document.write( "Question 492943: Jane drove to an appointment averaging 60 mph. On the return trip, her average speed was 28 mph due to traffic, and she took 24 minutes longer. How far did she travel to the appointment? \n" ); document.write( "
Algebra.Com's Answer #335224 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate (r) times Time (t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let d=distance to her appointment
\n" ); document.write( "And t=time required to travel to her appointment
\n" ); document.write( "Then d=60t
\n" ); document.write( "On the return trip:
\n" ); document.write( "d=28(t+24/60)=28(t+2/5) (Lets deal in hours for the time being )
\n" ); document.write( "So our equation to solve is:
\n" ); document.write( "60t=28(t+2/5)
\n" ); document.write( "60t=28t+56/5
\n" ); document.write( "32t=56/5
\n" ); document.write( "160t=56
\n" ); document.write( "t=56/160=7/20 hr or 21 min
\n" ); document.write( "d=60t=60*7/20 =21 mi\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "at 28 mph on the return trip, it would take her
\n" ); document.write( "21/28 = 3/4 hr or 45 min
\n" ); document.write( "45-21=24 min longer\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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