Four pipes fill a cistern in 15,20,30 and 60 hours respectively.The first was opened at 6 am,the second at 8 am,the third at 10 am and fourth at noon.when will the reservoir will be filled?

• ans) 2pm
  Pipe a will be opened for 6am-12pm -> 6hours ===> 6/15
  Pipe b will be opened for 8am-12pm -> 4hours ===> 4/20
  Pipe c will be opened for 10am-12pm -> 2hours ==> 2/30
  From noon, all pipes are opened...
  Up-to 12pm, the amount of the reservoir is 2/3
  6/15 + 4/20 + 2/30 = 40/60 = 2/3
  
  In 1 hour, the four pipes are going to fill 1/6th of reservoir
  1/15 + 1/20 + 1/30 + 1/60 = 10/60 = 1/6
  
  So, The remaining 1/3 will be filled by 4 pipes in 2 hours...
  So, the reservoir will be filled by 2PM...
• 10 years agoHelpfull: Yes(8) No(0)
• (1/15)*2 +(1/15+1/20)*2+(1/15+1/20+1/30)*2+(1/15+1/20+1/30+1/60)*x=1
  so that x=2;
  so that reservoir will be filled at 2 o'clock
• 9 years agoHelpfull: Yes(0) No(0)
• Loss matches = X
  Win matches = X+3
  2 points for + (-1) points for loss
  2*(X+3) + (-1)*X = 23
  X+6= 23
  X= 17.
  Therefore loss matches =17
  Now as per question
  total matches = loss matches + won matches
  (X )+ (X+3) = 2x + 3
  2(17)+3= 37.
  Hence 37 matches they played.
• 6 years agoHelpfull: Yes(0) No(0)