There is a cycle race going on in a circular track and 1/5th of the total in front a person and 5/6th of the total behind that person gives up the total number of participants. Total how many participants are there?

• 11 participants
• 15 years agoHelpfull: Yes(2) No(0)
• 31
• 14 years agoHelpfull: Yes(2) No(0)
• Assume there are x participants in the race.
  participants in front of a person wil b x-1 and
  that behind him wil b x-1.
  so
  1/5(x-1) + 5/6(x-1) = x
  we get,
  x=31
  
  
  
  
• 10 years agoHelpfull: Yes(2) No(2)
• 11
• 15 years agoHelpfull: Yes(0) No(0)
• 11 participants
  
• 15 years agoHelpfull: Yes(0) No(0)
• 1/6
• 15 years agoHelpfull: Yes(0) No(1)
• 31 participants
• 15 years agoHelpfull: Yes(0) No(0)
• let total participents =x
  x/5+5*x/6+1=x
  31x+30=30x
  x=30 answer
• 11 years agoHelpfull: Yes(0) No(2)
• Assume there are x participants in the race.In a round race,no: of
  participants in front of a person wil b x-1 an that behind him wil b
  x-1. i.e, 1/5(x-1) + 5/6(x-1) = x ; solving x = 31
• 10 years agoHelpfull: Yes(0) No(1)