At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.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, which of the following represents the speed of the faster train, in miles per hour?


total distance traveled in 3.5 hours is 287 miles.
let the speed of faster train be x miles per hour...slower is x-6 miles per hour..the relative speed would be sum of the avg speeds because they are moving in opp directions

total distance = speed * time
287=(2x-6)*3.5
82=2x-6....x=88/2=44

38
40
44
48
50

Dear ashetty
the most important thing in this Q is the formula.. when two vehicle comes to each other at differenet speed and meet each other after specific time: total destination = S1t + S2t
S1= the first vehicle speed S2= second vehicle speed t=the time they meet each other

regarding this Q: we have S1 slower train speed S2= faster train speed= S1+6
t= 1:30 p.m- 10 a.m = 3.5 hrs.
now ==> 3.5 s1 + 3.5 (s1+6) = 287 ==> 7S1= 266 ==> S1= 38 and S2= 38+6= 44

hope it helps u to solve all the same type of Q