A cistern can be filled by two pipes A and B in 7 1/2 h. Both of them are open for 2 1/2 h and then B is closed. A alone now require 6 2/3 h more to fill cistern. How long would A take to fill the cistern working alone?
[A] 8 h
[C] 9 h
[B] 7 h
[D] 10 h

• (A+B) 1 hr work=2/15 (A+B)'s work in 5/2 hrs=2/15*5/2=1/3
  Remaining Part=1-1/3=2/3
  since A fills 2/3 parts in 20/3 hrs
  so A fills 1 part in(20/3*3/2)hrs=10hrs
  Answer:-10hrs
• 11 years agoHelpfull: Yes(38) No(0)
• A n B both fill the cistern in 15/2 hours..so 2/15 part in 1 hour..
  Now,both are open for 5/2 hours=5/2(2/15)=1/3
  Remaining part=1-1/3=2/3
  So.A is fill 2/3 part by 2o/3 hours...
  So,how many time it take for filling 1/3 part?.=(1/3*3/2*20/3)=10/3 hours.
  So cistern can be fill in total=(10/3+20/3)=10 hr
  
  Ans is 10 hr
• 13 years agoHelpfull: Yes(19) No(1)
• a+b=2/15
  
  (a+b)*5/2 + a*(20/3) =1
  
  55a+15b=6,
  15a+15b=2
  
  So, a can do it in 10 hours.
• 10 years agoHelpfull: Yes(3) No(1)
• we know that A+B 's 1hr work =2/15 and then work done in 5/2 hrs is=5/2*2/15=1/3
  so remaining part=1-1/3=2/3
  now A alone require 20/3 hr to fill cistern ,so A alone will take=2/3*x=20/3 =>x=10
  So 10 hrs is the answer.
• 9 years agoHelpfull: Yes(0) No(0)
• (D) 10 hours
• 6 years agoHelpfull: Yes(0) No(0)