document.write( "Question 1034999: Solve. Round to the nearest tenth of necessary\r
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document.write( "Mr. Albert drove to his cottage in the mountains at a rate of 45 mi/hr. On the way back he drove at a rate of 50 mi/hr but took a route 15 miles shorter than the route by which he traveled to the cottage. How many total miles did he travel if it took him 1/2 hr longer to go than to return? \n" );
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Algebra.Com's Answer #649732 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Mr. Albert drove to his cottage in the mountains at a rate of 45 mi/hr. \n" ); document.write( " On the way back he drove at a rate of 50 mi/hr but took a route 15 miles shorter than the route by which he traveled to the cottage. \n" ); document.write( " How many total miles did he travel if it took him 1/2 hr longer to go than to return? \n" ); document.write( ": \n" ); document.write( "let d = the distance drove to the cottage \n" ); document.write( "The return trip was 15 mi shorter, therefore: \n" ); document.write( "(d-15) = return distance \n" ); document.write( ": \n" ); document.write( "write a time equation; time = dist/speed \n" ); document.write( "To time - return time = .5 hrs \n" ); document.write( " \n" ); document.write( "multiply equation by 450, cancel the denominators and you have \n" ); document.write( "10d - 9(d-15) = 450(.5) \n" ); document.write( "10d - 9d + 135 = 225 \n" ); document.write( "d = 225 - 135 \n" ); document.write( "d = 90 mi to the cottage \n" ); document.write( "then obviously, \n" ); document.write( "90 - 15 = 75 mi return home \n" ); document.write( ": \n" ); document.write( "\"How many total miles did he travel \" \n" ); document.write( "90 + 75 = 165 mi \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this out, find the actual time each way \n" ); document.write( "90/45 = 2.0 hrs \n" ); document.write( "75/50 = 1.5 hrs \n" ); document.write( "------------------- \n" ); document.write( "time dif: .5 hrs \n" ); document.write( " |