document.write( "Question 284297: Fred and Desi left Steamtown Mall at 9:00 am and began walking in opposite directions. At 1:00 pm that same day they were 20 miles apart. If Fred walks 0.5 mph slower than Desi, what is Desi's speed of walking?\r
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document.write( "A. 2.25 mph B. 2.75 mph C. 3.5 mph D. 5 mph E. None of these. \n" );
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Algebra.Com's Answer #854209 by ikleyn(53750) You can put this solution on YOUR website! . \n" ); document.write( "Fred and Desi left Steamtown Mall at 9:00 am and began walking in opposite directions. \n" ); document.write( "At 1:00 pm that same day they were 20 miles apart. If Fred walks 0.5 mph slower than Desi, \n" ); document.write( "what is Desi' speed of walking? \n" ); document.write( "A. 2.25 mph B. 2.75 mph C. 3.5 mph D. 5 mph E. None of these. \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " In the post by @mananth, the treatment is not adequate to the problem. \n" ); document.write( " Due to this reason, the solution and the answer are incorrect.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " I came to bring a correct solution.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "let desi's speed be x mph \n" ); document.write( "Fred's speed = x-0.5 mph \n" ); document.write( "In 4 hours Desi has walked 4x miles \n" ); document.write( "In 4 hours Fred has walked 4(x-0.5) miles = 4x-2 \n" ); document.write( "4x + (4x -2) = 20 \n" ); document.write( "8x = 22 \n" ); document.write( "x = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "CHECK. The distance apart in 4 hours, at 1:00 pm, is 4*2.75 + 4*(2.75 - 0.5) = 20 miles. ! Precisely correct !\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Solved correctly.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |