Two water taps together fill a tank in 9 3/8 hours.the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. find the time in which each tap can separately fill the tank.

• Larger tap fill in = 15hrs
  Smaller tap fill in = 25hrs
  
  Let x = No. of hours required to fill by longer tap.
  So, x+10 = No. of hours required to fill by smaller tap.
  And v = Volume of the tank.
  
  Portion of tank fill in 1hr by longer tap = v/x
  Portion of tank fill in 1hr by smaller tap = v/x+10
  Portion of tank fill in 1hr by both tap = v/(75/8)
  
  v/x + v/(x+10) = v/(75/8)
  
  1/x + 1/(x+10) = 8/75
  
  75(2x+10) = 8(x^2 +10x)
  150x+750 = 8x^2 + 80x
  8x^2 - 150x + 80x -750 = 0
  8x^2 - 70x - 750 = 0
  By simplifying,
  4x^2 - 35x - 375 = 0
  
  (i.e) b^2 -4ac = 7225>0
  x = 35+(square root (7225)) /8 (or) x = 35 - (square root (7225)) /8
  x = (35+85) /8
  x = 120/8
  [x = 15hrs] and [x+10 = 15+10 = 25hrs].
  
  Therefore,
  
  Larger tap can fill in = 15hrs
  Smaller tap can fill in= 25hrs
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