document.write( "Question 854941: a student bikes to school averaging 5 mph. he bikes home the exact same distance averaging 10 mph. what is his average speed for the round trip? \n" ); document.write( "

Algebra.Com's Answer #515009 by KMST(5328) $\"\"$ $\"About$

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If the school was 10 miles from the student's home,
\n" ); document.write( "it would take him $\"10miles%2F5mph=2hours\"$ to bike to school,
\n" ); document.write( "and $\"10miles%2F10mph=1hours\"$ to bike home.
\n" ); document.write( "The round trip would be $\"2%2A10miles=20miles\"$
\n" ); document.write( "and it would take $\"2hours%2B1hour=3hours\"$ .
\n" ); document.write( "the average speed would be
\n" ); document.write( " $\"20miles%2F3hours=highlight%286.3mph%29\"$ (rounded).
\n" ); document.write( "
\n" ); document.write( "The 10 mile distance was made up, but the calculation would be the same with any distance.
\n" ); document.write( "We could say, it is $\"d\"$ miles,
\n" ); document.write( "so the times in hours would be $\"d%2F5\"$ and $\"d%2F10\"$ .
\n" ); document.write( "The round trip would be $\"2d\"$ miles,
\n" ); document.write( "and it would take $\"d%2F5%2Bd%2F10=2d%2F10%2Bd%2F10=3d%2F10\"$ hours.
\n" ); document.write( "Them, the speed (in mph) would be calculated as
\n" ); document.write( " $\"2d%2F%283d%2F10%29=2d%2A%2810%2F3d%29=20d%2F30d=20%2F3=6.3\"$ .
\n" ); document.write( "The distance cancels out, and it was easier to understand and calculate by making $\"d=10miles\"$ . \n" ); document.write( "

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