A can do a piece of work in 24 days, B in
52 days and C in 64 days. All being to do
it together, but A leaves after 6 days and
B leaves 6 days before the completion of
the work. How many days did the work
last?

• *I think you have mistakenly written 32 as 52 for B(as per SSC Paper)* but still solving with your given data
  
  A takes 24 days
  B takes 52 days
  C takes 64 days
  So, Total work = LCM of (A,B,C)
  LCM (24,52,64)=2496 units
  So, efficiency of A= 2496/24 = 104 units
  efficiency of B= 2496/52 =48 units
  efficiency of C= 2496/64= 39 units
  efficiency of A+B+C = 191 units
  So they worked for 6 days before A leaves
  i.e. Work done =191*6 =1146 units
  Remaining Work = 2496-1146 =1350 units
  Now , B left C alone for 6 days. So, C alone did 6 days work
  i.e. Work Done=39*6=234 units
  Remaining Work = 1350-234= 1116 units ,which was done by both B and C
  So, Days taken by B+C=1116/(efficiency of B and C)=1116/87=12.82
  
  Total Days taken= 6+6+12.82=24.82= 25 (approx) Ans.
• 7 years agoHelpfull: Yes(0) No(0)