document.write( "Question 1034999: Solve. Round to the nearest tenth of necessary\r
\n" ); document.write( "\n" ); 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?
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Algebra.Com's Answer #649732 by ankor@dixie-net.com(22740)\"\" \"About 
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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?
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\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
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\n" ); document.write( "write a time equation; time = dist/speed
\n" ); document.write( "To time - return time = .5 hrs
\n" ); document.write( "\"d%2F45\" - \"%28%28d-15%29%29%2F50\" = .5
\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
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\n" ); document.write( "\"How many total miles did he travel \"
\n" ); document.write( "90 + 75 = 165 mi
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\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
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\n" ); document.write( "time dif: .5 hrs
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