document.write( "Question 55477: A passenger train can travel 325 mi in the same time a freight train takes to travel 200mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed of each.\r
\n" ); document.write( "\n" ); document.write( "Thanks in advance!
\n" ); document.write( "ac \n" ); document.write( "
Algebra.Com's Answer #37724 by stanbon(75887)\"\" \"About 
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A passenger train can travel 325 mi in the same time a freight train takes to travel 200mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed of each.
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\n" ); document.write( "Pass train DATA:
\n" ); document.write( "distance =325 mi ; time = x hrs ; rate= 325/x hr.
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\n" ); document.write( "Freight train DATA:
\n" ); document.write( "distance =200 mi ; time = x hrs ; rate = 200/x hr.
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\n" ); document.write( "EQUATION:
\n" ); document.write( "pass train rate - 25 mph = freight train rate
\n" ); document.write( "(325/x) - 25 = (200/x)
\n" ); document.write( "Multiply thru by x to get:
\n" ); document.write( "325 - 25x = 200
\n" ); document.write( "25x = 125
\n" ); document.write( "x=5 hrs
\n" ); document.write( "Rate of pass train = 325/5= 65 mph
\n" ); document.write( "Rate of freight train = 200/5 = 40 mph
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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