Rahul took part in a cycling race with several others on a circular track. If…
2023
Rahul took part in a cycling race with several others on a circular track. If 1/5 of the participants are ahead of him and 5/6 of the participants are behind him, find the total number of participants.
- A.
15
- B.
31
- C.
29
- D.
23
Attempted by 2 students.
Show answer & explanation
Correct answer: B
General relation for such “ahead + behind” fraction problems: with x total participants on a circular track, the (x − 1) other participants get split by two given fractions — one fraction ahead of the person, one behind. When the two fractions add up to more than 1 (as here), the two counts overlap around the loop, so their sum equals the full field size x, not just the (x − 1) other participants: fraction_ahead × (x − 1) + fraction_behind × (x − 1) = x.
Let x = total number of participants, so everyone besides Rahul numbers (x − 1).
Participants ahead of Rahul = 1/5 × (x − 1); participants behind Rahul = 5/6 × (x − 1).
Apply the general relation: 1/5(x − 1) + 5/6(x − 1) = x.
Take the LCM of 5 and 6 (= 30): 6/30(x − 1) + 25/30(x − 1) = x, that is, 31/30(x − 1) = x.
Cross-multiply: 31(x − 1) = 30x, so 31x − 31 = 30x, giving x = 31.
Cross-check: with x = 31, the other 30 participants split as 1/5 × 30 = 6 ahead and 5/6 × 30 = 25 behind; 6 + 25 = 31, matching the total field size exactly, confirming the total is 31.
