At 10 a.m two trains started traveling toward each other from stations 287 miles apart.they passed each other at 1.30p.m the same day.If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour,find speed of the faster train in miles per hour?

• 44 miles per hour
  
  Let the speed of faster train be x
  Speed of slower train = (x-6)
  
  Time taken to cross each other = 3hours 30 min = 7/2 hours
  Distance travelled = 287
  Relative speed = x+x-6 = 2x-6
  
  Speed = diatance / time
  2x - 6 = 287/7/2
  =82
  2x = 88
  x = 44
  
  Speed = 44 miles per hour
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