Ive been working on this one for two days Any help is apprec

I\'ve been working on this one for two days. Any help is appreciated.
Girl is hiking and standing on the middle of a train bridge over a river. She is 300 ft from point P which is also the direction the train is approaching from. She is 500 ft from point Q in the opposite direction of the train\'s approach. Heather can reach point P at the same time as the train will reach point P while she is running 10mph. She can reach point Q at the same time as the train will reach point Q while she is running 10mph. How fast in mph is the train traveling?
I\'ve tried various d=rt combinations, using the table format, but nothing is correct. I know that the girl can reach point P in 30 units of time or reach point Q in 50 units of time. The train can reach point Q in 50 units of time + the unknown amount of time it takes to reach the bridge. There is an unknown in the distance the train is from the bridge. I think I\'ve looked at this too long.

Solution

The train travels x ft while the girl travels 300 ft.
The Train travels x + 300 ft + 500 ft while the girls travels 500 ft.
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Info: 10mph = 10(mi/hr)(5280 ft/1 mi)(1 hr/60 min) = 880 ft/min
Time for the girl to arrive at P.
Distance = 300 ft ; Rate = 880 ft/min ; time = 300/880 = 0.341 minute
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Time for the girl to arrive at Q.
Distance = 500 ft ; rate = 880 ft/min ; time = 500/880 = 0.568 minute
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Train to P DATA:
Distance = x ft ; time = 0.341 min;
Rate = d/t = x/0.341 ft/min
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Train to Q DATA:
Distance = (800+x) ft ; time = 0.568; rate = (800+x)/0.568 ft.min
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EQUATION:
Train rate to P = train rate
x/0.341 = (800+x)/0.568
0.568x = 0.341*800 + 0.341x
0.227x = 272.8
x = 1201.762 ft
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Train Rate = x/0.341 = 1201.762/0.341 = 3524.229 ft/min
Info: 3524.229(ft/min)(1mi/5280 ft)(60 min/1 hr) = 40.048 mph
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Checking:
(800+1201.762)/0.568 = 3524.229 ft/min = 40.048 mph