document.write( "Question 996027: Mark made a business trip of
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document.write( "247.5 miles. He averaged
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document.write( "51 mph for the first part of the trip and
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document.write( "60 mph for the second part. If the trip took
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document.write( "4.5 hours, how long did he travel at each rate?
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Algebra.Com's Answer #614646 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Mark made a business trip of 247.5 miles. \n" ); document.write( " He averaged 51 mph for the first part of the trip and \n" ); document.write( " 60 mph for the second part. \n" ); document.write( " If the trip took 4.5 hours, how long did he travel at each rate? \n" ); document.write( ": \n" ); document.write( "let d = distance traveled at 51 mph \n" ); document.write( "the total distance is 247.5 mi therefore: \n" ); document.write( "(247.5-d) = distance traveled at 60 mph \n" ); document.write( ": \n" ); document.write( "Write a time equation; time = dist/speed \n" ); document.write( ": \n" ); document.write( "51mph time + 60 mph time = 4.5 hrs \n" ); document.write( " \n" ); document.write( "Least common factor of 51 and 60 is 1020, multiply equation by 1020 \n" ); document.write( "1020* \n" ); document.write( "Cancel the denominators and we have \n" ); document.write( "20d + 17(247.5-d) = 4590 \n" ); document.write( "20d + 4207.5 - 17d = 4590 \n" ); document.write( "20d - 17d = 4590 - 4207.5 \n" ); document.write( "3d = 382.5 \n" ); document.write( "d = 382.5/3 \n" ); document.write( "d = 127.5 mi at 51 mph \n" ); document.write( "then \n" ); document.write( "247.5 - 127.5 = 120 mi at 60 mph \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Confirm this answer, find the actual time at each speed \n" ); document.write( "127.5/51 = 2.5 hrs \n" ); document.write( "120/60 = 2 hrs \n" ); document.write( "----------------- \n" ); document.write( "total time 4.5 hrs \n" ); document.write( " \n" ); document.write( " |