document.write( "Question 1106450: Mr. Brown drove from his home to his school at the rate of 25 mph and returned by a different route at the rate of 30 mph. The route by which he returned was 5 miles longer than the route by which he went. The return trip took 10 minutes less than the trip out. Find the distance Mr. Brown traveled each way. \n" ); document.write( "
Algebra.Com's Answer #721448 by ankor@dixie-net.com(22740)\"\" \"About 
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Mr. Brown drove from his home to his school at the rate of 25 mph and
\n" ); document.write( " returned by a different route at the rate of 30 mph.
\n" ); document.write( " The route by which he returned was 5 miles longer than the route by which he went.
\n" ); document.write( " The return trip took 10 minutes less than the trip out.
\n" ); document.write( " Find the distance Mr. Brown traveled each way.
\n" ); document.write( ":
\n" ); document.write( "let t = travel time at 25 mph
\n" ); document.write( "10 min = 1/6 hr, therefore
\n" ); document.write( "(t-\"1%2F6\") = time at 30 mph
\n" ); document.write( ":
\n" ); document.write( "write a distance equation; dist = speed * time
\n" ); document.write( "return dist = to dist + 5 mi
\n" ); document.write( "30(t-\"1%2F6\") = 25t + 5
\n" ); document.write( "30t - 5 = 25t + 5
\n" ); document.write( "30t - 25t = 5 + 5
\n" ); document.write( "5t = 10
\n" ); document.write( "t = 10/5
\n" ); document.write( "t = 2 hrs is the travel time at 25 mph
\n" ); document.write( "therefore
\n" ); document.write( "25 * 2 = 50 mi to school
\n" ); document.write( "and
\n" ); document.write( "50 + 5 = 55 mi return home
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution by finding the return dist
\n" ); document.write( "30(2-\"1%2F6\")
\n" ); document.write( "30(1\"5%2F6\") = 55 mi
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