A, B and C can do a piece of work in 30, 20 and 10 days respectively. A is assisted by B on one day and by C on the next day, alternately. How long would the work take to finish?

• Let Total units of work are 60
  Therefore, A in one day will complete 60/30 units=2 units
  Similarly B in one day = 60/20=3 units
  & C in one day=60/10=6 units
  On first day A+B=2+3=5 units are completed
  On second day A+C=2+6=8 units completed
  Total units completed in 2days=5+8=13 units
  Total days required to complete all 60 units =60/13=4 which is equivalent to 8 days since 13 was 2 days work
  Still 8 units are remained because 13*4=52
  On 9th day A&B will complete 5 units
  Still 3 units are remained which on 10th day A&C will do in 3/8 day
  Hence total days =9 3/8 days.
• 7 years agoHelpfull: Yes(8) No(0)
• 9 3/13 days
  
  Total 2 day's work = 2 days's work by A + 1 day work by B + 1 day work by C = 2/30 + 1/20 + 1/10 =13/60
  So days require to finish the work= 2*(60/13)= 120/13 = 9 3/13
• 8 years agoHelpfull: Yes(0) No(3)
• A+B's 1dy wrk is 5/60; A+C's 1day woek is 8/60.
  So 2 days total work is 13/60; And total work to be done is 1.
  therefore by propotionality concepts,
  (13/60)*x=2*1;
  x=120/13;
  x=9 3/13days;
  i.e total time taken to finish the work is 9 3/13days;
  
• 8 years agoHelpfull: Yes(0) No(3)
• L.C.M of their work unit is 60. A,B,C work unit is 2,3,6 respectively.first day work unit is 5&2nd day work unit is 9.It's going alternatively.for 8 days they work for 56unit.4 unit of work is done in 2/7 day.Totaly the work is done by 8*(2/7)days.
• 7 years agoHelpfull: Yes(0) No(0)