document.write( "Question 81412: a passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passanger train is 25 mi/phr faster than the speed fo the freight train, find the speed of each. \n" ); document.write( "
Algebra.Com's Answer #58305 by stanbon(75887) ![]() You can put this solution on YOUR website! a passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passanger train is 25 mi/phr faster than the speed fo the freight train, find the speed of each. \n" ); document.write( "---------- \n" ); document.write( "Passenger train DATA: \n" ); document.write( "distance=25 mi ; rate = x mph ; time = d/r = 325/x hrs \n" ); document.write( "----------------- \n" ); document.write( "Freight train DATA: \n" ); document.write( "distance = 200 m ; rate = x-25 mph ; time = d/r = 200/(x-25) hr \n" ); document.write( "------------------- \n" ); document.write( "EQUATION: \n" ); document.write( "time = time \n" ); document.write( "325/x = 200/(x-25 \n" ); document.write( "Crsoo- multiply to get: \n" ); document.write( "325(x-25) = 200x \n" ); document.write( "325x-8125 = 200x \n" ); document.write( "125x = 8125 \n" ); document.write( "x = 65 mph ( speed of passenger train) \n" ); document.write( "x-25 = 40mph(speed of freight train)\r \n" ); document.write( "\n" ); document.write( "===================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |