A is twice as efficient as B and B is thrice as efficient as C, who can finish a job alone in 24 days. If A, B and C work on alternate days with A starting first, in how many days will the job finished?

• 6.66 days
  3 days work =5/12
  6 days work 10/12
  remaning=2/12
  remaining work will be completed by A in (2/12)*4=2/3 days
  so total days to complete the work =6+2/3 days
• 8 years agoHelpfull: Yes(10) No(2)
• As a:b=2:1,b:c=3:1(efficiency) time ratio to do the work=1:2,1:3 so a can complete in 4d,b can compete in 6d as mentioned c in 24d .a+b+c 3 day work=1/4+1/8+1/24=5/12.so 5/12*2=10/12 means a+b+c 6 days work=10/12.(5/6).
  remaining work=1-5/6=1/6(.16666) but a can do 1/4(.25)work in i day and remaining(1/6) is less than a"s 1 day work i.e (1/4). so on 7th day work is completed
• 8 years agoHelpfull: Yes(1) No(5)
• on 7th day work is finished as a days work of C=1/24 ,B=1/8 AND A=1/4
  SO 3 days work of A+B+C= 1/24+1/8+1/4=20/24 .
  NOW total work left is = 1-20/24 = 4/24 = 1/6
  so 1/6 is less than 1/4 so it will be completed on 7th day BY "A"
• 8 years agoHelpfull: Yes(1) No(1)
• 7 th day
• 7 years agoHelpfull: Yes(0) No(0)