| 99.0 | ~8,000 – 9,500 rank |
| 99.5 | ~4,000 – 5,500 rank |
| 99.8 | ~1,200 – 1,800 rank |
| 99.9 | Top 800 – 900 rank |
Read Also: JEE Syllabus