SOLUTION: A passenger train traveled 180 miles in the same amount of time it took a freight train to travel 120 miles. The rate of the freight train was 15 miles per hour slower than the rat

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Question 1071054: A passenger train traveled 180 miles in the same amount of time it took a freight train to travel 120 miles. The rate of the freight train was 15 miles per hour slower than the rate of the passenger train. Find the rate of the passenger train.

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
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Let "r" be the rate of the passenger train, in mph.

Then the rate of the freight train is (r-15) mph.

The time the passenger train spent for 180 miles trip was 180%2Fr hours.

The time the freight train spent for 120 miles trip was 180%2F%28r-1%29 hours.


The condition says that these two amounts of times are equal:


180%2Fr = 120%2F%28r-15%29   ("A passenger train traveled 180 miles in the same amount of time it took a freight train to travel 120 miles.")


Solve this equation. For it, multiply both sides by r*(r-15). You will get

180*(r-15) = 120r  --->  180r - 2700 = 120 r  --->  180r - 120 r = 2700  --->  60r = 2700  --->  r = 2700%2F60 = 45.


Answer. The passenger train rate is 45 mph. That for the freight train is 30 mph.

Solved.