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by MBA.Aspirant » Tue Nov 15, 2011 3:03 pm
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?

A. 38
B. 40
C. 44
D. 48
E. 50
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by shankar.ashwin » Tue Nov 15, 2011 4:16 pm
Since they travel in opposite direction, their relative speed is sum of speeds of both trains, X and (X+6)

Speed = Distance / time

Given time = 3.5 and distance = 287

Speed = 287/3.5 = 82.

2X + 6 = 82

2X = 76

X = 38.

Since faster is asked (X+6) = 44 C IMO
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by GMATGuruNY » Tue Nov 15, 2011 6:10 pm
MBA.Aspirant wrote: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?

A. 38
B. 40
C. 44
D. 48
E. 50
We can plug in the answers, which represent the speed of the faster train.

Answer choice C: 44 miles per hour
The speed of the slower train = 44-6 = 38 miles per hour.
Combined rate for the two trains = 44+38 = 82 miles per hour.
Distance covered by the two trains in 7/2 hours = 82(7/2) = 287 miles.
Success!

The correct answer is C.
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