A can do a work in 8 days, B can do a work in 7 days, C can do a work in 6 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work.

• A can do a work in 8 days, B can do a work in 7 days, C can do a work in 6 days.
  A works on the first day, B works on the second day and C on the third day resly.that is they work on alternate days. When will they finish the work.(which day will they finish the work)
  6 7/168 days 7 7/169 days 7 6/168 days 7/168 days 7 7/168 days
  
  After day 1, A finishes 1/8 of the work.
  
  After day 2, B finishes 1/7 more of the total work. (1/8) + (1/7) is finished.
  
  After day 3, C finishes 1/6 more of total work. Total finished is 73/168.
  
  So, after day 6, total work finished is 146/168.
  
  On day 7, A will work again and finish 1/8 = 21/168 more of total work.
  
  Total work completed at end of day 7 is (146/168) + (21/168) = 167/168.
  
  Work will be completed on day 8 when B is working. He must finish 1/168 of total remaining work. Since he takes 7 days to finish the total task, he will need 7/168 of the day.
  
  Total days required is 7 + (7/168).
• 10 years agoHelpfull: Yes(7) No(3)
• a-8 days, b- 7 days, c- 6 days
  
  (a+b+c)'s 3 days work= (1/8+1/7+1/6)
  = 73/168
  
  work done in 3 triplets of days=(2*73)/168
  = 73/84
  
  remaining work =(1-(73/84))
  =11/84
  
  now remaining work is done by (A+b)together= 11/84
  1/8 + B =11/84
  B = 11/84 - 1/8 =1/168
  
  1/168 work done by b in (7*1/168)= 7/168 days
  
  total time taken = 7 + 1/168= 7 7/168
• 11 years agoHelpfull: Yes(4) No(13)