Question

At 10 am two trains started travelling towards each other from stations 287 miles apart. they passed each other at 1:30 pm the same day.if the average speed of the fastest train exceeds the average speed of the slower train by 6 miles per hour.find the speed of the first train in miles per hour

Answer 1

as per given information:

=> Distance d= 287

=> fastest train = X+6

=> Slower train = X

time = Distance/ Speed

train start's at 10:00 am and passed each other at 1:30 pm 3.5 hrs [ 7/2 hrs ]

i.e as two trains are going opposite directions to each other we will add two speeds so...

287/ X+X+6= 7/2

286/ 2x +6 = 7/2

14x +42= 574

14x = 532

X= 38

ans is: first train = X+6

= 38+6

= 44