Rahul took part in a cycling race with several other participants on a…

Concept: In race problems that give one fraction 'ahead' and one fraction 'behind', first fix the REFERENCE SET each fraction acts on, write the ahead-count and behind-count as those fractions of that set, and then turn the problem's STATED verbal condition into a single equation in the unknown total; when the two fractions add to more than 1 the relation is not a physical split of the reference set but a numerical condition to be solved exactly as given. Here the reference set is the participants other than Rahul (x − 1), and the stated condition is that the ahead-count plus the behind-count equals the TOTAL number of participants x — one more than the x − 1 others themselves, which is exactly what makes the fractions 1/5 and 5/6 combine into a solvable equation.

1. Let x = total number of participants, so everyone besides Rahul numbers (x − 1).

2. Participants ahead of Rahul = 1/5 × (x − 1); participants behind Rahul = 5/6 × (x − 1).

3. Apply the stated condition: 1/5(x − 1) + 5/6(x − 1) = x.

4. Take the LCM of 5 and 6 (= 30): 6/30(x − 1) + 25/30(x − 1) = x, that is, 31/30(x − 1) = x.

5. Cross-multiply: 31(x − 1) = 30x, so 31x − 31 = 30x, giving x = 31.

Verification: with x = 31, the other 30 participants give 1/5 × 30 = 6 ahead and 5/6 × 30 = 25 behind; 6 + 25 = 31, matching the stated condition that these two counts add up to the total number of participants (not just the other 30) — confirming x = 31.