JAMB 2010 · UTME · Q2

Change the right side to base ten: $511_9 = 5 \times 81 + 9 + 1 = 415$.

The place values in base eight are 64, 8 and 1, so $6r7_8 = 384 + 8r + 7 = 391 + 8r$.

Set $391 + 8r = 415$: $8r = 24$, so $r = 3$, option C.