Three pipes A,B and C can fill a cistern in 7 hours.After working at it together for 2 hours C is closed. Now A and B require 10 hours more to fill it. How long will C take to fill the cistern alone?
[A] 12 h
[C] 14 h
[B] 13 h
[D] 15 h

• (A+B+C)'s 1hr work=1/7
  partly filled in 2hrs=2/7
  remaining work=5/7
  Hence A+B's 10hrs work=5/7
  (A+B)'s 1hr work=5/70
  c work alone=((A+B+C)'s 1hr work)-((A+B)'s 1hr work)
  =1/7-5/70
  =14hrs
• 13 years agoHelpfull: Yes(51) No(0)
• a+b+c=7HR, 1work in 7hr
  Then a+b+c in 2 hrs done 2/7 work
  Left wid 5/7 work
  5/7 work is done in 10 hrs by a+b,
  So, 1w done in 7/5*10=14 days by a+b,
  Now we knw a+b do in 14 days n a+b+c in 7 days
  By solving we get 14days alone by C.
  This is d oral way to solve a T n W questions :)
  
• 10 years agoHelpfull: Yes(2) No(1)
• (a+b+c)=1/7 hrs
  (a+b)*10=5/7
  (a+b+c)*2=2/7
  After solving c =1/14,
  So, c will take 14 hours.
• 10 years agoHelpfull: Yes(1) No(0)
• part filled by a+b+c in 1hr=1/7 ->part filled in 2 hr= 2/7
  hence 2/7+(part filled by a+b in 10 hr)=1 i.e tank is full
  therefore (part filled by a+b)=1/14
  c will fill in=1/7-1/14=1/14
  hence 14 hrs c will take.
• 10 years agoHelpfull: Yes(0) No(0)