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2026

JoSAA

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JEE Main 2026 Percentile vs Rank – Complete Conversion Table

99.9% = ~1,200 rank · 99% = ~12,000 · 95% = ~60,000 · 90% = ~1,20,000 · Based on 12 lakh candidates

How Percentile Converts to Rank

JEE Main 2026 rank is calculated from the NTA score (percentile) using the following formula:

Rank ≈ (1 − NTA Score / 100) × Total Unique Candidates

For JEE Main 2026, approximately 12 lakh unique candidates appeared (counting each person once across both sessions). NTA uses the best NTA score from Session 1 or Session 2 for rank computation. Ties in NTA score are broken using inter-se merit criteria defined by NTA.

Note: This formula gives an approximation. The actual rank distribution also accounts for category reservations, tie-breaking rules, and the exact total of unique candidates.

JEE Main 2026 Percentile to Rank Conversion Table

PercentileApprox RankCandidate ZoneBest College Option
1001–10Perfect ScoreAll IITs (via JEE Adv)
99.9~1,200Top 0.1%IIT zone / NIT CSE top
99.5~6,000Top 0.5%NIT Trichy CSE (Gen)
99.0~12,000Top 1%NIT Warangal CSE
98.0~24,000Top 2%NIT Calicut CSE
97.0~36,000Top 3%Mid NITs CSE
95.0~60,000Top 5%NIT branches / IIITs
90.0~1,20,000Top 10%GFTIs / CSAB NITs
85.0~1,80,000Top 15%CSAB / State
80.0~2,40,000Top 20%State counselling
75.0~3,00,000Top 25%Private colleges

* Rank estimates assume ~12 lakh unique candidates. College options are indicative for General category via JoSAA 2026.

Rank Required for Top Colleges

CollegeBranchApprox Closing Rank (Gen)
NIT TrichyCSE~3,500
NIT WarangalCSE~5,200
NIT SurathkalCSE~7,500
NIT CalicutCSE~10,000
IIIT HyderabadCSE~1,500
NIT JaipurCSE~18,000
NIT RaipurCSE~35,000
GFTI (NIFFT)Manufacturing~1,20,000
JEE Main
JEE Advanced
DASA / NRI

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Frequently Asked Questions – Percentile vs Rank

Common questions about JEE Main 2026 percentile to rank conversion and college eligibility

1
What rank corresponds to 90 percentile in JEE Main 2026?
With approximately 12 lakh unique candidates in JEE Main 2026, a 90 percentile score corresponds to an approximate rank of 1,20,000. The formula is: Rank ≈ (1 − percentile/100) × total candidates. So: (1 − 0.90) × 12,00,000 = 1,20,000.
2
Which NIT can I get with 95 percentile in JEE Main 2026?
95 percentile in JEE Main 2026 corresponds to approximately rank 60,000. At this rank, options include GFTIs and newer NITs via CSAB. For General category candidates, most top NITs (like NIT Trichy, Warangal) require under 25,000 rank for CSE. However, branches like Civil or Mechanical Engineering at NITs like NIT Calicut or NIT Jaipur may have closing ranks beyond 50,000–70,000.
3
How do I calculate my JEE Main rank from percentile?
Use the formula: Rank ≈ (1 − NTA Score/100) × Total Unique Candidates. For 2026, total unique candidates is approximately 12 lakh. Example: For 97 percentile → (1 − 0.97) × 12,00,000 = 36,000. Note: This is an approximation; actual rank depends on tied candidates and category reservations.
4
What is the difference between percentile and rank in JEE Main?
Percentile (NTA score) shows your relative performance within a session (0–100 scale). Rank is your absolute position among all unique candidates after the best-of-two-sessions rule is applied. Two candidates with the same percentile in different sessions may get different ranks after normalization.
5
What rank is needed for NIT vs GFTI admission in JEE Main 2026?
For top NITs (CSE): General rank under 10,000–15,000. For mid NITs (CSE): 15,000–40,000. For newer NITs (various branches): 40,000–1,50,000. For GFTIs: Up to 2,00,000+ depending on institute and branch. CSAB Special Round fills seats beyond JoSAA closing ranks, sometimes up to rank 2,50,000+.

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Related Pages

Marks vs PercentileJEE Main Cutoff 2026JEE Main 2026 ResultNIT Cutoff 2026JoSAA 2026 GuideCollege Predictor

Disclaimer

Rank estimates are approximate based on ~12 lakh unique candidates. Actual ranks depend on NTA normalization and tie-breaking rules. College cutoffs are indicative from JoSAA 2025.