There are m pipes from 1 to m..such that tge speed at which it fills a pool alone is equal to the time taken by other m-1 pipes(not for m=1)..suppose time taken by 9 pipe is 40min when filled alone..what would be the time taken if pipe no. 12 and 13 fill together..??

• using Efficiency consept..
  1/T9=1/40, given
  =>1/T9=1/T8 + 1/T7 +.......+ 1/T1 = 1/40
  =>1/T10=1/T9 + (1/T8 + 1/T7 +.......+ 1/T1) = 1/40+1/40
  =>1/T11=1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 = 1/20 + 1/20
  =>1/T12=1/T11 + 1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 =1/10 + 1/10
  =>1/T13=1/T12 + 1/T11 + 1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 =1/5 + 1/5
  now, time taken by pipe no. 12 & 13 will be given by
  1/T12 + 1/T13 = 1/5 + 2/5 =3/5
  Hence,time = 5/3 =1.66min
• 10 years agoHelpfull: Yes(77) No(6)
• let t1, t2...t8 r d time taken by the correponding pipes individually.
  9 pipes-> 40 min= 10th pipe seperately
  1/t1 +1/t2... +1/t9= 1/40
  10 pipes-> 1/40 +1/40= 1/20 =11th pipe sep.
  11 pipes-> 1/20+ 1/20=1/10 =12th pipe
  12 pipes-> 1/10+1/10=1/5 =13th pipe
  so 12th & 13th together= 1/10+ 1/5=3/10
  means 10/3=3.33 min.
  
  
  
• 10 years agoHelpfull: Yes(6) No(2)
• 12, 13 will take only 20.20 mintes to fill the pool.
• 10 years agoHelpfull: Yes(1) No(9)
• Does any body remember options of this question
  
• 9 years agoHelpfull: Yes(0) No(0)