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At 10 a.m. two trains started travelling toward each other from stations 287 miles apart. They passed each other at 1:30 pm the same day. They 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?

At 10 a.m. two trains started travelling toward each other from stations 287 miles apart. They passed each other at 1:30 pm the same day. They 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? Correct Answer 44

Let the average speed of slower and faster trains be S and (S + 6) miles/hr respectively. Time taken = 10 am to 1:30 pm = 3.5 hr Distance travelled by slower train = 3.5S Distance travelled by faster train = 3.5(S + 6) Now,  3.5S + 3.5(S + 6) = 287 or, 3.5S + 3.5S + 21 = 287 or, 7S = 287 - 21 or, 7S = 266 or, S = 266/7 = 38 miles/hr So, speed of the faster train, S + 6 = 38 + 6 =  44 miles/hr

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