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)![]() ![]() You can put this solution on YOUR website! 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- \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- \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- \n" ); document.write( "30(1 \n" ); document.write( " \n" ); document.write( " |