A can do a piece of work in 36 days,B can do it in 72 days,C can do it in 78 days.The three take turns in working i.e A works on the 1st day,B works on the 2nd day,C works on the 3rd day and so on.Which day will they finish the work?

1)17th day
2)54th day
3)55th day
4)19th day

• ans is 55
  (A+B+C)'s 3days work=1/36+1/72+1/78=51/936
  in next step we hav to multiply 51 with some number,so that it reaches to den(936)
  that number is 18;
  now work done in 18 days for each of A,B, and C(18*3=54days)(because they worked alternatively)=18*51/936=918/936;
  so remaining work to be done is=1-918/936=18/936;
  in the next day it is A's side; A,s 1 day work=1/36;
  from proportionality rule:
  1/36work---->1 day
  18/936work--->?days(say x)
  x=(18/936(1))/(1/36)=9/13;(appr=1 day)and previously there are 18 daysfor A,Band C(i.e)total days=(3*18)+9/13=55
• 12 years agoHelpfull: Yes(8) No(3)
• A work 100/36 =2.78
  B work 100/72 =1.38
  C work = 100/78 =1.28
  3 day work of A , B,C is 5.43
  1 day work = 5.43/3 =1.81
  time to complete total work =100/1.81 =55.2
  so ans = 55
• 10 years agoHelpfull: Yes(5) No(1)
• On 19Th Day..."A" Does 26 nits Of Work "B" Does 13 Units Of Work "C" Does 12 Units oF Work On Each Day...Total Work Is 936 Units So For 18 Days 918 Units Are Done N On 19TH Day "A" Dos Remaining 18 Units...So 19 Days...
• 13 years agoHelpfull: Yes(4) No(10)
• a+b+c three days work is 1/36+a/72+1/78=1/916.
  work done in 6 pairs of days(i.e 6*3=18 days)=(1/916)*18= 1/52.
  so on 19th day a can finish the work
• 13 years agoHelpfull: Yes(3) No(8)
• consider 6 unit of work has to be done.
  a=6/36
  b=6/72
  c=6/78
  x is no. of days require
  (6/36+6/72+6/78) * x = 6
  solve for x and get 18.35 which is 19 days
• 6 years agoHelpfull: Yes(0) No(1)