Question 115995
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
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Whew!!! That was some problem.
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Cheers,
Stan H.