10. At 10 a.m. two trains started travelling towards each other from stations 287 miles apart. They passed each other at 1:30pm on the same day. If the average speed of the faster train exceeds the average speed of the slower train by 6 miles per hour, the speed of the faster train in miles per hour is

• 44 mph
  
  Let the speed of the faster train= x mph and slower train=(x-6) mph
  Relative speed while travelling towards each other=x + (x-6) = (2x-6) mph
  After starting at 10:00 am, trains meet at 1:30 pm i.e. in 3.5 hrs or 7/2 hrs.
  So time taken when they meet = 287/(2x-6) = 7/2
  => 41*2 = 2x-6 or x=44
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