document.write( "Question 200669: roy drives to work at 70 mph and home that night along the same route at 30 mph.
\n" ); document.write( "what is his average speed ? \n" ); document.write( "

Algebra.Com's Answer #150913 by vleith(2983) $\"\"$ $\"About$

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Careful here. The kneejerk answer is (70+30)/2 = 50
\n" ); document.write( "But is not correct.
\n" ); document.write( "Why?
\n" ); document.write( "Sure the distance both way is the same, but the time required to cover that distance is less going 70 than it is at 30.\r
\n" ); document.write( "\n" ); document.write( "Let the distance be given as D
\n" ); document.write( "The total distance is 2D.
\n" ); document.write( " $\"Distance+=+AverageSpeed%2Atime\"$
\n" ); document.write( " $\"2D+%2F+time+=+AverageSpeed\"$ \r
\n" ); document.write( "\n" ); document.write( "So we need to find out how much time is spent on the trip each way
\n" ); document.write( " $\"timeToWork+=+D%2F70\"$
\n" ); document.write( " $\"timeHome+=+D%2F30\"$
\n" ); document.write( " $\"time+=+D%2F70+%2B+D%2F30\"$
\n" ); document.write( " $\"time+=+%2830D%2B70D%29%2F2100\"$
\n" ); document.write( " $\"time+=+100D%2F2100\"$
\n" ); document.write( " $\"time+=+D%2F21\"$
\n" ); document.write( "So
\n" ); document.write( " $\"2D+%2F+time+=+AverageSpeed\"$
\n" ); document.write( " $\"2D+%2F+%28D%2F21%29+=+AverageSpeed\"$
\n" ); document.write( " $\"42+=+AverageSpeed\"$ \r
\n" ); document.write( "\n" ); document.write( "Thus, becaue the return trip takes over twice as long as getting to work, the average speed is well less than the kneejerk answer.\r
\n" ); document.write( "\n" ); document.write( "Make sure you understand this concept. A problem just like this was on every SAT/GRE test I ever took.
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