National Institute of Technology, Rourkela (NIT Rourkela)
CGPA to percentage
Quick answer: At National Institute of Technology, Rourkela (NIT Rourkela), a CGPA of 8.5 is about 85%, using the official formula % = CGPA × 10. (official source) Last verified 2026-07-19.
National Institute of Technology, Rourkela (NIT Rourkela) — CGPA → %
How National Institute of Technology, Rourkela (NIT Rourkela) converts CGPA to percentage
National Institute of Technology, Rourkela (NIT Rourkela) (Odisha) uses the formula % = CGPA × 10 to convert a 10-point CGPA into a percentage for transcripts, job applications and higher studies. Enter your CGPA above to convert it instantly.
Worked example
Take a CGPA of 8.5 on the 10-point scale. Applying the official National Institute of Technology, Rourkela (NIT Rourkela) formula % = CGPA × 10 gives a percentage of about 85%. Change the CGPA in the calculator above and the percentage updates instantly using this same formula — no rounding shortcuts.
Why the formula matters
National Institute of Technology, Rourkela (NIT Rourkela) uses % = CGPA × 10 on its 10-point scale. Using the official conversion keeps your percentage defensible if an admissions office or employer asks how you calculated it.
Source & verification
NIT Rourkela states Equivalent Percentage = CGPA × 10, to be adopted notionally at the recipient's discretion. The institute follows a 7-point grade scale (Ex=10, A=9, B=8, C=7, D=6, P=5, F=2); the ×10 conversion is the officially notified rule regardless. Note on older records: an earlier NIT Rourkela conversion certificate used Percentage = 10 × CGPA − 5 (so e.g. CGPA 8.5 → 80% rather than 85%), and some older grade cards or third-party sources still cite that −5 form. It is superseded by the 2021 Senate-notified ×10 rule above.
Comparing universities? See the full CGPA to percentage directory or calculate your CGPA first with the CGPA calculator.