document.write( "Question 175451: A passenger train travels 392 mi in the same time that it takes a freight train to
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document.write( "travel 322 mi. If the passenger train travels 20 mi/h faster than the freight train,
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document.write( "find the speed of each train. \n" );
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Algebra.Com's Answer #130523 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A passenger train travels 392 mi in the same time that it takes a freight \n" ); document.write( " train to travel 322 mi. If the passenger train travels 20 mi/h faster than \n" ); document.write( " the freight train, find the speed of each train. \n" ); document.write( "; \n" ); document.write( "Let s = speed of the freight \n" ); document.write( "then \n" ); document.write( "(s+20) = speed of the passenger train \n" ); document.write( ": \n" ); document.write( "Since the problems states the times are equal, write a time equation \n" ); document.write( "Time = dist/speed \n" ); document.write( ": \n" ); document.write( "Pass time = freight time \n" ); document.write( " \n" ); document.write( "Cross multiply \n" ); document.write( "392s = 322(s+20) \n" ); document.write( ": \n" ); document.write( "392s = 322s + 6440 \n" ); document.write( ": \n" ); document.write( "392s - 322s = 6440 \n" ); document.write( ": \n" ); document.write( "70s = 6440 \n" ); document.write( "s = \n" ); document.write( "s = 92 mph is the freight \n" ); document.write( "then \n" ); document.write( "92 + 20 = 112 mph is the passenger train \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the times of each \n" ); document.write( "392/112 = 3.5 hrs \n" ); document.write( "322/92 = 3.5 hrs \n" ); document.write( " \n" ); document.write( " |