Two pipes a and b can fill a tank in 6hrs and 4 hrs respectively. if they are opened on alternate hour and if pipe a is opened first in how many hours, the tank shall be full?

• a's 1 hr work = 1/6
  b's 1 hr work = 1/4
  
  in 2 hrs a ll fill 2 * 1/6 = 1/3
  in 2 hrs b ll fill 2 * 1/4 = 1/2
  
  so 1/3 + 1/2 = 5/6
  Remaining part yet to be filled = 1 - 5/6 = 1/6
  
  So 2 + 2 = 4 hrs [A, B, A, B]
  Next turn is A's...
  A ll fill 1/6 part in 1hr...
  
  So 2 + 2 + 1 = 5hrs
  
  Ans : 5 hrs
  
• 10 years agoHelpfull: Yes(8) No(0)
• 1st hr tank is filled by a i.,e 1/6
  2nd hr tank is filled by b i.,e 1/4
  for 2hrs 1/6 + 1/4 = 5/12
  4hrs =2(5/12)=10/12
  5 hrs=10/12 + 1/6=12/12=1
  so,within 5Hrs tank will be filled.
• 10 years agoHelpfull: Yes(2) No(0)
• (A+B)'s 2 hour's work when opened =1/6+1/4=5/12
  (A+B)′s 4 hour's work=(5/12)∗2=5/6
  Remaining work = 1−(5/6)=1/6
  Now, its A turn in 5th hour
  1/6 work will be done by A in 1 hour
  Total time = 4+1=5hours
• 10 years agoHelpfull: Yes(1) No(0)
• Lcm of 6 and 4 is 12( total capacity of tank )
  So 1st hr...2( A's one hr work)
  2nd hr..3(B's one hr work )
  2
  3
  2
  So total 5 hrs
• 6 years agoHelpfull: Yes(0) No(0)