JEE Main Rank Predictor: Percentile to All India Rank
Type your JEE Main percentile (NTA score) to get a likely All India Rank band, worked out from the official number of candidates and calibrated on the official count at the JEE Advanced cut-off. Only have your marks? The marks mode first turns them into a percentile range — published tables disagree, so that band is wider.
The total NTA score on your scorecard (the better of your two sessions)
Estimate · 2026 data · confidence: medium-high
Confidence: medium-high — calibrated on the official count at the 93 percentile cut-off.
✓ At or above the 2026 UR (General) JEE Advanced cut-off of 93.4123549 (96,873 candidates).
Step-by-step solution
- N = 15,38,468 unique candidates (2026, NTA)
- Calibration c = 96,873 ÷ ((100 − 93.4123549) × N ÷ 100) = 0.9558
- (100 − 97.5) × 15,38,468 ÷ 100 = 38,462
- × 0.9558 = about 36,763
- Band ±10% → AIR 33,000 – 41,000 (rounded outward)
| Category | Cut-off |
|---|---|
| UR (General) | 93.4123549 |
| EWS | 82.4164528 |
| OBC-NCL | 80.9232583 |
| SC | 63.9172792 |
| ST | 52.0174712 |
How to use the JEE main rank predictor
- Choose “My percentile” and enter the total percentile from your NTA scorecard (or choose “My marks” for a rougher estimate).
- Pick the data year — 2026 (latest) or 2025.
- Read the All India Rank band and whether you are above the JEE Advanced qualifying cut-off for UR; the category cut-offs are listed under the result.
Formulas
| Rank estimate | AIR ≈ (100 − percentile) × N ÷ 100 × cN = unique candidates (2026: 15,38,468); c = calibration ≈ 0.956 from the official count at the UR cut-off. |
|---|---|
| Band | ±10% of the estimate (±15% above 99.9 percentile)Rounded outward to 2 significant figures. |
| Percentile (per shift) | 100 × (candidates in your shift with a raw score ≤ yours) ÷ (candidates in your shift)NTA Information Bulletin 2026, Appendix VII. |
Solved examples
Example 1: 99 percentile in 2026
- (100 − 99) × 15,38,468 ÷ 100 = 15,385
- × 0.9558 = about 14,705
- ±10%
Answer: AIR about 13,000 – 17,000
Example 2: 97.5 percentile in 2026
- (100 − 97.5) × 15,38,468 ÷ 100 = 38,462
- × 0.9558 = about 36,763
- ±10%
Answer: AIR about 33,000 – 41,000
Example 3: Check: the 2026 UR cut-off 93.4123549
- (100 − 93.4123549) × 15,38,468 ÷ 100 = 1,01,349
- NTA’s official count at this cut-off = 96,873
- c = 96,873 ÷ 1,01,349 = 0.9558
Answer: the calibration used above
Why the formula needs a calibration
A percentile tells you what share of your shift scored the same as you or less, so (100 − percentile)% of candidates are ahead. Multiplying by the number of candidates gives a first guess at the rank. But percentiles are worked out per session, while ranks are over unique candidates — and about 8 lakh candidates sat both sessions in 2026, keeping only their better score. In 2025 and 2026 the plain formula overshot the official count at the UR cut-off by about 4.5%, so we multiply by about 0.956.
The calibration is checked at one point, the 93.4 percentile cut-off. Near it the band is reliable; far above 99.9 or well below 90 percentile the error can grow, which the result says.
Why marks give a wide band
NTA never publishes marks next to percentiles for real shifts. Coaching and news sites publish tables, but they disagree by up to 6 percentile points around 100–120 marks, several are copies of each other, and in one 2025 analysis the 99th percentile needed anywhere from about 148 to 203 marks depending on the shift. We show the full range across the tables we checked instead of picking one.
JEE Advanced qualifying cut-off
In 2026 the UR (General) cut-off for JEE Advanced was an NTA score of 93.4123549, and 96,873 candidates were at or above it; about 2.5 lakh candidates qualified across all categories. The 2.5 lakh figure is the sum of category quotas — it does not mean 93 percentile is rank 2.5 lakh.
Ties in JEE Main ranks are broken by the NTA score in Maths, then Physics, then Chemistry, then the ratio of wrong to right answers. Age and application number are no longer used (from 2025).
Frequently asked questions
What rank is 99 percentile in JEE Main 2026?
About 13,000 to 17,000. With 15,38,468 candidates, (100 − 99) × 15,38,468 ÷ 100 = 15,385, and the calibration on NTA’s official count brings it to about 14,700. It is an estimate; your rank is on your NTA scorecard.
How many students appeared in JEE Main 2026?
15,38,468 unique candidates appeared in Paper 1 (13,04,653 in Session 1 and 10,34,330 in Session 2; 8,00,516 sat both), according to NTA’s result press release of 20 April 2026.
What is the JEE Advanced cut-off 2026?
UR (General) 93.4123549, EWS 82.4164528, OBC-NCL 80.9232583, SC 63.9172792 and ST 52.0174712 (NTA scores).
Can I find my rank from my marks?
Only roughly. NTA publishes no marks-to-percentile table and the percentile depends on your shift, so the tool shows the range across published tables and the rank band for that whole range.
Which session counts if I appeared twice?
The better of your two total NTA scores. Ranks are given only after Session 2.
Sources
All accessed 13 October 2026. Official documents first; secondary sites are named for comparison only — their tables are not copied.
Official
- NTA press release, JEE Main 2026 final result (20 Apr 2026) — candidate counts, 100-percentilers, cut-offs
- NTA press release, JEE Main 2026 Session 1 (16 Feb 2026)
- JEE Main 2026 Information Bulletin (percentile, tie-break)
- JEE Main 2026 percentile distribution by shift
- JEE Main 2025 result notice (NTA document, copy hosted by Allen)
Keep learning
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