document.write( "Question 410841: A man drive to a town at an average speed of 60 miles per hour. He returns by a road which is 10 miles shorter, at an average speed of 45 miles per hour. the return trip takes him he same time as the first trip. Find the length of each road. \n" ); document.write( "

Algebra.Com's Answer #288828 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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A man drive to a town at an average speed of 60 miles per hour.
\n" ); document.write( " He returns by a road which is 10 miles shorter, at an average speed of 45 miles per hour.
\n" ); document.write( "the return trip takes him he same time as the first trip.
\n" ); document.write( " Find the length of each road.
\n" ); document.write( ":
\n" ); document.write( "Let d = distance of the longer road
\n" ); document.write( "then
\n" ); document.write( "(d-10) = distance of the return road
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: Time = dist/speed
\n" ); document.write( ":
\n" ); document.write( "To time = Return time
\n" ); document.write( " $\"d%2F60\"$ = $\"%28d-10%29%2F45\"$
\n" ); document.write( "Cross multiply
\n" ); document.write( "60(d-10) = 45d
\n" ); document.write( "60d - 600 = 45d
\n" ); document.write( "60d - 45d = 600
\n" ); document.write( "d = $\"600%2F15\"$
\n" ); document.write( "d = 40 mi is the longer road, then obviously, 30 mi is the return road
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check this by finding the actual time of each trip, (they should be equal)
\n" ); document.write( "40/60 = .67 hr
\n" ); document.write( "30/45 = .67 hr also \n" ); document.write( "

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